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In John M. Lee's Introduction to Smooth Manifolds, there is a comment regarding the difficulty of extending the Constant Rank Theorem to the case where the codomain is a smooth manifold with nonempty boundary:

On the other hand, the situation is considerably more complicated for a map whose codomain is a manifold with boundary: since the image of the map could intersect the boundary in unpredictable ways, there is no way to put such a map into a simple canonical form without strong restrictions on the map.

I am curious about what can be said in this case. I can think of a smooth immersion whose image intersects the boundary uncountably many times, for example the map $F: \mathbb{R} \to \mathbb{H}^2$ given by

$$F(t) = \begin{cases}\left(t, e^{-1/t}\right) && t > 0\\ \left(t, 0\right) && t \leq 0\end{cases}$$

However, I don't know if this example violates the conclusions of the Constant Rank Theorem. I originally thought that it didn't, because the smooth charts $\left(\mathbb{R}, \text{id}\right)$ and $\left(\mathbb{H}^2, \psi\right)$ where $$\psi(x, y) = \begin{cases}\left(x, y - e^{-1/x}\right) && x > 0 \\ (x, y) && x \leq 0\end{cases}$$

allow us to represent $F$ in coordinates as $\hat{F}(u) = \left(\psi \circ F \circ \text{id}^{-1}\right)(u) = (u, 0)$ for all $u \in \mathbb{R}$.

But then I realized that $\left(\mathbb{H}^2, \psi\right)$ is not actually a smooth boundary chart, since $(1, 0)$ is mapped to $\left(1, -\frac{1}{e}\right)$ which is not even in $\mathbb{H}^2$.

So I do not know if there are smooth charts with respect to which the coordinate representation of $F$ does have the form $\hat{F}(u) = (u, 0)$, or whether possible counterexamples to the Constant Rank Theorem when applied to codomains with nonempty boundary would require more pathological functions.

Note that this question is similar but not a duplicate because that question is asking for the Constant Rank Theorem with the domain being a smooth manifold with boundary, while my question is about the codomain being a smooth manifold with boundary.

Tob Ernack
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1 Answers1

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Let $K \subset \mathbb{R}$ be the Cantor set. There is a nonnegative smooth function $f \colon \mathbb{R} \to \mathbb{R}$ such that $K = f^{-1}(0)$. Consider the following parametrization $F \colon \mathbb{R} \to \mathbb{H}^2$ of the graph of $f$ : $F(t) = (t, f(t))$, which is a smooth immersion. If $t \in K$ and $(U, \varphi)$, $(V, \psi)$, are charts such that $t \in U$ and $F(U) \subset V$, then the map $\hat{F} = \psi \circ F \circ \varphi^{-1}$ must send infinitely many points to $\partial \mathbb{H}^2$ and infinitely many points to $\operatorname{Int} \mathbb{H}^2$. Therefore, it cannot have the form $x \mapsto (x, 0)$.

vizietto
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