# Kernel of the tangent map

If $\varphi:U\subset \mathbb{R}^n \to \mathbb{R}^m$ is $C^1$, let $\mathrm{T}\varphi:\mathrm{T}U \to \mathrm{T}R^m$ be its tangent map. The inverse function theorem tells us that if $\ker(\mathrm{T}\varphi(x))$ is zero, $\varphi$ is injective in some neighborhood of $x$. If the kernel is non-zero, what can we say about $\varphi$ near $x$ provided we know the kernel? In particular, can we say anything about curves through $x$ whose tangents belong to this kernel?

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You can consider the Constant Rank Theorem: en.wikipedia.org/wiki/… –  Pierre-Yves Gaillard Sep 10 '10 at 4:55
You should be more concrete about what you want to know. Books have been written and careers built upon the study of singularities of smooth maps, so unless you are more specific it is hard to know what you are after! –  Mariano Suárez-Alvarez Sep 10 '10 at 13:12
@Mariano: I don't know if mmm is in this position, but it can be difficult for someone unfamiliar with a particular field to know if their question related to it is a simple one with a concrete answer, or a fundamental problem upon which books have been written. Perhaps the best response in such a case is to leave a comment saying "Your question is a fundamental problem in [name of field]. A good reference is [citation of textbook]." –  Rahul Sep 10 '10 at 15:22
@Rahul, but s/he is familiar with what he wants to know! "What can we say about X?" is quite non-descriptive about what he wants to know. –  Mariano Suárez-Alvarez Sep 10 '10 at 17:55
@Pierre-Yves, I would be happy if you would add that as an answer. –  mmm Sep 11 '10 at 2:10

As Wikipedia says:

"The inverse function theorem (and the implicit function theorem) can be seen as a special case of the constant rank theorem, which states that a smooth map with locally constant rank near a point can be put in a particular normal form near that point."

The Constant Rank Theorem is stated as Theorem (7.1) p. 47 of An Introduction to Differentiable Manifolds and Riemannian Geometry, Revised Second Edition, William M. Boothby, Academic Press. (This is the reference given by Wikipedia.)

Here is, for the reader's convenience, a statement of the Constant Rank Theorem.

Let $k,n$ and $r$ be positive integers, let $a$ be in $\mathbb R^n$, let $b$ be in $\mathbb R^k$, let $f$ be a smooth map from a neighborhood of $a$ to $\mathbb R^k$ sending $a$ to $b$, and let $\ell$ be the linear map from $\mathbb R^n$ to $\mathbb R^k$ sending $x$ to $(x_1,\dots,x_r,0,\dots,0)$. Assume that the rank of the tangent map to $f$ at $x$ is equal to $r$ for all $x$ in our neighborhood of $a$.

Then there is a diffeomorphism $g$ from a neighborhood of $a$ to a neighborhood of 0 in $\mathbb R^n$, and a diffeomorphism $h$ from a neighborhood of 0 in $\mathbb R^k$ to a neighborhood of $b$, such that the equality $f=h\circ\ell\circ g$ holds in some neighborhood of $a$.

EDIT OF MARCH 19, 2011

Here is a statement and a proof of the Constant Rank Theorem.

Theorem. Let $U$ be open in $\mathbb{R}^n$, let $a$ be a point in $U$, and let $f$ be $C^p$ map ($1\le p\le\infty$) of rank $r$ from $U$ to $\mathbb{R}^k$. Then there are open sets $U_1,U_2\subset\mathbb{R}^n$, $U_3\subset\mathbb{R}^k$ and $C^p$ diffeomorphisms $\varphi:U_1\to U_2$, $\psi:U_3\to U_3$ such that $a\in U_1$ and $(\psi\circ f\circ\varphi^{-1})(x)=(x_1,\dots,x_r,0,\dots,0)$ for all $x$ in $U_2$.

Proof. For $$x\in\mathbb{R}^r,\quad y\in\mathbb{R}^{n-r},\quad(x,y)\in U$$ write $$f(x,y)=(f_1(x,y),f_2(x,y)),\quad f_1(x,y)\in\mathbb{R}^r,\quad f_2(x,y)\in\mathbb{R}^{k-r}.$$

We can assume that $\partial f_1(x,y)/\partial x$ is invertible for all $(x,y)\in U$. Define $$\varphi:U\to\mathbb{R}^n,\quad(x,y)\mapsto(f_1(x,y),y).$$ By the Inverse Function Theorem, there are open sets $$U_1\subset\mathbb{R}^n,\quad U_4\subset\mathbb{R}^r,\quad U_5\subset\mathbb{R}^{n-r}$$ such that $a\in U_1\subset U$, $\varphi$ is a $C^p$ diffeomorphism from $U_1$ onto $U_2:=U_4\times U_5$, and $U_5$ is connected.

Then $f(\varphi^{-1}(x,y))=(x,g(x,y))$ for some $C^p$ map $g$ from $U_2$ to $\mathbb{R}^{k-r}$. As $\partial g/\partial y=0$, we can write $g(x)$ for $g(x,y)$, and it suffices to set $U_3:=U_4\times\mathbb{R}^{k-r}$ and $\psi(u,v):=(u,v-g(u))$ for $u\in U_4$ and $v\in\mathbb{R}^{k-r}.$

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While true, the constant rank condition is not usually satisfied in mmm's set-up, unless the map $\varphi$ already has maximal rank. Generically the map $\varphi$ won't have maximal rank in any neighbourhood of $x$ if its rank at $x$ is less than $\min(m,n)$. –  Robin Chapman Sep 11 '10 at 10:08