Equivariant extension of distributionson GLnbyBich-Ngoc Cao NguyenB.Sc., The University of Manitoba, 2009M.Sc., The University of British Columbia, 2011A THESIS SUBMITTED IN PARTIAL FULFILLMENT OFTHE REQUIREMENTS FOR THE DEGREE OFDOCTOR OF PHILOSOPHYinThe Faculty of Graduate and Postdoctoral Studies(Mathematics)THE UNIVERSITY OF BRITISH COLUMBIA(Vancouver)October 2020c Bich-Ngoc Cao Nguyen 2020The following individuals certify that they have read, and recommend tothe Faculty of Graduate and Postdoctoral Studies for acceptance, the thesisentitled:Equivariant extension of distributions on GLnsubmitted by Bich-Ngoc Cao Nguyen in partial fulfillment of the require-ments for the degree of Doctor of Philosophy in Mathematics.Examining Committee:Julia Gordon, MathematicsSupervisorWilliam Casselman, MathematicsSupervisory Committee MemberLior Silberman, MathematicsSupervisory Committee MemberKalle Karu, MathematicsUniversity ExaminerGordon Semeno↵, PhysicsUniversity ExaminerClifton Cunningham, Mathematics at the University of CalgaryExternal ExamineriiAbstractLet k be a non-Archimedean local field, and C1c (GLn) the space of locallyconstant compactly supported complex-valued functions on the general lin-ear group GLn over k. For every irreducible representation (⇡, V ) of GLn,the space Hom(C1c (GLn), V ⌦ eV ) is one-dimensional. This space is gener-ated by an element denoted by ⇣, which can be thought of as an integralagainst matrix coecients. In this thesis, we are interested in the so-called”extension problem” of ⇣.More explicitly, for 0 m n, GLn can be embedded into the spaceRm of all n ⇥ n matrices over k of rank at least m. If ⇣ lies in the imageof the induced map of this embedding, then we say that ⇣ can be extendedto rank at least m. For m = 0, the extension problem of ⇣ to rank at least0 has been completely answered in Tate’s thesis for n = 1, and by Moeglin,Vigne´ras, Waldspurger, and Minguez for general n. Our goal is to determinethe least value m for a given representation such that ⇣ can be extended torank at least m.A representation is said to appear in rank m if Hom(C1c (Rm), V ⌦ eV )is non-trivial. It is natural to conjecture that ⇣ extends to rank at leastm + 1 but does not extend to rank at least m where m is the highest rankless than n that ⇡ appears in. In this thesis, this conjecture is proved forspherical representations, by means of extending Satake transform to thespace of K-bi-invariant functions on Mn and obtaining a partial descriptionof the image of the rank filtration under this extended Satake transform.Some explicit computations for spherical representations of GL3 are in-cluded as motivating examples of the general case. There are also somesuggestive calculations for non-spherical representations of GL2.iiiLay SummaryWe study the so-called ”extension problem” that is a generalization of aclassical result from Tate’s thesis in non-Archimedean harmonic analysis.The objects of interest are maps on function spaces that are equipped withactions. When a smaller space is embedded inside a bigger space, one canask whether the map defined on the smaller space can be extended to oneon the bigger space and still remain compatible with the given action onthe bigger space. It is known that for certain actions, such extension to thewhole bigger space is not possible. We are interested in determining exactlyhow far one can extend the map before it breaks down. Our main result isthat for a given action, we can determine the largest subset of the biggerspace that the map can be extended to.ivPrefaceThis dissertation is original, unpublished, independent work by the author,B-N.C. Nguyen.vTable of ContentsAbstract . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iiiLay Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ivPreface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . vTable of Contents . . . . . . . . . . . . . . . . . . . . . . . . . . . . viAcknowledgements . . . . . . . . . . . . . . . . . . . . . . . . . . . ixDedication . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . x1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.1.1 The extension problem . . . . . . . . . . . . . . . . . 11.1.2 The guiding questions . . . . . . . . . . . . . . . . . . 31.1.3 Description of results . . . . . . . . . . . . . . . . . . 42 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92.1 The case n = 1 (Tate’s thesis) . . . . . . . . . . . . . . . . . 92.1.1 Unramified characters and Mellin transform . . . . . 112.2 The set-up for GLn . . . . . . . . . . . . . . . . . . . . . . . 122.2.1 A filtration of C1c (Mn) by descending rank . . . . . . 132.2.2 A filtration of C1c (Mn) by ascending rank . . . . . . 142.3 Review of classical results . . . . . . . . . . . . . . . . . . . . 152.3.1 Induced representations and duality . . . . . . . . . . 152.3.2 Representations of GLn . . . . . . . . . . . . . . . . . 162.3.3 The Zelevinsky classification of irreducible non-cuspidalrepresentations of GLn in segment notation . . . . . . 172.3.4 Segment notation of an irreducible unramified repre-sentation (⇡, V ) of GLn(k) . . . . . . . . . . . . . . . 20viTable of Contents2.3.5 Minguez’s results on segment notation of the uniqueirreducible quotient of ⌧ ⇥ ⇢ . . . . . . . . . . . . . . 202.4 Zeta integrals and L-functions . . . . . . . . . . . . . . . . . 213 The case n = 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 243.1 Summary of results for GL2 . . . . . . . . . . . . . . . . . . 243.2 The space HomG⇥G(C1c (R1), V ⌦ eV ) . . . . . . . . . . . . . 263.2.1 A description of C1c (R1) as an induced representation 263.2.2 The computation of HomG⇥G(C1c (R1), V ⌦ eV ) . . . . 273.2.3 An explicit generator of HomG⇥G(C1c (R1), V ⌦ eV ) . 343.3 The descending rank filtration for GL2 . . . . . . . . . . . . 373.3.1 An explicit form of dµ˙(x, y) on S\G⇥G . . . . . . . 373.3.2 An extension of ⇣R1⇡ from C1c (R1) to C1c (R1) . . . 383.4 Spherical representations and Satake transforms . . . . . . . 423.4.1 Extension of ⇣G⇡ using the ascending rank filtration . 424 Spectrum of C1c (Rm) . . . . . . . . . . . . . . . . . . . . . . . . 484.1 General results on C1c (Rm) . . . . . . . . . . . . . . . . . . . 484.1.1 A description of C1c (Rm) as an induced representation 484.1.2 The condition on (⇡, V ) so that HomGLn⇥GLn(C1c (Rm), V⌦eV ) 6= {0} . . . . . . . . . . . . . . . . . . . . . . . . . 494.1.3 One-dimensionality of HomGLn⇥GLn(C1c (Rm), V⌦eV )when it is non-trivial . . . . . . . . . . . . . . . . . . 534.2 The spherical part of the spectrum of C1c (Rm) . . . . . . . . 564.2.1 The case m = n . . . . . . . . . . . . . . . . . . . . . 564.2.2 The case m < n . . . . . . . . . . . . . . . . . . . . . 575 Extension problem and the local L-function: partial resultsfor general irreducible representations of GLn . . . . . . . . 625.1 The extension of ⇣GLn⇡ to ⇣Mn⇡ if LGLn(⇡, s) does not have apole at s = n12 . . . . . . . . . . . . . . . . . . . . . . . . 625.2 Poles of LGLn(⇡, s) if (⇡, V ) appears in a lower rank . . . . . 635.2.1 The case m = 0 . . . . . . . . . . . . . . . . . . . . . 655.2.2 The case m = 1 . . . . . . . . . . . . . . . . . . . . . 655.2.3 The case m = n 1 . . . . . . . . . . . . . . . . . . . 665.3 Counterexample: L-function fails to detect the rank . . . . . 676 The extension problem for spherical representations of GLn 696.1 Summary of results . . . . . . . . . . . . . . . . . . . . . . . 69viiTable of Contents6.2 Satake parameters of spherical representations that appear inlower ranks . . . . . . . . . . . . . . . . . . . . . . . . . . . . 726.3 Equivariant extension of ⇣GLn⇡ and Satake transforms . . . . 746.3.1 Satake transform on Mn . . . . . . . . . . . . . . . . 746.3.2 Cones in the co-character lattice and neighbourhoodsin Mn . . . . . . . . . . . . . . . . . . . . . . . . . . . 766.3.3 Spherical functions supported on lower rank . . . . . 776.3.4 Satake transforms of C1c (Rm)K⇥K . . . . . . . . . . 786.3.5 Vanishing property of S(f(FH⇤nm⇤ )) if H 6= ; . . . 816.3.6 Vanishing property of S(f(⇤nm⇤ )) . . . . . . . . . . 876.3.7 Equivariant extension of ⇣GLn⇡ . . . . . . . . . . . . . 887 The extension problem for GL3 . . . . . . . . . . . . . . . . . 927.1 Summary of results for GL3 . . . . . . . . . . . . . . . . . . 927.2 Representations that appear in low ranks . . . . . . . . . . . 937.2.1 Representations that appear in rank 1 . . . . . . . . . 947.2.2 Representations that appear in rank 2 . . . . . . . . . 957.3 Equivariant extension of ⇣GL3⇡ and Satake transforms . . . . 977.3.1 The Satake transform of some explicit functions inC1c (GL3)K⇥K . . . . . . . . . . . . . . . . . . . . . . 987.3.2 The Satake transform of some explicit functions inC1c (R2)K⇥K . . . . . . . . . . . . . . . . . . . . . . 1007.3.3 Equivariant extension of ⇣GL3⇡ for the representation|.| det3 . . . . . . . . . . . . . . . . . . . . . . . . . 1037.3.4 Equivariant extension of ⇣GL3⇡ for the representation|.|⇥ |.| 12 det2 . . . . . . . . . . . . . . . . . . . . . . 1068 The Steinberg representation of GL2 . . . . . . . . . . . . . . 108Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116AppendicesA Injectivity of Satake transform on C1c (M2)K⇥K . . . . . . . 119A.1 Satake images of elements in the basis of C1c (M)K⇥K interms of elementary symmetric polynomials . . . . . . . . . . 119A.2 Injectivity of Satake transform on C1c (M)K⇥K . . . . . . . 121viiiAcknowledgementsFirstly, I would like to express my sincere gratitude to my advisors, Prof.William Casselman and Prof. Julia Gordon, for their guidance and continu-ous support during my study. Their patience and immense knowledge havegreatly assisted this research. Their constant encouragement and friendli-ness have turned my Ph.D. years into an enjoyable journey during whichI have learnt more than just math from them. Without them, this thesiswork would not have been completed or written.I am also thankful to my committee members, whose comments andsuggestions help improve the manuscript. Finally, I would like to thankeveryone whom I have had the honour of knowing throughout my years ofstudy. Their presence in my life have defined who I am today and made thisaccomplishment possible. Thank you.ixDedicationThis thesis is dedicated to my beloved aunt, Phuong Cao, who has loved meand supported me unconditionally throughout the years.xChapter 1Introduction1.1 Introduction1.1.1 The extension problemThis thesis explores what we call an extension problem for GLn(k), where kis a non-Archimedean local field.Before stating the extension problem for general n, let us first con-sider the case n = 1. In [23], A. Weil remarked that the results of thelocal theory in Tate’s thesis [21] can be rephrased as saying that the spaceHomk⇥(C1c (k),) is one-dimensional for every smooth character of k⇥,and asked for a generalization of such a result to GLn(k). Before we discusssuch a generalization, note that we can think of the elements of the spaceHomk⇥(C1c (k),) as k⇥-equivariant distributions with values in C on whichk⇥ acts via . The key observation that can be derived from Tate’s thesisis that the generator of the one-dimensional space Homk⇥(C1c (k),) has avery di↵erent origin depending on whether is the trivial character or not.For trivial , this space is generated by the delta-distribution ⇤0 , which canbe thought of as a k⇥-invariant integral on the space C1c ({0}) of functionssupported at the origin. For all other , it comes from the k⇥-equivariantextension of a k⇥-invariant integral on the space C1c (k⇥) of the locally con-stant compactly supported functions on k⇥. More precisely, given a choiceof Haar measure d⇥x on k⇥, there is a linear functionalf 7!Zk⇥f(x)(x) d⇥x,defined on the space C1c (k⇥) of locally constant functions compactly sup-ported on k⇥. This linear functional has a k⇥-equivariant extension to thespace C1c (k) if and only if the character is non-trivial. It is this type ofequivariant extension that we are interested in; we call the question of itsexistence the extension problem. It becomes clear from Tate’s proof thatthis equivariant extension exists if and only if the local zeta-integral⇣(, s) :=ZO(x)|x|s d⇥x11.1. Introductiondoes not have a pole at s = 0. We review this theory in Section 2.1, andsee explicitly that the pole of the local L-function at s = 0 in this case isprecisely the obstruction to the existence of the solution to our extensionproblem.In order to explore this question for general n, we first need to state theextension problem correctly. Here we follow Weil’s paper [23], as well assubsequent literature where his question was largely answered, namely [6]and [15]. For GLn(k), the correct analogue of the space of k⇥-equivariantdistributions on k is the spaceHomGLn(k)⇥GLn(k)(C1c (Mn(k)), V ⌦ eV ),where Mn(k) is the space of all n ⇥ n matrices, (⇡, V ) is an irreducibleadmissible representation of GLn(k), and (e⇡, eV ) is its contragredient repre-sentation. Here GLn(k) ⇥ GLn(k) acts on C1c (Mn(k)) by ((g, g0) · f)(x) =f(g1xg0). The elements of this space, in analogy with the one-dimensionalcase, can be thought of as End(V )-valued distributions. Godement andJacquet’s study of zeta integrals in [6], in a sense, answered Weil’s question,as pointed out by Moeglin, Vigne´ras and Waldspurger (see [[15],3.III.7]).More precisely, exactly as in the case n = 1, it is not hard to see that thespaceHomGLn(k)⇥GLn(k)(C1c (GLn(k)), V ⌦ eV )is one-dimensional, and is generated by⇣GLn(k)⇡ = ⇡() :C1c (GLn(k))! V ⌦ eVf 7!✓⇡(f) : v 7!ZGLnf(g)⇡(g)v dg◆,which can be thought of as an integral against matrix coecients of ⇡.Moeglin, Vigne´ras and Waldspurger pointed out that when the Godement-Jacquet local L-function attached to ⇡ does not have a pole at s = n12 ,this integral naturally extends equivariantly to test functions in C1c (Mn(k)).They used this argument to prove that the spaceHomGLn(k)⇥GLn(k)(C1c (Mn(k)), V ⌦ eV )is at least one-dimensional, and conjectured that its dimension is exactlyone, which is a very special case of the conjectures on theta-correspondencethat they discuss. This conjecture (for GLn(k) ⇥ GLm(k)) was proved byMinguez ([13]); we will refer to this result asmultiplicity-one forMn(k). Thisthesis began as a search for a refinement of this multiplicity-one result, as21.1. Introductionwell as a more direct proof, as we explain below. While we have not arrivedat an alternative proof in full generality, here we present results towardsdetailed understanding of the space HomGLn(k)⇥GLn(k)(C1c (Mn(k)), V ⌦ eV ),especially for spherical representations. We also work out the cases of n = 2and n = 3 in full detail, which provides a good illustration of the generaldiculties.1.1.2 The guiding questionsA key ingredient in all the prior literature on the subject, e.g. [15], [13], and[10], is the filtration by the rank of the matrix. We introduce some notation:for 0 m n, let• Rm(k) be the subset of Mn(k) consisting of matrices of rank m,• Rm(k) – the subset of Mn(k) consisting of matrices of rank at leastm,• Rm(k) – the subset of Mn(k) consisting of matrices of rank at mostm, which is the closure of Rm(k) under the usual topology.Given an admissible representation ⇡ of G = GLn(k), we say that ⇡appears in lower rank if for some m < n, the spaceHomGLn(k)⇥GLn(k)(C1c (Rm(k)), V ⌦ eV ) is non-trivial. Minguez called suchrepresentations appearing on the boundary.We say that the ⇡⌦ ⇡˜-valued zeta-integral on G extends to Mn when thegenerator of the space HomGLn(k)⇥GLn(k)(C1c (GLn(k)), V ⌦ eV ) extends toan element of HomGLn(k)⇥GLn(k)(C1c (Mn(k)), V ⌦ eV ).While the main focus of [15], [13], and [10] is the multiplicity-one result,which is needed to establish theta-correspondence, here we ask questions ofa di↵erent kind, which are not addressed in the existing literature:1. Can we prove directly that if the L-function of ⇡ has a pole at s =n12 , then the ⇡ ⌦ ⇡˜-valued zeta-integral on G does not extend toMn(k)? This is, in fact, the converse to the statement of Moeglin,Vigne´ras, and Waldspurger [15]. Generally, how much can one learnabout the extension problem from the poles of the L-function? (Heuris-tically, as we explain below, one can think of the pole at s = n12 ofthe local L-function of ⇡ as an ’obstruction’ to extension).2. Furthermore, when the ⇡ ⌦ ⇡˜-valued zeta-integral on GLn(k) doesnot extend to Mn(k), what is the lowest rank that it does extend31.1. Introductionto? Namely, what is the smallest value of m such that this integralextends to C1c (Rm(k))? Can one read o↵ this m from the poles ofthe L-function attached to ⇡?3. Can we find an explicit generator of the spaceHomGLn(k)⇥GLn(k)(C1c (Mn(k)), V ⌦ eV )?In the case where the ⇡⌦⇡˜-valued zeta-integral onG does not extend toMn(k), we expect that it comes from an invariant integral of functionssupported on Rm(k) where m is the lowest rank that (⇡, V ) appearsin. Does multiplicity one hold for C1c (Rm(k)) for every m where0 m n?Let K be the maximal compact open subgroup GLn(O) of GLn(k). Atthe level of K-fixed vectors, our work is motivated by the following conjec-ture:Conjecture (Conjecture 3). Let H(GLn(k),K) be the spherical Hecke al-gebra. For a spherical representation (⇡, V ) of GLn(k),(i) HomH(GLn(k),K)(C1c (Rm(k))K⇥K , V K ⌦ eV K) ⇠=HomH(GLn(k),K)(C1c (Rm+1(k))K⇥K , V K⌦ eV K) if (⇡, V ) does not ap-pear in rank m;(ii) HomH(GLn(k),K)(C1c (Rm(k))K⇥K , V K ⌦ eV K) ⇠=HomH(GLn(k),K)(C1c (Rm(k))K⇥K , V K ⌦ eV K) if (⇡, V ) appears in rankm.In particular, we expect that multiplicity one holds at every rank, namely,for 0 m n,HomH(GLn(k),K)(C1c (Rm(k))K⇥K , V K ⌦ eV K) ⇠= C.1.1.3 Description of resultsIt turns out that even for GL2(k), the answers to most of these questionsare not straightforward. However, for GL2(k), we have an almost completeanswer to all of the above questions. For general n, we construct examplesillustrating the diculties in finding a complete answer. In particular, weshow that the answer to question (1) is, generally, negative. However, forspherical representations, the situation is much nicer, and the main part ofthe thesis is devoted to answering question (2) for spherical representations.41.1. IntroductionWe answer this question completely for almost all spherical representations,and construct examples illustrating the diculty of the remaining cases. ForGL3(k), our answer to question (2) for spherical representations is complete.We study the case n = 2 in Chapter 3, with the study of the extensionproblem for the Steinberg representation deferred to Chapter 8. We provethe following:Theorem (Theorem 9, Proposition 8). For a spherical irreducible admissi-ble representation (⇡, V ) of GL2(k), the generator of the spaceHomGL2(k)⇥GL2(k)(C1c (M2(k)), V ⌦ eV )”comes from” an invariant integral on the following space of functions:• C1c (GL2(k)) if the L-function of ⇡ has no pole at s = 12 ;• C1c (R1(k)) if the L-function of ⇡ has a pole at s = 12 but no pole ats = 12 ,• C1c ({0}) if the L-function of ⇡ has poles at both s = 12 and s = 12 .Here, ”comes from” means that it is the image of the corresponding in-tegral (which we construct in Section 3.2.3) under the maps described inSection 2.2.We believe that our method of producing the generator for the spaceHomGL2(k)⇥GL2(k)(C1c (M2(k)), V ⌦ eV ) should generalize to arbitrary n, butwe do not pursue this question here. We also note that this is the only placewhere the spaces C1c (Rm(k)) play a role in this thesis (we also saw nomention of them in the earlier literature).As a corollary, in the case of GL2(k), we obtain an alternative proofof the multiplicity-one result for all representations except for a particularfamily of irreducible principal series without Iwahori-fixed vectors. Thisalready illustrates the diculty one encounters for representations of higherdepth.The complete answer for question (2) for spherical representations ofGL3(k) is deferred to Chapter 7. This chapter uses the general methods wedeveloped for spherical representations, and can be used as an illustrationof the complicated proofs in Chapter 6.Then in Chapter 4, we give a description of the space C1c (Rm(k)) as aninduced representation, re-prove Minguez’s criterion for a representation ⇡to ‘appear’ in Rm(k), and develop a version of ’Satake isomorphism’ for thisspace.51.1. IntroductionIn Chapter 5, we develop some partial results about the extension prob-lem for a general representation, in particular, fill in the details in [15], whichyields the following result in Section 5.1:Theorem (Theorem 11). If LGLn(k)(⇡, s) has no pole at s = n12 , thenthe generator ⇣GLn(k)⇡ of HomGLn(k)⇥GLn(k)(C1c (GLn(k)), V ⌦ eV ) extends toMn(k).Heuristically, it appears that on the one hand, ⇡ appearing in rankm < nis an obstruction to extension of the ⇡ ⌦ ⇡˜-invariant integral to Mn(k); andon the other hand, the local L-function of ⇡ having a pole at s = n12 isalso an obstruction. Thus it seems natural to ask:(a) If the L-function of ⇡ has a pole at s = n12 , is ⇡ guaranteed to appearin lower rank?(b) Can the L-function of ⇡ detect the ranks in which ⇡ appears?To the best of our knowledge, these questions have not been studied. Theanswer to question (a) for spherical representations is positive, which weproved in Proposition 12(ii), and otherwise, we do not know.We find that the answer to question (b) is, in fact, negative. There aredistinct representations with the same L-function that appear in di↵erentranks (Example 5.3). Thus, the L-function alone is too coarse an invariantto capture the precise rank in which ⇡ may appear. However, we expectthat if ⇡ appears in rank m, then the L-function associated to ⇡ must havecertain poles. We discuss this question in Section 5.2, and make a preciseconjecture (Conjecture 2). Recall Minguez’s criterion for a representation ⇡to ‘appear’ in Rm(k) which we reprove as Proposition 10:Proposition ( Proposition 10 ). If ⇡ appears in rank m, then ⇡ is the uniqueirreducible quotient of IndGLn(k)Pm ⌧ ⌦ 1nm for some ⌧ 2 Irr(GLm(k)).Using the above proposition, we can determine the cuspidal support of⇡, but this does not give the full Zelevinsky classification datum of ⇡. (Wereview Zelevinsky classification in Section 2.3.3.) The main diculty is todetermine the segments of ⇡ given the segments of ⌧ and 1nm. In thecases where the segments of ⇡ can be deduced explicitly (for example, whenm = 0, 1, n 1), we prove that if ⇡ appears in rank m, then the L-functionassociated to ⇡ has poles at s = n12 , . . . ,n12 +(nm1) (and possiblymore). Moreover, we find that if ⇡ appears in lower rank, then this rank61.1. Introductionis not necessarily unique, that is, ⇡ can appear in multiple lower ranks(Example 3).However, if we restrict our attention to spherical representations, thenmore can be said about the poles of the L-function and the rank(s) that(⇡, V ) appears in. In Chapter 6, we focus on spherical representations, andprove our main theorem:Theorem. (Theorem 14)Assume that the extended Satake transform is injective on C1c (Mn(k))K⇥K(Hypothesis 1). If ⇡ is an irreducible spherical representation such that theL-function associated to ⇡ has a simple pole at s = n12 , then ⇡ appears ina unique rank r < n and ⇣GLn(k)⇡ extends to Rr+1(k) but does not extendto Rr(k).The proof of the above theorem is complicated and relies on an ex-tension of Satake transform on C1c (Rm(k))K⇥K and the assumption thatthe Satake transform is injective on C1c (Mn(k))K⇥K (see Hypothesis 1).We use downward induction on the rank m and show that the questionwhether ⇣GLn(k)⇡ extends to Rm(k) is equivalent to the question whetherthe evaluation map extends from the Satake image of C1c (Rm+1(k))K⇥Kto that of C1c (Rm(k))K⇥K . The natural next step is to describe the im-age of C1c (Rm(k))K⇥K under the Satake transform. Note that the Sa-take transform on C1c (Rm(k))K⇥K is defined through the extended Sa-take transform on C1c (Mn(k))K⇥K as introduced by L. La↵orgue [11]. Werecall that the extended Satake transform on C1c (Mn(k))K⇥K is realizedas an integral twisted by | det |s and then its value is defined via ana-lytic continuation (see Section 6.3.1). Observe that the Satake image ofa function in C1c (Rm(k))K⇥K is a symmetric rational function whose de-nominator isnQi=1(1 q n12 zi). For rank 1, we give a complete descriptionof the image in Proposition 13. For other ranks, it is dicult to give anice description of the Satake image of C1c (Rm(k))K⇥K due to the com-binatorial complexity of the numerators (see Example 8). However, tostudy the extension of ⇣GLn(k), we only need to study the image of theSatake transform of C1c (Rm(k))K⇥K under the evaluation map that sendsz1 = n12 , . . . , znm = n12 + (n m 1), as defined in Definition 15.Note that these values of zi’s are related to the ranks that ⇡ appears in (seeProposition 12).Thus, the L-function associated to ⇡ having a pole at s = n12 is pre-cisely the obstruction to the extension. Our approach is to show that underthe evaluation map, the numerators of the Satake image of C1c (Rm(k))K⇥K71.1. Introductionvanish if m > r and do not vanish if m = r. This proof relies on the as-sumption that s = n12 is a simple pole. Thus, when s = n12 is not asimple pole, we do not know the answer.We also show that the composition of the evaluation map and the Sataketransform on K-fixed functions on GLn(k) is the same as the compositionof the Satake transform on GLm(k) and the Harish-Chandra constant termmap (Proposition 11).8Chapter 2Preliminaries2.1 The case n = 1 (Tate’s thesis)In this section, we review A. Weil’s approach to Tate’s thesis with emphasison the general technique that can be applied to GLn (cf. [9], [21], [23]).Observe that n = 1 is the unique value of n where GLn is abelian,which allows many simplifications. Here, GL1 = k⇥, M1 = k, and R0 ={0}. Instead of considering the action of GL1⇥GL1 as introduced above, itsuces to consider the following action of a single copy of GL1 on k:Definition 1 (A (left) action of GL1 on k). Let g 2 GL1, and x 2 k. Then,g · x = gx.Moreover, irreducible representations of GL1 are one-dimensional. Thus,instead of (⇡⌦e⇡, V ⌦eV ), we can consider simply (,C), where is a smoothcharacter of GL1.Definition 2. Let dx be the usual Haar measure on k that is normalizedsuch that O has volume 1. Denote by d⇥x the Haar measure on k⇥ defined byd⇥x =dx|x| . Then, O⇥ has volume 1 with respect to d⇥x. For an irreduciblerepresentation (,C) of GL1, let⇣k⇥ : C1c (k⇥)! Cf 7!Zk⇥f(x)(x) d⇥x.Let (,C) be an irreducible representation of GL1. In this case, wehave the following short exact sequence of GL1-spaces:0! C1c (k⇥) ext! C1c (k) res! C1c ({0})! 