Why is the matrix in Dirac's bracket formula invertible?

Question

I am reading the book "Introduction to mechanics and symmetry" by J.Marsden and T.Ratiu and am experenced a problem.

Let $(P,\Omega)$ be a symplectic manifold, a submanifold $S\subset P$ is called a symplectic submanifold when $\omega:=i^*\Omega$ is a symplectic form on $S, i:S\rightarrow P$ being the inclusion. Assume that $\dim P=2n,\dim S=2k$. In a neighborhood of a point $z_0\in S$, choose coordinates $z^1,...,z^{2n}$ on $P$ such that $S$ is given by $$z^{2k+1}=0,...,z^{2n}=0,$$ and so $z^1,...,z^{2k}$ provide the local coordinates for $S$.

In the formulation of Dirac's bracket formula, it appears the inverse of the matrix defined by $$C^{ij}(z)=\{z^i,z^j\},\ i,j=2k+1,...,2n.$$

The author said that it is easy to see one can choose coordinates such that the matrix $C=\{C^{ij}\}$ is invertible. But I cannot see the reason.

Thank you if you can give a short proof or refer me to a good reference.

Chengbo

Answer 1

The issue is discussed, perhaps not completely clearly, in Henneaux and Teitelboim, Quantization of Gauge Systems. Princeton University Press, 1992. They prove, in chapter two, that the Dirac bracket is precisely the Poisson bracket of $S$ determined by the symplectic form $\Omega|_S$ which is the pullback of $\Omega$ from $P$. In fact, this approach is preferable over the use of Dirac brackets, as I understand the story, because pulling back differential forms is a more elementary operation than generating these elaborate Dirac brackets.