Open Collections

UBC Theses and Dissertations

UBC Theses Logo

UBC Theses and Dissertations

Spaces of homomorphisms and commuting orthogonal matrices. Higuera Rojo, Galo 2014

Your browser doesn't seem to have a PDF viewer, please download the PDF to view this item.

Item Metadata


24-ubc_2014_spring_higuerarojo_galo.pdf [ 419.02kB ]
JSON: 24-1.0167318.json
JSON-LD: 24-1.0167318-ld.json
RDF/XML (Pretty): 24-1.0167318-rdf.xml
RDF/JSON: 24-1.0167318-rdf.json
Turtle: 24-1.0167318-turtle.txt
N-Triples: 24-1.0167318-rdf-ntriples.txt
Original Record: 24-1.0167318-source.json
Full Text

Full Text

Spaces of homomorphisms and commutingorthogonal matricesbyGalo Higuera RojoA thesis submitted in partial fulfilment of the requirements forthe degree ofDoctor of PhilosophyinThe Faculty of Graduate and Postdoctoral Studies(Mathematics)The University of British Columbia(Vancouver)April 2014c© Galo Higuera Rojo, 2014AbstractIn this work we study the space of group homomorphisms Hom(pi,G) forspecial choices of pi and G. In the first part of this thesis, we enumerate anddescribe the path components for the spaces of ordered commuting k-tuplesof orthogonal and special orthogonal matrices respectively. This correspondsto choosing pi = Zk and G = O(n), SO(n). We also provide a lower boundon the number of components for the case G = Spin(n) for sufficiently largen.In the second part, we describe the space Hom(Γ, SU(2)), where Γ is agroup arising from a central extension of the form0→ Zr → Γ→ Zk → 0.The description of this space is good enough that, using some known results,it allows us to compute its cohomology groups.iiPrefaceAll of the results in this thesis are my own work under the supervision ofAlejandro Adem. The paper “On the space of commuting orthogonal matri-ces” is based on the work done in the first chapter and has been publishedin [Roj13]. I had a lot of great conversations with Juan Souto and he gaveme great ideas and suggestions on how to attack the problem.For the second chapter, I mostly worked on my own with guidance ofAlejandro but Jose´ Manuel Go´mez was of great help when I was trying tounderstand the space of homomorphisms from abelian groups into SU(2).iiiTable of ContentsAbstract . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iiPreface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iiiTable of Contents . . . . . . . . . . . . . . . . . . . . . . . . . . . . ivAcknowledgements . . . . . . . . . . . . . . . . . . . . . . . . . . . viIntroduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11 Spaces of commuting matrices. . . . . . . . . . . . . . . . . . . 61.1 Background and preliminaries . . . . . . . . . . . . . . . . . . 91.1.1 The space of commuting tuples . . . . . . . . . . . . . 101.1.2 The case G = O(n) . . . . . . . . . . . . . . . . . . . . 111.2 A general strategy to study Ck(G) . . . . . . . . . . . . . . . . 121.3 The connected components of Ck(O(n)) . . . . . . . . . . . . . 141.4 The connected components of Ck(SO(n)) . . . . . . . . . . . . 281.4.1 The proof of the recursive formula . . . . . . . . . . . . 341.5 Bounds using Stiefel-Whitney classes cannot be sharp . . . . . 381.6 A bound for the number of components of Ck(Spin(n)) . . . . 422 Spaces of homomorphisms from central extensions of freegroups by free groups into SU(2). . . . . . . . . . . . . . . . . 452.1 The spaces Hom(Zn ⊕ A, SU(2)) . . . . . . . . . . . . . . . . . 492.2 The reduction modulo 2 . . . . . . . . . . . . . . . . . . . . . 54iv2.3 Extensions with center of rank 1. . . . . . . . . . . . . . . . . 562.4 The general case. . . . . . . . . . . . . . . . . . . . . . . . . . 61Concluding remarks. . . . . . . . . . . . . . . . . . . . . . . . . . . 67Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69A Maximal abelian subgroups of O(n). . . . . . . . . . . . . . . . 72vAcknowledgementsFirstly, I would like to thank my supervisor, Alejandro Adem, for all ofhis support, suggestions, and the fruitful conversations he had with me, foralways being available, patient, and encouraging. It was a joy being hisstudent.I am also thankful to Juan Souto, he was great to talk to and he intro-duced me to the ideas and techniques that eventually led me to finish the firstpart of this thesis, he also taught some really fun classes. To Jose´ ManuelGo´mez, who in addition to patiently reading and criticizing anything I wouldwrite, gave me very useful pointers and suggestions for the work done in thesecond part of this work.I also want to thank Dale Rolfsen, for his support from the beginning ofmy PhD, when I was still trying to find my place at UBC, and his continuedinterest and help throughout my time here.Finally, I would like to thank CONACyT for their financial support duringmy studies, and Edgardo Rolda´n for his help when I was trying to simplifyone rather complicated formula.viIntroductionThe spaces of group homomorphisms Hom(pi,G) between a discrete group piand a Lie Group G are interesting for their relation with bundle theory andcan be quite complicated. The topology on Hom(pi,G) is given by consideringit as a subspace of Gk, where k is the number of elements in a generating setof pi. The group G acts naturally on these spaces by conjugation, the quotientspace Rep(pi,G) := Hom(pi,G)/G is called the representation space and itcan be identified with the moduli space of flat G−bundles over a manifoldwith fundamental group equal to pi. These moduli spaces of bundles areimportant in physics, they are related with a number of important quantumfield theories, such as Yang-Mills and Chern-Simons theories.In particular, when pi = Zk, the space Hom(Zk, G) can be identified withthe spaceCk(G) := {(g1, . . . , gk) ∈ Gn : gigj = gjgi for all i, j},of ordered commuting k-tuples in G. The fundamental group of (S1)k is Zkso the space Rk(G) := Rep(Zk, G) can be identified with the moduli spaceof flat G−bundles over the k−torus.The spaces Ck(G) were studied from the homotopical viewpoint by Ademand Cohen in [AC07], where they find a stable splitting (after one suspension)of the space Ck(G) for large class of Lie groups, including all closed subgroups1of GL(n,C). The splitting has the following form:Σ(Ck(G)) '∨1≤j≤kΣ∨(kj)Cj(G)/Sj(G) ,where Sj(G) is the subspace of Cj(G) of tuples with at least one entry equal tothe identity of G. The quotients Cj(G)/Sj(G) are highly singular spaces andare very complicated in general, so explicit computations using this techniquehave been limited. However, based on this splitting, Crabb in [Cra11] andindependently Baird, Jeffrey and Selick in [BJS11] were able to compute thestable homotopy type of Ck(SU(2)) explicitly.The spaces Ck(G) are in general not connected, even if G is connectedand simply connected. But note that if G is connected, then Rk(G) andCk(G) have the same number of path connected components.The number of path components of Ck(G) is known for some special casesof G, for example, in [AC07] it was shown that Ck(G) is connected when Gis one of U(q), SU(q) or Sp(q). In [TGS08] Torres-Giese and Sjerve foundthe number of connected components of Ck(SO(3)), and in [ACG13] thecalculations were done for the case when G is a central product of specialunitary groups. In [KS00], it was shown that the space C3(Spin(n)) is dis-connected for n ≥ 7, previously, a wrong assumption about the connectivityof these spaces led to a miscalculation of the number of vacuum states insupersymmetric Yang-Mills theories over spatial (S1)3 by Witten in [Wit82].In the first chapter of this dissertation, we will find the number of con-nected components of Ck(O(n)) and Ck(SO(n)) for all k and n. Here, thegroup O(n) is the group of orthogonal n× n real matrices, and SO(n) is thesubgroup of O(n) of matrices with determinant equal to one. It is interestingto note that O(n) is a disconnected group and that all of the Lie groups inthe examples above are connected. However, even though O(n) is not con-nected, our analysis will show that conjugating by an element of G preserves2the connected components of Ck(O(n)), so our results effectively computethe connected components of Rk(O(n)) and Rk(SO(n)).The proofs will be based in the approach described in Section 2.4, andalthough we use it here mainly to calculate connected components this tech-nique has been used to obtain a variety of results about the topology of thespaces of commuting tuples, some examples are [Bai07], [GPS12] and [PS13].We will also show that the components of Ck(O(n)) and Ck(SO(n)) sta-bilize for sufficiently large values of n, which allows us to calculate the com-ponents of Ck(O) and Ck(SO), where O and SO are the direct limits of O(n)and SO(n) respectively under the inclusion maps. At the end of the chapter,we use the calculations for SO(n) and Stiefel-Whitney classes to find a lowerbound to the number of components of Ck(Spin(n)) (for a sufficiently largen), where Spin(n) is the spinor group of dimension n, which is the connecteddouble cover of SO(n).Up to this point, we will have only been looking at spaces of homomor-phisms fixing the source group to be Zk and changing the target group. Inthe second chapter, we will explore what happens when we try to take a moregeneral group as a source group. Namely, we study the spaces of homomor-phisms Hom(Γ, SU(2)), where Γ is a central extension of the form0→ Zr → Γ→ Zn → 0.Notice that we have restricted the target group to be SU(2), this was one ofthe first groups to be chosen as a target when the spaces of commuting tuplesbegan to be studied systematically in [AC07]. Since we are generalizing thesource group, we should look at a target group for which we know enoughabout the space of commuting tuples to help us understand the case witha different source group. As mentioned before, the stable homotopy typeof Ck(SU(2)) was found in [Cra11] and [BJS11] so this makes SU(2) anideal candidate for our purposes. This is also what motivates the class ofsource groups we are considering, these extensions are close enough to the3free abelian case so that we can use what we know about Ck(SU(2)) togive a description of Hom(Γ, SU(2)) good enough to be able to compute itscohomology.The way we go about analyzing this space is by splitting it into twosubspaces and then studying each piece separately. We have a decompositionHom(Γ, SU(2)) = Hom(Γ/[Γ,Γ], SU(2)) ∪RΓ,where RΓ is the space of all the homomorphisms that do not factor throughthe abelianization of Γ. The first piece of this decomposition is analyzed inSection 2.1 using results by Adem and Go´mez in [AG11]. Section 2.3 dealswith RΓ in the case when r = 1 and finally, the general case for r is done inSection 2.4.In [PS13] it was shown that if K is a maximal compact subgroup ofG, then the space Ck(G) deformation retracts to Ck(K), this means thatour results about the number of connected components of Ck(O(n)) andCk(SO(n)) can be translated verbatim to Ck(GL(n,R)) and Ck(SL(n,R))since O(n) and SO(n) are the maximal compact subgroups of GL(n,R) andSL(n,R) respectively. More generally, in [Ber13], Bergeron proves that, givenany nilpotent group N then Hom(N,G) deformation retracts to Hom(N,K).Since the groups considered in Chapter 2 are nilpotent, this means that wecan also translate the results in Chapter 2 about Hom(Γ, SU(2)) to theircorresponding counterparts for Hom(Γ, SL(2,C)).Each chapter begins with a summary of the precise statements of themain results proved in that chapter, as well as how they are organized withinthe chapter. Some of the proofs in Chapter 1 rely heavily on a few well knownfacts of linear algebra, and for the sake of completeness and self-containmentof this text I have added their simple proofs as an appendix.I have also added a small epilogue with the concluding remarks, where wemention some of the possible directions in which one would want to generalizethe results of this thesis and state some of the complications that may arise.4Notation: Throughout this dissertation, the symbol Z2 will representthe multiplicative group with two elements {1,−1} and Z/2Z will be theadditive group of 2 elements {0, 1}. I will use |A| to denote the cardinalityof the set A and the symbol(nk)is the number of ways to choose k elementsfrom a set containing n elements.5Chapter 1Spaces of commuting matrices.In this chapter we we will enumerate the path connected components of thespace of ordered commuting k-tuples of orthogonal and special orthogonalmatrices respectively. We will also provide a lower bound on the numberof connected components of the space of commuting elements in the spinorgroups.The first section will serve as an introduction to the study of spaces ofhomomorphisms in general and will present some known results. The secondsection presents one of the current strategies used for the general study ofthe space of commuting elements in a Lie Group.In Section 1.3, we will use the strategy described in Section 1.2 to provethe main theorem in this chapter, which calculates the number of connectedcomponents of the space Ck(O(n)). The theorem is:Theorem 1.0.1. For each n, k ∈ N, the space Ck(O(n)) hasbn2 c∑j=0(2kn− 2j)connected components.Note that the formula no longer depends on n when n ≥ 2k − 1 since the6binomial coefficients vanish when n − 2j is large enough. So we get thefollowing corollary:Corollary 1.0.2. If n ≥ 2k − 1 then Ck(O(n)) has 22k−1 connected compo-nents.In fact, in this range, the inclusion O(n) ↪→ O(n+ 1) will induce an isomor-phism at the pi0 level. So if we let O = colim−→O(n) be the infinite orthogonalgroup, then Ck(O) has 22k−1 connected components.Although the group O(n) is not connected, our analysis will show thatconjugating by an element of G acts as the identity on pi0. As a result, wehave the following corollary:Corollary 1.0.3. Rk(O(n)) and Ck(O(n)) have the same number of compo-nents.So we effectively calculate the number of connected components of Rk(O(n))as well.At the end of the section, we also make some progress in the direction ofcomputing higher invariants of the spaces Ck(O(n)), in the form of provingthat they have(2kn)components homeomorphic to O(n)/Zn2 .The space Ck(SO(n)) is equal to the intersection of Ck(O(n)) with SO(n)k.In Section 1.4 we use this fact and Theorem 1.0.1 to calculate the number ofconnected components of Ck(SO(n)):Corollary 1.0.4. If n < 2k, the space Ck(SO(n)) has12kn−12∑j=0(2kn− 2j)components when n is odd, and it has12kn2∑m=0((2kn− 2j)+ (−1)n−2j2(2k − 1)(2k−1n−2j2)).7components when n is even. If n ≥ 2k − 1 it has 22k−k−1 components.Similar to the case of O(n), if we let SO = colim−→SO(n) we obtain thatCk(SO) has 22k−k−1 components. Since SO(n) is connected, Corollary 1.0.4also computes the number of connected components of the representationspaces Rk(SO(n)).To finish the section we extend the computation of the cohomology ofCk(SO(3)) done in [TGS08] to Ck(O(3)).In [AC07] Adem and Cohen found lower bounds for the number of compo-nents of Hom(pi,O(n)) for a general discrete finitely generated group pi. Theirapproach was to separate components using the Stiefel-Whitney classes. Inthe case pi = Zk this approach does not give a sharp result, in Section 1.5we will see that the Stiefel-Whitney classes are not enough to identify thecomponents of Ck(O(n)). In other words, we prove the following corollary:Corollary 1.0.5. Suppose k ≥ 3 and n ≥ 2k − 1. There exist homomor-phisms lying in different components of Ck(O(n)) with the same total Stiefel-Whitney class, and so the total Stiefel-Whitney class does not distinguish allthe components of Ck(O(n)).Even though the Stiefel-Whitney classes fail to distinguish the compo-nents completely, in Section 1.6 we will see how the second Steifel-Whitneyclass helps to find a bound for the components of Ck(Spin(n)). We will provethe following:Corollary 1.0.6. If n ≥ 2k − 1 then Ck(Spin(n)) has at least 22k−k−1−(k2)connected components.This bound is sharp enough to see that the space of commuting triples inSpin(n) is disconnected if n ≥ 7. This fact first appeared in [KS00].81.1 Background and preliminariesLet pi be a finitely generated group and G any Lie group. Consider the setHom(pi,G). This set can be given a topology as follows. Let Fk be the freegroup on k generators and let f : Fk → pi be any surjection. This inducesan inclusion f ∗ : Hom(pi,G)→ Hom(Fk, G) = Gk so we can give Hom(pi,G)the subspace topology. It is easy to see that this topology does not dependon the choice of k or f .In the next paragraph we explain the relation between these spaces andbundle theory; in particular, how the connected components of these spacescould help us understand principal G-bundles over Bpi.Given f ∈ Hom(pi,G) we can apply the classifying space functor to geta continuous map Bf ∈ Map∗(Bpi,BG). It is known that the classifyingspace functor is continuous as long as the compactly generated topology isused on the source, and so we can pass to connected components and get amap B0 : pi0(Hom(pi,G))→ [Bpi,BG]. Now recall that [Bpi,BG] classifies allprincipal G-bundles over Bpi, this means that understanding pi0(Hom(pi,G))and the map B0 may help us understand principal G-bundles over Bpi. Thisis of particular interest for the cases when Bpi is a compact manifold, forexample when pi = Zk.The spaces Hom(pi,G) have a G action induced by the conjugation actionon G. Since inner automorphisms induce mappings homotopic to the identityon BG we have that B0 factors through Rep(pi,G) := Hom(pi,G)/G. Somepeople prefer to study Rep(pi,G) instead but note that if G is connected thenRep(pi,G) has the same number of components as Hom(pi,G).91.1.1 The space of commuting tuplesIn the particular case when pi = Zk the space Hom(Zk, G) can be identifiedwith the space Ck(G) of ordered commuting k-tuples in Gk, i.e. the setCk(G) = {(g1, . . . , gk) ∈ Gk : gigj = gjgi for all i 6= j}.We do this by identifying f ∈ Hom(Zk, G) with (f(e1), . . . , f(ek)) ∈ Ck(G),where {e1, . . . , ek} is the standard basis for Zk. In the sequel, we will identifythese two spaces without explicitly mentioning it. In this case, Bpi = (S1)kand by studying Ck(G) we will be getting information about principal G-bundles over the k-torus.It is worth mentioning that the space Ck(G) may have many componentseven when G is connected and simply connected. Such an example wasgiven by Kac and Smilga in [KS00] where they showed that C3(Spin(7)) isnot connected. Here is a condition which guarantees the connectedness ofHom(Zk, G):Proposition 1.1.1. If every abelian subgroup of G is contained in a maximaltorus then Ck(G) is path connected.This proposition was proved in [AC07]. We will give an alternate proof ofthis proposition in the following section as an example of an easy applicationof our approach.As a consequence of this proposition we have the following corollary.Corollary 1.1.2. For G = SU(n), U(n), Sp(n) the spaces Ck(G) are pathconnected.This follows from the known fact that maximal abelian subgroups in thegiven groups are precisely the maximal tori (see for example [BtD95]).More recently, in [AG12], Adem and Go´mez have described all compactsimply connected Lie groups that have connected Ck(G) for all k as the finiteproducts of Sp(n)’s and SU(m)’s.10The final goal is to compute pi0(Ck(G)) for other kinds of groups G. Thisproblem was solved forG = SO(3) by Torres-Giese and Sjerve in [TGS08] andby Adem, Cohen and Go´mez for central products of special unitary groupsin [ACG13]. In the following sections we will present a complete solution tothis problem in the case when G = O(n), SO(n) generalizing the result in[TGS08], we also find a lower bound for the case G = Spin(n).1.1.2 The case G = O(n)Before restricting ourselves to studying Ck(O(n)) we state here what is knownabout the components of Hom(pi,O(n)) for a general finitely generated grouppi. These results are due to Adem and Cohen in [AC07] and they are obtainedby separating components of Hom(pi,O(n)) using the first and second Stiefel-Whitney classes.Theorem 1.1.3. There is a decomposition of Hom(pi,O(n)) into closed andopen subspacesHom(pi,O(n)) =∐w∈H1(pi,F2)Hom(pi,O(n))wwhere Hom(pi,O(n))w is the subset of Hom(pi,O(n)) of homomorphisms withfirst Stiefel-Whitney class equal to w.Theorem 1.1.4. If pi is a finitely generated discrete group andH2(pi/[pi, pi];F2)→ H2(pi;F2)is surjective, then|pi0(Hom(pi,O(n)))| ≥ |H1(pi;F2)||H2(pi;F2)|.We will show in Section 1.5 that not even the total Stiefel-Whitney classis able to identify the components of Ck(O(n)), which means that bounds11obtained in this fashion cannot be sharp in general.1.2 A general strategy to study Ck(G)In this section we will introduce a general strategy to approach the problemof studying the spaces Ck(G) for general G. This strategy has proven fruitfuland has been used to obtain a variety of results; some examples are [Bai07],[GPS12] and [PS13].The idea is to look at the following maps: for every abelian subgroup Aof G we have mapsΦA : G× Ak → Ck(G)defined by(g, a1, . . . , ak) 7→ (ga1g−1, . . . , gakg−1).The space G× Ak has an N(A)-action given byn · (g, a1, . . . , ak) = (gn−1, na1n−1, . . . , nakn−1)where N(A) is the normalizer of A in G. The map ΦA i s invariant on N(A)orbits so it induces a map on the quotient:Φ˜A : G×N(A) Ak → Ck(G).The space G×N(A)Ak is homeomorphic to G/A×W (A)Ak, where W (A) =N(A)/A is the Weyl group of A in G.Let SA = Im(ΦA). Note that this is just the orbit of Ak under theconjugation action of G on Ck(G). Then we have the following proposition,which is immediate from the definitions.Proposition 1.2.1. The following statements hold for all abelian subgroupsA and A′ in G:121. If A′ ⊂ A then SA′ ⊂ SA.2. SgAg−1 = gSAg−1 = SA.3. If f ∈ Ck(G) and A = Im(f) then f ∈ SA.From these properties we have the following direct corollary.Corollary 1.2.2. There is a decompositionCk(G) =⋃A∈ISAwhere I is a set of representatives of conjugacy classes of maximal abeliansubgroups of G.This way we have obtained a decomposition of Ck(G) into smaller spacesand we can use a “Mayer-Vietoris” approach to the problem by analyzingthe smaller pieces and their intersections.The relevant thing is that in general, even though the map Φ˜A : G/A×W (A)Ak → SA is not injective, we can use Φ˜A to translate information aboutG/A ×W (A) Ak to information about SA, which has the advantage that thespace G/A×W (A) Ak may be much easier to understand than Ck(G).Another good thing is that in many cases, for example when G is compact,the set I can be taken finite which significantly simplifies the analysis. For adetailed description of the “many cases” of the previous sentence, one shouldlook at [PS13].This approach however does require some understanding of the abeliansubgroups of G, or at least of the maximal conjugacy classes of abelian sub-groups. In the next sections we will apply this to compute the connectedcomponents in the cases G = O(n), SO(n), where we have a good under-standing of conjugacy classes of maximal abelian subgroups, or a so called“normal form” for abelian subgroups. We should emphasize that even thoughthis is mainly used here to compute connected components this strategy may13be useful to compute higher invariants of these spaces. For example in Propo-sition 1.3.13 we use this approach to compute the homeomorphism type ofsome of the components of Ck(O(n)).Note that if A is a maximal abelian subgroup then G×Ak is a manifold,on which path components and connected components agree. It follows thatthe path components and connected components of SA agree. If G is compactthis implies that the path components and connected components of Ck(G)also agree.As an example of an easy application of this approach we give an alternateproof of Proposition 1.1.1.Proof of Proposition 1.1.1. The hypothesis clearly implies that G is con-nected. Since all maximal tori are conjugate and every abelian subgroupis contained in a maximal torus by hypothesis, we can take I from Corollary1.2.2 to be I = {T}, where T is any maximal torus. Then Ck(G) = ST =ΦT (G× T k) and so it is connected since G× T k is connected.1.3 The connected components of Ck(O(n))In this section we will apply the strategy described in the previous sectionto enumerate the connected components of the space Ck(O(n)) for all k andn. Our goal is to prove Theorem 1.0.1, which we restate here:Theorem 1.0.1. For each n, k ∈ N, the space Ck(O(n)) hasbn2 c∑j=0(2kn− 2j)connected components.The strategy to prove Theorem 1.0.1 is to obtain a decomposition ofCk(O(n)) into closed and open disjoint subspaces, and then count the number14of components of each of these subspaces. Theorem 1.0.1 follows directly fromthe next theorem:Theorem 1.3.1. There is a decomposition of Ck(O(n)) into closed and opensubspaces:Ck(O(n)) =bn2 c∐j=0Uj,and for each j, the subspace Uj has( 2kn−2j)connected components.We will construct the subspaces Uj using the techniques described in theprevious section. The rest of this section is devoted to the proof of Theorem1.3.1.Consider the subgroup Aj = (S1)j × (Z2)n−2j ⊂ O(n) where the corre-spondence is(θ1, . . . , θj, t1, . . . , tn−2j)↔Mθ1 0 . . . 