Elliptic Curves Spring 2017 Problem Set #10

Transcription

1 Ellptc Curves Sprng 2017 Problem Set #10 Descrpton These problems are related to the materal covered n Lectures Instructons: Solve one of Problems 1 3, then fll out the survey problem 4. For ths problem set you are not allowed to consult outsde sources other than your favorte (ntroductory) algebra textbook. You may refer to lecture notes and references n the syllabus, and you may dscuss the problems wth fellow students. The frst to spot each non-trval typo/error wll receve 1-3 ponts of extra credt. Problem 1. Congruence subgroups (100 ponts) Let Γ(1) := SL 2 (Z) denote the modular group and H := {τ : m τ > 0} Q { } the extended upper half plane. The dagram below depcts a fundamental regon F for H / Γ(1) n H, along wth nne of ts translates. Each translate γf s labeled by γ, where γ s expressed n terms of the generators S = and T = for Γ(1). The colored labels ρ,, wthn the regon labeled by γ ndcate the ponts γρ, γ, and γ, respectvely (where ρ = e 2π/3 ). Note that the red, black, and blue colored along the top of the dagram are all the same pont, but there are three dstnct translates of on the real axs (at 1, 0, 1), each of whch les n two translates of F (ths llustrates a key pont: translates of F may overlap at ponts whose stablzers act non-trvally, namely, the ponts, ρ, ). The regon F ncludes the arc from to ρ along the unt crcle and the lne from ρ to along the magnary axs, but no other ponts on ts boundary other than ; the translates of these have been colored and orented accordngly. T 1 I T ρ ρ ρ ρ ρ ρ (ST ) 2 ρ ρ S T S ST S ST (T S) 2 T ST ρ ρ -3/2-1 -1/2 0 1/2 1 3/2 1

2 We recall the followng subgroups of SL 2 (Z), defned for each nteger N 1: and correspondng modular curves Γ(N) := {γ SL 2 (Z) : γ ( 1 0 ) mod N}, 0 1 Γ 1 (N) := {γ SL 2 (Z) : γ ( 1 ) mod N}, 0 1 Γ 0 (N) := {γ SL 2 (Z) : γ ( 0 ) mod N}, X(N) := H / Γ(N), X 1 (N) := H / Γ 1 (N), X 0 (N) := H / Γ 0 (N). For any congruence subgroup Γ we call the Γ(1)-translates of n H or H / Γ cusps. (a) Determne the ndex of Γ(2) n Γ(1), and the number of Γ(2) cusp orbts. Then gve a connected fundamental regon for H / Γ(2) by lstng a subset of the translates of F n the dagram above and dentfy the cusps that le n your regon. Compute the genus of X(2) by trangulatng your fundamental regon and applyng Euler s formula V E + F = 2 2g. Be careful to count vertces and edges correctly ntally specfy vertces and edges as H -ponts n the dagram (e.g. ST ρ), then determne whch vertces and edges are Γ(2)-equvalent (note that edges whose end ponts are equvalent need not be equvalent). Do the same for Γ 0 (2) and X 0 (2). (b) For each of the followng congruence subgroups, determne ts ndex n Γ(1), the number of cusp orbts, and a set of cusp representatves: Γ 0 (3), Γ 1 (3), Γ(3). (c) Prove that for each nteger N 1 we have an exact sequence 1 Γ(N) SL 2 (Z) SL 2 (Z/NZ) 1. Show that n general one cannot replace SL 2 wth GL 2 n the sequence above (so your proof for SL 2 needs to use more than the fact that Z Z/NZ s surjectve). (d) Derve formulas for the ndex [ Γ(1) : Γ] for Γ = Γ(N), Γ 1 (N), Γ 0 (N) and any N 1. Use the Euler functon φ(n) := #(Z/NZ) where approprate. For any congruence subgroup Γ, let ν 2 ( Γ) and ν 3 ( Γ) count the number of SL 2 (Z) translates of and ρ, respectvely, that le n a fundamental regon of H for Γ and are fxed by some γ Γ other than ±I. Let ν ( Γ) be the number of cusp-orbts for Γ. (e) For Γ = Γ(p), Γ 1 (p), Γ 0 (p) derve formulas for ν 2 ( Γ), ν 3 ( Γ), ν ( Γ), where p s a prme (hnt: show that for any δ SL 2 (Z), f γ SL 2 (Z) {±I} stablzes δ then t has trace 0, and f t stablzes δρ then t has trace ±1). Let Γ(N), Γ 1 (N), Γ 0 (N) denote the mages of the groups Γ(N), Γ 1 (N), Γ 0 (N) n PSL 2 (Z) := SL 2 (Z)/{±I} respectvely, and for any congruence subgroup Γ wth mage Γ n PSL 2 (Z) defne ( [ Γ(1) : Γ] f I Γ µ( Γ) := [ Γ(1) : Γ] = [ Γ(1) : Γ]/2 f I 6 Γ Usng the Remann-Hurwtz genus formula one can prove that for any congruence subgroup Γ the genus of the modular curve X Γ := H / Γ s gven by the formula µ( Γ) ν 2 ( Γ) ν 3 ( Γ) ν ( Γ) g(x Γ ) = For convenence we may wrte g( Γ) for g(x Γ ). 2

