Two Special Cases of the DynamicalMordell-Lang Conjecture in PositiveCharacteristicbyKristina NelsonB.Sc., The University of British Columbia, 2015A THESIS SUBMITTED IN PARTIAL FULFILLMENT OFTHE REQUIREMENTS FOR THE DEGREE OFMASTER OF SCIENCEinThe Faculty of Graduate and Postdoctoral Studies(Mathematics)THE UNIVERSITY OF BRITISH COLUMBIA(Vancouver)April 2017c© Kristina Nelson 2017AbstractWe prove the positive characteristic version of the Dynamical Mordell-LangConjecture in two novel cases. Let p be a prime andK a field of characteristicp > 0. Let k ∈ N, α ∈ Gkm(K) and V ⊂ Gkm be a variety. Let ϕ : Gkm → Gkmbe a group endomorphism defined over K. We knowϕ(x1, x2, ..., xk) = (xa1,11 xa1,22 · · ·xa1,kk , ..., xak,11 xak,22 · · ·xak,kk )for some ai,j ∈ Z. In the case where the matrix of exponents, (ai,j) is similarto a single Jordan block, we show that the set S = {n ∈ N : ϕn(α) ∈ V } is afinite union of arithmetic progressions. When the dimension k = 3, we showS is a finite union of arithmetic progressions for any group endomorphismϕ.iiPrefaceThe topic and general ideas behind this thesis were the suggestion of myadvisor, Prof. Dragos Ghioca. Examples 1.1 was previously published in [2]and Examples 5.1 and 5.2 appeared in [3]. The remainder of this thesis isthe original, unpublished work of the author.iiiTable of ContentsAbstract . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iiPreface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iiiTable of Contents . . . . . . . . . . . . . . . . . . . . . . . . . . . . ivAcknowledgements . . . . . . . . . . . . . . . . . . . . . . . . . . . v1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42.1 Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42.2 Lemmas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63 Single Jordan Block . . . . . . . . . . . . . . . . . . . . . . . . 153.1 Proof of Proposition 3.1 . . . . . . . . . . . . . . . . . . . . . 153.2 Theorem 1.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . 174 Dimension 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 195 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 236 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27ivAcknowledgementsI would like to thank my advisor, Prof. Dragos Ghioca, not only for hisguidance with regard to this thesis, but also for his wisdom and adviceregarding my adventures in academia and mathematics.Thank you also, dad, for telling me seven years ago I just had to go toUBC and take Science One.vChapter 1IntroductionLet K be a characteristic zero field, X a quasiprojective variety, α ∈ X apoint, ϕ an endomorphism of X defined over K, and V ⊂ X a subvariety.As is usual in arithmetic dynamics we write ϕn(α) = ϕ ◦ ϕ ◦ · · · ◦ ϕ(α)to denote n applications of the endomorphism. Given this, the dynamicalMordell-Lang Conjecture predicts that the setS := {n ∈ N : ϕn(α) ∈ V } (1.1)is a finite union of arithmetic progressions. Here, as in the remainder of thisnote, a singleton set is considered an arithmetic progression with commondifference zero between its terms.The dynamical Mordell-Lang Conjecture has been shown to hold undercertain conditions [1, 4, 5, 6, 9]. For a full survey of recent progress onthe dynamical Mordell-Lang Conjecture readers may review [2]. A naturalextension of the conjecture is to positive characteristic K, however in thiscase counterexamples immediately arise where S cannot be written as anyfinite union of arithmetic progressions. See Example 1.1 below, and Chapter5 for further examples. In the following, as in the rest of this note, N0 denotesthe set of non-negative integers.Example 1.1. Let K = Fp(t). Let α = (1, t, 1, 1 − t) ∈ G4m, and defineϕ : G4m → G4m byϕ(x1, x2, x3, x4) = (x1x2, x2, x3x4, x4).Thenϕn(α) = (tn, t, (1− t)n, (1− t)).Let f(x1, x2, x3, x4) = x1 + x3 − 1. We take V = Z(f) = {β ∈ G4m(K) :f(β) = 0} and show in this case the set S of (1.1) is not a finite union of1Chapter 1. Introductionarithmetic progressions. Let n = mpj for m coprime to p, and j ∈ N0. Thenf(ϕn(α)) = tmpj+ (1− t)mpj − 1= tmpj+m∑i=0(mi)(−t)ipj − 1.If m > 1 then f(ϕn(α)) contains a non-zero term of the form m(−t)pj ,contributed by the i = 1 term of the sum. Given this, it is easy to seef(ϕn(α)) = 0 occurs if and only if m = 1. Thus the set of n ∈ N such thatϕn(α) is in Z(f) is {pj : j ∈ N0}. Note that this is not a finite union ofarithmetic progressions.So, the dynamical Mordell-Lang Conjecture as stated above fails inpositive characteristic. However, motivated by results of Moosa and Scanlon[8], the following positive characteristic version of Dynamical Mordell-Langconjecture was proposed [2, Conjecture 13.2.0.1]. Here, as in the remainderof this note, [m] denotes the set of integers {1, ...,m}.Conjecture 1.2. (Ghioca-Scanlon). Let K be a field of characteristic p > 0,X a quasiprojective variety, V a subvariety, α ∈ X(K) a point and ϕ anendomorphism of X defined over K. Then the set S = {n ∈ N : ϕn(α) ∈ V }is a finite union of arithmetic progressions, along with finitely many sets ofthe form