Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|improve this question
2  
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. –  Theo Buehler Mar 3 '11 at 0:00
2  
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". –  Ralph Mar 3 '11 at 0:34
1  
Soule has made some integral calculations for $SL_n(\mathbb Z)$ for $n=3,4$, which is not too far away from $GL_n(\mathbb Z)$. –  Jim Conant Mar 3 '11 at 0:42
2  
@Jim: Soule's paper "The cohomology of $SL_3(\mathbf{Z})$" also contains the integral cohomology of $GL_3(\mathbf{Z}) = SL_3(\mathbf{Z}) \times \mathbf{Z}/2\mathbf{Z}$. –  Ralph Mar 3 '11 at 1:25
1  
@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 –  François Brunault Mar 3 '11 at 9:50
show 1 more comment

2 Answers 2

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.

share|improve this answer
    
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). –  Ralph Jun 24 '11 at 6:40
    
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. –  John Rognes Jun 24 '11 at 19:23
    
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). –  Ralph Jun 24 '11 at 21:42
add comment

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!

share|improve this answer
    
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 –  Hugo Chapdelaine Jun 26 '11 at 16:00
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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