# Submanifold of matrix space

One can identify the space $$M_{\mathbb{R}}(n,n)$$ of real $$n \times n$$ matrices as $$\mathbb{R}^{n^2}$$. Consider the subset $$S:=\{ A \in M_{\mathbb{R}}(n,n) : det(A)=1 \}$$ and show it is a smooth submanifold of $$\mathbb{R}^{n^2}$$. How is the tanget space defined for elements of $$S$$?

What I have tried so far: So, one way to show that $$S$$ is a submanifold is to use a proposition that states S $$\subset \mathbb{R}^n$$ is a submanifold of dimension $$k$$ iff for each $$x \in S$$, there exists an open set $$U$$ subset $$\mathbb{R}^n$$ with $$x \in U$$ and a smooth map $$\phi: U \rightarrow \mathbb{R}^{n-k}$$ such that $$S \cap U= \phi^{-1}(\{0\})$$ and for each $$y \in S \cap U$$ the map $$D\phi: \mathbb{R}^n \rightarrow \mathbb{R}^{n-k}$$ is surjective.

One thing that sets me of, is topology on $$M_{\mathbb{R}}(n,n)$$ respectively $$S$$. $$S$$ should as a subset of $$M_{\mathbb{R}}(n,n)$$ have the induced topology. My Problem is, that I don't know what the topology on $$M_{\mathbb{R}}(n,n)$$ is. So I can't determine what the induced topology on $$S$$ is.

The only thing that come to mind is, that if $$M_{\mathbb{R}}(n,n)$$ is the "whole" space, then $$M_{\mathbb{R}}(n,n)$$ has to be an open set.

So if I consider $$U=M_{\mathbb{R}}(n,n)$$, then for every $$x \in S$$, $$x$$ is also in $$U$$.

Now, since $$S$$ defined by using $$det(A)=1$$. The idea would be to let $$\phi(A):=det(A)-1$$. Then we get $$S \cap U=\phi^{-1}(\{ 0 \})$$. What is left is, to show that $$\phi: U \rightarrow \mathbb{R}^{1}$$ is smooth. But since the $$det$$ is smooth, we also got that. Since $$\phi: U \rightarrow \mathbb{R}$$ I also get that $$S$$ should have dimension $$n^2 -1$$.

The last part is to show that $$D\phi$$ is surjective. For that, I need to show that $$D\phi$$ as a matrix has rank $$1$$.

From analysis I remember that if $$f: \mathbb{R}^p \rightarrow \mathbb{R}$$, then $$Df=(\frac{\partial f(x)}{\partial x_1} ... \frac{\partial f(x)}{\partial x_p} )$$.

I now need to show that this row vector has rank 1, i.e. is not zero. That's where I am stuck. Is my approach thus far correct? How can I finish this proof (show that $$\phi$$ is surjective)?

So far your approach looks good, though n.b. we only need $$D\phi$$ to be surjective on $$S$$.
Hint First restrict $$\phi$$ to the open subset $$\operatorname{GL}(n, \Bbb R) := M_{\Bbb R}(n, n) \setminus \det^{-1}(0)$$ of invertible matrices---this is no problem for us, as $$S \subset \operatorname{GL}(n, \Bbb R)$$. Since $$\phi$$ and $$\det$$ differ by a constant, $$D\phi = D\det$$.
To show that $$D\det$$ has rank $$1$$, it's enough to show for each $$A \in S$$ that there is some $$B \in T_A \operatorname{GL}(n, \Bbb R)$$ such that $$D_A\det \cdot B \neq 0$$, but it's not much more work just to compute $$D\det$$ explicitly: Use the entries $$A_i^j$$ as global coordinates on $$\operatorname{GL}(n, \Bbb R)$$, expand by minors along the $$ith$$ column, and apply Cramer's Rule to compute $$\frac{\partial}{\partial A_i^j} \det A = (\det A) (A^{-1})_j^i .$$
It follows that $$\boxed{D_A\det \cdot B = (\det A) \operatorname{tr} (A^{-1} B)} .$$
• I haven't really figured out what sets in $M_{\mathbb{R}}(n,n)$ are open. So I don't see why $GL(n,\mathbb{R})$ is an open set. Apr 14 at 5:01
• The space $\operatorname{GL}(n, \Bbb R)$ is (more or less by definition) the preimage of the open set $\Bbb R \setminus \{0\} \subset \Bbb R$ under the continuous map $\det$. Apr 14 at 5:04