Let $\phi:M\rightarrow N$ be a smooth map between smooth manifolds, $v \in \operatorname{Vect}(M)$. Let $\{x^\mu\}$ and $\{y^i\}$ be local coordinates on $M$ and $N$ respectively.

How can I show that the components of the pushforward vector $\phi_*v$ in local coordinates is given by the following: $$(\phi_*v)^i(y) = \frac{\partial y^i}{\partial x^\mu}v^\mu(x)$$

Differential geometry in general is rather new to me so any help would be appreciated.

  • $\begingroup$ You could show what each of these do to a function defined near $y$. $\endgroup$ – Matthew Leingang Apr 6 '18 at 13:14
  • 1
    $\begingroup$ Here's a sloppy (physicist) sketch : $(\phi_*v) = \xi^i \frac{\partial}{\partial y^i}$. By definition, for any $f \in C^{\infty}(N)$, $$ (\phi_*v) f = v(f \circ \phi) = v^i \frac{\partial}{\partial x^i} (f \circ \phi) = v^i \frac{\partial (f \circ \phi)}{\partial x^i} = v^i \frac{\partial f}{\partial y^j} \frac{\partial y^j }{\partial x^i} = \Big( v^i \frac{\partial y^j }{\partial x^i} \frac{\partial}{\partial y^j} \Big) f $$ So $\xi^i = v^j \frac{\partial y^i }{\partial x^j} $ $\endgroup$ – kelvinn aja Apr 6 '18 at 13:23
  • $\begingroup$ So if I follow through with this and arrive at $\xi^i = v^\mu\frac{\partial y^i}{\partial x^\mu}$ does it follow that $(\phi_{*}v)^{i}(y) = \xi^{i}(y) = \frac{\partial y^i}{\partial x^\mu}v^\mu(x)$? $\endgroup$ – FridgeRacer Apr 10 '18 at 4:23

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