Constant Rank Theorem for Manifolds with Boundary

Question

I'm trying to answer problem 4-3 from Lee's Introduction to Smooth Manifolds, 2nd edition. The problem says:

Formulate and prove a version of the rank theorem for a map of constant rank whose domain is a smooth manifold with boundary.

Lee himself gave a hint in another question. Here is what I have so far:

I'm assuming I have a smooth map of constant rank $F: \mathbb{H}^m\rightarrow\mathbb{R}^n$. Let's say $\mathrm{rank}(F)=k$. Then I extend $F$ to a smooth map $\tilde{F}:\mathbb{R}^m\rightarrow\mathbb{R}^n$. I can shrink the domain to a small enough neighborhood of $\mathbf{0}$, $U$, such that there is a projection $\pi:\mathbb{R}^n\rightarrow\mathbb{R}^k$ with $\pi\tilde{F}_{|U}$ a submersion. The actual constant-rank theorem then says we have charts $(A,\alpha)$ and $(B,\beta)$ with $$ \beta\circ\pi\tilde{F}\circ\alpha^{-1}:\alpha(A)\rightarrow\beta(B)$$

If I write the coordinates of $\mathbb{H}^m$ as $(x,y)$ and the coordinates of of $\alpha(A)$ as $(a,t)$, then this map looks like $$ \beta\circ\pi\tilde{F}\circ\alpha^{-1}(a,t) = a$$

I can switch the order so the projection is the last map applied: $$ \pi\circ(\beta\times\mathrm{id})\circ\tilde{F}\circ\alpha^{-1}\equiv\beta\circ\pi\tilde{F}\circ\alpha^{-1}$$

Which finally means $$(\beta\times\mathrm{id})\circ\tilde{F}\circ\alpha^{-1}(a,t)=(a,S(a,t))$$ Where $S:\alpha(A)\rightarrow\mathbb{R}^{(n-k)}$.

This is pretty similar to how the constant-rank theorem is proved, but now I'm stuck. The map above, restricted to $\alpha(A\cap\mathbb{H}^m)$, has rank $k$. But that does not mean $S$ is independent of $y$. That's because $\alpha$ is not necessarily a boundary chart on $A\cap\mathbb{H}^m$. I also don't see any way to make $\alpha$ a boundary chart (something similar is proved in Lee's book, but crucially uses that $F$ is an immersion).

Can someone give me a hint as to how to finish this? Ideally I'd get to a place where (with possibly different charts) $$ \tilde{\beta}\circ F\circ\tilde{\alpha}^{-1}(a,t) = (a,0)$$ and $\tilde{\alpha}$ is a boundary chart.

Bonus question: where do I use Lee's assumption (from the hint) that $\ker dF_p\not\subseteq T_p\partial\mathbb{H}^m$?

Answer 1

The result can be proved by following the hints given by Lee and carefully studying the proof of the Rank Theorem for the boundaryless case.

First, by considering local coordinates, we may assume that $M$ is an open subset of $\mathbb{H}^{m}$ containing $p = 0$ and $N$ is an open subset of $\mathbb{R}^{n}$ containing $F(p) = 0$. Now extend $F$ to a map $\widetilde{F}$ taking values in $\mathbb{R}^{n}$ defined on an open subset $\widetilde{M}$ of $\mathbb{R}^{m}$. We can ensure that $\widetilde{F}$ maps into $N$ by shrinking $\widetilde{M}$ if necessary.

Note that $\text{d}F_{0} = \text{d}\widetilde{F}_{0}$. If $\widetilde{F}(x, y) = (Q(x, y), R(x, y))$ then we can ensure that the matrix $(\partial Q_{i}/\partial x_{j})$ at $(0, 0)$ has rank $r$ by possibly permuting coordinates. The condition $\text{ker}(\text{d}F_{p})\not\subseteq T_{p}(\partial M)$ ensures that this can be done by permuting only the first $m-1$ coordinates in the domain (so that we're still working with boundary charts).

Define $\varphi : \widetilde{M}\rightarrow\mathbb{R}^{m}$ by $\varphi(x, y) = (Q(x, y), y)$. Since matrix $D\varphi(0, 0)$ is non-singular, we may use the Inverse Function Theorem to obtain connected neighborhoods $U_{0}$ of $(0, 0)$ and $\widetilde{U}_{0}$ of $\varphi(0, 0)$ such that $\varphi : U_{0}\rightarrow \widetilde{U}_{0}$ is a diffeomorphism. By shrinking $U_{0}$ and $\widetilde{U}_{0}$ if necessary, we may assume that $\widetilde{U}_{0}$ is an open cube. An important point to note here is that the definition of $\varphi$ ensures that it restricts to a diffeomorphism from $U_{0}\cap\mathbb{H}^{m}$ onto $\widetilde{U}_{0}\cap\mathbb{H}^{m}$. In other words, $\varphi|_{U_{0}\cap\mathbb{H}^{m}}$ is a boundary chart for $M$.

It can be shown that $F\circ(\varphi^{-1}|_{\widetilde{U}_{0}\cap\mathbb{H}^{m}})(x, y) = (x, R(x, y))$ for some smooth function $R : \widetilde{U}_{0}\cap\mathbb{H}^{m}\rightarrow \mathbb{R}^{n-r}$. Using the constant rank condition for $F$ (not $\widetilde{F}$!) and the fact that $\widetilde{U}_{0}$ is a cube, we conclude that $R$ is independent of $y$. The proof can now be completed by suitable choosing a coordinate chart in some neighborhood of $F(p)$ in $N$.

Answer 2

Lee's Introduction to Smooth Manifolds deals with the case of local immersions for manifolds with boundary in Theorem 4.15. So let us suppose that $F:\mathbb{H}^m \rightarrow \mathbb{R}^n$ has $\mathrm{rank}(F)=k<m$. For $\mathbb{H}^m=\{(x^1,\dots,x^m)\in \mathbb{R}^m, x^m\geq 0\}$, the assumption $\ker dF_p\not\subseteq T_p\partial\mathbb{H}^m$ in Lee's answer to the original question means that $dF_p(\partial/\partial x^m+a_1\partial/\partial x^1+\dots+a_{m-1}\partial/\partial x^{m-1})=0$ for some numbers $a_i$. The search for $k$ linearly independent tangent vectors in the image can therefore be restricted to $dF_p(\partial/\partial x^i), i<m$. Let's suppose that $k$ is the rank of $(\partial F_i/\partial x^j)_p, 1\leq i, j \leq k$. The coordinate change $(x^1,\dots,x^m) \rightarrow (F_1,\dots,F_k,x^{k+1},\dots,x^m)$ produces another boundary chart $x^m\geq 0$ for which the rest of the proof works as in the case of manifolds without a boundary.