Why are pull backs always defined for 1 forms but not push forwards?

Question

So I know that given a differentiable function $\mathcal{F}: \mathcal{M} \rightarrow \mathcal{N}$, we can define the push forward map: $$ \mathcal{F}_*: T_p(\mathcal{M}) \rightarrow T_{\mathcal{F}(p)}(\mathcal{N}) $$

However, the pull back map for vectors would require the inverse function $\mathcal{F}^{-1}$ which may not necessarily exist and this is why we cannot define a pull back for vector fields.

Now coming to forms, I do not understand why the pull back is always defined while the push forward is not? That is, given that I know $\langle \omega, v \rangle$, shouldn't I be able to define $\langle \mathcal{F}_* (\omega), \mathcal{F}_*(v)\rangle$? Why is this not possible?

Answer 1

Your question really comes down to sets, and functions, and compositions of functions.

Let's just do 1-forms.

A differential 1-form $\omega$ on $\mathcal M$ is a certain kind of function whose domain is the set $T \mathcal M$: it inputs a tangent vector on $M$ and outputs a number, i.e. it outputs an element of the set $\mathbb R$: $$(1) \qquad \omega : T \mathcal M \to \mathbb R $$ A differential 1-form $\rho$ on $\mathcal N$ similarly inputs a tangent vector on $N$ and outputs a number: $$(2) \qquad \rho : T \mathcal N \to \mathbb R $$

Now consider a smooth function $\mathcal F : \mathcal M \to \mathcal N$ with corresponding pushforward map $$(3) \qquad \mathcal F_* : T \mathcal M \to T \mathcal N $$ How can we chain these function diagrams together? What choices to we have for composing these functions?

Well, first of all, we can easily compose (2) and (3) to get $$T \mathcal M \xrightarrow{ \mathcal F_*} T \mathcal N \xrightarrow{\rho} \mathbb R $$ and that's the definition of the pullback $\mathcal F^*(\rho) : T \mathcal M \to \mathbb R$.

But how do you propose composing (1) and (3)? There's simply no composition to be made. And that's why the "pushforward" of forms is not defined. Whether or not $\mathcal F$ has an inverse is irrelevant.

Answer 2

I just want to add some intuition here for you. You should think of pullback of forms in terms of substitution (as you did with substitutions in integrals in your second calculus course). For example, if $f\colon\Bbb R\to \Bbb R^2$, say $f(t)=(t\cos t, t^2)$, and $\omega$ is a $1$-form on $\Bbb R^2$, say $\omega = P\,dx + Q\,dy$, then $$f^*\omega = f^*(P\,dx + Q\,dy) = P(t\cos t,t^2)d(t\cos t) + Q(t\cos t,t^2) d(t^2)$$ is what you get merely by substituting $x=t\cos t$, $y=t^2$, and using $dg=g'(t)\,dt$ as you did in substitutions in integrals. Indeed, the whole premise of substitution is that you're pulling back and doing the integral in a new coordinate system.