This is a general question about ranks of differential forms.
I read in a book the phrase "symplectic form has constant rank..." I understand that the symplectic form is a nondegenerate differential 2-form. But what is the rank of a symplectic form?
In general, what is the rank of a differential form? When I think rank, I think about the dimension of the range of a matrix. Is there some matrix associated with a differential form? Or does rank in this context refer to tensor rank?
I also understand the notion of the rank of a differentiable map between manifolds. If $f:M \rightarrow N$ is differentiable and $p\in M$, the rank is the dimension of the image of the derivative:
$\mathrm{rank}(f)_p = \mathrm{dim}(\mathrm{image}(df_p))$,
where $df_p:T_p M \rightarrow T_{f(p)} N$.
Any two-form gives a (skew-symmetric) bilinear form $TM_p \times TM_p \rightarrow \mathbb{R}$.
The rank of the two-form at $p$ is the rank of this bilinear form.
What's the rank of a bilinear form?
A bilinear map $B: V \times V \rightarrow \mathbb{R}$ is equivalent to a linear map $V \rightarrow V^*$ given by $v \mapsto B(v,\cdot)$. The rank of this linear map is the rank of the bilinear form.
Some further comments: Non-degenerate means full rank, i.e. that the linear maps $TM_p \rightarrow TM^*_p$ you get are isomorphisms. So a symplectic form is always constant rank.