$\DeclareMathOperator\SL{SL}\DeclareMathOperator\Sp{Sp}$Suppose that $F/Q$ is a number field.

Using automorphic forms, Borel computed the ($R$-) stable cohomology of $\SL_n(O_F)$, and as a result, computed $K_i(O_F)\otimes Q$. Nowadays, using work of Suslin, Voevodksy, Rost, etc, one has an almost "complete" understanding of the actual integral groups $K_i(Z)$, say, modulo Vandiver's conjecture. This does not directly give the stable cohomology of $\SL_n$, because one set of groups is computing homotopy and the other is computing cohomology, but let us not worry about that distinction for the moment.

Borel also computed the ($R$-) stable cohomology of $\Sp_{2n}(O_F)$. My question, loosely speaking, is whether one can describe the stable integral cohomology of $\Sp_{2n}(Z)$ in as detailed away as the algebraic K-groups of $Z$ "describe" the integral cohomology of $\SL_n(Z)$.

Let me summarize some of what I have found out (following up some of the answers below), mostly though emails from experts.

For affine objects, which certainly includes $Z$, K-theory is about the monoidal category $P(A)$ of projective finitely generated A-modules, and Hermitian K-theory is about the monoidal category $P(A)_h$ of objects in $P(A)$ equipped with a non-degenerate symmetric (or skew-symmetric) form. One of the issues with computing or working with such a theory over $Z$ is that irritating issues arise in characteristic $2$, as one might expect with quadratic forms present. It seems that one might be in good shape to understand the groups $K^h_i(Z[1/2])$. For usual K-theory, there is an excision formula relating $K_i(Z)$ to $K_i(Z[1/2])$ and $K_i(F_2)$. The latter group is "easy" (or at least was computed by Quillen).

Of interest to me in $K_i(Z)$ are the Soulé classes. The next thing I will be thinking about is whether Soule classes can give rise to elements in the stable cohomology of $\Sp_2n(Z)$.

Andy said some very interesting things, but I will probably be awarding the bounty to Oscar, since the paper he linked to was more directly relevant to what I was trying to find out. (Sorry Andy...but it looks like you have lots of reputation anyway!)

  • $\begingroup$ Oh, I think Oscar absolutely deserves it. Congrats Oscar! Is this the first bounty that anyone's won? $\endgroup$ Nov 9, 2009 at 6:35
  • $\begingroup$ Also, shoot me an email if you learn anything interesting about the stable cohomology of Sp_2g(Z). My knowledge of algebraic k-theory is painfully classical -- essentially what is useful for geometric group theory and surgery theory -- but Sp_2g(Z) is one of my favorite groups and I'm always interested in learning new things about it. $\endgroup$ Nov 9, 2009 at 6:38
  • $\begingroup$ There has been progress in this general area, see especially the work of Feng--Galatius--Venkatesh: arxiv.org/abs/2007.15078 They also refer to the important nine-author work on hermitian K-theory (the third part being probably especially relevant). There also has been recent work by Schlichting and Huzney Parvez Sarwar, e.g: homepages.warwick.ac.uk/~masiap/research/H3Sp.pdf $\endgroup$ Jun 30, 2022 at 6:23

3 Answers 3


The algebraic K-groups of Z are to the homology of SLn(Z) as the Hermitian K-groups of Z are to the homology of Sp2g(Z). There is a paper by Berrick and Karoubi here in which they discuss, and make some calculations of, the Hermitian K-theory of Z and Z[1/2].


It is known that $H_2$ of $Sp_{2g}(Z)$ is $Z$ for $g$ at least 4 and $Z+(Z/2Z)$ for $g=3$. The latter calculation plus references for the first calculation can be found in the following paper.

  • MR0372056 (51 #8273) Stein, Michael R. The Schur multipliers of ${\rm Sp}_{6}(Z)$, ${\rm Spin}_{8}(Z)$, ${\rm Spin}_{7}(Z)$, and $F_{4}(Z)$. Math. Ann. 215 (1975), 165–172.

Also, it is easy to see that $Sp_{2g}(Z)$ is perfect for $g$ at least 2, so $H_1(Sp_{2g}(Z);Z)=0$.

One can then calculate $H^2$ and $H^3$ integrally using the universal coefficients theorem together with their rational values.

  • 1
    $\begingroup$ Mike published a bunch of wonderful papers in the 70's working out analogues of the classical algebraic K-groups (ie K_0, K_1, and K_2) for Chevalley groups over general rings. They contain a number of very nice calculations that don't seem to be as well-known as they should be. For instance, a consequence of his work is that H_2 of Sp_{2g}(Z/LZ) is 0 as long as L is not divisible by 4 (I do not know if it is known what happens when L is divisible by 4). For L a prime, this was originally due to Steinberg. $\endgroup$ Nov 4, 2009 at 21:51
  • $\begingroup$ Continuing my previous comment. Anyway, I needed these types of results about a year ago and I consulted a number of experts. None of them knew about Mike's calculations! I ended up finding them after spending untold hours browsing through mathscinet. The paper I referred to above is one of the later papers in this sequence. $\endgroup$ Nov 4, 2009 at 22:09

I don't know what $H_3(Sp_{2g}(Z))$ is, but I want to give a reference for a related fact. Let $Mod_g$ be the mapping class group. The action on $H_1$ of the surface preserves the algebraic intersection form and gives a representation $Mod_g\to Sp_{2g}(Z)$ which is well-known to be surjective. Rationally, the stable cohomology of $Sp_{2g}(Z)$ injects into the stable cohomology of $Mod_{g}$ and gives "one half" of the stable cohomology of $Mod_{g}$.

Anyway, I do know a reference for some stable integral homology calculations for the mapping class group. Namely, in The low-dimensional homotopy of the stable mapping class group, Ebert proves that in a stable range we have $H_3$ of the mapping class group equal to $Z/12Z$ and $H_4$ equal to $Z^2$. He does this by using the fact that Madsen and Weiss's work identifies the infinite loop space $(Mod_{\infty})^+$ with a space whose first few homotopy groups are known. Ebert then builds with his bare hands the first few stages of the Postnikov tower for this space.

Presumably this either has been done or could be done for $Sp_{2g}(Z)$ by a sufficiently motivated homotopy theorist.


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