How do I remove the specific assumptions in the proof of the rank theorem in the special case of injective differential?

Question

I have the following proof in my notes. It is a particular case of the more general rank theorem.

Theorem (Rank theorem for injective differential). Suppose $M$ is a smooth manifold of dimension $m$, and that $N$ is a smooth manifold of dimension $n$. Suppose $F : M \to N$ is smooth. Let $p \in M$. If $dF_p$ is injective, then there are charts $(U, \varphi)$ of M around p and $(V,\psi )$ of N around $F(p)$ such that $F(U) \subseteq V$ and for all $x \in\varphi(U)$, and

$\psi\circ F \circ \varphi^{−1}(x) = (x, 0_{n−m})$

Proof

We prove the theorem in the case that $M$ is an open subset of $\Bbb R^m$, $N$ open subset of $\Bbb R^n$, $p = 0_m$ and $F(p) = 0_n$ By using appropriate charts around p and F(p), we can prove the general case.

Suppose $dF_p$ is injective. Then $m \le n$. Define $Q: M \to \Bbb R^m$ and $R: M \to \Bbb R^{n−m}$ by $F(x) = (Q(x), R(x))$ for $x \in M$.

Since $DF(p)$ is injective, its matrix has an $m × m$ invertible submatrix. We can do this by possibly exchanging coordinates ensuring that $DQ(p)$ is such a matrix; we will assume this from now on. Because $DQ(p)$ is invertible, then due to the inverse function theorem there are nbhs $U$ of $p$ and $\tilde V$ of $Q(p)$ such that $\varphi := Q|_U : U \to\tilde V$ is a diffeomorphism

Define $V := \tilde V \times \Bbb R^{n−m}$.This is an open nbhd of $F(p) = 0_n$, because $Q(p) = 0_m$.

Define :$ \psi:V \to \Bbb R^n: \psi(v; w) = (v, w − R \circ \varphi^{−1}(v))$, for $v \in \tilde V$ and $w \in \Bbb R^{n−m}$.

This is a diffeomorphism onto its image, with inverse given by

$\psi^{−1}(s, t) = (s, t + \varphi^{−1}(s))$, for $s \in \Bbb R^m $ and $t \in \Bbb R^{n−m}$ such that $(s, t) \in \psi(V )$.

And for all $x \in \varphi(U)$,

$\psi\circ F \circ \varphi^{−1}(x) =\psi(x,R\circ\varphi^{-1}(x))= (x, 0_{n−m}) \tag{*}$

I have a couple of doubts

1. How does one come up a priori with the idea to define $\psi$ like that? It is supposed to be a chart of $ N$ around $F(p)$, so how does one know that it has that form?

2. How do I prove the general case ? (that is removing the assumptions: $M$ is an open subset of $\Bbb R^m$, $N$ open subset of $\Bbb R^n$, $p = 0_m$ and $F(p) = 0_n$ but still considering injective derivative)I am having trouble writting it down. Moreover why are they using charts in the proof above if they are not working with general manifolds?

Thanks for your help.

Answer 1

The proof is awkwardly written and obscures the central ideas. We assume that we’re in Euclidean soace and the derivative at $p=0$ is $\begin{bmatrix} I\\0\end{bmatrix}$. Consider the map $g(x,y)=F(x)+(0,y)$. The inverse function theorem gives a local inverse $\tau$ for $g$. Then look at $(\tau\circ g)(x,0)=(\tau\circ F)(x)$.

Answer 2

See Product manifolds in Dieudonné, Vol. 3: Reduction of proofs to open sets in $\Bbb{R}^n$ for some indications of how to handle such reductions.

Just to be super explicit for this question, suppose you prove the rank theorem for injective differentials in your given special case. For completely general $F:M\to N$,

• fix a point $p\in M$ about which we’re going to prove things.

• fix an arbitrary chart $(V_0,\psi_0)$ in $N$ around $F(p)$, and consider the translation $\tau_1:\Bbb{R}^n\to\Bbb{R}^n$ given by $\tau_1(y)=y-\psi_1(F(p))$ (translations are obviously diffeomorphisms). Now, define $V_1=V_0$, and $\psi_1=\tau_1\circ\psi_0$. Then, $(V_1,\psi_1)$ is a chart for $N$ around the point $F(p)$, and furthermore, $\psi_1(F(p))=0\in\Bbb{R}^n$.

