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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
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@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

1 Answer 1

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."

See http://en.wikipedia.org/wiki/Derivative_rule_for_inverses#Constant_rank_theorem

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
    
@Robin Chapman : Your comment has been split into 5 comments. I mentioned the Const. Rk Thm in a comment to the question, and the OP kindly asked me to add it as an answer. I agree that my answer has not even an atom of originality, I'm sure there are more clever things to say about this, and I'd read with pleasure any further comment or answer you, or other users, may make. [I don't know if I'm the only one to see 5 comments instead of one, but otherwise, it would be clearer for the reader if you fixed that.] –  Pierre-Yves Gaillard Sep 11 '10 at 10:48
    
Pierre-Yves, I'm a bit sceptical about how useful the constant rank theorem is. As far as I can see it won't be applicable in very many circumstances, save for the maximal rank case where it reduces to the familiar characterization of immersions and submersions.... –  Robin Chapman Sep 11 '10 at 11:55
    
@Robin Chapman - If I understand you correctly, you want statements which hold GENERICALLY. That was not the way I understood the question. Why don't you explain this genericity issue in an answer? Same thing for "the familiar characterization of immersions and submersions" that you mention. [I just thought it couldn't hurt to state the Constant Rank Theorem, but I know that it doesn't answer all the questions.] –  Pierre-Yves Gaillard Sep 11 '10 at 13:39

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