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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

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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.

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  • $\begingroup$ 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 $\endgroup$ – kevint Feb 16 at 21:51
  • $\begingroup$ @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 $\phi^{-1}$. $\endgroup$ – Rhys Steele Feb 16 at 21:59
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    $\begingroup$ @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. $\endgroup$ – Rhys Steele Feb 16 at 22:00
  • $\begingroup$ Understood. I have included an extract of the notes where this exercises are asked to be proved at the bottom of the page. $\endgroup$ – kevint Feb 16 at 22:12
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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.

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