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Problem 4-3 in J.M. Lee's introductory text about smooth manifolds, asks to formulate and prove a version of the constant rank theorem for a map of constant rank whose domain is a smooth manifold with boundary. That is, show that,

If $F:M\rightarrow N$ is smooth, $N$ with empty boundary, $F$ of constant rank $r$, then, for every point $p$ in $M$, $F$ has a local representation of the form $\tilde{F}(x)=(x_1, ...,x_r,0,...,0)$

Lee gives a hint: After extending $F$ (the interesting case is when $p\in\partial M$), follow the proof of the regular constant rank theorem, until you have to make use of the constant rank hypothesis. The problem may be that the extension has higher rank. Lee's hint is to modify the map so that it has constant rank.

I don't see how to do this.

(If it's a silly question, I'm sorry, I haven't slept in over 24hs.)

Bill
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1 Answers1

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Since this question has remained open for a while, I thought I should think more carefully about what's involved. @Ted Shifrin's suggestion is the right way to start, but after thinking about it for a while, I decided that the problem really needs a better hint than the one I put in the book. So here's a (hopefully) better suggestion to get you started. (I've also added this to my list of corrections on my ISM web page.)

First, to get a good result, you'll have to add the assumption that $\ker dF_p\not\subseteq T_p\partial M$. After choosing smooth coordinates, you can assume $M \subseteq \mathbb H^m$ and $N\subseteq\mathbb R^n$, and extend $F$ to a smooth function $\widetilde F$ on an open subset of $\mathbb R^m$.

Now, as Ted suggested, assuming that $F$ has constant rank $r$, show that there is a coordinate projection $\pi\colon\mathbb R^n\to\mathbb R^r$ such that $\pi\circ \widetilde F$ is a submersion, and apply the rank theorem to $\pi\circ \widetilde F$ to find new coordinates in which $\widetilde F$ has a coordinate representation of the form $(x,y) \mapsto (x,R(x,y))$. Then use the rank condition to show that $R|_M$ is independent of $y$.

Jack Lee
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    I can't see how the end of the proof works here, namely when you get back a boundary chart by using the original coordinates on $M$ and set up the chart on $N$ accordingly. I think this is where the assumption $\ker dF_p\not\subseteq T_p\partial M$ should be used, having in mind the projection $F:\mathbb{R}\times[0,+\infty)\to\mathbb{R}$, $F(x,y)=y$, for which the part of the proof you described works without the assumption, but for which the rank theorem does not apply. Sorry for the late question, I will ask this as a separate question if you think it is better. – Balloon Aug 08 '21 at 14:32