3 (f) Use your answers to (d) and (e) to gve asymptotc approxmatons for g( Γ) for Γ = Γ(p), Γ 1 (p), Γ 0 (p) and ncreasng prmes p that have an exact leadng term (so of the form f(p) + O(g(p)) for some functons f and g wth g = o(f)). Conclude that the set of prmes p for whch g( Γ) takes any fxed value s fnte. Modular curves of genus 0 and 1 are of partcular nterest because we can use these curves to obtan nfnte famles of ellptc curves over Q (or a number feld) that have partcular propertes, for example, a torson pont of order p. By Faltngs Theorem, over a number feld a curve of genus g 2 has only a fnte number of ratonal ponts. (g) For Γ = Γ(p), Γ 1 (p), Γ 0 (p) determne the prmes p for whch g( Γ) = 0, and the prmes p for whch g( Γ) = 1. You may use Sage to check your answers (and gan ntuton), but your proofs must stand on ther own. To create the congruence subgroups Γ(N), Γ 1 (N), Γ 0 (N) n Sage use Gamma(N), Gamma1(N), Gamma0(N), respectvely. The returned objects support ndex(), nu2(), nu3(), cusps(), and genus() methods that you may fnd useful. Problem 2. Non-congruence subgroups of fnte ndex (100 ponts) Recall that a congruence subgroup s a subgroup of Γ(1) = SL 2 (Z) that contans Γ(N) for some N 1. Every congruence subgroup s a fnte ndex subgroup of SL 2 (Z). In ths problem you wll prove that the converse does not hold; there exst fnte ndex subgroups of SL 2 (Z) that are not congruence subgroups. Let PSL (Z) := SL 2 (Z)/{±I}, let α be the mage of S = 1 0 n PSL 2 (Z), and let β be the mage of ST = 0 1 n PSL 2 (Z). 1 1 (a) Let Z 2,3 be the fntely presented group wth generators x, y satsfyng the relatons 2 x = y 3 = 1 (and no others). Prove that the map Z 2,3 PSL 2 (Z) defned by x 7 α and y 7 β s an somorphsm. You may fnd the dagram from Problem 1 helpful. Part (a) mples that for any fnte group H = ha, b wth a = 2 and b = 3 we have a surjectve group homomorphsm SL 2 (Z) PSL 2 (Z) H, where the frst map s quotent map SL 2 (Z) PSL 2 (Z) and the second s the composton of the somorphsm PSL 2 (Z) Z 2,3 and the surjectve homomorphsm Z 2,3 H defned by x 7 a, y 7 b. The kernel Γ H of such a homomorphsm s a fnte ndex subgroup of SL 2 (Z). Our strategy s to show that for many fnte groups H = ha, b ths kernel cannot contan Γ(N) for any nteger N and s therefore not a congruence subgroup. To smplfy matters, we wll focus on cases where H s a smple group, meanng that H s a non-trval group that contans no normal subgroups other than the trval group and tself. Every non-trval fnte group G has a composton seres 1 = G 0 / G 1 / / G k 1 / G k = G n whch each G s a normal subgroup of G +1 and each quotent G +1 /G s smple. The quotents G +1 /G are called the smple factors of G (analogous to prme factors of an nteger). Ths composton seres s not unque, but the Joran-Hölder theorem states that the smple factors G +1 /G that appear n any composton seres for G are unque up to somorphsm (and occur wth the same multplcty). 3