m∑j=1cjpkjnj : nj ∈ N0 for each j ∈ [m] , (1.2)for some cj ∈ Q, and kj ∈ N.To describe the cases of this conjecture under consideration in this note,we introduce the following notation. In the case where X = Gkm for somek ∈ N and ϕ is a group endomorphism of Gkm, we have that ϕ is given byϕ(x1, x2, ..., xk) = (xa1,11 xa1,22 · · ·xa1,kk , ..., xak,11 xak,22 · · ·xak,kk ) (1.3)for some ai,j ∈ Z, and all (x1, x2, ..., xk) ∈ Gkm. Let A be a matrix with(A)i,j = ai,j ∈ Z. Then we denote the endomorphism of (1.3) with ϕA = ϕ.Since the matrix of ϕA◦ϕB is simply A·B, we have in general that ϕnA = ϕAn .2Chapter 1. IntroductionConjecture 1.2 has been proven in [2], in the case where X = Gkm andϕ = ϕA for a diagonalizable matrix A (see Proposition 4.1). Conjecture 1.2has also been proven by [3] when X = Gkm, ϕ is any regular self-map, andthe variety V is a curve. We prove the conjecture in the following two novelcases.Theorem 1.3. Let K be a field of characteristic p > 0. Let the groupendomorphism ϕA : Gkm → Gkm be defined byϕ(x1, ..., xk) = (xa1,11 xa1,22 · · ·xa1,kk , ..., xak,11 xak,22 · · ·xak,kk )for some ai,j ∈ Z, and assume that the matrix A = (ai,j) is similar to asingle Jordan block. Let V ⊂ Gkm be a variety, and α ∈ Gkm(K) a point.Then the setS = {n ∈ N : ϕnA(α) ∈ V }is a finite union of arithmetic progressions.Theorem 1.4. Let K be a field of characteristic p > 0. Let ϕ : G3m → G3mbe a group endomorphism, let V ⊂ G3m be a variety, and α ∈ G3m(K) apoint. Then the setS = {n ∈ N : ϕn(α) ∈ V }is a finite union of arithmetic progressions.As we can see, in these cases the more complicated sets of (1.2) do noteven arise.3Chapter 2Preliminaries2.1 NotationThroughout the remainder of this note, K will be a field of characteristicp > 0. We set X = Gkm for k ∈ N and suppose ϕ is a group endomorphismof Gkm. A function field E over field F is a finitely generated field extension,E/F , of positive transcendence degree. We will often consider the casewhere K is a function field over Fp.In this note an arithmetic progression N is a subset of N of the formN = {b+ d · i : i ∈ N0},for some b ∈ N and d ∈ N0. Note that N may be a singleton set if d = 0,otherwise N is infinite.Given polynomials f1, ..., fm ∈ K[x1, ..., xk] we use Z(f1, ..., fm) to de-note their zero locus in Gkm. That is, the points of Z(f1, ..., fm) in K are{β ∈ Gkm(K) : fi(β) = 0, ∀ i ∈ [m]}.For the sake of brevity, we invent the following notation. Given V ⊂ Gkma variety, α = (α1, ..., αk) ∈ Gkm(K), A ∈ Mk×k(Z), and N , we let S =S(V, α,A,N ) = {n ∈ N : ϕnA(α) ∈ V }. Given a matrix A ∈ Mk×k(Z), wesay ‘S(A) is a finite union of arithmetic progressions’ ifS = S(V, α,A,N ) = {n ∈ N : ϕnA(α) ∈ V }is indeed a finite union of arithmetic progressions for any choice of varietyV ⊂ Gkm, α ∈ Gkm(K) and arithmetic progression N .Definition 2.1. We say a sequence (an)∞n=0 is preperiodic if there existsN, d ∈ N such that for all n > N , an = an−d.42.1. NotationGiven a point α of quasiprojective variety X and an endomorphism ϕ :X → X, we say α is a preperiodic point of ϕ if the sequence (ϕi(α))∞i=1 ispreperiodic.Definition 2.2. An absolute value on the field F is a map | · |v : F → Rsuch that:• |x|v ≥ 0 for all x ∈ F and |x|v = 0 if and only if x = 0.• |xy|v = |x|v|y|v for all x, y ∈ F .• |x+ y|v ≤ |x|v + |y|v for all x, y ∈ F .Furthermore, we say that an absolute value | · |v on F is non-archimedean if• |x+ y|v ≤ max(|x|v, |y|v) for all x, y ∈ F .In this paper F will usually be a field of positive characteristic. Everyabsolute value on a field of positive characteristic is non-archimedian [7,Chapter 1, Section 1].Definition 2.3. Field F is said to be a Product Formula Field if the fol-lowing conditions are satisfied:• There exists a set of absolute values on F , MF .• For any fixed x ∈ F×, |x|v 6= 1 holds for only finitely many | · |v ∈MF .• There exists a function MF → N with its value at | · |v ∈ MF denotedby Nv, such that the following product formula holds:∏|·|v∈Mk|x|Nvv = 1 for all x ∈ F×. (2.1)In this note we rely on the fact that any function field F is a productformula field, see [7, Chapter 2] for background and details. By replacingevery | · |v in the standard set of absolute values with | · |Nvv we may assumethe set is normalized with Nv = 1 for all | · |v. For simplicity we make thisthe definition of MF .Definition 2.4. Let MK be a normalized, standard set of absolute valueson K.We also have the following fact from the theory of absolute values onfunction fields.Fact 2.5. If K is a function field over the field of constants Fp, and|x|v = 1 for all | · |v ∈MK , then x ∈ Fp. See [7, Chapter 2, Section 5].52.2. Lemmas2.2 LemmasThe following results will be used by Chapters 3 and 4, and will help usreduce certain instances of the problem to other already solved cases.On Arithmetic ProgressionsProposition 2.6.1. For all i ∈ [`], let Si be a finite union of arithmetic progressions.Then ∩`i=1Si is also a finite union of arithmetic progressions.2. The complement of a finite union of arithmetic progressions is anotherfinite