# Pull-back of differential forms

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?

• By exterior algebra you mean one forms ? Mar 1, 2021 at 15:31
• 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. Mar 1, 2021 at 16:26
• Okay, thanks for the clarification ! Mar 1, 2021 at 20:33

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.

• Thanks a lot! I understand now (physicist's notation haha). Mar 1, 2021 at 16:29

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."