Let $f: \mathbb R^n \to \mathbb R^m$ be smooth and let $x_0 \in \mathbb R^n$ be such that $\operatorname{rank}{(J_f(x_0))} = k $. Then there exists a neighboudhood of $x_0$ and diffeomorphisms $\phi, \psi$ such that

$$ \phi \circ f \circ \psi = (x_1, \dots, x_n ) \mapsto (x_1, \dots, x_k, 0 \dots, 0)$$

on this neighbourhood.

This is the constant rank theorem.

It seems to me that this is saying that any smooth map can be written as a projection onto some of its coordinates on some neighbourhood.

Is this really what this is saying? Please could someone help me get the intuition behind this theorem?

  • 1
    $\begingroup$ As Areaman's answer states this is an incorrect statement of the theorem, for example for $f:\Bbb R \to \Bbb R$ with $f(x)=x^2$ restricted to any neighborhood of zero does not look like the zero map locally. $\endgroup$ – PVAL-inactive Aug 23 '15 at 2:24

Shouldn't the rank be constant on a neighborhood of that point to make this conclusion? It is sufficient to check at that point if the rank is maximal, but otherwise the rank can "jump" up suddenly, meaning you would need more coordinates. It can't jump down suddenly, because the determinant function is smooth. It takes some time for a nonzero continuous function to become zero, but a function that is zero can "instantly" become non-zero.

Otherwise your intuition seems more or less correct. Sometimes a useful term is locally: "The constant rank theorem says that a smooth function with locally constant rank is locally a linear map of that same rank, up to diffeomorphism."

  • 5
    $\begingroup$ Just a supplement for the OP: The rank of the Jacobian of a smooth (or even just continuously differentiable) function is lower semicontinuous. This explains why the rank can't jump down, as AreaMan says, and this holds regardless of what the rank is (i.e. maximal or not). $\endgroup$ – Kyle Aug 24 '15 at 12:51

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