Vanishing Ranges for the Cohomology of Finite Groups of Lie Type

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1 for the Cohomology of Finite Groups of Lie Type University of Wisconsin-Stout, University of Georgia, University of South Alabama

2 The Problem k - algebraically closed field of characteristic p > 0 G - reductive algebraic group over k (split over F p ) q = p r for an integer r 1 G(F q ) - associated finite Chevalley group

3 The Problem k - algebraically closed field of characteristic p > 0 G - reductive algebraic group over k (split over F p ) q = p r for an integer r 1 G(F q ) - associated finite Chevalley group Question What is the least positive i such that H i (G(F q ), k) 0?

4 Quillen, 1972 H i (GL n (F q ), k) = 0 for 0 < i < r(p 1). No claim of sharpness.

5 Quillen, 1972 H i (GL n (F q ), k) = 0 for 0 < i < r(p 1). No claim of sharpness. For a split reductive G, there exists some constant C depending only on the root system, such that H i (G(F q ), k) = 0 for 0 < i < r C

6 Quillen, 1972 H i (GL n (F q ), k) = 0 for 0 < i < r(p 1). No claim of sharpness. For a split reductive G, there exists some constant C depending only on the root system, such that H i (G(F q ), k) = 0 for 0 < i < r C For SL 2 : C = p 1 2 ; no other values given. Not sharp in general (cf. Carlson, 1983)

7 More Friedlander, 1975 Vanishing ranges for special orthogonal and symplectic groups.

8 More Friedlander, 1975 Vanishing ranges for special orthogonal and symplectic groups. Hiller, 1980 Vanishing ranges for G simply connected for all types.

9 More Friedlander, 1975 Vanishing ranges for special orthogonal and symplectic groups. Hiller, 1980 Vanishing ranges for G simply connected for all types. Friedlander and Parshall, 1986 B(F q ) GL n (F q ) H i (B(F q ), k) = 0 for 0 < i < r(2p 3)

10 More Friedlander, 1975 Vanishing ranges for special orthogonal and symplectic groups. Hiller, 1980 Vanishing ranges for G simply connected for all types. Friedlander and Parshall, 1986 B(F q ) GL n (F q ) H i (B(F q ), k) = 0 for 0 < i < r(2p 3) H r(2p 3) (B(F q ), k) 0

11 Notation G - simple and simply connected Φ - root system T - torus B - Borel subgroup (corresponding to negative roots) U - its unipotent radical W - Weyl group X (T ) - weights X (T ) + - dominant weights For λ X (T ) +, H 0 (λ) := ind G B (λ) - the costandard module For λ X (T ) +, V (λ) := H 0 (λ ) - the Weyl or standard module w 0 - longest word W λ := w 0 (λ)

12 The Strategy Step 1: H i (G(F q ), k) H i (G, ind G G(F q) (k))

13 The Strategy Step 1: H i (G(F q ), k) H i (G, ind G G(F q) (k)) Step 2: Filter ind G G(F q) (k) by H0 (λ) H 0 (λ ) (r)

14 The Strategy Step 1: H i (G(F q ), k) H i (G, ind G G(F q) (k)) Step 2: Filter ind G G(F q) (k) by H0 (λ) H 0 (λ ) (r) Step 3: For r = 1, relate H i (G, H 0 (λ) H 0 (λ ) (1) ) to H i (G 1, H 0 (λ))

15 The Strategy Step 1: H i (G(F q ), k) H i (G, ind G G(F q) (k)) Step 2: Filter ind G G(F q) (k) by H0 (λ) H 0 (λ ) (r) Step 3: For r = 1, relate H i (G, H 0 (λ) H 0 (λ ) (1) ) to H i (G 1, H 0 (λ)) Step 4: Use root combinatorics and a description of H i (G 1, H 0 (λ)) to obtain vanishing information.

16 The Strategy Step 1: H i (G(F q ), k) H i (G, ind G G(F q) (k)) Step 2: Filter ind G G(F q) (k) by H0 (λ) H 0 (λ ) (r) Step 3: For r = 1, relate H i (G, H 0 (λ) H 0 (λ ) (1) ) to H i (G 1, H 0 (λ)) Step 4: Use root combinatorics and a description of H i (G 1, H 0 (λ)) to obtain vanishing information. Step 5: Use inductive arguments to get vanishing for higher r.

17 The Filtration Proposition As a G-module, ind G G(F q) (k) has a filtration with factors of the form H 0 (λ) H 0 (λ ) (r) with multiplicity one for each λ X (T ) +.

18 The Filtration Proposition As a G-module, ind G G(F q) (k) has a filtration with factors of the form H 0 (λ) H 0 (λ ) (r) with multiplicity one for each λ X (T ) +. Proof: Consider k[g] as a G G-module structure via the left and right regular actions. As a G-module via G G G, Known Fact: k[g] has a filtration by H 0 (λ) H 0 (λ ).