00 Mθ2. . .... Mθj...t10. . . 0. . . 0 tn−2j,whereMθ =(cos θ − sin θsin θ cos θ)is the 2× 2 matrix corresponding to a clock-wise rotation of an angle θ.For each j = 1, . . . , bn2 c we have mapsΦj : O(n)× Akj → Ck(O(n))15given byΦj(g, a1, . . . , ak) = (ga1g−1, . . . , gakg−1).It is well known that the Aj’s are representatives of conjugacy classes ofmaximal abelian subgroups of O(n). Or in other words, we have that ifSj = Im(Φj) thenCk(O(n)) =bn2 c⋃j=0SjThe sets Sj are certainly not disjoint, but we will produce our sets Ujusing the maps Φj.Notation: For the remainder of this section we let (e1, . . . , en) be thestandard ordered basis for Rn. We will denote the span of the vectorsv1, . . . , vl by 〈v1, . . . , vl〉.Definition 1.3.2. Given a = (a1, . . . , ak) ∈ Akj we can write each coordinatelike:am = (θm,1, . . . , θm,j, tm,1 . . . , tm,n−2j),and given i ∈ {1, . . . , n− 2j} we letri(a) := (t1,i, . . . tk,i) ∈ Zk2.The vector ri(a) is the vector of eigenvalues of the common eigenvector e2j+ifor the am’s. We also define the following subset of Akj :Pj := {a ∈ Akj : ri(a) 6= ri′(a) whenever i 6= i′}and we letUj := Φj(O(n)× Pj).The geometric description of the sets Uj is as follows: recall that everyrepresentation ρ : A → O(n) from a finitely generated abelian group Asplits up as a direct sum of orientable 2-dimensional representations and line16bundles. This follows from the fact that commuting complex matrices aresimultaneously diagonalizable. The set Uj consists of the representations thathave a direct sum decompositions in which j summands are 2-dimensionalorientable representations but do not have a direct sum decompositions inwhich j + 1 summands are 2-dimensional orientable representations.These Uj’s are the sets involved in Theorem 1.3.1. We shall prove thatthey are closed and disjoint, and that they have the claimed number ofconnected components.Note that for every g ∈ O(n) and every j we have gUjg−1 = Uj so thesesets are invariant under the conjugation action on Ck(O(n)).We begin the proof of Theorem 1.3.1 with the following proposition.Proposition 1.3.3. Let a ∈ Pj, b ∈ Pj′ and g ∈ O(n) such that gag−1 = b.Then j = j′ and {r1(a), . . . , rn−2j(a)} = {r1(b), . . . , rn−2j(b)}.Since the proof is slightly technical we will first introduce some prelimi-nary definitions and examples that will clarify the proof.Definition 1.3.4. Let a = (a1, . . . , ak) ∈ Akj , where each coordinate isai = (θi,1, . . . , θi,j, ti,1 . . . , ti,n−2j).We will say a is m-ordered if m ∈ {0, . . . , j} is such that:1. For every l ∈ {m+1, . . . , j} and i ∈ {1, . . . , k} it holds that θi,l ∈ {0, pi}.2. For every l ∈ {1, . . . ,m} there exists an i ∈ {1, . . . , k} such that θi,l /∈{0, pi}.Part (1) of this definition means that the only nonzero entries of eachai below row 2m are ±1 in the diagonal. Part (2) of this definition meansthat the matrices (a1, . . . , ak) have no common eigenvectors in the span of(e1, . . . , e2m). Note that an element of Ajk is m-ordered for at most onem ∈ {0, . . . , j}, but it may be that it is not m-ordered for any of them.Below are some examples to further clarify this definition:17Example 1.3.5. Let k = 3, n = 5 and j = 2. Leta = ((pi/2, 0, 1), (pi/4, pi,−1), (pi/6, 0, 1)) ∈ A32,which corresponds to the triple of matrices0 −11 01 00 11,1√2−1√21√21√2−1 00 −1−1,√32−1212√321 00 11.The triple a is 1-ordered. Letb = ((pi, pi/2, 1), (0, pi/3,−1), (0, 0,−1)) ∈ A32,which corresponds to the triple of matrices−1 00 −10 −11 01,1 00 112−√32√3212−1,1 00 11 00 1−1.The triple b is not m-ordered for any m since e1 and e2 are common eigenvec-tors of the 3 matrices and there are nonzero entries away from the diagonalbelow their corresponding rows. Letc = ((pi/2, pi,−1), (pi, pi/3,−1), (pi/6, 0, 1)) ∈ A32,which corresponds to the triples of matrices:0 −11 0−1 00 −11 ,−1 00 −112−√32√3212−1 ,√32−1212√321 00 11 .The triple c is 2-ordered.18Looking at b we see that not all elements x ∈ Akj are m-ordered for somem. However, we can always conjugate x by a permutation matrix to getx′ ∈ Akj which is m-ordered for some m. In the case of the example abovewe can conjugate b to get the tripleb′ =0 −11 0−1 00 −11 ,12−√32√32121 00 1−1 ,1 00 11 00 1−1 ,which is 1-ordered. Notice that we only have to permute some of the first 2jvectors of the standard basis and so this process does not alter the value ofany ri, since these only involve the last n− 2j rows of the matrices. This issummarized in the following lemma:Lemma 1.3.6. Given a ∈ Akj there exists a permutation matrix g ∈ O(n)such that a′ = gag−1 ∈ Akj is m-ordered for some m and ri(a) = ri(a′) for alli ∈ {1, . . . , n− 2j}.We need one more definition before going into the proof of Proposition1.3.3:Definition 1.3.7. Let a = (a1, . . . , ak) ∈ Akj be m-ordered. Then we definethe eigenmatrix of a to be the (n − 2m) × k matrix X with entries ±1determined by the equations:ai(e2m+l) = Xl,ie2m+l.Example 1.3.8. If a, b′ and c are as in Example 1.3.5 the the eigenmatrixof a is 1 −1 11 −1 11 −1 1 ,19the eigenmatrix of b′ is−1 1 1−1 1 11 −1 −1 ,and the eigenmatrix of c is(1 −1 1).Notice that if a ∈ Akj is m-ordered and X is its eigenmatrix then thevectors r1(a), . . . , rn−2j(a) are the last n − 2j rows of X. Also notice thatthe first 2(j −m) rows of X appear by pairs since they come from rotationsof angles 0 or pi. If a ∈ Pj then all of the ri(a) are different, which meansthat they are the only rows of X that appear an odd number of times. Sowe have the following lemma:Lemma 1.3.9. Let a ∈ Pj be m-ordered and let X be its eigenmatrix, thenthe set {r1(a), . . . , rn−2j(a)} is the set of rows of X that appear an odd numberof times in X.Now we are ready to prove Proposition 1.3.3.Proof of Proposition 1.3.3. Let a = (a1, . . . , ak) ∈ Pj and b = (b1, . . . , bk) ∈Pj′ . By Lemma 1.3.6 we may assume without loss of generality that a ism-ordered and b is m′-ordered.Let E±i be the ±1-eigenspace of ai and let F±i be the ±1-eigenspace ofbi. Since gaig−1 = bi we know that g(E±i ) = F±i for all i. This implies thatg(k⋂i=1(E+i ⊕ E−i ))=k⋂i=1(F+i ⊕ F−i ).Since a is m-ordered we know thatk⋂i=1(E+i ⊕ E−i ) = 〈e2m+1, . . . , en〉20and since b is m′-ordered we know thatk⋂i=1(F+i ⊕ F−i ) = 〈e2m′+1, . . . , en〉.This implies that m = m′ and that g preserves 〈e2m+1, . . . , en〉. In otherwords, g has the form: (? 00 U)where U is a (n− 2m)× (n− 2m) non-degenerate matrix.Let X be the eigenmatrix of a and Y be the eigenmatrix of b. We claimthat X and Y have the same rows up to permutation. This would finish theproof since we know by Lemma 1.3.9 that the sets {r1(a), . . . , rn−2j(a)} and{r1(b), . . . , rn−2j(b)} are the sets of rows that appear an odd number of timesin X and Y respectively.If (up,l) are the entries of U then for l ∈ {1, . . . , n− 2m} we havege2m+l =n−2m∑p=1up,le2m+p.For i ∈ {1, . . . , k} we can calculate:bige2m+l = ai(n−2m∑p=1up,le2m+p)=n−2m∑p=1up,lYp,ie2m+pgaie2m+l = g(Xl,ie2m+l) =n−2m∑p=1up,lXl,ie2m+p.Since big = gai then up,lYp,i = up,lXl,i for every p,i and l. In particular,if up,l 6= 0 then the p-th row of Y equals the l-th row of X. Since g isan isomorphism this implies that for every p ∈ {1, . . . , n − 2m} there is anl ∈ {1, . . . , n − 2m} such that ul,p 6= 0, so each row on X appears at least21once in Y .Now suppose the rows Xl1,•, . . . , Xlq ,• are equal, with li 6= li′ if i 6= i′. Weknow that the dimension of g(〈el1 , . . . , elq〉) is q, which means that the spacegenerated by the columns u•,l1 , . . . , u•,lq has dimension q. This means thatthere are at least q values of p for which there is an r such that up,lr 6= 0,i.e. the row Xl1,• appears at least q times in X. Since X and Y have thesame number of rows this shows that X and Y have the same rows up topermutation.As we said before, by Lemma 1.3.9 the rows that appear an odd numberof times in the matrix X are {r1(a), . . . , rn−2j(a)}. Similarly the rows thatappear an odd number of times in Y are {r1(b), . . . , rn−2j′(b)}. We haveproved that X and Y have the same rows so the proposition follows.Proposition 1.3.10. The spaces Uj are all closed and satisfyCk(O(n)) =bn2 c∐j=0Uj.Proof. The condition ri(a) 6= ri′(a) is closed in Akj for each pair (i, i′). SoPj is an intersection of closed subsets of Akj which is compact. This showsthat Pj is compact. Since Uj = Φj(O(n)×Pj) then Uj is compact, and henceclosed as required.Now we will show that the Uj’s cover Ck(O(n)). Since the Aj’s are repre-sentatives of maximal conjugacy classes of abelian subgroups of O(n) we canconjugate any commuting tuple into some Akj . And since the Uj’s are invari-ant under conjugation it is enough to show that they cover every Akj . Sup-pose then that a = (a1, . . . , ak) ∈ Akj with ai = (θi,1, . . . , θi,j, ti,1 . . . , ti,n−2j).If (t1,i, . . . tk,i) 6= (t1,i′ , . . . tk,i′) whenever i 6= i′ then a ∈ Pj. Otherwise, thereexist i 6= i′ such that (t1,i, . . . tk,i) = (t1,i′ , . . . tk,i′), which means that we canconjugate a into Akj+1 by simply reordering the rows to get θi,j+1 = 0 or pi forall i. We can repeat the process until we find a number j′ and a g ∈ O(n) for22which gag−1 ∈ Pj′ . We finish this procedure because Pbn2 c = Akbn2 c. It followsthat a ∈ Uj′ and shows thatCk(O(n)) =bn2 c⋃j=0Uj.To show that the union is disjoint suppose x ∈ Uj ∩ Uj′ . Then x =gag−1 = hbh−1 for some g, h ∈ O(n), a ∈ Pj and b ∈ Pj′ . This impliesh−1ga(h−1g)−1 = b and so j = j′ by Proposition 1.3.3.Since there are finitely many Uj’s it follows from Proposition 1.3.10 thateach Uj is open and closed in Ck(O(n)) and so each Uj is a union of com-ponents. All that is left to do then is count the components of Uj for eachj.It is important in what follows to understand the action of the Weyl groupW (Aj) = N(Aj)/Aj on Aj by conjugation, where N(Aj) is the normalizer ofAj in O(n). We explain this action in the following paragraphs.Recall that Aj admits a N(Aj) conjugation action. Since Aj acts by theidentity this induces an action of W (Aj) = N(Aj)/Aj on Aj.If we think about a matrix in O(n) as a linear map on Rn, then con-jugating corresponds to changing the basis on which we write the map. If(e1, . . . , en) is the standard basis for Rn and c ∈ N(Aj) then ce2j+1 is a unitvector on which every element of Aj acts as ±1, but the only unit vectorswith that property are {±e2j+1, . . . ,±en} so ce2j+1 must be one of them. Thesame goes for e2j+2, . . . , en and so c permutes the set {±e2j+1, . . . ,±en}. Asimilar analysis gives that c maps the plane generated by {e2l−1, e2l} isomet-rically to some other plane of that form for l = 1, . . . , j2 . And so the actionby conjugation on Aj is given in coordinates as above by:c · (θ1, . . . , θj, t1, . . . , tn−2j) = ((−1)s1θσ(1), . . . , (−1)sjθσ(j), tξ(1), . . . , tξ(n−2j))23where σ ∈ Σj and ξ ∈ Σn−2j are permutations and sl = 0, 1. We now proceedto count the components of Uj.The action of N(Aj) on Aj induces an action on O(n)× Akj given byc · (g, a1, . . . , ak) = (gc−1, ca1c−1, . . . , ca1c−1).Since Aj acts on the coordinate Akj by the identity this induces an action ofW (Aj) = N(Aj)/Aj on O(n)/Aj × Akj .From the paragraphs above it follows that if c ∈ N(Aj) and a ∈ Akj then(r1(cac−1), . . . , rn−2j(cac−1)) = (rξ(1)(a), . . . , rξ(n−2j)(a))for some ξ ∈ Σn−2j depending only on [c] ∈ W (Aj). This implies that theW (Aj) action restricts to Pj and to O(n)/Aj×Pj. Recall that Φj is constanton the orbits of this action, so it induces a surjective mapΦ˜j : O(n)/Aj ×W (Aj) Pj → Uj.Lets call the projection mapsq : O(n)× Pj → O(n)/Aj × Pjandp : O(n)/Aj × Pj → O(n)/Aj ×W (Aj) Pj.We want to count the components of Uj by counting the components ofO(n)/Aj ×W (Aj) Pj and then showing that Uj has the same number of com-ponents.Proposition 1.3.11. The space O(n)/Aj ×W (Aj) Pj has( 2kn−2j)connectedcomponents, indexed by the subsets {r1, . . . , rn−2j} ⊂ Zk2 of size n − 2j. Infact, given (g, a) ∈ O(n)× Pj the connected component containing q(p(g, a))is the one corresponding to the set {r1(a), . . . , rn−2j(a)}.24Proof. The set Lj := {(r1, . . . , rn−2j) ∈ (Zk2)n−2j : rm 6= rl for all m 6= l}is a discrete set with 2k!