I am trying to understand the push forward of a vector field by going through a specific calculation.

Consider $f: \mathbb{R}^3 \rightarrow \mathbb{R}^2$ given by

$f(x,y,z) = (x+y+7, z-x-5)$

and the vector field

$X= - x^3 \partial_x + 2y^2\partial_y + xy\partial_z$ on $\mathbb{R}^3$.

I wish to find the push forward of vectors $X_{(0,0,0)}$ and $X_{(1,-1,1)}$ w.r.t. $f$.

For a $C^{\infty}$ function $f: M \rightarrow N$ between manifolds $M,N$ and $X_p \in T_pM$, the push forward is given by $f_{\star}X_{f(p)}:C^{\infty}(f(p)) \rightarrow \mathbb{R}$ with $g \rightarrow X_p(g \circ f)$.

Work so far

To work out the push forward in local coordinates, I try to apply the following theorem:

For $C^{\infty}$ function $f: M \rightarrow N$, $(x^1, ..., x^m)$ local coordinates for $p \in M$ and $(y^1, ..., y^m)$ local coordinates for $f(p) \in N$. Then if

$X_p = \sum^{m}_{\mu=1} A^{\mu} \frac{\partial}{\partial x^\mu} |_p$ is in $T_pM$ then

$f_{\star}X_{f(p)} = \sum^{m}_{\mu=1} \sum^{n}_{\nu=1} A^{\mu} \frac{\partial}{\partial x^\mu} |_p (y^\nu \circ f) \frac{\partial}{\partial y^\nu} |_{f(p)}$

What follows is my attempt at plugging in the details of the question:

Say we denote the local coordinates for $\mathbb{R}^3$ as $(x,y,z)$ and the ones for $\mathbb{R}^2$ as $(s,t)$ and expand the sums from above (bear with me), it will look like

$f_{\star}X_{f(p)} = -x^3 \frac{\partial}{\partial x} |_p (s \circ f) \frac{\partial}{\partial s}|_{f(p)} - x^3 \frac{\partial}{\partial x} |_p (t \circ f) \frac{\partial}{\partial t}|_{f(p)} \\ + 2y^2 \frac{\partial}{\partial y} |_p (s \circ f) \frac{\partial}{\partial s}|_{f(p)} + 2y^2 \frac{\partial}{\partial y} |_p (t \circ f) \frac{\partial}{\partial t}|_{f(p)} \\ + xy \frac{\partial}{\partial z} |_p (s \circ f) \frac{\partial}{\partial s}|_{f(p)} + xy \frac{\partial}{\partial z} |_p (t \circ f) \frac{\partial}{\partial t}|_{f(p)}$

Is this correct so far? I am not sure how to add the remaining details and at what point one would plug in the points itself. Any help is appreciated.

  • $\begingroup$ Isn’t the answer simply $\mathrm{d}f[X]$ evaluated at the appropriate points? Moreover, $f_\star$ is linear, so that should tell you what the pushforward of $X_{(0,0,0)}$ ought to be at a glance. $\endgroup$ – amd Apr 23 at 23:41
  • $\begingroup$ @amd I got lost in the notation, but yes, in the end it is fairly straight forward. I think I have the answer now, hope to add it here later. And the pushforward of the latter should be the zero vector. $\endgroup$ – casimir Apr 25 at 14:58
  • $\begingroup$ I find the notation somewhat cumbersome myself, but the underlying ideas are fairly straightforward. $\endgroup$ – amd Apr 25 at 17:27

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