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

I am 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 $$x^{'\mu} = x^{\mu} \bullet \phi$$.

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]

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• Might Mathematics be better suited for this math question? – Kyle Kanos Feb 18 at 12:10
• please do not cross-post questions, it is considered an abuse of the SE sites and is looked down upon. Just choose one site and if it doesn't get an answer after ~1 week, try at a different site. – Kyle Kanos Feb 18 at 12:44
• Math mods: Please merge. – Qmechanic Feb 19 at 0:35

I'm not sure what you mean by "proof" of a definition. Note that you can't really "push forward" a form. You can only push forward a vector at point $$p$$. Your book's "push forward" $$\phi_*$$ of a form is really the pull-back of the form along the inverse map $$\phi^{-1}$$.

If $$\phi: M\to N$$ takes $$x\mapsto z(x)$$ then, by definition, the pushforward in coordinate language of
$$X= \left.X^\mu \frac{\partial}{\partial x^\mu}\right\vert_p \in TM_p$$ is $$\phi_* X= \left.X^\mu \frac{\partial z^\nu}{\partial x^\mu} \frac{\partial}{\partial z^\nu}\right\vert_{\phi(p)} \in TN_{\phi(p)}$$ Take care that you can't push forward a vector field unless $$\phi_* X$$ is 1-1.This is why the book says "diffeomorphism" rather than a general map. You can, however, always pull back a form even when $$\phi$$ is not 1-1.

If, for example, $$\eta=\eta_\mu(z) dz^\mu \in \Lambda^1 (T^*N)$$ and the map $$\phi: M\to N$$ takes $$x\mapsto z(x)$$ then $$\phi^* \eta= \eta_\mu(z(x)) d(z^\mu(x))= \eta_\mu(z(x)) \frac{\partial z^\mu}{\partial x^\nu}dx^\nu\in \Lambda^1 (T^* M)$$

There is no real "proof" here, just the use of the chain-rule $$dz^\mu = \frac{\partial z^\mu}{\partial x^\nu}dx^\nu$$ to transcribe into a specific coordinate system the statement of the definition. You can, however, use the explicit formula for the push-forward of a vector to check that this recipe is consistent with the coordinate free langauge

I think, conventionally the pull-back is defined as adjoint to push-forward. So if you have a manifold $$\bar{\mathcal{M}}$$, and manifold $$\mathcal{M}$$ (possibly the same manifold). You then need a map $$\Phi:\bar{\mathcal{M}}\to\mathcal{M}$$, such that $$x^{\left(i\right)}=\Phi^{\left(i\right)}\left(\bar{x}\right)$$.

Based on this you can define a push-foward $$\phi: T_\bar{p} \bar{\mathcal{M}}\to T_{\Phi\left(\bar{p}\right)}\mathcal{M}$$, such that

$$\phi\left(\bar{A}^i \bar{\partial}_i\right)=\bar{A}^i \frac{\partial \Phi^{(k)}}{\partial \bar{x}^{(i)}} \partial_k$$

Now you can also define forms on both manifolds, i.e. $$\omega \in T_p \mathcal{M}^*$$, $$\omega: T_p \mathcal{M}\to\mathbb{R}$$, and same for $$\bar{\omega} \in T_p \bar{\mathcal{M}}^*$$. It is convenient to use the following notation for the action of the form $$\omega$$ one the vector $$A$$:

$$\langle \omega | A\rangle = \omega\left(A\right)=\omega_i A^i$$

You can then ask what is the result of applying the form to the push-forwarded vector:

$$\langle \omega | \phi \bar{A}\rangle = \omega_k \frac{\partial \Phi^{(k)}}{\partial \bar{x}^{(i)}}\bar{A}^i$$

Finally you can define the adjoint to the push-forward, the pull-back, as:

$$\langle \omega | \phi \bar{A}\rangle = \langle \phi^* \omega | \bar{A}\rangle = \omega_k \frac{\partial \Phi^{(k)}}{\partial \bar{x}^{(i)}}\bar{A}^i$$

Where the induced pull-back is $$\phi^*: T_{\Phi\left(\bar{p}\right)}\mathcal{M}^* \to T_{\bar{p}}\mathcal{\bar{M}}^*$$, and $$\phi^*\left(\omega_i dx^i\right)=\omega_i \frac{\partial\Phi^{(i)}}{\partial\bar{x}^{(k)}} d\bar{x}^k$$

Often there is abuse of notation, where one says $$x^{(i)}=x^{(i)}\left(\bar{x}\right)$$ (dropping $$\Phi$$)