Compute the contraction of a 1-form with a vector field Question: Let $\alpha$ be the $1$-form on $\mathbb{R}^3$ given by $\alpha=zdy-ydz$ and let $\mathbb{X}$ be the vector field on $\mathbb{R}^3$ given by $\mathbb{X}=(0,y,-z)$. Compute $i_\mathbb{X}\alpha$
Answer: $i_{(0,y,-z)}(zdy-ydz)=i_{(0,y,-z)}(zdy)-i_{(0,y,-z)}(ydz)=zy-y(-z)=2yz.$
Is this answer correct? I feel that I must have done something wrong as I feel there should be more to the question.
I am not familiar with taking the contraction of a one form.
Can you just use the relation: $i_vf^{(j)}=v^j$?
Any feedback would be greatly appreciated.
 A: Your computation of $i_{\mathbb{X}}\alpha$ is correct. Let me explain some general facts that work on any manifold (at least locally), but keep it restricted to $\mathbb{R}^3$.
First note that a one-form $\alpha$ is a function on vector fields. Any one-form can be written as a linear combination of $dx$, $dy$, and $dz$ where the coefficients are functions of $x$, $y$, and $z$ - another way of saying this is that $\{dx, dy, dz\}$ is a basis for the one-forms. By linearity, in order to understand the value of a one-form on a vector field, it is enough to know the values of $dx$, $dy$, and $dz$ on a vector field. If $V = (v^1, v^2, v^3)$ is a vector field (here the $v^i$ are functions of $x$, $y$, and $z$), then 
$$dx(v^1, v^2, v^3) = v^1\qquad dy(v^1, v^2, v^3) = v^2\qquad dz(v^1, v^2, v^3) = v^3.$$
Because of these relationships, one can specify $dx$, $dy$, and $dz$ as the dual basis of the standard basis of vector fields.
What does any of this have to do with contraction? Well, in the case of one-forms $i_{\mathbb{X}}\alpha = \alpha(\mathbb{X})$; i.e. contraction by a vector field is nothing but evaluation of the form on the vector field. Note, in the case of $k$-forms where $k > 1$, contraction is not the same as evaluation.
A: $\beta$'s irrelevance notwithstanding, your computation of $\iota_{\mathbb X}\alpha$ is correct. I'm not sure what your final question is supposed to mean.
