Modular Representation Theory of Endomorphism Rings

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1 Modular Representation Theory of Endomorphism Rings Lehrstuhl D für Mathematik RWTH Aachen University Bad Honnef, April 2010

2 Alperin s Weight Conjecture Assumptions: Let G be a finite group and k be an algebraically closed field with characteristic p > 0.

3 Alperin s Weight Conjecture Assumptions: Let G be a finite group and k be an algebraically closed field with characteristic p > 0. Definition and Remark Let X be an indecomposable kg-module. 1 Let H G. Then X is called H-projective, if X (X H ) G. 2 Any minimal subgroup Q of G such that X is Q-projective is called a vertex of X. 3 Vertices are p-subgroups of G. The set of vertices of X is a conjugacy-class of G.

4 Alperin s Weight Conjecture Theorem (Green Correspondence) Let Q G be a p-subgroup and N G (Q) H G. Then there is a one-to-one-correspondence between {X : X indecomposable kg-module with vertex Q} = g {Y : Y indecomposable kh-module with vertex Q} = given via 1 X H = f(x) Z, where each indecomposable direct summand of Z has a vertex in {Q g H : g G \ H}, 2 Y G = g(y ) W, where each indecomposable direct summand of W has a vertex in {Q g Q : g G \ H}. f

5 Alperin s Weight Conjecture Definition A weight for G is a tupel (Q,S), where Q is a p-subgroup of G and S is a simple kn G (Q)-module with vertex Q. In this case, we call S a weight module and the Green correspondent g(s) of S in G a weight Green correspondent (WGC).

6 Alperin s Weight Conjecture Definition A weight for G is a tupel (Q,S), where Q is a p-subgroup of G and S is a simple kn G (Q)-module with vertex Q. In this case, we call S a weight module and the Green correspondent g(s) of S in G a weight Green correspondent (WGC). Weights are understood up to the natural equivalence induced by conjugation of G. (Q,S) is a weight if and only if S is a simple and projective kn G (Q)/Q-module. Any WGC is (isomorphic to) a direct summand of the permutation module k G P, where P Syl p(g).

7 Alperin s Weight Conjecture Conjecture (Alperin s Weight Conjecture, 1987) The number of simple kg-modules (up to isomorphism) is equal to the number of weights for G (up to equivalence).

8 Endomorphism Ring of k G P Let Y := k G P = n i=1 Y i be a decomposition of k G P into indecomposable direct summands and put E := End kg (Y ). For ϕ E, ϕ(y) denotes the image of y Y under ϕ. The Hom-Functor Hom kg (Y, ) : mod-kg mod-e induces a decomposition of E E into the PIMs P i := Hom kg (Y,Y i ) (Fitting Correspondence). Of course, hd(p i ) is a simple E-module, but soc(p i ) is in general not simple.

9 Computational Experiments With the computer algebra programes GAP, MeatAxe:

10 Computational Experiments With the computer algebra programes GAP, MeatAxe: Construct Y = k G P, Y = n i=1 Y i, E E, E E = n i=1 P i. Analyze hd(y i ), soc(y i ), hd(p i ), soc(p i ). Determine character table, the Cartan- and decomposition matrix of E.

11 Computational Experiments With the computer algebra programes GAP, MeatAxe: Construct Y = k G P, Y = n i=1 Y i, E E, E E = n i=1 P i. Analyze hd(y i ), soc(y i ), hd(p i ), soc(p i ). Determine character table, the Cartan- and decomposition matrix of E. Observation For almost all groups analyzed so far: {simple constituents of soc(e E )} = {simple kg-modules} = {weights for G}

12 Computational Experiments With the computer algebra programes GAP, MeatAxe: Construct Y = k G P, Y = n i=1 Y i, E E, E E = n i=1 P i. Analyze hd(y i ), soc(y i ), hd(p i ), soc(p i ). Determine character table, the Cartan- and decomposition matrix of E. Observation For almost all groups analyzed so far: { simple constituents of soc(e E )} / = {simple kg-modules} / = = { weights for G} / Exception: M 11 in characteristic 3.

