24

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. :)

John Thompson
  • 1,067
  • 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

2 Answers2

30

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.

Henry T. Horton
  • 18,318
  • 5
  • 60
  • 77
  • 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 Sep 27 '13 at 21:04
  • @HenryT.Horton Can I ask why is $\mathbb{R}^{n^2}$ a smooth manifold? – Joe Feb 17 '16 at 22:13
  • @Joe It is Hausdorff and second countable (known facts about topological spaces of the type $\mathbb{R}^{m}$). Also, each point has an open neighborhood $U$ homeomorphic to $\mathbb{R}^{n^{2}}$: just take $U=\mathbb{R}^{n^{2}}$ and the identity map as the homeomorphism. Hence, it is a topological manifold. Notice that this homeomorphism covers the whole $U=\mathbb{R}^{n^{2}}$, so an atlas would be $(id, \mathbb{R}^{n^{2}})$. The changes of basis are vacuously smooth. Done. – soap Feb 21 '17 at 19:00
  • @Soap and the $nxn$ matrices can be put in correspondence with $\mathbb{R}^{n^{2}}$ since we think about them as elements of a vector space? i.e just scalar multiplication and adding components? the matrix is thus essential all one long vector. – user123124 Feb 05 '19 at 18:49
  • 1
    @Maxed Yes, that's it – soap Feb 06 '19 at 13:48
22

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.

Aru Ray
  • 1,712