4 (b) Prove that f a fnte smple group S s a quotent of G (meanng S = G/K for some K / G) then S s a smple factor of G but that the converse does not hold n general. (c) Prove that f a fnte group G s the drect product of non-trval groups H 1,..., H n then the factors of G are precsely the factors of the H (counted wth multplcty). Conclude that f S s a smple quotent of G then t s a quotent of one of the H. (d) Prove part (c) of Problem 1. e 1 e r (e) Let N = p 1 p r > 1 by wth p 1,..., p r dstnct prmes. Prove that every smple quotent of SL 2 (Z/NZ) s a smple quotent of SL 2 (Z/p e Z) for some. (f) Usng the fact that PSL 2 (Z/pZ) s a non-abelan smple group for prmes p 5, show that the smple factors of SL 2 (Z/p e Z) are: a cyclc group of order 2, PSL 2 (Z/pZ), and 3e 3 cyclc groups of order p, and that n partcular, PSL 2 (Z/pZ) s the unque non-abelan smple factor of SL 2 (Z/p e Z), for all prmes p 5. (g) Usng the fact that the alternatng group A n s a non-abelan smple group for all n 5, prove that A n s not a quotent of SL 2 (Z/NZ) for any N and any n > 5. (h) Usng Sage, fnd elements a of order 2 and b of order 3 that generate A 9 and lst them n cycle notaton. You don t need to wrte down a proof that they generate A 9 but you should verfy ths n Sage. To create A 9 use A9=AlternatngGroup(9), and to check whether a and b generate A 9 use A9.subgroup([a,b]) == A9. In fact, A n s generated by an element of order 2 and an element of order 3 for all n 9 (see [1]), but you are not asked to prove ths. It follows from the dscusson after (a) that there s a surjectve homomorphsm SL 2 (Z) A 9 that sends ±α to a and ±β to b. The kernel Γ of ths homomorphsm s a fnte ndex subgroup of SL 2 (Z). () Prove that Γ s not a congruence subgroup. We now want to construct a short lst of generators for Γ. The frst step s to convert the representaton of A 9 wth generators a and b of orders 2 and 3 nto a fntely presented group that s a fnte quotent of Z 2,3 specfed by relatons. To do ths use the Sage command: H=A9.subgroup([a,b]).as_fntely_presented_group().smplfed() Ths may take a few seconds. The second step s to plug S and ST nto all the relatons n the fnte presentaton of H you created above. Important: Sage may swap the roles of a and b when t constructs the fnte presentaton check the relatons to see f ths happened (f you see a3 and b 2 n the lst of relatons rather than a2 and b 3 then you know they were swapped). Assumng a 2 and b 3 are the frst two relatons, you can use G=SL(2,Integers()); S=G([0,-1,1,0]); T=G([1,1,0,1]) for n range(2,len(h.relatons())): prnt H.relatons()[].subs(a=S,b=S*T) to get a lst of matrces n SL 2 (Z) that, together wth S and T generate Γ. Note that the length of the lst you get wll depend on your choce of a and b, but shouldn t be more than 10 or 20 matrces (n fact one can do t wth 4). (j) Record the number of matrces n your lst above (not ncludng S and T ), and a smallest and largest matrx n your lst accordng to the L -norm (maxmum of absolute values of matrx entres). 4

5 Problem 3. Polycyclc presentatons (100 ponts) Let α~ = (α 1,..., α k ) be a sequence of generators for a fnte abelan group G, and let G = hα 1,..., α be the subgroup generated by α 1,..., α. The subnormal seres 1 = G 0 / G 1 / / G k 1 / G k = G, s a polycyclc seres: each G 1 s a normal subgroup of G and each of the quotents G /G 1 = hα G 1 s a cyclc group (whch we don t requre to have prme order, but one can always further decompose the seres so that they are). Every fnte solvable group admts a polycyclc seres, but we restrct ourselves here to abelan groups (wrtten multplcatvely). When G s the nternal drect product of the cyclc groups hα, we have G /G 1 = hα and call α~ a bass for G, but ths s a specal case. For abelan groups, G /G 1 s somorphc to a subgroup of hα, but t may be a proper subgroup, even when G s cyclc. The sequence r(α~) = (r 1,..., r k ) of relatve orders for α~ s defned by r = G : G 1, and satsfes r = mn{r : αr G 1 }. We necessarly have r α, but equalty typcally does Q not hold (α~ s a bass precsely when r = α for all ). In any case, we always have r = G, thus computng the r determnes the order of G. (a) Let α~ = (α 1,..., α k ) be a sequence of generators for a fnte abelan group G, wth relatve orders r(α~) = (r 1,..., r k ). Prove that every β G can be unquely represented n the form α x k β = ~x α~ = α x 1 1 k, where each x Z satsfes 0 x < r. Show that f β = α r, then x j = 0 for j. By analogy wth the case r = 1, we call ~x the dscrete logarthm of β wth respect to α~ (but note that the dscrete logarthm of the dentty element s now the zero vector). The vector ~x can be convenently encoded as an nteger x n the nterval [0, G 1] va X Y x = x N, N = r j, 1 k 1 j< and we may smply wrte x = log α β to ndcate that x s the nteger encodng the vector ~ ~x = log α β. Note that x = bx/n c mod r, so t s easy to recover ~x from ts encodng x. ~ (b) Desgn a generc group algorthm that, gven a sequence of generators α~ = (α 1,..., α k ) for a fnte abelan group G, constructs a table T wth entres T [0],..., T [ G 1] wth the property that f T [n] = β, then n = log α ~ β. Your algorthm should also output the relatve orders r, and the ntegers s for whch T [s ] = α r. Ths allows us to compute a polycyclc presentaton for G, whch conssts of the sequence α~, the relatve orders r(α~) = (r 1,..., r k ), and the vector of ntegers s(α~) = (s 1,..., s k ). Wth ths presentaton n hand, we can effectvely smulate any computaton n G wthout actually performng any group operatons (.e. calls to the black box). Ths can be very useful when the group operaton s expensve. 5

