We have the exterior 1-form on $\mathbb{R}^3$ given by $\alpha=x dy\wedge dz-ydz\wedge dx\in\Omega^2(\mathbb{R}^3)$. I'm trying to get a explicit expression of $\alpha$ restricted to the surface $xy=1$.
I get that applied to a vector $(v_1,v_2,v_3)\in T_pM$ with $p=(p_1,p_2,p_3)$ it should be $v_3(p_1v_2-p_2v_1)$, but I'm not sure I understood it well.
First we parametrice the surface as follows, $$\left\{\begin{matrix}x&=&x\\y&=&\frac{1}{x}\\z&=&z\end{matrix}\right.$$ so we get that, $$\left\{\begin{matrix}dx&=&dx\\dy&=&-\frac{1}{x^2}dx\\dz&=&dz\end{matrix}\right.$$ so in $xy=1$ it is, $$x dy\wedge dz+ydz\wedge dx=-\frac{1}{x}dx\wedge dz-\frac{1}{x}dz\wedge dx=0$$