0, (2.1)which gives the following long exact sequence in cohomology:0! Homk⇥(C1c ({0}),C) res⇤! Homk⇥(C1c (k),C) ext⇤!Homk⇥(C1c (k⇥),C)! . . .(2.2)92.1. The case n = 1 (Tate’s thesis)First, by Schur’s Lemma, we have that Homk⇥(C1c ({0}),C) ⇠= C if is triv-ial, and {0} otherwise. On the other hand, we have that Homk⇥(C1c (k⇥),C) ⇠=C for every irreducible representation (,C) of GL1. Specifically, the spaceHomk⇥(C1c (k⇥),C) is generated by ⇣k⇥ as defined in Definition 2.Lemma 1. The map ext⇤ is surjective if and only if is a non-trivialcharacter.Proof. Suppose that there exists a k⇥-equivariant lift ⇣k 2 Homk⇥(C1c (k),C)of ⇣k⇥ . It follows from ⇣k being an extension of ⇣k⇥ that for f 2 C1c (k⇥),⇣k(f) = ⇣k⇥ (f). Namely, if f 2 C1c (k⇥), then⇣k(f) =Zk⇥f(x)(x) d⇥x.In particular, for the characteristic function chO⇥ of O⇥,⇣k(chO⇥) =(1 if is unramified,0 otherwise.By [[3], Lemma 7.4 p.52], the space C1c (k) is spanned by characteristicfunctions of the form cha+$mO, for a 2 k and m 2 Z. Thus, every functionf 2 C1c (k) can be written as A chO +f 0, where A 2 C, f 0 2 C1c (k⇥). Hence,a necessary and sucient condition for the existence of ⇣k is that ⇣k(chO)is well-defined, since by linearity,⇣k(f) = A⇣k(chO) + ⇣k(f0),and the latter is well-defined for f 0 2 C1c (k⇥). Observe that:chO⇥ = chO ch$O = chO $1 · chO = (1$1) · chO .Thus, if ⇣k(chO) is well-defined, then:⇣k(chO⇥) = (1 ($))⇣k(chO).Observe that being trivial is the unique case where ⇣k(chO) cannot bedefined since ⇣k(chO⇥) = 1 while the right hand side is 0. In the othercases, we may set:⇣k(chO) =(0 if is ramified,(1 ($))1 if is non-trivial and unramified.Thus, a k⇥-equivariant lift ⇣k 2 Homk⇥(C1c (k),C) of ⇣k⇥ exists if and onlyif is non-trivial.102.1. The case n = 1 (Tate’s thesis)Using the long exact sequence (2.2), we conclude that:Corollary 1. For every smooth character (,C), the space Homk⇥(C1c (k),C)is one-dimensional.Moreover, we can, in fact, give an explicit description of a generator ofHomk⇥(C1c (k),C). For = 1, the space Homk⇥(C1c (k),C1) is generatedby the -distribution on {0}, which we denote by ⇣k1 for consistency withthe notation in later sections. More explicitly,⇣k1 :C1c (f)! Cf 7! f(0).On the other hand, for 6= 1, the space Homk⇥(C1c (k),C) is generatedby:⇣k :C1c (f)! Cf 7!Zk⇥f(x)(x) d⇥x.Recall the local L-function for each representation of k⇥:L(, s) =(1 if is ramified,(1 ($)qs)1 if is non-trivial and unramified.Observe that being trivial is the unique case where L(, s) has a pole ats = 0, and this pole is precisely the obstruction to the surjectivity of ext⇤.We will later see that a similar phenomenon also occurs for a general GLn.2.1.1 Unramified characters and Mellin transformIn this section, we revisit the question about the extension of ⇣k⇥ where is an unramified character of GL1 using Mellin transform. The followingwell-known argument is included here for completeness since the methodwill be generalized later to the case of spherical representations of GLn inTheorem 14.First, since is unramified, we can consider just the subspaces of O⇥-fixed vectors in the short exact sequence (2.1) and obtain:0! C1c (k⇥)O⇥ ext! C1c (k)O⇥ res! C1c ({0})O⇥ ! 0. (2.3)Next, we recall the following definition of the Mellin transform.112.2. The set-up for GLnDefinition 3 (Mellin transform on C1c (k)O⇥). LetS :C1c (k)O⇥ ! 11 zC[z±1]f 7!Zk⇥f(x)|x|s d⇥x,which converges absolutely if <(s) 0. The Mellin transform S(f) off 2 C1c (k)O⇥ , which is a special case of Satake transform, is defined tobe the analytic continuation of the above integral to s = 0. More explicitly,as a formal series,S(f) =Xi2Zf($i)zi.It is well-known that C1c (k⇥)O⇥ ⇠= C[z±1] (Satake isomorphism) andC1c (k)O⇥ ⇠= 11zC[z±1]. By applying the Mellin transform S on C1c (k⇥)O⇥and C1c (k)O⇥in the short exact sequence (2.3), we obtain the followingshort exact sequence:0! C[z±1]! 11 zC[z±1]! C! 0. (2.4)Moreover, for f 2 C1c (k)O⇥ , via the Mellin transform, the action of $ on ftranslates to multiplication by z; namely, S($ ·f) = zS(f). For an arbitraryf 2 C1c (k)O⇥ , the function (1$1) ·f generally does not lie in C1c (k)O⇥ .However, for chO 2 C1c (k)O⇥ , we have that(1$1) · chO = chO⇥ ,which lies in C1c (k)O⇥. Using this particular function, we observe that theextension of ⇣k⇥ depends on the non-vanishing ofS((1$1) · chO) = (1 z)S(chO) 2 11 zC[z±1].We will prove an analogue of this for spherical representations of GLn inTheorem 14 under the assumption that the extended Satake transform isinjective (Hypothesis 1).2.2 The set-up for GLnLet k be a non-Archimedean local field. Denote by Mn the topological spaceof all n⇥n matrices with entries in k, and GLn the topological group of n⇥ninvertible matrices with entries in k. We first define an action of GLn⇥GLnon Mn as follows:122.2. The set-up for GLnDefinition 4 (A (left) action of GLn⇥GLn on Mn). Let (g, g0) 2 GLn⇥GLn,and x 2 Mn. Then, (g, g0) · x = gxg01.For 0 m n, denote by Rm (resp. Rm, and Rm) the topologicalspace of matrices in Mn of rank m (resp. rank at most m, and rank at leastm). Then, under the action of GLn⇥GLn, the space Mn decomposes inton+ 1 distinct orbits based on the rank of the matrix and Mn =S0mnRm.In particular, Mn = Rn, GLn = Rn, and each Rm and Rm are alsoGLn⇥GLn-spaces under the above defined action of GLn⇥GLn.Let X be Rm, Rm or Rm for some m, considered as a GLn⇥GLn-space with the GLn⇥GLn-action defined above. Denote by C1c (X) thespace of C-valued locally constant compactly supported functions on X.Then, the action of GLn⇥GLn on X induces the following (left) action onthe function space:Definition 5. The (left) action of GLn⇥GLn on C1c (X) is given by:((g, g0) · f)(x) = f((g1, g01) · x) = f(g1xg0),for (g, g0) 2 GLn⇥GLn and f 2 C1c (X).Given an irreducible admissible representation (⇡, V ) of GLn and itscontragredient (e⇡, eV ), consider the following action of GLn⇥GLn on V ⌦ eV :Definition 6. The (left) action of GLn⇥GLn on V ⌦ eV is given by:(g, g0) · (v ⌦ ev0) = ⇡(g)v ⌦ e⇡(g0)ev0,for (g, g0) 2 GLn⇥GLn, and v ⌦ ev0 2 V ⌦ eV .We recall the following natural filtrations of C1c (Mn).2.2.1 A filtration of C1c (Mn) by descending rankFor every 1 m n, we have the following short exact sequence ofGLn⇥GLn-spaces:0! C1c (Rm) ext! C1c (Rm) res! C1c (Rm1)! 0, (2.5)where ext : C1c (Rm) ! C1c (Rm) is the extension by zero map, andres : C1c (Rm) ! C1c (Rm1) is the restriction map. Note that thesemaps are well-defined on the space of locally constant compactly supportedfunctions because Rm is open in Rm and Rm1 is closed in Rm. Note132.2. The set-up for GLnalso that ext and res are both GLn⇥GLn-homomorphisms. Then, to eachshort exact sequence of the form (2.5), we may apply the contravariant func-tor HomGLn⇥GLn(, V ⌦ eV ), and obtain the following long exact sequencein cohomology:0! HomGLn⇥GLn(C1c (Rm1), V ⌦ eV ) res⇤! HomGLn⇥GLn(C1c (Rm), V ⌦ eV )ext⇤! HomGLn⇥GLn(C1c (Rm), V ⌦ eV )! . . .(2.6)2.2.2 A filtration of C1c (Mn) by ascending rankFor every 0 m < n, since Rm+1 is open in Rm and Rm is closed inRm, we also have the following short exact sequence of GLn⇥GLn-spaces:0! C1c (Rm+1) ! C1c (Rm) ! C1c (Rm)! 0, (2.7)where the first map is the extension by zero map, and the second map is therestriction map.This yields the following long exact sequence in cohomology:0! HomGLn⇥GLn(C1c (Rm), V ⌦ eV ) ! HomGLn⇥GLn(C1c (Rm), V ⌦ eV )! HomGLn⇥GLn(C1c (Rm+1), V ⌦ eV )! . . .(2.8)We observe that this exact sequence was studied by Kudla [8] and Minguez[13]. Everywhere below, when we say that a distribution on C1c (Rm+1)extends to C1c (Rm), we mean that it is an image of an element ofHomGLn⇥GLn(C1c (Rm), V ⌦ eV ).Note that the next term in the long exact sequence (2.8) isExt1GLn⇥GLn(C1c (Rm), V ⌦ eV ).If we can show that Ext1GLn⇥GLn(C1c (Rm), V ⌦ eV ) is trivial, then a dis-tribution on C1c (Rm+1) extends to C1c (Rm). However, the Ext com-putation turns out to be dicult and if Ext1GLn⇥GLn(C1c (Rm), V ⌦ eV ) isnon-trivial, we would need to study the connecting homomorphism fromHomGLn⇥GLn(C1c (Rm+1), V ⌦ eV ) to Ext1GLn⇥GLn(C1c (Rm), V ⌦ eV ) in or-der to determine if a distribution extends. Thus, we will not pursue thisdirection and instead, study the extension problem for spherical representa-tions of GLn using the extended Satake transform.142.3. Review of classical results2.3 Review of classical results2.3.1 Induced representations and dualityLet G be a reductive p-adic group. Here, we shall review some basic resultsand set notation for the representations of G. For a detailed exposition ofthe representation theory of reductive p-adic groups, see [1], [5], [18].Let H be a closed subgroup of G. Let µH be a left Haar measure onH. Let H : H ! R⇥+ be the unique group homomorphism such that for allf 2 C1c (H), we have that:ZHf(xh) dµH(x) = H(h)ZHf(x) dµH(x).We call H the modulus character of H.Remark 1. As we sometimes refer to results in [3], it is important to notethat the definition of a modulus character given above is di↵erent from thaton page 29 [3]. More precisely, the value of the modulus character under thedefinition given here is the reciprocal of that in [3]. We choose to use theabove definition of a modulus character since it is more commonly used inliterature.Let (⌧,W ) be an admissible representation ofH. We use IndGH ⌧ , c-IndGH ⌧ ,and indGH ⌧ to denote the unnormalized induction, unnormalized compactinduction, and normalized induction of (⌧,W ) from H to G, respectively.More explicitly,indGH ⌧ ⇠= IndGH ⌧12H ,where H is the modulus character of H.For an admissible representation (⇡, V ) of G, we say that (⇡, V ) has acomposition series if there exists an ascending chain of G-invariant subspacesof V :{0} =: V`+1 ( V` ( . . . ( V1 ( V0 := V,such that Vj/Vj+1 is irreducible. The subquotients Vj/Vj+1 are called theJordan-Holder factors of the composition series. We denote by JH(⇡) theset of Jordan-Holder factors of (⇡, V ).We recall the following well-known result, which shall be used later on.Lemma 2 ([13], Lemma 1.3 p.5). Let X be a locally profinite space, X 0 aclosed subspace of X. Suppose that a locally profinite group G acts continu-ously on X and G ·X 0 = X. Let G0 be the stabilizer of X 0 in G. Denote by152.3. Review of classical results⇡ the natural representation of G in the space C1c (X) consisting of locallyconstant compactly supported C-valued functions on X, and ⇢ the naturalrepresentation of G0 in C1c (X 0). Then⇡ ⇠= c-IndGG0 ⇢.We also recall the following classical results on duality of admissiblerepresentations.Proposition 1 ([1], Proposition 6 p.14). Let (⇡, V ) and (⇢,W ) be admissiblerepresentations of G. Then,HomG(V,fW ) ⇠= HomG(W, eV ).Proposition 2 ([3] Duality Theorem p.32). Let H be a closed subgroup ofG and (⌧,W ) an admissible representation of H. Then,^c-IndGH ⌧ ⇠= IndGH e⌧H .Proposition 3 ([1], Proposition 29 p.14). Let P = LN be a parabolic sub-group of G and (⌧,W ) an admissible representation of L extended triviallyto P . Then,i^ndGP ⌧ ⇠= indGP e⌧ .2.3.2 Representations of GLnLet n = (n1, . . . , nr) be a partition of n, that is, n1+ . . .+nr = n. Denote byPn (resp. Pn) the parabolic subgroup of GLn consisting of blocks of upper(resp. lower) triangular matrices in GLn corresponding to n. We writeLn and Nn (resp. Nn) for the corresponding Levi subgroup and unipotentradical in Pn (resp. Pn). We have that:Pn = LnNn = NnLn, and Ln ⇠= GLn1 ⇥GLn2 ⇥ . . .⇥GLnr .The minimal parabolic subgroup corresponding to the partition n = (1, . . . , 1)shall be denoted by B, and we write T and N for the Levi subgroup and theunipotent subgroup of B, respectively. For maximal parabolic subgroups,that is, when n has the form (m,nm), we shall use the subscriptm insteadof n to denote the partition. For example, we will write Pm instead of Pn.Since we will frequently make use of the modulus character of parabolicsubgroups Pm and B, we include here the explicit form of their modulus162.3. Review of classical resultscharacters.B :B = TN ! C⇥diag(t1, . . . , tn)u 7!nYj=1|tj |n2j+1Pm :Pm = LmNm ! C⇥hg1 00 g4iu 7! | det(g1)|nm| det(g4)|m.(2.9)Let (⇡j , Vj) be an admissible representation of GLnj . Then, we may viewthe tensor product (⇡1⌦. . .⌦⇡r, V1⌦. . .⌦Vr) as an admissible representationof Pn where Nn acts trivially. We will use the notation ⇡1⇥. . .⇥⇡r to denotethe representation indGLnPn ⇡1 ⌦ . . .⌦ ⇡r.Let (⇡, V ) be an admissible representation of GLn. LetV (Nn) := {v ⇡(u)v | v 2 V, u 2 Nn}.Then, VNn := V/V (Nn) is the Jacquet module of (⇡, V ) with respect to Nn,and (⇡Nn , VNn) is an admissible representation of Ln. This gives rise to thenormalized Jacquet functor rn,n from the category Rep(GLn) of admissiblerepresentations of GLn to the category Rep(Ln) of admissible representa-tions of Ln:rn,n :Rep(GLn)! Rep(Ln)(⇡, V ) 7! (⇡Nn12Pn, VNn).Let Irr(GLn) be the set of equivalence classes of irreducible representa-tions of GLn. We will write 1n to denote the trivial representation of GLn,and simply 1 for the trivial character of GL1 = k⇥. For a one-dimensionalrepresentation detn and any admissible representation (⇡, V ) of GLn, wewill write (⇡, V ) to denote the twist of ⇡ by , namely, the representation( detn)⌦ ⇡.2.3.3 The Zelevinsky classification of irreduciblenon-cuspidal representations of GLn in segmentnotationIn this section, we will review the Zelevinsky classification of irreduciblenon-cuspidal representations of GLn in segment notation (cf. [24], [14]).We recall the following definition of a segment as given in [[24], 3.1 p.180]:172.3. Review of classical resultsDefinition 7 (A segment). Let ⇢ be a cuspidal representation of GLn and` 2 N. One sets: = {⇢, |.|⇢, . . . , |.|`1⇢ =: ⇢0}.We call a segment and often denote it by = [⇢, ⇢0]. In the case where⇢ = ⇢0, then we may shorten the notation into = [⇢].Remark 2. As we will refer to results in [14], it is important to note that[14] uses a di↵erent notation to represent a segment . More explicitly,as defined on page 8 [14], let ⇢ be a cuspidal representation of GLn, b, e 2Z, b e. Then, a segment = {|.|b⇢, |.|b+1⇢, . . . , |.|e⇢}is often denoted by = {b, e}⇢. However, under this notation, we donot have a unique way of writing as {b, e}⇢. For example, the segment{|.|, |.|2, |.|3} may be denoted by either {0, 2}|.| or {1, 3}1. To avoid confu-sion, we will always follow the notation of a segment as introduced in [24]and write {|.|, |.|2, |.|3} as [|.|, |.|3].Proposition 4 ([14], Proposition 4.1 p.9). To each segment = [⇢, |.|`1⇢],⇢ being a cuspidal representation of GLn/`, ` 2 N such that ` divides n, onecan associate irreducible representations hi and hit of GLn such that:(1) hi (resp. hit) is the unique irreducible submodule (resp. quotient) of⇢⇥ |.|⇢⇥ . . .⇥ |.|l1⇢.(2) r(n/`,...,n/`),n(hi) = ⇢ ⌦ |.|⇢ ⌦ . . . ⌦ |.|l1⇢, (resp. r(n/`,...,n`),n(hit) =|.|l1⇢⌦ . . .⌦ |.|⇢⌦ ⇢).The representations hi and hit are also characterized by the property (2).Example 1 ([24], Example 3.2 p.181). Consider the segment = [|.|n12 , |.|n12 ]for n 2. Then, hi is the trivial representation 1n of GLn since:r(1,...,1),n(1n) = |.|n12 ⌦ . . .⌦ |.|n12 ,which, by Proposition 4, characterizes 1n uniquely as hi. Note that thecomputation of r(1,...,1),n(1n) can be done explicitly as follows: recall thatV (N) = {v ⇡(u)v | u 2 N},182.3. Review of classical resultswhich is trivial in this case since ⇡(u)v = v for all u 2 N and v 2 V . Thus,VN := V/V (N) = 1⌦ 1⌦ . . .⌦ 1,as a representation of T . Finally, we have thatr(1,...,1),n(1n) := 12B VN ,which yields the desired result. In this case, hit is the Steinberg represen-tation of GLn that fits in the short exact sequence below:0! 1n ! |.|n12 ⇥ . . .⇥ |.|n12 ! hit ! 0.Let 1 = [⇢1, ⇢01] and 2 = [⇢2, ⇢02] be two segments. One says that 1and 2 are linked if 1 6✓ 2, 2 6✓ 1, and 1 [2 is also a segment. If1 and 2 are linked and ⇢2 = |.|a⇢1 for some a 2 N, then we say that 1precedes 2. We refer to a multi-set of segments as a multi-segment.Theorem 1 ( ([24],Theorem 4.2 p.184), ([14], Theorem 4.2 p.10)). Thefollowing conditions are equivalent:(1) For every 1 i, j r, the segments i and j are not linked.(2) h1i ⇥ . . .⇥ hri is irreducible.(3) h1it ⇥ . . .⇥ hrit is irreducible.We have the following classification of irreducible non-cuspidal represen-tations of GLn in terms of segments:Theorem 2 ([14], Theorem 4.3 p.10). (1) Let (1, . . . ,r) be a multiseg-ment. Suppose that, if i < j, i does not precede j. Then, therepresentation h1it ⇥ . . . hrit admits a unique irreducible quotient,which is denoted by h1, . . . ,rit. The multiplicity of h1, . . . ,rit inJH(h1it ⇥ . . . hrit) is exactly 1.(2) The representations h1, . . . ,rit and h01, . . . ,0rit are equivalent ifand only if the sequences (1, . . . ,r) and (01, . . . ,0r) are equal up toa rearrangement.(3) Every irreducible representation of GLn can be written in the formh1, . . . ,rit.For the dual version of the above theorem for irreducible submodules,see [[24], Theorem 6.1 p.188] or [[14], Theorem 4.6 p.21].192.3. Review of classical results2.3.4 Segment notation of an irreducible unramifiedrepresentation (⇡, V ) of GLn(k)We first recall the classification of irreducible unramified representations ofGLn(k). The following theorem is due to Zelevinsky, which is quoted herein the form that appeared in [17].Theorem 3 ([17], Theorem 9.10 p.195). Let 1, . . . ,n be n unramifiedcharacters of k⇥, which are ordered to satisfy the ”do not precede” condition.Then, the representation h[1], . . . , [n]it is an unramified representation ofGLn(k). Furthermore, every irreducible admissible representation of GLn(k)is equivalent to h[1], . . . , [n]it for some 1, . . . ,n.Remark 3. We often denote the unramified characters i as |.|si wheresi 2 C. Using this notation, if 1, . . . ,n are ordered to satisfy the ”do notprecede” condition, then sj 6= si + 1 for i < j. Equivalently, if sj = si + 1then j < i.2.3.5 Minguez’s results on segment notation of the uniqueirreducible quotient of ⌧ ⇥ ⇢In this section, we will review some Minguez’s results in [14]. The followingtheorem will be used later to show that an induced representation of theform ⌧ 1nm has a unique irreducible quotient.Theorem 4 ([14], Theorem 5.1 p.15, Corollary 5.5, p.16). Let 1, . . . ,rbe segments which are not linked, and satisfy the following condition:If i 6= j, then either i = j or i \j = ;. (2.10)Let ⇢ = h1, . . . ,rit (resp. h1, . . . ,ri), and ⇡ 2 Irr(Gn). Then, ⇡ ⇥ ⇢(resp. ⇢ ⇥ ⇡) has a unique irreducible submodule (resp. quotient), and itappears with multiplicity 1 in JH(⇡ ⇥ ⇢) (resp. JH(⇢⇥ ⇡)).Next, we recall some of Minguez’s results on the segment notation ofthe unique irreducible quotient of ⌧ ⇥ ⇢ where ⌧ is irreducible and ⇢ issupercuspidal. To simplify the notation, we will adapt the result to ourparticular case where ⇢ = |.|n12 as a representation of GL1(k), which wewill use later on. For the exact result in its full generality, refer to [[14],Section 7 p.21].Let m = (1, . . . ,r) be the multisegment corresponding to ⌧ . Letj(1), . . . ,j(t) the segments of m in decreasing order, where j(s) has the202.4. Zeta integrals and L-functionsform [|.|bj(s) , |.|n12 ] for 1 s t. Note that m may not necessarily haveany such j(s). For every integer v between 1 and r, we define inductivelythe integer i(v) to be the smallest integer di↵erent from i(1), . . . , i(v 1),such that i(v) ends in |.|n32 and precedes j(tv+1), and let i(v) = r + 1if such integer does not exist. We let `(1), . . . , `(u) be the indices such that`(s) ends in |.|n32 , but `(s) is not one of the i(v)’s defined above. For`(s) = [|.|b(s), |.|n32 ], let +`(s) = [|.|b(s), |.|n12 ]. Define:!0 = h[|.|n12 ],1, . . . ,rit,!s = h1, . . . ,+`(s), . . . ,rit, for 1 i u.Proposition 5 ([14], Corollary 7.3 p.21). The unique irreducible quotientof ⌧ ⇥ |.|n12 is !1 if u > 0 and !0 if u = 0.2.4 Zeta integrals and L-functionsIn this section, we recall some basic definitions and results on zeta inte-grals and L-functions for representations of GLn. In particular, we reviewthe connection between the L-function and the segment notation of an irre-ducible representation of GLn. For a more detailed exposition, see [ [6],§3],and [[22],§2.5 ].Let (⇡, V ) be an admissible representation of GLn. For f 2 C1c (Mn),ev0 2 eV and v 2 V , we consider the integral⇣Mn(f, ev0, v, s) = ZGLnf(g)hev0,⇡(g)vi| det(g)|s dg.We recall the following important result of Godement and Jacquet:Theorem 5 ([6],Theorem 3.3 p.30). Let (⇡, V ) be an admissible irreduciblerepresentation of GLn.(1) There exists s0 2 R so that for <(s) > s0, f 2 C1c (Mn), ev0 2 eV andv 2 V , the integral⇣Mn(f, ev0, v, s) = ZGLnf(g)hev0,⇡(g)vi| det(g)|s dgconverges absolutely.212.4. Zeta integrals and L-functions(2) There exists LGLn(⇡, s) such that (LGLn(⇡, s))1 2 C[qs] and for everyf 2 C1c (Mn), ev0 2 eV and v 2 V ,⇣Mn⇣f, ev0, v, s+ n12 ⌘LGLn(⇡, s)2 C[qs, qs].Remark 4 ([22], 2.5.3 p.274). For an irreducible cuspidal representation(⇡, V ) of GLn (n 2), we have that:LGLn(⇡, s) = 1.Next, using the segment notation of (⇡, V ), Wedhorn computes the L-function LGLn(⇡, s) using the following inductive relations [[22], 2.5.3 p.273-274]:• If (⇡, V ) = (,C) is a representation of k⇥, then from Tate’s thesis:Lk⇥(, s) =(1 if is ramified,(1 ($)qs)1 if is unramified.• If ⇡ = hit, where = [⇢, |.|`1⇢] for some ` 2 N and an irreduciblecuspidal representation ⇢, thenLGLn(⇡, s) = LGLn/`(⇢, s+ ` 1).• If ⇡ = h1, . . . ,rit, thenLGLn(⇡, s) =rYi=1L(hiit, s).Using the above inductive relations and Remark 4, we recall that follow-ing well-known result, which we include the proof here for completeness.Lemma 3. The L-function LGLn(⇡, s) associated to an irreducible repre-sentation (⇡, V ) of GLn has a pole at s = e if and only if there existsa segment in the multisegment corresponding to ⇡, which has the form[|.|b, |.|e] for some b e.Proof. Let ⇡ = h1, . . . ,rit. ThenLGLn(⇡, s) =rYi=1L(hiit, s).222.4. Zeta integrals and L-functionsSo, there exists a segment i such that s = e is a pole of L(hiit, s).Suppose that i = [⇢, |.|`1⇢] for some cuspidal representation ⇢ of GLj(j n) and ` 2 N. ThenLGLj`(hiit, s) = LGLj (⇢, s+ ` 1).By Remark 4, if j 2, then LGLj (⇢, s + ` 1) = 1 which does not havea pole. So, j = 1 and ⇢ = is a character of k⇥. Furthermore, is anunramified character since Lk⇥(, s + ` 1) = 1 if is ramified. Thus, = |.|b for some b. Then,LGLjl(hiit, s) = Lk⇥(|.|b, s+ ` 1) = (1 qbs`+1)1,which has a pole at s = b `+1. So, e = b+ `1 and |.|`1⇢ = |.|e, whichcompletes the proof.As an immediate consequence of Theorem 3 and Lemma 3, we have thefollowing observation:Corollary 2. Let (⇡, V ) be an irreducible unramified representation of GLn(k).Then, its L-function LGLn(⇡, s) has exactly n poles (counting multiplicities).Moreover, the values of the poles and its multiplicities determine the repre-sentation (⇡, V ) uniquely.23Chapter 3The case n = 2In this chapter, we study the case n = 2 in detail. Some results obtainedhere are a special case of the more general results that hold for an arbitraryn, whose proofs will be presented in the later chapters. However, somemethods here are specific to the case n = 2: in particular, the descriptionof C1c (R1) as an induced representation is slightly di↵erent from the onewe use in general later, and the construction of a semi-invariant integralon R1 is specific to GL2 (though we believe it should generalize). Further,in Section 3.3 we use the descending rank filtration to obtain an alternateapproach to the multiplicity-one result in this case. Thus, Sections 3.2, 3.3are specific to GL2. In contrast, Section 3.4 is designed to be an example ofthe proof of the general theorem from Chapter 6. Finally, we see that theSteinberg representation of GL2 ‘appears in R1’; this is the only representa-tion that is not spherical for which we are able to obtain a complete answerto the extension problem, but we defer that discussion to the last chaptersince it relies on all the methods developed in this thesis, and at the sameis the least likely to generalize.For convenience, in this chapter, we will use G := GL2, and M := M2,B for the Borel subgroup of G, and T and N for the Levi subgroup and theunipotent subgroup of B, respectively.We start with a quick summary of the results of this chapter.3.1 Summary of results for GL2We start with the spectrum of C1c (R1): in Section 3.2.1, we show thatas a G ⇥ G-representation, C1c (R1) is compactly induced from the trivialrepresentation of the centralizer of the matrix [ 1 00 0 ] in G ⇥ G, which wedenote by S (see Proposition 6). (In later sections, we will use inductionin stages and think of C1c (Rm) as a parabolically induced representation,but for GL2 this description suces and is actually more convenient). Thisgroup S is not unimodular; we compute its modulus character in Lemma 4,and give an expression for the semi-invariant measure on R1 ' S\G⇥G in(3.4).243.1. Summary of results for GL2In Section 3.2.2, we prove our first result, Theorem 6, which gives a com-plete list of representations appearing inR1 – the space HomG⇥G(C1c (R1), V ⌦ eV )is non-trivial if and only if ⇡ is one of the following:• the one-dimensional representation | det |,• Steinberg representation, which is the irreducible quotient of IndGB |.|⌦ |.|1,• the irreducible principal series |.| 12 ⇥ , where 6= |.| 12 , |.| 32 (i.e. theirreducible principal series with the first component of its Satake pa-rameter equal to 12).For the representations on this list, the space HomG⇥G(C1c (R1), V ⌦ eV )is one-dimensional. In Section 3.2.3 we use the canonical pairing betweenthe Jacquet modules VN and VN and a semi-invariant measure µ˙ on S\G⇥Gto give an explicit formula for a generator of this space for all representationson this list (see Theorem 7). This formula is particularly nice when Jacquetmodule of ⇡ is 1-dimensional, namely, when ⇡ = |.