$f: X \to Y$ is a smooth map and $\omega$ is a $p$-form on $Y$, what is $\omega[f(x)]$?

Question

I am reading Guillemin and Pollack's Differential Topology Page 163:

If $f: X \to Y$ is a smooth map and $\omega$ is a $p$-form on $Y$, define a $p$-form $f^*\omega$ on $X$ as follows: $$f^*\omega(x) = (df_x)^*\omega[f(x)].$$

So my question is, what is $\omega[f(x)]$?

I have been trying to make this question self-contained, but here is the whole background: Definition of pullback. In that question, I have shown $(df_x)^*T(v_1, \dots, v_p) = T \circ df_x (v_1, \dots, v_p)$ and $\omega$ is $(df_x)^*T$. But I want the missing piece about what that $\omega[f(x)]$ is equal to, to make sense of the definition of $f^* \omega (x)$.

Answer 1

You can write your differential form (at least locally) as a sum of "pure forms", i.e. forms of the type $$\omega_\alpha(y) = \alpha(y)dx^{i_1}\wedge\ldots\wedge dx^{i_p}$$ with $\alpha$ a smooth function. Then $$f^*\omega_\alpha(x) = (df_x)^*[(\alpha\circ f)(x)dx^{i_1}\wedge\ldots\wedge dx^{i_p}]$$ In other words, $\omega[f(x)]$ means that you precompose the "weight functions" of the pure forms with $f$.

Answer 2

$\omega[f(x)]$ is the value of the form $\omega$ at the point $f(x)$. It is a vector in the $p$-th exterior power of the cotangent space at $f(x)$. (Remember that a $p$-form is a smooth function assigning to each point of a manifold such a vector.) I guess you were just confused by the use of square brackets, which were, I think, intended just to make it easier to read; round brackets would have been OK too, and it seems they didn't cause you any problems with the entirely analogous expresson $f^*\omega(x)$ on the left side of the equation.