union of arithmetic progressions.Part (1) of Proposition 2.6 follows from the fact that the intersectionof two arithmetic progressions is another arithmetic progression or theempty set. Part (2) follows from the fact that the complement of a singlearithmetic progression is a finite (possibly empty) union of arithmeticprogressions.Lemma 2.7. Let X be a quasiprojective variety, V ⊂ X a subvariety, ϕ :X → X an endomorphism and α ∈ X(K) a point. Suppose the sequence(ϕn(α))∞n=1 is preperiodic (in the sense of definition 2.1). Then {n ∈ N :ϕn(α) ∈ V } is a finite union of arithmetic progressions.Proof. Because (ϕn(α))∞n=1 is preperiodic we can find N, d ∈ N such thatϕn(α) = ϕn−d(α) for all n > N . For all i ∈ [N ], letMi := {i} be a singletonarithmetic progression. For all i ∈ [d], let Ni = {(N + i) + d · k : k ∈ N0}.Then we have defined N · d arithmetic progressions, and their union is N.For i ∈ [N ], let βi = ϕi(α). For i ∈ [d], let γi = ϕN+i(α), and note thatϕn(α) = γi for any n ∈ Ni. Thus {n ∈ N : ϕn(α) ∈ V } is a finite union ofthe Mi for which βi ∈ V , and the Ni for which γi ∈ V . On the Jordan Normal FormFact 2.8. We briefly outline several definitions and results from the theoryof generalized eigenvectors and the Jordan normal form of a matrix. LetA ∈Mk×k(C).62.2. Lemmas1. Definition. A matrix J is said to be in Jordan normal form if it is ofthe form:J =J1 . . .JN ,for square Ji =λi 1λi. . .. . . 1λi ,for some λi ∈ C. The Ji matrices are said to be Jordan blocks.2. Definition. Let B be a basis and B be the matrix whose columns arethe vectors of B. Then B is an ordered canonical basis B for A ifB−1AB is in Jordan normal form.3. Definition. The vector x is a generalized eigenvector of A, witheigenvalue λ and rank m, if (A− λI)mx = 0 and (A− λI)m−1x 6= 0.4. A basis is canonical if and only if it is composed completely of disjointJordan chains, that is, sets of the form {(A−λI)kx : k ∈ {0, ...,m−1}}where x is a generalized eigenvector of A with eigenvalue λ and rank m.5. There exists an ordered canonical basis of A. Furthermore, if A hasinteger entries and integer eigenvalues, then it is possible to choosethe canonical basis such that all basis vectors are in Qk.6. Each Jordan chain of A corresponds to a Jordan block in the Jordannormal form of A, and the number of vectors in a Jordan chain equalsthe size of the corresponding block.Lemma 2.9. Suppose every eigenvalue of A ∈ Mk×k(Z) is in Z, and letJ ∈ Mk×k(Z) be its Jordan normal form. Then S(A) is a finite union ofarithmetic progressions, if S(J) is.72.2. LemmasProof. Suppose S(J) is a finite union of arithmetic progressions, in otherwords, suppose S(V ′, α′, J,N ′) is a finite union of arithmetic progressionsfor any choice of variety V ′ ⊂ Gkm, α′ ∈ Gkm(K) and arithmetic progressionN ′. Let V ⊂ Gkm be a variety, α ∈ Gkm(K) a point and N an arithmeticprogression, we show S(V, α,A,N ) is a finite union of arithmetic progres-sions as well.By Fact 2.8 part 5 there exists invertible B ∈ Mk×k(Q) such that A =B−1JB. At the cost of scaling B and B−1, we may assume B ∈ Mk×k(Z).Because the inverse is obtained through gaussian elimination, B−1 is inMk×k(Q), and there exists non-zero c ∈ Z such that cB−1 ∈ Mk×k(Z). LetC = c · Ik and D = CB−1 ∈ Mk×k(Z). At the cost of replacing K witha finite extension, we can find γ ∈ Gkm(K), such that ϕC(γ) = α and letβ = ϕB(γ). Then we have:ϕnA(α) =ϕAn(α) = ϕB−1JnB(α) = ϕB−1JnB(ϕC(γ))=ϕCB−1JnB(γ) = ϕD(ϕJnB(γ)) = ϕDJn(β).Then ϕD(ϕJn(β)) ∈ V if and only if ϕJn(β) ∈ ϕ−1D (V ). Because ϕD isan endomorphism of Gkm we have that the last set is again a variety, sayW = ϕ−1D (V ). So we have shown ϕnA(α) ∈ V if and only if ϕnJ(β) ∈ W andthe lemma follows easily. Lemma 2.10. Let N ∈ N. Let A ∈Mk×k(Z), let JN be the Jordan normalform of AN and assume JN ∈ Mk×k(Z). Then S(A) is a finite union ofarithmetic progressions if S(JN ) is.Proof. Assume S(JN ) is a finite union of arithmetic progressions. LetV ⊂ Gkm be a variety, α ∈ Gkm(K) a point and N an arithmetic progression.We show show S(V, α,A,N ) is a finite union of arithmetic progressions aswell.We call the sequence of points (ϕA(α), ϕ2A(α), ϕ3A(α), ...) the orbit of ϕAat α. Note that the orbit of ϕA at α can be written as the finite union ofthe orbits of ϕAN at α,ϕA(α), ..., ϕN−1A (α). Thus S(V, α,A,N ) is a finiteunion of arithmetic progressions if S(V, ϕiA(α), AN ,N ) is a finite union ofarithmetic progressions for all i ∈ {0, ..., N − 1}. Since S(V, ϕiA(α), AN ,N )is a finite union of arithmetic progressions by Lemma 2.9, we are done. 82.2. LemmasLemma 2.11. Let N ∈ N and a ∈ Z. Let Ja and JaN be k×k single Jordanblock matrices with a and aN on the diagonal, respectively. That is:Ja =a 1 · · · 00 a. . . 0....... . ....0 0 0 a , JaN =aN 1 · · · 00 aN. . . 