| det2 or when ⇡ is theSteinberg representation.Next, in Section 3.2.3, we use these explicit formulas and also the ex-plicit description of the semi-invariant measure on R1 to see which of theseV ⌦ eV -valued distributions on R1 ‘extend across 0’, that is, extend toR1. It turns out that for all representations on this list, the generatorof HomG⇥G(C1c (R1), V ⌦ eV ) extends to R1 (see Theorem 8). We notethat this gives an almost complete alternative proof of multiplicity-one forM2. This is the only place in the thesis where we study the descending rankfiltration – it seems that getting suciently explicit information about theinvariant integrals for general n would be too complicated.Finally, we switch to the traditionally studied ascending rank filtration,and explore the question of the extension of the ⇡ ⇥ ⇡˜-valued integral ⇣G⇡on G to the space C1c (R1). We note that the only representations forwhich this question is interesting are the representations ‘appearing in R1’on the above list, and the trivial representation. We use Section 3.4 toillustrate the general method we developed in Chapter 6 for the spheri-cal principal series representations on the list. In particular, in Chapter6, we rely on an extension of the Satake transform on C1c (GLn)K⇥K tothe space C1c (Mn)K⇥K of K-biinvariant functions in C1c (Mn), as definedby L. La↵orgue [11]. For spherical representations, instead of studying thespace HomG⇥G(C1c (M), V ⌦ eV ), we only focus on the homomorphisms be-tween K-fixed vectors. For G = GL2, we are able to explicitly computethe image of C1c (R1)K⇥K under this extension of the Satake map (see253.2. The space HomG⇥G(C1c (R1), V ⌦ eV )Proposition 7). Using this description, we prove that ⇣G⇡ does not extend toC1c (R1)K⇥K for all the irreducible spherical representations on our list.On the other hand, for the trivial representation, the zeta-integral (whichis just the linear functional f 7! RG f(g) dg) extends to C1c (R1)K⇥K , andthen fails to extend to C1c (M)K⇥K (see Proposition 8).Finally, we use similar methods (which become much more combinato-rially complicated when working with Iwahori-fixed vectors) to study theSteinberg representation in Chapter 8. There we obtain that the zeta-integral on G does not extend to R1 for the Steinberg representation atthe Iwahori-fixed level.3.2 The space HomG⇥G(C1c (R1), V ⌦ eV )In this section, we want to study the space HomG⇥G(C1c (R1), V ⌦ eV ). Notethat the computation done in this section can also be generalized to thecase of a general n with some modifications, which we do not pursue in thisthesis.3.2.1 A description of C1c (R1) as an induced representationWe use the fact that R1 is a single orbit under G ⇥ G-action on M . Let = [ 1 00 0 ], and let S be the stabilizer of in G⇥G:S = {(s, s0) 2 G⇥G | ss01 = }.Proposition 6. As representations of G⇥G, we have the following iso-morphism:c-IndG⇥GS 1S ⇠= C1c (R1).Proof. Define the following linear operator:F : C1c (R1)! c-IndG⇥GS 1S(f : R1 ! C) 7! Ff :G⇥G! C(g, g0) 7! f(g1g0)We first prove that Ff 2 c-IndG⇥GS 1S . Let (s, s0) 2 S, (g, g0) 2 G⇥G.It follows from the homeomorphism S\G⇥G ⇠= R1 as G⇥G-spaces thatf is compactly supported on R1 if and only if Ff is compactly supportedmodulo S. Moreover, f is locally constant if and only if there exists an open263.2. The space HomG⇥G(C1c (R1), V ⌦ eV )subgroup K 0 of G⇥G such that f 2 C1c (R1)K0 , and therefore, Ff is fixedby K 0 as well.Finally, we have:Ff((s, s0)(g, g0)) = Ff(sg, s0g0) = f(g1s1s0g0) = f(g1g0) = Ff(g, g0).Define the following inverse linear operator:G : c-IndG⇥GS 1S ! C1c (R1)(f : G⇥G! C) 7! Gf :R1 ! Cr 7! f(g, g0), where g1g0 = r.Such pair (g, g0) exists because G⇥G acts transitively on R1. To show thatGf is well-defined, suppose that there exist (g, g0), (h, h0) 2 G⇥G such thatg1g0 = r = h1h0. Then, (hg1, h0g01) 2 S, andf(g, g0) = f((hg1, h0g01)(g, g0)) = f(h, h0).It is clear that G = F1. It remains to show that F is a morphism ofG⇥G-representations. Let (g, g0), (x, y) 2 G⇥G,((g, g0) · Ff)(x, y) = (Ff)(xg, yg0) = f(g1x1yg0)= ((g, g0) · f)(x1y) = F((g, g0) · f)(x, y)Thus, C1c (R1) ⇠= c-IndG⇥GS 1S as G⇥G-representations.Remark 5. This characterization of C1c (R1) is di↵erent from that obtainedbelow from Lemma 12 for a general n. If we apply Lemma 12, we would getC1c (R1) ⇠= IndG⇥GB⇥B C1c (k), where the action of B ⇥ B on C1c (k) is givenby:(⇣hb1 b20 b4i,hb01 0b03 b04i⌘· f)(x) = f(b11 xb01),for f 2 C1c (k).3.2.2 The computation of HomG⇥G(C1c (R1), V ⌦ eV )Our goal is to find a criteria for a representation (⇡, V ) to intertwine withC1c (R1). Namely, we want to find all (⇡, V ) such thatHomG⇥G(C1c (R1), V ⌦ eV ) 6= {0}.273.2. The space HomG⇥G(C1c (R1), V ⌦ eV )Let (⇡, V ) be an irreducible admissible representation of G. In order tocompute HomG⇥G(C1c (R1), V ⌦ eV ), observe that:HomG⇥G(C1c (R1), V ⌦ eV ) ⇠= HomG⇥G(eV ⌦ V, ^C1c (R1)).As shown above in Proposition 6, C1c (R1) ⇠= c-IndG⇥GS 1S , so, by the[[3], Duality Theorem p.56], ^C1c (R1) ⇠= IndG⇥GS S where S is the moduluscharacter of S. We start by computing S explicitly.Lemma 4. 1. Explicitly, the set S is the subset of B ⇥ B consisting ofpairs of matrices with the same element in the upper-left corner:S =n⇣⇥ s1 s20 s4⇤,hs1 0s03 s04i⌘2 G⇥Go.2. The modulus character of S is:S :S ! C⇥⇣⇥ s1 s20 s4⇤,hs1 0s03 s04i⌘7!s04s4 .Proof. 1. Suppose that s = [ s1 s2s3 s4 ] and s0 =hs01 s02s03 s04iare such that (s, s0) 2 S.Then, from the equality = ss01, we get:[ 1 00 0 ] =1s01s04 s02s03hs1s04 s1s02s3s04 s3s02i,) s1s04s01s04 s02s03= 1, s1s02 = 0, s3s04 = 0, s3s02 = 0.Therefore, s3 = s02 = 0, and s1 = s01. The converse holds by reversingthese implications.2. For (x, y) 2 S, we denote by (x, y) = (txnx, tyny) the correspondingdecomposition in TN ⇥TN . Using this decomposition, we may definethe following linear functional on the space C1c (S): 7!ZT⇥TZN⇥N(txnx, tyny) dµT⇥T (tx, ty)dµN⇥N (nx, ny).One can verify that the functional defined above is left S-invariant,and therefore, is a left Haar integral. Next, we compute the modulus283.2. The space HomG⇥G(C1c (R1), V ⌦ eV )of S. Given (s, s0) =⇣⇥ s1 s20 s4⇤,hs1 0s03 s04i⌘2 S, we may use the followingdecomposition:⇣⇥ s1 s20 s4⇤,hs1 0s03 s04i⌘=0BB@⇥ s1 00 s4 ⇤| {z }tsh1 s2s110 1i| {z }ns,hs1 00 s04i| {z }ts0h1 0s03s014 1i| {z }ns01CCA ,Then,ZS((x, y)(s, s0)) d(x, y)=ZT⇥TZN⇥N(txsnxs, tys0nys0) dµT⇥T (tx, ty)dµN⇥N (nx, ny)=ZT⇥TZN⇥N(txtst1s nxtsns, tyts0t1s0 nyts0ns0) dµT⇥T (tx, ty)dµN⇥N (nx, ny).Using the isomorphism N ⇥N ! k ⇥ k which maps [ 1 a0 1 ] , ⇥ 1 0b 1 ⇤ 7!(a, b) and identifying the Haar measures on both spaces, we have that:ZN⇥N(t1s nxts, t1s0 nyts0) dµN⇥N (nx, ny)=Zk⇥k(⇣h1 s11 as40 1i,h1 0s1bs014 1i⌘dµk⇥k(a, b)=s04s4 Zk⇥k([ 1 a0 1 ] ,⇥1 0b 1⇤dµk⇥k(a, b)=s04s4 ZN⇥N(nx, ny) dµN⇥N (nx, ny).Hence, ZS((x, y)(s, s0)) d(x, y) =s04s4 ZS(x, y)d(x, y),and we obtain the desired modulus character of S.Now, given a representation (⇡, V ), we want to fix the central characterwhich allows us to obtain a nice description of the subspace of C1c (R1) thatintertwines with V ⌦ eV (see Lemma 6). More precisely, let Z(G⇥G) be thecentre of G⇥G, and ↵ the central character of our irreducible representation293.2. The space HomG⇥G(C1c (R1), V ⌦ eV )(⇡, V ). Then, the central character of V ⌦ eV has the form ! := ↵⌦↵1. Forthe purpose of computing HomG⇥G(C1c (R1), V ⌦ eV ), only the subrepresen-tation of IndG⇥GS S on which the centre of G⇥G acts by !1 matters (seeLemma 5). Moreover, this subrepresentation is, in fact, a principal seriesrepresentation of G⇥G, as we shall see in Lemma 6.Definition 8. Let ! be a character of Z(G⇥G) and I the identity elementof G. Define the space(IndG⇥GS S)! = {f 2 IndG⇥GS S | (aI, a0I) · f = !(a, a0)f, 8(a, a0) 2 k⇥ ⇥ k⇥}.Lemma 5. Let V be an irreducible admissible representation of G withcentral character ↵, and let ! = ↵⌦ ↵1. ThenHomG⇥G(eV ⌦ V, IndG⇥GS S) = HomG⇥G(eV ⌦ V, (IndG⇥GS S)!1).Proof. It suces to show that if ⇠ 2 HomG⇥G(eV ⌦V, IndG⇥GS S), then ⇠(eV ⌦V ) ✓ (IndG⇥GS S)!1 , and by linearity it suces to show this containmentfor the images of simple tensors. Note that the central character of eV ⌦ Vis !1. Then, for ⇠ 2 HomG⇥G(eV ⌦ V, IndG⇥GS S), ev0 ⌦ v 2 eV ⌦ V , and(aI, a0I) 2 Z(G⇥G), we have that:(aI, a0I) · ⇠(ev0 ⌦ v) = ⇠((aI, a0I) · (ev0 ⌦ v))= ⇠(!1(a, a0)(ev0 ⌦ v)) = !1(a, a0)⇠(ev0 ⌦ v),where the last equality follows from linearity of ⇠, since !(a, a0) is a scalar.Hence, ⇠(ev0 ⌦ v) 2 (IndG⇥GS S)!1 , and the lemma follows.As above, let ! = ↵⌦ ↵1 be a character of Z(G⇥G). Next, we wouldlike to obtain a nice description of the space (IndG⇥GS S)!1 . We define arepresentation of B ⇥B as follows:Definition 9. Given a character ↵ of k⇥, define the one-dimensional rep-resentation ↵ by:↵ :B ⇥B ! C⇥⇣hb1 b20 b4i,hb01 0b03 b04i⌘7! ↵✓b01b1◆ b1b04b01b4 .Lemma 6. Let ↵ be a character of k⇥, and let ! = ↵ ⌦ ↵1 be the corre-sponding character of Z(G⇥G). As G⇥G-representations,(IndG⇥GS S)!1 ⇠= IndG⇥GB⇥B ↵.303.2. The space HomG⇥G(C1c (R1), V ⌦ eV )Proof. Both (IndG⇥GS S)!1 and IndG⇥GB⇥B ↵ are spaces consisting of lo-cally constant C-valued functions on G⇥G on which G⇥G acts by righttranslation. Thus, it suces to check that they coincide as sets. Letf 2 (IndG⇥GS S)!1 . We need to prove that for (b, b0) 2 B ⇥B,f(b, b0)(x, y) = ↵(b, b0)f(x, y).Let b =hb1 b20 b4iand b0 =hb01 0b03 b04i. Then, we can write (b, b0) = (1, z0)(s, s0),where z0 is the diagonal matrix b01b1I, and (s, s0) =✓hb1 b20 b4i,b1 0b1b03b01b1b04b01◆2 S.Recall that !1 = ↵1 ⌦ ↵. Then, we have:f(b, b0)(x, y)= f((1, z0)(s, s0)(x, y)) = f((s, s0)(x, y)(1, z0))= ↵✓b01b1◆f((s, s0)(x, y)) = ↵✓b01b1◆S(s, s0)f(x, y)= ↵✓b01b1◆ b1b04b01b4 f(x, y) = ↵(b, b0)f(x, y),and thus f 2 IndG⇥GB⇥B ↵.Conversely, let f 0 2 IndG⇥GB⇥B ↵. Observe that S ✓ B⇥B and ↵ |S= Sby Lemma 4, so that f 0 2 IndG⇥GS S . Moreover, for any (aI, a0I) 2 Z(G⇥G),we have that:(aI, a0I) · f 0(x, y) = f 0((x, y)(aI, a0I)) = f 0((aI, a0I)(x, y))= ↵✓a0a◆f 0(x, y) = !1(a, a0)f 0(x, y).Thus, f 0 2 (IndG⇥GS S)!1 .Remark 6. To obtain a similar result for a general n, we would need to fix astandard parabolic subgroup P of GLn and an irreducible representation Pof P . In the case n = 2, the computation is greatly simplified because GL2only has a unique standard parabolic subgroup (up to conjugation), namely,the Borel subgroup B. Moreover, S\(B ⇥ B) ⇠= k⇥, so instead of an irre-ducible representation of B, which is a pair of characters, it suces to fixone character of k⇥.Combining Lemmas 5 and 6, we get:HomG⇥G(eV ⌦ V, (IndG⇥GS S)!1) ⇠= HomG⇥G(eV ⌦ V, IndG⇥GB⇥B ↵). (3.1)313.2. The space HomG⇥G(C1c (R1), V ⌦ eV )Now we can pass to Jacquet modules, and apply Frobenius reciprocity.We have that:HomG⇥G(eV ⌦ V, (IndG⇥GS S)!1) ⇠= HomG⇥G(eV ⌦ V, IndG⇥GB⇥B ↵)⇠= HomT⇥T ((eV ⌦ V )N⇥N ,↵)⇠= HomT⇥T (eVN ⌦](eVN ),↵),where the last isomorphism is obtained from VN⇠=](eVN ) by [[5], Corollary4.2.5 p.42].Theorem 6. Let (⇡, V ) be an irreducible admissible representation of G.Then, the space HomG⇥G(C1c (R1), V ⌦ eV ) is:• one-dimensional if (⇢, V ) is |.|det2, StG, |.| 12⇥, where 6= |.| 12 , |.| 32 ,• trivial otherwise.Proof. We use the classification of irreducible representation (⇡, V ) ofG, andin some cases, the explicit knowledge of their Jacquet modules, to computeHomT⇥T (eVN ⌦](eVN ),↵). As a representation of T ⇥ T , we can denote ↵as (↵1|.|⌦ |.|1)⌦ (↵|.|1 ⌦ |.|). We have the following cases:• Supercuspidal representations (⇡, V ): Then, (e⇡, eV ) is also supercusp-idal, and hence, eVN ⇠= {0}. Thus, HomT⇥T (eVN ⌦](eVN ),↵) = {0}.• One dimensional representations (⇡, V ) ⇠= ( det2,C): In this case,↵ = 2, (e⇡, eV ) ⇠= (1 det2,C), and its Jacquet module is eVN ⇠=1 ⌦ 1.HomT⇥T (eVN ⌦](eVN ),↵) ⇠=HomT⇥T(1 ⌦ 1)⌦ ( ⌦ ), (2|.|⌦ |.|1)⌦ (2|.|1 ⌦ |.|) ,which, by inspection, is isomorphic to C if = |.|, and {0} otherwise.• Special representations (⇡, V ) ⇠= StG:In this case, ↵ = 2, (e⇡, eV ) ⇠= 1 StG, and eVN ⇠= 1|.| ⌦ 1|.|1.Therefore,HomT⇥T (eVN ⌦](eVN ),↵) ⇠=HomT⇥T(1|.|⌦ 1|.|1)⌦ (|.|1 ⌦ |.|), (2|.|⌦ |.|1)⌦ (2|.|1 ⌦ |.|) ,which is isomorphic to C if = 1, and {0} otherwise.323.2. The space HomG⇥G(C1c (R1), V ⌦ eV )• Irreducible principal series representations (⇡, V ) ⇠= 1 ⇥ 2 where112 6= |.|±1. Then, the central character ↵ = 12, the contragredi-ent (e⇡, eV ) ⇠= 11 ⇥ 12 , and we have two sub-cases.– If 1 6= 2, theneVN ⇠= (11 |.| 12 ⌦ 12 |.| 12 ) (12 |.| 12 ⌦ 11 |.| 12 ).Thus, eVN ⌦](eVN ) is a direct sum of 4 one-dimensional summands,and we study when each summand of eVN ⌦](eVN ) is isomorphic to↵ = (11 12 |.|⌦ |.|1)⌦ (12|.|⌦ |.|).Summand of eVN ⌦](eVN ) Summand is isomorphic to ↵ if(11 |.|12 ⌦ 12 |.|12 )⌦ (1|.| 12 ⌦ 2|.| 12 ) 2 = |.| 12(11 |.|12 ⌦ 12 |.|12 )⌦ (2|.| 12 ⌦ 1|.| 12 ) 1 = 2 = |.| 12 (excluded since 1 6= 2)(12 |.|12 ⌦ 11 |.|12 )⌦ (1|.| 12 ⌦ 2|.| 12 ) 1 = 2 = |.| 12 (excluded since 1 6= 2)(12 |.|12 ⌦ 11 |.|12 )⌦ (2|.| 12 ⌦ 1|.| 12 ) 1 = |.| 12Since 1 ⇥ 2 ⇠= 2 ⇥ 1 for 12 6= |.|±1, we have thatHomT⇥T (eVN ⌦](eVN ),↵) ⇠= Cif (⇡, V ) ⇠= |.| 12 ⇥ 2 and 2 6= |.| 12 , |.| 32 .– If 1 = 2, then eVN is a 2-dimensional indecomposable represen-tation of T , which is the non-split extension in the following shortexact sequence:0! 11 |.|12 ⌦ 11 |.|12 ! eVN ! 11 |.| 12 ⌦ 11 |.| 12 ! 0.Thus, eVN ⌦](eVN ) has a one-dimensional quotient (11 |.| 12 ⌦11 |.|12 ) ⌦ (1|.| 12 ⌦ 1|.| 12 ), which is isomorphic to ↵ if andonly if 1 = |.| 12 . Hence,HomT⇥T (eVN ⌦](eVN ),↵) ⇠= Cif (⇡, V ) ⇠= |.| 12 ⇥ |.| 12 .333.2. The space HomG⇥G(C1c (R1), V ⌦ eV )3.2.3 An explicit generator of HomG⇥G(C1c (R1), V ⌦ eV )In this section, we will construct an explicit generator of HomG⇥G(C1c (R1), V⌦eV ) for almost all irreducible representations (⇡, V ) of G such thatHomG⇥G(C1c (R1), V ⌦ eV ) ⇠= C,except for the irreducible principal series representation |.| 12 ⇥ |.| 12 . More ex-plicitly, we construct an explicit generator when (⇡, V ) is one of the followingrepresentations:|.| det2, StG, |.| 12 ⇥ where 6= |.|± 12 , |.| 32 . (3.2)Note that this section is separate from the rest of the thesis and while it ispossible to generalize, we do not pursue this approach for n > 2.The main reason why we exclude the representation |.| 12 ⇥ |.| 12 in thissection is because this particular construction of the generator involves theJacquet module of (⇡, V ). If (⇡, V ) is such that HomG⇥G(C1c (R1), V ⌦ eV ) ⇠=C, then it is an irreducible subspace of |.| 12 ⇥ where is a character of k⇥.The representation |.| 12 ⇥ |.| 12 is the unique case that appears in rank 1 andits Jacquet module VN is a two dimensional indecomposable representationof T . Furthermore, recall that we are interested in this construction mainlyto determine if ext⇤ is surjective for the case of (⇡, V ) being the Steinbergrepresentation, and the surjectivity of ext⇤ for the representation |.| 12 ⇥ |.| 12can be proved later via Ext computation (see Lemma 9).Observe that in order to construct an explicit generator of HomG⇥G(C1c (R1), V⌦eV ), it suces to construct a linear functional C1c (R1) ⇥ eV ⇥ V ! C. By[[3], Proposition 3.4 p.30], there exists a unique (up to scalar) positive semi-invariant measure µ˙ on S\G⇥G such that for every f 2 (C1c (S\G⇥G), S),we have that:ZS\G⇥Gf(xg, yg0)dµ˙(x, y) =ZS\G⇥Gf(x, y)dµ˙(x, y), 8(g, g0) 2 G⇥G.We will make use of the existence of this positive semi-invariant measure inorder to define a linear functional on C1c (R1)⇥ eV ⇥ V .Before defining such a linear functional, observe that if (⇡, V ) is suchthat HomG⇥G(C1c (R1), V ⌦ eV ) ⇠= C and (⇡, V ) 6= |.| 12 ⇥ |.| 12 , then eVN ⌦ VNhas a unique one-dimensional direct summand Q on which T ⇥ T acts via(1|.| 12 ⌦ |.|1) ⌦ (|.| 12 ⌦ |.|). We define a pairing h., .i,Q on V ⌦ eV asfollows: for v 2 V and ev0 2 eV , denote by vN (resp. ev0N ) the image of v343.2. The space HomG⇥G(C1c (R1), V ⌦ eV )(resp. ev0) under the quotient map V ! VN (resp. eV ! eVN ). Then, letev0N ⌦ vN be the image of ev0N ⌦ vN under the quotient map from eVN ⌦ VNto Q. Note that on Q, we have a natural pairing BQ. Let:hev0, vi,Q := BQ(ev0N ⌦ vN ).We make the following two important observations:• When (⇡, V ) is either the Steinberg representation or |.| det2, thepairing hev0, vi,Q defined above is simply the canonical pairing hev0, vion eV ⌦ V . This is a consequence of [[5], Proposition 4.2.3 p.41], andthe fact that eVN ⌦ VN is exactly one-dimensional in those cases.• For (s, s0) 2 S, then we shall see later that:he⇡(s)ev0,⇡(s0)vi,Q = S(s, s0)hev0, vi,Q.In fact, this is the motivating reason why we define the pairing oneV ⌦ V this way.Using the above defined pairing, we first prove the following auxiliary lemma:Lemma 7. For every f 2 C1c (R1), ev0 2 eV , and v 2 V , define:fev0,v :G⇥G! C(x, y) 7! he⇡(x)ev0,⇡(y)vi,Qf(x1y).Then, fev0,v is in (C1c (S\G⇥G), S).Proof. We want to show that for (s, s0) 2 S, then fev0,v(sx, s0y) = S(s, s0)fev0,v(x, y).We have that:fev0,v(sx, s0y) = he⇡(sx)ev0,⇡(s0y)vi,Qf(x1s1s0y) = he⇡(sx)ev0,⇡(s0y)vi,Qf(x1y),since s1s0 = . So, to obtain the desired equality, it suces to showthat he⇡(sx)ev0,⇡(s0y)vi,Q = S(s, s0)he⇡(x)ev0,⇡(y)vi,Q. Let s = ⇥ s1 s20 s4 ⇤ ands0 =hs1 0s03 s04i. Since s0 =h1 0s03s11 1i hs1 00 s04i, we have that:(⇡(s0y)v)N = (⇡(h1 0s03s11 1i)⇡(hs1 00 s04i)⇡(y)v)N= (⇡(hs1 00 s04i)⇡(y)v)N (sinceh1 0s03s11 1i2 N and N acts trivially on VN )= ⇡N (hs1 00 s04i)(⇡(y)v)N .353.2. The space HomG⇥G(C1c (R1), V ⌦ eV )Similarly, we get (e⇡(sx)ev0)N = e⇡N (⇥ s1 00 s4 ⇤)(e⇡(x)ev0)N . So,he⇡(sx)ev0,⇡(s0y)vi,Q = he⇡(⇥ s1 00 s4 ⇤)e⇡(x)ev0,⇡(h s1 00 s04 i)⇡(y)vi,QSince on Q, the group T ⇥ T acts via (1|.| 12 ⌦ |.|1) ⌦ (|.| 12 ⌦ |.|), weget:he⇡(sx)ev0,⇡(s0y)vi,Q = s04s4 he⇡(x)ev0,⇡(y)v)i,Q = S(s, s0)he⇡(x)ev0,⇡(y)vi,Q,and we obtain the desired equality.Thus, the following linear functional is well-defined:⇣R1 :C1c (R1)⇥ eV ⇥ V ! C(f, ev0, v) 7! ZS\G⇥Ghe⇡(x)ev0,⇡(y)vi,Qf(x1y) dµ˙(x, y). (3.3)In summary, we have provedTheorem 7. 1. Let (⇡, V ) be the Steinberg representation or the one-dimensional representation |.| det2 of GL2. Then the spaceHomG⇥G(C1c (R1), V ⌦ eV ) is one-dimensional, generated by⇣R1⇡ :C1c (R1)! End(V )sm ⇠= V ⌦ eVf 7! v 7!ZS\G⇥Gf(x1y)⇡(x1y)v dµ˙(x, y)!,where End(V )sm is the space of smooth endomorphisms of V .2. Let (⇡, V ) be the irreducible principal series |·| 12⇥, with 6= |·| 12 , |·| 32 .Then the space HomG⇥G(C1c (R1), V ⌦ eV ) is one-dimensional, gener-ated by ⇣R1 defined by the formula (3.3), where Q is the 1-dimensionalquotient of the Jacquet module eVN ⌦ VN on which T ⇥ T acts via(1|.| 12 ⌦ |.|1)⌦ (|.| 12 ⌦ |.|).Remark 7. Observe that if (⇡, V ) is a spherical representation that appearsin rank 1, then it is the irreducible quotient of |.| 12 ⇥ , and its associatedL-function LG(⇡, s) has a pole at s = 12 . Moeglin, Vigne´ras, and Wald-spurger noted that for such (⇡, V ), the residue of ⇣M⇡ at s = 12 generatesthe space HomG⇥G(C1c (M), V ⌦ eV ) (cf. [[15], III.7 p.65]). On the otherhand, it follows from Proposition 8 and the long exact sequence (2.8) that thegenerator of the space HomG⇥G(C1c (M), V ⌦ eV ) comes from the invariantintegral on R1. Thus, in fact, Equation (3.3) is a realization of the residueof ⇣M⇡ at s = 12 as an integral over S\(G⇥G).363.3. The descending rank filtration for GL23.3 The descending rank filtration for GL2In this section, we will use the descending rank filtration for GL2 to showthat if (⇡, V ) is a representation that appears in either rank 0 or 1, thenthe space HomG⇥G(C1c (R1), V ⌦ eV ) is one-dimensional (Theorem 9). Ourapproach is to first realize an explicit measure dµ˙(x, y) on S\G⇥G (Section3.3.1). Then, through computation, we show that the generator ⇣R1⇡ ofHomG⇥G(C1c (R1), V ⌦ eV ), which we constructed in 3.2.3, can be extendedto ⇣R1⇡ via meromorphic continuation (Theorem 8). Note that while thisproves the surjectivity of the map ext⇤ for a representation (⇡, V ) that islisted in (3.2), it was mainly needed to prove the surjectivity in the caseof Steinberg representation. For all representations (⇡, V ) which appear inrank 1 except for StG, we can simply compute the groupExt1GLn⇥GLn(C1c ({0}), V ⌦ eV )in order to show that ext⇤ is surjective (Lemma 9).Note that we will only use the descending rank filtration in this sectionfor GL2 and nowhere else, and while it can be generalized to an arbitraryn, we do not pursue this approach. Thus, this section can be read indepen-dently from the rest of the thesis.3.3.1 An explicit form of dµ˙(x, y) on S\G⇥GAs we will need to do computation on S\G⇥G in the upcoming section inorder to study the convergence of certain integrals, it will be convenient torealize the measure dµ˙(x, y) in a more concrete form.Let K be the maximal compact subgroup of G. Using the decompositionG = BK = BK, we may write (g, g0) = (b, b0)(k, k0) where g, g0 2 G, k, k0 2K, b 2 B, b0 2 B. Furthermore, if b =hb1 b20 b4iand b0 =hb01 0b03 b04i, then:(b, b0) =⇣b,hb1 0b03b1b011 b04i⌘⇣1,hb01b11 00 1i⌘.Note that⇣b,hb1 0b03b1b011 b04i⌘2 S.Therefore, we define a linear functional on the space (C1c (S\G⇥G), S)as follows: 7!Zk⇥ZK⇥K(k, [ t 00 1 ] k0)d(k, k0)d⇥t. (3.4)373.3. The descending rank filtration for GL2Next, using a similar argument to that given in [[3], Corollary 7.6 p.55], weshow that the above linear functional is a realization of the positive semi-invariant measure dµ˙(x, y) on S\G⇥G.Lemma 8. Haar measure on K⇥K⇥k⇥ can be normalized so that for anyf 2 (C1c (S\G⇥G), S),ZS\G⇥Gf(x, y) dµ˙(x, y) =Zk⇥ZK⇥K(k, [ t 00 1 ] k0)dkdk0d⇥t.Proof. We have that:(K ⇥ ⇥ k⇥ 00 1 ⇤K) \ S = n⇣⇥ s1 s20 s4 ⇤ , h s1 0s03 s04 i⌘ | s1, s4, s04 2 O⇥, s2, s03 2 Oo ,which is a compact subgroup on which the restriction of S is trivial. Therestriction of functions is an isomorphism from C1c (S\G⇥G, S) toC1c (((K ⇥⇥k⇥ 00 1⇤K) \ S)\(K ⇥ ⇥ k⇥ 00 1 ⇤K)), 1).Thus, dµ˙(x, y) restricts to a semi-invariant measure on C1c (((K⇥⇥k⇥ 00 1⇤K)\S)\(K ⇥ ⇥ k⇥ 00 1 ⇤K)), 1), which coincides with the restriction of any Haarmeasure on K ⇥ ⇥ k⇥ 00 1 ⇤K up to a scalar multiple. Thus, we may normalizethe Haar measure on K⇥ ⇥ k⇥ 00 1 ⇤K such that the desired equality holds.3.3.2 An extension of ⇣R1⇡ from C1c (R1) to C1c (R1)In this section, (⇡, V ) is assumed to be one of those representations for whichwe have constructed ⇣R1⇡ , which are explicitly listed in (3.2). We shall showthat there exists a lift of ⇣R1⇡ from HomG⇥G(C1c (R1), V ⌦ eV ) to the spaceHomG⇥G(C1c (R1), V ⌦ eV ) via meromorphic continuation. In particular,this shows that the map ext⇤ in the sequence (2.6) is surjective for theSteinberg representation.For a fixed s 2 C, denote by ⇣R1s the following map defined via formalzeta integral:⇣R1s :C1c (R1)⇥ eV ⇥ V ! C(f, ev0, v) 7! he⇡(x)ev0,⇡(y)vi|.|s,Qf(x1y) dµ˙(x, y).Recall that the Steinberg representation is the irreducible subspace of |.| 12 ⇥ |.| 12 ,which corresponds to = |.| 12 , that is, s = 12 . So, if ⇣R1 12is well-defined,then it will give rise to the lift of ⇣R1⇡ for the Steinberg representation. In383.3. The descending rank filtration for GL2order to show that ⇣R1 12is well-defined, we consider the following formalzeta integral:⇣R1 :C1c (R1)⇥ eV ⇥ V ⇥ C! C(f, ev0, v, s) 7! ZS\G⇥Ghe⇢(x)ev0, ⇢(y)vi|.|s,Qf(x1y) dµ˙(x, y).Next, we show that ⇣R1 is well-defined at all complex values s except fors = 12 . Note that the value s =12 corresponds to the representation |.|12⇥|.| 12 ,which we already exclude in the construction of ⇣R1 . The computation usedto prove the next theorem is similar to that in the proof of [[3], Proposition26.1 p.163]. The main idea is to reduce the computation of integrals onS\G⇥G to those on k⇥, and then use the result from Tate’s thesis.Theorem 8. For ⇣R1 as defined above, there exists s0 2 R such that theintegral ⇣R1(f, ev0, v, s) converges absolutely and uniformly in vertical stripsin the region <(s) > s0 for all f 2 C1c (R1), v 2 V, ev0 2 eV . The inte-gral represents a rational function in qs+12 . Moreover, the meromorphiccontinuation of ⇣R1 to the whole complex s-plane has a pole at s = 12 .Proof. Let f 2 C1c (R1), v 2 V , and ev0 2 eV . Denote by K0 a compactopen subgroup of G such that f is K0-bi-invariant, v 2 V K0 and ev0 2 eV K0 .Let be a set of representatives of (K ⇥K)/(K0 ⇥K0); in particular, isfinite. We will reduce the integral on S\G⇥G to a finite sum of integralson k⇥. First, using the explicit form of dµ˙(x, y), we have that:⇣R1(f, ev0, v, s)=ZS\G⇥Ghe⇡(x)ev0,⇡(y)vi|.|s,Qf(x1y) dµ˙(x, y)=Zk⇥ZK⇥Khe⇡(k)ev0,⇡([ t 00 1 ] k0)vi|.|s,Qf(k1 [ t 00 1 ] k0) d(k, k0)d⇥t=Zk⇥X(ki,k0j)2ZK0⇥K0he⇡(kik)ev0,⇡([ t 00 1 ] k0jk0)vi|.|s,Qf(k1k1i [ t 00 0 ] k0jk0) d(k, k0)d⇥t.Since K0 was chosen so that f is K0-bi-invariant, v 2 V K0 and ev0 2 eV K0and T ⇥T acts on Q via (1|.| 12 ⌦ |.|1)⌦ (|.| 12 ⌦ |.|), for = |.|s, we get393.3. The descending rank filtration for GL2that:Zk⇥X(ki,k0j)2ZK0⇥K0he⇡(kik)ev0,⇡([ t 00 1 ] k0jk0)vi|.|s,Qf(k1k1i [ t 00 0 ] k0jk0) d(k, k0)d⇥t= µ(K0 ⇥K0)Zk⇥X(ki,k0j)2|t|s 12 he⇡(ki)ev0,⇡(k0j)vi|.|s,Qf(k1i [ t 00 0 ] k0j) d⇥t= µ(K0 ⇥K0)X(ki,k0j)2he⇡(ki)ev0,⇡(k0j)vi|.|s,Q✓Zk⇥f(k1i [ t 00 0 ] k0j)|t|s12 d⇥t◆,where the last equality holds because is finite. Thus, it suces to showthat for (ki, k0j) 2 , the integralZk⇥f(k1i [ t 00 0 ] k0j)|t|s12 d⇥thas the desired properties. For (ki, k0j) 2 , define:fki,k0j :k ! Ct 7! f(k1i [ t 00 0 ] k0j).Note that fki,k0j 2 C1c (k), and since for 2 C1c (k), the integralZk⇥(t)|t|s 12 d⇥tconverges absolutely and uniformly in vertical strips in the region <(s) > s0for some s0 2 R; it represents a rational function in qs+ 12 , and its mero-morphic continuation to the whole s-plane only has a pole at s = 12 , whichcompletes the proof.Observe that as a corollary of Theorem 8, the map ext⇤ in the se-quence (2.6) is surjective for the Steinberg representation. To show thatext⇤ is surjective for the remaining representations (⇡, V ) for which thespace HomG⇥G(C1c (R1), V ⌦ eV )) is non-trivial, we will compute the groupExt1G⇥G(C1c ({0}), V ⌦ eV ).Lemma 9. Let (⇡, V ) be |.| det2 or |.| 12 ⇥ , where 6= |.| 12 , |.| 32 . Then,Ext1G⇥G(C1c ({0}), V ⌦ eV ) ⇠= {0}.403.3. The descending rank filtration for GL2Proof. If (⇡, V ) ⇠= (|.|det2), then the group Ext1G⇥G(C1c ({0}), V ⌦ eV ) clas-sifies G⇥G-isomorphism classes of two-dimensional representations (⌧,C2)of G⇥G that fit in the short exact sequence:0! (|.| det2,C)! (⌧,C2)! (12,C)! 0.An analogous computation to that in the proof of [[2], Proposition 4.5.8p.482] shows that the above short exact sequence splits, and ⌧(g, g0) =h| det2(gg01)| 00 1iis the unique such representation of G⇥G up to isomor-phism.On the other hand, if (⇡, V ) ⇠= |.