0
$\begingroup$

I currently have a very limited understanding of this topic and would be very grateful if someone could work me through a solution to this question:

Let $\phi : M \to N$ be a diffeomorphism of manifolds. For a vector field $X$ on $M$ define the push-forward vector field $Z = \phi_* X$ on $N$ by $$Z|_y = \mathrm{d}\phi_x(X|_x)$$ where $x = \phi^{-1}(y)$. Show that for any function $f:N \to \mathbb{R}$ $$(\phi_*X)\cdot f = (X \cdot (f \circ \phi))\circ \phi^{-1}.$$

Where do I start? I don't really have a great understanding of vector fields to begin with, let alone how to manipulate the algebra to get to this solution. Thanks in advance.

$\endgroup$
1
$\begingroup$

By definition, $(\phi_*X).f=df(y).(\phi_*X)(y)$

$=df(y).d\phi(\phi^{-1}(y))(X(\phi^{-1}(y)))$.

$X.(f\circ \phi)(x)=d(f\circ\phi)(x).X(x)=df(\phi(x)).d\phi(x)(X(x))$.

This implies that $X.(f\circ \phi)(\phi^{-1}(y))=df(\phi(\phi^{-1}(y)).d\phi(\phi^{-1}(y))(X(\phi^{-1}(y)))$

$=df(y).d\phi(\phi^{-1}(y))(X(\phi^{-1}(y))$.

The both expressions are equal.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.