I'm relatively new to contact forms, and to differential forms in generals; please forgive if this is too simple:
I want to show that if $w$ is a contact form (say for a 3-manifold $M^3$), then $dw$ , the exterior derivative of $w$ is nowhere zero on the contact planes. The doubt I have is that my result seems to depend on using basis vectors of the contact plane, and I don't know if this is enough to guarantee the result for general vectors in the contact plane. Here is what I have so far (all evaluated at a generic point $p$ in $M^3$):
Let $w$ be a contact form for $M^3$. Then we have that $ w \wedge dw \neq 0$. Let $X$ be transverse to the contact plane $\eta_p$ at $p$ and let {$Y_p,Z_p$} be basis for $\eta_p$. Then we have : $$w \wedge dw (X,Y,Z)=w(X)dw(Y,Z)-w(Y)dw(X,Z)+w(Z)dw(X,Y) $$ (##).
Now, $w$ is zero on the contact planes b definition ( since the contact planes are defined as the kernel of the form.), so that the second and third term above are zero, and since $w$ is a contact form, we have $w \wedge dw \neq 0$. Since I am assuming $X$ is transverse to the plane, we have $w(X) \neq 0$, so we must then have $dw(Y,Z)$ --the restriction of $w$ to the contact plane -- non-zero.
Is this correct? Thanks.
Your argument is fine. If $dw$ was zero on the contact plane at $p$ this means that $dw(Y,Z)=0$ for any two $Y,Z$ in the contact plane, so it suffices to produce a single counterexample like you did. That being said your argument covers almost every combination of $Y,Z$ any way - the only other possibility is when $Y = cZ$, and of course $\omega(Z,cZ)=0$ for any differential form $\omega$ so there's not really any information here.