most recent 30 from http://mathoverflow.net 2013-05-19T07:51:53Z http://mathoverflow.net/feeds/question/57183 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/57183/how-to-compute-the-cohomology-of-the-general-linear-group-with-integral-entries Hugo Chapdelaine 2011-03-02T23:38:50Z 2011-06-25T23:55:38Z

<p>Q: So how does one compute the cohomology groups $H^*(GL_n(\mathbf{Z}),\mathbf{Z})$?</p> <p>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.</p> <p>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 $G=GL_n(\mathbf{Z})$, $E=\mathbf{R}^n-0$.</p> <p>By the way, does $E/G$ have a geometrical description?</p>

http://mathoverflow.net/questions/57183/how-to-compute-the-cohomology-of-the-general-linear-group-with-integral-entries/68660#68660 John Rognes 2011-06-23T21:34:10Z 2011-06-23T21:34:10Z

<p>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 $GL(Z) = GL_\infty(Z)$.</p> <p>The Serre spectral sequence implies that there is little difference between the case of $GL(Z)$ and $SL(Z)$.</p> <p>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). </p> <p>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.)</p> <p>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.</p>

http://mathoverflow.net/questions/57183/how-to-compute-the-cohomology-of-the-general-linear-group-with-integral-entries/68669#68669 Agol 2011-06-23T22:41:45Z 2011-06-23T23:03:36Z

<p>The quotient $E/G$ is non-Hausdorff, I'm not sure there will be a nice geometric description. </p> <p>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! </p>