. Algebraic tori and a computational inverse Galois problem. David Roe. Jan 26, 2016

Transcription

1 Algebraic tori and a computational inverse Galois problem David Roe Department of Mathematics University of Pittsburgh Jan 26, 2016

2 Outline 1 Algebraic Tori 2 Finite Subgroups of GL n (Z) 3

3 Tori over R When you hear torus, you probably think Today: an algebraic version Define three basic tori over R: U, with U(R) = {z C : z z = 1}, G m, with G m (R) = R, S, with S(R) = C Theorem (cf [1, Thm 2]) Ẹveryalgebraictorusover R isaproductofthesetori

4 Algebraic tori G m is the variety defined by xy 1: for any ring R its points are the units R U is the variety defined by x 2 + y 2 1; after tensoring with C can factor as (x + iy)(x iy) 1 Both are in fact groupschemes: the set of points has a group structure Definition An algebraictorus over a field K is a group scheme, isomorphic to (G m ) n after tensoring with a finite extension Can also give T( K) plus a continuous action of Gal( K/K) on it

5 Character lattices Definition Ṭhe characterlattice of T is X (T) = Hom K(T, G m ), X (T) is a free rank-n Z-module with a Gal( K/K) action Can take {χ i : (z 1,, z n ) z i } as a basis for X (G n m) X (G m ) = Z with trivial action, X (U) = Z with conjugation acting as x x, X (S) = Zv Zw with conjugation exchanging v and w Theorem Thefunctor T X (T) definesacontravariantequivalenceof categories K-Tori Gal( K/K)-Lattices

6 Finding tori Goal 1 Createadatabaseofalgebraictoriover p-adicfields (wwwlmfdborg) 2 Usetostudystructureofalgebraicgroups, p-adic representationtheoryandlocallanglands, especiallyfor exceptionalgroups Some will apply to other fields and to Galois representations

7 Strategy We break up the task of finding tori into two pieces: 1 For each dimension n, list all finite groups G that act (faithfully) on Z n For fixed n, the set of G is finite 2 For each G and p, list all Galois extensions L/Q p with Gal(L/Q p ) G For fixed G and p, the set of L is finite Moreover, when p does not divide G, this question is easy

8 Finite Subgroups of GL n (Z) With a choice of basis, a faithful action of G on Z n is the same as an embedding G GL n (Z) Two G-lattices are isomorphic if and only if the corresponding subgroups are conjugate within GL n (Z) Two G-lattices are isogenous if the corresponding subgroups are conjugate within GL n (Q) G m U and S are isogenous but not isomorphic, since ( 1 0 and ( ) are conjugate in GL n(q) but not in GL n (Z) 0 1 )

9 Previous Computations CARAT [2] Up to dimension 6, the software package CARAT lists all of the finite subgroups of GL n (Z), up to Z- and Q-conjugacy IMF GAP Library [4] The group theory software package GAP has a library for maximal finite subgroups where the corresponding lattice is irreducible as a G-module The Q-classes are known for n 31, the Z-classes for n 11 and n {13, 17, 19, 23}

10 Indecomposible subgroups A G-lattice is indecomposible if it does not split as a direct sum of G-submodules For example, X (S) is not irreducible, since a + b is a stable submodule, as is a b But it is indecomposible: the sum of these submodules has index 2 For n > 6, work remains to recover a list of indecomposible subgroups Note that the decomposition into indecomposible submodules is NOT unique

11 Interlude: p-adic fields For each prime p, define v p : Q Z { } by v p (p k α) = k when α is relatively prime to p Set x p = p v p(x), and Q p as the completion Z p = {x Q p : v(x) 0} and P p = {x Z p : v(x) > 0} is the unique maximal ideal in Z p, with quotient F p (residue field) A uniformizer is an element of valuation 1, ie p u Q p Z p p Z and Z p F p (1 + P p ) For example, is an element of Q 5

12 Interlude: p-adic extensions Algebraic extensions of Q p are much richer than those of R Let K/Q p be a finite extension There is a unique extension of v to a valuation v K : K Q { } L/K is unramified if the image of v K is the same as v L There is a unique unramified extension of each degree (comes from the residue field) L/K is totallyramified if the corresponding extension of residue fields is trivial A totally ramified extension is tame if [L : K] is prime to p These are obtained by adjoining roots of uniformizers A totally ramified extension is wild if [L : K] is a power of p Any extension L/K can be split as L/L t /L u /K, with L u /K unramified, L t /L u tame and L/L t wild

13 Number of Subgroups (up to GL n (Z)-conjugacy) Dimension Real Unramified Tame adic adic adic adic Local All Note that each subgroup corresponds to multiple tori, since there are multiple field extensions with that Galois group

14 Order of Largest Subgroup Dimension Real Unramified Tame adic adic adic adic Irreducible Weyl A 1 G 2 B 3 F 4 B 5 2 E 6 Dim Largest Irreducible Subgroup (E 7 ) (E 8 ) (B 31 )

15 Inverse Galois Problem Classic Problem: determine if a finite G is a Galois group Depends on base field: every G is a Galois group over C(t) Most work focused on L/Q: S n and A n, every solvable group, every sporadic group except possibly M 23, Generic polynomials f G (t 1,, t r, X) are known for some (G, K): every L/K with group G is a specialization Computational Problem Give an algorithm to find all of the field extensions of K = Q p with a specified Galois group

