A (Failed) Attempt to Generalize Rank Theorems

Lee Smooth Manifolds

I… hate this problem… Spent two weeks to realize I the main theorem problem statement below just doesn’t work. The boundary coordinate has to be non-critical. Bye, nothing here of value to see. -Brandon on June 28th 2026, 12:34 AM from Coda room 46

We aim to prove the

Main Theorem: Let $M$ be a $m$-dimensional smooth manifold with boundary and $N$ be a $n$-dimensional smooth manifold without boundary. Let $F:M\to N$ is a smooth map with constant rank $r$. For each $p\in M$ there exist smooth charts $(U,\varphi )$ for $M$ centered at $p$ and $(V,\psi )$ for $N$ centered at $F(p)$ such that $F(U)\subseteq V$, in which $F$ has a coordinate representation of the form

$$\begin{array}{c}\hat{F}(x_1,x_2,\dots ,x_r,x_{r+1},\dots ,x_m)=(x_1,x_2,\dots ,x_r,0,\dots ,0)\end{array}\text{\hspace{1em}(∼)}$$

We first consider the case where $\partial M=\varnothing$. We can simplify the problem one level by treating $M$ and $N$ as open subsets of ${\mathbb{R}}^m$ and ${\mathbb{R}}^n$, respectively, and finding a coordinate representation resembling (~) at an arbitrary point $p\in M$. As $F$ has rank $r$ at $p$, WLOG assume the upper left $r\times r$ submatrix of $DF(p)$ is nonsingular. For simplicity we relabel ${\mathbb{R}}^m$ as $(x,y)$ where $x\in {\mathbb{R}}^r$ and $y\in {\mathbb{R}}^{m-r}$, and $F$ as $F(x,y)=(Q(x,y),R(x,y))$ where $Q:{\mathbb{R}}^m\to {\mathbb{R}}^r$ and $R:{\mathbb{R}}^m\to {\mathbb{R}}^{n-r}$. WLOG, assume $p=(0,0)$ and $F(p)=(0,0)$.

As in the implicit function theorem, we consider the function $\sigma (x,y)=(Q(x,y),y)$, such that

$$D\sigma (0,0)=\left(\begin{array}{cc}\frac{\partial Q^i}{\partial x^j}(0,0)& \frac{\partial Q^i}{\partial y^j}(0,0)\\ 0& {\delta }_i^j\end{array}\right)$$

which is clearly nonsingular. By the inverse function theorem, we can take connected neighborhoods $U\subset M$ of $p$ and $V$ of $F(p)$ such that $\sigma$ is a diffeomorphism from $U$ to $V$. Taking restrictions if necessary, assume $U$ is an open cube. The smooth inverse ${\sigma }^{-1}$ must then be of the form ${\sigma }^{-1}(x,y)=(A(x,y),y)$ where $A:{\mathbb{R}}^m\to {\mathbb{R}}^r$ satisfies $x=Q(A(x,y),y)$. On $V$, the map $F\circ {\sigma }^{-1}:{\mathbb{R}}^m\to {\mathbb{R}}^n$ satisfies $F\circ {\sigma }^{-1}(x,y)=(x,R(A(x,y),y))$. Take $\tilde{R}(x,y)=R(A(x,y),y)$ for simplicity. Here, we can note that $F$ must have rank $r$ on $V$ as composition with a diffeomorphism leaves rank unchanged. However, for $v\in V$,

$$DF(v)=\left(\begin{array}{cc}{\delta }_i^j& 0\\ \frac{\partial {\tilde{R}}^i}{\partial x^j}& \frac{\partial {\tilde{R}}^i}{\partial y^j}\end{array}\right)$$

$DF(v)$ can have rank $r$ only if $\partial {\tilde{R}}^i/\partial y^j=0$ for all relevant $(i,j)$. As $U$ is a cube, $V$ has a straightforward geometry and if we take $S(x)=\tilde{R}(x,y)$, $F\circ {\sigma }^{-1}(x,y)=(x,S(x))$ for all $(x,y)\in V$. We chose $(U,\sigma )$ as our chart on the domain to simplify the first $r$ coordinates; we now choose a chart in the codomain to clean up the remaining $n-r$ dimensions. Consider the set $\tilde{V}\subset N$ such that $\tilde{V}=\{(u,v)\in N,u\in {\mathbb{R}}^r,v\in {\mathbb{R}}^{n-r}|(u,0)\in V\}$. For any $(x,y)\in V$, $F\circ {\sigma }^{-1}(x,y)\in N\cap (\{x\}\times {\mathbb{R}}^{n-r})$, and thus $F\circ {\sigma }^{-1}(V)\subset \tilde{V}$. Define the function $\psi :\tilde{V}\to {\mathbb{R}}^n$ given by $\psi (u,v)=(u,v-S(u))$. This is clearly smooth as its inverse is simply ${\psi }^{-1}(u,v)=(u,v+S(u))$. Equipped with the charts $(U,\sigma )$ and $(\tilde{V},\psi )$, for any $(x,y)\in V=\sigma (U)$, we have