0....... . ....0 0 0 aN .Then S(Ja) is a finite union of arithmetic progressions if S(JaN ) is.Proof. We first prove the minimal polynomial of JNa is (x−aN )k. The matrixJNa is triangular with aN along the diagonal; thus this is its only eigenvalueand the minimal polynomial of JNa is p(x) = (x − aN )m for some m ∈ [k].Let q(x) = (x − a)k be the minimal polynomial of Ja. Because p(JNa ) = 0we must have that q(x) | p(xN ). Let ξ0, ..., ξN−1 be the N th roots of unity,so that p(xN ) =∏N−1j=0 (x− aξj)m. Then we have:(x− a)k ∣∣ N−1∏j=0(x− aξj)m,and it follows that m = k, as desired. Because the minimal polynomialof JNa is (x − aN )k, its Jordan form must contain a Jordan block of size k,corresponding to eigenvalue aN . But then the Jordan form of JNa is preciselyJaN , and the claim follows from Lemma 2.10. Simplifying CasesClaim 2.12. To prove S(A) is a finite union of arithmetic progressions,it is sufficient to show S(Z(f), α,A,N ) is a finite union of arithmetic pro-gressions, for any f ∈ K[x], α ∈ Gkm(K) and arithmetic progression N , byProposition 2.6.With the following three technical lemmas we develop tools to be usedby both Chapters 3 and 4. Our setting will be as follows: let s ∈ [k], letA ∈Mk×k(Z), and defineA˜s ∈Mk−1×k−1(Z) (2.2)to be the submatrix of A formed by removing the sth row and column.Similarly, given any α = (α1, ..., αk) ∈ Gkm, let α˜s be the vector with the sthentry removed. Using a circumflex to denote the missing element, we haveα˜s = (α1, ..., α̂s, ..., αk).92.2. LemmasLemma 2.13. Let K be a function field over Fp. Let A ∈Mk×k(Z), s ∈ [k],and suppose every element of the sth row is zero, except possibly As,s. LetA˜s be as in (2.2) and assume S(A˜s) is a finite union of arithmetic progres-sions. Suppose α = (α1, ..., αk) ∈ Gkm satisfies αs ∈ Fp. Then S(V, α,A,N )is a finite union of arithmetic progressions, for any variety V ⊂ Gkm andarithmetic progression N .Proof. Since s ∈ [k] is fixed, we drop the superscript s from our notation,writing A˜ := A˜s and α˜ := α˜s.Let h ∈ K[x1, ..., xk]. By Claim 2.12, it is sufficient to showS(Z(h), α,A,N ) is a finite union of arithmetic progressions.Let m be the order of αs in Fp×, and let (A)i,j denote the value of(A)i,j modulo m. Because the reduced matrix An ranges over a finite setand An is a linear function of An−1, we see the sequence (An : n ∈ N) ispreperiodic, in the sense of definition 2.1.By An’s preperiodicity, we can find arithmetic progressions {Ni}`i=1such that ∪`i=1Ni = N and An is a fixed matrix (modulo m) for all n ∈ Ni.Note that:ϕnA(α1, ..., αs, ..., αk) =(α(An)1,11 · · ·α(An)1,ss · · ·α(An)1,kk ,. . .α01 · · ·α(An)s,ss · · ·α0k,. . .α(An)k,1k · · ·α(An)k,ss · · ·α(An)k,kk).Given this, for each i ∈ [`] and j ∈ [k], choose any n ∈ Ni and use it todefineβi,j := α(An)j,ss .This is independent to the particular n ∈ Ni chosen. For each i we alsodefine:hi(x1, ..., x̂s, ...xk) := h(x1βi,1, ..., βi,s, ...xkβi,k).By our assumption that (A)s,i is zero for all i 6= s, we have (A˜)n = (˜An).102.2. LemmasIt follows easily that:hi(ϕnA˜(α˜)) = h(ϕnA(α)), (2.3)for all n ∈ Ni. The set S(Z(hi), α˜, A˜,Ni) is a finite union of arithmeticprogressions by assumption, and so by (2.3) S(Z(h), α,A,Ni) is a finiteunion of arithmetic progressions as well. The lemma follows. Let s ∈ [k]. In the following two lemmas we will express polynomials ofK[x1, ..., xk] with the following notation:h(x1, ..., xk) =d∑i=0xishi(x1, ..., x̂s, ..., xk), (2.4)where d is the degree of h in xs.Lemma 2.14. Let A ∈ Mk×kZ, s ∈ [k], and suppose every element of thesth column of A is zero, except possibly As,s. Suppose S(A˜s) is a finite unionof arithmetic progressions (with A˜s as in (2.2)). Let h be as in (2.4) andsuppose that d = 0. Then S(Z(h), α,A,N ) is a finite union of arithmeticprogressions as well, for any α ∈ Gkm(K) and arithmetic progression N .Proof. Since s ∈ [k] is fixed, we drop the superscript s from our notation,writing A˜ := A˜s and α˜ := α˜s.Let hi and d be as defined in (2.4). We have d = 0, so leth˜(x1, ..., x̂s, ..., xk) := h(x1, ..., xk).By our assumption that (A)i,s is zero for all i 6= s, we have (A˜)n = (˜An). Itfollows easily thath(ϕnA(α)) = h˜(ϕnA˜(α˜)). (2.5)By assumption we have that S(Z(h˜), α˜, A˜,N ) is a finite union of arithmeticprogressions, and so by (2.5) we get that S(Z(h), α,A,N ) is a finite unionof arithmetic progressions as well. Lemma 2.15. Let A ∈ Mk×k(Z), s ∈ [k], and suppose every element ofthe sth column of A is zero, except possibly As,s. Suppose S(A˜s) is a finiteunion of arithmetic progressions, and that S(Z(h), α,A,N ) is as well, forany α ∈ Gkm(K), arithmetic progression N and polynomial h (of the form(2.4)) satisfying hd(ϕnA˜s(α˜s)) 6= 0 for all n ∈ N . Then S(A) is a finite unionof arithmetic progressions as well.112.2. LemmasProof. Since s ∈ [k] is fixed, we drop the superscript s from our notation,writing A˜ := A˜s and α˜ := α˜s.Assume S(Z(h), α,A,N ) is a finite union of arithmetic progressionswhenever h’s leading term in xs, hd, is non-zero at ϕnA˜(α˜) for all n ∈ N .Claim 2.16. Consider S(Z(h), α,A,N ). If h’s leading term hd(ϕnA˜(α˜)) isidentically zero for all n ∈ N , then proving S(Z(h), α,A,N ) is a finite unionof arithmetic progressions reduces to proving the same for S(Z(h′), α,A,N ),for a polynomial h′ whose degree in xs is strictly less