(2k−n+2j)! points, and O(n) has 2 components. SincePj ∼= (S1)jk × Lj then O(n) × Pj has exactly 2 × 2k!(2k−n+2j)! components. Itremains to see how q and p glue these components.If j = n2 then Pj = Akj∼= (S1)jk is connected. If {e1, . . . , en} is thestandard basis of Rn then matrix which sends e1 to −e1 and is the identityon 〈e1〉⊥ has determinant −1 and is in the normalizer of Aj. Then the setO(n)/Aj ×W (Aj) Pj is connected and so it has( 2kn−2j)=(2k0)= 1 componentsas required.If j 6= n2 then Aj contains at least one element of determinant −1 andso O(n)/Aj is connected. This means that p glues components by pairsand so O(n)/Aj × Pj has 2k!(2k−n+2j)! components. Notice that these com-ponents are indexed by the set Lj. Recall that for all a ∈ Pj and [c] ∈W (Aj) we have that (r1(a), . . . , rn−2j(a)) = (rξ(1)(cac−1), . . . , rξ(n−2j)(cac−1))for some ξ ∈ Σn−2j. By definition of Pj this shows that each W (Aj)-orbit inO(n)/Aj×Pj intersects exactly (n−2j)! connected components. This showsthat O(n)/Aj ×W (Aj) Pj has2k!(2k−n+2j)!(n−2j)! =( 2kn−2j)connected componentsas required, indexed by the subsets {r1, . . . , rn−2j} ⊂ Zk2 of size n− 2j.To finish the proof of our main theorem it remains to show the followingproposition.Proposition 1.3.12. The subspace Uj has( 2kn−2j)connected components, in-dexed by the subsets {r1, . . . , rn−2j} ⊂ Zk2 of size n− 2j.Proof. From Proposition 1.3.11 it is enough to show that if(g, a), (h, b) ∈ O(n)× Pjare such that Φj(g, a) = Φj(h, b), then{r1(a), . . . , rn−2j(a)} = {r1(b), . . . , rn−2j(b)}25(i.e. the map Φ˜j is injective on pi0). This is a direct application of Proposition1.3.3 by recalling that by the definition of Φj the fact that Φj(g, a) = Φj(h, b)implies thath−1ga(h−1g)−1 = b.This completes the proof of Theorem 1.3.1 and in turn proves Theorem1.0.1.In conclusion, this analysis shows that the components of Ck(O(n)) areindexed by the subsets of A ⊂ Zk2 such that |A| ≡ n(mod2) and |A| ≤ n.The inclusion O(n) ↪→ O(n+ 1) induces a mappi0(Ck(O(n)))→ pi0(Ck(O(n+ 1))).Using our description of the components this map is very easy to describe, itsends the component corresponding to A ⊂ Zk2 to the component correspond-ing to the subset A4{(1, . . . , 1)}, where 4 denotes the symmetric differenceof sets.From this it follows easily that if n ≥ 2k − 1 the inclusion induces abijection on pi0. This shows that|pi0(Ck(O))| = 22k−1as stated in Corollary 1.0.2.Proposition 1.3.3 shows that conjugating by an element ofO(n) acts as theidentity on pi0(Ck(O(n))). This shows that Rk(O(n)) has the same number ofconnected components as Ck(O(n)). This fact was stated in Corollary 1.0.3.We finish the section by identifying the homeomorphism type of some ofthe components of Ck(O(n)).Proposition 1.3.13. Each component of U0 is homeomorphic to O(n)/Zn2 .26Proof. The space P0 is discrete, since it is a subspace of (A0)k which is dis-crete (∼= Znk2 ). Let p ∈ P0. With arguments similar to the ones used inProposition 1.3.3, it is easy to see that the stabilizer of p under the conjuga-tion action of O(n) is exactly A0, the group with ±1 as diagonal entries andzeros everywhere else. This group is clearly isomorphic to Zn2 .The set Φ0(O(n)× {p}) is the orbit of p in Ck(O(n)) under the conjuga-tion action of O(n) and so it is homeomorphic to O(n)/Z(p) = O(n)/Zn2 . Allwe have to argue now is that this is the full component of Ck(O(n)) corre-sponding to {r1(p), . . . , rn(p)}. For this, we look at the following commutingdiagram:O(n)× P0Φ0 //U0O(n)/A0 ×W0 P0Φ˜088pppppppppppp.We know that the map Φ˜0 sends components of O(n)/A0 ×W0 P0 ontocomponents of U0, so it is enough to show that the image of O(n) × {p}under the quotient map is a component of O(n)/A0 ×W0 P0.Note that all the elements of Ak0 are 0-ordered by default, and can beidentified with their corresponding eigenmatrix. By definition, P0 is pre-cisely the subset of Ak0 consisting of elements that have eigenmatrix withoutrepeating rows. The way in which W (A0) acts on Ak0 is by permuting therows of the eigenmatrices, so we see that W (A0) acts freely on P0, whichis discrete. This implies that O(n)/A0 × {p} is homeomorphic to its imagein O(n)/A0 ×W (A0) P0 and that the image of O(n)/A0 × {p} is one of thecomponents of O(n)/A0 ×W (A0) P0. This finishes the proof.Proposition 1.3.14. All of the components of U[n2 ] are homeomorphic toeach other. In particular, they are all homeomorphic to the component con-taining the trivial homomorphism.27Proof. Note that if n is even U[n2 ] has only one component so there is nothingto prove. If n is odd U[n2 ] has 2k components indexed by the elements of Zk2.In this case A[n2 ] = P[n2 ]∼= (S1)[n2 ] × Z2. Let t = (t1, . . . , tk) ∈ Zk2. The mapΦj restricts to a surjectionΦj : O(n)× (S1)[n2 ]k × {t} → KtwhereKt is the component of Ck(O(n)) corresponding to t. Let 1 = (1, . . . , 1) ∈Zk2. We have a homeomorphismft : O(n)× (S1)k[n2 ] × {t} → O(n)× (S1)k[n2 ] × {1}.(g, (θij), t) 7→ (g, (tjθij), 1)This homeomorphism induces a homeomorphism between Kt and K1.1.4 The connected components of Ck(SO(n))In this section we will use what we know so far to count the components ofthe space Ck(SO(n)). Given our enumeration of the components of Ck(O(n))we can reduce the problem to a combinatorial problem regarding matriceswith ±1 entries. We then proceed to solve the combinatorial problem byshowing that the numbers we are looking for satisfy a recurrence relation.We will also calculate the cohomology ring of Ck(O(3)) with coefficients ina field with characteristic not equal to 2. This is a simple application of theresults in [Bai07] and [TGS08].Let’s begin by seeing how to translate our problem to a combinatorialone. Note thatCk(SO(n)) = SO(n)k ∩ Ck(O(n))and since SO(n) is closed and open in O(n) then Ck(SO(n)) is a union ofcomponents of Ck(O(n)).28Let Bj = Uj ∩ SO(n)k, then it follows from Theorem 1.3.1 thatCk(SO(n)) =bn2 c∐j=0Bj,and that each Bj is a union of components. Like before, it is enough tocalculate the number of components of Bj.Similar to the case of O(n) we do it by looking at Φ˜−1j (Bj) which, as weknow, has the same number of components than Bj. Note thata = (θ1, . . . , θj, t1, . . . , tn−2j) ∈ Aj ∩ SO(n)if and only ifn−2j∏i=1ti = 1.Recall that Pj ∼= (S1)kj × Lj whereLj = {(r1, . . . , rn−2j) ∈ (Zk2)n−2j : rm 6= rl for all m 6= l}.Then we have that Φ−1(Bj) ∼= O(n)× (S1)kj ×Mj, where Mj is the followingset:Mj := {(r1, . . . , rn−2j) ∈ Lj :n−2j∏i=1tl,i = 1 for all l = 1, . . . , k}.In the definition above, we must remember that ri = (t1,i, . . . , tk,i) ∈ Zk2.Also, remember that the components of O(n)/Aj ×W (Aj) Pj are indexed bythe sets {r1, . . . , rn−2j} ⊂ Zk2 of size n − 2j. The component correspondingto {r1, . . . , rn−2j} is in Φ˜−1j (Bj) if and only ifn−2j∏i=1tl,i = 129for all l = 1, . . . , k.We are interested in the cardinality of Mj so let (r1, . . . , rn−2j) ∈Mj andconsider putting the ri’s as rows in a matrix, this translates the problemof counting the components of Φ˜−1j (Bj) to solving the following countingproblem, in which m = n− 2j:Question 1.4.1. How many matrices of size m× k with entries in Z2 haveall their rows different and the product of all the entries in any column equalto 1?Lets call the number of such matrices f(m), recall that k is fixed fromthe start. Then the number components of Bj isf(m)m! . This is because thecomponents are indexed by the unordered set of rows not the ordered tupleof rows. We will now focus on finding f(m) and begin by showing that f isdefined recursively by:Proposition 1.4.2. The function f satisfies the following recurrence rela-tion:f(m) =2k!(2k − (m− 1))!− (m− 1)× (2k − (m− 2))× f(m− 2)whenever m ≤ 2k. Clearly f(m) = 0 if m > 2k.Proof. We first choose the first (m − 1) rows of our matrix. We want themto be different so there are 2k!(2k−(m−1))! ways of doing this. The property thatthe product of the entries in each column has to be 1 determines the lastrow uniquely. However, it may be that the last row turns out to be equal tosome other row, so we must subtract these cases.Note that if a matrix has exactly two rows equal and the product of allthe entries of each column is 1, then the product of all the entries in eachcolumn of the non-repeating rows is 1 as well. So if we are counting thematrices which have exactly two rows equal, one of which is the last one, wehave f(m−2) choices for the non-repeated rows, then 2k−(m−2) options for30the row we are repeating and (m−1) options for the position of the repeatedrow.It is clear from the definition that f(1) = 1 (vector with all entries 1)and f(0) = 1 (the empty matrix). Since the recurrence relation is two-stepthese values determine f(m) for all m. It is also clear from the definition(and the inductive formula) that f(2) = 0. Next we give formulas for f(m)in the even and odd case.Proposition 1.4.3. If m ≤ 2k is odd thenf(m) =(2k − 1)!(2k −m)!If 0 ≤ m ≤ 2k−1 thenf(2m) =12k(2k!(2k − 2m)!+(−1)m(2k − 1)2k−1!(2m)!m! (2k−1 −m)!)The proof that these formulas satisfy the recurrence relation will be givenat the end of the section in its own subsection. This is because the proof islong and purely computational, and it would distract us from our final goalof finding a formula for the components of Bj.As we said earlier we have the relation|pi0(Bj)| =f(n− 2j)(n− 2j)!.Using the explicit formula for f given in Proposition 1.4.3 we get an explicitformula for the number of components of Bj.Proposition 1.4.4. If n is odd, then|pi0(Bj)| =12k(2kn− 2j),31and if n is even, then|pi0(Bj)| =12k((2kn− 2j)+ (−1)n−2j2(2k − 1)(2k−1n−2j2)).Also recall thatCk(SO(n)) =bn2 c∐j=1Bj.Summing over j we get the formulas for number of connected components ofCk(SO(n)) as stated in Corollary 1.0.4, which we will restate here.Corollary 1.0.4. If n ≤ 2k, the space Ck(SO(n)) has12kn−12∑j=0(2kn− 2j)components when n is odd, and it has12kn2∑m=0((2kn− 2j)+ (−1)n−2j2(2k − 1)(2k−1n−2j2)).components when n is even. If n ≥ 2k − 1 it has 22k−k−1 components.The fact that the number of components stabilizes after n = 2k−1 followsboth from the fact that the ones for O(n) stabilize at that point, and from theformulas given above. The value to which they stabilize is easily computedfrom either of the formulas.The components of Ck(SO(3)) had been counted by Torres Giese andSjerve in [TGS08] and independently by Adem, Cohen and Go´mez in [ACG13]32using two different approaches. Our formula in the case n = 3 reads:|pi0(Ck(SO(3)))| =((2k3)+(2k1))12k=(2k3)12k+ 1=(2k − 1)(2k−1 − 1)3+ 1.Which is precisely the result in both references above. So our approach givesthe known answer in the case n = 3.Also in [TGS08], Torres-Giese and Sjerve calculated the homeomorphismtype of the components of Ck(SO(3)) which do not contain the identity. Inthis case we only have B0 and B1, the latter being the component of thetrivial homomorphism. The space B0 is a union of components of U0 and inProposition 1.3.13 we found the homeomorphism type of the components ofU0. It can easily be seen that our result for n = 3 agrees with the result in[TGS08]. The following example shows that we can now find the cohomologyring of Ck(O(3)) with coefficients in any field with characteristic 6= 2.Example 1.4.5. The cohomology ring H∗(Ck(O(3));F ). In this case we onlyhave U0 and U1 which have(2k3)and 2k components respectively. Proposition1.3.13 tells us that each component of U0 is homeomorphic to O(3)/Z32 andProposition 1.3.14 tells us that all the components of U1 are homeomorphicto the component containing the element 1 = (1, . . . , 1) ∈ Ck(O(3)). Letscall that component Rk,3.The rational cohomology groups of Rk,3 were computed in [TGS08] aswell as the mod − 2 cohomology ring. The ring structure with coefficientsin a field of characteristic relatively prime to |W | can be computed using([Bai07], Theorem 4.3):Theorem 1.4.6. Let G be a connected, compact Lie group, let T be a max-33imal torus in G and let F be a field with characteristic relatively prime to|W |, where W is the Weyl Group. Then H∗(Rk,3, F ) ∼= H∗(G/T × T k, F )W .This theorem applies to connected groups, but note thatRk,3 ⊂ Ck(SO(3))so we can use G = SO(3) to compute the cohomology. In this case G/T ∼= S2soH∗(G/T × T k, F ) ∼= ΛF (y1, . . . , yk)⊗F [x]< x2 >.Where |x| = 2 and |yi| = 1. The Weyl group is W = Z2 and acts by changingthe sign on each yi and x. So it follows that if F has characteristic 6= 2 wehaveH∗(Rk,3, F ) ∼= (ΛevenF (y1, . . . , yk))⊕(ΛoddF (y1, . . . , yk)⊗ x).So the cohomology ring H∗(Ck(O(3)), F ) is the direct sum of 2k copies ofH∗(Rk,3;F ) and(2k3)copies of the cohomology of O(3)/Z32 ∼= S3/Q8.1.4.1 The proof of the recursive formulaIn this subsection we will prove Proposition 1.4.3, we will show that theformulas given there satisfy the recurrence relation:f(m+ 2) =2k!