| 12 ⇥ ⇠= IndGB |.|⌦|.|12 , then (e⇡, eV ) ⇠=|.| 12 ⇥ 1 ⇠= IndGB 1⌦ 1|.|12 andV ⌦ eV ⇠= IndG⇥GB⇥B(|.|⌦ |.| 12 )⌦ (1⌦ 1|.| 12 ).Using [[4], Theorem A.12 p.927], we may pass to Jacquet modules, and get:Ext1G⇥G(C1c ({0}), V ⌦ eV ) ⇠= Ext1G⇥G ⇣1G⇥G, IndG⇥GB⇥B(|.|⌦ |.| 12 )⌦ (1⌦ 1|.| 12 )⌘⇠= Ext1T⇥T⇣1T⇥T , (|.|⌦ |.| 12 )⌦ (1⌦ 1|.| 12 )⌘.Observe that for any character , we have that(|.|⌦ |.| 12 )⌦ (1⌦ 1|.| 12 ) 6⇠= 1T⇥T .Thus, it follows from the observation that T⇥T ⇠= (k⇥)4 and [[2], Proposition4.5.8 p.482] that for any character of k⇥,Ext1T⇥T (1T⇥T , (|.|⌦ )⌦ (1⌦ 1|.|1)) = {0},as desired.Using the above results, we obtain a complete description for the spaceHomG⇥G(C1c (R1), V ⌦ eV ):Theorem 9. Let (⇡, V ) be an irreducible admissible representation of G.Then, HomG⇥G(C1c (R1), V ⌦ eV ) is:• one-dimensional if (⇡, V ) is |.| det2,12, StG, |.| 12 ⇥ , where 6=|.| 12 , |.| 32 ,• trivial otherwise.413.4. Spherical representations and Satake transformsProof. Recall the following exact sequence:0! HomG⇥G(C1c ({0}), V ⌦ eV ) res⇤! HomG⇥G(C1c (R1), V ⌦ eV )ext⇤! HomG⇥G(C1c (R1), V ⌦ eV )! Ext1G⇥G(C1c ({0}), V ⌦ eV )! . . .We proceed by cases. Observe that if (⇡, V ) is not one of the followingrepresentations: |.| det2,12, StG, |.| 12 ⇥ , where 6= |.| 12 , |.| 32 , then bothHomG⇥G(C1c ({0}), V ⌦ eV ) and HomG⇥G(C1c (R1), V ⌦ eV ) are trivial. So,in those cases, HomG⇥G(C1c (R1), V ⌦ eV ) ⇠= {0}.If (⇡, V ) is the trivial representation, then HomG⇥G(C1c ({0}), V ⌦ eV ) ⇠=C, but HomG⇥G(C1c (R1), V ⌦ eV ) ⇠= {0}. Thus, HomG⇥G(C1c (R1), V ⌦eV ) ⇠= C.If (⇡, V ) is one of the following representations:|.|det2, StG, |.| 12⇥ where 6= |.| 12 , |.| 32 , then HomG⇥G(C1c ({0}), V⌦eV ) ⇠= {0} and HomG⇥G(C1c (R1), V⌦eV ) ⇠= C. In this case, for all but the Steinberg representation,Ext1G⇥G(C1c ({0}), V ⌦ eV ) ⇠= {0},which shows that ext⇤ is surjective. For the Steinberg representation, wehave constructed an explicit generator for HomG⇥G(C1c (R1), V ⌦ eV ), andTheorem 8 shows that such a generator can be lifted to an element ofHomG⇥G(C1c (R1), V⌦eV ). Hence, in all cases where HomG⇥G(C1c (R1), V⌦eV ) ⇠= C, the map ext⇤ is surjective, and HomG⇥G(C1c (R1), V ⌦ eV ) ⇠= C,as desired.3.4 Spherical representations and Sataketransforms3.4.1 Extension of ⇣G⇡ using the ascending rank filtrationLet H be the spherical Hecke algebra H(G,K). In this section, we studythe extension problem of ⇣G⇡ using the ascending rank filtration. More ex-plicitly, we focus on the case where (⇡, V ) is a spherical representationof GL2 such that ⇣G⇡ does not extend H-equivariantly to an element inHomH(C1c (M)K⇥K , V K⌦ eV K). From the ascending rank filtration, we mayview the extension of ⇣G⇡ from C1c (G)K⇥K to C1c (M)K⇥K as a two-step pro-cess: first from C1c (G)K⇥K to C1c (R1)K⇥K , and then from C1c (R1)K⇥Kto C1c (M)K⇥K . Thus, we want to know at which step the extension failsto exist, and in the case of GL2, it is equivalent to finding out if ⇣G⇡ extends423.4. Spherical representations and Satake transformsH-equivariantly to an element in HomH(C1c (R1)K⇥K , V K ⌦ eV K). Notethat as a consequence of Theorem 11, which we will prove in Section 5.1,and the classification of spherical representations, if (⇡, V ) is an irreduciblespherical representation for which ⇣G⇡ does not extend equivariantly to an el-ement in HomH(C1c (M)K⇥K , V K⌦ eV K), then (⇡, V ) is one of the following:12, |.| det2, or |.| 12 ⇥ |.|t where t 6= 12 , 32 .In this section, we use the definitions of Chapter 6 and notation definedthere.The Satake transform of C1c (R1)K⇥KWe discuss the definition of the extension of Satake transform to C1c (Mn)K⇥Kin Section 6.3.4. For now, the reader can either refer to that section, orthink of a naive definition, by summing formal power series. Our goal inthis section is just to do the formal calculations for GL2, to provide the firstguiding example for Chapter 6. Since this section is intended as an exampleof the more general calculations, we also use the notation for test functionsthat we develop in Section 6.3.3: we will deal with functions f⇤, where ⇤is a cone in the co-character lattice; such a function should be understoodas a (formal) sum of characteristic functions of the double cosets KaK,parametrized by the co-characters 2 ⇤ (see Section 6.3.2 for detail). Withthese conventions, we describe the image of C1c (R1)K⇥K under the Sataketransform.Proposition 7. The Satake transform gives a surjective morphism of Heckemodules. Namely,S(C1c (R1)K⇥K)⇣ C[z±11 , z±12 ]W"1 z1z2(1 q 12 z1)(1 q 12 z2)#First, we compute the image under the Satake transform of the charac-teristic functions of the setsKh$⇤2O$⇤2iK,in C1c (R1)K⇥K , where ⇤2 2 Z. Note that these functions project to a basisof C1c (R1)K⇥K under restriction to R1, and their images under the Sataketransform will be used later in the proof of the above proposition.Lemma 10. Using the notation introduced in (6.2), let f⇤1⇤be the charac-teristic function of the setKh$⇤2O$⇤2iK,433.4. Spherical representations and Satake transformswhich is an element in C1c (R1)K⇥K , where ⇤2 2 Z. Then,S(f⇤1⇤) =1 z1z2(1 q 12 z1)(1 q 12 z2)(z1z2)⇤2 .Note that in the above lemma, we denote the characteristic function ofthe setKh$⇤2O$⇤2iK,by f⇤1⇤to be consistent with the notation that will be introduced in (6.2).In this case, the function defined in Lemma 10 corresponds to f⇤1⇤where⇤ = (⇤2) 2 ⇤1⇤2 .Proof. Observe that by (6.3), we have thatf⇤1⇤= f(⇤1⇤ ) + f(F{↵1}⇤1⇤ ) .In this case, (F{↵1}⇤1⇤) = {(⇤2,⇤2)} and (⇤1⇤) = {(1,⇤2) | 1 > ⇤2}.Thus, by applying the Satake transform and using the definition of f⇤1⇤, wehave thatS(f⇤1⇤) = S(f(⇤2,⇤2)) +X2(⇤1⇤ )S(f),We proceed to compute S(f), where = (1,⇤2) for 1 > ⇤2 (when 2 (⇤1⇤)), or = (⇤2,⇤2) (when 2 (F{↵1}⇤1⇤)) using Macdonald’sformula, which will be recalled as Theorem 12 in Section 6.3.1.For = (1,⇤2) where 1 > ⇤2, µM = 1 because the Levi componentM of the centralizer of $ is trivial and12 ($) = (q12 )1(q12 )⇤2 ,[w]($) =(z11 z⇤22 , if w = ()z12 z⇤21 , if w = (12)Thus,S(f(1,⇤2)) =z1 q1z2z1 z2 (q12 z1)1(q12 z2)⇤2 +z2 q1z1z2 z1 (q12 z2)1(q12 z1)⇤2 .On the other hand, for = (⇤2,⇤2) which lies in (F{↵1}⇤1⇤), µM =11+q1because the Levi component M of the centralizer of $ is G, and12 ($) = 1,[w]($) = (z1z2)⇤2 , 8w 2W.443.4. Spherical representations and Satake transformsThus,S(f(⇤2,⇤2)) =11 + q1✓z1 q1z2z1 z2 (z1z2)⇤2 +z2 q1z1z2 z1 (z1z2)⇤2◆= (z1z2)⇤2 .Now we can compute the Satake tranform S(f⇤1⇤).S(f⇤1⇤) = S(f(⇤2,⇤2)) +1X1=⇤2+1S(f(1,⇤2))= (z1z2)⇤2 +1X1=⇤2+1✓z1 q1z2z1 z2 (q12 z1)1(q12 z2)⇤2 +z2 q1z1z2 z1 (q12 z2)1(q12 z1)⇤2◆= (z1z2)⇤2 +z1 q1z2z1 z2 (q 12 z2)⇤2 · (q12 z1)⇤2+11 q 12 z1+z2 q1z1z2 z1 (q 12 z1)⇤2 · (q12 z2)⇤2+11 q 12 z2= 1 +z1 q1z2z1 z2 ·q12 z11 q 12 z1+z2 q1z1z2 z1 ·q12 z21 q 12 z2!(z1z2)⇤2=1 z1z2(1 q 12 z1)(1 q 12 z2)(z1z2)⇤2 .Using the above Lemma, we now prove Proposition 7, which is a specialcase of Proposition 13. Note that in general it is dicult to describe theexplicit image of C1c (Rm)K⇥K under the Satake transform.Proof of Proposition 7. Recall that the space C1c (R1)K⇥K is generated byC1c (G)K⇥K and characteristic functions of sets of the formKh$⇤2O$⇤2iK,where ⇤2 2 Z. Moreover, S(C1c (G)K⇥K) ⇠= C[z±11 , z±12 ]W and by Lemma 10,S chK$⇤2O$⇤2K!=1 z1z2(1 q 12 z1)(1 q 12 z2)(z1z2)⇤2 .Thus, we have thatS(C1c (R1)K⇥K) ✓ C[z±11 , z±12 ]W"1 z1z2(1 q 12 z1)(1 q 12 z2)#.453.4. Spherical representations and Satake transformsIn particular, it follows fromS✓chKhO1iK◆=1 z1z2(1 q 12 z1)(1 q 12 z2)that the Satake map S is surjective andS(C1c (R1)K⇥K)⇣ C[z±11 , z±12 ]W"1 z1z2(1 q 12 z1)(1 q 12 z2)#.Extension of ⇣G⇡ to ⇣R1⇡Proposition 8. Let (⇡, V ) be a spherical representation of GL2 such that ⇣G⇡does not extend H-equivariantly to an element in HomH(C1c (M)K⇥K , V K⌦eV K). Then,(i) if (⇡, V ) is |.|det2 or |.| 12⇥|.|t where t 6= 12 , 32 , then ⇣G⇡ does not extendH-equivariantly to an element in HomH(C1c (R1)K⇥K , V K ⌦ eV K);(ii) if (⇡, V ) is 12, then ⇣G⇡ extends H-equivariantly to an element inHomH(C1c (R1)K⇥K , V K ⌦ eV K).Note that the proof of the above proposition is a particular case of thatof Theorem 14. However, unlike the result for general n which relies onHypothesis 1 in Section 6.3.1, this result for GL2 does not since we can provethat the extended Satake transform is injective on C1c (M)K⇥K throughexplicit computations (see Appendix A).Proof of (i). Recall the following short exact sequence:0! C1c (G)K⇥K ! C1c (R1)K⇥K ! C1c (R1)K⇥K ! 0,which gives the following sequence after the application of the Satake trans-form:0! C[z±11 , z±22 ]W ! C[z±11 , z±12 ]W"1 z1z2(1 q 12 z1)(1 q 12 z2)#! Q1 ! 0,where Q1 is the quotient module. Let (⇡, V ) be |.| det2 or |.| 12 ⇥ |.|t wheret 6= 12 , 32 . Observe that for these (⇡, V ), the map ⇣G⇡ on C1c (G)K⇥K corre-sponds to the composition of the evaluation map eval on C[z±11 , z±22 ]W which463.4. Spherical representations and Satake transformssends z1 7! q 12s and z2 7! qts where t 6= 12 , 32 , and the evaluation ats = 0. To show that the map ⇣G⇡ does not extend H-equivariantly to anelement in HomH(C1c (R1)K⇥K , V K⌦ eV K), we provide an explicit elementf 2 C1c (R1)K⇥K such that eval(S(f)) is undefined at s = 0.Consider the function chKhO1iK2 C1c (R1)K⇥K . Using the explicitcomputation in the proof of Proposition 7, we have thateval(S(chKhO1iK)) =1 q 12t2s(1 qs)(1 q 12ts).Since t 6= 12 , it follows that eval(S(chKhO 1 iK)) is undefined at s = 0.Therefore, the map ⇣G⇡ does not extend H-equivariantly to an element inHomH(C1c (R1)K⇥K , V K ⌦ eV K).Proof of (ii). Let (⇡, V ) be the trivial representation 12 of GL2. Then,the evaluation map eval on C[z±11 , z±22 ]W , that corresponds to ⇣G⇡ , sendsz1 7! q 12s, and z2 7! q 12s. We will show that ⇣G⇡ extends H-equivariantlyto an element in HomH(C1c (R1)K⇥K , V K⌦ eV K) by showing that the sameevaluation map is well-defined on S(C1c (R1)K⇥K . By Proposition 7, sinceS(C1c (R1)K⇥K)⇣ C[z±11 , z±12 ]W"1 z1z2(1 q 12 z1)(1 q 12 z2)#it suces to show that the evaluation of 1z1z2(1q 12 z1)(1q 12 z2)is well-defined ats = 0. We have thateval 1 z1z2(1 q 12 z1)(1 q 12 z2)!=1 q2s(1 qs)(1 q1s) =1 + qs1 q1s ,which is well-defined at s = 0. Thus, the evaluation map is well-defined onS(C1c (R1)K⇥K) and the map ⇣G⇡ extends H-equivariantly to an elementin HomH(C1c (R1)K⇥K , V K ⌦ eV K).47Chapter 4Spectrum of C1c (Rm)4.1 General results on C1c (Rm)4.1.1 A description of C1c (Rm) as an induced representationRecall that one of the objectives is to determine which irreducible admissiblerepresentations (⇡, V ) of GLn give HomGLn⇥GLn(C1c (Rm), V ⌦ eV ) 6= {0}.One of the key tools in computing Hom spaces is Frobenius Reciprocitywhich requires a realization of C1c (Rm) as an induced representation fromsome closed subgroup of GLn⇥GLn. To obtain such a realization, let Xmbe the closed subset of Rm consisting of rank m matrices of the form [ t 00 0 ]where t 2 GLm. Note that we can also choose Xm = {⇥1m 00 0⇤} if we wantto fix a particular representative. Then, Rm = (GLn⇥GLn) · Xm, andLemma 2 would allow us to describe C1c (Rm) as an induced representationfrom the stabilizer subgroup of Xm. Note that this stabilizer is not a point-wise stabilizer but rather the set of elements in GLn⇥GLn that fix Xm asa set. Thus, our first task is to compute the stabilizer of Xm.Lemma 11. For 0 m n, the stabilizer of Xm under the action ofGLn⇥GLn is Pm ⇥ Pm.Proof. Clearly, any (p, p0) 2 Pm ⇥ Pm of the form⇣⇥ p1 p20 p4⇤,hp01 0p03 p04i⌘stabi-lizes Xm. Now, let (g, g0) 2 GLn⇥GLn such that (g, g0) · Xm = Xm. Wemay write g = [ g1 g2g3 g4 ] and g0 =hg01 g02g03 g04iwhere g1, g01 2 Mm, g2, g02 2 Mm,nm,g3, g03 2 Mnm,m and g4, g04 2 Mnm. Then, for every t 2 GLm,(g, g0) · [ t 00 0 ] 2 Xm , 9t0 2 GLm, g [ t 00 0 ] g01 =⇥t0 00 0⇤) g [ t 00 0 ] =⇥t0 00 0⇤g0 )hg1t 0g3t 0i=hg01 g020 0i, g3 = 0, g02 = 0.Moreover, since rank g1 + rank g4 = rank g = n, we get that rank g1 = mand rank g4 = n m. So, g1 2 GLm and g4 2 GLnm. Similarly, we getg01 2 GLm, g04 2 GLnm, and the statement of the lemma follows.484.1. General results on C1c (Rm)By Lemma 2, we get that C1c (Rm) ⇠= IndGLn⇥GLnPm⇥Pm C1c (Xm) as GLn⇥GLn-representations.Definition 10. We denote by (µm, C1c (GLm)) the representation of Pm ⇥Pm on C1c (GLm) given by:(µm(p, p0)f)(x) = f(p11 xp01),where p =⇥ p1 p20 p4⇤, p0 =hp01 0p03 p04iwith p1, p01 2 GLm, p4, p04 2 GLnm, p2 2Mm,nm, p03 2 Mnm,m, and f 2 C1c (GLm), x 2 GLm.Note that C1c (Xm) ⇠= (µm, C1c (GLm)) as Pm⇥Pm-representations. Weobtain the following characterization of C1c (Rm) and its contragredient:Lemma 12. For 0 m n, C1c (Rm) ⇠= IndGLn⇥GLnPm⇥Pm µm and ^C1c (Rm) ⇠=IndGLn⇥GLnPm⇥Pm eµmPm⇥Pm as GLn⇥GLn-representations.This specific choice of realization is particularly useful as it allows us toapply Frobenius Reciprocity in the computation of the Hom spaces. Notethat the above characterization of C1c (Rm) as a parabolically induced repre-sentation of GLn⇥GLn is the same up to notation di↵erences as that givenby Minguez [[13], p.7].4.1.2 The condition on (⇡, V ) so thatHomGLn⇥GLn(C1c (Rm), V ⌦ eV ) 6= {0}Let (⇡, V ) be an irreducible representation of GLn. We say that ⇡ appearsin rank m as a quotient if the space HomGLn⇥GLn(C1c (Rm), V ⌦ eV ) is non-trivial. The goal of this section is to give a necessary and sucient conditionfor (⇡, V ) to appear in rank m as a quotient. To that end, we first considerthe two extreme cases where m = 0 or m = n. In these cases, the Homspaces can be computed using elementary arguments, and a generator foreach space can be given explicitly. For 0 < m < n, we need to make use ofFrobenius Reciprocity and Jacquet modules.The case m = 0Observe thatR0 = {0}, so C1c (R0) ⇠= C with the trivial action of GLn⇥GLn.We have that:494.1. General results on C1c (Rm)Lemma 13. Let (⇡, V ) be an irreducible admissible representation of GLn.Then,HomGLn⇥GLn(C1c (R0), V ⌦ eV ) ⇠=(C if (⇡, V ) ⇠= (1n,C),{0} otherwise.Proof. Since C1c (R0) and V⌦eV are both irreducible as GLn⇥GLn-representations,the lemma follows from [[1], Schur’s Lemma p.19].The case m = nNote that Rn = GLn, and we recall the following well-known result:Lemma 14 ([15], III.3 p.59). Let (⇡, V ) be an irreducible admissible repre-sentation of GLn. Then,HomGLn⇥GLn(C1c (Rn), V ⌦ eV ) = HomGLn⇥GLn(C1c (GLn), V ⌦ eV ) ⇠= C.The case 0 < m < nProposition 9. Let (⇡, V ) be an irreducible admissible representation ofGLn. Then, for 0 < m < n,HomGLn⇥GLn(C1c (Rm), V ⌦ eV ) ⇠= HomLm⇥Lm(µm1Pm⇥Pm , geVNm ⌦ eVNm).Proof. For 0 < m < n, we make use of the characterization of C1c (Rm) as aparabolically induced representation of GLn⇥GLn. By Lemma 12, we havethat:HomGLn⇥GLn(C1c (Rm), V ⌦ eV ) ⇠= HomGLn⇥GLn(IndGLn⇥GLnPm⇥Pm µm, V ⌦ eV )⇠= HomGLn⇥GLn(eV ⌦ eeV , ^IndGLn⇥GLnPm⇥Pm µm),where the last isomorphism is obtained by applying [[4], Proposition A.11p.926]. Then, it follows from [[3], Proposition 2.9 p.24] and [[3], DualityTheorem p.56] that:HomGLn⇥GLn(eV ⌦ eeV , ^IndGLn⇥GLnPm⇥Pm µm) ⇠= HomGLn⇥GLn(eV ⌦ V, IndGLn⇥GLnPm⇥Pm fµmPm⇥Pm).Using [[4], Theorem A.12 p.927], we obtain the following isomorphism:HomGLn⇥GLn(eV⌦V, IndGLn⇥GLnPm⇥Pm fµmPm⇥Pm) ⇠= HomLm⇥Lm(eVNm⌦VNm , fµmPm⇥Pm).504.1. General results on C1c (Rm)Then, by applying [[4], Proposition A.11 p.926] again, we get:HomGLn⇥GLn(C1c (Rm), V ⌦ eV ) ⇠= HomLm⇥Lm(µm1Pm⇥Pm , geVNm ⌦ gVNm).By [[5], Corollary 4.2.5 p.42], we also have that gVNm ⇠= eVNm , which allowsus to further reduce it to:HomGLn⇥GLn(C1c (Rm), V ⌦ eV ) ⇠= HomLm⇥Lm(µm1Pm⇥Pm , geVNm ⌦ eVNm),as desired.First, observe that from this reduction, we can conclude the following:Corollary 3. Let (⇡, V ) be an irreducible admissible cuspidal representationof GLn. Then, for 0 < m < n,HomGLn⇥GLn(C1c (Rm), V ⌦ eV ) ⇠= {0}.Proof. Since (⇡, V ) is cuspidal, so is (e⇡, eV ), and for cuspidal representations,eVNm ⇠= {0} for any proper parabolic subgroup Pm (that is, when 0 < m < n).Thus, we may restrict our attention to those (⇡, V ) that are quotientsof some parabolically induced representation of GLn. For this, we first useone further isomorphism, that is, by tensoring both representations withPm⇥Pm and using the fact that Pm⇠= gPm , we get a convenient descriptionin terms of their contragredients:HomGLn⇥GLn(IndGLn⇥GLnPm⇥Pm µm, V ⌦ eV ) ⇠= HomLm⇥Lm(µm, ^(Pm eVNm)⌦ (Pm eVNm)),(4.1)where Lm is the Levi subgroup of Pm and Pm. In order to see the advan-tage of doing this extra step, observe that Lm ⇠= GLm⇥GLnm, and thesubgroup n⇣⇥ 1m 00 g4⇤,h1m 00 g04i⌘| g4, g04 2 GLnmo(4.2)of Lm acts trivially on C1c (GLm), which is clear from the definition of µm.So, a necessary condition for HomGLn⇥GLn(C1c (Rm), V ⌦ eV ) 6= {0} is thatthe subgroup in (4.2) also acts trivially on (PmgeVNm) ⌦ (Pm eVNm). Next,we prove the following lemma, whose proof uses arguments in the proofs of[[13], Lemma 2.2, Corollary 2.3, Theorem 2.4 p.7-9] :514.1. General results on C1c (Rm)Proposition 10. Let (⇡, V ) be an irreducible admissible representation ofGLn. Then, HomGLn⇥GLn(C1c (Rm), V ⌦ eV ) 6= {0} if and only if thereexists an irreducible admissible representation (⌧,W ) of GLm such thatHomGLn(IndGLnPm⌧ ⌦ 1nm,⇡) 6= {0}.Proof. First, we show that HomGLn⇥GLn(C1c (Rm), V ⌦ eV ) 6= {0} if andonly if there exists an irreducible admissible representation (⌧,W ) of GLmsuch that HomLm(⌧⌦1nm, ^(Pm eVNm)) 6= {0}. By Equation (4.1), the spaceHomGLn⇥GLn(C1c (Rm), V ⌦ eV ) being non-trivial is equivalent to the exis-tence of a non-trivial intertwining operator of Lm ⇥ Lm-representations be-tween (µm, C1c (GLm)) and^(Pm eVNm)⌦ (Pm eVNm). So, there exists a non-trivial irreducible sub-representation of the image of such operator. Sinceit is irreducible, it must have the form Y ⌦ eY where Y is an irreduciblerepresentation of Lm. Furthermore, since Lm ⇠= GLm⇥GLnm, we havethat Y ⇠= ⌧ ⌦ , where ⌧ is an irreducible representation of GLm, and isan irreducible representations of GLnm. As discussed before, a necessarycondition for HomGLn⇥GLn(C1c (Rm), V ⌦ eV ) 6= {0} is that the subgroup in(4.2) also acts trivially on Y ⌦ eY . Hence, ⇠= 1nm. Then,HomLm⇥Lm((⌧ ⌦ 1nm)⌦ (e⌧ ⌦ 1nm), ^(Pm eVNm)⌦ (Pm eVNm)) 6= {0}, HomLm((⌧ ⌦ 1nm), ^(Pm eVNm)) 6= {0}.Conversely, suppose that HomLm((⌧ ⌦ 1nm), ^(Pm eVNm)) 6= {0}. SinceHomGLm⇥GLm(C1c (GLm), ⌧ ⌦ e⌧) ⇠= C, we can construct a non-trivial in-tertwining operator in HomLm⇥Lm(µm, (⌧ ⌦ 1nm) ⌦ (e⌧ ⌦ 1nm)) by let-ting GLnm⇥GLnm act trivially. Thus, by composing with the inclu-sion map, we obtain a non-trivial GLn⇥GLn-map from (µm, C1c (GLm)) to^(Pm eVNm) ⌦ (Pm eVNm). Hence, HomGLn⇥GLn(C1c (Rm), V ⌦ eV ) 6= {0} ifand only if there exists an irreducible admissible representation (⌧,W ) ofGLm such that HomLm(⌧ ⌦ 1nm, ^(Pm eVNm)) 6= {0}.Next, we show that HomLm(⌧⌦1nm, ^(Pm eVNm)) ⇠= HomGLn(IndGLnPm ⌧⌦1nm, V ). By Proposition 1,HomGLn(IndGLnPm⌧ ⌦ 1nm, V ) ⇠= HomGLn(eV , ^IndGLnPm ⌧ ⌦ 1nm).Then, using [[3], Duality Theorem p.56], we have that:HomGLn(eV , ^IndGLnPm ⌧ ⌦ 1nm) ⇠= HomGLn(eV , IndGLnPm (e⌧ ⌦ 1nm)Pm).524.1. General results on C1c (Rm)Next, it follows from [[4], Theorem A.12 p.927] that:HomGLn(eV , IndGLnPm (e⌧ ⌦ 1nm)Pm) ⇠= HomLm(eVNm , (e⌧ ⌦ 1nm)Pm).Finally, by applying Proposition 1 again, we get:HomLm(eVNm , (e⌧ ⌦ 1nm)Pm) ⇠= HomLm(⌧ ⌦ 1nm, ^(Pm eVNm)).Thus, the statement of the proposition follows.Remark 8. We note that Proposition 10 is consistent with the finding forthe case of m = 0 and m = n using elementary methods. Indeed,• For m = 0, by Proposition 10 in Section 4.1.2,HomGLn⇥GLn(C1c (R0), V⌦eV ) 6= {0}, HomGLn(IndGLnGLn 1n,⇡) 6= {0}.However, IndGLnGLn 1n is precisely the trivial representation of GLn, which,by Schur’s Lemma, means that ⇡ has to be the trivial representationas well.• For m = n, by Proposition 10 in Section 4.1.2,HomGLn⇥GLn(C1c (Rn), V ⌦ eV ) 6= {0}if and only if there exists an irreducible admissible representation (⌧,W )of GLn such that HomGLn(IndGLnGLn⌧,⇡) 6= {0}. Since IndGLnGLn ⌧ = ⌧ , wemay choose ⌧ = ⇡, which shows that the Hom space is non-trivial (andin fact, one-dimensional) for every irreducible admissible representa-tion (⇡, V ) of GLn.4.1.3 One-dimensionality of HomGLn⇥GLn(C1c (Rm), V ⌦ eV )when it is non-trivialLet (⇡, V ) be an irreducible admissible representation of GLn such thatHomGLn⇥GLn(C1c (Rm), V ⌦ eV ) 6= {0}. Equivalently, there exists an irre-ducible admissible representation (⌧,W ) of GLm such thatHomGLn(IndGLnPm⌧ ⌦ 1nm,⇡) 6= {0}.Since (⇡, V ) is irreducible, it follows that (⇡, V ) is an irreducible quotientof IndGLnPm ⌧ ⌦ 1nm. We first show that in fact, it is the unique such quo-tient. Note that while the uniqueness of ⇡ as an irreducible quotient of534.1. General results on C1c (Rm)IndGLnPm ⌧ ⌦ 1nm is mentioned throughout this thesis, it is only used in thissection when we prove that the space HomGLn⇥GLn(C1c (Rm), V ⌦ eV ) is onedimensional. Moreover, note that ⇡ being the unique irreducible quotient ofIndGLnPm ⌧ ⌦ 1nm is not the Langlands classification unless ⌧ happens to bean irreducible tempered representation.Lemma 15. Let (⇡, V ) be an irreducible admissible representation of GLnsuch that there exists an irreducible admissible representation (⌧,W ) of GLmsuch that HomGLn(IndGLnPm⌧ ⌦ 1nm,⇡) 6= {0}. Then, (⇡, V ) is the uniqueirreducible quotient of IndGLnPm ⌧ ⌦ 1nm, and it appears with multiplicity 1in JH(IndGLnPm ⌧ ⌦ 1nm).Proof. Let ⌧ be an irreducible representation of GLm. We write IndGLnPm⌧ ⌦ 1nmin terms of normalized induction:IndGLnPm ⌧ ⌦ 1nm ⇠= indGLnPm (⌧ ⌦ 1nm) 12Pm⇠= |.|nm2 ⌧ ⇥ |.|m2 detnm,where Pm is the modulus character in (2.9). We have that |.|m2 detnm isthe unique irreducible submodule of indGLnmBnm |.|m2 12Bnm , where:Bnm = |.|nm1 ⌦ |.|nm3 ⌦ . . .⌦ |.|n+m+1,) |.|m2 12Bnm = |.|n+2m+12 ⌦ |.|n+2m12 ⌦ . . .⌦ |.|n12 .So, in segment notation, as a submodule,|.|m2 detnm = h[|.|n12 (nm1), |.|n12 ]i.By [[14], Corollary 5.5 p.16], IndGLnPm ⌧ ⌦ 1nm ⇠= |.|nm2 ⌧ ⇥ |.|m2 detnmhas a unique irreducible quotient, which has to be ⇡, and ⇡ appears withmultiplicity 1 in JH(IndGLnPm ⌧ ⌦ 1nm).Using the uniqueness of (⇡, V ) as an irreducible quotient of IndGLnPm ⌧ ⌦ 1nm,we can, in fact, show that the space HomGLn(IndGLnPm⌧ ⌦ 1nm,⇡) is one-dimensional.Lemma 16. Let (⇡, V ) be the unique irreducible quotient of IndGLnPm ⌧ ⌦ 1nm.Then,HomGLn(IndGLnPm⌧ ⌦ 1nm,⇡) ⇠= C.544.1. General results on C1c (Rm)Proof. Suppose that dimCHomGLn(IndGLnPm⌧ ⌦ 1nm,⇡) 2. Consider theJordan-Holder series for IndGLnPm ⌧ ⌦ 1nm:0 = Y`+1 ( Y` ( . . . ( Y0 = IndGLnPm ⌧ ⌦ 1nm,that is, Yj/Yj+1 2 JH(IndGLnPm ⌧ ⌦ 1nm) is irreducible for 0 j `. Then,for every j, we have the following short exact sequence of GLn-spaces:0! Yj+1 ! Yj ! Yj/Yj+1 ! 0,which yields the following long exact sequence in cohomology:0! HomGLn(Yj/Yj+1, V ) ! HomGLn(Yj , V ) ! HomGLn(Yj+1, V )! . . .Since ⇡ is the unique irreducible quotient, Y0/Y1 ⇠= ⇡. Thus, for j = 0, weget:0! C ! HomGLn(IndGLnPm ⌧ ⌦ 1nm, V ) ! HomGLn(Y1, V )! . . .Thus, since we are assuming that dimCHomGLn(IndGLnPm⌧⌦1nm,⇡) 2, wehave that HomGLn(Y1, V ) 6= {0}. Let r 1 be the smallest value such thatHomGLn(Yr, V ) 6= {0}, and HomGLn(Yr+1, V ) = {0}. Then, by the long ex-act sequence above, HomGLn(Yr/Yr+1, V ) 6= {0}. Moreover, since Yr/Yr+1 isirreducible, Yr/Yr+1 ⇠= ⇡. This is a contradiction to ⇡ appearing with multi-plicity 1 in JH(IndGLnPm ⌧ ⌦1nm). Thus, HomGLn(IndGLnPm ⌧ ⌦ 1nm,⇡) ⇠= C,as desired.Note that in the proof of Proposition 10, we have in fact shown thatHomGLn⇥GLn(C1c (Rm), V ⌦ eV ) ⇠= HomGLn(IndGLnPm ⌧ ⌦ 1nm,⇡).Thus, we get the following corollary:Corollary 4. Let (⇡, V ) be an irreducible admissible representation of GLnsuch that HomGLn⇥GLn(C1c (Rm), V ⌦ eV ) is non-trivial. Then,HomGLn⇥GLn(C1c (Rm), V ⌦ eV ) ⇠= C.Note that while Proposition 10 gives a necessary and sucient conditionfor a representation (⇡, V ) to appear in rank m, it is generally dicultto determine the rank(s) that a particular irreducible representation (⇡, V )may appear in. However, for spherical representations, we can obtain somepartial results on the ranks (see Proposition 12 in Section 6.2). Moreover,recall that just like in Section 3.4 for GL2, we will eventually make use ofthe Satake transform to prove our main result which only makes sense when(⇡, V ) is a spherical representation. Thus, in the next section, we will studythe spherical part of the spectrum of C1c (Rm) and later in Chapter 6, wewill restrict our attention to the class of spherical representations.554.2. The spherical part of the spectrum of C1c (Rm)4.2 The spherical part of the spectrum of C1c (Rm)Let H(GLn,K) be the spherical Hecke algebra and consider the action ofH(GLn,K) on the space C1c (Rm)K⇥K given by:( · f)(x) =ZGLn(g)f(gx) dg,where 2 H(GLn,K), f 2 C1c (Rm)K⇥K , and x 2 Rm. We would like tostudy the following questions:• What are the eigenfunctions and eigenvalues of C1c (Rm)K⇥K underthe action above?• For which spherical representations (⇡, V ) of GLn, is the spaceHomH(GLn,K)(e⇡K ⌦ ⇡K , C1c (Rm)K⇥K)non-trivial?First, we recall the known results using Satake isomorphism for the casem = n to obtain the complete answers in that case. Then, we study the casem < n using a commutative diagram (see Proposition 11). Note that theHom-space in the second question is di↵erent from the Hom-space that hasbeen studied for the extension problem. Namely, for the extension problem,we study Hom-spaces of the form:HomGLn⇥GLn(C1c (X),⇡ ⌦ e⇡)where X is Rm or Rm, which are studied in Chapters 5. A conjectureon a necessary and sucient condition on the representation (⇡, V ) to be asubquotient of C1c (Rm)K⇥K is given in Conjecture 10.4.2.1 The case m = nFirst, we review the known results given by the Satake isomorphism for thespecial case when m = n. Let (s1, . . . , sn) 2 Cn, z = (z1, . . . , zn) 2 Cnwhere zi = qsi , and z : B ! C⇥ the character defined by:z(an) =nYi=1|ai|si , where a = diag(a1, . . . , an).564.2. The spherical part of the spectrum of C1c (Rm)Consider the normalized parabolically induced representation indGLnB z,which is spherical. Its subspace (indGLnB z)K of K-fixed vectors is one-dimensional, and it is generated by fz, where fz : GLn ! C is definedby:fz(ank) = (12Bz)(a).Since (indGLnB z)K is one-dimensional, for every 2 C1c (GLn)K⇥K , wehave that · fz is a multiple of fz, and we may define (S)(z) 2 C by( · fz) = (S)(z)fz. Thus, we obtain the following map:S :C1c (GLn)K⇥K ! C[z±11 , . . . , z±nn ]W 7! (S)(z) :=ZGLn(x)fz(x) dx.Theorem 10 (Satake isomorphism, Theorem 3 p.253 [19]). The map S isan isomorphism of algebras.Using the Satake isomorphism, we obtain the following answers:• The eigenvectors of C1c (GLn)K⇥K are the functions fz, where z 2 Cn,and its eigenvalues are (S)(z), where 2 C1c (GLn)K⇥K ;• For every spherical representations (⇡, V ) of GLn, the spaceHomH(GLn,K)(e⇡K ⌦ ⇡K , C1c (Rn)K⇥K)is non-trivial.4.2.2 The case m < nIn this section, we would like to obtain a similar characterization of theeigenvectors of C1c (Rm)K⇥K as that for C1c (GLn)K⇥K . We first recallsome definitions of maps.Definition 11. Consider the following map, which is motivated by theHarish-Chandra constant term map (§6 p.174 [16]):HS :C1c (GLn)K⇥K ! C1c (GLm)Km⇥Km 7!