than d. To see this,defineh′(x1, ..., xk) =d−1∑i=0xishi(x1, ..., x̂s, ..., xk),and note h′(ϕnA(α)) = h(ϕnA(α)) for all n ∈ N .Claim 2.17. Proving S(Z(h), α,A,N ) is a finite union of arithmetic progres-sions reduces to proving the same for S(Z(h), α,A,N ′) where hd(ϕnA˜(α˜)) isidentically zero for all n ∈ N ′. To see this, note that by assumption,{n ∈ N : hd(ϕnA˜(α˜)) = 0}is a finite union of arithmetic progressions. Thus by Proposition 2.6, we canfind a collection of arithmetic progressions such that their union is N , andhd(ϕnA˜(α˜)) is either identically zero or non-zero on each. If it is non-zero weare done by the assumption. The claim follows.Finally, let g ∈ K[x1, ..., xk], α ∈ Gkm(K) and arithmetic progression Nbe arbitrary. That S(Z(g), α,A,N ) is a finite union of arithmetic progres-sions now follows easily by alternatively applying Claims 2.16 and 2.17, andby Lemma 2.14. Asymptotic BoundDefinition 2.18. Given functions f, g : N→ R, we say that f = o(g) if forevery > 0, there exists an N ∈ N such that n > N implies |f(n)| < |g(n)|.Lemma 2.19. Let K be a function field over Fp. Let N be an arithmeticprogression. Let g1, ..., gk be functions N → K×, and | · |w ∈ MK . Leth ∈ K[x1, ..., xk], and write h(x1, ..., xk) =∑di=0 xi1hi(x2, ..., xk) where d isthe degree of h in x1. Suppose the following conditions hold:(a) |g1|w is monotonically increasing to infinity, and log(|gi|u) =122.2. Lemmaso(log(|g1|w)) for all i > 1 and any | · |u ∈MK .(b) d > 0, and hd(g2(n), ..., gk(n)) is non-zero for all n ∈ N .(c) There exist only finitely many | · |u ∈ MK such that|hd(g2(n), ..., gk(n))|u > 1 for some n ∈ N .Given this, the setS = {n ∈ N : h(g1(n), ..., gk(n)) = 0}is finite.Proof. Let yn,i = gi(n). We begin with a remark on assumption (a). Forany | · |u ∈MK , `2, ..., `k ∈ N, α ∈ K× and b > 0 we have:log(|α|u) + `2 log(|g2(n)|u) + ...+ `k log(|gk(n)|u)< b log(|g1(n)|w)for sufficiently large n ∈ N . Thus:|α · y`2n,2 · · · y`kn,k|u < |yn,1|bw,and in particular for any i ∈ [d], there exists an Ni ∈ N such that:|hi(yn,2, ..., yn,k)|u < |yn,1|bw (2.6)for all n > Ni.Suppose for contradiction that there exists infinite set I ⊂ N such thath(yn,1, ..., yn,k) = 0 for all n ∈ I. Note first that at any point (y1, ..., yk) ∈Gkm we have:|h(y1, ..., yk)|w ≥∣∣∣yd1∣∣∣w·∣∣∣∣∣|hd(y2, ..., yk)|w −∣∣∣∣∣d−1∑i=1yi−d1 hi(y2, ..., yk)∣∣∣∣∣w∣∣∣∣∣ .Thus we must have:|hd(yn,2, ..., yn,k)|w =∣∣∣∣∣d−1∑i=1yi−dn,1 hi(yn,2, ..., yn,k)∣∣∣∣∣wfor all n ∈ I. By (2.6) there exists N such that for all n > N in I:|hd(yn,2, ..., yn,k)|w =∣∣∣∣∣d−1∑i=1yi−dn,1 hi(yn,2, ..., yn,k)∣∣∣∣∣w< |yn,1|−1/2w . (2.7)132.2. LemmasBy throwing out at most finitely many elements from I, we may assume(2.7) and |yn,1|1/2w > 1 hold for all n ∈ I. Let W be the set of places| · |u ∈MK where |hd(g2(n), ..., gk(n))|u > 1 for some n. Let s = |W|, whichis finite by assumption (c).By (b) hd(yn,2, ..., yn,k) is non-zero and so the product formula, (2.1),holds: ∏u∈MK|hd(yn,2, ..., yn,k)|u = 1. (2.8)From (2.7) and (2.8), we know s > 0 and moreover that for each n ∈ I theremust exist place | · |wn ∈ W such that:1 < |yn,1|12sw < |hd(yn,2, ..., yn,k)|wn .But W is finite, so a single place, | · |w′ , must appear infinitely many times.We can find an infinite subset I ′ ⊂ I such that|yn,1|12sw < |hd(yn,2, ..., yn,k)|w′ ,for all n ∈ I ′. This contradicts (2.6), and so the original set I cannotexist. Remark 2.20. In Lemma 2.19, we singled out x1 and g1, but by relabelling,the results apply for xs and gs, for any s ∈ [k].14Chapter 3Single Jordan BlockWe prove Theorem 1.3. In this case X = Gkm, ϕ is a group endomorphismof X, and the matrix A ∈ Mk×k(Z), corresponding to ϕ, is similar to asingle Jordan block.Proposition 3.1. Let K be a field of characteristic p > 0. Let a ∈ Z andlet Ja ∈Mk×k(Z) be a single Jordan block, that is:Ja =a 1 0 · · · 00 a 1 · · · 0.......... . ....0 0 0 a 10 0 0 0 a .Then S(Ja) is a finite union of arithmetic progressions.3.1 Proof of Proposition 3.1We show S = S(V, α, Ja,N ) is a finite union of arithmetic progressions,where V ⊂ Gkm is a variety, α ∈ Gkm(K) a point and N an arithmeticprogression. As usual we write α as α = (α1, ..., αk). Since V, α, and ϕJaare defined over a finitely generated subfield of K, we may assume, withoutloss of generality, that K is a function field over Fp.Claim 3.2. It is sufficient to prove the proposition in the case where a > 0.Proof. If a = 0, then (Ja)n = 0 for all n ≥ k, so the image ϕnJa(α) = (1, ..., 1)for all n ≥ k. It follows easily that S(V, α, Ja,N ) = {n ∈ N : ϕnJa(α) ∈ V }is a finite union of arithmetic progressions.If a < 0, then by Lemma 2.11 it is sufficient to show S(Ja2) is a finiteunion of arithmetic progressions, where Ja2 is the matrix consisting of asingle Jordan block with a2 > 0 on the diagonal. 