6 (c) Let α~, r(α~), and s(α~) be a polycyclc presentaton for a fnte abelan group G. Gven ntegers x = log α ~ β and y = log α ~ γ, explan how to compute the nteger z = log α ~ βγ usng r(α~) and s(α~), wthout performng any group operatons. Also explan how to compute the nteger w = log α ~ β 1. As a sde beneft, the algorthm you desgned n part (b) gves a more effcent way to enumerate the class group cl(d) than we pused n Problem Set 9, snce the class number h(d) s asymptotcally on the order of D (ths s a theorem of Segel). But frst we need to fgure out how to construct a set of generators for G. We wll do ths usng prme forms. These are forms f = (a, b, c) for whch a s prme and a < b a (but we do not requre a c, so prme forms need not be reduced). Prme forms correspond to prme deals whose norm s prme (degree-1 prmes). Recall that magnary quadratc orders O are determned by ther dscrmnant D, whch can always be wrtten n the form D = u 2 D K, where D K s the dscrmnant of the maxmal order O K and u = [O K : O] s the conductor of O. (d) Let a be a prme. Prove that f a dvdes the conductor then there are no prme forms of norm a, and that otherwse there are exactly 1+( D a ) prme forms of norm a, where ( D a ) s the Kronecker symbol. 1 Wrte a program that ether outputs a prme form (a, b, c) wth b 0 or determnes that none exsts. When D s fundamental, we can generate cl(d) usng prme forms of norm at most p D /3; ths follows from the bound proved n Problem Set 9 and the fact that the maxmal order O K s a Dedeknd doman (so deals can be unquely factored nto prme deals). We can stll generate cl(d) wth prme forms when D s non-fundamental, but boundng the prmes nvolved s slghtly more complcated, so we wll restrct ourselves to fundamental dscrmnants for now. (e) Implement the algorthm you desgned n part p(b), usng the program from part (d) to enumerate the prme forms of norm a D /3 n ncreasng order by a. Use the prme forms as generators, but use a table lookup to dscard prme forms that are already present n your table so that your α all have relatve orders r > 1 (warnng: prme forms need not be reduced: be sure to reduce them before makng any comparsons). For the group operaton, you can create bnary quadratc forms n Sage usng BnaryQF([a,b,c]), and then compose forms f and g usng h=f*g. Use h.reduced form() to get the reduced form. You wll only be usng ths code on small examples, so don t worry about effcency; you wll only be graded on your answers to part (f) (whch you can solve mostly by hand, wth some help from Sage). (f) Run your algorthm on D = 5291, and then run t on the frst fundamental dscrmnant D < N, where N s the frst fve dgts of your student ID. Don t lst all the elements of cl(d), just gve the reduced forms for the elements of α~ and the nteger vectors r(α~) and s(α~). Santy check your results by verfyng that you at least get the rght class number for D (you can check ths n Sage usng NumberFeld(x**2-D, t ).class number()). 1 Thus ( D ) s 0 f D s even, 1 f D 1 mod 8, and 1 f D 5 mod 8. Note that we refer to a as the 2 norm of the form (a, b, c), snce the correspondng deal has norm a. 6

7 Problem 4. Survey Complete the followng survey by ratng each of the problems you attempted on a scale of 1 to 10 accordng to how nterestng you found the problem (1 = mnd-numbng, 10 = mnd-blowng ), and how dffcult you found t (1 = trval, 10 = brutal ). Also estmate the amount of tme you spent on each problem to the nearest half hour. Problem 1 Problem 2 Problem 3 Interest Dffculty Tme Spent Also, please rate each of the followng lectures that you attended, accordng to the qualty of the materal (1= useless, 10= fascnatng ), the qualty of the presentaton (1= epc fal, 10= perfecton ), the pace (1= way too slow, 10= way too fast, 5= just rght ) and the novelty of the materal (1= old hat, 10= all new ). Date Lecture Topc Materal Presentaton Pace Novelty 4/24 Remann surfaces and modular curves 4/26 The modular equaton Please feel free to record any addtonal comments you have on the problem sets or lectures, n partcular, ways n whch they mght be mproved. References [1] I.M.S. Dey and J. Wegold, Generators for alternatng and symmetrc groups, J. Australan Mathematcal Socety 12 (1971),

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