2
$\begingroup$

Studying for an exam, and trying to get my head around the concepts of push forwards.

The question I'm attempting to answer is:

"Give an example of a continuously differentiable diffeomorphism F and a continuously differentiable vector field X, such that the push forward of X is continuous but not differentiable." The answer given in the back of the book is:

On the real numbers, take

$F(x) = $ $x^2$ if $x \ge 0$; $−x^2$ if $x \gt 0$,

and $X(x) = 1$.

(Apologies for the awful formatting - I'll edit again in future when I work out how to use MathJaX to get piecewise functions!)

I'd really appreciate if someone could explain why this example satisfies the question.

  • 0
    Thanks for being so responsive.2012-11-14

1 Answers 1

1

If $F: M \to N$ is a diffeomorphism and $X$ is a vector field on $M$ then the pushforward vector field $dF~X$ is defined by $(dF~X)(y) = dF\vert_{F^{-1}(y)} X_{F^{-1}(Y)}$. In your case $dF\vert_x$ is multiplication by $2|x|$ and the inverse of any $y \in \mathbb R$ is given by $\sqrt{|y|}$ if $y > 0$ and $-\sqrt{|y|}$ if $y < 0$.

Therefore $(dF~X)(y) = 2\sqrt{|y|}$, which is continuous but not differentiable at the origin.