153.1. Proof of Proposition 3.1Remark 3.3. For n ≥ k we have:Jna =an nan−1(n2)an−2 · · · ( nk−1)an−(k−1)0 an nan−1 · · · ( nk−2)an−(k−2).......... . ....0 0 0 an nan−10 0 0 0 an ,and so:ϕnJa(α1, ..., αk) = (αan1 αnan−12 · · ·α( nk−1)an−(k−1)k , αan2 · · ·α( nk−2)an−(k−2)k , ..., αank ).We now proceed to prove Proposition 3.1 via induction on the dimensionk. By Claim 2.12, we may assume V = Z(f) for f ∈ K[x1, ..., xk], and byClaim 3.2 that a > 0.Base CaseLet k = 1. The set Z(f) is either empty, all of Gkm, or a finite set of points,and in the first two cases the claim trivially holds – so consider only thelast. Let β ∈ Z(f) be arbitrary, it is sufficient to show {n ∈ N : ϕnJa(α) = β}is a finite union of arithmetic progressions. If ϕnJa(α) = β for only onen ∈ N then we are done, otherwise α is a preperiodic point of ϕJa and weare done by lemma 2.7.Thus the set of n ∈ N such that ϕnJa(α) ∈ Z(f) is a finite union ofarithmetic progressions, as desired.Inductive StepNow assume the inductive hypothesis for dimension k − 1, that is, as-sume S(V ′, α′, J ′a,N ′) is a finite union of arithmetic progressions forany J ′a ∈ Mk−1×k−1(Z), α′ ∈ Gkm(K), variety V ′ ⊂ Gkm and arithmeticprogression N ′.If |αk|v ≤ 1 for all | · |v ∈ MK , then αk ∈ Fp by Fact 2.5, and soby Lemma 2.13 (with s = k) and the induction hypothesis we are done.Instead, assume we can find | · |v ∈MK such that |αk|v > 1.Let the maximum power of x1 found in f(x1, ..., xk) be d. Then for somepolynomials fi ∈ K[x2, ..., xk] we have:f(x1, ..., xk) =d∑i=0xi1fi(x2, ...xk).163.2. Theorem 1.3We now invoke Lemma 2.19. Recall we have:ϕnJa(α1, ..., αk) = (αan1 αnan−12 · · ·α( nk−1)an−(k−1)k , αan2 · · ·α( nk−2)an−(k−2)k , ..., αank ).Given this, define the gi from Lemma 2.19 so that ϕnJa(α) =(g1(n), ..., gk(n)). In particular we have (for example) that:g1(n) = αan1 αnan−12 · · ·α( nk−1)an−(k−1)k .Let | · |v be the place | · |w of the lemma. Then because |αk|v > 1 we have(nk − i)an−(k−i) log(|αj |u) = o((nk − 1)an−(k−1) log(|αk|v)),for any place | · |u ∈MK , all j ∈ [k] and 1 ∈ [k] with i > 1. Assumption (a)of the lemma follows easily. We let f be the polynomial h of the lemma. ByLemma 2.14 we may assume d > 0, while by Lemma 2.15 (with s = 1), wemay assume fd(ϕnAk−1(α2, ..., αk)) is non-zero for all n ∈ N . Together thesegive us assumption (b). Finally, assumption (c) follows by noting there areonly finitely many places where either a coefficient of hd, or an αi has normgreater than 1.Thus Lemma 2.19 completes the proof. 3.2 Theorem 1.3We now prove Theorem 1.3. For convenience, the theorem is reproducedbelow.Theorem 1.3. Let K be a field of characteristic p > 0. Let the groupendomorphism ϕA : Gkm → Gkm be defined byϕ(x1, ..., xk) = (xa1,11 xa1,22 · · ·xa1,kk , ..., xak,11 xak,22 · · ·xak,kk )for some ai,j ∈ Z, and assume that the matrix A = (ai,j) is similar to asingle Jordan block. Let V ⊂ Gkm be a variety, and α ∈ Gkm(K) a point.Then the setS = {n ∈ N : ϕnA(α) ∈ V }is a finite union of arithmetic progressions.173.2. Theorem 1.3Proof. Let A ∈Mk×k(Z), and suppose A has Jordan form:Ja =a 1 0 · · · 00 a 1 · · · 0.......... . ....0 0 0 a 10 0 0 0 a ,for some a ∈ C. By Proposition 3.1 and Lemma 2.9, in order to proveTheorem 1.3, it is sufficient to show that a must be in Z. Note that Tr(A) =Tr(Ja) = ka and Det(A) = Det(Ja) = ak. Thus both ka and ak are in Z.We see that a is both a rational number and an algebraic integer, and so wemust have a ∈ Z. 18Chapter 4Dimension 3The goal of this section is to prove Theorem 1.4. We will need the followingresult, which appears in [2, Proposition 13.3.0.2].Proposition 4.1. Let K be a field of characteristic p > 0, α ∈ Gkm(K),V ⊂ Gkm be a variety and N be an arithmetic progression. Let A ∈Mk×k(Z)be diagonalizable. ThenS(V, α,A,N ) = {n ∈ N : ϕnA(α) ∈ V }is a finite union of arithmetic progressions. Our proof of Theorem 1.4 will rely on the following lemma.Lemma 4.2. Let K be a function field over Fp. Let J ∈ M3×3(Z) be amatrix in Jordan form, let V ⊂ G3m be a variety, α ∈ G3m(K) a point andN an arithmetic progression. Then the setS = {n ∈ N : ϕnJ(α) ∈ V }is a finite union of arithmetic progressions.Proof. We show S(V, α, J,N ) is a finite union of arithmetic progressions,where α is written α = (α1, α2, α3). By Proposition 3.1 and Proposition4.1, it is sufficient to consider the case where matrix J is composed of twoJordan blocks. Without loss of generality we write:J =a 1 00 a 00 0 b ,for a, b ∈ Z. If a = 0 thenJ2 =0 0 00 0 00 0 b2 ,19Chapter 4. Dimension 3Thus J2 is already diagonal and we are done by Lemma 2.10 and Theorem4.1. Thus we may assume a 6= 0. In this case we claim J2 has Jordan form:J ′ =a2 1 00 a2 00 0 b2 .This follows as J2 has eigenvalues b2 and a2 with algebraic multiplicity 1and 2 respectively, and is not diagonalizableBy the previous paragraph, and Lemma 2.10, it is enough to show S(J ′)is a finite union of arithmetic progressions. Thus we may now assume a > 0and b ≥ 0.By Claim 2.12 it is sufficient to consider V = Z(f) for polynomialf = f(x1, x2, x3). If b = 0, then let g(x1, x2) = f(x1, x2, 1) and notef(ϕnJ(α)) = g(ϕnJa((α1, α2))), where Ja = ( a 10 a ), so we are done by Proposi-tion 3.1.By the above, we may assume a > 0 and b > 0. We split into two casesdepending on the relative size of a and b.Case 1: 0 < a < b. Note