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My professor gave me the definition: Assuming $\phi : M \rightarrow N $ a smooth map between smooth manifolds M and N.

$\phi^*$ : $\Lambda(N) \rightarrow \Lambda(M)$ the pullback defined by:

($\phi^*\omega)_x$ ( $v^1,...,v^p$) = $\omega_{\phi(x)}$ ($T_x \phi v^1,...,T_x \phi v^p$)

Where I think the $\Lambda(N)$ and $\Lambda(M)$ are exterior algebra on manifold N and M.

So I knew the definition of the pullback defined on the dual spaces.

But there I don't see what does $T_x \phi v^1$ means?

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  • $\begingroup$ By exterior algebra you mean one forms ? $\endgroup$
    – Flewer47
    Mar 1, 2021 at 15:31
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    $\begingroup$ I think it's more general. $\Lambda (M)$ = $ \bigoplus \Lambda^p (M)$ where the sum is from p=0 to dim(M), and $\Lambda^p (M)$ are the p-forms. $\endgroup$ Mar 1, 2021 at 16:26
  • $\begingroup$ Okay, thanks for the clarification ! $\endgroup$
    – Flewer47
    Mar 1, 2021 at 20:33

2 Answers 2

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This is somewhat abuse of notation for the pushforward. Let $\phi : M \to N$, then for a $k$-form $\omega$ in $\Lambda^k TN$, we have that, $$(\phi^*\omega)_x(X_1,\dots, X_k) = \omega_{\phi(x)}(\mathrm d\phi_x(X_1), \dots,\mathrm d\phi_x(X_k))$$

which is a form taking vectors on $M$, where the pushforward is $\mathrm d\phi_x : T_xM \to T_{\phi(x)} N$. If one thinks of a tangent vector as a derivation on functions, then $\mathrm d\phi_x(X)(f) = X(f\circ\phi)$.

Another way to see the pushforward: choosing a chart $U$ around $x$ and $V$ around $\phi(x)$, the pushforward is a map (using the same notation) $\phi : U \to V$ and then, $$\frac{\partial}{\partial u^i} \mapsto \frac{\partial \phi^j}{\partial u^i} \frac{\partial}{\partial v^j}.$$


In practice this is how you compute it hands on: take for example $\omega = uv^3 \mathrm du \wedge \mathrm v$ with $u,v$ coordinates on $\mathbb R^3$ and we have a map $\phi : \mathbb R^3 \to \mathbb R^2$ given by $(x,y,z) \mapsto (x^2+yz,e^{xyz}).$

Then we have $\phi^* \omega = (x^2+yz)(e^{xyz})^3 \mathrm d(x^2+yz) \wedge \mathrm d(e^{xyz})$, where $\mathrm d(x^2+yz)$ for example is interpreted in the usual way, when taking the exterior derivative of a degree zero form. You can expand this out and you'll get a form in $\mathrm dx, \mathrm dy, \mathrm dz$.

You should convince yourself from the formal definition that in practice it means just substituting in when you have a smooth map like this.

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  • $\begingroup$ Thanks a lot! I understand now (physicist's notation haha). $\endgroup$ Mar 1, 2021 at 16:29
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Here $T_x\phi$ probably denotes the differential of $\phi$ at $x$. Since it is applied to $v^i$ you should think of $v^i$ as a tangent vector to $M$ at $x$. I.e. we take $v^1,\ldots, v^n\in T_xM$ and $T_x\phi(v^i)\in T_{\phi(x)}N$.

Some people prefer this notation as it is more "functorial."

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