✓ : p1 7! | det(p1)|(nm)2ZNmZGLnm⇣uhp1 00 p4i⌘du dp4◆,where Km is the maximal compact subgroup of GLm.574.2. The spherical part of the spectrum of C1c (Rm)Remark 9. Note that in order to check that HS is well-defined, we need toverify two conditions:• the integral that represents (p1) needs to be convergent, and• the function is compactly supported on GLm.Both of these conditions are satisfied due to the following observations: sinceK is compact, the set {p 2 Pm | 9k 2 K, pk 2 supp()} is also compact.Moreover, the decomposition of Pm as NmLm is unique. Thus, if the set ofp is compact, so are the sets of u and ` obtained from such decomposition ofp. See also [[3], Lemma p.163] where the same map is considered for GL2,but the proof is clearly general.Next, we introduce the following definition of the map evalnm:Definition 12. For every 0 m n, we define the evaluation map evalnmwhere we twist all zi’s by qs, and evaluate the last n m variables atparticular values:evalnm :C[z±1i ]! C[z±11 , . . . , z±1m ][qs]P (z1, z2, . . . , zn) 7! P (z1qs, . . . , zmqs, qn12 s, qn12 +1s, . . . , qn12 +(nm1)s)We extend the evaluation map naturally to the space of rational functionsin the variables zi.Let ev : C(qs)! C be the map s 7! 0.Proposition 11. The following diagram is commutative.C1c (GLn)K⇥KS //HS✏✏C[z±11 , . . . , z±1n ]Wnev evalnm✏✏C1c (GLm)Km⇥KmS // C[z±11 , . . . , z±1m ]WmProof. Let 2 C1c (GLn)K⇥K . From the definition of Satake isomorphism,we have that:S()(z) =ZGLn(x)fz(x) dx=ZKZPm(pk)fz(pk) dp dk (by [[3], Equation (7.6.2) p.5] )=ZKZPm(p)fz(p) dp dk (since and fz are K-fixed)=ZPm(p)fz(p) dp,584.2. The spherical part of the spectrum of C1c (Rm)with the normalization that µ(K) = 1. Since Pm = NmLm, we have that:ZPm(p)fz(p) dp =ZNmZLm (ul) fz(ul)1Pm(ul) dl du.Since fz is Nm-left invariant, and Lm ⇠= GLn⇥GLnm, we obtain:ZNmZLm (ul) fz (ul) 1Pm(ul) dl du=ZNmZGLmZGLnm⇣uhp1 00 p4i⌘fz⇣hp1 00 p4i⌘1Pm⇣hp1 00 p4i⌘dp1 dp4 du=ZNmZGLmZGLnm⇣uhp1 00 p4i⌘fz⇣hp1 00 p4i⌘| det(p1)|(nm)| det(p4)|m dp1 dp4 du.Denote by z0 = (z1, . . . , zm) the tuple of its first m components of z. Ifhp1 00 p4i= u diag(a1, . . . , an)hk1 00 k4i, thenfz(hp1 00 p4i) =nYi=1|ai|n2i+12 +si ,fz0(p1) =mYi=1|ai|m2i+12 +si .Thus,fz⇣hp1 00 p4i⌘| det(p1)|(nm) = mYi=1|ai|n2i+12 +sinYi=m+1|ai|n2i+12 +si!mYi=1|ai|(nm)=mYi=1|ai|n2i+12 +si(nm)nYi=m+1|ai|n2i+12 +si=mYi=1|ai|m2i+12 +simYi=1|ai|(nm)2nYi=m+1|ai|n2i+12 +si= fz0(p1)| det(p1)|nm2nYi=m+1|ai|n2i+12 +si .Thus,S()(z) =ZNmZGLmZGLnm⇣uhp1 00 p4i⌘fz0(p1)| det(p1)|nm2nYi=m+1|ai|n2i+12 +si+m dp1 dp4 du.594.2. The spherical part of the spectrum of C1c (Rm)Next, the evaluation ev evalnm setszm+1 = qn12 , . . . , zn = qn2n+12 m = qn12 +(nm1).In other words, ev evalnm maps zi = qn2i+12 m for m+ 1 i n, andwith a suitable normalization on GLnm so that µ(Knm) = 1, we get:ev evalnm S()(z) =ZNmZGLmZGLnm⇣uhp1 00 p4i⌘fz0(p1)| det(p1)|nm2 dp1 dp4 du=ZGLm(HS)(p1)fz0(p1) dp1,by Definition 11. Note that it is precisely the Satake transform of HS() 2C1c (GLm)Km⇥Km , and the proof is complete.Remark 10. It is important that the evaluation map evalnm evaluatesexactly at zm+1 = qn12 s, . . . , zn = qn12 +(nm1)s so that the diagramin Proposition 11 is commutative. This is because in the Definition 11 of themap HS, we integrate over the GLnm-block of Pm, which corresponds tothe variables zm+1, . . . , zn. An evaluation map at n m di↵erent variableszi’s would simply correspond to a map HS where is integrated over theLevy subgroup of a di↵erent parabolic subgroup. Thus, the statement ofProposition 11 can be modified to work with an evaluation map at any nmvariables if we modify the Definition 11 of HS accordingly. As a result, wewill later define evalnm with permutation in Definition 15 in Section 6.3.1.Using the commutative diagram in Proposition 11, and the fact thatevalnm is surjective and S is an isomorphism, we obtain the following corol-lary:Corollary 5. The map HS : C1c (GLn)K⇥K ! C1c (GLm)Km⇥Km, as de-fined in Definition 11, is surjective.The above corollary motivates us to make the following conjecture:Conjecture 1. (1) The eigenvectors of C1c (Rm)K⇥K are the functions fz,where z 2 Cn has the form z = (z1, . . . , zm, qn12 , . . . , qn12 +(nm1)),and its eigenvalues are (S)(z), where 2 C1c (GLn)K⇥K ;(2) An irreducible spherical representation (⇡, V ) of GLn is a subquotientof C1c (Rm)K⇥K if and only if the Satake parameters {s1, . . . , sn} asso-ciated to ⇡ contains the sequence {n12 , n12 1, . . . , n12 (nm1)}.604.2. The spherical part of the spectrum of C1c (Rm)A partial result on part (2) of the above conjecture is given in Proposi-tion 12. Recall that in order to study the extension problem in later chapters,we need to consider (⇡, V ) for whichHomH(GLn,K)(C1c (Rm)K⇥K ,⇡K ⌦ e⇡K)is non-trivial, in which case, we say that ⇡ appears in rank m as a quotient.Note that if (⇡, V ) is an irreducible quotient of IndGLnPm ⌧⌦1nm, then (⇡, V )appears in rank m as a quotient; however, it might not be spherical. Ifinstead we take spherical subquotient of IndGLnPm ⌧ ⌦1nm, then it might notappear in rank m as a quotient (see Example 2 below). In Proposition 12,we give a criterion for determining the lowest rank m in which ⇡ appears asa quotient in terms of Satake parameters of ⇡.Example 2. In GL2 with m = 1, let ⌧ = 1. Then, IndGL2P11 ⌦ 1 hasthe Steinberg representation StGL2 as an irreducible quotient and the trivialrepresentation 12 as a subspace. StGL2 is not spherical but it does appear inrank 1 as a quotient, whereas 12 is spherical, but it appears in rank 0, andnot in rank 1 as a quotient.61Chapter 5Extension problem and thelocal L-function: partialresults for general irreduciblerepresentations of GLn5.1 The extension of ⇣GLn⇡ to ⇣Mn⇡ if LGLn(⇡, s) doesnot have a pole at s = n12In this section, we will show that if the L-function associated to a repre-sentation (⇡, V ) does not have a pole at s = n12 , then it does imply asolution to the extension problem at the highest rank.Theorem 11. Let (⇡, V ) be an irreducible admissible representation of GLnsuch that the associated L-function LGLn(⇡, s) does not have a pole at s =n12 . Then, ⇣GLn⇡ can be extended GLn⇥GLn-equivariantly to a map ⇣Mn⇡in HomGLn⇥GLn(C1c (Mn), V ⌦ eV ).Proof. Let (⇡, V ) be an irreducible admissible representation of GLn suchthat the associated L-function LGLn(⇡, s) does not have a pole at s = n12 .We will construct explicitly a lift ⇣Mn⇡ of ⇣GLn⇡ in HomGLn⇥GLn(C1c (Mn), V⌦eV ). First, observe that we have a map:HomGLn⇥GLn(C1c (Mn)⇥ eV ⇥ V,C)! HomGLn⇥GLn(C1c (Mn)⌦ (eV ⌦ V ),C),where given an element f in HomGLn⇥GLn(C1c (Mn)⇥ eV ⌦V,C) , the imagef 0 of f in HomGLn⇥GLn(C1c (Mn) ⌦ (eV ⌦ V ),C) is the map obtained viathe universal property of tensor products. Furthermore, we also have thefollowing isomorphisms of GLn⇥GLn-modules:HomGLn⇥GLn(C1c (Mn)⌦ (eV ⌦ V ),C) ⇠= HomGLn⇥GLn(C1c (Mn),Hom(eV ⌦ V,C))⇠= HomGLn⇥GLn(C1c (Mn), V ⌦ eV ).625.2. Poles of LGLn(⇡, s) if (⇡, V ) appears in a lower rankThus, in order to construct ⇣Mn⇡ in HomGLn⇥GLn(C1c (Mn), V ⌦eV ), it sucesto construct an element in HomGLn⇥GLn(C1c (Mn)⇥ eV ⇥ V,C) and look atits image in HomGLn⇥GLn(C1c (Mn), V ⌦ eV ) via the above maps.Given the generator ⇣GLn⇡ , we have the following element ⇣GLn0 in thespace HomGLn⇥GLn(C1c (GLn)⇥ eV ⇥ V,C):⇣GLn0 :C1c (GLn)⇥ eV ⇥ V ! C(f, ev0, v) 7! ZGLnhev0,⇡(g)vif(g) dg.Note that ⇣GLn⇡ corresponds to an element ⇣GLn0 of HomGLn⇥GLn(C1c (GLn)⇥eV ⇥ V,C), and thus, we may prove the existence of a lift for ⇣GLn⇡ by show-ing that ⇣GLn0 can be lifted to an element ⇣Mn0 in HomGLn⇥GLn(C1c (Mn)⇥eV ⇥ V,C). Following the definition of zeta integrals, for s 2 C, consider theformal zeta integral of the form:⇣Mns :C1c (Mn)⇥ eV ⇥ V ! C(f, ev0, v) 7! ZGLnhev0,⇡(g)vif(g)| det g|s dg.The integral ⇣Mns converges absolutely for <(s) 0, for all f 2 C1c (Mn), ev0 2eV and v 2 V by part (1) of Theorem 5. Thus, as a function of s, we mayanalytically continue ⇣Mn : C1c (Mn) ⇥ eV ⇥ V ⇥ C ! C to a meromorphicfunction on the whole s-plane. Note that if ⇣Mn does not have a pole ats = 0, then ⇣Mn0 is an extension of ⇣GLn0 .Next, we will show that ⇣Mn has a pole at s = 0 if and only if theassociated L-function of (⇡, V ) has a pole at s = n12 . By Theorem 5 (2),we have that for every f 2 C1c (Mn), ev0 2 eV and v 2 V ,⇣Mn(f, ev0, v, s+ n12 )LGLn(⇡, s)✓ C[qs, qs].So, ⇣Mn has a pole at s = 0 if and only if LGLn(⇡, s) also has a pole ats = n12 . Thus, the statement of the lemma follows.5.2 Poles of LGLn(⇡, s) if (⇡, V ) appears in a lowerrankLet (⇡, V ) be an irreducible representation of GLn such that for somem < n,we have that HomGLn⇥GLn(C1c (Rm), V ⌦ eV ) ⇠= C. We have the following635.2. Poles of LGLn(⇡, s) if (⇡, V ) appears in a lower rankconjecture about the poles of the L-function attached to ⇡ in this case:Conjecture 2. For m < n, if HomGLn⇥GLn(C1c (Rm), V ⌦ eV ) ⇠= C, thenthe L-function attached to ⇡ has a pole ats = n 12,n 12+ 1, . . . ,n 12+ (nm 1).By Proposition 10, if HomGLn⇥GLn(C1c (Rm), V ⌦ eV ) ⇠= C, then ⇡ is theunique irreducible quotient of IndGLnPm ⌧ ⌦ 1nm ⇠= |.|mn2 ⌧ ⇥ |.|m2 detnmfor some ⌧ 2 Irr(GLm). Thus, the L-function of ⇡ is closely related to the L-functions of |.|mn2 ⌧ and of |.|m2 detnm since ⇡ and |.|mn2 ⌧ ⇥ |.|m2 detnmhave the same cuspidal support. Observe that:|.|m2 detnm = h[|.|n12 ], . . . , [|.|n12 (nm1)]it.Thus, the L-function of |.|m2 detnm has poles ats = n 12,n 12+ 1, . . . ,n 12+ (nm 1).However, it is unclear that as the unique irreducible quotient of |.|mn2 ⌧ ⇥|.|m2 detnm, ⇡ would inherit these poles as well. We have the followingobservation:Lemma 17. If |.|mn2 ⌧ ⇥ |.|m2 detnm is irreducible, then the L-functionattached to ⇡ has a pole ats = n 12,n 12+ 1, . . . ,n 12+ (nm 1).Proof. Suppose that |.|mn2 ⌧⇥ |.|m2 detnm is irreducible, which means thatit is precisely ⇡. Then:LGLn(⇡, s) = LGLm(|.|mn2 ⌧, s)LGLnm(|.|m2 detnm, s).As discussed above, the L-function of |.|m2 detnm has exactly the poles ats = n 12,n 32, . . . ,n 12+ (nm 1).Thus, so does the L-function attached to ⇡.We can also prove the conjecture for the case of a spherical representation(⇡, V ). Namely,645.2. Poles of LGLn(⇡, s) if (⇡, V ) appears in a lower rankLemma 18. If (⇡, V ) is an irreducible spherical representation of GLn suchthat HomGLn⇥GLn(C1c (Rm), V ⌦ eV ) ⇠= C, then the L-function attached to⇡ has a pole ats = n 12,n 12+ 1, . . . ,n 12+ (nm 1).Proof. The lemma follows from part (1) of Proposition 12, which will beproved later in Section 6.2.For a general irreducible representation (⇡, V ) of GLn, by examiningsome special cases where we can compute the L-function of ⇡ explicitly, weobtain some partial results.5.2.1 The case m = 0As discussed before, the only representation which appears in rank 0 is thetrivial representation (1n,C), which is the unique irreducible quotient of therepresentation indGLnB 12B. Thus:1n = h[|.|n12 ], [|.|n12 1], . . . , [|.|n12 (n1)]it,LGLn(1n, s) =nYj=1(1 qsn12 +(j1))1.Thus, LGLn(1n, s) has a pole ats = n 12,n 12+ 1, . . . ,n 12+ (n 1),and the conjecture is true for m = 0.5.2.2 The case m = 1By Proposition 10, if HomGLn⇥GLn(C1c (R1), V ⌦ eV ) ⇠= C, then ⇡ is theunique irreducible quotient of ⇥ |.| 12 detn1 for some character . Insegment notation, we have that:⇥ |.| 12 detn1 = h[]i ⇥ h[|.|n32 , |.|n12 ]i.By Theorem 1, the representation h[]i ⇥ h[|.|n32 , |.|n12 ]i is irreducible ifthe segment [] is not linked with h[|.|n32 , |.|n12 ]i, that is, if 6= |.|n+12or |.|n12 . Thus, the cases we are interested in are when = |.|n+12 or655.2. Poles of LGLn(⇡, s) if (⇡, V ) appears in a lower rank|.|n12 . Furthermore, observe that if ⇡ is the unique irreducible quotient of⇥|.| 12 detn1, then it is the unique irreducible submodule of |.| 12 detn1⇥,and we may use [[14], Theorem 6.6 p.20] to determine the multisegmentcorresponding to ⇡.By [[14], Theorem 6.6 (2) p.20], the multisegment corresponding to ⇡ is:⇡ =(h[|.|n+12 ], [|.|n12 ], . . . , [|.|n32 ]it if = |.|n+12 ,h[|.|n12 ], . . . , [|.|n12 , |.|n32 ]it if = |.|n12 .Observe that in either cases, each segment of |.| 12 detn1 that ends in |.|sfors = n 12,n 12, . . . ,n 12+ (n 2)still remains ending in |.|s as a segment of ⇡. Thus, the L-function attachedto ⇡ has a pole ats = n 12,n 12, . . . ,n 12+ (n 2),and the conjecture holds for m = 1.5.2.3 The case m = n 1By Proposition 10, if HomGLn⇥GLn(C1c (Rn1), V ⌦ eV ) ⇠= C, then ⇡ is theunique irreducible quotient of IndGLnPn1 ⌧ ⌦ 1 ⇠= |.|12 ⌧ ⇥ |.|n12 . Let m =(1, . . . ,r) be the multisegment corresponding to ⌧ . By Proposition 5,the unique irreducible quotient ⇡ of |.| 12 ⌧ ⇥ |.|n12 is either !0 or !1, where:!0 = h[|.|n12 ],1, . . . ,rit,!1 = h1, . . . ,+l(1), . . . ,rit,where +l(1) has the form [|.|b(1), |.|n12 ] (refer to Section 2.3.5 for the no-tation). Observe that in either case, the multisegment corresponding to ⇡contains a segment which ends in |.|n12 . Thus, by Lemma 3, the L-functionattached to ⇡ has a pole at s = n12 , and the conjecture holds form = n1.It is unclear for other ranks how the segments of ⇡ are determined fromthe segments of |.|mn2 ⌧ ⇥ |.|m2 detnm. However, we expect that eachsegment of |.|m2 detnm that ends in |.|s fors = n 12,n 12+ 1, . . . ,n 12+ (nm 1)will remain ending in |.|s as a segment of ⇡, which will lead to the L-functionof ⇡ having the wanted poles.665.3. Counterexample: L-function fails to detect the rank5.3 Counterexample: L-function fails to detectthe rankAs we discussed in the introduction, heuristically one might hope that the L-function of ⇡ might detect the rank in which ⇡ appears. However, beyond thecase of n = 2, this is not so. Here we construct an example of two irreduciblerepresentations with the same L-function that appear in di↵erent ranks.An example of two irreducible representations with the sameL-function that appear in di↵erent ranksLet (⇡1, V1) be the irreducible (normalized) parabolically induced repre-sentation |.| 12 StGL2 ⇥1. In segment notation, (⇡1, V1) can be written ash[1, |.|], [1]it. On the other hand, let (⇡2, V2) be the irreducible representa-tion of GL3 corresponding to the segment notation h[], [|.|], [1]it, where is a ramified character. More concretely, (⇡2, V2) is the irreducible paraboli-cally induced representation ⇥|.| 12 det2. Then, using Wedhorn’s definitionof the L-function of a representation given its multisegment as described inSection 2.4, we get:LGL3(⇡1, s) = LGL3(⇡2, s) =(1 qs)(1 q1s)1 .First, observe that ⇥|.| 12 det2 ⇠= IndGL3P1 |.|⌦12, and thus, HomGL3(IndGL3P1 |.|⌦12,⇡2) ⇠= C. By Proposition 10, it follows that HomGL3⇥GL3(C1c (R1),⇡2 ⌦f⇡2) ⇠= C, and (⇡2, V2) appears in rank 1.We now show that (⇡1, V1) appears in rank 2 by producing an explicit ir-reducible representation ⌧ 2 Irr(GL2) such that HomGL3(IndGL3P2 ⌧⌦1,⇡1) ⇠=C. Consider ⌧ = |.| 12 ⇥ |.| 12 = h[|.| 12 ], [|.| 12 ]it 2 Irr(GL2). Then, we use [[14],Corollary 7.3 p.21] to compute the segment notation of the unique irre-ducible quotient of |.| 12 ⌧ ⇥ |.| = h[1], [1]it ⇥ h[|.|]it. We get that the uniqueirreducible quotient of |.| 12 ⌧ ⇥ |.| has the segment notation h[1, |.|], [1]it,which is precisely that of (⇡1, V1). Hence, (⇡1, V1) appears in rank 2.Next, we will show explicitly that (⇡1, V1) does not appear in rank 1, thatis, HomGL3⇥GL3(C1c (R1),⇡1⌦f⇡1) is trivial. If HomGL3⇥GL3(C1c (R1),⇡1⌦f⇡1) ⇠= C, then there exists (⌧,C) 2 Irr(GL1) such that HomGL3(IndP1 ⌧ ⌦12,⇡1) ⇠= C. Observe that:|.| 12 StGL2 ⇥1 ⇠= 1⇥ |.|12 StGL2⇠= IndGL3P1 |.|1 ⌦ |.| StGL2 .675.3. Counterexample: L-function fails to detect the rankWe compute the Hom space HomGL3(IndP1 ⌧ ⌦ 12, IndGL3P1 |.|1 ⌦ |.| StGL2).By Frobenius reciprocity,HomGL3(IndP1 ⌧ ⌦ 12, IndGL3P1 |.|1 ⌦ |.| StGL2) ⇠= HomL1((IndP1 ⌧ ⌦ 12)N1 , |.|1 ⌦ |.| StGL2).We have that the Jordan-Holder composition factors of (IndP1 ⌧ ⌦12)N1 are⌧ ⌦ 12 and |.|⌦ (IndGL2B ⌧ |.|1 ⌦ 1). For every ⌧ , there does not exist a mapfrom either factors to |.|1⌦ |.| StGL2 . Thus, HomL1((IndP1 ⌧⌦12)N1 , |.|1⌦(|.| StGL2)) is trivial. Hence, (⇡1, V1) does not appear in rank 1.Alternatively, we can compute explicitly the segment notation of theunique irreducible quotient of IndP1 ⌧ ⌦ 12 ⇠= |.|1⌧ ⇥ |.|12 det2 for every⌧ . Observe that the unique irreducible quotient of |.|1⌧ ⇥ |.| 12 det2 is theunique irreducible submodule of |.| 12 det2⇥|.|1⌧ by [[14], Proposition 7.1p.21]. We have that |.| 12 det2 = h[|.|], [1]it in segment notation. So, byusing [[14], Theorem 6.6 (2) p.20], we have that the segment notation of theunique irreducible quotient of |.|1⌧ ⇥ |.| 12 det2 is:8><>:h[|.|], [1], [1]it if ⌧ = |.|,h[|.|], [|.|1,1]it if ⌧ = 1,h[|.|], [1], [|.|1⌧ ]it otherwise.Since (⇡1, V1) = h[1, |.|], [1]it, which is none of the above, it follows that(⇡1, V1) does not appear in rank 1.Thus, we have produced an example of two irreducible representations ofGL3 which have the same L-function, but appear in di↵erent rank filtration.So, an L-function of a representation may not be a fine enough invariant todistinguish this.68Chapter 6The extension problem forspherical representations ofGLnIn this chapter, we restrict our attention to the class of spherical irreduciblerepresentations of GLn and study the Hom-space between subspaces of K-fixed vectors, namely,HomH(GLn,K)(C1c (Rm)K⇥K ,⇡K ⌦ e⇡K).Note that we only study the Hom-space above instead of the full spaceHomGLn⇥GLn(C1c (Rm),⇡⌦ e⇡) that was introduced in Section 2.2, despitethat C1c (Rm) is not generated by its K-fixed vectors.6.1 Summary of resultsIn Section 6.2, we first study the connection between the ranks that (⇡, V )appears in as a quotient and the set {s1, . . . , sn} and obtain the followingpartial result:Proposition (Proposition 12). (i) If (⇡, V ) appears in C1c (Rm) as a quo-tient, then (⇡, V ) is the unique irreducible quotient of a representationof the form |.|s1 ⇥ |.|s2 ⇥ . . .⇥ |.|sn such that the sequence {n12 , n12 1, . . . , n12 (nm 1)} is contained in {s1, . . . , sn};(ii) If (⇡, V ) is the unique irreducible quotient of a representation of theform |.|s1⇥|.|s2⇥. . .⇥|.|sn where n12 , n12 1, . . . , n12 (n r 1) is the sequence of maximal length that is contained in {s1, . . . , sn}, then(⇡, V ) appears in C1c (Rr) as a quotient for r < n.Thus, using the set {s1, . . . , sn}, one can determine the lowest rank thata spherical representation (⇡, V ) appears in as a quotient. In the case that(⇡, V ) appears in multiple low ranks as a quotient, we give a conjecture on696.1. Summary of resultshow to determine all the ranks that (⇡, V ) appears in as a quotient usingan inductive method based on the set {s1, . . . , sn}.The rest of the chapter is devoted to studying the equivariant extensionof ⇣GLn⇡ to lower ranks. In Sections 6.3.1 and 6.3.2, we recall the definitionof Satake transform on Mn as introduced by L. La↵orgue in [11], the Mac-donald’s formula and we develop some notation that will be used to studyspherical functions supported on lower rank. In Section 6.3.3, we character-ize the functions in C1c (Rm)K⇥K that project to a basis of C1c (Rm)K⇥Kunder the restriction map (Lemma 20). In 6.3.4, we define the ideal Jm inC[z±1i ]W by:S(C1c (Rm)K⇥K) =JmnQi=1(1 q n12 zi).For m = 1, we obtain an explicit description of J1 in terms of J2 with oneadditional generator (Proposition 13). However, it is dicult in general togive a nice description of Jm due to the combinatorial complexity of thenumerators (see Example 8). Nonetheless, we observe that for the extensionof ⇣GLn⇡ , it suces to study the image of Jm under the map ev evalnmthat substitutes z1 = n12 , . . . , znm = n12 +(nm 1). Let Im be theideal in C[znm+1, . . . , zn]W defined by:Im = hnYi=nm+1(1 q n12 (nm)zi)i.Our main technical result of this chapter is the following description of theimage of Jm under the map evalnm.Theorem (Theorem 13). For 0 m n,ev evalnm(Jm) ✓ Im.The proof of Theorem 13 is contained in Sections 6.3.5 and 6.3.6, and itrelies on explicit computation of the Satake transform of a test function usingMacdonald’s formula. Lastly, in Section 6.3.7, we prove the main result,which provides a complete answer to the extension problem of ⇣GLn⇡ for ageneric spherical representation (⇡, V ) that appears in low rank. However,note that this result depends on Hypothesis 1 in Section 6.3.1 that theextended Satake transform is also injective on C1c (Mn)K⇥K .Theorem (Theorem 14). Assume Hypothesis 1 in Section 6.3.1 and let(⇡, V ) be a spherical irreducible admissible representation of GLn, which is706.1. Summary of resultsan irreducible quotient of |.|s1 ⇥ . . . |.|sn. Suppose that the value n12 appearsexactly once in the set {s1, . . . , sn}, and let r be the length of the longestdecreasing sequence of consecutive integers in {s1, . . . , sn} starting at n12 .Then,(1) ⇡ appears in rank r as a quotient;(2) for every m > r, ⇣GLn⇡ extends GLn⇥GLn-equivariantly to an elementof HomH(GLn,K)(C1c (Rm)K⇥K , V K ⌦ eV K);(3) the map ⇣GLn⇡ does not extend GLn⇥GLn-equivariantly to an elementof HomH(GLn,K)(C1c (Rr)K⇥K , V K ⌦ eV K).Our proof relies on Theorem 13, which show that under the map eval,the ideal Jm vanishes if m > r and does not vanish if m = r. The conditionthat the value n12 appears exactly once in the set {s1, . . . , sn} is to ensurethat the L-function LGLn(⇡, s) has a simple pole at s = n12 . Thus, whenthe pole at s = n12 is not simple, we do not know the answer. In general,we have the following conjecture:Conjecture 3. For a spherical representation (⇡, V ),(i) HomH(GLn,K)(C1c (Rm)K⇥K , V K⌦eV K) ⇠= HomH(GLn,K)(C1c (Rm+1)K⇥K , V K⌦eV K) if (⇡, V ) does not appear in rank m;(ii) HomH(GLn,K)(C1c (Rm)K⇥K , V K⌦eV K) ⇠= HomH(GLn,K)(C1c (Rm)K⇥K , V K⌦eV K) if (⇡, V ) appears in rank m.In particular, we expect that multiplicity one holds at every rank, namely,for 0 m n,HomH(GLn,K)(C1c (Rm)K⇥K , V K ⌦ eV K) ⇠= C.For example, for the representation (⇡, V ) = |.| ⇥ |.| 12 det2 of GL3 whichappears in ranks 3, 2, and 1. From the conjecture, we expect that(i) HomH(GL3,K)(C1c (GL3)K⇥K , V K ⌦ eV K) ⇠= C is generated by ⇣GL3⇡ ;(ii) HomH(GL3,K)(C1c (R2)K⇥K , V K ⌦ eV K) ⇠= C is generated by ⇣R2⇡ ;(iii) HomH(GL3,K)(C1c (R1)K⇥K , V K⌦eV K) and HomH(GL3,K)(C1c (M3)K⇥K , V K ⌦ eV K)are both one-dimensional, and each is generated by ⇣R1⇡ .Statement (i) is known from Lemma 14, and statement (ii) is a consequenceof Proposition 17, which relies on Hypothesis 1 and the exact sequence (2.8).Statement (iii) is purely conjectural.716.2. Satake parameters of spherical representations that appear in lower ranks6.2 Satake parameters of sphericalrepresentations that appear in lower ranksWe recall from Theorem 3 that if (⇡, V ) is a spherical irreducible admis-sible representation of GLn, then it is the unique irreducible quotient of arepresentation of the form |.|s1 ⇥ |.|s2 ⇥ . . .⇥ |.|sn .Proposition 12. (i) If (⇡, V ) appears in C1c (Rm) as a quotient, then(⇡, V ) is the unique irreducible quotient of a representation of the form|.|s1 ⇥ |.|s2 ⇥ . . .⇥ |.|sn such that the sequence {n12 , n12 1, . . . , n12 (nm 1)} is contained in {s1, . . . , sn};(ii) If (⇡, V ) is the unique irreducible quotient of a representation of theform |.|s1⇥|.|s2⇥. . .⇥|.|sn where n12 , n12 1, . . . , n12 (n r 1) is the sequence of maximal length that is contained in {s1, . . . , sn}, then(⇡, V ) appears in C1c (Rr) as a quotient for r < n.Proof of (i). Since (⇡, V ) appears in rank m, by Proposition 10, there existsan irreducible admissible representation (⌧,W ) of GLm such that (⇡, V ) isthe unique irreducible quotient of IndGLnPm ⌧ ⌦ 1nm ⇠= |.|nm2 ⌧ ⇥ |.|m2 detnm. Since ⇡ is spherical, so is ⌧ and by Theorem 3, |.|nm2 ⌧ is anirreducible quotient of a representation of GLm of the form |.|s1 ⇥ . . .⇥ |.|smwhere si 6= sj+1 if i j. On the other hand, |.|m2 detnm is an irreduciblequotient of |.|n12 ⇥ |.|n12 1 ⇥ . . .⇥ |.|n12 (nm1). Through a compositionof quotient maps, we have that:|.|s1 ⇥ . . .⇥ |.|sm ⇥ |.|n12 ⇥ |.|n12 1 ⇥ . . .⇥ |.|n12 (nm1) ! |.|nm2 ⌧ ⇥ |.|m2 detnm ! ⇡.Thus, (⇡, V ) is the unique irreducible quotient of a representation of the form|.|s1⇥|.|s2⇥. . .⇥|.|sn where the sequence {n12 , n12 1, . . . , n12 (nm1)}is contained in {s1, . . . , sn}.Proof of (ii). We will show that (⇡, V ) appears in rank r as a quotient byusing Proposition 10 and constructing an explicit irreducible representation⌧ of GLr such that ⇡ is the unique irreducible quotient of IndGLnPr⌧ ⌦ 1nr.Consider the longest sequence of indices i1, . . . , inr such that sij = n12 (n j 1) and ij is chosen to be minimal. We re-order the si’s as follows:let s01, . . . , s0m be those si’s which are not among the particular values sijabove. We order the s0i’s so that if s0k = si and s0l = sj where i > j, thenk > l, that is, we preserve the order among the s0i’s from before. Then,the multisegment ([|.|s01 ], [|.|s02 ], . . . , [|.|s0m ]) also satisfies the ”do not precede”726.2. Satake parameters of spherical representations that appear in lower rankscondition. Thus, by Theorem 2, there exists an irreducible representation⌧ 2 Irr(GLr) such that:⌧ = h[|.|s01 ], [|.|s02 ], . . . , [|.|s0r ]it.On the other hand, observe that:h[|.|si1 ], [|.|si2 ], . . . , [|.|sinr ]it = h[|.|n12 ], [|.|n12 1], . . . , [|.|n12 (nr1)]it⇠= |.| r2 detnr.By [[14],Theorem 5.1 p.15], the representation ⌧ 0⇥ |.| r2 detnr has a uniqueirreducible quotient, denoted by ⇡0. We will show that ⇡0 ⇠= ⇡. Recall thatwe have the following composition of quotient maps:|.|s01 ⇥ . . .⇥ |.|s0r ⇥ |.|si1 ⇥ . . .⇥ |.|sinr ! ⌧ 0 ⇥ |.