Proving that the general linear group is a differentiable manifold

Question

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. :)

Answer 1

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.

Answer 2

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.