13 Computational Experiments From this, the following questions and ideas arise: Generalize analysis to more general modules than k G P, e.g. k G Q i for certain p-subgroups Q i of G. Which role does the Hom-functor, and especially its evaluation at a simple kg-module play? Characterize families of groups for which the observation hold.

14 M 11 in characteristic 3 p = 3, G = M 11, P Syl 3 (G), G = k G P hd soc E E k G P hd soc

15 M 11 in characteristic 3 Generalize kp G to kg P kg Q, where Q = 3. Y := (kp G kg Q )/(Defect 0, multiplicities)

16 M 11 in characteristic 3 Y := (kp G kg Q )/(Defect 0, multiplicities) Y hd soc Y hd soc

17 M 11 in characteristic 3 Y := (kp G kg Q )/(Defect 0, multiplicities) Y hd soc Y hd soc E E Y hd soc E E Y hd soc

18 Hom-Functor F := Hom kg (Y, ) evaluated at simple kg-module - what can happen?

19 Hom-Functor F := Hom kg (Y, ) evaluated at simple kg-module - what can happen? In general: F(S) for a simple kg-module S not simple. For example: A 6 in characteristic 5: F(8) = 1 a 1 b. Even soc(f(s)) may not be simple. For example: M 11 in characteristic 3. Then F(10 3 ) =

20 Hom-Functor Assumptions for this section:

21 Hom-Functor Assumptions for this section: Consider some kg-module Y such that {S : S hd(y ) simple }/ = = {T : T soc(y )}/ =. Let Y = n i=1 Y i be a decomposition into indecomposable direct summands. Assume: Head of Y i is simple for all 1 i n. Y i = Yj i = j. Set E := End kg (Y ).

22 Hom-Functor Assumptions for this section: Remark Consider some kg-module Y such that {S : S hd(y ) simple }/ = = {T : T soc(y )}/ =. Let Y = n i=1 Y i be a decomposition into indecomposable direct summands. Assume: Head of Y i is simple for all 1 i n. Y i = Yj i = j. Set E := End kg (Y ). There is a partial order on {Y i : 1 i n}, given via: Y i Y j, if and only if there is a surjection ϕ : Y i Y j.

23 Hom-Functor Notation: Fix S = hd(y i ) for some 1 i n. Then i ϕ : Y i S may be considered as an element of F(S) = Hom kg (Y,S) E.

24 Hom-Functor Notation: Lemma Fix S = hd(y i ) for some 1 i n. Then i ϕ : Y i S may be considered as an element of F(S) = Hom kg (Y,S) E. (a) Let ψ k F(S) be simple E-submodule. Then ψ is of the form ψ = i ϕ for some 1 i n such that hd(y i ) = S. (b) The following are equivalent. (i) Y i is maximal w.r.t. for some 1 i n. (ii) i ϕ k is a simple E-submodule of F(S).

25 Hom-Functor Theorem Let S be a simple socle constituent of Y and assume (possibly after renumbering) that hd(y k ) = S for all 1 k l for some l n and that hd(y k ) = S for all l + 1 k n. (a) The following are equivalent: (i) The E-submodule soc(f(s)) has exactly r non-isomorphic constituents. (ii) Among {Y k : 1 k l} there are exactly r maximal elements with respect to. (b) Let ψ k E E be simple E-submodule. Then there is 1 i n such that ψ = E i ϕ.

26 Hom-Functor Corollary Assume that for each simple module occurring in the head of Y there is exactly one maximal element with respect to.

27 Hom-Functor Corollary Assume that for each simple module occurring in the head of Y there is exactly one maximal element with respect to. Then soc(f(s)) is a simple E-module for each simple constituent S of hd(y ). Moreover, {S : S hd(y ) simple}/ = {T : T soc(e) simple}/ = S soc(f(s)) is a bijection.

