Rank of vector bundle homomorphism is almost constant

Question

Let $M$ be a smooth manifold, $\pi : E \to M$ and $\pi' : E' \to M$ smooth vector bundles and $f : E \to E'$ a smooth bundle map, i.e. $\pi' \circ f = \pi$ and the restriction of $f$ to each fibre is a linear map $f_x : \pi^{-1}(\{x\}) \to \pi'^{-1}(\{x\})$ ($x \in M$).

In my lectures I have stumbled across the following statement which I do not see how to prove:

There is an open dense set $M_{\mathrm{reg}} \subseteq M$ such that for all $x \in M_{\mathrm{reg}}$ the function $x \mapsto \dim \mathrm{im}(f_x)$ is constant.

Do you have any suggestions on how to tackle this problem? Thanks in advance.

EDIT: this statement becomes true if one replaces "constant" with "locally constant". See below for a proof.

Answer 1

I believe that the claim is false:

Let $M$ be the real line, and both $E$ and $E'$ be the trivial line bundle over the real line.

Let $f(x, t) = (x, g(x)t)$, where $$ g(x) = \begin{cases} 0 & x \ge 0\\ \exp(-1/x^2) & x < 0 \end{cases} $$

Then $rank(f_x) = 1$ for $x < 0$, and $rank(f_x) = 0$ for $x \ge 0$.

This same kind of construction can be carried out on the circle if you want $M$ compact, but it's a bit messier.

Answer 2

In fact, I found a proof for the slightly weaker statement

The set $A := \{ x \in M \ | \ \exists O \subseteq M \ \mathrm{open}: x \in O \ \mathrm{and} \ x \mapsto \mathrm{rank}(f_x|_O) \mathrm{\ is \ constant}\}$ is dense and open in $M$. Or, in other words, the rank function is locally constant on a dense and open subset of $M$.

For the sake of completeness, I will provide the proof here. I'm also pretty confident that this was the intended statement. Note: for John Hughes' counter example (thank you) we find $A = \mathbb{R} \setminus \{0\}$ which is clearly open and dense in $\mathbb{R}$.

1. Denote by $r : M \to \mathbb{N}_0$ the rank map $x \mapsto \mathrm{rank}(f_x)$. Claim 1: $r$ is lower-semicontinuous.

   We have to show that $r^{-1}((p, \infty))$ is open for all $p \in \mathbb{R}$. Since $r$ is integer-valued, its maximum value $m \in \mathbb{N}_0$ is attained at some point $s \in M$. Clearly, if $p \geq m$, then $r^{-1}((p, \infty))$ is the empty set and thus open. Also, if $p < 0$, then $r^{-1}((p, \infty))$ is the whole base manifold $M$ and thus open. Let $0 \leq p < m$ and $x_0 \in r^{-1}((p, \infty))$ be arbitrary with $r(x_0) =: k > p$. By the definition of vector bundles it is possible to choose an open set $V \subseteq M$ containing $x_0$ such that $E$ and $E'$ are simultaneously trivialized over $V$. Hence, we identify $\pi^{-1}(V)$ with $V \times \mathbb{R}^a$ and $\pi'^{-1}(V)$ with $V \times \mathbb{R}^b$ where $a,b$ are the ranks of the vector bundles. Furthermore, under this identification, the submersions $\pi, \pi'$ take the form $$ \pi: V \times \mathbb{R}^a \to V, \ (x,v) \mapsto x $$ and $$ \pi' : V \times \mathbb{R}^b \to V, \ (x,w) \mapsto x. $$ That $f$ is assumed to be a bundle homomorphism implies the following: First, from $\pi' \circ f = \pi$ it follows that $f$ takes locally the form $$ f : V \times \mathbb{R}^a \to V \times \mathbb{R}^b, \ (x,v) \mapsto (x, A_x v) $$ with $A_x \in \mathbb{R}^{b \times a}$ being the matrix representation of $f_x$. Second, since $f$ is smooth and thus continuous, the map $x \mapsto A_x$ is continuous on $V$. Now, since $r(x_0) = k > p \geq 0$ there exists a $k \times k$-minor $B_{x_0}$ of $A_{x_0}$ such that $\det(B_{x_0}) > 0$. Using continuity of $x \mapsto \det(B_{x})$ we find an open neighborhood $U \subseteq V$ of $x_0$ such that $\det(B_x) > 0$ for all $x \in U$. But this means that $r(x) \geq k$ for all $x \in U$ and thus $U \subseteq r^{-1}((p, \infty))$.

2. Claim 2: The set $A$ from above is dense and open. It follows directly from the definition of $A$ that $A$ is open. To see that it is dense, choose $x_0 \in M$ together with an open neighborhood $U \subseteq M$ arbitrarily. One needs to prove that $U \cap A \neq \emptyset$. Let $m \in \mathbb{N}_0$ be the maximum value of $r|_U$, so $r(x) \leq m$ for all $x \in U$. It is attained at some point $s \in U$. By 1. there is an open set $\tilde U \subseteq U$ containing $s$ such that $r(x) \geq m$ for all $x \in \tilde U$. So, finally $r(x) = m$ for all $x \in \tilde U$ and thus $s \in U \cap A$.