(2k − (m+ 1))!− (m+ 1)× (2k −m)× f(m), (1.1)described in Proposition 1.4.2 with initial values f(1) = 1 and f(0) = 1.Since this recurrence relation only depends on the terms two steps beforethere is a unique solution given those initial values. We will separate ouranalysis into two cases depending on the parity of m. We start with the oddcase.Proposition 1.4.7. If m ≤ 2k is odd, thenf(m) =(2k − 1)!(2k −m)!34satisfies the recursive equation (1.1) and has initial value f(1) = 1.Proof. The fact that f(1) = 1 is obvious. Here are the calculations that showthat the recurrence relation is satisfied:f(m+ 2) =2k!(2k − (m+ 1))!− (m+ 1)× (2k −m)× f(m)=2k!(2k − (m+ 1))!− (m+ 1)× (2k −m)×(2k − 1)!(2k −m)!=(2k − 1)!(2k − (m+ 1))!(2k − (m+ 1))=(2k − 1)!(2k − (m+ 2))!Now we move on to deal with the even case, which is a little more com-plicated.Proposition 1.4.8. If m ≤ 2k−1, thenf(2m) =12k(2k!(2k − 2m)!+(−1)m(2k − 1)2k−1!(2m)!m! (2k−1 −m)!)satisfies the recurrence relation (1.1) and has initial value f(0) = 1.Proof. First we check the initial value:f(0) =12k(2k!2k!+(2k − 1)2k−1!2k−1!)=12k(2k)= 1as required.35We will prove the recurrence relation by parts. Letg(m) =(−1)m(2k − 1)2k−1!(2m)!m! (2k−1 −m)!andh(m) =2k!(2k − 2m)!,thenf(2m) =12k(h(m) + g(m))Claim 1. The function g satisfies the recurrence relation:g(m+ 1) = −(2m+ 1)(2k − 2m)g(m).Here is the calculation:g(m+ 1) =(−1)m+1(2k − 1)2k−1!(2m+ 2)!(m+ 1)! (2k−1 −m− 1)!= −(−1)m(2k − 1)2k−1!(2m)!(2m+ 1)(2m+ 2)(2k−1 −m)m!(m+ 1) (2k−1 −m− 1)!(2k−1 −m)= −(−1)m(2k − 1)2k−1!(2m)!(2m+ 1)(2)(2k−1 −m)m! (2k−1 −m)!= −g(m)(2m+ 1)(2k − 2m).Claim 2. The function h satisfies the recurrence relation:h(m+ 1) =(2k)(2k!)(2k − 2m− 1)!− (2m+ 1)(2k − 2m)h(m).36Here is the calculation:h(m+ 1) =2k!(2k − 2(m+ 1))!=2k!(2k − 2m− 1)(2k − 2m)(2k − 2m)!=(2k!(2k − 2m)!)(2k(2k − 2m)− (2m+ 1)(2k − 2m))=(2k)(2k!)(2k − 2m− 1)!− (2m+ 1)(2k − 2m)(2k!(2k − 2m)!)=(2k)(2k!)(2k − 2m− 1)!− (2m+ 1)(2k − 2m)h(m).And now we can finally prove the recurrence relation for f :f(2m+ 2) =12k(h(m+ 1) + g(m+ 1))=12k((2k)(2k!)(2k − 2m− 1)!− (2m+ 1)(2k − 2m) (h(m) + g(m)))=2k!(2k − 2m− 1)!− (2m+ 1)(2k − 2m)(12k(h(m) + g(m)))=2k!(2k − (2m+ 1))!− (2m+ 1)(2k − 2m)f(2m).This completes the proof of Proposition 1.4.3.Remark 1.4.9. In the odd case, there is a way to avoid the recurrence relationand calculate f directly as follows. Let D be the set of m× k matrices with37entries in Z2 and that have all their rows different. Let ρ : D → Zk2 bethe function that sends a matrix to the vector containing the product of itscolumns. Note that f(m) is precisely the cardinality of ρ−1(1, . . . , 1). If mis odd, then changing the signs of all the entries in a column of a matrixwill change the sign of the product of that column. This implies that thecardinality of ρ−1(x) does not depend on x ∈ Zk2 and sof(m) =|D||Zk2|=2k!2k(2k −m)!,which is exactly the formula in Proposition 1.4.3. However, this doesn’t workin the case when m is even since changing the sign of all the entries in onecolumn does not change the sign of the product of the entries in the column.Hence, we are forced to solve the problem as we did, by solving the recurrencerelation. The fact that, in this case, the fibres of ρ don’t all have the samesize is captured in the formula by the strange correction term that does notappear in the odd case, I have yet to find the combinatorial meaning of thisterm.1.5 Bounds using Stiefel-Whitney classes can-not be sharpIn this section we will prove that the Stiefel-Whitney classes cannot distin-guish all of the components of Ck(O(n)).As described in Section 1.1 we have a mapB : Ck(O(n))→Map∗(BZk, BO(n)).The total Stiefel-Whitney class (SW class for short) is a locally constant38functionw : Map∗(BZk, BO(n))→ H∗(Zk;Z/2Z),so we get a functionwB : Ck(O(n))→ H∗(Zk;Z/2Z).This function is locally constant so it is constant on connected components.In [AC07] Adem and Cohen obtained the bounds stated in Theorem 1.1.4 forpi0(Ck(O(n))) by analyzing the first and second SW classes of the elementsin Ck(O(n)). A natural question following that is if the total SW classdistinguishes all the components of Ck(O(n)), in other words, is the mapwB0 injective? We will use our calculations and some results from [ACC03]to explain why the answer of this question is “No”.Given a representation ρ : Zk → O(n) we will write w(ρ) when we reallymean wB0(ρ), this should not cause any confusion. Recall that the total SWclass has the formw(ρ) = 1 + w1(ρ) + · · ·+ wn(ρ)where wi(ρ) ∈ H i(Zk;Z/2Z) is the i-th SW class of ρ. Also recall thatany representation ρ : Zk → O(n) splits as a direct sum of 2-dimensionalorientable representations and line bundles. In Lemma 3.1 of [ACC03] it isshown that twice any line bundle gives an SO(2) representation and thatSO(2) representations have trivial total SW class.Let ρ ∈ Ck(O(n)). We can writeρ = Θ1 ⊕ · · · ⊕Θj ⊕ θ1 ⊕ · · · ⊕ θn−2jwhere Θi is an SO(2) representation and the θi’s are non-repeating line bun-dles. This is the same as saying ρ ∈ Uj. In fact, given the description ofthe components of Uj we know that the component in which ρ lies is indexed39by the set {w1(θ1), . . . , w1(θn−2j)} under the bijection between H1(Zk;Z/2Z)and Zk2 induced by the standard basis {e1, ..., ek} ⊂ H1(Zk;Z/2Z).The total SW class splits sums as products, we know w(Θi) = 1 and theonly possible non-zero SW class of a line bundle is the first, so we havew(ρ) = w(θ1) · · ·w(θn−2j) = (1 + w1(θ1)) · · · (1 + w1(θn−2j)).The total SW class w(ρ) is of the form 1 +α for α ∈ H>0(Zk;Z/2Z). Weknow that H∗(Zk;Z/2Z) ∼= ΛZ/2Z(e1, . . . , ek) so H>0(Zk;Z/2Z) has cardinal-ity 22k−1. Corollary 1.0.2 tells us that if n is large enough then Ck(O(n)) hasexactly the same number of components. So if we want to show that wB0 isnot injective it is enough to show that there exists an α such that 1+α is notthe SW class of any representation. The following follows from ([ACC03],Proposition 3.4):Proposition 1.5.1. Let ρ : Zk → O(n) be a sum of line bundles. If w1(ρ) =w2(ρ) = 0 then w(ρ) = 1.This tells us that if k ≥ 3 then there are many α ∈ H>0(Zk;Z/2Z) suchthat 1 +α is not the SW class of any representation. Hence we’ve shown thefollowing:Corollary 1.5.2. If k ≥ 3 and n ≥ 2k−1 then the map wB0 is not injective,or in other words, the total Stiefel-Whitney class does not distinguish all thecomponents of Ck(O(n)).We can actually find explicit pairs of different components with equalSW class. By our description of components, this is the same as finding 2different subsets {x1, . . . , xi} and {y1, . . . , yi′} of H1(Zk;Z/2Z) such thati∏j=1(1 + xj) =i′∏j=1(1 + yj). (1.2)40This correspondence occurs because we can easily find line bundles whichhave as first SW class any specified element in H1(Zk,Z/2Z). So if we letθ1, . . . , θi be line bundles such that w1(θj) = xj and θ′1, . . . , θ′i′ be line bundlessuch that w1(θ′j) = yj and constructρ = θ1 ⊕ · · · ⊕ θiand similarly ρ′, then ρ and ρ′ lie in different components of Ck(O(n)) sincethe subsets {x1, . . . , xi} and {y1, . . . , yi′} are different, but w(ρ) = w(ρ′) byequation 1.2.Note that if x ∈ H1(Zk;Z/2Z) then x2 = 0 and so (1 + x)2 = 1. So it isactually enough to find a subset A ⊂ H1(Zk;Z/2Z) such that∏x∈A(1 + x) = 1.And then for every B ⊂ A the components corresponding to B and A − Bwill have the same SW class. The following proposition shows that A =H1(Zk;Z/2Z) works.Proposition 1.5.3. If k ≥ 3 then∏x∈H1(Zk;Z/2Z)(1 + x) = 1.Proof. The proof is by induction on k. Recall that we know thatH1(Zk;Z/2Z) ∼=ΛZ/2Z(e1, . . . , ek). The case k = 3 is a direct computation, which we will omit.41Suppose the proposition is true for k and let A = H1(Zk;Z/2Z)∪ {0}, then:∏x∈H1(Zk+1;Z/2Z)(1 + x) =∏x∈A(1 + x)∏x∈A(1 + x+ ek+1)=∏x∈A(1 + x+ ek+1)=∏x∈A(1 + x) + ek+1∑x∈A∏y∈A−{x}(1 + y)= 1 + ek+1∑x∈A(1 + x) = 1.Where the last sum is equal to zero since each generator appears 2k−1 timesand 1 appears 2k times.This shows that if k ≥ 3 and n ≥ 2k then for every subsetB ⊂ H1(Zk;Z/2Z)the components of Ck(O(n)) corresponding to B and to H1(Zk;Z/2Z) − Bhave the same SW class.1.6 A bound for the number of componentsof Ck(Spin(n))In his section we will use the second SW class and our results about thecomponents of Ck(SO(n)) to find a lower bound for the number of connectedcomponents of Ck(Spin(n)) for sufficiently large values of n. We will proveCorollary 1.0.6, which we restate here:Corollary 1.0.6. If n ≥ 2k − 1 then Ck(Spin(n)) has at least 22k−k−1−(k2)connected components.Here the Lie group Spin(n) is the universal cover of SO(n), we will callthat covering map s : Spin(n)→ SO(n).42In [KS00], Kac and Smilga showed that the space of commuting triples inSpin(n) is disconnected if n ≥ 7. Notice that this fact can be deduced fromour bound as well. So this can be consider a generalization of their result.It is known that a homomorphism ρ : Zk → SO(n) lifts to Spin(n) if andonly if w2(ρ) = 0. In other words, in the following sequence:Ck(Spin(n))s∗ // Ck(SO(n))w2 // H2(Zk;Z/2Z) ,we have that Im(s∗) = w−12 (0). Let K = w−12 (0). Since K is the continuousimage of Ck(Spin(n)) it follows that |pi0(Ck(Spin(n)))| ≥ |pi0(K)|. Corollary1.0.6 then follows from the next proposition:Proposition 1.6.1. If n ≥ 2k − 1 then |pi0(K)| ≥ 22k−k−(k2).Proof. As we said before, w2 is locally constant so it induces a mapw2 : pi0(Ck(SO(n)))→ H2(Zk;Z/2Z)and pi0(K) is the inverse image of 0 under this map.The cohomology group H2(Zk;Z/2Z) is a Z/2Z-vector space. If n islarge enough we can give pi0(Ck(SO(n))) a structure of a Z/2Z-vector spaceas described in Proposition 1.6.2 in such a way that w2 is linear. Then|Im(w2)| × |Ker(w2)| = |pi0(Ck(SO(n)))|.The proposition then follows from the inequality:|Im(w2)| ≤ |H2(Zk;Z/2Z)| = 2(k2),and the equalities: |Ker(w2)| = |pi0(K)|, |pi0(Ck(SO(n)))| = 22k−1−k.Proposition 1.6.2. If n ≥ 2k− 1 then pi0(Ck(SO(n))) admits a structure ofZ/2Z-vector space for which w2 is linear.43Proof. Recall that for n ≥ 2k−1 the map SO(n) ↪→ SO(n+1) induces anisomorphism on pi0 and so we get an isomorphism between pi0(Ck(SO(n))) andpi0(Ck(SO)). We will define the Z/2Z-vector space structure on pi0(Ck(SO)).Given x ∈ Ck(SO) we denote the component containing x by [x]. Letα, β ∈ Ck(SO). We know that α factors through some SO(n) and β factorsthrough some SO(m) so we can define α + β to be the compositionZkα⊕β// SO(n)⊕ SO(m) // SO(n+m) // SOGiven [α], [β] ∈ pi0(Ck(SO)) we can define [α] + [β] := [α + β]. Strictlyspeaking α+β depends on the choice of m and n but by