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Q: So how does one compute the cohomology groups $H^*(GL_n(\mathbf{Z}),\mathbf{Z})$?

First note that $H^*(GL_n(\mathbf{Z}),\mathbf{Z})$ is isomorphic to $H_B^*(Y/GL_n(\mathbf{Z}),\mathbf{Z})$ (Betti cohomology) where $Y$ is any contractible space on which $GL_n(\mathbf{Z})$ acts freely. Maybe one should first ask to compute the cohomology with rational coefficients and then deal with the torsion separately.

Secondly, note that $GL_n(\mathbf{Z})$ acts on $\mathbf{R}^n-\{0\}$. Unfortunately it does not act discontinuously on $\mathbf{R}^n-\{0\}$ so its quotient by $GL_n(\mathbf{Z})$ will be quite messy. Nevertheless it might be possible to use some version of the Leray spectral sequence on $$ G\rightarrow E\rightarrow E/G $$ where $G=GL_n(\mathbf{Z})$, $E=\mathbf{R}^n-0$.

By the way, does $E/G$ have a geometrical description?

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    $\begingroup$ A better candidate for $E$ would certainly be the symmetric space associated to $\operatorname{GL}_{n}(\mathbb{R})$, i.e., the symmetric positive definite matrices. Serre has made extensive calculations of the cohomology of discrete subgroups of Lie groups (e.g. here springerlink.com/content/0171m21753248642), but I think mostly with real coefficients. $\endgroup$
    – Theo Buehler
    Commented Mar 3, 2011 at 0:00
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    $\begingroup$ To my knowledge not much is known for general n. There are some results by Ash (see his homepage: www2.bc.edu/~ashav). You also may have a look at the book "Knudson: Homology of Linear Groups". The stable rank ($n = \infty$) has been computed by Borel in the paper "Stable real cohomology of arithmetic groups". $\endgroup$
    – Ralph
    Commented Mar 3, 2011 at 0:34
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    $\begingroup$ Soule has made some integral calculations for $SL_n(\mathbb Z)$ for $n=3,4$, which is not too far away from $GL_n(\mathbb Z)$. $\endgroup$
    – Jim Conant
    Commented Mar 3, 2011 at 0:42
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    $\begingroup$ @Jim: Soule's paper "The cohomology of $SL_3(\mathbf{Z})$" also contains the integral cohomology of $GL_3(\mathbf{Z}) = SL_3(\mathbf{Z}) \times \mathbf{Z}/2\mathbf{Z}$. $\endgroup$
    – Ralph
    Commented Mar 3, 2011 at 1:25
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    $\begingroup$ @Hugo : See also P. Elbaz-Vincent, H. Gangl et C. Soulé, "Quelques calculs de la cohomologie de GL_N(Z) et de la K-theorie de Z" (in French), math.uiuc.edu/K-theory/0581 $\endgroup$
    – François Brunault
    Commented Mar 3, 2011 at 9:50

2 Answers 2

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There are homological stability results (due to Ruth Charney and Hendrik Maazen around 1979, if I recall correctly) saying that $H_*(GL_n(Z); Z) \to H_*(GL_{n+1}(Z); Z)$ is about $n/2$-connected. So in a range of degrees increasing to infinity with n you might just ask about the (co-)homology of $GL(Z) = GL_\infty(Z)$.

The Serre spectral sequence implies that there is little difference between the case of $GL(Z)$ and $SL(Z)$.

For the rational result, Armand Borel computed $H^*(SL(Z); Q)$ in his paper (MR0387496) "Stable real cohomology of arithmetic groups", in Ann. Sci. \'Ecole Norm. Sup. (1974).

For integral results, Bill Dwyer and Steve Mitchell compute $H^*(GL(Z); Z)$ in their paper (MR1633505) "On the $K$-theory spectrum of a ring of algebraic integers", in $K$-Theory 14 (1998). See 1.5 and section 10 of their paper. They assume the now proven Lichtenbaum--Quillen conjecture (Voevodsky for $p=2$, Rost, Voevodsky, Weibel? for $p$ odd.)

In both cases the results are more general, and suffice to compute the cohomology of $GL(R)$ and the (rational) algebraic K-theory of R for R any ring of integers in a number field.

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  • $\begingroup$ Did Dwyer-Mitchell really consider integral coefficients ? For, if my understanding of the topic is right, there is a close connection between large torsion in the integral (co)homology of $GL(\mathbb{Z})$ and $K_*(\mathbb{Z})$ and the latter is related to Vandiver's conjecture on irregular primes. (I think there is also a paper of Soule´ that estimates such torsion). $\endgroup$
    – Ralph
    Commented Jun 24, 2011 at 6:40
  • $\begingroup$ No, you are right, they work with $Z/\ell$-coefficients. The answer for $R = O_F[1/\ell]$, with $F$ a number field, involves a matrix of maps $BU \to BU$ determined by the Iwasawa module of $F$, and this is how the Bernoulli numbers enter for $F = Q$. There was a 1997 Univ. of Washington Ph.D. thesis "Torsion in the Homology of the General Linear Group for a Ring of Algebraic Integers" by Prashanth Adhikari (probably supervised by Mitchell) that elaborated on this. I'm not sure that it was published. $\endgroup$
    – John Rognes
    Commented Jun 24, 2011 at 19:23
  • $\begingroup$ John, thanks for clarification. The paper of Soule´ I meant is arxiv.org/pdf/math/9812171v1. (the result has been generalized by Soule´ to the rings of integers of arbitrary number fiedls in math.uiuc.edu/K-theory/0603/cdn1.pdf). $\endgroup$
    – Ralph
    Commented Jun 24, 2011 at 21:42
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The quotient $E/G$ is non-Hausdorff, I'm not sure there will be a nice geometric description.

There's a standard way to get $Y$. The symmetric space for $GL(n,\mathbb{R})$ is the symmetric space $Q$ of positive definite symmetric matrices of determinant $>0$, isomorphic to $GL(n,\mathbb{R})/O(n,\mathbb{R})$. Then $GL(n, \mathbb{Z})$ acts discretely on this space, but torsion elements have fixed points. Also, the torsion elements of $GL(n,\mathbb{Z})$ map non-trivially to $GL(n,\mathbb{Z}/p)$ for some prime $p$. One may take a $K(GL(n,\mathbb{Z}/p),1)=X$, then $GL(n,\mathbb{Z}/p)$ and therefore $GL(n,\mathbb{Z})$ acts on the universal cover $\tilde{X}$. Now, take the diagonal action of $GL(n,\mathbb{Z})$ on $Q\times \tilde{X}$. This action is free and discrete. Of course, this assumes that you have a nice way to construct $X$, which must be infinite dimensional!

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  • $\begingroup$ Well $K(GL(n,Z/p),1)=X$ will be a CW-complex of infinite dimension and the only way I know to construct it is with the usual killing cells technique which is kind of tautological $\endgroup$
    – Hugo Chapdelaine
    Commented Jun 26, 2011 at 16:00

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