13
$\begingroup$

The Steinberg representation is a remarkable irreducible representation of a reductive algebraic group over a finite field or local field, or a group with a BN-pair. It is analogous to the 1-dimensional sign representation ε of a Coxeter or Weyl group that takes all reflections to –1. The Steinberg representation is the Alvis–Curtis dual of the trivial 1-dimensional representation.

Question: What are the analogs of the Steinberg representation for sporadic simple groups ? (If there are any ideas for other finite groups beyond standard - it also welcome).

Wikipedia suggests: Some of the sporadic simple groups act as doubly transitive permutation groups so have a BN-pair for which one can define a Steinberg representation, but for most of the sporadic groups there is no known analogue of it.

However googling does not lead me to an answer even about that "some of sporadic ... " .


Further question (cohomological construction) :

At Mathoverflow D.Pasechnik answering S.Lentner question: (weak?) BN-Pair / Tits System for Sporadic Groups ? writes: Instead of starting from a weak BN-pair, one can weaken Tits' axioms from his "Local approach to buildings" to develop a theory dealing with sporadics.

That may give a way to answer my question since it is known that Steinberg representation is realized in the cohomology group of the Bruhat–Tits building.

Question 2 Are there similar cohomological constructions of "Steinberg" representation for sporadic groups ?


Further question (mod-p reduction):

In the remarkable (must&pleasure to read - imho) survey "The Steinberg representation" BAMS 1987 J. E. Humphreys takes modular representation point view on the Steinberg representation. The point is: for G(F_p) "St" can be reduced mod "p". Moreover the reduction is quite remarkble - it preserves irreducibility and some other properties (quote: it is also a "principal indecomposable" representation determining a "block" by itself. Moreover, the character of the representation vanishes at all elements of G having order divisible by p) - that follows from Brauer-Nesbit theory.

Question 3 If there any mod-p properties of "St" for sporadic ? The positive answer would have a strange conclusion - that some sporadics might be considered as kind of groups defined over "F_p" for some "p".


Motivation (one of):

As described here: MO271067 I would hope to have some bijection(s) between irreducible representations of sporadic groups (and other groups also) and their conjugacy classes. Distinguishing some representations as "Steinberg like" would be quite helpful.

For example for Mathiew group M11 looking on the character table (e.g. here page 3 or Magma):


Class | 1 2 3 4 5 6 7 8 9 10

Size | 1 165 440 990 1584 1320 990 990 720 720

Order | 1 2 3 4 5 6 8 8 11 11

X.1 + 1 1 1 1 1 1 1 1 1 1

X.2 + 10 2 1 2 0 -1 0 0 -1 -1

X.3 0 10 -2 1 0 0 1 Z1 -Z1 -1 -1

X.4 0 10 -2 1 0 0 1 -Z1 Z1 -1 -1

X.5 + 11 3 2 -1 1 0 -1 -1 0 0

X.6 0 16 0 -2 0 1 0 0 0 Z2 Z2#2

X.7 0 16 0 -2 0 1 0 0 0 Z2#2 Z2

X.8 + 44 4 -1 0 -1 1 0 0 0 0

X.9 + 45 -3 0 1 0 0 -1 -1 1 1

X.10 + 55 -1 1 -1 0 -1 1 1 0 0

Z1 = i * sqrt(2) Z2 = (-1+i * sqrt(11))/2

One can easily guess to map pair of conjugacy classes of order 8 to pair of irreps X3,X4 and classes of order 11 to X.6 X.7 because

1) that are only 4 complex classes/irreps - that distiguishes 4 classes and 4 irreps

2) orders 8, 11 of classes correspond to degree of rationality of those irreps - thus we get: two classes of order 8 <-> two irreps X3,X4; two classes of order 11 <-> two irreps X6,X7;

Bonus Question What irrep is Steinberg for M11 and what conjugacy class might correspond to it ?


More generally one may ask is there an analog of Alvis-Curtis duality for sporadics ? Is there splitting of classes and characters to semisimple/unipotent for sporadics ? Analogs of parabolic subgroups ? Etc... (i.e. can one extend properties from Lie groups to sporadics ? )


EDIT: From the comments it becomes clear that there is lot of research generalizing the Steinberg representation. Still it is not clear for the direction suggested in Wikipedia - from double transitive action to BN-pair , and hence to Steinberg reprsentation.