our description of theconnected components of Hom(Zk, SO) we know that [α + β] only dependson the non repeating line bundles in the decomposition of α ⊕ β, and thatdoes not depend on any choices so the operation is well defined.Another way to think about the same operation is the symmetric differ-ence of sets. We know that each component of Ck(SO) corresponds uniquelyto some subset of Zk2. Given two such subsets A,B ⊂ Zk2 we can defineA+B := A4B, where 4 represents the symmetric difference.This gives Ck(SO) the structure of an abelian group on which everyelement is 2 torsion. In other words, this gives Ck(SO) the structure ofa Z/2Z-vector space.It remains to see that w2 : Ck(SO)→ H2(Zk;Z/2Z) is linear.Given [α], [β] ∈ pi0(Ck(SO)) we know thatw2([α] + [β]) = w2([α + β]) = w2(α⊕ β)= w1(α)w1(β) + w2(α) + w2(β)= w2(α) + w2(β) = w2([α]) + w2([β])Here we remember that w1 is zero for SO representations. This finishesthe proof of Proposition 1.6.2 and completes the proof of Proposition 2Spaces of homomorphisms fromcentral extensions of freegroups by free groups intoSU(2).In the previous chapter we talked about spaces of the form Hom(pi,G) forthe particular case when pi = Zk and then we specialized to the cases G =O(n), SO(n), Spin(n). In this chapter, we will explore what happens whenwe try to take a slightly more general source group pi. However, this comesat the cost of having to specialize G to a simpler group, in this case, SU(2).We will study the space of group homomorphisms Hom(Γ, SU(2)), whereΓ is a central extension of a free abelian group by a free abelian group. Weprovide a detailed analysis of this space capable of computing its cohomol-ogy and path components in terms of the cohomology class classifying theextension.Historically, SU(2) has been the first interesting candidate for explicit cal-culations when talking about spaces of homomorphisms, so it makes sense tostart there when we are trying to generalize the source group. The cohomol-45ogy of the space Ck(SU(2)) = Hom(Zk, SU(2)) was first studied in [AC07]for k = 2, 3 and in [Cra11] and [BJS11] for general k. In [AC07] Adem andGo´mez studied topological invariants of the space Hom(A,G), where A is anyabelian group and G is in a class of compact Lie groups containing SU(2),in particular they obtain the number of its connected components. We willuse these results to do our analysis.Throughout this chapter, Γ will be a central extension of the form0→ Zr → Γ→ Zk → 0.These extensions are classified by elements of H2(Zk;Zr), so by r-tuples ofthe formω = (ω1, . . . , ωr) ∈ H2(Zk;Zr) ∼=(H2(Zk;Z))r,whereωl =∑1≤i<j≤kβli,je∗i ∧ e∗j ∈ H2(Zk;Z).In this description, {ei}ki=1 are the standard generators of Zk, {e∗i }ki=1 are thedual generators of the cohomology ring H∗(Zk;Z) and βli,j ∈ Z. The groupΓ corresponding to the class ω is given in terms of generators and relationsbyΓ = 〈E1, . . . , Ek, X1, . . . , Xr : [Ei, Ej] =r∏l=1Xβli,jl , Xi is central〉.We chose this type of groups in an attempt to generalize the source groupfrom the abelian case in a way where we can still exploit our knowledge ofthe spaces Ck(SU(2)). These groups proved to be good candidates as thenext step following the understanding of the spaces of commuting tuples.One fundamental difference between Hom(Γ, SU(2)) and Ck(SU(2)) isthat the latter is always path connected and the former may not. In fact,there are two kinds of elements of Hom(Γ, SU(2)): the ones that factor46though the abelianization of Γ and the ones that do not. This gives a de-composition into disjoint subspacesHom(Γ, SU(2)) = Hom(Γ/[Γ,Γ], SU(2)) ∪RΓ,where RΓ is the space of all the homomorphisms that do not factor throughthe abelianization. The space Hom(Γ/[Γ,Γ], SU(2)) is compact as it is aclosed subspace of SU(2)k+r, and it will be shown that RΓ is also compact,so this is actually a decomposition as coproduct.The abelianization of Γ can be obtained from the homology exact se-quence of the extensionH2(Zk)φ// H1(Zr) // H1(Γ) // H1(Zk) // 0 .Since the last nonzero map splits we see thatH1(Γ) ∼= Γ/[Γ,Γ] ∼= Zk ⊕ coker(φ),whereφ : H2(Zk)→ H1(Zr)is given in matrix notation by:φ =β11,2 β11,3 · · · β1k−1,k......βr1,2 βr1,3 · · · βrk−1,k .The cokernel of this matrix can be computed using its Smith normal form tobecoker(φ) ∼= Z/a1Z⊕ · · · ⊕ Z/at ⊕ Zr−t.Whereai =di(φ)di−1(φ),47di(φ) is the greatest common divisor of the i× i minors of φ and t is the rankof φ viewed as a transformation between Q-vector spaces.In [AG11], Adem and Go´mez determine the number of path connectedcomponents of Hom(Zm⊕A,G) in terms of the orbits of the Weyl group actionon Hom(A, T ), where T is a maximal torus of G. In the case G = SU(2)their techniques can be used to find the homeomorphism type of each ofthose components. This will be discussed in Section 2.1, where we prove thefollowing theorem:Theorem 2.0.3. Let A = Z/q1Z ⊕ · · · ⊕ Z/qnZ be a finite group, whereq1, . . . , qr are all even and qr+1, . . . , qn are all odd, then the space Hom(Zk ⊕A, SU(2)) has 2r path connected components homeomorphic to Ck(SU(2))and q1q2···qn−2r2 path connected components homeomorphic to S2 × (S1)k.This theorem completely describes the subspace Hom(Γ/[Γ,Γ], SU(2)).The rest of the chapter is devoted to the understanding of RΓ. In Section 2.3we explain that this subspace only depends on the reduction modulo 2 of ω.Namely, we prove:Proposition 2.0.4. Let Γ be the quotient of Γ induced by Zr → (Z/2Z)r asin the diagram0 // Zr //Γ //Zk //00 // (Z/2Z)r // Γ // Zk // 0.Let RΓ the subspace of Hom(Γ, SU(2)) of homomorphisms that do not factorthough the abelianization of Γ, thenRΓ ∼= RΓ.In Section 2.3 we cover the case when the kernel of the extension has rank1 (r = 1) and show that, in this case, RΓ is either empty or 2k−2 copies of48SO(3) depending on the rank of the reduction of ω modulo 2.Section 2.4 deals with the case when the kernel has rank greater than1 (r > 1). We explain how the space RΓ is a disjoint union of subspaceshomeomorphic to RΓ(v) , where v ∈ (Z/2Z)r and Γ(v) is a group like the onescovered in Section 2.3. This means that RΓ is a disjoint union of copies ofSO(3) in this general case as well. For the final statement we use an auxiliaryfunction t which is defined in Section 2.4.Theorem 2.0.5. Let Γ be the central extension corresponding to the classω ∈ H2(Zk;Zr), then RΓ is homeomorphic to t(ω)× 2k−2 copies of SO(3).Since the cohomologies of Ck(SU(2)), S2×(S1)k and SO(3) are all known,and our description of Hom(Γ, SU(2)) is in terms of these spaces, we can usethese results to calculate the cohomology groups of Hom(Γ, SU(2)) and, inparticular, its connected components.2.1 The spaces Hom(Zn ⊕ A, SU(2))In this section, we will use the results from [AG11] and [BJS11] to com-pute the cohomology of the spaces Hom(Zn ⊕A, SU(2)), where A is a finiteabelian group. When A is empty this is the space of ordered commutingn−tuples in SU(2). In [AC07] Adem and Cohen found a stable splitting forthe space of commuting tuples, and then Crabb in [Cra11] and Baird, Jeffreyand Selick in [BJS11] independently found the homotopy type of the directsummands. This description can be used to explicitly calculate the cohomol-ogy of the space of commuting elements in SU(2). In [AG11], Adem andGo´mez calculated the number of path-connected components of the spaceHom(Zn⊕A,G) for G in a class of compact lie groups containing SU(2). Inthe case of SU(2), we will show how their techniques can be used to calculatenot only the number of path-connected components but also the cohomologygroups of all these components.49We begin by recalling the known results for the case when A is empty.The stable splitting of the commuting n−tuples in SU(2) given in [BJS11]is as follows:Σ(Cn(SU(2))) ' Σ(n∨m=1(nm)ΣS(mL)), (2.1)whereΣS(mL) 'S3 if m = 1,S2 ∨ (RP 4/RP 2) if m = 2,ΣRP 2 ∨ (RP k+2/RP k−1) if m > 2.(2.2)Although an explicit formula for the cohomology groups of Cn(SU(2))was not given in [BJS11], one can directly get it from (2.1) and (2.2). Thecohomology groups are:Hk(Cn(SU(2));Z) ∼=Z if k = 0,0 if k = 1,Z(n2) if k = 2,Zn ⊕ (Z/2Z)2n−1−n−(n2) if k = 3,Z(nk) ⊕ (Z/2Z)(nk−1)+(nk−2) if k ≥ 4 even,Z(nk−2) if k ≥ 5 odd.(2.3)Now we move on to the case when A is not empty. For this case, we willuse the approach that Adem and Go´mez used in [AG11]. For this sectiononly, we will use the shorthand notation X := Hom(Zn ⊕ A, SU(2)) andG := SU(2).If A has a set of generators of size i then there is an inclusion X ⊂Hom(Zn+i, G) induced by the surjection Zn+i → Zn ⊕ A. The conjugationaction of G on Hom(Zn+i, G) restricts to a G action on X. It was shown in[AG11] that the isotropy groups of the G-action on X are connected and ofmaximal rank, that is, for every x ∈ X the isotropy group Gx is connectedand we can find a maximal torus Tx in G such that Tx ⊂ Gx.50Pick a maximal torus T in G, let NG(T ) be its normalizer and W =NG(T )/T be the Weyl group of G. Consider the mapφ : G×XT → X(g, x) 7→ gx.As shown in [AG11], the fact that G acts with maximal rank isotropy impliesthat this map is surjective. There is a right NG(T )-action on G×XT givenby (g, x) · n = (gn, n−1x) and φ is invariant under this action, so we get aninduced map on the quotientϕ : G×NG(T ) XT = G/T ×W XT → X[g, x] 7→ gx.In this case XT = Hom(Zn ⊕ A, T ) = T n × Hom(A, T ). The followinglemma tells us that the map ϕ is injective in a large subset of its domain.Lemma 2.1.1. Let f ∈ Hom(Zn⊕A, T ). Assume that there exists x ∈ Zn⊕Asuch that f(x) is a regular1 element in G, then ϕ−1(ϕ([g, f ])) = {[g, f ]} forall g ∈ G.Proof. Let g ∈ G and assume there exists [g′, f ′] ∈ G/T ×W Hom(Zn⊕A, T )such that ϕ([g′, f ′]) = ϕ([g, f ]). This implies that gfg−1 = g′f ′g′−1 and, inparticular, that gf(x)g−1 = g′f ′(x)g′−1, sof(x) = (g−1g′)f ′(x)(g−1g′)−1 ∈ (g−1g′)T (g−1g′)−1.Since f(x) is regular this implies that T = (g−1g′)T (g−1g′)−1 and henceg−1g′ ∈ NG(T ), and since (g, f) · (g−1g′) = (g′, f ′) this proves that [g′, f ′] =[g, f ] as required.1Recall that an element in a Lie Group is regular if it is in exactly one maximal torus.51Remark 2.1.2. The Lemma is true for a general compact Lie group G, butnote that in our case, G = SU(2), the hypothesis on f can be stated as “theimage of f is not contained in {±1}” since ±1 are the only 2 elements ofSU(2) that are not regular.By [AG11, Corollary 3.4] the number of path connected components of Xis equal to the number of different orbits of the action of W on Hom(A, T ).Notice that Hom(A, T ) is a finite set. If o ⊂ Hom(A, T ) is a W -orbit wewill denote the component corresponding to o as Xo. If 1 is the trivialhomomorphism then X{1} is homeomorphic to Cn(SU(2)).In our case the W = Z/2Z-action on T = S1 is given by complex conju-gation. The only fixed points of this action are {±1}. Given f ∈ Hom(A, T )there are only two possibilities: either W = Z/2Z acts trivially on f or itacts freely on f . Since the only fixed points of the W -action on T are {±1}then W will act trivially on f precisely when f(A) ⊂ {±1}. In this case theW -orbit of f is only {f} and we have the surjectionϕ : G/T ×W (Tn × {f})→ X{f}which is just a shift by f ofϕ : G/T ×W (Tn × {1})→ X{1}∼= Cn(SU(2))and so X{f} is homeomorphic to Cn(SU(2)).If f(A) 6⊂ {±1} then W acts freely on f and its orbit is {f, f¯}. In thiscase, W acts freely on G/T ×T n×{f, f¯} interchainging the two components,hence G/T ×W (T n × {f, f¯}) ∼= G/T × T n. Furthermore, since the image off contains regular elements of G the mapϕ : G/T ×W (Tn × {f, f¯})→ X{f,f¯}is a homeomorphism by Lemma 2.1.1 together with the fact that ϕ is sur-52jective. This shows that X{f,f¯} is homeomorphic to G/T × Tn. We can nowsubstitute G = SU(2) and T = S1 to get that X{f,f¯} is homeomorphic toS2 × (S1)n. We have proved then the following lemma:Lemma 2.1.3. Let T ⊂ SU(2) be a maximal torus, W = Z/2Z the Weylgroup of SU(2) and o ⊂ Hom(A, T ) a W -orbit, then the component ofHom(Zn⊕A, SU(2)) corresponding to o is is homeomorphic to Cn(SU(2)) ifo is a point and it is homeomorphic to S2 × (S1)n if o is not a point.The only thing left to determine is how many W -orbits there