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We know that the the general linear group is defined as the set $\{A\in M_n(R): \det A \neq 0\}$. I have a homework on how to prove that it is a smooth manifold. So far my only idea is that we can think of each matrix, say $A$, in that group as an $n^2-$dimensional vector. So i guess that every neighborhood of $A$ is homeomorphic to an open ball in $\mathbb{R}^{n^2}$ (However, i don't know how to prove this.)

Now, I'm asking for help if anyone could give me a hint on how to prove that the general linear group is a smooth manifold since I really don't have an idea on how to do this. (By the way, honestly, I don't really understand what a $C^{\infty}-$smooth structure means which is essential to the definition of a smooth manifold.) Your help will be greatly appreciated. :)

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What properties must be fulfilled for the GL to be a smooth manifold? Maybe it helps to consider the cases $\det{A}>0$ and $\det{A}<0$? – vanguard2k Jul 26 '12 at 13:30
I'm not sure with this but GL must have a smooth structure, that is, it must have a maximal atlas. Hmmm, for those cases, I still don't have an idea on what to do with them. :( – John Thompson Jul 26 '12 at 13:38
up vote 8 down vote accepted

Construct a map $f:M_n(\mathbb{R}) \rightarrow \mathbb{R}$ by taking each matrix to its determinant, where $M_n(\mathbb{R})$ is the set of all $n \times n$ matrices. $f^{-1}(\mathbb{R}\backslash\{0\})=GL_n(\mathbb{R})$, and $\mathbb{R}\backslash\{0\}$ is an open subset of $\mathbb{R}$. Therefore, $GL_n(\mathbb{R})$ is an open subset of $M_n(\mathbb{R})$. I'll leave the rest to you.

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Thanks for the hint! Maybe all I need is to sit this one down and focus to get the right proof. :) – John Thompson Jul 26 '12 at 14:17

Here are some hints:

  • $\mathbb{R}^{n^2}$ is trivially a smooth manifold.

  • The determinant map $$\det: \mathbb{R}^{n^2} \longrightarrow \mathbb{R},$$ which we define by considering elements of $\mathbb{R}^{n^2}$ as $n \times n$ matrices, is continuous (it is a polynomial in the entries of the matrix). Then $$\mathrm{GL}(n; \mathbb{R}) = \mathrm{det}^{-1}(\mathbb{R}\setminus\{0\})$$ is an open subset of $\mathbb{R}^{n^2}$.

  • Now show that an open subset of a smooth manifold is itself a smooth manifold with the obvious smooth structure.

If you need more clarification, let me know.

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I have to go for now, but I will add some explanation on smooth structures when I am back. – Henry T. Horton Jul 26 '12 at 13:53
Okay, for clarification, is it the standard smooth structure that you're talking about for GL? – John Thompson Jul 26 '12 at 14:04
@JohnThompson he's talking about the smooth structure of $M_n(\mathbb{R})$ – Pax Kivimae Sep 27 '13 at 21:04
@HenryT.Horton Can I ask why is $\mathbb{R}^{n^2}$ a smooth manifold? – Joe Feb 17 at 22:13

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