# Action of a 1-form on the push-forward of a vector

I am currently a physicist studying differential geometry I am trying to proof the expression below.

Given that for a map $$\phi$$ : $$M$$ $$\to$$ $$M$$ the pull-back $$\phi$$*$$\omega$$ $$\in$$ $$T^\ast_p M$$ of a 1-form $$\omega$$ $$\in$$ $$T^\ast_p M$$ is defined by :

($$\phi$$*$$\omega$$)$$(v)$$ = $$\omega$$($$\phi_{*}v$$) where $$v$$ $$\in$$ $$T_{p}M$$.

How would we proof this in a coordinate basis $$dx^{\mu}_{p}$$, $$\phi^{*}\omega$$ has components:

$$(\phi^{*}\omega)_{\nu} = \frac{\partial x^{'\mu}}{\partial x^{v}}\omega_{\mu}$$

where $$\mathbf{\omega} = \omega_{\mu}dx^{\mu}_{\phi(p)}$$ and $$x^{'\mu} = x^{\mu} \bullet \phi$$.

EDIT and also prove that if $$\phi$$ is a diffeomorphism, then the push-forward is $$\phi$$*$$\omega$$ $$\in$$ $$T^{\ast}_{\phi(p)} M$$ of a 1-form $$\omega$$ $$\in$$ $$T^{\ast}_{p} M$$ is defined by:

$$(\phi_{*}\omega)(v) = \omega(\phi^{*}v)$$ for any $$v \in T^{\ast}_{\phi(p)} M$$. Prove that in the coordinate basis $$dx^{\mu}_{\phi(p)}, \phi_{*}\omega$$ has components :

$$(\phi_{*}\omega)_{\nu} = \frac{\partial x^{\mu}}{\partial x^{'v}}\omega_{\mu}$$.

To clarify things please find the extract of the notes I am reading: extract

Thanks

We want to show that $$\phi^* \omega = \frac{\partial (x')^\mu}{\partial x^\nu} \omega_\mu dx^\nu$$ where I use the usual summation convention and have written $$\omega = \omega_\mu dx^\mu$$ in local coordinates.

Since the pull-back is linear, it will be enough to check that $$(\phi^* dx^\mu)_\nu = \frac{\partial (x')^\mu}{\partial x^\nu}$$ where my subscript refers to the coordinate representation with respect to coordinates $$x^\nu$$.

Now $$(\phi^* dx^\mu)_\nu = (\phi^* dx^\mu)\bigg(\frac{\partial}{\partial x^\nu} \bigg) = dx^\mu \bigg( \phi_* \frac{\partial}{\partial x^\nu} \bigg) = \bigg( \phi_* \frac{\partial}{\partial x^\nu} \bigg)_\mu.$$ Finally, we have $$\bigg( \phi_* \frac{\partial}{\partial x^\nu} \bigg)_\mu = \bigg( \phi_* \frac{\partial}{\partial x^\nu}\bigg)(x^\mu) = \frac{\partial}{\partial x^\nu}(x^\mu \circ \phi) = \frac{\partial (x')^\mu}{\partial x^\nu}$$ which gives the desired result.

• Thanks Rhys. I have edited the question we also have to prove the pushforward of the 1-form $\omega$ and we are explicitly told the proof should show subscripts p and $\phi_{p}$ on all basis vectors $\frac{\partial}{\partial x^{\mu}}$ and 1-forms $dx^{\mu}$. Is an analogous proof to what you done for the second part still applies ? And is it OK to ignore the indices p and $\phi(p)$? Thanks – kevint Feb 16 at 21:51
• @kevint If I'm honest, the notation in your edit is a bit confusing to me, I think there are some typos. However, it seems like your definition of the push-forward will just be the pull-back by $\phi^{-1}$ and then this will follow by the first result by replacing $\phi$ with $\phi^{-1}$. – Rhys Steele Feb 16 at 21:59
• @kevint By the way, in future you shouldn't really edit questions to ask further things after you receive an answer. Instead just ask a new question! Since you're a new user, I don't mind answering here this time. – Rhys Steele Feb 16 at 22:00
• Understood. I have included an extract of the notes where this exercises are asked to be proved at the bottom of the page. – kevint Feb 16 at 22:12

I find this all easier to understand if I write $$\phi$$ as a map between two different manifolds (where the two might coincidentally be the same manifold).

Also, it all comes from abstract linear algebra: If $$A_*: V \rightarrow W$$ is a linear map (i.e., a pushforward), then there is a natural dual map $$A^*: W^* \rightarrow V^*$$ (i.e., a pullback), such that for any $$v \in V$$ and $$\ell \in W^*$$, $$(A^*\ell)(v) = \ell(A_*v).$$ If you choose bases for $$V$$ and $$W$$ and use the dual bases for $$V^*$$ and $$W^*$$, then you can write $$A_*$$ and $$A^*$$ as matrices.

Now you can apply this to the differential of a map $$\phi: M \rightarrow N$$, which is a linear map $$\phi_*: T_pM \rightarrow T_{\phi(p)}N$$, which is analogous to the map $$A_*$$ above and defined as follows: Given $$v \in T_pM$$, there exists a curve $$c: (-\delta,\delta) \rightarrow M$$ such that $$c(0) = p$$ and $$c'(0) = v$$. You can compose $$\phi$$ with $$c$$ to get a curve in $$N$$ and define $$\phi_*v = \left.\frac{d}{dt}\right|_{t=0}\phi(c(t)) \in T_{\phi(p)}N.$$ Since $$\phi_*: T_pM \rightarrow T_{\phi(p)}N$$ is a linear map like $$A_*$$ above, there is a dual map $$\phi^*: T_{\phi(p)}^*N \rightarrow T_p^*M$$.

When you push forward a vector field $$v$$, you're just applying the linear map $$\phi_*$$ to $$v(p)$$ for each $$p \in M$$. Notice that if $$\phi$$ is either not injective or not surjective, $$\phi_*v$$ is not a vector field on $$N$$. Similarly, the pullback of a differential form $$\omega$$ on $$N$$ is simply applying $$\phi^*$$ to $$\omega(\phi(p))$$ for each $$p$$. Notice that, contrast to the pushforward, the pullback of a smooth differential form on $$N$$ is a smooth differential form on $$M$$.

Since everything above was defined without using local coordinates, we now know they don't depend on any choice of coordinates.

If you now choose local coordinates on $$M$$ near $$p \in M$$ and on $$N$$ near $$\phi(p)$$, then you get a basis of $$T_pM$$ by holding all but one coordinate on $$M$$ fixed and differentiating the curve with respect to the remaining coordinate. You can do the same using the coordinates on $$N$$. You can now write $$\phi_*: T_pM \rightarrow T_{\phi(p)}N$$ as a matrix, just as for $$A_*$$ above. Using the corresponding dual bases, you can write $$\phi^*: T^*_{\phi(p)}N \rightarrow T^*_pM$$ as matrices, just as described for $$A^*$$. Now you can check that the matrices for $$\phi_*$$ and $$\phi^*$$ are essentially the Jacobian matrix of partial derivatives of $$\phi$$ written with respect to the local coordinates on $$M$$ and $$N$$.

Finally, to minimize confusion, I recommend never talking about the pushforward of a differential form or the pullback of a vector field. If $$\phi$$ is a diffeomorphism, then there is a pushforward of vector fields on $$N$$ by $$\phi^{-1}_*$$ and a pullback of differential forms on $$M$$ by $$\phi^{-1}$$. Such precision in language makes it much less likely you'll get confused or make mistakes.