are of eachtype. We do this in the next Lemma.Lemma 2.1.4. Let A = Z/q1Z⊕ · · · ⊕ Z/qkZ, where q1, . . . , qr are all evenand qr+1, . . . , qk are all odd, then Hom(A, T ) has 2r trivial W -orbits andq1q2···qk−2r2 nontrivial W -orbits.Proof. Since every orbit has either one or two points it is enough to find thefixed points of the W -action. Notice Hom(A, T ) = Z/q1Z⊕· · ·⊕Z/qkZ wherewe identify Z/qZ ⊂ S1 = T as the q-th roots of the unity. The W -action iscomplex conjugation and the only fixed points of this action are the ones onwhich every entry is either 1 or −1. The only entries that could have a valueof −1 are the first r. This proves that there are exactly 2r trivial orbits.The remaining q1q2 · · · qk− 2r points are paired up in nontrivial orbits of size2.Putting together Lemmas 2.1.3 and 2.1.4 we get the main theorem of thissection.Theorem 2.1.5. Let A = Z/q1Z ⊕ · · · ⊕ Z/qkZ be a finite group, whereq1, . . . , qr are all even and qr+1, . . . , qk are all odd, then the space Hom(Zn ⊕A, SU(2)) has 2r path connected components homeomorphic to Cn(SU(2))and q1q2···qk−2r2 path connected components homeomorphic to S2 × (S1)n.53This theorem, together with the formulas (2.3), can calculate the co-homology groups of Hom(Zn ⊕ A, SU(2)) explicitly for all n and A. As acorollary, we know exactly how many connected components these spaceshave.Corollary 2.1.6. Let A = Z/q1Z ⊕ · · · ⊕ Z/qkZ, where q1, . . . , qr are alleven and qr+1, . . . , qk are all odd, then the space Hom(Zn ⊕ A, SU(2)) hasq1q2···qk+2r2 path connected components.This corollary could have been deduced from [AG11, Corollary 3.4] and asimple calculation like the one in Lemma 2.1.4, but Theorem 2.1.5 gives anexplicit description of the homeomorphism type of every component.2.2 The reduction modulo 2In the last section we completely described the subspace of Hom(Γ, SU(2))of homomorphisms that factor through the abelianization of Γ, from nowon we will be focused on describing RΓ, the space of homomorphisms withnon-abelian image.In this section, we will show that if ω is the classifying class of the ex-tension, then RΓ only depends on the reduction of ω modulo 2. This followsfrom the easy but crucial fact about SU(2), as we will explain.Proposition 2.2.1. The center of any non-abelian subgroup of SU(2) is{±1}.Recall that Γ is given in terms of generators and relations by:Γ = 〈E1, . . . , Ek, X1, . . . , Xr : [Ei, Ej] =r∏l=1Xβli,jl , Xi is central〉.Given a homomorphism f : Γ → SU(2) we let Ai = f(Ei) and Bi = f(Xi).From the relations in Γ we know that the matrices Bi are in the center of54f(Γ). In the particular case when f ∈ RΓ, we know that f(Γ) is a non-abeliansubgroup of SU(2) and hence its center is {±1} by Proposition 2.2.1. Thisimplies that if f ∈ RΓ, then f factors through the quotient Γ induced byZr → (Z/2Z)r:0 // Zr //Γ //Zk //00 // (Z/2Z)r // Γ // Zk // 0.The cohomology class in H2(Zk; (Z/2Z)r) classifying Γ is simply the reduc-tion of ω modulo 2. We summarize this in the following proposition:Proposition 2.2.2. Let Γ be the quotient of Γ induced by Zr → (Z/2Z)r andlet RΓ the subspace of Hom(Γ, SU(2)) of homomorphisms that do not factorthough the abelianization of Γ, thenRΓ ∼= RΓ.Note that this proposition is not saying that Hom(Γ, SU(2)) is homeo-morphic to Hom(Γ, SU(2)), since Γ and Γ may have different abelianizations.We will use ω to denote the reduction of ω modulo 2. Proposition 2.2.2 hasthe two following direct corollaries.Corollary 2.2.3. If Γ′ is the central extension corresponding to the classω′ ∈ H2(Zk;Zr), and ω′ = ω, then RΓ′ is homeomorphic to RΓ.Proof. This follows from the fact that Γ ∼= Γ′ whenever ω = ω′.Corollary 2.2.4. If ω = 0, then RΓ is empty.Proof. In this case, Γ is the trivial extension and hence abelian. This impliesRΓ = ∅.55This means that for the purposes of studying RΓ we can restrict ourselvesto the simpler case of extensions of the form0→ (Z/2Z)r → Γ→ Zk → 0.2.3 Extensions with center of rank 1.In this section we will finish the description of Hom(Γ, SU(2)) for the specialcase when Γ is a central extension of the form0→ Z→ Γ→ Zk → 0.The cohomology class of this extension has the formω =∑1≤i<j≤kβi,je∗i ∧ e∗j ∈ H2(Zk;Z).Recall that we have the decompositionHom(Γ, SU(2)) = Hom(Γ/[Γ,Γ], SU(2)) ∪RΓ,and that by the previous section RΓ ∼= RΓ. We will begin by describingHom(Γ/[Γ,Γ], SU(2)) more explicitly.The abelianization of Γ in this case isΓ/[Γ,Γ] ∼= Zk ⊕ Z/mZ,where m is the maximum common divisor of the βi,j’s. We can use Theorem2.1.5 to give an explicit description of the space Hom(Γ/[Γ,Γ], SU(2)). Thereare 3 cases that depend on the value of m:56Corollary 2.3.1. If m = 0, thenHom(Γ/[Γ,Γ], SU(2)) ∼= Ck+1(SU(2))if m 6= 0 and m is even, thenHom(Γ/[Γ,Γ], SU(2)) ∼= Ck(SU(2))⊔Ck(SU(2))⊔m−22⊔i=1S2 × (S1)kif m is odd, thenHom(Γ/[Γ,Γ], SU(2)) ∼= Ck(SU(2))⊔m−12⊔i=1S2 × (S1)k .All that is left then is to describe RΓ. We have seen in Proposition2.2.2 that RΓ is determined by reduction of ω modulo 2. So, by the theskew-symmetric bilinear form ω ∈ H2(Zk;Z/2Z), and such bilinear formsare determined by their rank. Furthermore, the following proposition followsdirectly from [RV96, Proposition 4.1]:Proposition 2.3.2. There exists a Z-module basis g1, . . . , gk of Zk such thatω = g∗1 ∧ g∗2 + · · ·+ g∗2m−1 ∧ g∗2mwhere 2m ≤ n.This means that by changing basis on Zk we may assume thatω = e∗1 ∧ e∗2 + · · ·+ e∗2m−1 ∧ e∗2m. (2.4)The number 2m is called the rank of ω.We are now ready for the main theorem of this section.57Theorem 2.3.3. Let ω be the reduction of ω modulo 2 and let 2m be therank of ω, then the space RΓ is homeomorphic to the disjoint union of 2k−2copies of SO(3) if m = 1, and it is empty otherwise.Before proving Theorem 2.3.3 we must prove a preliminary result aboutthe space of anti-commuting pairs in SU(2).Proposition 2.3.4. The spaceY := {(A1, A2) ∈ SU(2)× SU(2) : [A1, A2] = −Id}of anti-commuting pairs in SU(2) is homeomorphic to SO(3).Proof. The pairs of anti-commuting elements in SU(2) are actually uniqueup to conjugation, this can be seen using [BFM02, Proposition 4.1.1] or byan easy direct calculation. In other words, the (diagonal) conjugation actionof SU(2) on Y is transitive, which implies thatY ∼= SU(2)/Zwhere Z is the stabilizer of any element (A1, A2) ∈ Y . This stabilizer must be{±Id} by Proposition 2.2.1 since A1 and A2 do not commute, and thereforeY ∼= SU(2)/{±1} ∼= SO(3)as claimed.Now we are ready to prove Theorem 2.3.3.Proof of Theorem 2.3.3. We first prove that if m ≥ 2 then RΓ is empty.We may assume ω is as in (2.4) and m > 1 (in particular this implies thatk ≥ 4). Assume for a contradiction that f ∈ RΓ. Let Ai = f(Ei), since weare assuming (2.4) we know that A1 commutes with A3 and A4, and thatthese two do not commute with each other. By Proposition 2.2.1 this implies58that A1 is equal to ±Id, however, [A1, A2] = −Id which is a contradiction.This proves that if m > 1, then RΓ is empty.Now we assume that m = 1 and that ω = e∗1 ∧ e∗2. Given f ∈ RΓ we letAi = f(Ei) ∈ SU(2), for each i > 2 we have that Ai commutes with A1 andA2, and these two do not commute, which means that Ai = ±Id. In otherwords we have that:RΓ ∼= {(A1, A2,±Id, . . . ,±Id) ∈ SU(2)k : [A1, A2] = −Id}∼= {(A1, A2) ∈ SU(2)2 : [A1, A2] = −Id} × {±Id}k−2∼=2k−2⊔i=1SO(3).The last homeomorphism follows from Proposition 2.3.4.Notice that this in particular shows that RΓ is compact, so we have thatthe decompositionHom(Γ, SU(2)) ∼= Hom(Γ/[Γ,Γ], SU(2)) unionsqRΓis in fact a decomposition as a coproduct. This decomposition, Theorem2.3.3, Corollary 2.3.1 and equation (2.3) are all the pieces we need to find thecohomology groups of the space Hom(Γ, SU(2)) in the case we are consideringnow. For the easy particular case when k = 2 we can use all this to get aneat description of the space Hom(Γ, SU(2)):Corollary 2.3.5. Let Γ be the extension corresponding to the cohomologyclass βe∗1 ∧ e∗2 ∈ H2(Z2;Z), thenHom(Γ, SU(2)) ∼={Hom(Γ/[Γ,Γ], SU(2)) if β is even,Hom(Γ/[Γ,Γ], SU(2))⊔SO(3) if β is odd.59Note that, in this case, the abelianization of Γ is simply Z2 ⊕ Z/βZ, sowe can use Corollary 2.3.1 and (2.3) to easily find the cohomology groups ofHom(Γ, SU(2)) in this case. In particular, we can calculate the number ofpath connected components.Corollary 2.3.6. Let Γ be the extension corresponding to the cohomologyclass βe∗1 ∧ e∗2 ∈ H2(Z2;Z), then|pi0(Hom(Γ, SU(2)))| =1 if β = 0,β+22 if β 6= 0 is even,β−12 + 2 if β is odd.As an example, we calculate the cohomology groups of the space of ho-momorphisms from the Heisenberg group to SU(2):Example 2.3.7. The Heisenberg group H fits in the extension0→ Z→ H → Z2 → 0which corresponds to the class e∗1 ∧ e∗2 ∈ H2(Z2,Z). In terms of generatorsand relations it is given by:H = 〈E1, E2, X : [E1, E2] = X,X is central〉.The abelianization of H is Z2 and so we have a decompositionHom(H,SU(2)) ∼= C2(SU(2))⊔SO(3).60The cohomology groups in this case are given by:H i(Hom(H,SU(2));Z) ∼=Z⊕ Z if i = 0,0 if i = 1,Z⊕ Z/2Z if i = 2,Z⊕ Z⊕ Z if i = 3,Z/2Z if i = 4,0 if i ≥ 5.The quotient SU(2)/{±Id} is homeomorphic to SO(3), and the quotientmap SU(2)→ SO(3) induces a map ρ : Hom(H,SU(2))→ Hom(Z2, SO(3)).The reason ρ lands on Hom(Z2, SO(3)) is because if f ∈ RH then f(X) =±Id, which implies (ρf)(X) = Id so ρf factors through the abelianiza-tion of H. It was independently shown in [ACG13], [Roj13] and [TGS08]that the space Hom(Z2, SO(3)) has exactly two components, the componentcontaining the trivial homomorphism and one component homeomoprhic toSO(3)/(Z/2Z)2. We know that C2(SU(2)) is connected so ρ(C2(SU(2))) iscontained in the component of Hom(Z2, SO(3)) which contains the trivialhomomorphism. The component of Hom(H,SU(2)) that is homeomorphicto SO(3) gets mapped under ρ to the component of Hom(Z2, SO(3)) that ishomeomorphic to SO(3)/(Z/2Z)2, ρ is simply the quotient map.2.4 The general case.We will now go back to the general case and investigate Hom(Γ, SU(2)) whenΓ is a central extension of the form0→ Zr → Γ→ Zk → 061with classifying cohomology classω = (ω1, . . . , ωr) ∈ H2(Zk;Zr) ∼=(H2(Zk;Z))r,whereωl =∑1≤i<j≤kβli,je∗i ∧ e∗j ∈ H2(Zk;Z).Here, {ei}ki=1 are the standard generators of Zk, {e∗i }ki=1 are the generators ofthe cohomology ring H∗(Zk;Z) and βli,j ∈ Z. Such group Γ is given in termsof generators and relations byΓ = 〈E1, . . . , Ek, X1, . . . , Xr : [Ei, Ej] =r∏l=1Xβli,jl , Xi is central〉.As before, we have a decomposition into subspacesHom(Γ, SU(2)) = Hom(Γ/[Γ,Γ], SU(2)) ∪RΓ,and it is enough to understand each of those subspaces. As mentioned in thebeginning of this chapter, the abelianization of Γ can be obtained via theexact sequenceH2(Zk)φ// H1(Zr) // H1(Γ) // H1(Zk) // 0 ,we obtain that H1(Γ) ∼= Γ/[Γ,Γ] ∼= Zk ⊕ coker(φ), whereφ : H2(Zk)→ H1(Zr)62is given in matrix notation by:φ =β11,2 β11,3 · · · β1k−1,k......