16 Database of p-adic Fields Jones and Roberts [3] have created a database of p-adic fields Lists all L/Q p with a given degree, including non-galois; Includes up to degree 10; Gives Galois group and other data about the extension; Biggest table is [L : Q 2 ] = 8, of which there are 1823 We need G in degree up to 96 (tame) or 14, 60, 144, 144 (wild, p = 7, 5, 3, 2 resp) Their database solves the problem for small G, but most of the target G fall outside it

17 Structure of p-adic Galois groups The splitting of L/K into unramified, tame and wild pieces induces a filtration on Gal(L/K) We can refine this filtration to G G 0 G 1 G 2 G r = 1 For every i, G i G; G/G 0 = F is cyclic, and L G 0 /K is maximal unramified; G 0 /G 1 = τ is cyclic, order prime to p and FτF 1 = τ p ; For 0 < i < r, G i /G i+1 F k i p Finding such filtrations on an abstract group is not difficult

18 Inductive Approach For tame extensions: lift irreducible polynomials from residue field for unramified, then adjoin n th roots of p u Thus, it suffices to solve: Problem Fix a Galois extension L/K, set H = Gal(L/K) and suppose G is an extension of H: 1 A G H 1, with A F k p Find all M/L st M/K Galois and Gal(M/K) G

19 Interlude: Local Class Field Theory Let M/L/Q p with [M : L] = m and Γ = Gal(M/L) Theorem (Local Class Field Theory [6, Part IV]) H 2 (Γ, M ) = u M/L 1 m Z/Z u M/L : Γ ab = Ĥ 2 (Γ, Z) Ĥ0 (Γ, M ) = L / Nm M/L M Themap M Nm M/L M givesabijectionbetween abelianextensions M/L andfiniteindexsubgroupsof L Pauli [5] gives algorithms for finding a defining polynomial of the extension associated to a given norm subgroup Upshot Since A = F k p abelian, can use LCFT to find possible M/L in terms of subgroups of L

20 A Mod-p Representation Given 1 A G H 1 and L/K, let V = (1 + P L )/(1 + P L ) p, an F p [H]-module Since A = Gal(M/L) has exponent p, it corresponds to a subgp N (1 + P L ) p and L /N (1 + P L )/(N (1 + P L )) Let W = (N (1 + P L ))/(1 + P L ) p, a subspace of V M/K is Galois iff W is stable under H = Gal(L/K) The MeatAxe algorithm finds such subrepresentations For each W, check V/W A as F p [H]-modules The corresponding M/K are candidates for Gal(M/K) G

21 Extension Classes There may be multiple extensions 1 A G H 1 yielding the same action of H on A Use group cohomology to distinguish them Choosing a section s : H G, define a 2-cocycle by (g, h) s(g)s(h)s(gh) 1 A Get bijection H 2 (H, A) {1 A G H 1}/ Two approaches to picking out G: 1 Just compute Gal(M/K), 2 Try to find the extension class, given W

22 A Conjecture on the Fundamental Class Conjecture Let N L correspondto M/L underlcft andset G = Gal(M/K), H = Gal(L/K) and A = Gal(M/L) Thentheimageof u L/K underthenaturalmap istheextensionclassfor H 2 (H, L ) H 2 (H, L /N) H 2 (H, A) 1 Gal(M/L) Gal(M/K) Gal(L/K) 1 If this conjecture holds, can compute a 2-cocycle representing u L/K and use it for each W

23 Summary of Algorithm Data: G G 0 G 1 G 2 G r = 1 Result: List of all Galois F/Q p with Gal(F/Q p ) G Find tame extensions L 1 /Q p with Gal(L 1 /Q p ) G/G 1 ; for 0 < i < r do Find class σ i of 1 G i /G i+1 G/G i+1 G/G i 1; for each L = L i do Compute a 2-cocycle representing u L/Qp ; Find all stable submodules W with L /W G i /G i+1 ; for each W do if u L/Qp σ i H 2 (L/Q p, L /W) then Add the M/L matching W to the list of L i+1 ; end end end end

24 Future Work Thanks References Future Work 1 Flesh out details of algorithm and implement it, 2 Extend group theoretic analysis to dimension 7 and 8, 3 Compute additional data for each torus: cohomology groups, embeddings into induced tori, Moy-Prasad filtrations, conductors, component groups of Néron models 4 Put data online at wwwlmfdborg

25 Future Work Thanks References Thank you for your attention!

26 Future Work Thanks References References [1] B Casselman Computationsinrealtori, Representation theory of real groups, Contemporary Mathematics 472, AMS (2007) [2] C Cid, J Opgenorth, W Plesken, T Schulz CARAT wwwbmathrwth-aachende/carat/ [3] J Jones, D Roberts Adatabaseoflocalfields, J Symbolic Comput 41 (2006), [4] G Nebe, W Pleskin, M Pohst, B Souvignier Irreduciblemaximalfinite integralmatrixgroups GAP Library [5] S Pauli Constructingclassfieldsoverlocalfields, J Théor Nombres Bordeaux 18 (2006), [6] J-P Serre Localfields Springer, New York, 1979