How One Forms act on elements of $T_x\left(M \right)$

Question

I know that a one-form is a map $\varphi: TM \to \mathbb{R}$ that restricts to a linear map on each tangent space of the manifold $M$. I need to think about the kernel of such a one-form, but I'm a bit lost as to how the one-form actually acts on elements of the tangent space.

To be more concrete, I have a one-form $\omega= dz-ydx$ (so that $M=\mathbb{R}^3$ I think) and need to figure out its kernel to prove $\omega$ in fact gives a contact structure on $\mathbb{R}^3$. So, how does $\omega$ act on something of the form $$ v_1\frac{\partial}{\partial x}+v_2\frac{\partial}{\partial y} + v_3\frac{\partial}{\partial z} $$ which afaik is how elements of $T_x(\mathbb{R}^3)$ look?

Answer 1

Let $\omega=dz-ydx\in\Lambda^1((\mathbb R^3 )^*)$, where $x,y,z$ are global coordinates of $\mathbb R^3$.
The space $\Lambda^1((\mathbb R^3)^*)$ is defined as $\{\omega:T\mathbb R^3:=\bigsqcup_{p\in\mathbb R^3}T_p\mathbb R^3\to\mathbb R\text{ s.t. }\omega \text{ is alternating}\}.$
Take a generic element of the tangent space $X=\underbrace{f_1\partial_x+f_2\partial_y+f_3\partial_z}_{f_1,f_2,f_3\in\mathcal C^{\infty}(\mathbb R^3)}$ and apply $\omega$: $$\omega(X)=\omega(f_1\partial_x+f_2\partial_y+f_3\partial_z)=\omega(f_1\partial_x)+\omega(f_2\partial_y)+\omega(f_3\partial_z)=\\=f_1(dz-ydx)(\partial_x)+f_2(dz-ydx)(\partial_y)+f_3(dz-ydx)(\partial_z)=\\=f_1[dz(\partial_x)-ydx(\partial_x)]+f_2[dz(\partial_y)-ydx(\partial_y)]+f_3[dz(\partial_z)-ydx(\partial_z)]$$ now observe that $dx,dy,dz$ is the (global) dual basis of $\partial_x,\partial_y,\partial_z$, so by definition $$f_1[0-y\cdot 1]+f_2[0-y\cdot 0]+f_3[1-y\cdot 0]=\\=-y\cdot f_1=0.$$

Answer 2

On $\mathbb R^3$, the definitions are very simple. \begin{align*} dx(a\partial_x+b\partial_y+c\partial_z) &= a,\\ dy(a\partial_x+b\partial_y+c\partial_z) &= b,\\ dz(a\partial_x+b\partial_y+c\partial_z) &= c.\\ \end{align*} In other words, $d(\text{variable})$ returns the coefficient of $\frac{\partial}{\partial(\text{variable})}$.

In general, if you are working on a manifold $M$ with a coordinate patch $(x^1,\dots,x^n)$, and vector fields $\partial_{x^1},\dots,\partial_{x^n}$, then $dx^i(\partial_{x^j}) = \delta^i_j$, where $\delta^i_j$ is the Kronecker delta symbol, equal to $1$ when $i = j$, and $0$ otherwise.