# Differential forms and how to show one is smooth

A class I am current taking makes some use of differential forms. In particular, we are asked to show that a certain differential form is smooth. I know in general that a differential $k$-form, $\omega$ is smooth if for smooth vector fields $X_1,\ldots,X_k$, $\omega_p(X_1(p),\ldots,X_k(p))$ is smooth in $p$. Is there a good way to visualize this in local coordinates, and is there a general easy way to prove that a differential form is smooth?

Also, and I believe related, is the question of how these vector fields interact with the differential form. ie is it true that if we have a $2$-manifold (in local coordinates) $X_1(p)=a_{11}(p)\frac{\partial}{\partial x_1}+a_{12}(p)\frac{\partial}{\partial x_2}$ and $X_2(p)=a_{21}(p)\frac{\partial}{\partial x_1}+a_{22}(p)\frac{\partial}{\partial x_2}$, does if $\omega_p=f(p) dx_1\wedge dx_2$, what would the formula for be for $\omega_p(X_1(p),X_2(p))$. I think this question is simple yet would illustrate to me how these things work.

I have looked up references for differential forms, but none seem to answer the questions about them that I seem to have. Any reference however would also be greatly appreciated.

Thank you for any help.

Smoothness of a form is the same as the smoothness of its coefficients in some (equivalently, any) system of coordinates. Let's verify this on the $2$-form example in your post. First, suppose that $f$ is a smooth function. Then for any smooth vector fields $X_1,X_2$ we have $$\omega_p(X_1(p),X_2(p)) = f(p)(a_{11}(p)a_{22}(p)-a_{12}(p)a_{21}(p))\tag 1$$ which is smooth. Note that (1) is calculated using the duality relations
$$(dx_1\wedge dx_2)\left(\frac{\partial}{\partial x_1},\frac{\partial}{\partial x_2}\right)=1$$ $$(dx_1\wedge dx_2)\left(\frac{\partial}{\partial x_2},\frac{\partial}{\partial x_1}\right)=-1$$ $$(dx_1\wedge dx_2)\left(\frac{\partial}{\partial x_1},\frac{\partial}{\partial x_1}\right)=0$$ $$(dx_1\wedge dx_2)\left(\frac{\partial}{\partial x_2},\frac{\partial}{\partial x_2}\right)=0$$ That is, the result of plugging basis vectors into a basis form is $1$ if the vector indices match the form indices exactly; it is $(-1)^\sigma$ if they match up to $\sigma$ transpositions, and is $0$ in other cases.

Conversely, suppose that the form $\omega_p$ is smooth. Plug in the coordinate vector fields to obtain $$\omega_p\left(\frac{\partial}{\partial x_1},\frac{\partial}{\partial x_2}\right)= f(p)$$ and conclude that the coefficient $f$ is smooth.

For general $k$-forms the argument is exactly the same.

How one proves that a form is smooth depends on how the form was constructed. The usual way to construct a form involves some of the following steps:

• taking a scalar function ($0$-form)
• differentiating a form
• multiplying two forms
• pulling a form back by a smooth map

All of the above operation preserve $C^\infty$ smoothness: if your ingredients are smooth, so is the result.

Sometimes the most nontrivial part of proving that a form is smooth is to prove that it is well-defined, that is, different choices of local coordinates give the same form. The Fubini-Study form is an example of such situation. For simplicity take $n=1$. The Fubini-Study form is the 2-form defined on $\mathbb C\mathbb P^1=\mathbb C\cup\{\infty\}$ by means of two charts: $$\omega = \partial\bar\partial \ln(1+|z|^2),\quad z\ne \infty$$ $$\omega = \partial\bar\partial \ln(1+|1/z|^2),\quad z\ne 0$$ One then has to check that the definitions are consistent on the overlap $z\ne 0,\infty$. Indeed, the difference of Kähler potentials on the overlap is $$\ln(1+|z|^2)-\ln(1+|1/z|^2)= \ln (|z|^2)$$ which is a harmonic function. Since $\partial\bar\partial\ln (|z|^2)=0$, the form is well-defined. Smoothness is not an issue: it's a local property, to be checked on each coordinate chart separately. And it's clear that on each chart we differentiate a $C^\infty$ function.