3
$\begingroup$

Let $M$ be an smooth manifold manifold of dimension $m$ and let $N$ be a smooth manifold of dimension $n$, and let $F:M\rightarrow N$ be a smooth map. Then the set $W=\left\{ p\in M:F\mbox{ has full rank at }p\right\}$ is open.

Here is my attempt at the proof.

Suppose without loss of generality that $m<n$. Let $\left(U,\varphi\right)$ be a chart at $p$ and let $\left(V,\psi\right)$ be a chart at $F\left(p\right)$ such that $F\left(U\right)\subseteq V$. By identifying linear maps with matrices (with respect to the standard bases for $\mathbb{R}^{n}$ and $\mathbb{R}^{m}$), the map $G:\varphi\left(U\right)\rightarrow M\left(n\times m,\mathbb{R}\right)$ defined by $x\longmapsto d\left(\psi\circ F\circ\varphi^{-1}\right)_{x}$ is continuous. For each $n\times m$ matrix $A$, let $m_A$ be the set of all invertible $m\times m$ submatrices of $A$. It is a standard result in linear algebra that $A$ has full rank if and only if $m_A$ is nonempty. The map $H:M(n\times m,\mathbb{R})\rightarrow\mathbb{R}$ defined by $A\longmapsto\sum_{S\in m_{A}}\left|\mbox{det}S\right|$ is continuous. It follows that the composition $H\circ G\circ\varphi$ is continuous. Hence, $\left(H\circ G\circ\varphi\right)^{-1}\left(\mathbb{R}-\left\{ 0\right\} \right)$ is an open subset of $U$. Taking the union of all such sets will result in $W$. Therefore $W$ is open.

Is it correct?

$\endgroup$
  • $\begingroup$ This looks like the same basic idea as this proof. $\endgroup$ – Michael Lee Jun 16 '17 at 0:15
4
$\begingroup$

The proof is correct, although I would suggest doing a few things slightly differently. Set $k = \min(m,n)$ and let $p \in W$. Choose a coordinate system around $p$ and consider your map $G$. At $\varphi(p)$, the matrix $G(\varphi(p))$ has rank $k$. Hence, there exists some $k \times k$ submatrix whose determinant is non-zero. Denote by $H \colon M_{n \times m}(\mathbb{R}) \rightarrow \mathbb{R}$ the function which returns the determinant of the specific $k \times k$ submatrix which works for $G(\varphi(p))$. Then $H$ is continuous and $(H \circ G \circ \varphi)^{-1}(\mathbb{R} \setminus \{ 0 \})$ is an open neighborhood around $p$ in which $\operatorname{rank} G = k$ is full.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.