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

Question

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]

Answer 1

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

Answer 2

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 $\Phi$)