$\endgroup$
11
  • $\begingroup$ Tiep has determined which groups have a Steinberg like character $\endgroup$ Jul 30, 2017 at 13:40
  • $\begingroup$ @GeoffRobinson Is this: "p-Steinberg Characters of Finite Simple Groups" sciencedirect.com/science/article/pii/S0021869397967910 Abstract: LetGbe a finite group andpa prime divisor of |G|. Ap-Steinberg character ofGis an irreducible character χ ofGsuch that χ(x) = ± |CG(x)|pfor everyp′-elementx ∈ G. A conjecture of W. Feit states that if a finite simple groupGhas ap-Steinberg character thenGis a finite simple group of Lie type in characteristicp. In this paper we prove this conjecture, using the classification of finite simple groups. $\endgroup$ Jul 30, 2017 at 16:21
  • $\begingroup$ @GeoffRobinson that somehow contradicts Wikipedia suggesting that there are analogs for some sporadics... Probably Tiep made too restrictive condition... $\endgroup$ Jul 30, 2017 at 16:22
  • 1
    $\begingroup$ Tiep required exactly the congruence condition he specified, which iis one natural choice. There are analogues of the Steinberg character ( obtained from the p-subgroup complex) for every finite group. These were studied extensively by P.J. Webb and J. Thevenaz, following earlier work of K.S. Brown. However, these are not necessarily irreducible, and are usually virtual characters, vanishing in all p-singular elements. There is a book by S.D. Smith paying particular attention to related cohomological constructions for sporadic groups. $\endgroup$ Jul 30, 2017 at 18:10
  • $\begingroup$ A nice article about generalizing buildings to arbitrary finite groups (or to the other primes) is Subgroup complexes by Peter Webb, pp. 349-365 in: ed. P. Fong, The Arcata Conference on Representations of Finite Groups, AMS Proceedings of Symposia in Pure Mathematics 47 (1987). This leads to the generalized Steinberg characters mentioned by Geoff in his last comment. BTW for the general linear groups and a prime $p$ NOT its characteristic, one should look at the Benson complex, i.e., the elementary abelian $p$-subgroups that contain ALL central elements of order $p$ of their centralizers. $\endgroup$
    – j.p.
    Jul 30, 2017 at 19:51

1 Answer 1

17
$\begingroup$

The approach that I have taken to generalizing the Steinberg module to finite groups other than groups of Lie type is that, in general, the object we should consider is a chain complex, rather than just a module. For every finite group G and for every prime p there is a finite chain complex of projective modules, canonically defined up to isomorphism, that in the case of a group with a split (B,N) pair in characteristic p, is the Steinberg module, shifted in degree so that it appears in degree equal to the dimension of the building. I like to call this canonically defined complex the `Steinberg complex' of G at the prime p, and it exists for every finite group and every prime.

The Steinberg complex is shown to exist and is defined for the first time (without calling it by that name) in P.J. Webb, A split exact sequence of Mackey functors, Commentarii Math. Helv. 66 (1991), 34-69, doi: 10.1007/BF02566635. It arises from Theorem 2.7.1 there, applied to Brown's simplicial complex of non-identity p-subgroups of G, or to any complex G-homotopy equivalent to this complex. This includes Quillen's complex of non-identity elementary abelian p-subgroups, Bouc's complex of p-radical subgroups, as well as Benson's complex which has been discussed recently. Benson's complex was introduced in an exercise at the end of section 6.6 of his book Representations and Cohomology II, and it is discussed by Smith in his book Subgroup Complexes. My theorem states that over a $p$-local ring (such as a field of characteristic p) the augmented chain complex of Brown's complex is the direct sum of a contractible complex and a complex of projective modules. This implies that if we remove all contractible summands from the augmented chain complex of Brown's complex we obtain a complex of projective modules, unique up to isomorphism

After my initial paper, other proofs have been given of my result. The following papers both contain proofs: P. Symonds, The Bredon cohomology of subgroup complexes, J. Pure Appl. Algebra 199 (2005), 261-298, doi: 10.1016/j.jpaa.2004.12.010. S. Bouc, Resolutions de foncteurs de Mackey, Proc Symposia in Pure Math. 63 (1998), 31-83.

The fact that the Steinberg complex exists has as a consequence the method of computing group cohomology described in my paper, A local method in group cohomology, and used quite extensively by various people, including Adem and Milgram.

Work on the structure of the Steinberg complex in the situation where its homology is not projective was done by Dan Swenson in his thesis. You can download a copy at http://www-users.math.umn.edu/~webb/PhDStudents/index.html

In 1986 I wrote a survey with the title Subgroup Complexes that appeared in vol 47 of Proc. Symposia in Pure Math, before I had proved my theorem about the Steinberg complex. In that survey I gave values of the Lefschetz module of the Steinberg complex for certain groups, including some sporadic groups. This is the alternating sum of the projective modules in the Steinberg complex, so it is a virtual projective module, and it could be called the generalized Steinberg module. I just want to point out that there are many errors in that article. The value I gave for the generalized Steinberg module of M_12 is completely wrong, and the value for A_7 should have a minus sign. These and other errors are listed on my web site, at http://www-users.math.umn.edu/~webb/Publications/SubgroupComplexes.errors

I hope this helps.

$\endgroup$
1
  • 3
    $\begingroup$ Thank you very much ! Welcome to Mathoverflow ! Great answer ! $\endgroup$ Aug 3, 2017 at 4:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.