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)?


1 Answer 1


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 $$\boxed{D_A\det \cdot B = (\det A) \operatorname{tr} (A^{-1} B)} .$$

  • $\begingroup$ 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. $\endgroup$
    – Philip
    Apr 14 at 5:01
  • $\begingroup$ 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 $\det$. $\endgroup$ Apr 14 at 5:04

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