Some questions about the proof of the General Linear Group being a manifold.

Question

I understand the idea behind proving that GL(n,$\mathbb{R}$) is a smooth manifold by first using the fact that it is isomorphic to $\mathbb{R}^{n^{2}}$ and using the continuity of the determinant function to show that it is an open subset of a manifold. However, I have a couple of questions about details.

First, the author mentions that because $\mathbb R^{m\times n}\simeq \mathbb R^{mn}$ we can give $\mathbb R^{m\times n}$ the same topology. What exactly will this topology look like? What will be an open set in $\mathbb R^{m\times n}$ and what would the metric look like? How could we measure "distance" between two matrices?

Second, why is the determinant function a continuous function? I guess I may be able to prove this once I understand what an open set in $\mathbb R^{m\times n}$ is. If somebody can help me or direct me to a source where this question is answered I would appreciate it. Thank you.

Answer 1

The space of $m\times n$ matrices with real entries can be naturally identified with $\Bbb R^{mn}$ because you can assign an arbitrary $n\times m-$matrix $A$ the $mn$-tuple $$(A_{11},A_{12},\ldots,A_{1n},A_{21},A_{22},\ldots A_{2n},\ldots,A_{m1},\ldots,A_{mn}),$$ so you can endow $\Bbb R^{m\times n}$ with a metric space structure by setting the distance function $$d(A,B)=\left(\sum_{i,j}(A_{ij}-B_{ij})^2\right)^{1/2}.$$

The determinant function is a polynomial, as has already been mentioned, namely, $$\det(A)=\sum_{\tau\in S_n}\DeclareMathOperator{\sign}{sign} \sign(\tau)\prod_{i=1}^nA_{1,\tau(i)}$$ where $S_n$ is the symmetric group on $\{1,\ldots,n\}$. This implies $\det$ is continuous.

Answer 2

I think this should be a valid argument * for why $Gl(n; \mathbb R)$ is a manifold: the determinant is a continuous function of the entries of a matrix, and in particular, a smooth map from $M_n \mathbb R \rightarrow \mathbb R$. Then $Gl(n; \mathbb R)$ is the inverse image of the open set $\mathbb R- 0$, so that it is an open subset of $M_n \mathbb R$, and an open subset of a manifold is itself a manifold. As Grumpy and I commented, the determinant is continuous in $M_n \mathbb R$, because it is a linear function of the coefficients of a matrix.

Re $\mathbb R^{mn}$ and $\mathbb R{m \times n}$, while I don't know the textbook you are using, I believe the topology is the product topology $(\mathbb R^m \times...\mathbb R^m)$, which is the product topology. In this case, an $n \times m$ matrix is just the element of $\mathbb R^{mn}$ given by $(a_{11},a_{12},..,a_{1n};a_{21}...,a_{2n};...a_{mn})$.