# Proving that the general linear group is a differentiable manifold

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. :)

• What properties must be fulfilled for the GL to be a smooth manifold? Maybe it helps to consider the cases $\det{A}>0$ and $\det{A}<0$? Jul 26, 2012 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. :( Jul 26, 2012 at 13:38

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.

• I have to go for now, but I will add some explanation on smooth structures when I am back. Jul 26, 2012 at 13:53
• Okay, for clarification, is it the standard smooth structure that you're talking about for GL? Jul 26, 2012 at 14:04
• @JohnThompson he's talking about the smooth structure of $M_n(\mathbb{R})$
– Pax
Sep 27, 2013 at 21:04
• @HenryT.Horton Can I ask why is $\mathbb{R}^{n^2}$ a smooth manifold?
– Joe
Feb 17, 2016 at 22:13
• @Maxed Yes, that's it
– Soap
Feb 6, 2019 at 13:48

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.

• Thanks for the hint! Maybe all I need is to sit this one down and focus to get the right proof. :) Jul 26, 2012 at 14:17