Why does a two-form which annihilates tangent vectors of a submanifold imply that the restriction of the two-form to the submanifold vanishes?

Question

In the definition of isotropic and Lagrangian submanifolds of a symplectic manifold, it seems that there are two equivalent definitions.

1. The symplectic two-form annihilates tangent vectors (denoted $T_i$) of the submanifold $\gamma$, i.e., \begin{equation} \omega(T_1,T_2)=0. \end{equation}

2. The restriction of the two-form to the submanifold, $\gamma$ vanishes. \begin{equation} \omega|_{\gamma}=0. \end{equation}

Why are these two definitions equivalent, and does such an equivalence hold for any two-form?

Answer 1

The equivalence of these two properties has nothing to do with symplectic geometry.

Suppose $S$ is a submanifold of a manifold $M$, and let $i_S : S \to M$ be the inclusion. Then for any differential $k$-form $\alpha$ on $M$, the restriction of $\alpha$ to $S$, denoted $\alpha|_S$, is the by definition the $k$-form on $S$ given by $\alpha|_S := i_S^*\alpha$. Acting on vector fields $X_1, \dots, X_k$ on $S$, we have

$$(\alpha|_S)(X_1, \dots, X_k) = (i_S^*\alpha)(X_1, \dots, X_k) = \alpha((i_S)_*X_1, \dots, (i_S)_*X_k).$$

Note that $(i_S)_* = i_{TS}$ where $i_{TS} : TS \to TM$ is the inclusion map. So by identifying $X_j$ with its image in $TM$, we see that

$$(\alpha|_S)(X_1, \dots, X_k) = \alpha(X_1, \dots, X_k).$$

It follows that $\alpha|_S = 0$ if and only if $\alpha(X_1, \dots, X_k) = 0$ for all vector fields $X_1, \dots, X_k$ on $S$.