• fix an arbitrary chart $(U_0,\phi_0)$ in $M$ around $p$. and consider the translation $\tau_0:\Bbb{R}^m\to\Bbb{R}^m$ given by $\tau(x)=x-\phi_0(p)$ (obviously a diffeomorphism). Now, define $U_1=U_0\cap F^{-1}(V_0)$ (so that we can compose things later) and $\phi_1=\tau_0\circ\phi_0|_{U_1}$. Then, $(U_1,\phi_1)$ is still a chart for $M$ around the point $p$, and furthermore, $\phi_1(p)=0\in\Bbb{R}^m$.

• Now, define the local representative $f:=\psi_1\circ F\circ\phi_1^{-1}:\phi_1[U_1]\subset\Bbb{R}^m\to \psi_1[V_1]\subset\Bbb{R}^m$. This map is $F$ composed on the domain and target by charts, so $f$ is a smooth map between open subsets of cartesian spaces; also all our translations and stuff ensure that $f(0_{\Bbb{R}^m})=0_{\Bbb{R}^n}$.

• We are now in the perfect position to apply our special case to the map $f$. We find that there exist diffeomorphisms $\widetilde{\phi}:\widetilde{U}_1\subset\phi_1[U_1]\to \widetilde{U}_2\subset\Bbb{R}^m$ and $\widetilde{\psi}\subset V_1\subset\psi_1[V_1]\to\widetilde{V}_2\subset\Bbb{R}^n$, where $\widetilde{U}_1,\widetilde{V}_1$ are respective open neighborhoods of $0$, and $\widetilde{U}_2,\widetilde{V}_2$ are open, such that $\widetilde{\psi}\circ f\circ\widetilde{\phi}^{-1}$ has the desired canonical form $x\mapsto (x,0)$.

• now, re-track everything. The desired chart about $p$ is $(U,\phi)$, where $U=\phi_1^{-1}[\widetilde{U}]$, and $\phi:=\widetilde{\phi}\circ \phi_1|_U$. Likewise, the desired chart about $F(p)$ is $(V,\psi)$ where $V=\psi_1^{-1}[\widetilde{V}]$ and $\psi:=\widetilde{\psi}\circ\psi_1|_{V}$. A little unwinding shows that $\psi\circ F\circ \phi^{-1}=\widetilde{\psi}\circ f\circ \widetilde{\phi}^{-1}: x\mapsto (x,0)$.

So, you see the reason nobody ever says all of this is because it is first of all annoying (I really hope I don’t have any typos above), because all we’re doing is suitably shrinking open sets on the domain/target, and composing on the domain/target by appropriate diffeomorphisms. Both of these ‘operations’ do not change whether or not we have a chart for our manifold. That’s why we simply say ‘without loss of generality’. Also, the notation very quickly gets out of hand.

But really, I strongly urge you to convince yourself of these arguments. Making these simplifications should be the first thing in any complicated proof. In fact, the proof also made another simplifying assumption about the derivative having such and such block format. More precisely, this means we need to compose $\psi_1$ and $\phi_0$ by suitable invertible linear transformations (which encapsulate the appropriate row/column operations)… but I’m not going to type that out as well. Anyway, see also Trying to understand the statement of Rudin's Rank Theorem.

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“Another” way of saying all of this is:

1. the theorem holds for $F$ if and only if for all open $U_0\subset M,V_0\subset N$, it holds for $F|_{U_0\to V_0}:U_0\to V_0$. This is then equivalent to the theorem holding for all $U,V$ which are domains of chart maps (because chart domains are open in the manifolds, and conversely for every open set in the manifold, and every point in the open set, there exist chart domains completely contained in the open set and containing the given point). In other words, we have literally reduced our daunting task to the case where $M,N$ are shrinked to chart-domains.
2. the theorem holds for a map if and only if it holds for any map obtained by composing on the domain and target by diffeomorphisms.
3. by putting statements (1) and (2) together, it suffices to prove the theorem for $\psi\circ F\circ \phi^{-1}$, which is a map between open subsets. This is why you can reduce to the case that $M\subset \Bbb{R}^m,N\subset\Bbb{R}^n$ are open.
4. by using (2) again, we can compose the domain and target by translations (since they’re diffeomorphisms). This allows us to reduce to the case that $F(0)=0$.
5. using (2) again, we can compose the domain and target by linear isomorphisms (i.e row/column reduction), which are of course diffeomorphisms, and suppose furthermore $DF_{0_{\Bbb{R}^m}}(x)=(x,0_{\Bbb{R}^{n-m}})$.

Now, step (5) is the case that you prove things in, because we have cleaned up all the distracting garbage. But as explained by steps (1)-(4), proving step (5), which at first glance seems like a special case, is sufficient to deduce the general case.