0

Already seen a bunch of questions about proving that $(GL_{n}(\mathbb{R}),||-||)$ is a manifold, with the norm $||-||:M_{n}(\mathbb{R})\longrightarrow \mathbb{R}$ with $||(a_{i,j})||=\sqrt{A^tA}=\sqrt{\sum_{j=1}^{n}(\sum_{i=1}^{n}a_{i,j}^{2})}$, ($M_{n}(\mathbb{R})\cong\mathbb{R}^{n^2}$). None of them give any hints on how to prove it is a topological space ANII (or second countable) and that it is locally Euclidean.

For the first property, i have to prove that there exists a base which is countable, i.e. there exists a countable set $\{\mathcal{U}_{i}\}_{i\in I}$, $\mathcal{U}_{i}\subset GL_{n}(\mathbb{R})$ open, such that every open set in $GL_{n}(\mathbb{R})$ is written as $\mathcal{O}\subset GL_{n}(\mathbb{R}), \mathcal{O}=\cup_{i\in I}\mathcal{U}_{i}$. Is that true?

Also, i have no idea on how to see it is locally Euclidean.

I supose we have to use at any point the norm defined previously, although I have no idea how.

Any help will work, thanks.

HornyPigeon54
  • 479

2 Answers2

1

$GL_n(\Bbb R)$ is an open subset of $M_n(\Bbb R)\cong \Bbb R^{n^2}$, namely the inverse image of $\Bbb R\setminus\{0\}$ under the continuous $\det$ map.

Hagen von Eitzen
  • 374,180
0

Hint: Think of each element of $GL_n$, an $n \times n$ matrix, as a list of $n^2$ numbers (read left-to-right, top-to-bottom, for instance). Now you can see that $GL_n$ is just a (very large) subset of $X = \Bbb R^{n^2}$. That helps a lot with proving things. Because you can take "all balls in $X$ whose centers have rational-number coordinates, and whose radii are rational, and which contain no matrices of determinant zero" as your collection of open sets to use.

John Hughes
  • 93,729
  • BTW, the approach described by @Hagenvoneitzen is the one almost universally used in practice, but I have the sense you wanted to see how one might do things without invoking any more powerful machinery, so I suggested that collection of open sets. – John Hughes Jan 01 '21 at 15:00