28 Indecomposable Liftable Modules in Cyclic Blocks For which families of groups does observation hold?

29 Indecomposable Liftable Modules in Cyclic Blocks For which families of groups does observation hold? Joint work with Gerhard Hiss.

30 Indecomposable Liftable Modules in Cyclic Blocks For which families of groups does observation hold? Joint work with Gerhard Hiss. Assumptions: Let B be cyclic kg-block with defect group Q. Question 3 : Which indecomposable B-modules are liftable?

31 Indecomposable Liftable Modules in Cyclic Blocks For which families of groups does observation hold? Joint work with Gerhard Hiss. Assumptions: Let B be cyclic kg-block with defect group Q. Question 3 : Which indecomposable B-modules are liftable? Kupisch 1969: Description of indecomposable B-modules in terms of paths on the Brauer tree:

32 Indecomposable Liftable Modules in Cyclic Blocks Paths of Type I: Brauer tree, with multiplicity m = 3 of exceptional vertex: A 2 A 1 χ 0 E χ 1 1 E 2 E 3 B 1 B 2 χ Λ χ 3

33 Indecomposable Liftable Modules in Cyclic Blocks Paths of Type I: Choose path: A 2 A 1 χ 0 E χ 1 1 E 2 E 3 B 1 B 2 χ Λ χ 3

34 Indecomposable Liftable Modules in Cyclic Blocks Paths of Type I: Choose marking on path and multiplicity t = 2 m = 3: A 2 A 1 χ 0 E χ 1 1 E 2 E 3 B 1 B 2 t=2 χ Λ χ 3

35 Indecomposable Liftable Modules in Cyclic Blocks Paths of Type I: Choose marking on path and multiplicity t = 2 m = 3: A 2 A 1 χ 0 E χ 1 1 E 2 E 3 B 1 B 2 t=2 χ Λ χ 3 From this, we get the following indecomposable kg-module: A 1 E 3 E 2

36 Indecomposable Liftable Modules in Cyclic Blocks Paths of Type I: Choose marking on path and multiplicity t = 2 m = 3: A 2 A 1 χ 0 E χ 1 1 E 2 E 3 B 1 B 2 t=2 χ Λ χ 3 A 1 A2 E1 E3 B 1 E 2 B 2 E 3 E 2

37 Indecomposable Liftable Modules in Cyclic Blocks Paths of Type I: Choose marking on path and multiplicity t = 2 m = 3: A 2 A 1 χ 0 E χ 1 1 E 2 E 3 B 1 B 2 t=2 χ Λ χ 3 A 1 A2 E1 E3 B 1 E 2 B 2 E 3 E 2 Note: The module Y is of Type I if and only if, soc(y ) and hd(y ) have no common constituents.

38 Indecomposable Liftable Modules in Cyclic Blocks Paths of Type II: Brauer tree, with multiplicity m = 3 of exceptional vertex: A 2 A 1 χ 0 E χ 1 1 E 2 E 3 B 1 B 2 χ Λ χ 3

39 Indecomposable Liftable Modules in Cyclic Blocks Paths of Type II: Choose path: A 2 A 1 χ 0 χ E 1 1 E 2 E 3 B 1 B 2 χ Λ χ 3

40 Indecomposable Liftable Modules in Cyclic Blocks Paths of Type II: Choose marking on path and multiplicity t = 2 m = 3: A 2 A 1 χ 0 χ E 1 1 E 2 E 3 B 1 B 2 t=2 χ Λ χ 3 From this, we get the following indecomposable kg-module:

41 Indecomposable Liftable Modules in Cyclic Blocks Paths of Type II: Choose marking on path and multiplicity t = 2 m = 3: A 2 A 1 χ 0 χ E 1 1 E 2 E 3 B 1 B 2 t=2 χ Λ χ 3 From this, we get the following indecomposable kg-module: E 2 B 2 A 1 E 2

