Proving $\theta$ is zero if and only if $\theta X=0$

Question

A one form $\theta$ is zero if and only if $\theta X=0$ for all $\mathscr{X}(M)$. A vector field is zero if and only if $\theta X=0$ for all $\theta\in\mathscr{X}^* (M)$

$\rightarrow$ If $\theta=0$ then $\theta X=X(0)=0$.

$\leftarrow$ If $\theta X=0$ then $\theta X=X(\theta)=0$ implies that the specific $\theta$ must be zero once $X$ is arbitrary vector field.

$\rightarrow$ If $X=0$ then $\theta X=0$.

$\leftarrow$ If $ X=0$ then $\theta X=X(\theta)=0$ implies that the specific $\theta$ must be zero once $\theta$ is arbitrary one form field.

Question:

Are my proofs correct? This seems to be too straightforward. If not can you explain how to prove the assertions?

Thanks in advance!

Edit:

In the comments, it was suggested to write in local coordinates. So consider $\zeta=\{x_1,...,x_n\}$ on a neighborhood $U$ of $p$.

Then $\theta(X)=\theta_p(\sum X_p(x_i)\partial_i)=\sum(X_p(x_i)\theta_p(\partial_i))=\sum dx_i(X_p)\theta_p(\partial_i)$

The last expression is only 0 if $\theta(\partial_i)$ is zero for every coordinate that is if $\theta=0$.

Answer 1

I'm not going to help with the direction $\theta = 0 \Rightarrow \theta X = 0 \forall X$. You should write out each detail yourself to be very sure you understand why in fact it is true.

So let us consider the direction $\forall X \theta X = 0 \Rightarrow \theta = 0$. This is equivalent to the contrapositive, if $\theta \neq 0$ then there exists some $X$ such that $\theta X \neq 0$. (Why the contrapositive? Because it can be easier to construct something to show existence.)

What does $\theta = 0$ mean? It means that in all local coordinate charts, all its coordinate functions equal zero. So if $\theta \neq 0$, then there exists at least one coordinate chart $x_i$ where $\theta = \theta_i dx^i$ and at least one $\theta_i \neq 0$.

Now can you build a vector field $X$ such that $\theta X \neq 0$?

Take a bump function $\phi$ supported on the open set associated to the coordinates $x_i$. Define $X$ by $X_i = \phi\theta_i\partial_i$ in the coordinate set and $X = 0$ out of it. Then the function $\theta X$ is also supported on the open set associated to the coordinates and can be computed in those coordinates: $\theta X = \sum \theta_i dx^i(\sum \phi\theta_j \partial_j ) = \phi \sum \theta_i^2 > 0$ at at least one point, as there is at least one $\theta_i \neq 0$.