βr1,2 βr1,3 · · · βrk−1,k .The cokernel of this matrix can be computed using its Smith normal form tobecoker(φ) ∼= Z/a1Z⊕ · · · ⊕ Z/at ⊕ Zr−t.Whereai =di(φ)di−1(φ),di(φ) is the greatest common divisor of the i × i minors of φ and t is therank of φ viewed as a transformation between Q-vector spaces. After calcu-lating the abelianization, we can use the results of Section 2.1 to describeHom(Γ/[Γ,Γ], SU(2)). All that is left to do is to describe RΓ.We have seen in Section 2.2 that RΓ is homeomorphic to RΓ, where Γ isthe reduction of Γ induced by Zr → (Z/2Z)r. We can break down RΓ intopieces by considering the restriction mapres : Hom(Γ, SU(2))→ Hom((Z/2Z)r, SU(2))induced by the inclusion (Z/2Z)r ⊂ Γ. Since the only elements of order 2in SU(2) are ±Id the space Hom((Z/2Z)r, SU(2)) can be identified withHom((Z/2Z)r,Z/2Z) ∼= (Z/2Z)r. Since this is a discrete set, the inverseimages under res of the elements in (Z/2Z)r induce a decomposition of thespace Hom(Γ, SU(2)) into closed and open subspaces. This decompositioninduces a decomposition of RΓ into closed and open subspaces by defining,for each v ∈ (Z/2Z)r, the subsetsR(v)Γ:= RΓ ∩ res−1(v).63We have argued then the following Proposition:Proposition 2.4.1. There is a disjoint union decompositionRΓ =⊔v∈(Z/2Z)rR(v)Γ.The description of these subsets can be made explicit as follows: a homo-morphism f ∈ Hom(Γ, SU(2)) is a choice of matrices Ai for i ∈ {1, . . . , k}and Bi = ±Id for i ∈ {1, . . . , r} such that for all i, j:[Ai, Aj] =r∏l=1Bβli,jl (2.5)For every v = (v1, . . . , vr) ∈ (Z/2Z)r we can describe R(v)ΓbyR(v)Γ= {f ∈ RΓ : Bi = (−1)viId}.Given Proposition 2.4.1, to understand RΓ, we only need to understandR(v)Γfor each v ∈ (Z/2Z)r. Recall that we can view v as a homomorphismv : (Z/2Z)r → Z/2Zso v induces a mapv∗ : H2(Zk; (Z/2Z)r)→ H2(Zk;Z/2Z).Definition 2.4.2. Given v ∈ (Z/2Z)r we define Γ(v) to be the extensioncorresponding to the cohomology classω(v) := v∗(ω) ∈ H2(Zk;Z/2Z).64Explicitly, the cohomology class ω(v) is given byω(v) =∑1≤i<j≤nβ(v)i,j e∗i ∧ e∗j ∈ H2(Zn;Z/2Z),where the coefficients β(v)i,j are given in terms of the coefficients of ω, and vasβ(v)i,j =r∑l=1vlβli,j ∈ Z/2Z.Note that if f ∈ R(v)Γ, then the relations (2.5) simplify to:[Ai, Ai] =r∏l=1(−1)vlβli,jId= (−1)∑rl=1 vlβli,jId= (−1)β(v)i,j Id,which are precisely the relations required for a homomorphism in RΓ(v) . Thisproves:Theorem 2.4.3. If Γ(v) is as in Definition 2.4.2 then R(v)Γ is homeomorphicto RΓ(v).Also, note that RΓ(v) can be calculated using Theorem 2.3.3 since Γ(v) is acentral extension with kernel of rank 1. The spaceRΓ(v) will be homeomorphicto 2k−2 copies of SO(3) if the rank of ω(v) is 2 and it will be empty otherwise.Let T (ω) : (Z/2Z)r → 2N be the function defined as:T (ω)(v) := rank(v∗(ω)) = rank(ω(v)).And let t : H2(Zk;Zr)→ N be the function defined as:t(ω) = |T (ω)−1(2)|.65By putting together Theorems 2.3.3, 2.4.3, and Proposition 2.4.1 we finallyobtain:Theorem 2.4.4. The space RΓ is homeomorphic to t(ω) × 2k−2 copies ofSO(3).Note that in particular this means that RΓ is compact, so the decompo-sitionHom(Γ, SU(2)) = Hom(Γ/[Γ,Γ], SU(2)) unionsqRΓis a coproduct. Our description of the cofactors is all in terms of Ck(SU(2)),SO(3), and S2×(S1)k, and the cohomology groups of all of these is known, sowe are able to compute the cohomology groups of the space Hom(Γ, SU(2)).66Concluding remarks.In this short epilogue I will mention some of the possible directions in whichone might try to extend this work, and I will explain some of the complica-tions that arise.The first thing one might try to do is get more information about thehigher invariants for Ck(O(n)). One can try to do something similar to whatwe did to find the connected components, since we have a decomposition ofCk(O(n)) as a disjoint union of the Uj’s, we can concentrate on understandingeach Uj independently. We can then look at the surjective mapsΦ˜j : O(n)/Aj ×W (Aj) Pj → Ujto try to get information about Uj. The first step would be to understand thevalue of the invariant we are looking for on O(n)/Aj ×W (Aj) Pj. I anticipatethis can potentially be done, since the space O(n) × Pj is understood andwe understand the W (Aj) action on it. What I think will be the harderproblem is to know what Φ˜j does to the invariant. This was the hard partin the case of the connected components as well. However, I do think thisapproach can work since one can get a good idea of the subspace whereΦ˜j is not injective and maybe, with a careful analysis of this subspace anddepending on the invariant we are looking for, that could be enough. Noticethat since Ck(SO(n)) is just a union of components of Ck(O(n)), and wehave determined exactly which ones, so anything we can get for O(n) we willhave for SO(n) as well.67Another thing one might try to do is get the exact number of componentsof Ck(Spin(n)). For this one could try first to describe the components ofCk(SO(n)) which have second Stiefel-Whitney class equal to zero. I thinkthis is a problem that is combinatorial in nature. And secondly, one wouldneed to understand how many components of Ck(Spin(n)) are mapped toeach of those components under the mapped induced by the covering map.There are two directions one could try to go when trying to extend theresults in Chapter 2. The first one is to try to extend the target group. Goodcandidates for this might be SU(p) for some prime p. The big difference hereis that Proposition 2.2.1 is not true anymore for p > 2, and that simple factwas used all over the chapter to analyze RΓ. One possibility to compensatethis could be to stratify the space RΓ by the rank of the centralizer of theimage of the homomorphisms and analyze the strata one by one. I mustadmit I do not know how hard this could get.The second direction is to try to generalize the source group. Goodcandidates for this are groups that are obtained by a sequence of extensionsby free abelian groups. For example, one first step could be to look at thespace Hom(Γ′, SU(2)) where Γ′ fits in an extension of the form0→ Zn → Γ′ → Γ→ 0where Γ is a group like the ones considered in Chapter 2. An even lessrestrictive option would be to try to study Hom(N,SU(2)) for a nilpotentgroup N , and someone more ambitious might want to try to directly look atHom(N,SU(p)). The idea here would be to somehow induct on the degreeof nilpotency of the group, but I don’t have a concrete idea of how this mightwork.68Bibliography[AC07] Alejandro Adem and Frederick R. Cohen. Commuting elementsand spaces of homomorphisms. Math. Ann., 338(3):587–626, 2007.[ACC03] A. Adem, D. Cohen, and F. R. Cohen. On representations andK-theory of the braid groups. Math. Ann., 326(3):515–542, 2003.[ACG13] Alejandro Adem, F. R. Cohen, and Jose´ Manuel Go´mez. Commut-ing elements in central products of special unitary groups. Proc.Edinb. Math. Soc. (2), 56(1):1–12, 2013.[AG11] A. Adem and J.M. Gomez. On the structure of spaces of commut-ing elements in compact lie groups. Configuration Spaces: Geome-try, Topology and Combinatorics, Publ.Scuola Normale Superiore,55(3):805–813, 2011.[AG12] Alejandro Adem and Jose´ Manuel Go´mez. Equivariant K-theory ofcompact Lie group actions with maximal rank isotropy. J. Topol.,5(2):431–457, 2012.[Bai07] Thomas John Baird. Cohomology of the space of commuting n-tuples in a compact Lie group. Algebr. Geom. Topol., 7:737–754,2007.[Ber13] Maxime Bergeron. The topology of nilpotent representationsin reductive groups and their maximal compact subgroups.arXiv:1310.5109v1, 10 2013.69[BFM02] Armand Borel, Robert Friedman, and John W. Morgan. Almostcommuting elements in compact Lie groups. Mem. Amer. Math.Soc., 157(747):x+136, 2002.[BJS11] Thomas Baird, Lisa C. Jeffrey, and Paul Selick. The space ofcommuting n-tuples in SU(2). Illinois J. Math., 55(3):805–813(2013), 2011.[BtD95] Theodor Bro¨cker and Tammo tom Dieck. Representations of com-pact Lie groups, volume 98 of Graduate Texts in Mathematics.Springer-Verlag, New York, 1995. Translated from the Germanmanuscript, Corrected reprint of the 1985 translation.[Cra11] M. C. Crabb. Spaces of commuting elements in SU(2). Proc. Edinb.Math. Soc. (2), 54(1):67–75, 2011.[Gal00] Jean H. Gallier. Geometric Methods and Applications, For Com-puter Science and Engineering, volume 38. Springer, 2000.[Gan77] F.R. Gantmacher. The Theory of Matrices, volume I. AMSChelsea, 1977.[GPS12] Jose´ Manuel Go´mez, Alexandra Pettet, and Juan Souto. On thefundamental group of Hom(Zk, G). Math. Z., 271(1-2):33–44, 2012.[KS00] V. G. Kac and A. V. Smilga. Vacuum structure in supersymmetricYang-Mills theories with any gauge group. In The many faces ofthe superworld, pages 185–234. World Sci. Publ., River Edge, NJ,2000.[PS13] Alexandra Pettet and Juan Souto. Commuting tuples in reduc-tive groups and thir maximal compact subgroups. Geometry andTopology, 17:2513–2593, 2013.70[Roj13] G. Higuera Rojo. On the space of commuting orthogonal matrices.Journal of Group Theory, Volume 0, Issue 0, ISSN (Online) 1435-4446, ISSN (Print) 1433-5883, 2013.[RV96] Zinovy Reichstein and Nikolaus Vonessen. Rational central simplealgebras. Israel J. Math., 95:253–280, 1996.[TGS08] E. Torres Giese and D. Sjerve. Fundamental groups of commutingelements in Lie groups. Bull. Lond. Math. Soc., 40(1):65–76, 2008.[Wit82] E. Witten. Constraints on supersymmetry breaking. NuclearPhysics, 202:253–316, 1982.71Appendix AMaximal abelian subgroups ofO(n).In this short appendix we explain how every abelian subgroup A ⊂ O(n) canbe conjugated inside one of the form Aj = (S1)j × (Z2)n−2j ⊂ O(n). Thisproves that the subgroups Aj are representatives of the conjugacy classes ofmaximal abelian subgroups of O(n) as required for the work in Chapter 1.This is purely a fact of linear algebra and it follows from the so called “normalform” for orthogonal matrices. I provide these proofs for completeness andstate the results in ways that are useful for the applications in Chapter 1,but none of this is new in any way. Similar results can be found in variousbooks in linear algebra such as [Gan77] or [Gal00].We will use the following well known fact about abelian subgroups ofU(n)Proposition A.0.5. If A ⊂ U(n) is an abelian subgroup then all the elementsof A can be simultaneously diagonalized by an orthonormal basis.Given a ∈ O(n) we know we can diagonalize a over the complex numbers.Given an eigenvalue λ of a let aλ be the eigenspace of a corresponding toλ. If λ is a non-real eigenvalue of a, then λ is also an eigenvalue of a and72has the same multiplicity as λ. If v ∈ aλ, then v ∈ aλ. This means we canchose an orthonormal basis of {v1, v1, . . . , vj, x1, . . . , xn−2j} where the pairsvl, vl correspond to pairs of non-real eigenvalues λl, λl and the xl’s are realeigenvectors for the real eigenvalues. From this basis we can construct a newreal orthonormal basis by letting yl =vl−vl√2iand y′l =vl+vl√2for l = 1, . . . , j.This gives a new set of real vectors {y1, y′1 . . . , y′j, x1, . . . , xn−2j} and it isroutine to check they are orthonormal and so they form an orthonormalbasis for Rn.If λl = cos θl + i sin θl we have the followingayl =avl − avl√2i=(cos θl + i sin θi)vl − (cos θl − i sin θi)vl√2i= cos θl(vl − vl√2i) + sin θl(vl + vl√2)= cos θlyl + sin θly′lSimilarly we can show ay′l = − sin θlyl + cos θly′l. Which shows that if welet p ∈ O(n) be the matrix which has as columns (y1, y′1 . . . , y′j, x1, . . . , xn−2j)thenp−1ap =cos θ1 − sin θ1 . . .sin θ1 cos θ1 0. . .... cos θj − sin θjsin θj cos θj0 ±1. . .. . . ±173This is known as the canonical form of an orthogonal matrix. Notice thatwe got p out of the diagonalization of a over C. We have then the followingfact.Proposition A.0.6. If A ⊂ O(n) is an abelian subgroup then there is anorthogonal decomposition W1 ⊕ · · · ⊕Wj ⊕ V1 ⊕ · · · ⊕ Vn−2j ∼= Rn such thatevery Wl has dimension 2, every Vl has dimension one, every element of Aacts on each Wl by a rotation (possibly trivial) and every element of A actsby ±1 on each Vl.Proof. By Corollary A.0.5 we can find an orthonormal basis {v1, . . . , vn} ofCn which diagonalizes all the elements in A. In other words each vi is aneigenvector for each a ∈ A. Since A ⊂ O(n) we can further assume that thebasis has the form {v1, v1, . . . , vj, vj, x1, . . . , xn−2j} where x1, . . . , xn−2j are allreal vectors. We constuct the basis {y1, y′1, . . . , yj, y′j, x1, . . . , xn−2j} as avobeand let Wl = 〈yl, y′l〉 and Vl = 〈xl〉.Consider the subroup Aj = (S1)j × (Z2)n−2j ⊂ O(n), where the inclusionis given as in Section 1.3. From Proposition A.0.6 we get the following twoCorollaries.Corollary A.0.7. For every abelian subgroup A ⊂ O(n) there exists p ∈O(n) and j ∈ N such that p−1Ap ⊂ Aj.Corollary A.0.8. If (a1, ..., ak) ∈ Ck(O(n)), then there exists p ∈ O(n) andj ∈ N such that p−1alp ∈ Aj for all l = 1, . . . , k.Proof. Let A = 〈a1, ..., ak〉 and apply Corollary A.0.7.74


Citation Scheme:


Citations by CSL (citeproc-js)

Usage Statistics



Customize your widget with the following options, then copy and paste the code below into the HTML of your page to embed this item in your website.
                            <div id="ubcOpenCollectionsWidgetDisplay">
                            <script id="ubcOpenCollectionsWidget"
                            async >
IIIF logo Our image viewer uses the IIIF 2.0 standard. To load this item in other compatible viewers, use this url:


Related Items