42 Indecomposable Liftable Modules in Cyclic Blocks Paths of Type II: Choose marking on path and multiplicity t = 2 m = 3: A 2 A 1 χ 0 χ E 1 1 E 2 E 3 B 1 B 2 t=2 χ Λ χ 3 From this, we get the following indecomposable kg-module: E 2 A 1 E 3 E2 E3 B2 E 2

43 Indecomposable Liftable Modules in Cyclic Blocks Paths of Type II: Choose marking on path and multiplicity t = 2 m = 3: A 2 A 1 χ 0 χ E 1 1 E 2 E 3 B 1 B 2 t=2 χ Λ χ 3 From this, we get the following indecomposable kg-module: E 2 A 1 E 3 E2 E3 B2 Note: The module Y is of Type II if and only if, soc(y ) and hd(y ) have common constituents. E 2

44 Indecomposable Liftable Modules in Cyclic Blocks Theorem Let τ be a path of Type I on the Brauer tree of B. Then the indecomposable module X constructed from τ is liftable if and only if τ is as in the following figure and one of the following cases occurs: χ 0 E χ 0 1 E 1 (a) χ 1 is not the exceptional vertex. Then the character of a lift of X equals χ 1.

45 Indecomposable Liftable Modules in Cyclic Blocks Theorem Let τ be a path of Type I on the Brauer tree of B. Then the indecomposable module X constructed from τ is liftable if and only if τ is as in the following figure and one of the following cases occurs: χ 0 E χ 0 1 E 1 (a) χ 1 is not the exceptional vertex. Then the character of a lift of X equals χ 1. (b) χ 1 is the exceptional vertex, which has multiplicity m. Each of E 0 and E 1 occurs t times as composition factor of X for some 1 t m. The character of any lift of X is a sum of t distinct exceptional characters.

46 Indecomposable Liftable Modules in Cyclic Blocks χ 0 E 0 χ 1 E 1 Note: If X of Type I, then X is uniserial in both cases. The head of X is either isomorphic to E 0 or E 1. If the head of X is isomorphic to E 0, the successor of E 1 around χ 1 equals E 0, and if the head of X is isomorphic to E 1, the successor of E 0 around χ 1 equals E 1.

47 Liftable Indecomposable Modules in Cyclic Blocks Theorem Let τ be a path of Type II on the Brauer tree. Then the indecomposable module X constructed from τ is liftable if and only if τ satisfies one the following three cases: Case 1: The vertex χ 0 is a leaf of the Brauer tree. χ 0 E 0 χ 1 χ s E s t χλ

48 Liftable Indecomposable Modules in Cyclic Blocks Theorem Let τ be a path of Type II on the Brauer tree. Then the indecomposable module X constructed from τ is liftable if and only if τ satisfies one the following three cases: Case 1: The vertex χ 0 is a leaf of the Brauer tree. χ 0 E 0 χ 1 χ s E s t χλ Case 2: Either E u0 is in the head or in the socle of X. If E u0 is in the head of X, the successor of E 0 around χ 0 is E u0. If E u0 is in the socle of X, the successor of E u0 around χ 0 is E 0. χ u0 χ E 0 u0 E 0 χ 1 χ s E s t χλ

49 Liftable Indecomposable Modules in Cyclic Blocks Theorem Let τ be a path of Type II on the Brauer tree. Then the indecomposable module X constructed from τ is liftable if and only if τ satisfies one the following three cases: Case 3: The successor of E u0 around χ 0 is E 0 and E u0 is in the socle of X. χ u0 E u0 χ 0 E 0 χ 1 χ s E s t χλ χ d0 E d0

50 Liftable Indecomposable Modules in Cyclic Blocks Theorem Let τ be a path of Type II on the Brauer tree. Then the indecomposable module X constructed from τ is liftable if and only if τ satisfies one the following three cases: Case 3: The successor of E u0 around χ 0 is E 0 and E u0 is in the socle of X. χ u0 E u0 χ 0 E 0 χ 1 χ s E s t χλ χ d0 E d0 The character of any lift of X equals χ 0 + χ 1 + χ s + µ, where µ is a sum of t distinct exceptional characters.

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