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How could I prove that $\mathrm{GL}(m, \mathbb{R})$ is a differential manifold of dimension $m^2$?

$\mathrm{GL}(m, \mathbb{R})$ is the set of all non-singular $m\times m$ matrices in $\mathbb{R}$

Thank you

Thomas Andrews
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Cherax
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1 Answers1

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It can clearly be represented to be the complement in $\mathbb R^{m^2}$ of the zeros of the "determinant" polynomial. So it is an open subset of $\mathbb R^{m^2}$, and thus can inherit the trivial differential structure on $\mathbb R^{m^2}$.

This space is obviously not connected, since there is no path from the matrices with positive determine to the matrices with negative determinants.

Thomas Andrews
  • 177,126
  • I'm not sure to understand what you said.

    As hint I have that if I consider the map $$\Phi: gl(m,\mathbb{R})\rightarrow Gl(m,\mathbb{R}) : A \rightarrow e^A = \sum_{0}^{\infty} A^n/n!$$

    then $\Phi$ is a diffeomorphism of a neighborhood of $0\in gl(m,\mathbb{R})$ onto a neighborhood of $I\in Gl(m,\mathbb{R})$.

    Using group multiplication to move neighborhoods around I should have the answer, but don't know how.

    Thanks

     – Cherax May 18 '15 at 20:27