19 The Filtration Proposition As a G-module, ind G G(F q) (k) has a filtration with factors of the form H 0 (λ) H 0 (λ ) (r) with multiplicity one for each λ X (T ) +. Proof: Consider k[g] as a G G-module structure via the left and right regular actions. As a G-module via G G G, Known Fact: k[g] has a filtration by H 0 (λ) H 0 (λ ). New action: G G G Id Fr r G G Key Step: Show that ind G G(F q) (k) k[g] with this new action.

20 Vanishing Criteria Theorem Let m be the least positive integer such that there exists ν X (T ) + with H m (G, H 0 (ν) H 0 (ν ) (r) ) 0. Let λ X (T ) + be such that H m (G, H 0 (λ) H 0 (λ ) (r) ) 0. Suppose H m+1 (G, H 0 (ν) H 0 (ν ) (r) ) = 0 for all ν < λ that are linked to λ. Then (i) H i (G(F q ), k) = 0 for 0 < i < m; (ii) H m (G(F q ), k) 0; (iii) if, in addition, H m (G, H 0 (ν) H 0 (ν ) (r) ) = 0 for all ν X (T ) + with ν λ, then H m (G(F q ), k) = H m (G, H 0 (λ) H 0 (λ ) (r) ).

21 The New Goal Study H i (G, H 0 (λ) H 0 (λ ) (r) ) Ext i G (V (λ)(r), H 0 (λ)).

22 The New Goal Study H i (G, H 0 (λ) H 0 (λ ) (r) ) Ext i G (V (λ)(r), H 0 (λ)). Lemma If H i (G r, H 0 (λ)) ( r) admits a good filtration, then H i (G, H 0 (λ) H 0 (λ ) (r) ) Ext i G (V (λ)(r), H 0 (λ)) Hom G (V (λ), H i (G r, H 0 (λ)) ( r) ) Hypothesis holds for r = 1 and p > h (the Coxeter number). For r > 1,???

23 Root Combinatorics Consider H i (G 1, H 0 (λ)). When can this be non-zero?

24 Root Combinatorics Consider H i (G 1, H 0 (λ)). When can this be non-zero? By block considerations, we need λ = pµ + w 0 for some µ X (T ) and w W. Further

25 Root Combinatorics Consider H i (G 1, H 0 (λ)). When can this be non-zero? By block considerations, we need λ = pµ + w 0 for some µ X (T ) and w W. Further Theorem (Andersen-Jantzen 1986, Kumar-Lauritzen-Thomsen 1999) For p > h, H i (G 1, H 0 (λ)) ( 1) = where u = Lie(U). {ind G i l(w) B (S 2 (u ) µ) if λ = w 0 + pµ 0 otherwise, Note: since p > h and λ is dominant, µ must also be dominant.

26 Root Combinatorics - Continued Say λ = pµ + w 0, Hom G (V (λ), H i (G 1, H 0 (λ)) ( 1) ) Hom G (V (λ), ind G B S i l(w) 2 (u ) µ) Hom B (V (λ), S i l(w) 2 (u ) µ) Therefore, λ µ = (p 1)µ + w 0 is a sum of i l(w) 2 positive roots.

27 Root Combinatorics - Continued Say λ = pµ + w 0, Hom G (V (λ), H i (G 1, H 0 (λ)) ( 1) ) Hom G (V (λ), ind G B S i l(w) 2 (u ) µ) Hom B (V (λ), S i l(w) 2 (u ) µ) Therefore, λ µ = (p 1)µ + w 0 is a sum of i l(w) 2 positive roots. Lemma Assume that p > h. Assume H i (G, H 0 (λ) H 0 (λ ) (1) ) 0. Then i (p 1) µ, α 1, where α is the longest root. Note: µ, α 1

28 Generic Vanishing Theorem Theorem Assume that p > h. Then (a) H i (G, H 0 (λ) H 0 (λ ) (r) ) = 0 for 0 1 case requires working with Ext i G (V (λ)(r), H 0 (ν)) for possibly distinct λ, ν.

29 Generic Vanishing Theorem Theorem Assume that p > h. Then (a) H i (G, H 0 (λ) H 0 (λ ) (r) ) = 0 for 0 1 case requires working with Ext i G (V (λ)(r), H 0 (ν)) for possibly distinct λ, ν. ( ) While better than previous bounds in most cases: r p 1 2, these are still not sharp in general.

30 Finding Sharp Bounds For λ = pµ + w 0, if µ, α 2, then i 2(p 1) 1 = 2p 3.

31 Finding Sharp Bounds For λ = pµ + w 0, if µ, α 2, then i 2(p 1) 1 = 2p 3. In general, the sharp bound will lie between p 2 and 2p 3.

32 Finding Sharp Bounds For λ = pµ + w 0, if µ, α 2, then i 2(p 1) 1 = 2p 3. In general, the sharp bound will lie between p 2 and 2p 3. If it is strictly less than 2p 3, µ is necessarily a fundamental dominant weight. And in fact certain specific such weights.

33 Finding Sharp Bounds For λ = pµ + w 0, if µ, α 2, then i 2(p 1) 1 = 2p 3. In general, the sharp bound will lie between p 2 and 2p 3. If it is strictly less than 2p 3, µ is necessarily a fundamental dominant weight. And in fact certain specific such weights. The task is then to study H i (G, H 0 (λ) H 0 (λ ) (1) ) for those weights and see whether it in fact is non-zero. In part, use Kostant Partition Functions.

