The stable cohomology of SL(F_p)

Back by popular demand: an actual mathematics post!

Today’s problem is the following: compute the cohomology of \mathrm{SL}(\mathbf{F}_p) for a (mod-p) algebraic representation.

Step 0 is to say what this problem actually is. It makes sense to talk about certain algebraic representations of \mathrm{SL}_n(\mathbf{F}_p) as n varies (for example, the standard representation or the adjoint representation, etc.). For such representations, one can prove stability phenomena for the corresponding cohomology groups. But my question is whether one can actually compute these groups concretely.

The simplest case is the representation \mathbf{L} = \mathbf{F}_p and here one has a complete answer: these cohomology groups are all zero in higher degree, a computation first done by Quillen and which is closely related to the fact that the K_n(\mathbf{F}_p) \otimes \mathbf{F}_p = 0.

Most of the references I have found for cohomology computations of special linear groups in their natural characteristic consider the case were p is very large compared to n, but let me remind the reader that we are exactly the opposite situation. One of the few references is a paper of Evens and Friedlander from the ’80s which computes some very special cases in order to compute K_3(\mathbf{Z}/p^2 \mathbf{Z}).

Note, however, that p should still be thought of as “large” compared to the partition which defines the corresponding stable local system(s).

In order to get started, let us make the following assumptions:

ANZATZ: There exists a space X with a \mathrm{SL}(\mathbf{Z}_p) pro-cover such that:

1. The corresponding completed cohomology groups with \mathbf{F}_p coefficients are \mathbf{F}_p for i = 0 and vanish otherwise.
2. If \mathbf{L} is the mod-p reduction of (an appropriately chosen) lattice in a (added non-trivial irreducible) algebraic representation of \mathrm{SL}(\mathbf{Z}_p), then H^i(X,\mathbf{L}) = 0 for i small enough compared to the weight of \mathbf{L}.

Some version of this is provable in some situations and it may be generally true, but let us ignore this for now. (One explicit example is given by the locally symmetric space for \mathbf{SL}(\mathbf{Z}[\sqrt{-2}]) and taking the cover corresponding to a prime \mathfrak{p} of norm p satisfying certain global conditions.) The point is, this anzatz allows us to start making computations. From the first assumption, one deduces by Lazard that

H^i(X(p),\mathbf{F}_p) = \wedge^i M,

where M is the adjoint representation. But now one has a Hochschild-Serre spectral sequence:

H^i(\mathrm{SL}(\mathbf{F}_p), \mathbf{L} \otimes \wedge^j M) \Rightarrow 0.

The point is now that one can now start to unwind this (even knowing nothing about the differentials) and make some conclusions, for example:

1. H^1(\mathrm{SL}(\mathbf{F}_p),\mathbf{L}) = 0.
2. H^2(\mathrm{SL}(\mathbf{F}_p),\mathbf{L}) = H^0(\mathrm{SL}(\mathbf{F}_p),\mathbf{L} \otimes M).

In particular, the first cohomology always vanishes, and the second cohomology is non-zero only for the adjoint representation where it is one dimensional. (One can see the non-trivial class in H^2 in this case coming from the failure of the tautological representation to lift mod p^2.) Note of course I am not claiming that the first cohomology vanishes for all representations, but only the “algebraic” ones, and even then with p large enough (compared to the weight). Note also that one has to be careful about the choice of lattices, but that is somehow built into the stability — for n fixed, the dual of M is given by trace zero matrices in M_n(\mathbf{F}_p) and so (from the cohomology side) “\mathbf{L} = M” is the correct object to consider rather than its dual since the dual is not stable even in degree zero. But I think you can secretly imagine that p is big enough and the weight small enough so that you can choose n so that all these representations are actually irreducible).

The first question is whether 1 & 2 are known results — I couldn’t find much literature on these sort of questions (they are certainly consistent with the very special cases considered by Evens and Friedlander).

The second question is what about degrees bigger than 2? For H^3 things start getting a little murkier, but it seems possible that H^3 always vanishes. Beyond that (well even before that) I am just guessing. But one might hope to even come up with a guess the the answer which is consistent with the spectral sequence above.

Added: Some of the discussions in the comments below contain some minor inaccuracies, but in back and forth conversations with Will via email he has formulated a pretty convincing conjectural answer to my questions (and also my secret unasked questions). Hopefully I will come back to this post later when these are all proved!

This entry was posted in Mathematics, Uncategorized and tagged , , , , , . Bookmark the permalink.

3 Responses to The stable cohomology of SL(F_p)

  1. Will Sawin says:

    The obvious guess is that H^{2d} (SL(F_p),L)=(L tensor Sym^d M)^{SL(F_p)}. This is because your spectral sequence looks like the Leray spectral sequence for a torus bundle on a base space where the total space has no cohomology and we want to calculate the cohomology of the base, and the natural way to do this is to have the base be the classifying space of the torus.

    In fact it seems that, if the characteristic of F_p were zero, knowing the cohomology of X(p) as a dg-algebra would tell you that it is a torus by rational homotopy theory, and the base would have to really be a classifying space.

    That makes me think it should be possible to calculate the cohomology explicitly by thinking about the spectral sequence, maybe using the algebra structure and the p large assumption at some point. (Maybe we get stuck at a certain point if p is too small…)

    The next thing to compute is E_{2}^{3,0} (where E_2 is the first page of your spectral sequence). This has a differential going to E_2^{1,1}, which you know vanishes, and then E_{3}^{3,0} has a differential going to E_{3}^{0,2}, which is the cokernel of the differential E_{2}^{2,1} => E_2^{0,2}. With your prexisting calculations, we see that this is H^0 (SL(F_p), L tensor M tensor M) => H^0( SL(F_p), L tensor wedge^2 M). The map is obtained from cup product with the map you wrote down, so it comes from the natural map M tensor M => wedge^2 M that arises from cup product in the formula you get from Lazard, which I guess is the obvious such map. If so then this splits in characteristic > 2 at least, and so H^3 vanishes in characteristic greater than two.

    By some big inductive argument maybe you get H^i has a simple formula for i at most p, and the first really interesting cohomology (i.e. not stable in p and n simultaneously) shows up in degree p+1.

    • Brilliant! This is the second version of my comment. In the first, I was going to say that Yes, this was exactly my first guess too (although for more prosaic reasons), and that it seemed to work in degrees 1,2 and 3, but then it went pear shaped in degree 4. And my “reasoning” was that things went wrong when L was the trivial representation, because then Sym^2(M) had SL(F_p) invariants. But then when writing down this comment I realized that I was talking rubbish, because it’s not Sym^2(M) but Wedge^2(M) which has the invariant form. I first did this computation 4 years ago and should probably have checked it again when I wrote this post. Your suggestions look promising!

      • Will Sawin says:

        Thanks! I want to believe that having invariants of the representation won’t kill you even if it happens for higher i, because the cup product structure helps you. What I mean is that one should be able to calculate the differentials for higher j values from the differentials for lower j values and the cup product structure of the cohomology (specifically because the cohomology is generated in degree 1, or at least it looks like it is). But I don’t know how to justify this yet.

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Google photo

You are commenting using your Google account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s