Integral of symplectic form?

Question

It is often said that "differential forms are used for integration".

Typically people like to talk about the integral $\int_M \omega$ of a differential form $\omega$, and exterior derivative, one of the most important operation of differential form, is defined such that Stokes' theorem $\int_{\partial M} \omega = \int_M \mathrm{d} \omega$ can hold.

Symplectic manifold is a manifold equipped with a symplectic 2-form $\omega$. Therefore, according to the former consideration, it should be interesting to talk about integral of the symplectic form on a two-dimensional submanifold of a symplectic manifold.

However, it seems that people in the field of symplectic geometry is quite indifferent on this. Why?

Answer 1

This is only true if you are very pedantic about what you mean by "symplectic geometry". Namely, if you include symplectic topology, then one of the most important invariants of a symplectic manifold requires integrating the symplectic form over $2$-dimensional submanifolds. To be more precise, one considers maps $u:\Sigma\to M$ for $\Sigma$ a Riemann surface, and then integrates $\int_\Sigma u^*\omega$. If $u$ is an embedding, then this is clearly the same as integrating over the submanifold $u(\Sigma)\subseteq M$.

The reason why this is done, is because one wants to "count" the number of pseudo-holomorphic maps $\Sigma\to M$, up to reparameterisation. To be able to do this, one has to restrict to maps with finite energy, i.e. $\int_\Sigma u^*\omega<\infty$. Then the "moduli space" of solutions can be constructed, and compactified (finite energy is crucial to ensure compactification). If this moduli space has expected dimension $0$, then the compactification results in a finite number of points, which can be counted in some appropriate sense. The numbers which are obtained in this way are called the Gromov-Witten invariants of the symplectic manifold.

Answer 2

While I love the answer of Quaere Verum and cant give enough +1s on it, here is a few more "simple" facts about the symplectic form $\omega$ on a symplectic manifold $(M,\omega)$ of dimension $2n$:

First of all, note that a $2$-form on a $2n$ dimensional manifold (unless $n=1$) is not really meant to be integrated. But let me explain why closed stll matters and how we can still integrate using this fact:

$\omega^n$ is a volume form, called the symplectiv volume form. It gives you a "canonical" volume form on your manifold compatible with the structure. If in addition $M$ is closed (i.e. compact with no boundary), then you know that all the even cohomology groups $H^{2k}(M)\neq 0$. Use the previous fact to proof this!

Hamiltonian vector fields, moment maps - they all make use of the symplectic form. It usually involves the Cartan formula, which for the symplectic form just reads $\mathcal{L}_X\omega=d(i_{X}\omega)$, where $d$ is exeterior differentiation and $i$ is the contraction/interior product with a vector field $X$.

Last but not least, we also have so called Kähler manifolds, where the symplectic form can be seen as an "intermediate" form between the Riemannian metric $g$ and the (almost-)complex structure $J$ on M, i.e. we have $$ \omega(\cdot,\cdot)=g(\cdot,J\cdot). $$ This construction would define you a non-degenerate $2$-form for any choice of $J$ and $g$, however, it is not necessarily closed.