34 Finding Sharp Bounds From the above description of G 1 -cohomology, we can more precisely say: dim H i (G, H 0 (λ) H 0 (λ ) (1) ) = u W( 1) l(u) P i l(w) (u λ µ) 2 P k (ν) is the dimension of the ν weight space in S k (u ). Alternatively, P k (ν) is the number of distinct ways that ν can be expressed as a sum of exactly k positive roots.

35 Φ of Type A n, n 2 Theorem Assume that p > n + 1. (a) (Generic case) If p > n + 2 and n > 3, then (i) H i (G(F p ), k) = 0 for 0 < i < 2p 3; (ii) H 2p 3 (G(F p ), k) = k. (b) If p = n + 2, then (i) H i (G(F p ), k) = 0 for 0 < i < p 2; (ii) H p 2 (G(F p ), k) = k k.

36 Φ of Type A n, n 2 - Continued Theorem Assume that p > n + 1. (c) If n = 2 and 3 divides p 1, then (i) H i (G(F p ), k) = 0 for 0 5, then (i) H i (G(F p ), k) = 0 for 0 < i < 2p 6; (ii) H 2p 6 (G(F p ), k) = k.

37 Φ of Type A n - r > 1 Theorem Assume that p > 2(n + 1). Suppose further that p > 11 whenever n = 4. Then (a) H i (G(F p ), k) = 0 for 0 < i < r(2p 3); (b) H r(2p 3) (G(F p ), k) 0.

38 Φ of Type B n, n 3 - Under Construction Guess Assume that p > 2n. (a) (Generic case) If n 7, n = 6 with p 13, or n = 5 with p 11, 13 then (i) H i (G(F p ), k) = 0 for 0 < i < 2p 3; (ii) H 2p 3 (G(F p ), k) 0. (b) If n = 3, 4, then (i) H i (G(F p ), k) = 0 for 0 < i < 2p 8; (ii) H 2p 8 (G(F p ), k) 0. (c) If n = 5 and p = 11, then (i) H i (G(F p ), k) = 0 for 0 < i < 2p 7; (ii) H 2p 7 (G(F p ), k) 0. (c) If n = 5, 6 and p = 13, then (i) H i (G(F p ), k) = 0 for 0 < i < 2p 5; (ii) H 2p 5 (G(F p ), k) 0. Note: For the small n cases, the vanishing ranges can are known.

39 Φ of Type C n, n 1 Theorem Assume that p > 2n. Then (a) H i (G(F q ), k) = 0 for 0 < i < r(p 2); (b) H r(p 2) (G(F q ), k) 0. Note: The only dominant weight with H p 2 (G, H 0 (λ) H 0 (λ ) (1) ) 0 is λ = (p 2n)ω 1, where ω 1 is the first fundamental dominant weight.

40 Φ of Type D n, n 4 Theorem Assume that p > 2n 2. Then (a) H i (G(F p ), k) = 0 for 0 1.

41 Φ of Type E 6 Theorem Assume that p > 12. (a) If p 13, 19, then (i) H i (G(F p ), k) = 0 for 0 < i < 2p 3. (ii) H 2p 3 (G(F p ), k) 0. (b) If p = 13, then (i) H i (G(F p ), k) = 0 for 0 1.

42 Φ of type E 7 Theorem Assume that p > 18. (a) If p 19, 23, then (i) H i (G(F p ), k) = 0 for 0 < i < 2p 3. (ii) H 2p 3 (G(F p ), k) 0. (b) If p = 19, then (i) H i (G(F p ), k) = 0 for 0 1.

43 Φ of type E 8 Theorem Assume that p > 30. (a) H i (G(F q ), k) = 0 for 0 < i < r(2p 3). (b) H r(2p 3) (G(F q ), k) 0. Note: The root lattice equals the weight lattice here. More generally, this should be the answer whenever Φ is simply laced and the group G is of adjoint type (as opposed to simply connected).

44 Φ of type F 4 - Under Construction Assume that p > 12. Know: H i (G(F p ), k) = 0 for 0 < i < 2p 9. Guess: H 2p 9 (G(F p ), k) = 0 and H 2p 8 (G(F p ), k) 0

45 Φ of type G 2 Theorem Assume that p > 6. Then (i) H i (G(F p ), k) = 0 for 0 < i < 2p 8. (ii) H 2p 8 (G(F p ), k) 0.

DECOMPOSITION OF TENSOR PRODUCTS OF MODULAR IRREDUCIBLE REPRESENTATIONS FOR SL 3 (WITH AN APPENDIX BY C.M. RINGEL)

DECOMPOSITION OF TENSOR PRODUCTS OF MODULAR IRREDUCIBLE REPRESENTATIONS FOR SL 3 (WITH AN APPENDIX BY CM RINGEL) C BOWMAN, SR DOTY, AND S MARTIN Abstract We give an algorithm for working out the indecomposable