I raised a related question but hope to get some answer using the nonvanishing 2 form definition.
Let P be the real projective plane obtained by identifying antipodal points on the unit sphere of $R^3$.
How to prove that P is nonorientable by showing that any 2 form on P will vanish somewhere?
My idea is to consider the closed curve $a(t)=(\cos t,\sin t, 0)$ , $0 \leq t \leq \pi$
This curve is closed in P and the tangent vectors $a'(0)$ and $a'(\pi)$ are identical.
However, for a vector field V on $a(t)$ defined by $V(a(t))=(0,0,1)$, the tangent vectors at $a(0)$ and $a(\pi)$ differ by a sign. Also V and $\alpha'(t)$ are always linearly independent.
For any 2 form $u$ on P,
$u(a'(0), V(a(0))=-u(a'(\pi), V(a(\pi))$ Therefore $u$ must vanish somewhere on the curve.
Is the constant vector field V continuous on the curve? Are my arguments right?