ϕnJ(α1, α2, α3) = (αan1 αnan−12 , αan2 , αbn3 ). Inthis case, the factor of |αbn3 |v dominates the value of |f(ϕnA(α))|v. Howeverwe must first eliminate the cases where α3 has small norm, or x3 does notappear in f .Let Ja = ( a 10 a ); S(Ja) is a finite union of arithmetic progressions byProposition 3.1. If |α3|v ≤ 1 for all | · |v ∈MK , then by Fact 2.5 and Lemma2.13 (with s = 3) we are done. Therefore we now assume |α3|v > 1 for some| · |v ∈MK .Let f(x1, x2, x3) =∑di=0 xi3fi(x1, x2). By Lemma 2.14 (with s = 3),we may assume d > 0. By Lemma 2.15 (with s = 3), we may assumefd(ϕnJa(α1, α2)) is non-zero for all n ∈ N .Finally, we apply Lemma 2.19 with s = 3 (using Remark 2.20). We letg1(n) = αan1 αnan−12 , g2(n) = αan2 and g3(n) = αbn3 . Condition (a) of thelemma follows from the fact that(ni)an−i log(|αj |u) = o (bn log(|α3|v)) ,20Chapter 4. Dimension 3for any place | · |u ∈ MK , i ∈ {0, 1} and j ∈ {1, 2} (here we use that0 < a < b implies in particular that 1 < b). Letting f be polynomial h fromthe lemma, (b) follows from the previous paragraph, and (c) can be seen bynoting there are only finitely many places where either a coefficient of fd oran αi has norm greater than 1. This concludes Case 1 of the proof.Case 2: 0 < b ≤ a. In this case the factor of |αnan−12 |v dominates.Let D =(a 00 b); S(D) is a finite union of arithmetic progressions byProposition 4.1. Thus if |α2|v ≤ 1 for all | · |v ∈ MK , then by Fact 2.5and Lemma 2.13 (with s = 2) we are done. So suppose |α2|v > 1 for some| · |v ∈MK .Let f(x1, x2, x3) =∑di=0 xi1fi(x2, x3). By lemma 2.14 (with s = 1)we may assume d > 0, and by Lemma 2.15 (with s = 1) we may assumefd(ϕnD((α1, α3))) is non-zero for all n ∈ N .We apply Lemma 2.19 with s = 1. We let the gi and polynomial h be asin Case 1. It is easy to check, by analogous arguments to those of Case 1,that conditions (a), (b) and(c) still hold. Thus by Lemma 2.19 S(V, α,A,N )is a finite union of arithmetic progressions. Theorem 1.4 (reproduced below for the reader’s convenience) is now aneasy corollary of Lemma 4.2.Theorem 1.4. Let K be a field of characteristic p > 0. Let ϕ : G3m → G3mbe a group endomorphism, let V ⊂ G3m be a variety, and α ∈ G3m(K) apoint. Then the setS = {n ∈ N : ϕn(α) ∈ V }is a finite union of arithmetic progressions.Proof. Let V, α and ϕ be as in the lemma statement. Let A ∈ M3×3(Z)be the matrix associated with ϕ, so that ϕ = ϕA. Since V, α, and ϕA aredefined over a finitely generated subfield of K, we may assume, withoutloss of generality, that K is a function field over Fp.We prove S(V, α,A,N ) is a finite union of arithmetic progressions.Recall that Q is a perfect field, that is, any polynomial irreducible over Q21Chapter 4. Dimension 3is separable. Let pA ∈ Z[x] be the degree three characteristic polynomialof A. If pA were irreducible over Q then it would have three distinct rootsand we would be done by Proposition 4.1.Suppose instead pA were reducible over Q, then we can write pA(x) =q(x)(x− r), for some r ∈ Q and q ∈ Q[x]. If q were irreducible over Q thenit would have two distinct non-rational roots, and so pA would have threedistinct roots. Thus we may assume q is reducible over Q. Then both of q’sroots must be in Q. Since pA is monic, all three of its roots – that we havejust seen are in Q – are in fact in Z. Then by Lemma 2.9 and Lemma 4.2we have the claim. 22Chapter 5ExamplesConjecture 1.2 has been proven by [3] when X = Gkm, ϕ is any regularself-map, and the variety V is a curve. Combining this with Theorem 1.4,Conjecture 1.2 has now been shown for the case where X = Gkm(K) fork ∈ {1, 2, 3} and ϕ is a group endomorphism. These cases are in a senseeasier because, as we saw in the three dimensional case, the set S containsnone of the more complicated sets of equation (1.2). As we saw in Example1.1, and will see in the following two examples, this property no longerholds in dimension ≥ 4.In both of the following examples, let the ambient field K be Fp(t).Example 5.1. Let p > 2. Let α = (1, t, 1, 1 + t, 1, 1 − t) ∈ G6m(K), anddefineA =1 1 0 0 0 00 1 0 0 0 00 0 1 1 0 00 0 0 1 0 00 0 0 0 1 10 0 0 0 0 1ThenϕnA(α) = (tn, t, (1 + t)n, (1 + t), (1− t)n, (1− t)).Let f(x1, x2, x3, x4, x5, x6) = x3 + x5 − 2x1 − 2. We claim the set S = {n ∈N : ϕnA(α) ∈ Z(f)}, equals{pn1 + pn2 : n1, n2 ∈ N0}. (5.1)It is easy to check (5.1) is contained in S, so we prove the reverse contain-ment. Suppose f(ϕnA(α)) = 0 and let n = `pa for ` coprime to p. Then:f(ϕnA(α)) = (1 + tpa)` + (1− tpa)` − 2tpa` − 2=`−1∑i=1(`i)((tpa)i + (−tpa)i)− tpa` + (−tpa)`.23Chapter 5. ExamplesIt follows immediately that ` must be even, and so we have:f(ϕnA(α)) =`−1∑i=1(`i)((tpa)i + (−tpa)i) . (5.2)If ` = 2 we are done, so suppose ` > 2. Then from the i = 2 term of (5.2)we see that ` ≡ 1 (mod p). Since ` − 1 6= 0 we can write ` − 1 = pbm forsome m coprime to p. Then ` ≡ 1 (mod p) implies b ∈ N (?). Given this,we have n = pa(pbm+ 1) and:f(ϕnA(α)) = (1 + tpa)pbm+1 + (1− tpa)pbm+1 − 2tpa(pbm+1) − 2=m∑i=0(mi)((tpa+b)i + (−tpa+b)i)− 2+ tpam∑i=0(mi)((tpa+b)i − (−tpa+b)i)− 