# Computation of the push forward of vectors

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.

• 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. – amd Apr 23 at 23:41
• @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. – casimir Apr 25 at 14:58
• I find the notation somewhat cumbersome myself, but the underlying ideas are fairly straightforward. – amd Apr 25 at 17:27