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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 "almost freely" on $\mathbf{R}^n-\{0\}$ (\mathbf{R}^n-\{0\}$. Unfortunately it acts does not act discontinuously and if one passes to a suitable subgroup it acts freely). Unfortunately (or fortunately!) on$\mathbf{R}^n-\{0\}$is not contractible since it is homotopic to so its quotient by$S^{n-1}$. GL_n(\mathbf{Z})$ will be quite messy. Nevertheless it might be possible to use some version of the Leray spectral sequence for the "almost fibration" 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 nice geometrical description?

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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 "almost freely" on $\mathbf{R}^n-\{0\}$ (it acts discontinuously and if one passes to a suitable subgroup it acts freely). Unfortunately (or fortunately!) $\mathbf{R}^n-\{0\}$ is not contractible since it is homotopic to $S^{n-1}$. Nevertheless it might be possible to use the Leray spectral sequence for the "almost fibration" $$G\rightarrow E\rightarrow E/G$$ where $G=GL_n(\mathbf{Z})$, $E=\mathbf{R}^n-0$.

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

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