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Hugo Chapdelaine
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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!) it does not act discontinuously on $\mathbf{R}^n-\{0\}$ is not contractible since it is homotopic toso 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?

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?

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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Hugo Chapdelaine
  • 7.6k
  • 2
  • 28
  • 70

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 fibration"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?

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 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?

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?

Source Link
Hugo Chapdelaine
  • 7.6k
  • 2
  • 28
  • 70
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