| r2 detnr ! ⇡0.Thus, ⇡0 is the unique irreducible quotient of |.|s01 ⇥ . . . ⇥ |.|s0r ⇥ |.|si1 ⇥. . . ⇥ |.|sinr . Moreover, it follows from Theorem 2 that ⇡0 ⇠= ⇡. Let ⌧ :=⌧ 0|.|nr2 2 Irr(GLr). Then,IndGLnPr ⌧ ⌦ 1nr ⇠= ⌧ 0 ⇥ |.|m2 detnr ! ⇡,which shows that ⇡ appears in rank r, as desired.As an immediate consequence of Proposition 12(ii), if the charactersassociated to (⇡, V ) are distinct, then (⇡, V ) appears in at most one lowrank r < n as a quotient, and this rank is uniquely determined by thecharacters. However, in general, a representation (⇡, V ) may appear in morethan one low ranks as a quotient (see Example 3), in which case, we canonly determine the lowest rank r < n that (⇡, V ) appears in, but not theranks above r. In this case, we can obtain a complete spectrum of C1c (Rm)only in the case of GL3 through explicit computations ( see Chapter 7).We also observe that the indices i for which si take on the particularvalues:n 12,n 12 1, . . . , n 12 (nm 1)need not be consecutive. For example, consider the representation (⇡, V ) =|.| 32 det2⇥|.| det3 of GL5 which corresponds to h|.|2, |.|2, |.|, |.|,1it. Wecannot find consecutive indices si’s that take on the values 2, 1, and 0.Corollary 6. The following representations appear in C1c (Rm) as a quo-tient:736.3. Equivariant extension of ⇣GLn⇡ and Satake transforms• all irreducible representations (⇡, V ) for which the longest decreasingsequence of consecutive values of si’s starting atn12 has length m;• some irreducible representations (⇡, V ) for which the longest decreasingsequence of consecutive values of si’s starting atn12 has length morethan m.We have a conjecture that given an irreducible spherical representation(⇡, V ), it is possible to determine all the ranks that (⇡, V ) appears in as aquotient using an inductive method.Conjecture 4. To find all the ranks that (⇡, V ) appears in as a quotient, wefirst find the longest decreasing sequence of consecutive values of si’s startingat n12 , which will determine the lowest rank that (⇡, V ) appears in. If thelength of the longest such sequence is n r, then (⇡, V ) appears in rank r asa quotient. Then, we exclude the si’s that give those values in the sequencefrom consideration. Among the remaining si’s, we find the longest decreasingsequence of consecutive values of si’s starting atn12 , which will determinethe second lowest rank that (⇡, V ) appears in. We continue inductively untilwe can no longer find a sequence starting at n12 or we exhaust all the si’s.For example, consider the representation (⇡, V ) = |.|⇥ |.| 12 det2 of GL3in Example 3. (⇡, V ) is the unique irreducible quotient of |.|⇥ |.|⇥ 1 wheres1 = s2 = 1 and s3 = 0. The longest decreasing sequence of consecutivevalues of si’s starting at 1 has length 2 (s1 = 1 and s3 = 0), which means that(⇡, V ) appears in rank 1 as a quotient. Excluding these values of si’s fromconsideration, we only have s2 = 1 left at this stage. The longest decreasingsequence of consecutive values of si’s starting at 1 is simply s2 = 1, whichcorresponds to (⇡, V ) appearing in rank 2 as well. Since we have exhaustedall the si’s, we get that (⇡, V ) appears in rank 1 and 2 as a quotient, whichagrees with our explicit computation in Chapter 7.6.3 Equivariant extension of ⇣GLn⇡ and Sataketransforms6.3.1 Satake transform on MnWe need to extend the definition of Satake transform from the sphericalHecke algebra of GLn to the space of K-bi-invariant functions on Mn. In[11], L. La↵orgue defined such an extension and proved that it is a formalseries “of divergence n12 ” in the function field case [[11], Proposition746.3. Equivariant extension of ⇣GLn⇡ and Satake transformsIII.23 p.94). We note that the proofs do not depend on the fact that heis working with a function field since the integral which defines the Sataketransform gives the same formal series when it is broken down by valuation.We recall from [[11], Definition III.16 p.88] the notion of ‘divergence’: a seriesP a has divergence d if there exists a polynomial P such that |a| P (deg())qdeg()d˙. In particular, this means that a series of divergence d inz1, . . . , zn will converge absolutely when it is evaluated at (z1qs, . . . , znqs)for <(s) suciently large.First, we recall various definitions of the Satake transform on the Heckealgebra. Let f 2 C1c (GLn)K⇥K . From [[7] Equation (3.4) p.6], the Sataketransform S(f) can be defined as a function on T by the integralS(f)(t) = 12 (t)ZNf(tu) du.Using the notation that will be introduced in Section 6.3.2, we observethat the Hecke algebra is generated by characteristic functions of the formchK$K with 2 X⇤(T )++, where X⇤(T )++ is the lattice of dominant co-characters, and we can restrict our attention to the Satake transform of suchfunctions. To shorten notation, letc(z) =Q>0(1 q1($))Q>0(1 ($)) , (6.1)and for w 2W , let w · z = (zw(j)).We have the Macdonald’s formula that allows us to compute the imageof chK$K under the Satake transform:Theorem 12 (Macdonald’s formula, Theorem 4.1.2 p.52 [12]). For an un-ramified character of the maximal split torus T , and f = ch(K$K) with 2 X⇤(T )++,S(f) = µM · 12 ($) Xw2Wc(w · z) · [w]($)!,where µM is the volume of the Levi component M that centralizes $.Let f 2 C1c (Mn)K⇥K . For s 2 C, consider the integral:12 (t)ZNf(tu)| det(t)|s du.756.3. Equivariant extension of ⇣GLn⇡ and Satake transformsIt follows from [[11], Proposition III.23 p.24] that the above integral is aformal series of divergence n12 , which will converge absolutely when it isevaluated at (z1qs, . . . , znqs) for suciently large <(s). Thus, the valueof the integral at s = 0 may be defined through analytic continuation, whichwe call the Satake transform of f at t, when it exists.Moreover, recall the following results from [Theorem 3 p. 253 [19]],[Theorem 6 p.397 [20]], and [11]:S(C1c (GLn)K⇥K) ⇠= C[z±1i ]W ,S(C1c (Mn)K⇥K)⇣1nQi=1(1 q n12 zi)C[z±1i ]W .For the rest of the thesis, we rely on the following hypothesis on the injec-tivity of the Satake transform of C1c (Mn)K⇥K :Hypothesis 1.S(C1c (Mn)K⇥K) ⇠=1nQi=1(1 q n12 zi)C[z±1i ]W .We do not have a proof but we believe that the hypothesis for the func-tion field case is a consequence of the Plancherel’s formula obtained byLa↵orgue [[11] Corollary IV.37 p.132]. For the p-adic field case, we includethe proof of the hypothesis for GL2 through explicit computations in Ap-pendix A.6.3.2 Cones in the co-character lattice and neighbourhoodsin MnWe will need a system of subsets of Mn(O) which, roughly speaking, providea system of K-biinvariant neighbourhoods of the sets Rm for various m. Itappears convenient to label such subsets by cones in the co-character latticeX⇤(T ).First, let us develop the notation for cones. Given 2 X⇤(T )++ adominant co-character, we denote by ⇤ the set {µ 2 X⇤(T ) | µ }. Thiscan be thought of a set of lattice points in the cone C := {µ 2 X⇤(T )⌦R |µ }. As we will soon need to talk about the sets of lattice points lyingon faces of cones, we use the superscript to record the dimension of thecorresponding cone. Using this notation, we write ⇤n0 for the set of dominantco-characters of GLn. Let ↵1, . . . ,↵n1 be the simple roots of GLn. For a766.3. Equivariant extension of ⇣GLn⇡ and Satake transformspolyhedral cone C ⇢ X⇤(T ) ⌦ R bounded by hyperplanes orthogonal toroots, we denote its faces by FHC, where H is a set of simple roots that areorthogonal to the face. More explicitly, the face FHC consists of all 2 Csuch that h↵`,i = 0 for ↵` 2 H. Similarly, let FH⇤ = FHC \X⇤(A). Wewill also refer to the sets FH⇤ as “faces” even though strictly speaking, theseare sets of integral points on the faces. Denote by (FH⇤) the set of latticepoints that lie in the interior of the face FH⇤ with the additional conventionthat for the zero-dimensional face, which is a single point, its interior is itselfrather than empty.Definition 13. Given a cone ⇤ ⇢ X⇤(T ) (of any dimension), denote byM⇤ the subset of Mn(O) \GLn(F ) defined byM⇤ =G=(1,...,n)2⇤K$1O...$nOK.We let f⇤ be the characteristic function of M⇤.For every m, 0 < m n, we have the embedding GLnm⇥GLm ,! GLn(as a Levi subgroup with the (nm)⇥ (nm)-block in the top left corner),and the corresponding embedding of co-character lattices X⇤(TGLnm) X⇤(TGLm) ,! X⇤(TGLn). If ⇤ = (⇤nm+1, . . . ,⇤n) 2 ⇤m0 , we denote by⇤nm⇤ the subset⇤nm⇤ := ⇤nm⇤nm+1 ⇤ = {(1, . . . ,nm,⇤nm+1 . . . ,⇤n) 2 ⇤n0}. (6.2)Observe that we have a decomposition of ⇤nm⇤ into a disjoint union of theinteriors of its faces:⇤nm⇤ =[H✓{↵1,...,↵nm}FH⇤nm⇤. (6.3)Note thatF;⇤nm⇤is simply the interior of ⇤nm⇤ , which we denote by(⇤nm⇤ ).6.3.3 Spherical functions supported on lower rankThe spherical Hecke algebra C1c (GLn)K⇥K has a basis that we denote by{f}2⇤n0 of the characteristic functions of the double cosets K$K, where runs over the dominant cone ⇤n0 in the co-character lattice X⇤(T ) (thisis Cartan decomposition). We need to establish similar decomposition forrank m matrices, 0 m n. As in the case of Cartan decomposition forGLn, it follows from Gaussian row reductions.776.3. Equivariant extension of ⇣GLn⇡ and Satake transformsLemma 19. We have the disjoint union decompositionRm =Gnm+1···nK24 0 ... 0$nm+1...$n35K,where the number of zero entries on the diagonal is nm.For each m between 0 and n1, the space C1c (Rm+1)K⇥K embeds intoC1c (Rm)K⇥K with the quotient isomorphic to C1c (Rm)K⇥K . We will needa section – a collection of elements of C1c (Rm)K⇥K that projects onto abasis of C1c (Rm)K⇥K modulo C1c (Rm+1)K⇥K .Using the notation introduced in (6.2), for ⇤ 2 ⇤m0 , denote by f⇤nm⇤the characteristic function of the setK2664$⇤nm+1O...$⇤nm+1O$⇤nm+1...$⇤n3775K.Note that this is a function in C1c (Rm)K⇥K .Lemma 20. The functions f⇤nm⇤for ⇤ 2 ⇤m0 lie in C1c (Rm)K⇥K andproject to a basis of C1c (Rm)K⇥K under restriction to Rm.Proof. This is a direct consequence of Lemma 19.6.3.4 Satake transforms of C1c (Rm)K⇥KDefinition 14. For 0 m n, let Jm be the ideal in C[z±1i ]W defined by:S(C1c (Rm)K⇥K) =JmnQi=1(1 q n12 zi). (6.4)Note that J0 = C[z±1i ]W , and Jn = hnYi=1(1 q n12 zi)i.Observe that by using the following short exact sequence:0! C1c (Rm+1)K⇥K ! C1c (Rm)K⇥K ! C1c (Rm)K⇥K ! 0,we may describe Jm in terms of Jm+1 if we find the Satake images of func-tions in C1c (Rm)K⇥K that project to a basis of C1c (Rm)K⇥K under the786.3. Equivariant extension of ⇣GLn⇡ and Satake transformsrestriction map. By Lemma 20, such functions have the form f⇤nm⇤for⇤ 2 ⇤m0 . For J1, we can obtain an explicit description in terms of J2 withone additional generator. However, for more general Jm, such descriptionappears to be very dicult to get (see Example 8 in the next chapter). Re-call that our main goal is to prove Theorem 14, and for that, it suces forus to study the image of Jm under ev eval, which we obtain in Theorem 13.Proposition 13. We have that:J1 = J22664 1nQi=1zinQi=1(1 q n12 zi)3775 .Proof. As discussed previously, it suces to compute S(f⇤n1⇤) for ⇤ 2 ⇤10since such functions lie in C1c (R1)K⇥K and project to a basis of C1c (R1)K⇥Kunder the restriction map. Moreover, observe that for every ⇤ 2 ⇤10 = Z,note that $⇤is a central element in GLn, andS(f⇤n1⇤) = S⇣chK$⇤K ⇤f⇤n11⌘= S (chK$⇤K)S(f⇤n11 ) = nYi=1zi!⇤S(f⇤n11 ).Thus, it suces to compute S(f⇤n11 ). We have that:S(f⇤n11 ) = S(chMn(O)(1,$1) · chMn(O)) = (1nYi=1zi)S(chMn(O)).Recall the following computation done by Tamagawa [Theorem 6 p.397 [20]]:S(chMn(O)) =1nQi=1(1 q n12 zi).Thus, we get:S(f⇤n11 ) =1nQi=1zinQi=1(1 q n12 zi).796.3. Equivariant extension of ⇣GLn⇡ and Satake transformsDefinition 15. For every 0 m n, we define the evaluation map evalnmwhere we shift each zi by qs and evaluate the first n m variables atparticular values:evalnm :C[z±1i ]! C[z±1nm+1, . . . , z±1n ][qs]P (z1, z2, . . . , zn) 7! P (qn12 s, qn12 +1s, . . . , qn12 +(nm1)s, znm+1qs, . . . , znqs)We extend the evaluation map naturally to the space of rational functions inthe variables zi. Let ev : C(qs) ! C be the map s 7! 0, which is the samemap as defined in Definition 12.Remark 11. Let P be a symmetric polynomial. Denote by evali1,...,inm theevaluation map that sends zi1 7! qn12 s, zi2 7! qn12 +1s, . . . , zinm 7!qn12 +(nm1)s, zinm+1 7! zinm+1qs, . . . , zn 7! znqs. Consider theWeyl element w that maps j 7! ij for 1 l n m and fixes j other-wise. Then,evalnm(P )(z) = evali1,...,inm(P )(w · z).Thus, since evaluating a symmetric polynomial at nm specific values givesthe same image up to permutation, we often identify evalnm(P )(z) withevali1,...,inm(P )(w · z) to shorten notation.For 0 m n, let Im be the ideal in C[znm+1, . . . , zn]W defined by:Im = hnYi=nm+1(1 q n12 (nm)zi)i.Using the above notation, we obtain the description for the image ofC1c (Rm)K⇥K under the Satake transform. Note that this is the main tech-nical result that does not rely on Hypothesis 1. To facilitate understandingof this theorem, we also include explicit computations for the case of GL3in Chapter 7. Thus, Chapter 7 can be read in parallel as an illustration ofthis proof.Theorem 13. For 0 m n,ev evalnm(Jm) ✓ Im.The rest of this section is devoted to this theorem. Recall the definitionsand notations introduced in Section 6.3.2. Observe that by Lemma 20,806.3. Equivariant extension of ⇣GLn⇡ and Satake transformsit suces to show that ev evalnm(f⇤nm⇤) 2 Im for every dominant co-character ⇤ 2 ⇤m0 . We first write f⇤nm⇤as a sum over the open faces of⇤nm⇤ . It follows from (6.3) thatf⇤nm⇤=XH✓{↵1,...,↵nm}f(FH⇤nm⇤ ) .Applying the Satake transform and using the definition of f⇤nm⇤, we havethatS(f(FH⇤nm⇤ )) =X2(FH⇤nm⇤ )S(f),and from Theorem 12, the Satake transform formula for S(f) contains acoecient µM , which depends on the stabilizer subgroup of in W . Sinceall ’s that lie in the same face have the same µM , we use the notationµM(FH⇤nm⇤ )instead of µM . We think of S(f(FH⇤nm⇤ )) as the sum of themain term coming from (⇤nm⇤ ) and the sum over the interiors of the faces,which vanish under ev evalnm as we will now show.6.3.5 Vanishing property of S(f(FH⇤nm⇤ )) if H 6= ;Proposition 14. Let H 6= ; be a subset of {↵1, . . . ,↵nm}, and ` thesmallest index such that ↵` lies in H. LetP(FH⇤nm⇤ ) = nYi=1(1 q n12 zi)!S(f(FH⇤nm⇤ )).Then,eval`(P(FH⇤nm⇤ )) = 0.The rest of this section is devoted to the proof of Proposition 14. Recallfrom Theorem 12 that each S(f) is a summation over the Weyl elementsw 2 W . Thus, we may rearrange the terms in S(f(FH⇤nm⇤ )) by collectingall the terms with the same Weyl element. For each w 2W , let:Sw(f(FH⇤nm⇤ )) = µM(FH⇤nm⇤ )X2(FH⇤nm⇤ )12 ($)[w])$= µM(FH⇤nm⇤ )X`+1...nm⇤nm+1X1>...>`1`+112 ($)[w]($).816.3. Equivariant extension of ⇣GLn⇡ and Satake transformsNote that the j ’s for 1 j ` 1 are distinct and we have a summationover each j . For the sum over `1 . . . nm ⇤nm+1, we mean thateither j > j+1 if ↵j 62 H or j = j+1 if ↵j 2 H, in which case, we havea single summation over j+1 instead of two summations over j and j+1.Using the notation above, we have that:S(f(FH⇤nm⇤ )) =Xw2Wc(w · z)Sw(f(FH⇤nm⇤ )),where c is defined in (6.1).First, we study the possible factors that appear in the denominator ofSw(f(FH⇤nm⇤ )) by doing explicit computations. We observe that these fac-tors come from two possible sources: factors of the form (zw(i) zw(j))coming from the denominator of c(w · z), and factors obtained from thegeometric series represented by the sum over j ’s. We split the summa-tions over j ’s into two parts: those over j ’s for 1 j ` 1 and thoseover j ’s where ` + 1 j n m . The motivation behind this splittingis that eventually we want to evaluate S(f(FH⇤nm⇤ )) at an `-tuple of val-ues (z1, . . . , z`) = (qn12 , . . . , qn12 +(`1)). The summations over j ’s for1 j ` 1 give rise to factors of the denominator which we will call type1 (see Lemma 21), which can lead to the vanishing of P(FH⇤nm⇤ ) , whereasthe summations over j ’s where `+1 j nm give rise to factors of type2 (see Lemma 22), which cannot contribute to the vanishing of P(FH⇤nm⇤ ) .Note that the image under ev eval`1 of the factors of Type 1 is a con-stant, whereas ev eval`1 of the factors of Type 2 is a function in at leasttwo variables.Lemma 21. The inner sum in the expression for Sw simplifies as:X1>...>`1`+112 ($)[w]($) =M1(w · z)`1Qj=1(1 q j(mj)2jQh=1zw(h)), (6.5)where M1(w · z) is a monomial in z.Proof. We have that:[w]($) =nmYj=1zjw(j)nYj=nm+1z⇤jw(j),12 ($) =nmYj=1qj(n+12j)2nYj=nm+1q⇤j (n+12j)2 .826.3. Equivariant extension of ⇣GLn⇡ and Satake transformsThus, by definition,X1>...>`1`+112 ($)[w]($) =nYj=nm+1⇣qn+12j2 zw(j)⌘⇤j X`1`+1⇣qn+12(`1)2 zw(`1)⌘`1 0@. . . X1>2⇣qn12 zw(1)⌘11A .Observe that this is a nested geometric series. We proceed by induction onh where 1 h `1. We will show that the common ratio of the geometricseries over h as we simplify the inner geometric series over 1 up to h1 isqh(nh)2hQj=1zw(j). Note that the factors of this common ratio come from twopossible sources:• (R1) the factor (q n+12h2 zw(h))h coming from 12 ($)[w]($);• (R2) the numerator of the expression obtained by simplifying the geo-metric series over 1 up to h1, which comes from the common ratioof the geometric series over h1.For the base case, observe that the innermost summation over 1 is a geo-metric series with common ratio qn12 zw(1) coming from type (R1). At thisstage, we do not have factors of the common ratio coming from type (R2)yet. We have that:X1>2⇣qn12 zw(1)⌘1=⇣qn12 zw(1)⌘2+11 q n12 zw(1),whose numerator gives the factor of type (R2) for the summation over 2.Thus, in combination with the factor qn32 zw(2) coming from type (R1), weget that the summation over 2 is a geometric series with common ratioqn2zw(1)zw(2).Next, we describe the induction step from h1 to h. Suppose that aftersimplifying the summations over 1, . . . ,h2, we have that the summationover h1 is a geometric series with common ratio q(h1)(n(h1))2h1Qj=1zw(j).Then, after simplifying the geometric series over h1, we have that thecommon ratio over h contains the factor qn+12h2 zw(h) from type (R1) and836.3. Equivariant extension of ⇣GLn⇡ and Satake transformsthe factor q(h1)(n(h1))2h1Qj=1zw(j) from type (R2). Therefore, the commonratio over h is qh(nh)2hQj=1zw(j), as desired.Now we are ready to compute Sw.Lemma 22. We have thatSw(f(FH⇤nm⇤ )) = Q1(w · z) ·M2(w · z)nQj=1(1 q j(mj)2njQi=1zw(i)), (6.6)where nj ` + 1, 0 n n m ` 1, M2 is a monomial in z, andQ1(w · z) is the right hand side of equation (6.5).Observe that the di↵erence between the factors in the denominator in(6.5) and those in the denominator in (6.6) is the number of zi’s that appearin each factor. Specifically, in (6.6), there are at least ` + 1 of zi’s in eachfactor, whereas in (6.5), there are at most ` 1.Proof. Recall thatSw(f(FH⇤nm⇤ )) =X`+1...nm⇤nm+10@ X1>...>`1`+11/2($)[w]($)1A .Similar to the previous lemma, this lemma also follows from the observationthat the summation over j where `+1 j nm is a geometric series withcommon ratio qj(nj)2jQh=1zw(h). The condition 0 n nm`1 reflectsthat we may not have a summation over each j for `+ 1 j nm.Definition 16. We defineP(FH⇤nm⇤ ),w =nYi=1(1 q n12 zi)c(w · z)Sw(f(FH⇤nm⇤ )).Then, by definition,P(FH⇤nm⇤ ) =Xw2WP(FH⇤nm⇤ ),w. (6.7)846.3. Equivariant extension of ⇣GLn⇡ and Satake transformsNote that for each w 2W , the function P(FH⇤nm⇤ ),w is a rational functionin zi, z1i ’s, whereas the function P(FH⇤nm⇤ ) is a polynomial in zi’s.Let D2(w · z) be the denominator of the right hand side of (6.6), andM(w · z) the product M1(w · z)M2(w · z).Lemma 23. The only non-vanishing terms of ev eval`1(P(FH⇤nm⇤ )) arethe summands ev eval`1(P(FH⇤nm⇤ ),w) that corresponds to the Weyl ele-ments w 2W satisfying w(j) = j for 1 j ` 1. Moreover,ev eval`1(P(FH⇤nm⇤ ),w)=C`1 · c,`1(w · z) · ev eval`1(M(w · z)) ·nQi=`+1(1 q n12 (`1)zi)ev eval`1(D2(w · z)) ,(6.8)where C`1 is a constant, and c,`1(w · z) is defined recursively by Equa-tion (6.9) below.Proof. We proceed by induction on h for 0 h `1. Namely, we describethe factors that cancel in the numerator and the denominator as we evaluateev evalh1(P(FH⇤nm⇤ ),w) at zh = qn12 +(h1). For h = 0, that is, beforethe evaluation, the statement follows from Lemma 22 with c,0 = c.We emphasize that the denominator of P(FH⇤nm⇤ ),w contains the fol-lowing factors:(D1) factors in the denominator coming from Equation (6.5), which wecalled factors of Type 1. Note that these factors will completely cancelwhen we apply eval`1;(D2) factors of D2(w·z) coming from Equation (6.6), which we called factorsof Type 2. These factors will not cancel when we apply eval`1;(D3) the denominator of c(w ·z), which contains factors of the form (zw(i)zw(j)).The numerator of P(FH⇤nm⇤ ),w contains the following non-constant factors:(N1) factors of M(w · z) coming from Equation (6.6). These factors willnot cancel when we apply eval`1;(N2) the productnQi=1(1 q n12 zi), coming from Definition 16. This productwill later cancel with the product coming from the denominator ofc(w · z);856.3. Equivariant extension of ⇣GLn⇡ and Satake transforms(N3) the numerator of c(w · z), which contains factors of the form (zw(i)qzw(j)).We describe the induction step from h = 0 to h = 1 in detail. First,observe that we have (1 q n12 z1) in the numerator coming from (N2).Thus, if the denominator does not contain the factor (1 q n12 z1), thenev eval1(P(FH⇤nm⇤ ),w) = 0. By inspection, one only obtains the factor ofthis kind from (D1) if and only if w(1) = 1. When w(1) = 1, in the denom-inator of ev eval1(P(FH⇤nm⇤ ),w), we factor out qn12 from the factors oftype (D3) that contain z1, and get the productnQi=2(1 q n12 zi). We cancelout this product with the remaining factors of type (N2). As a result, nofactors of type (N2) are left. Moreover, when w(1) = 1, in the numeratorof ev eval1(P(FH⇤nm⇤ ),w) , we factor out qn12 from the factors of type(N3) that contain z1. We get the productnQi=2(1 q n12 1zi), which we nowlabel as (N2) again. We denote the ratio of factors of type (N3) and (D3)by c,1(w · z).Similarly, we carry out the induction step from h1 to h. At this stage,we have the factor (1q n12 (h1)zh) in the numerator of ev evalh1(P(FH⇤nm⇤ ),w)coming from (N2). If (1 q n12 (h1)zh) is not a factor of the denomi-nator of ev evalh1(P(FH⇤nm⇤ ),w), then ev evalh(P(FH⇤nm⇤ ),w) = 0. If(1 q n12 (h1)zh) is a factor of the denominator, then it has to come fromthe type (D1) when w(h) = h. In this case, we factor out qn12 +(h1) fromthe factors of type (N3) and (D3) that contain zh, and get:ev evalh(c,h1(w · z)) =nQi=h+1(1 q n12 hzi)nQi=h+1(1 q n12 (h1)zi)c,h(w · z). (6.9)We let the productnQi=h+1(1 q n12 hzi) become the new factors of type(N2).Proof of Proposition 14. Observe that ev eval`(P(FH⇤nm⇤ )) can be obtainedby evaluating ev eval`1(P(FH⇤nm⇤ )) at z` = qn12 +(`1). By Lemma 23,the only non-vanishing terms of ev eval`1(P(FH⇤nm⇤ )) are the summands866.3. Equivariant extension of ⇣GLn⇡ and Satake transformsev eval`1(P(FH⇤nm⇤ ),w) for the Weyl elements w 2W satisfying w(j) = jfor 1 j ` 1, and each summand has the form (6.8). Observe that(1 qn12 +(`1)z`)is not a factor of the denominator of ev eval`1(P(FH⇤nm⇤ ),w) but it is afactor of the numerator of ev eval`1(P(FH⇤nm⇤ ),w) of type (N2). There-fore, ev eval`(P(FH⇤nm⇤ ),w) = 0 for all w 2W , which completes the proof.6.3.6 Vanishing property of S(f(⇤nm⇤ ))Proposition 15. LetP(⇤nm⇤ ) = nYi=1(1 q n12 zi)!S(f(⇤nm⇤ )).Then,ev evalnm(P(⇤nm⇤ ))= Cnm0B@ Xw2W,w(j)=j1jnmc,nm(w · z)nYi=nm+1z⇤iw(i)1CA · nYi=nm+1(1 q n12 (nm)zi),where Cnm is a constant.Proof. This proposition follows the same computation done in the proof ofLemma 21.We have a corollary from the calculation done in the previous proof thatwill be used later.Corollary 7. Consider the function f⇤nm0⇤which lies in C1c (Rm)K⇥K .Then,ev evalnm⇣P⇤nm0⇤⌘= Cnm0B@ Xw2W,w(j)=j1jnmc,nm(w · z)1CA nYi=nm+1(1 q n12 (nm)zi),wherePw2W,w(j)=j1jnmc,nm(w · z) is a non-zero constant.876.3. Equivariant extension of ⇣GLn⇡ and Satake transformsProof. The form of ev evalnm(P⇤nm0⇤) follows directly from Proposition 15.To show thatPw2W,w(j)=j1jnmc,nm(w · z) is a non-zero constant, letW 0 := {w 2W | w(j) = j, 1 j nm} ⇠= Sm,which permutes the zi’s for nm+ 1 i n.Recall that the factors in the denominator ofPw2W 0c,nm(w ·z) have theform (zw(i) zw(j)). We will first show that (zw(i) zw(j)) is also a factorof the numerator ofPw2W 0c,nm(w · z). We can pair the c,nm(w · z)’saccording to the cosets of Sm/(ij). Namely, the two terms c,nm(w · z) inthe same pair will have all common factors in the numerators except one,that is, one has (zw(i)q1zw(j)) and the other has (zw(j)q1zw(i)). Thus,when zw(i) = zw(j), the sum of each pair will be zero, and (zw(i) zw(j))is a factor of the numerator ofPw2W 0c,nm(w · z). In particular, the de-nominator ofPw2W 0c,nm(w ·z) is identically 1, andPw2W 0c,nm(w · z) is infact a polynomial. However, the polynomials in the numerator and the de-nominator ofPw2W 0c,nm(w · z) originally have the same degree. Therefore,Pw2W,w(j)=j1jnmc,nm(w · z) is a non-zero constant.6.3.7 Equivariant extension of ⇣GLn⇡In this section, we study the extension properties of the generator ⇣GLn⇡ inHomH(GLn,K)(C1c (GLn)K⇥K , V K ⌦ eV K). We prove the following:Theorem 14. Assume Hypothesis 1 in Section 6.3.1, and let (⇡, V ) bea spherical irreducible admissible representation of GLn, which is an irre-ducible quotient of |.|s1⇥ . . . |.|sn. Suppose that the value n12 appears exactlyonce in the set {s1, . . . , sn}, and let r be the length of the longest decreasingsequence of consecutive integers in {s1, . . . , sn} starting at n12 . Then,(1) ⇡ appears in rank r as a quotient;(2) for every m > r, ⇣GLn⇡ extends H(GLn,K)-equivariantly to an elementof HomH(GLn,K)(C1c (Rm)K⇥K , V K ⌦ eV K);(3) the map ⇣GLn⇡ does not extend H(GLn,K)-equivariantly to an elementof HomH(GLn,K)(C1c (Rr)K⇥K , V K ⌦ eV K).886.3. Equivariant extension of ⇣GLn⇡ and Satake transformsProof of (1). Since {n12 , n12 1, . . . , n12 (n r 1)} is the sequenceof maximal length {s1, . . . , sn} contained in {s1, . . . , sn}, it follows fromProposition 12(ii) that (⇡, V ) appears in C1c (Rr) as a quotient.Proof of (2). We prove (2) by downward induction on m. When m = n > r,then the statement is vacuously true. Suppose that the statement holds form+1. We want to show that the statement also holds for m, where m > r,that is, ⇣GLn⇡ can be extended H(GLn,K)-equivariantly to an element inHomH(GLn,K)(C1c (Rm)K⇥K , V K ⌦ eV K).Recall the following exact sequence:0!C1c (Rm+1)K⇥K ext! C1c (Rm)K⇥K res! C1c (Rm)K⇥K ! 