2tpa(pbm+1).If m = 1 we are done, and m = 0 is impossible, so suppose m > 1. Thenit is easy to see the i = 1 term of the second sum, 2mtpatpa+b, must cancelwith some term of the first sum. We have:2mtpa+b+pa = −2(mi)tipa+b,and so pa+b + pa = ipa+b. From this we can deduce b = 0, contradicting (?).Recall Conjecture 1.2 allowed S to contain finitely many arithmetic pro-gressions, as well as finitely many sets of the formm∑j=1cjpkjnj : nj ∈ N0 for each j ∈ [m] , (5.3)for some cj ∈ Q, and kj ∈ N. Since {pn1 + pn2 : n1, n2 ∈ N0} cannot bewritten as a simple geometric series (or an arithmetic progression), Example5.1 shows that the sum to m in (5.3) is necessary. With our next examplewe show that the cj are indeed sometimes necessarily in Q.Example 5.2. Let p > 2. Let α = (1, tp−1, 1, (1 − t)p−1) ∈ G4m(K), anddefineA =1 1 0 00 1 0 00 0 1 10 0 0 1 .24Chapter 5. ExamplesThen ϕnA(α) = (tn(p−1), tp−1, (1− t)n(p−1), (1− t)p−1). Let f(x1, x2, x3, x4) =tx1 + (1− t)x3 − 1. We claim S = {n ∈ N : ϕnA(α) ∈ Z(f)} is equal to{pjp− 1 −1p− 1 : j ∈ N0}. (5.4)It is easy to check (5.4) is contained in S, so we prove the reverse contain-ment. As we saw in Example 1.1, m ∈ N satisfiestm + (1− t)m − 1 = 0if and only if m = pj for some j ∈ N0. So suppose f(ϕnA(α)) = 0 and then,defining m = (p− 1)n+ 1, we have:f(ϕnA(α)) = (tn(p−1))t+ ((1− t)n(p−1))(1− t)− 1= tm + (1− t)m − 1.Thus n = pj−1p−1 for some j ∈ N0.25Chapter 6ConclusionThe approach used in Chapters 3 and 4 to show Conjecture 1.2 for thesimilar to a Jordan block case and dimension 3 case will always yield a set Sthat is a finite union of arithmetic progressions. However, as the exampleshave shown, more complicated sets (as in equation (1.2)) already begin toappear in dimension 4. Thus new ideas will be required as we press on tohigher dimensions.26Bibliography[1] Jason P. Bell, Dragos Ghioca, and Thomas J. Tucker. The DynamicalMordell-Lang problem for e´tale maps. American Journal of Mathemat-ics, 132(6):1655–1675, 2010.[2] Jason P. Bell, Dragos Ghioca, and Thomas J. Tucker. The Dynami-cal Mordell–Lang Conjecture, volume 210 of Mathematical Surveys andMonographs. American Mathematical Soc., 2016.[3] Dragos Ghioca. The Dynamical Mordell-Lang Conjecture in positivecharacteristic. arXiv:1610.00367, 2016.[4] Dragos Ghioca and Thomas J. Tucker. Periodic points, linearizing maps,and the Dynamical Mordell–Lang problem. Journal of Number Theory,129(6):1392–1403, 2009.[5] Dragos Ghioca, Thomas J. Tucker, and Michael E. Zieve. Intersectionsof polynomial orbits, and a Dynamical Mordell–Lang Conjecture. In-ventiones mathematicae, 171(2):463–483, 2008.[6] Dragos Ghioca, Thomas J. Tucker, Michael E. Zieve, et al. Lin-ear relations between polynomial orbits. Duke Mathematical Journal,161(7):1379–1410, 2012.[7] Serge Lang. Fundamentals of Diophantine Geometry. Springer Science& Business Media, 2013.[8] Rahim Moosa and Thomas Scanlon. F-structures and integral points onsemiabelian varieties over finite fields. American Journal of Mathematics,pages 473–522, 2004.[9] Junyi Xie. Dynamical Mordell–Lang Conjecture for birational polyno-mial morphisms on A2. Mathematische Annalen, 360(1-2):457–480, 2014.27
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Two special cases of the Dynamical Mordell-Lang Conjecture in positive characteristic Nelson, Kristina 2017
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Title | Two special cases of the Dynamical Mordell-Lang Conjecture in positive characteristic |
Creator |
Nelson, Kristina |
Publisher | University of British Columbia |
Date Issued | 2017 |
Description | We prove the positive characteristic version of the Dynamical Mordell-Lang Conjecture in two novel cases. Let p be a prime and K a field of characteristic p>0. Let k ∈ ℕ, and let G denote the multiplicative group of K, of dimension k. Let α be an element of G, and V a variety contained in G. Let φ: G →G be a group endomorphism defined over K. We know φ(x₁,x₂,...,xk)=(x₁^a1,1 x₂^a1,2 ··· xk^a1k , ... , x₁^ak1 x₂^ak2 ··· xk^akk), for some integer exponents aij. In the case where the matrix of exponents, ( aij ) is similar to a single Jordan block, we show that the set S = { n ∈ ℕ double : φ^n(α) ∈ V } is a finite union of arithmetic progressions. When the dimension k = 3, we show S is a finite union of arithmetic progressions for any group endomorphism φ. |
Genre |
Thesis/Dissertation |
Type |
Text |
Language | eng |
Date Available | 2017-04-12 |
Provider | Vancouver : University of British Columbia Library |
Rights | Attribution-NonCommercial-NoDerivatives 4.0 International |
DOI | 10.14288/1.0343589 |
URI | http://hdl.handle.net/2429/61213 |
Degree |
Master of Science - MSc |
Program |
Mathematics |
Affiliation |
Science, Faculty of Mathematics, Department of |
Degree Grantor | University of British Columbia |
GraduationDate | 2017-05 |
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UBCV |
Scholarly Level | Graduate |
Rights URI | http://creativecommons.org/licenses/by-nc-nd/4.0/ |
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