$$\psi \circ F\circ {\sigma }^{-1}(x,y)=(x,S(x)-S(x))=(x,0)$$

this also shows $\psi$ is centered at $F(p)$ as $\psi (0,0)=\psi \circ F\circ {\sigma }^{-1}(0,0)=(0,0)$. Our proof for $\partial M=\varnothing$ is complete.

If $\partial M$ is nonempty but $p\in \text{Int }M$, we can simply apply the empty boundary case to the open submanifold $\text{Int }M$ to prove the theorem. Thus, it suffices to address the case $p\in \partial M$. Again, for simplicity, we can exploit a mild abuse of notation and identify $M$ as an open subset of ${\mathbb{H}}^m$ and $N$ as an open subset of ${\mathbb{R}}^n$. WLOG, assume $p=0$ and $F(p)=0$.

Let $\hat{F}$ be an arbitrary smooth extension of $F$ to an open neighborhood of $p$ in ${\mathbb{R}}^n$. By restricting the domain if necessary, assume $\hat{F}$ is at least rank $r$ on its domain. We first show one can modify $\hat{F}$ to be of constant rank on its domain. As in the original proof, reorder the coordinates such that the upper-left $r\times r$ submatrix of $\hat{F}(p)$ is nonsingular, and relabel ${\mathbb{R}}^m$ as $(x,y)$ and $F(x,y)=(Q(x,y),R(x,y))$. Defining $\sigma (x,y)=(Q(x,y),y)$ and using the inverse function theorem, we obtain a neighborhood $U$ of $p$ in ${\mathbb{R}}^n$ such that $\sigma$ on $U$ is a diffeomorphism (from now on, we consider $\sigma$ to only be defined on $U$).

The proof splits into two cases. The first case is that the boundary coordinate of ${\mathbb{H}}^m$ is not one of the $r$ first coordinates of $\hat{F}$.

We solve for the subcase where $F$ is a submersion or immersion on $M$. Take an arbitrary extension $\tilde{F}$ of $F$ defined on a neighborhood of $p$ in ${\mathbb{R}}^m$. By smoothness of the Jacobian and determinant, there is a neighborhood $U\subset {\mathbb{R}}^m$ of $p$ such that $\tilde{F}$ is also constant-rank on $U$ with the same rank of $F$ (we can only make this argument because $F$ is full-rank). Using the open rank theorem with $\tilde{F}:U\to N$ gives charts $(A,\varphi )$ and $(B,\psi )$ where $p\in A\subset U$ and $\tilde{F}(A)\subset B\subset N$ and the coordinate representation of $\tilde{F}$ is the projection map. The issue here is that $\varphi$ might not cleanly restrict to a boundary chart containing $p$. In the case of an immersion, this is easy to rectify; because $\dim N\ge \dim M$, we can define $\tilde{\varphi }={\varphi }^{-1}\times Id({\mathbb{R}}^{n-m})$ such that $\tilde{\varphi }$ maps $B$ to ${\mathbb{R}}^n$. By replacing $\psi$ with $\tilde{\psi }=\tilde{\varphi }\circ \psi$, we get, for any $x\in A$,

$$\tilde{\psi }\circ \tilde{F}(x)=\tilde{\varphi }\circ \psi \circ \tilde{F}\circ \varphi \circ {\varphi }^{-1}(x)=\tilde{\varphi }\circ {\pi }_{\varphi (A)\to \psi (B)}\circ {\varphi }^{-1}(x)={\pi }_{M\to \tilde{\psi }(N)}(x)$$

Thus, the charts $(A,Id)$ and $(B,\tilde{\psi })$ gives us our desired projection.

For submersions we have to reach deeper into the open rank theorem construction. Note that the chart $(A,\varphi )$ constructed using our open rank theorem proof is such that the

We now solve for the general case $p\in \partial M$. First, reorder the coordinates such that the top left $r\times r$ submatrix of $DF(p)$ is nonsingular.