0By applying the Satake transform to C1c (Rm+1)K⇥K and C1c (Rm)K⇥K ,we obtain the following short exact sequence:0! Jm+1nQi=1(1 q n12 zi)! JmnQi=1(1 q n12 zi)! Qm ! 0 (6.10)where Jm and Jm+1 are the ideals as defined in (6.4), and Qm is the quotientmodule. By the inductive hypothesis, we have the following commutativediagram:C1c (GLn)K⇥K _✏✏⇣GLn⇡''⇠= // C[z±1i ]Wev evalxx _✏✏C1c (Rm+1)K⇥K _✏✏⇣Rm+1⇡// V K ⌦ eV K ⇠= C Jm+1nQi=1(1q n12 zi) _✏✏eval⇤ooC1c (Rm)K⇥K⇠= //?88JmnQi=1(1q n12 zi)?eewhere eval : C[z±1i ]W ! C(qs) is the evaluation map zi 7! qsis and ev :C(qs) ! C is the evaluation map at s = 0. Note that the isomorphism inthe top row is the well-known Satake isomorphism, whereas the isomorphismin the bottom row is a consequence of Hypothesis 1 in Section 6.3.1.Observe that from the above commutative diagram, in order to show theexistence of an H(GLn,K)-equivariant extension of ⇣GLn⇡ inHomH(GLn,K)(C1c (Rm)K⇥K , V K ⌦ eV K),896.3. Equivariant extension of ⇣GLn⇡ and Satake transformsit suces to show that the map eval⇤ is well-defined from JmnQi=1(1q n12 zi)to C.Moreover, since eval⇤ is the extension of the composition ev eval, it sucesto show that ev eval is well-defined on JmnQi=1(1q n12 zi). Thus, we need to showthat every function in eval(S(C1c (Rm)K⇥K)) is well-defined at s = 0.Let f 2 C1c (Rm)K⇥K , and Pf (zi) =nQi=1(1q n12 zi)S(f) 2 Jm. Observethat the denominator of eval(S(f)) has exactly one factor (1 qs) becausethe value n12 appears exactly once in the set {s1, . . . , sn}. Thus, if we canshow that eval(Pf ) contains the factor (1 qs), then we are done. Letz0i = ziqs. Then,ev(Pf (z0i)) = Pf (zi).Since ev eval(Pf (zi)) = 0, we have that ev(Pf (qsis)) = 0. It followsfrom Pf (qsis) 2 C(qs) and ev(Pf (qsis)) = 0 that Pf (qsis) containsa factor of (1 qs). Thus, if ev eval(Jm) = 0, then every function ineval(S(C1c (Rm)K⇥K)) is well-defined at s = 0.Next, we will show that ev eval(Jm) = 0. Since m > r (that is, nm <n r), there exists an index inm+1 such that zinm+1 = qn12 +nm. ByTheorem 13, since ev evalnm(Jm) ✓ Im, it follows that ev evalnm+1(Jm) =0. Therefore, for (⇡, V ) appearing in rank r, ev eval(Jm) = 0. Thus,the composition ev eval is well-defined on JmnQi=1(1q n12 zi), which shows thatthe map eval⇤ can be extended to JmnQi=1(1q n12 zi). Therefore, ⇣GLn⇡ extendsH(GLn,K)-equivariantly to an element of HomH(GLn,K)(C1c (Rm)K⇥K , V K⌦eV K).Proof of (3). To show that ⇣GLn⇡ does not extend H(GLn,K)-equivariantlyto an element in HomH(GLn,K)(C1c (Rr)K⇥K , V K ⌦ eV K), we show that thecomposition ev eval from JrnQi=11q n12 zito C cannot be defined. Similarlyto the proof of (2), we consider only K-fixed vectors and use the Sataketransform to obtain the following sequence:0! Jr+1nQi=1(1 q n12 zi)! JrnQi=1(1 q n12 zi)! Qr ! 0 (6.11)906.3. Equivariant extension of ⇣GLn⇡ and Satake transformswhere Jr and Jr+1 are the ideals as defined in (6.4), and Qr is the quotientideal. It follows from (2) that there exists an H(GLn,K)-equivariant exten-sion of ⇣GLn⇡ in HomH(GLn,K)(C1c (Rr+1)K⇥K , V K⌦ eV K), and therefore, anextension eval⇤ of ev eval from Jr+1nQi=11q n12 zito C. To show that ev eval can-not be defined from JrnQi=11q n12 zito C, it suces to show that ev eval(Jr) 6=0. This is because the value n12 appears in the set {s1, . . . , sn} exactly once,which means the denominator of every eval(S(f)) for f 2 C1c (Rr)K⇥Kcontains a factor (1 qs). Thus, if ev eval(Jr) 6= 0, then for somef 2 C1c (Rr)K⇥K , we have that ev eval(Pf ) 6= 0, and thus, eval(S(f))is undefined at s = 0.To show that ev eval(Jr) 6= 0, we proceed by contradiction and as-sume that ev eval(Jr) = 0. Recall the function f⇤nm0⇤2 C1c (Rm)K⇥K asdefined in 6.3.2. By Corollary 7, we have thatev evalnm⇣P⇤nm0⇤⌘=0B@ Xw2W,w(j)=j1jnmc,nm(w · z)1CA nYi=nm+1(1q n12 (nm)zi).Thus, since ev evalnm(P⇤nm0⇤) = 0, there exists an index inr+1 such thatzinr+1 = qn12 +(nr). This contradicts the assumption that r is the lengthof the longest decreasing sequence of consecutive integers in {s1, . . . , sn}starting at n12 . Therefore, ev eval(Jr) 6= 0, and the map eval⇤ cannot bedefined from JrnQi=1(1q n12 zi)to C, which completes the proof.91Chapter 7The extension problem forGL3In this chapter, we study the case n = 3 in detail by applying the generalresults of Chapter 4 and Chapter 6. The goals of this chapter are the follow-ing: first, we want to have concrete examples to illustrate some dicultiesthat were not present for n = 2 in Chapter 3, but manifest themselveswhen we study the case of an arbitrary n. Secondly, we carry out some ex-plicit computation to facilitate understanding of the more technical resultsin Chapter 6. We start with a quick summary of the results in this chapter.7.1 Summary of results for GL3In Section 7.2, we obtain a complete list of irreducible spherical represen-tations (⇡, V ) that appear in the spherical part of the spectrum of C1c (R1)and C1c (R2) as a quotient (see Lemma 24 and Lemma 25). The methodused in the proofs of these lemmas is combinatorial in nature and it showsthe complexity of the problem for general n. From those results, we get anexample of a representation that appears in multiple low ranks as a quo-tient (Example 3). For GL3, the representation |.|⇥ |.| 12 det2 is the uniquesuch that appears in both ranks 1 and 2 as a quotient. The representa-tions appearing in multiple low ranks are among those to which the resultsof Theorem 14 do not apply since their Satake parameters {s1, s2, . . . , sn}contain the value n12 with multiplicity greater than 1.Then, in Section 7.3.2, we carry out some explicit computation withspherical functions in C1c (R2)K⇥K . In Example 8, we compute the Satakeimage of the characteristic function of the setKh$O$1iK,which is denoted by f⇤1⇤where ⇤ = (1, 0) 2 ⇤20 using the notation intro-duced in Section 6.3.2. Observe that S(f⇤1⇤) is complicated, which shows927.2. Representations that appear in low ranksthe diculty in finding a nice description for S(C1c (Rm)K⇥K) in general.Then, also in Example 8, we show that despite its complexity, when weevaluate the the numerator of S(f⇤1⇤) at z1 = q1, the image is containedin the ideal h(1 z2)(1 z3)i. This is a particular example of Theorem 13.Its proof follows the same argument as in the proof of Theorem 13, and itsinclusion here mainly serves as an example to demonstrate the main ideabehind the proof of Theorem 13. We also compute the Satake image of thecharacteristic function of the setKhO11iK,in Example 9, since it will be used later.Next, we provide a particular instance of Theorem 14 in Section 7.3.3.In Proposition 16, we show that for the representation (⇡, V ) = (|.|det3,C)that only appears in rank 1, ⇣GL3⇡ extends equivariantly to C1c (R2)K⇥K ,but does not extend to C1c (R1)K⇥K . The proof of Proposition 16 followsexactly the same outline as that of Theorem 14 and also depends on Hy-pothesis 1 in Section 6.3.1. Lastly, in Section 7.3.4, we study the extensionof ⇣GL3⇡ where ⇡ = |.|⇥ |.|12 det2 is the representation that appears in mul-tiple low ranks. Note that Theorem 14 does not apply directly in this casebecause the Satake parameters of (⇡, V ) contain the value 1 with multi-plicity 2. However, a similar argument to that in the proof of Theorem 14also shows that ⇣GL3⇡ does not extend to C1c (R2)K⇥K . This is becausehaving a pole at s = 1 with multiplicity is only an obstruction to provingthe extension of ⇣GL3⇡ .7.2 Representations that appear in low ranksIn this section, we classify the irreducible spherical representations of GL3that appear in lower ranks. Our main tools are Proposition 10, Proposi-tion 12 , and the classification of spherical representations of GL3.The approach that we will use is the following: by Proposition 10, ⇡is a quotient of a representation of the form ⌧ ⇥ |.|m2 det3m for some⌧ 2 Irr(GL3) if and only if ⇡ appears in rank m. On the other hand, recallthat by Theorem 3, every irreducible unramified representation of GL3 is theunique irreducible quotient of a representation of the form |.|s1 ⇥ |.|s2 ⇥ |.|s3where sj 6= si + 1 whenever i < j. As we shall see in general in Section 6.2Proposition 12 (i), the condition that ⇡ appears in a certain low rank gives arestriction on the values of si. For example, if (⇡, V ) appears in rank 1, then{0, 1} is a subset of {s1, s2, s3}. In this section, we examine case by case the937.2. Representations that appear in low rankspossible values of si to determine if ⇡ is a quotient of a representation of theform ⌧ ⇥ |.|m2 det3m for some ⌧ 2 Irr(GL3).The main diculty in generalizing this complete analysis of the spaceC1c (Rm) to an arbitrary n is the combinatorial nature of the relationshipbetween the si’s and the irreducible representation ⇡. Namely, if si = si+1+1, then the induced representation |.|si ⇥ |.|si+1 is reducible, and we need toconsider its unique irreducible quotient.7.2.1 Representations that appear in rank 1Lemma 24.HomGL3⇥GL3(C1c (R1), V ⌦ eV ) ⇠=(C if (⇡, V ) is |.| det3 or |.|t ⇥ |.| 12 det2 ,{0} otherwise.where t 6= 1, 2.Proof. By Proposition 12 (i), if (⇡, V ) appears in C1c (R1), then (⇡, V ) is theunique irreducible quotient of a representation of the form |.|s1 ⇥ |.|s2 ⇥ |.|s3where sj 6= si+1 whenever i < j and {0, 1} is a subset of {s1, s2, s3}. First,we consider the case where s1 = 1 and s2 = 0, that is, ⇡ is the uniqueirreducible quotient of |.|⇥ 1⇥ |.|s3 . We have two possibilities: s3 = 1 i.e.the segments [1] and [|.|s3 ] are linked, and s3 6= 1 i.e. they are not linked.Observe that we have the following quotient map(|.|⇥ 1)⇥ |.|s3 ! |.| 12 det2 ⇥ |.|s3 ! ⇡.In segment notation, |.| 12 det2 ⇥ |.|s3 = h[1, |.|]i ⇥ h[|.|s3 ]i. The di↵erencebetween the two cases is whether the segment [|.|s3 ] precedes the segment[1, |.|], which determines if |.| 12 det2⇥|.|s3 is irreducible and if |.| 12 det2⇥|.|s3is equivalent to |.|s3 ⇥ |.| 12 det2. We will see that the equivalence between|.| 12 det2 ⇥ |.|s3 and |.|s3 ⇥ |.| 12 det2 is the key to finding an irreduciblerepresentation ⌧ of GL1 such that HomGL3(⌧ ⇥ |.|12 det2,⇡) ⇠= C, which byProposition 10 would tell us that ⇡ appears in rank 1.• When s3 6= 1, that is, the segments [1, |.|] and [|.|s3 ] are not linked,it follows from Theorem 1 that |.| 12 det2 ⇥ |.|s3 = h[1, |.|]i ⇥ h[|.|s3 ]iis irreducible. Thus, ⇡ = |.| 12 det2 ⇥ |.|s3 is the unique irreduciblequotient of |.|⇥ 1⇥ |.|s3 . Moreover, by Theorem 2, |.| 12 det2⇥ |.|s3 ⇠=|.|s3 ⇥ |.| 12 det2, and thusHomGL3(|.|s3 ⇥ |.|12 det2,⇡) ⇠= C.947.2. Representations that appear in low ranksIt follows from Proposition 10 that ⇡ appears in rank 1.• When s3 = 1, we have that ⇡ = h[|.|], [1], [|.|1]it ⇠= 13. We will showthat 13 does not appear in rank 1 by showing that there does notexist an irreducible representation ⌧ of GL1 such that 13 is the uniqueirreducible quotient of ⌧ ⇥ |.| 12 det2. Note that if such ⌧ exists, then⌧ = |.|1. However, the unique irreducible quotient of |.|1⇥ |.| 12 det2,denoted by h[|.|1], [|.|], [1]it is not equivalent to h[|.|], [1], [|.|1]it = 13by Theorem 2 because the segment [|.|1] precedes the segment [1].Thus, no such ⌧ exists and 13 does not appear in rank 1.The case where s2 = 1 and s3 = 0 can be studied similarly. We summarizeall the possibilities in the following table:Cases |.|s1 ⇥ |.|s2 ⇥ |.|s3 (⇡, V ) ⌧ ⇥ |.| 12 det2s1 = 1, s2 = 0 |.|⇥ 1⇥ |.|s3 13 if s3 = 1 (none)|.| 12 det2⇥|.|s3 if s3 6= 1 |.|s3 ⇥ |.| 12 det2s2 = 1, s3 = 0 |.|s1 ⇥ |.|⇥ 1 |.| det3 if s1 = 2 |.|2 ⇥ |.| 12 det2|.|s1 ⇥ |.| 12 det2 if s1 6= 2 |.|s1 ⇥ |.| 12 det27.2.2 Representations that appear in rank 2Lemma 25.HomGL3⇥GL3(C1c (R2), V ⌦ eV ) ⇠=(C if (⇡, V ) is |.|2 det3 or ⌧ ⇥ |.|,{0} otherwise.where ⌧ = |.|t det2 where t 6= 52 ,12 , or ⌧ = |.|s ⇥ |.|t where s, t 6= 1, 2.Proof. In this case, Proposition 12 (i) states that if (⇡, V ) appears in C1c (R2),then (⇡, V ) is the unique irreducible quotient of a representation of the form|.|s1 ⇥ |.|s2 ⇥ |.|s3 where sj 6= si + 1 whenever i < j and {1} is a subset of{s1, s2, s3}. We will consider the case s1 = 1 in detail and the other caseswhere s2 = 1 or s3 = 1 can be done similarly.Suppose that s1 = 1. We need to consider each of the following possibili-ties: whether the segment [|.|s2 ] is linked with [|.|], and whether the segment[|.|s2 ] is linked with [|.|s3 ]. We have the following cases:• If the segment [|.|s2 ] is linked with [|.|], then s2 = 0. We have twosubcases:957.2. Representations that appear in low ranks– If the segment [|.|s2 ] = [1] is linked with [|.|s3 ], then s3 = 1.Thus, ⇡ = h[|.|], [1], [|.|1]it ⇠= 13. In this case, ⇡ is the uniqueirreducible quotient of |.|⇥ |.| 12 det2. However, by Theorem 2,|.|⇥ |.| 12 det2 6⇠= |.| 12 det2 ⇥ |.| because the segment [|.|1,1]corresponding to |.| 12 det2 precedes the segment [|.|]. Thus, 13does not appear in rank 2 because we cannot find an irreduciblerepresentation ⌧ of GL2 such that 13 is the unique irreduciblequotient of ⌧ ⇥ |.|.– If the segment [|.|s2 ] = [1] is not linked with [|.|s3 ], that is, s3 6=1, then we have the following quotient map(|.|⇥ 1)⇥ |.|s3 ! |.| 12 det2 ⇥ |.|s3 .Note that by Theorem 1, the representation |.| 12 det2 ⇥ |.|s3 =h[1, |.|]i ⇥ h[|.|s3 ]i is irreducible, because the segments [1, |.|] and[|.|s3 ] are not linked if s3 6= 2 and s3 6= 1. Thus, ⇡ = |.| 12 det2⇥|.|s3 is the unique irreducible quotient of |.|⇥ 1⇥ |.|s3 . However,since s3 6= 1, ⇡ is not the unique irreducible quotient of a rep-resentation of the form ⌧ ⇥ |.| where ⌧ 2 Irr(GL2). Therefore, byProposition 10, ⇡ does not appear in rank 2.• If the segment [|.|s2 ] is not linked with [|.|], that is, s2 6= 0, then wehave the following subcases:– If the segment [|.|s2 ] is linked with [|.|s3 ], that is, s3 = s21, thenwe have the following quotient map|.|⇥ (|.|s2 ⇥ |.|s21)! |.|⇥ |.|s2 12 det2.By Theorem 1, the representation |.| ⇥ |.|s2 12 det2 = h[|.|]i ⇥h[|.|s21, |.|s2 ]i is irreducible, because the segments h[|.|]i and h[|.|s21, |.|s2 ]iare not linked if s2 6= 3 and s2 6= 0. Thus, ⇡ = |.|⇥ |.|s2 12 det2is the unique irreducible quotient of |.|⇥ |.|s2 ⇥ |.|s21. Moreover,by Theorem 2, |.|⇥ |.|s2 12 det2 ⇠= |.|s2 12 det2 ⇥ |.|, and thus,HomGL3(|.|s212 det2 ⇥ |.|,⇡) ⇠= C.It follows from Proposition 10 that ⇡ appears in rank 2.– If the segment [|.|s2 ] is not linked with [|.|s3 ], that is, s3 6= s2 1,then the representation⇡ = |.|⇥ |.|s2 ⇥ |.|s3 = h[|.|]i ⇥ h[|.|s2 ]i ⇥ h[|.|s3 ]i967.3. Equivariant extension of ⇣GL3⇡ and Satake transformsis irreducible because the segments [|.|], [|.|s2 ] and [|.|s3 ] are pair-wise not linked. Moreover, by Theorem 2, |.| ⇥ |.|s2 ⇥ |.|s3 ⇠=(|.|s2 ⇥ |.|s3)⇥ |.|. Thus,HomGL3((|.|s2 ⇥ |.|s3)⇥ |.|,⇡) ⇠= C.Note that |.|s2 ⇥ |.|s3 is an irreducible representation of GL2 be-cause s3 6= s2 1 and s2 6= s3 1. It follows from Proposition 10that ⇡ appears in rank 2.We summarize all the possibilities in the following table:Cases |.|s1 ⇥ |.|s2 ⇥ |.|s3 (⇡, V ) ⌧ ⇥ |.|s1 = 1|.|⇥ 1⇥ |.|s3 1 det3 if s3 = 1 (none)|.| 12 det2⇥|.|s3 if s3 6= 1 (none)|.|⇥ |.|s2 ⇥ |.|s3 if s2 6= 0 |.|⇥ |.|s2 12 det2 if s3 = s2 1 |.|s2 12 det2⇥|.||.|⇥ |.|s2 ⇥ |.|s3 if s3 6= s2 1 |.|s2 ⇥ |.|s3 ⇥ |.|s2 = 1|.|2 ⇥ |.|⇥ |.|s3 |.| det3 if s3 = 0 (none)|.| 32 det2⇥|.|s3 if s3 6= 0 |.| 32 det2⇥|.| if s3 = 1|.|s1 ⇥ |.|⇥ |.|s3 if s1 6= 2 |.|s1 ⇥ |.| 12 det2 if s3 = 0 |.| 12 det2⇥|.| if s1 = 1|.|s1 ⇥ |.|⇥ |.|s3 if s3 6= 0 |.|s1 ⇥ |.|s3 ⇥ |.|s3 = 1|.|s1 ⇥ |.|2 ⇥ |.| |.|2 det3 if s1 = 3 |.| 52 det2⇥|.||.|s1 ⇥ |.| 32 det2 if s1 6= 3 (none)|.|s1 ⇥ |.|s2 ⇥ |.| if s2 6= 2 |.|s1 12 det2⇥|.| if s1 = s2 + 1 |.|s1 12 det2⇥|.||.|s1 ⇥ |.|s2 ⇥ |.| if s1 6= s2 + 1 |.|s1 ⇥ |.|s2 ⇥ |.|Example 3 (A representation that appears in multiple low ranks ). By theabove classification, the representation ⇡ = |.|⇥|.| 12 det2 of GL3 appears bothin rank 1 and 2. In fact, for GL3, ⇡ is the unique representation that appearsin multiple low ranks. Observe that in this case, the L-function LGL3(⇡, s)has a double pole at s = 1, whereas for any other representation of GL3that appears in one low rank, its associated L-function has a simple pole ats = 1.7.3 Equivariant extension of ⇣GL3⇡ and SataketransformsIn this section, we use the notation introduced in 6.3.977.3. Equivariant extension of ⇣GL3⇡ and Satake transforms7.3.1 The Satake transform of some explicit functions inC1c (GL3)K⇥KIn this section, we will compute the explicit images of some particular func-tions in C1c (GL3)K⇥K under the Satake transform, which will be used inthe computations in the next section. The main tool that we use is theMacdonald’s formula, which is stated as Theorem 12 in Section 6.3.1.Example 4. For f(1,1,0) 2 C1c (GL3)K⇥K which is the characteristic func-tion of the setKh$1$1iK, where 1 > 1,we have thatS(f(1,1,0)) =c(z)(qz1)1z2 + c((23) · z)(qz1)1z3 + c((12) · z)(qz2)1z1+c((123) · z)(qz2)1z3 + c((132) · z)(qz3)1z1 + c((13) · z)(qz3)1z2.Proof. The lemma follows from Theorem 12, where µM(1,1,0) = 1 becausethe Levi component M of the centralizer of $(1,1,0) is trivial, and12 ($(1,1,0)) = q1 ,[w]($(1,1,0)) =8>>>>>>>>><>>>>>>>>>:z11 z2, if w = ()z11 z3, if w = (23)z12 z1, if w = (12)z12 z3, if w = (123)z13 z1, if w = (132)z13 z2, if w = (13)Example 5. For f(1,1,0) 2 C1c (GL3)K⇥K which is the characteristic func-tion of the setKh$$1iK,we have thatS(f(1,1,0)) = 11 + q1 ((c(z) + c((12) · z))qz1z2 + (c((23) · z) + c((132) · z))qz1z3+(c((123) · z) + c((13) · z))qz2z3).987.3. Equivariant extension of ⇣GL3⇡ and Satake transformsProof. In this case, µM(1,1,0) =11+q1 because the Levi component M(1,1,0)of the centralizer of $(1,1,0) is isomorphic to GL2⇥{1}, and12 ($(1,1,0)) = q,[w]($(1,1,0)) =8><>:z1z2, if w = (), (12)z1z3, if w = (23), (132)z2z3, if w = (123), (13).Example 6. For f(0,0,0) = chK , we have thatS(f(0,0,0)) = 1.Proof. In this case, µM(0,0,0) =1(1+q1)(1+q1+q2) because the centralizer of$(0,0,0) is GL3, 12 ($(0,0,0)) = 1, and [w]($(0,0,0)) = 1 for all w 2W .Example 7. For f(1,0,0) 2 C1c (GL3)K⇥K which is the characteristic func-tion of the setKh$111iK, where 1 > 0,we have thatS(f(1,0,0)) =(z1 q1z2)(z1 q1z3)(z1 z2)(z1 z3) z11 +(z2 q1z1)(z2 q1z3)(z2 z1)(z2 z3) z12 +(z3 q1z1)(z3 q1z2)(z3 z1)(z3 z1) z13 .Proof. The lemma follows from Theorem 12, where µM(1,0,0) =11+q1 be-cause the Levi component M(1,0,0) of the centralizer of $(1,0,0) is isomor-phic to GL2, and12 ($(1,0,0)) = q1 ,[w]($(1,0,0)) =8><>:z11 , if w = (), (23)z12 , if w = (12), (123)z13 , if w = (13), (132)997.3. Equivariant extension of ⇣GL3⇡ and Satake transforms7.3.2 The Satake transform of some explicit functions inC1c (R2)K⇥KWe will illustrate the complexity of the image of C1c (R2)K⇥K under theSatake transform by computing the image of some particular functions inC1c (R2)K⇥K . In Example 8, we show that the image of a test functionunder the Satake transform can be complicated. However, after evaluatingat z1 = q1, the image of the numerator is simple and always lies in aparticular ideal. Then, in Example 9, we also compute the Satake image ofa test function that turns out to be nice, and it will be used later in theproof of Proposition 17.Example 8. Using the notation introduced in (6.2), let ⇤ = (1, 0) 2 ⇤20,and f⇤1⇤the characteristic function of the setKh$O$1iK,which is an element in C1c (R2)K⇥K . Then,S(f⇤1⇤) =✓11 + q1+qz11 qz1◆c(z)qz1z2 +✓11 + q1+qz21 qz2◆c((12) · z)qz1z2+✓11 + q1+qz11 qz1◆c((23) · z)qz1z3 +✓11 + q1+qz31 qz3◆c((132) · z)qz1z3+✓11 + q1+qz21 qz2◆c((123) · z)qz2z3 +✓11 + q1+qz31 qz3◆c((13) · z)qz2z3.LetP⇤1⇤= (1 qz1)(1 qz2)(1 qz3)S(f⇤1⇤).Then,ev eval1(P⇤1⇤) = (1 z2)(1 z3)(z2 + z3).Observe that even though the expression for S(f⇤1⇤) is complicated,ev eval1(P⇤1⇤) is relatively simple. More importantly, ev eval1(P⇤1⇤) 2h(1 z2)(1 z3)i, which is a key ingredient in our study of the extensionproblem of ⇣GL3⇡ . We will also see that the image of P⇤1⇤under ev eval1is determined by the image of P(⇤1⇤ ) , which turns out to be a generalphenomenon for any cone ⇤nm⇤ in the case of GLn.Proof. We first decompose the function as a sum of functions over the inte-riors of the faces of the cone using (6.3).f⇤1⇤= f(⇤1⇤ ) + f(F{↵1}⇤1⇤ ) .1007.3. Equivariant extension of ⇣GL3⇡ and Satake transformsNote that in this case, (F{↵1}⇤1⇤) = {(1, 1, 0)} and (⇤1⇤) = {(1, 1, 0) |1 > 1}. Thus, by applying the Satake transform and using the definitionof f⇤1⇤, we have thatS(f⇤1⇤) = S(f(1,1,0)) +X2(⇤1⇤ )S(f) = S(f(1,1,0)) +1X1=2S(f(1,1,0)).We will compute the summands of S(f⇤1⇤) that contain c(z) explicitly.From Examples 4 and 5, we have that:11 + q1c(z)qz1z2 +1X1=2c(z)(qz1)1z2 =11 + q1c(z)qz1z2 + c(z)(qz1)21 qz1 z2=✓11 + q1+qz11 qz1◆c(z)qz1z2.Other summands of S(f⇤1⇤) corresponding to di↵erent w 2 W can beobtained similarly.Next, to show thatev eval1(P⇤1⇤) = (1 z2)(1 z3)(z2 + z3),we observe that from the definition of P⇤1⇤, it follows thatP⇤1⇤= P(1,1,0) + P(⇤1⇤ ) .Since f(1,1,0) 2 C1c (GL3)K⇥K , we have that S(f(1,1,0)) 2 C[z±11 , z±12 , z±13 ]W .Hence, P(1,1,0) 2 h(1 qz1)(1 qz2)(1 qz3)i and ev eval1(P(1,1,0)) = 0.Therefore,ev eval1(P⇤1⇤) = ev eval1(P(⇤1⇤ )),and we may restrict our attention to ev eval1(P(⇤1⇤ )). Since each S(f) isa summation over w 2W , for every w 2W , we may defineP(⇤1⇤ ),w = (1 qz1)(1 qz2)(1 qz3)X2(⇤1⇤ )µM 12 ($)[w]($).Then, it follows from the definition of P(⇤1⇤ ) thatP(⇤1⇤ ) =Xw2WP(⇤1⇤ ),w.1017.3. Equivariant extension of ⇣GL3⇡ and Satake transformsObserve that if (1 qz1) is not a factor of the denominator ofX2(⇤1⇤ )µM 12 ($)[w]($),then (1 qz1) is a factor of the numerator of P(⇤1⇤ ),w, in which case,ev eval1(P(⇤1⇤ ),w) = 0. By inspection, (1 qz1) is a factor of the denomi-nator for w = () and w = (23). We have that:ev eval1(P(⇤1⇤ ),()) =(1 z2)(1 z3)(z2 q1z3)z2 z3 z2,ev eval1(P(⇤1⇤ ),(23)) =(1 z2)(1 z3)(z3 q1z2)z3 z2 z3.Thus,ev eval1(P(⇤1⇤ )) = ev eval1(P(⇤1⇤ ),()) + ev eval1(P(⇤1⇤ ),(23))= (1 z2)(1 z3)(z2 + z3),as desired.Next, we compute the Satake transform of another function in C1c (R2)K⇥K ,which will be used later in the proof of Proposition 17.Example 9. Using the notation introduced in Section 6.3.2, let ⇤ = (0, 0) 2⇤20, and f⇤1⇤the characteristic function of the setKhO11iK,which is an element in C1c (R2)K⇥K . Then,S(f⇤1⇤) =1 q(z1z2 + z1z3 + z2z3) + (q + q2)z1z2z3(1 qz1)(1 qz2)(1 qz3) .Proof. Observe that by (6.3), we have thatf⇤1⇤= f(⇤1⇤ ) + f(F{↵1}⇤1⇤ ) .1027.3. Equivariant extension of ⇣GL3⇡ and Satake transformsNote that in this case, (F{↵1}⇤1⇤) = {(0, 0, 0)} and (⇤1⇤) = {(1, 0, 0) |1 > 0}. Thus, by applying the Satake transform and using the computa-tions done in Examples 6 and 7, we have thatS(f⇤1⇤) = S(f(0,0,0)) +X2(⇤1⇤ )S(f)= S(f(0,0,0)) +X1>0S(f(1,0,0))= 1 +(z1 q1z2)(z1 q1z3)(z1 z2)(z1 z3) ·z11 z1 +(z2 q1z1)(z2 q1z3)(z2 z1)(z2 z3) ·z21 z2+(z3 q1z1)(z3 q1z2)(z3 z1)(z3 z1) ·z31 z3= 1 +q(z1 + z2 + z3) (q + q2)(z1z2 + z1z3 + z2z3) + (q + q2 + q3)z1z2z3(1 qz1)(1 qz2)(1 qz3)=1 q(z1z2 + z1z3 + z2z3) + (q + q2)z1z2z3(1 qz1)(1 qz2)(1 qz3) ,which completes the proof.7.3.3 Equivariant extension of ⇣GL3⇡ for the representation|.| det3In this section, we study the extension problem of ⇣GL3⇡ for the representation|.| det3, which is a specific instance of Theorem 14. By the classification in7.2, |.| det3 only appears in rank 1.Note that since (|.| det3,C) is the unique irreducible quotient of |.|2 ⇥|.| ⇥ 1, in this case, we have that s1 = 2, s2 = 1 and s3 = 0. Thus, themap eval : C[z±1i ]W ! C in the proof of Theorem 14, sends P (z1, z2, z3) 7!P (q2s, q1s, qs). In particular, we have that:eval1 :C[z±1i ]W ! C[z±11 , z±13 ]W [qs]P (z1, z2, z3) 7! P (z1qs, q1s, z3qs),eval2 :C[z±1i ]W ! C[z±11 ][qs]P (z1, z2, z3) 7! P (z1qs, q1s, qs).(7.1)The map eval1 in (7.1) agrees with eval1 in Definition 15 if we identifyC[z±11 , z±13 ]W [qs] with C[z±12 , z±13 ]W [qs] via z1 7! z2. Thus, we use eval1in both cases to denote the evaluation map of one of the zi at q1 (seeRemark 11).1037.3. Equivariant extension of ⇣GL3⇡ and Satake transformsProposition 16. Assume Hypothesis 1 in Section 6.3.1, and let (⇡, V ) bethe representation (|.| det3,C) of GL3 that only appears in rank 1. Then,(1) ⇣GL3⇡ extends H(GL3,K)-equivariantly to an element ofHomH(GL3,K)(C1c (R2)K⇥K , V K ⌦ eV K);(2) ⇣GL3⇡ does not extend H(GL3,K)-equivariantly to an element ofHomH(GL3,K)(C1c (R1)K⇥K , V K ⌦ eV K).Proof of (1). Recall the following exact sequence:0!C1c (GL3)K⇥K ext! C1c (R2)K⇥K res! C1c (R2)K⇥K ! 0.Applying the Satake transform to C1c (GL3)K⇥K and C1c (R2)K⇥K , weobtain the following short exact sequence:0! C[z±1i ]W !J23Qi=1(1 qzi)! Q2 ! 0,where J2 is the ideal defined in (6.4), and Q2 is the quotient module. Wehave the following commutative diagram:C1c (GL3)K⇥K _✏✏⇣GL3⇡''⇠= // C[z±1i ]Wev evalxx _✏✏C1c (R2)K⇥K? // V K ⌦ eV K ⇠= C J23Qi=1(1qzi)?oowhere eval : C[z±1i ]W ! C[qs] sends P (z1, z2, z3) 7! P (q2s, q1s, qs)and ev : C[qs]! C is the evaluation map at s = 0. Note that the isomor-phism in the top row is the well-known Satake isomorphism. We will showthat the composition of maps ev eval extends to a well-defined map eval⇤on J23Qi=1(1qzi).In order to do that, we show that every function in eval(S(C1c (R2)K⇥K))is well-defined at s = 0, for which it suces to show thateval(S(C1c (R2)K⇥K)) ✓1(1 q1s)(1 q1s)C[qs].1047.3. Equivariant extension of ⇣GL3⇡ and Satake transformsFirst, we will show that ev eval(J2) = 0. By Theorem 13, we have thatev eval1(J2) ✓ h(1 z1)(1 z3)i,which results in ev eval2(J2) = 0. Therefore, eval(J2), which can be ob-tained by evaluating ev eval2(J2) at z1 = q2, is also identically zero.Next, let f 2 C1c (R2)K⇥K and it remains to show thateval(S(f)) 2 1(1 q1s)(1 q1s)C[qs].Let Pf (z1, z2, z3) =3Qi=1(1 qzi)S(f) 2 J2. For z0i = ziqs, we have thatev(Pf (z01, z02, z03)) = Pf (z1, z2, z3).Thus, it follows from ev eval(Pf (z1, z2, z3)) = 0 that ev(Pf (q2s, q1s, qs)) = 0.Moreover, since Pf (q2s, q1s, qs) is a polynomial in qs, it followsfrom ev(Pf (q2s, q1s, qs)) being zero that Pf (q2s, q1s, qs) con-tains a factor of (1 qs). Finally, since the denominator of eval(S(f))has the form (1 qs)(1 q1s)(1 q1s), we get that eval(S(f)) 21(1q1s)(1q1s)C[qs]. Thus, the map ev eval extends to a well-definedmap eval⇤ on J23Qi=1(1qzi).Lastly, we use the map eval⇤ to define ⇣R2⇡ . For f 2 C1c (R2)K⇥K , let⇣R2⇡ (f) = eval⇤(S(f)).It follows from eval⇤ being an extension of eval that ⇣R2⇡ is an extension of⇣GL3⇡ , which completes the proof.Proof of (2). We consider the short exact sequence0!C1c (R2)K⇥K ext! C1c (R1)K⇥K res! C1c (R1)K⇥K ! 