I'm looking into symplectic forms which requires the differential 2-form to be closed and non-degenerate. I know that in Euclidean space, a differential 2-form $\omega:\mathbb{R}^n\times \mathbb{R}^n\rightarrow \mathbb{R}$ is non-degenerate if $\omega(v,w)=0$ implies $w=0$ for all $v\in\mathbb{R}^n$. However, what would it mean for a differential 2-form on a manifold to be non-degenerate? Would it simply be the same definition but $\omega(v,w)=0$ implies $w=0$ for all $v\in M$ where M is the manifold?
Also, how do you show a 2-form is non-degenerate, for example, in the standard symplectic in $\mathbb{R}^2$, if $v=\begin{pmatrix}a\\b \end{pmatrix}$ and $w=\begin{pmatrix}c\\d \end{pmatrix}$ then for $\omega (v,w)=0$, $\omega(v,w)=ad-bc=0$ but why does that imply $c=d=0$? Sorry if this is a stupid question.
Given a real vector space $V$, a bilinear form $\Omega:V\times V\rightarrow \mathbb{R}$ is non-degenerate if $\Omega(v,w)=0$ for all $w\in V$ implies $v=0$.
A differential $2$-form $\omega$ on a manifold $M$ gives at each point $p\in M$ a bilinear form on the tangent space $T_{p}M$: $$\omega_{p}:T_{p}M\times T_{p}M\rightarrow \mathbb{R}.$$ We say that $\omega$ is non-degenerate if $\omega_{p}$ is non-degenerate for all $p\in M$.
As for the standard symplectic form $\Omega$ on $\mathbb{R}^{2}$, assume that $v=(a,b)$ is such that $\Omega(v,w)=0$ for all $w\in\mathbb{R}^{2}$. This implies that $$ad-bc=0$$ for all $c,d\in\mathbb{R}$. In particular for $d=1$ and $c=0$, which yields $a=0$. Setting $d=0$ and $c=1$ gives $b=0$.