0.Applying the Satake transform to C1c (R2)K⇥K and C1c (R2)K⇥K , weobtain the following exact sequence:0! J23Qi=1(1 qzi)! J13Qi=1(1 qzi)! Q1 ! 0 (7.2)1057.3. Equivariant extension of ⇣GL3⇡ and Satake transformswhere J2 and J1 are the ideals as defined in (6.4), and Q1 is the quotientmodule. We have the following commutative diagram:C1c (R2)K⇥K _✏✏⇣R2⇡%%⇠= // J23Qi=1(1qzi)eval⇤zz _✏✏C1c (R1)K⇥K? // V K ⌦ eV K ⇠= C J13Qi=1(1qzi)?oowhere the maps ⇣R2⇡ and eval⇤ are defined in the proof of part (1). In thiscase, the isomorphism in the top row is a consequence of Hypothesis 1 inSection 6.3.1. To show that the map ⇣GL3⇡ cannot be extended H(GL3,K)-equivariantly to an element in HomH(GL3,K)(C1c (R1)K⇥K , V K ⌦ eV K), weshow that eval⇤ is not well-defined on J13Qi=1(1qzi)by providing an explicitelement P in J1 whose image under eval is undefined at s = 0.Consider the function f⇤10⇤2 C1c (R1)K⇥K . More explicitly, f⇤10⇤is thecharacteristic function of the setKhO O1iK.By Corollary 7, we have that ev eval2⇣P⇤10⇤⌘= 1 q1z1. Therefore,ev eval(P⇤10⇤) = 1 q3 6= 0, which shows that eval⇤ is not well-defined onJ13Qi=1(1qzi). Thus, the map ⇣GL3⇡ does not extend H(GL3,K)-equivariantly toan element in HomH(GL3,K)(C1c (R1)K⇥K , V K ⌦ eV K).7.3.4 Equivariant extension of ⇣GL3⇡ for the representation|.|⇥ |.| 12 det2Recall that ⇡ = |.| ⇥ |.| 12 det2 is the unique representation of GL3 thatappears in multiple low ranks. In segment notation, ⇡ = h[|.|], [|.|], [1]it andits L-function has a double pole at s = 1 and a simple pole at s = 0. In thiscase, eval is the map which sends S(f)(z1, z2, z3) 7! S(f)(q1s, q1s, qs).Proposition 17. Assume Hypothesis 1 in Section 6.3.1, and let (⇡, V ) bethe representation |.| ⇥ |.| 12 det2 of GL3 that appears in both ranks 1 and2. Then, ⇣GL3⇡ does not extend H(GL3,K)-equivariantly to an element ofHomH(GL3,K)(C1c (R2)K⇥K , V K ⌦ eV K).1067.3. Equivariant extension of ⇣GL3⇡ and Satake transformsProof. Observe that the map eval⇤ is well-defined if and only if every func-tion in the image of eval is well-defined at s = 0, that is,eval(S(C1c (R2)K⇥K) ✓11 q1sC[qs, qs].We will show that eval⇤ is not well-defined on S(C1c (R2)K⇥K) by providingan explicit function f 2 C1c (R2)K⇥K such that eval(S(f)) 62 11q1sC[qs, qs].Consider the function f⇤20⇤2 C1c (R2)K⇥K . More explicitly, f⇤20⇤is thecharacteristic function of the setKhO11iK.From Example 9, we obtain the image of f⇤20⇤under the Satake trans-form:S(f⇤20⇤) =1 q(z1z2 + z1z3 + z2z3) + (q + q2)z1z2z3(1 qz1)(1 qz2)(1 qz3) .Thus,eval(S(f⇤20⇤)) =1 q12s 2q2s + q13s + q3s(1 qq1s)(1 qqs)=qs q2s q12s + 1(1 qs)(1 q1s) .Since eval(S(f⇤20⇤)) is undefined at s = 0, the map eval⇤ is undefined. Thus,the map ⇣GL3⇡ does not extend H(GL3,K)-equivariantly to an element inHomH(GL3,K)(C1c (R2)K⇥K , V K ⌦ eV K).107Chapter 8The Steinberg representationof GL2In this section, we will use the same notation in Chapter 3. We will showthat for the Steinberg representation, there does not exist a GLn⇥GLn-equivariant map ⇣M⇡ such that ⇣M⇡ iMG = ⇣G⇡ . The approach used in thissection is inspired by and follows the same outline as that in Section 3.4.The main di↵erence is that since the Steinberg representation StG does nothave K-fixed vectors but only I-fixed vectors, we will work with the spaceC1c (G)I⇥I of Iwahori-bi-invariant functions instead. The diagram that weaim to construct in this section is the following:C1c (G)I⇥IZI,G // _iMG✏✏⇣G⇡((ZI,G(C1c (G)I⇥I) _iZ✏✏evalG⇡uu //M2(C((z1, z2)))V I ⌦ eV I ⇠=M2(C)C1c (M)I⇥IZI,M //⇣M⇡66ZI,M (C1c (M)I⇥I)evalM⇡ii //M2(C((z1, z2)))We first define some notations that will be used in this section. Fors1, s2 2 C, let s1,s2 be the character of B defined by:s1,s2 :B ! C⇥hb1 b20 b4i7! |b1|s1 |b4|s2 .denote by (⇡s1,s2 , Vs1,s2) the normalized induced representation indGB s1,s2 .The Steinberg representation can be realized as a G-invariant subspace ofindGB s1,s2 where s1 =12 and s2 = 12 . Observe that for all s1, s2 2 C,the I-invariant subspace V Is1,s2 is two-dimensional, spanned by the function108Chapter 8. The Steinberg representation of GL2f1s1,s2 and fws1,s2 , defined by:f1s1,s2 :G = BI tBwI ! C fws1,s2 : G! Cbi 7! 12B(b)s1,s2(b), bi 7! 0,bwi 7! 0, bwi 7! 12B(b)s1,s2(b).Note that to simplify the formulas, in this section, we normalize the Haarmeasure on G such that µG(I) = 1.The map ⇣G⇡s1,s2 : C1c (G)I⇥I ! V Is1,s2 ⌦]Vs1,s2IDefine the following map:⇣G⇡s1,s2 :C1c (G)I⇥I ! EndC(V Is1,s2)f 7!✓f 0 7!✓x 7!ZGf(g)f 0(xg) dg◆◆.Wemay represent ⇣G⇡s1,s2 as a 2-by-2 matrix with respect to the basis {f1s1,s2 , fws1,s2}of V Is1,s2 . More explicitly,⇣G⇡s1,s2 :C1c (G)I⇥I !M2(C)f 7!(⇣G⇡s1,s2(f)(f1s1,s2 ))(1) (⇣G⇢s1,s2(f)(fws1,s2 ))(1)(⇣G⇡s1,s2(f)(f1s1,s2 ))(w) (⇣G⇡s1,s2(f)(fws1,s2 ))(w)= RG f(g)f1s1,s2(g) dgRG f(g)fws1,s2(g) dgRG f(g)f1s1,s2(wg) dgRG f(g)fws1,s2(wg) dg.Denote by am,n matrices of the form⇥$m 00 $n⇤and bm,n matrices of theform⇥0 $m$n 0⇤. Let chIam,nI (resp. chIbm,nI) the characteristic function ofIam,nI (resp. Ibm,nI). Recall that C1c (G)I⇥I is generated by chIam,nI andchIbm,nI subject to the Bernstein relation. We have the following explicitform for the image of these generators under ⇣G⇡s1,s2 :Lemma 26. For m n, and a_ := ⇥$ 00 $1 ⇤, let := s1,s2. Then,⇣G⇡s1,s2 (chIam,nI) = 12 (am,n)(am,n) 01q111(a_)s1,s2 (am,n)+(1q1)[w]1(a_)1[w]1(a_) [w](am,n) [w](am,n)⇣G⇡s1,s2 (chIan,mI) = 12 (am,n)[w](am,n)q(1q1)1s1,s2 (a_)11(a_) (am,n)+q(1q1)1[w]1(a_) [w](am,n)0 (am,n)⇣G⇡s1,s2 (chIbm,nI) = 12 (am,n)0 q(am,n)[w(am,n)q(1q1)11(a_)(am,n)+q(1q1)1[w]1(a_) [w](am,n)⇣G⇡s1,s2 (chIbn,mI) = 12 (am,n)(1q1)1(a_)11(a_) (am,n)+(1q1)[w]1(a_)1[w]1(a_) [w](am,n) [w](am,n)q1(am,n) 0109Chapter 8. The Steinberg representation of GL2Proof. The computation done here follows exactly the same steps as thosein the proof of Lemma 27, and is omitted.The map ZI,G : C1c (G)I⇥I !M2(C((z1, z2)))Let:f1z1,z2 :G = BI tBwI ! C fwz1,z2 : G! Cbi =hb1 b20 b4ii 7! 12B(b)z⌫(b1)1 z⌫(b4)2 , bi 7! 0,bwi 7! 0, bwi =hb1 b20 b4iwi 7! 12B(b)z⌫(b1)1 z⌫(b4)2 .Define:ZI,G :C1c (G)I⇥I !M2(C((z1, z2)))f 7! RG f(g)f1z1,z2(g) dgRG f(g)fwz1,z2(g) dgRG f(g)f1z1,z2(wg) dgRG f(g)fwz1,z2(wg) dg.Consider the following evaluation map:evalGs1,s2 :ZI,G(C1c (G)I⇥I)! V Is1,s2 ⌦]Vs1,s2I ⇠=M2(C)z1 7! qs1 , z2 7! qs2 .Note that this also justifies that evalG⇡ is well-defined on ZI,G(C1c (G)I⇥I).First, we can define evalG⇡ on the subspace of M2(C((z1, z2))) on which it iswell-defined. By construction, evalGs1,s2 ZI,G = ⇣G⇡s1,s2 . If ZI,G(C1c (G)I⇥I isnot contained in the the domain of evalGs1,s2 , that would imply that ⇣⇡Gs1,s2isnot well-defined on some function of C1c (G)I⇥I , which would be a contra-diction.For clarity and future reference, we include here the image of the gener-ators under ZI,G:110Chapter 8. The Steinberg representation of GL2Lemma 27. For m n,ZI,G(chIam,nI) = qmn2 zm1 zn2 0(1q1) zm1 zn2zn1 zm21z11 z2zn1 zm2,ZI,G(chIan,mI) = qmn2"zn1 zm2 (1q)zn11 zm+12 z11 z21z11 z20 zm1 zn2#,ZI,G(chIbm,nI) = qmn2"0 qzm1 zn2zn1 zm2 (q1)zm1 zn2zn11 zm+121z11 z2#,ZI,G(chIbn,mI) = qmn2"(q11) zn1 zm2 zm11 zn+121z11 z2zn1 zm2q1zm1 zn2 0#.The map ZI,M : C1c (M)I⇥I !M2(C((z1, z2)))Define:ZI,M :C1c (M)I⇥I !M2(C((z1, z2)))f 7! RG f(g)f1z1,z2(g) dgRG f(g)f1z1,z2(wg) dgRG f(g)fwz1,z2(g) dgRG f(g)fwz1,z2(wg) dg.ZI,M is well-defined, and by definition, ZI,M iMG = ZI,G. We will rely onthe following hypothesis for the main result:Hypothesis 2. ZI,M is injective.Thus, assuming Hypothesis 2, we have the following commutative dia-gram of G⇥G-spaces where the first maps in each row are isomorphisms:C1c (G)I⇥IZI,G // _iMG✏✏⇣G⇡s1,s2))ZI,G(C1c (G)I⇥I) _iZ✏✏evalGs1,s2uu //M2(C((z1, z2)))V Is1,s2 ⌦]Vs1,s2I ⇠=M2(C)C1c (M)I⇥IZI,M // ZI,M (C1c (M)I⇥I) //M2(C((z1, z2)))We find an explicit element in the image of ZI,M :Lemma 28. We have that:ZI,M (chIhO 00 1iI) =24 11q 12 z1 0(1q1)q12 z1(1q12 z1)(1q12 z2)11q12 z235 .111Chapter 8. The Steinberg representation of GL2Proof. This proof is purely computational using the results from Lemma 27and the following observation that:ZI,M (chIhO 00 1iI) =1Xm=0ZI,M (chIam,0I).Using the above results, we obtain the following conclusion:Lemma 29. For s1 =12 or s2 =12 , there does not exist a GLn⇥GLn-equivariant map ⇣M⇡s1,s2 : C1c (M)I⇥I ! V Is1,s2⌦]Vs1,s2Isuch that ⇣M⇢s1,s2 iMG =⇣G⇢s1,s2 .Proof. Suppose that for s1 =12 or s2 =12 , there exists such a GLn⇥GLn-equivariant map ⇣M⇡s1,s2 : C1c (M)K⇥K ! V Ks1,s2 ⌦]Vs1,s2Ksatisfying ⇣M⇢s1,s2 iMG = ⇣G⇢s1,s2. Since ZI,M is an isomorphism onto its image, we may considerthe composition ⇣M⇡s1,s2 Z1I,M from ZI,M (C1c (M)I⇥I) to V Is1,s2⌦]Vs1,s2I ⇠= C.We now show that ⇣M⇡s1,s2 Z1I,M iZ = evalGs1,s2 . We have that:evalGs1,s2 = ⇣G⇡s1,s2 Z1I,G = (⇣G⇡s1,s2 iMG ) Z1I,G = ⇣G⇡s1,s2 Z1I,M ZI,M iMG Z1I,G= ⇣G⇡s1,s2 Z1I,M ZI,G Z1I,G = ⇣G⇡s1,s2 Z1I,M .Then, we have the following commutative diagram:C1c (G)I⇥IZI,G // _iMG✏✏⇣G⇡s1,s2))ZI,M (C1c (G)I⇥I) _iZ✏✏evalGs1,s2uu // C((z1, z2))V Is1,s2 ⌦]Vs1,s2I ⇠=M2(C)C1c (M)I⇥IZI,M //⇣M⇡s1,s255ZI,M (C1c (M)I⇥I)⇣M⇡s1,s2Z1K,Mii // C((z1, z2))Observe that the map ⇣M⇡s1,s2 Z1I,M is determined uniquely by its value onz1 and z2. It follows from ⇣M⇡s1,s2 Z1I,M iZ = evalGs1,s2 that via ⇣M⇡s1,s2 Z1I,M ,112Chapter 8. The Steinberg representation of GL2we also have that z1 7! qs1 and z2 7! qs2 . By Lemma 28,⇣M⇡s1,s2 (chIhO 00 1iI) = ⇣M⇡s1,s2 Z1K,M ZK,M (chM2(O))= ⇣M⇡s1,s2 Z1I,M0@24 11q 12 z1 0(1q1)q12 z1(1q12 z1)(1q12 z2)11q12 z2351A=24 11q 12s1 0(1q1)q12s1(1q12s1 )(1q12s2 )11q12s235 .However, for s1 =12 or s2 =12 , this map is undefined, which is a contradic-tion. Thus, the lemma follows.Lemma 30. For (⇡, V ) being the Steinberg representation, the map ⇣G⇡ doesnot extend equivariantly to a map ⇣M⇡ .Proof. For (⇡, V ) being the Steinberg representation, (⇡, V ) can be realizedas the G-invariant subspace of (⇡ 12 , 12 , V 12 , 12 ). Moreover, VI is the one-dimensional subspace of V I12 , 12, spanned by the function f112 , 12 q1fw12 , 12.By definition,⇣G⇡ :C1c (G)I⇥I ! EndC(V I)f 7!✓f 0 7! (x 7!ZGf(g)(f 0)(xg) dg)◆.With respect to the basis {f112 , 12 q1fw12 , 12}, we may identify EndC(V I)with C, and obtain:⇣G⇡ :C1c (G)I⇥I ! Cf 7!ZGf(g)(f112 , 12 q1fw12 , 12)(g) dg.Define:evalG⇡ :ZI,G(C1c (G)I⇥I)! C[ a1 a2a3 a4 ] 7! evalG12 , 12⇣[ a1 a2a3 a4 ]h1q1i[ 1 0 ]⌘= evalG12 , 12(a1 q1a2).Then, by definition, evalG⇡ ZI,G = ⇣G⇡ . Suppose that ⇣M⇡ is a G ⇥ G-equivariant extension of ⇣G⇡ . We can construct the map ⇣M⇡ Z1I,M such113Chapter 8. The Steinberg representation of GL2that ⇣M⇡ Z1I,M iZ = evalG⇢ . Then, ⇣M⇡ Z1I,M is also defined uniquely bythe same definition as evalG⇡ . We have the following diagram:C1c (G)I⇥IZI,G // _iMG✏✏⇣G⇡''ZI,G(C1c (G)I⇥I) _iZ✏✏evalG⇡uu //M2(C((z1, z2)))V I ⌦ eV I ⇠= CC1c (M)I⇥IZI,M //⇣M⇡77ZI,M (C1c (M)I⇥I)⇣M⇡ Z1I,Mii //M2(C((z1, z2)))Then,⇣M⇡ (chIhO 00 1iI) = ⇣M⇡ Z1I,M ZI,M (chIhO 00 1 iI)= ⇣M⇡ Z1I,M0@24 11q 12 z1 0(1q1)q12 z1(1q12 z1)(1q12 z2)11q12 z2351A= evalG12 , 12 11 q 12 z1!,which is undefined. Thus, no such ⇣M⇡ exists.We make the following important observations:• Observe that for irreducible representations (⇡, V ) with K-fixed vec-tors that can be embedded in (⇢s1,s2 , Vs1,s2) for some s1, s2 2 C, if⇣M⇡s1,s2 does not exist, then ⇣M⇡ does not exist as an immediately con-sequence. However, for the Steinberg representation StG which hasI-fixed vectors and can be embedded into (⇡ 12 , 12 , V 12 , 12 ), the non-existence of ⇣MStG does not follow from the non-existence of ⇣M⇡ 12 , 12.This is because V K12 , 12is already one-dimensional, and if (⇡, V ) hasK-fixed vectors, then V K = V K12 , 12. On the other hand, V I12 , 12is two-dimensional, and StIG is only one-dimensional, so we need to treat itwith care. It is possible that the map ⇣M⇡ 12 , 12into V I12 , 12⌦ V^ 12 , 12Idoesnot exist, but the map ⇣M⇡ into VI ⌦ eV I exists such as in the case of(⇡, V ) = |.| StG and (⇡ 32 ,12, V 32 ,12).114Chapter 8. The Steinberg representation of GL2• One can also show that for spherical representations of G such thatHomGLn⇥GLn(C1c (R1), V⌦eV ) ⇠= C, the space HomGLn⇥GLn(C1c (M), V⌦eV ) ⇠= C is one-dimensional using the above results. For the irreduciblerepresentations of the form (⇡ 12 ,t, V 12 ,t), the result is an immediateconsequence of Lemma 29. For (⇡, V ) being either 12 or |.| det2,then V I is the proper one-dimensional subspace of V Is1,s2 spanned byf1s1,s2+fws1,s2 , and a similar argument as in the proof of Lemma 30 willsuce.• It is well-known that for unramified principal series representations(⇡, V ), then ⇡(chM2(O)) is the local L-factor, as proven by Tamagawafor GLn in [20]. Note that as a consequence of the computation doneabove, we have, in fact, showed that for the Steinberg representation(⇡, V ), then ⇡(chIhO 00 1iI) is precisely the local L-factor.115Bibliography[1] Joseph Bernstein. Representations of p-adic groups. Notes by K. E.Rumelhart. Harvard Uni., 1992.[2] Daniel Bump. Automorphic forms and representations, volume 55 ofCambridge Studies in Advanced Mathematics. Cambridge UniversityPress, Cambridge, 1997.[3] Colin J. Bushnell and Guy Henniart. The local Langlands conjecture forGL(2), volume 335 of Grundlehren der Mathematischen Wissenschaften[Fundamental Principles of Mathematical Sciences]. 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Birkha¨user Boston, Boston,MA, 2003.116Bibliography[10] Stephen S. Kudla and Stephen Rallis. On first occurrence in the localtheta correspondence. In Automorphic representations, L-functions andapplications: progress and prospects, volume 11 of Ohio State Univ.Math. Res. Inst. Publ., pages 273–308. de Gruyter, Berlin, 2005.[11] Laurent La↵orgue. Noyaux du transfert automorphe de Langlands etformules de Poisson non line´aires. Jpn. J. Math., 9(1):1–68, 2014.[12] I.G. Macdonald. Spherical Functions on a Group of P-adic Type. Publi-cations of the Ramanujan Institute. Ramanujan Institute for AdvancedStudy in Mathematics, University of Madras, 1971.[13] Alberto Mı´nguez. Correspondance de Howe explicite: paires duales detype II. Ann. Sci. E´c. Norm. Supe´r. (4), 41(5):717–741, 2008.[14] Alberto Mı´nguez. Sur l’irre´ductibilite´ d’une induite parabolique. J.Reine Angew. Math., 629:107–131, 2009.[15] Colette Mœglin, Marie-France Vigne´ras, and Jean-Loup Waldspurger.Correspondances de Howe sur un corps p-adique, volume 1291 of LectureNotes in Mathematics. Springer-Verlag, Berlin, 1987.[16] Calvin C. Moore, editor. Harmonic analysis on homogeneous spaces.American Mathematical Society, Providence, R.I., 1973.[17] Dipendra Prasad and A. Raghuram. Representation theory ofGL(n) over non-archimedean local fields. 2007. Retrieved fromhttp://www.math.tifr.res.in/ dprasad/ictp2.pdf on May 24th, 2016.[18] Franc¸ois Rodier. Repre´sentations de GL(n, k) ou` k est un corps p-adique. In Bourbaki Seminar, Vol. 1981/1982, volume 92 of Aste´risque,pages 201–218. Soc. Math. France, Paris, 1982.[19] Ichiroˆ Satake. Theory of spherical functions on reductive algebraicgroups over p-adic fields. Inst. Hautes E´tudes Sci. Publ. Math., (18):5–69, 1963.[20] Tsuneo Tamagawa. On the ⇣-functions of a division algebra. Ann. ofMath. (2), 77:387–405, 1963.[21] John Torrence Tate, Jr. Fourier Analysis in number fields and Hecke’szeta-functions. ProQuest LLC, Ann Arbor, MI, 1950. Thesis (Ph.D.)–Princeton University.117Bibliography[22] Torsten Wedhorn. The local Langlands correspondence for GL(n) overp-adic fields. In School on Automorphic Forms on GL(n), volume 21of ICTP Lect. Notes, pages 237–320. Abdus Salam Int. Cent. Theoret.Phys., Trieste, 2008.[23] Andre´ Weil. Fonction zeˆta et distributions. Seminaire Bourbaki, 9:523–531, 1964-1966.[24] A. V. Zelevinsky. Induced representations of reductive p-adic groups.II. On irreducible representations of GL(n). Ann. Sci. E´cole Norm.Sup. (4), 13(2):165–210, 1980.118Appendix AInjectivity of Sataketransform on C1c (M2)K⇥KIn this appendix, we will prove the injectivity of the extended Satake trans-form on M2. The proof relies on expressing the explicit images of elementsin the basis of C1c (M2)K⇥K in terms of elementary symmetric polynomi-als using Macdonald’s formula. Then, by studying the leading term of theimages, we show that no linear combination of the basis elements can bemapped to zero under the Satake transform. Note that the injectivity of theextended Satake transform on C1c (M2)K⇥K combined with the computationby Tamagawa [Theorem 6 p.397 [20]] shows thatS(C1c (M2)K⇥K) ⇠=1(1 q 12 z1)(1 q 12 z2)C[z±11 , z±12 ]W ,which is a particular case of Hypothesis 1. Moreover, according to Cassel-man’s unpublished notes, the idea of studying the leading term of the imagesis, in fact, Satake’s.To shorten notation, we will use M := M2 and G := GL2.A.1 Satake images of elements in the basis ofC1c (M)K⇥K in terms of elementary symmetricpolynomialsSince the image of C1c (M)K⇥K under the Satake transform lies in the spaceof symmetric rational functions in z1 and z2, we may represent it in terms ofelementary symmetric polynomials. Let E1 = q12 (z1+z2) and E2 = z1z2. Wechoose the lexicographic order that E1 > E2 in order to determine leadingterm of a polynomial in E1 and E2.Observe that as a result of Lemma 20 in Section 6.3.3, a basis forC1c (M)K⇥K consists of characteristic functions of sets of the form:Kh$⇤1$⇤2iK, Kh$⇤2O$⇤2iK, or K⇥O O ⇤K.119A.1. Satake images of elements in the basis of C1c (M)K⇥K in terms of elementary symmetric polynomialsThus, we first study the images of these basis elements under the Sataketransform in terms of E1 and E2.Basis elements of C1c (M)K⇥K that project to a basis of C1c (G)K⇥KThe basis elements of C1c (M)K⇥K that project to a basis of C1c (G)K⇥Kare simply the basis elements of C1c (G)K⇥K itself. Namely, they are char-acteristic functions of double cosets:Kh$⇤1$⇤2iK,where ⇤1 ⇤2. We denote these functions by f(⇤1,⇤2). Recall from Macdon-ald’s formula that if ⇤1 = ⇤2, then:S(f(⇤2,⇤2)) = (z1z2)⇤2 = E⇤22 ,and if ⇤1 > ⇤2, then:S(f(⇤1,⇤2)) = q12 (⇤1⇤2)(z1z2)⇤2 z⇤1⇤2+11 z⇤1⇤2+12z1 z2 q1z1z2z⇤1⇤211 z⇤1⇤212z1 z2!.Note that we do not need to have an explicit expression for S(f(⇤1,⇤2)) interms of E1 and E2 and it suces for us to just keep track of the leadingterm of S(f(⇤1,⇤2)).Lemma 31. The leading term of S(f(⇤1,⇤2)) is q12 (⇤1⇤2)E⇤1⇤21 E⇤22 .Proof. Observe that:z⇤1⇤2+11 z⇤1⇤2+12z1 z2 = z⇤1⇤21 + z⇤1⇤211 z2 + . . .+ z1z⇤1⇤212 + z⇤1⇤21= (z⇤1⇤21 + z⇤1⇤22 ) + (z1z2)(z⇤1⇤221 + z⇤1⇤222 ) + . . .In terms of elementary symmetric polynomials, the leading term of the ex-pansion of (zn1 + zn2 ) is En1 . Thus, the leading term ofz⇤1⇤2+11 z⇤1⇤2+12z1 z2is E⇤1⇤21 . Similarly, the leading term ofq1z1z2z⇤1⇤211 z⇤1⇤212z1 z2is q1E⇤1⇤221 E2. Thus, the statement of the lemma follows.120A.2. Injectivity of Satake transform on C1c (M)K⇥KBasis elements of C1c (M)K⇥K that project to a basis ofC1c (R1)K⇥KThe basis element of C1c (M)K⇥K that project to a basis of C1c (R1)K⇥Kare the characteristic functions of the setsKh$⇤2O$⇤2iK.Using the notation in (6.2), the above functions are denoted by f⇤1(⇤2). FromLemma 10, we have thatS(f⇤1(⇤2)) =1 z1z2(1 q 12 z1)(1 q 12 z2)(z1z2)⇤2 =(1 E2)E⇤221 E1 + qE2 .Basis element of C1c (M)K⇥K that projects to a basis ofC1c ({0})K⇥KThe only basis element of C1c (M)K⇥K that projects to a basis of C1c ({0})K⇥Kis the characteristic function of the setK⇥O O ⇤K,which we denote by f⇤2 . It follows from the computation by Tamagawa[Theorem 6 p.397 [20]] thatS(f⇤2) = 1(1 q 12 z1)(1 q 12 z2)=11 E1 + qE2 .A.2 Injectivity of Satake transform on C1c (M)K⇥KWe first prove an auxiliary lemma about the form of the Satake image of anarbitrary element in C1c (M)K⇥K .Lemma 32. Let f 2 C1c (M)K⇥K . Then,S(f) = a1 E1 + qE2 +X⇤22Zb⇤2(1 E2)E⇤221 E1 + qE2 +X⇤22Zc⇤2E⇤22 +X(⇤1,⇤2)2Z2⇤1>⇤2d(⇤1,⇤2)S(f(⇤1,⇤2)),where only finitely many coecients a, b⇤2 , c⇤2 and d(⇤1,⇤2) are non-zero.121A.2. Injectivity of Satake transform on C1c (M)K⇥KProof. Let f 2 C1c (M)K⇥K . We first write f as a linear combination of thebasis elements in C1c (M)K⇥K , and then divide the summands into groupsof those that project to basis elements of C1c ({0})K⇥K , C1c (R1)K⇥K andC1c (G)K⇥K . Among those that are basis elements of C1c (G)K⇥K , we alsosplit into two summations depending on whether ⇤1 = ⇤2 or not. Moreexplicitly, we have thatf = af⇤2 +X⇤22Zb⇤2f⇤1(⇤2)+X⇤22Zc⇤2f(⇤2,⇤2) +X(⇤1,⇤2)2Z2⇤1>⇤2d(⇤1,⇤2)f(⇤1,⇤2),where only finitely many coecients a, b⇤2 , c⇤2 and d(⇤1,⇤2) are non-zero.Thus, its image under the Satake transform follows from the computationsin A.1.Using the above lemma and by studying the leading term, we obtain thefollowing result:Proposition 18. The map S : C1c (M)K⇥K ! 11E1+qE2C[E1, E±12 ] is in-jective.Proof. Let f 2 C1c (M)K⇥K be such that S(f) = 0. By Lemma 32, itsimage S(f) under the Satake transform is:a1 E1 + qE2 +X⇤22Zb⇤2(1 E2)E⇤221 E1 + qE2 +X⇤22Zc⇤2E⇤22 +X(⇤1,⇤2)2Z2⇤1>⇤2d(⇤1,⇤2)S(f(⇤1,⇤2)),(A.1)where only finitely many coecients a, b⇤2 , c⇤2 and d(⇤1,⇤2) are non-zero.First, we show that d(⇤1,⇤2) = 0 for all (⇤1,⇤2) 2 Z2 by contradiction.Suppose that there exists (⇤1,⇤2) 2 Z2 with ⇤1 > ⇤2 such that d(⇤1,⇤2) 6= 0.We first consider all ordered pairs (⇤1,⇤2) where d(⇤1,⇤2) 6= 0 and ⇤1 ⇤2 ismaximal. Then, we can find among those a unique pair (`⇤1, `⇤2) where `⇤2 ismaximal. Thus, (`⇤1, `⇤2) is a pair such that d(`⇤1,`⇤2) 6= 0, and `⇤1 `⇤2 and `⇤2are maximal (in that order of importance). By Lemma 31, the leading termof S(f(`⇤1,`⇤2)) is q12 (`⇤1`⇤2)E`⇤1`⇤21 E`⇤22 . It follows from the maximality of `⇤1 `⇤2and `⇤2, that the leading term ofX(⇤1,⇤2)2Z2d(⇤1,⇤2)S(f(⇤1,⇤2))122A.2. Injectivity of Satake transform on C1c (M)K⇥Kis d(`⇤1,`⇤2)q12 (`⇤1`⇤2)E`⇤1`⇤21 E`⇤22 . Moreover, observe that if we express S(f) as asingle fraction with the denominator (1 E1 + qE2), then the numeratorsof the first summand and those in the first summation in (A.1) do notcontain E1, and the leading terms in the numerators of those in the secondsummation in (A.1) have the form c⇤2E1E⇤22 . Thus, over the commondenominator (1 E1 + qE2), the leading term in the numerator of S(f)is still d(`⇤1,`⇤2)q12 (`⇤1`⇤2)E`⇤1`⇤2+11 E`⇤22 because `⇤1 `⇤2 + 1 2 (Recall that`⇤1 > `⇤2, so `⇤1 `⇤2 1). It follows from S(f) = 0 that d(`⇤1,`⇤2) = 0, which isa contradiction. Thus, d(⇤1,⇤2) = 0 for all (⇤1,⇤2) 2 Z2 where ⇤1 > ⇤2, andS(f) has the form:a1 E1 + qE2 +X⇤22Zb⇤2(1 E2)E⇤221 E1 + qE2 +X⇤22Zc⇤2E⇤22 , (A.2)where only finitely many coecients a, b⇤2 and c⇤2 are non-zero.Next, we show that c⇤2 = 0 for all ⇤2 2 Z by contradiction. Supposethat there exists ⇤2 2 Z such that c⇤2 6= 0. Let `⇤2 be maximal such thatc`⇤2 6= 0. So, c`⇤2E`⇤22 is the leading term ofP⇤22Zc⇤2E⇤22 . Observe that thenumerators of the first summand and those in the first summation in (A.2)do not contain E1. Therefore, if we express S(f) as a single fraction withthe denominator (1E1 + qE2), then the leading term in the numerator ofS(f) is c`⇤2E1E`⇤22 . Since S(f) = 0, we get c`⇤2 = 0, which is a contradiction.Therefore, we have thatS(f) = a1 E1 + qE2 +X⇤22Zb⇤2(1 E2)E⇤221 E1 + qE2=11 E1 + qE20@(a+ b0 b1) + X⇤22Z\{0}(b⇤2 b⇤21)E⇤221A .Since S(f) = 0, we get that a + b0 b1 = 0 and b⇤2 b⇤21 = 0 for all⇤2 2 Z\{0}. It follows from the recurrence relation on b⇤2 that b⇤2 = b0 if⇤2 > 0 and b⇤2 = b1 if ⇤2 < 1. Since only finitely many coecients b⇤2can be non-zero, we have that b⇤2 = 0 (and therefore, a = 0) for all ⇤2 2 Z.Thus, f = 0, and the map S is injective on C1c (M)K⇥K .123
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Equivariant extension of distributions on GLn Nguyen, Bich-Ngoc Cao 2020
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Title | Equivariant extension of distributions on GLn |
Creator |
Nguyen, Bich-Ngoc Cao |
Publisher | University of British Columbia |
Date Issued | 2020 |
Description | Let k be a non-Archimedean local field, and Cc∞(GLn) the space of locally constant compactly supported complex-valued functions on the general linear group GLn over k. For every irreducible representation (𝜋,V) of GLn, the space Hom(Cc∞(GLn), V ⊗ Ṽ) is one-dimensional. This space is generated by an element denoted by 𝜁, which can be thought of as an integral against matrix coefficients. In this thesis, we are interested in the so-called "extension problem" of 𝜁. More explicitly, for 0≤m≤n, GLn can be embedded into the space R≥m of all n×n matrices over k of rank at least m. If 𝜁 lies in the image of the induced map of this embedding, then we say that 𝜁 can be extended to rank at least m. For m=0, the extension problem of 𝜁 to rank at least 0 has been completely answered in Tate's thesis for n=1, and by Moeglin, Vignéras, Waldspurger, and Minguez for general n. Our goal is to determine the least value m for a given representation such that 𝜁 can be extended to rank at least m. A representation is said to appear in rank m if Hom(Cc∞(R≥m), V ⊗ Ṽ) is non-trivial. It is natural to conjecture that 𝜁 extends to rank at least m+1 but does not extend to rank at least m where m is the highest rank less than n that 𝜋 appears in. In this thesis, this conjecture is proved for spherical representations, by means of extending Satake transform to the space of K-bi-invariant functions on Mn and obtaining a partial description of the image of the rank filtration under this extended Satake transform. Some explicit computations for spherical representations of GL₃ are included as motivating examples of the general case.There are also some suggestive calculations for non-spherical representations of GL₂. |
Genre |
Thesis/Dissertation |
Type |
Text |
Language | eng |
Date Available | 2020-10-24 |
Provider | Vancouver : University of British Columbia Library |
Rights | Attribution-NonCommercial-NoDerivatives 4.0 International |
DOI | 10.14288/1.0394801 |
URI | http://hdl.handle.net/2429/76357 |
Degree |
Doctor of Philosophy - PhD |
Program |
Mathematics |
Affiliation |
Science, Faculty of Mathematics, Department of |
Degree Grantor | University of British Columbia |
GraduationDate | 2020-11 |
Campus |
UBCV |
Scholarly Level | Graduate |
Rights URI | http://creativecommons.org/licenses/by-nc-nd/4.0/ |
AggregatedSourceRepository | DSpace |
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