Just as the zero section in $T^*\mathbb{R}^N$ (equipped with the standard symplectic form) is the "model" / "quintessential" Lagrangian submanifold, does $T^*\mathbb{R}^N$ have a "model" co-isotropic submanifold?

## 1 Answer

Let $(M, \Omega)$ be a presymplectic manifold and let $E \longrightarrow M$ be the **characteristic bundle** of $(M, \Omega)$, defined fiberwise by
$$E_x = \{v \in T_x M : \Omega_x(v, -) = 0\}.$$
If $E^\ast$ denotes the dual bundle to $E$, then we have the following classification theorem.

Theorem.There exists a symplectic structure $\omega$ on a neighborhood $U$ of the zero section of $E^\ast$. $(M, \Omega)$ is coisotropically embedded in $(U, \omega)$ as the zero section, and furthermore all coisotropic embeddings of $(M, \Omega)$ into a symplectic manifold areneighborhood equivalentto this embedding.

The presymplectic form $\Omega$ specifies what the pullback of $\omega$ to $M$ is. For a Lagrangian embedding, $\Omega \equiv 0$ and hence $E^\ast = T^\ast M$, and we recover the fact that the zero section of $T^\ast M$ is the canonical model for a Lagrangian embedding of $M$ into a symplectic manifold.

For the details, see the original paper of Gotay:

Mark J. Gotay,

On coisotropic imbeddings of presymplectic manifolds. Proc. Amer. Math. Soc. 84 (1982), 111-114.

here. What is directly relevant to your question is Chapter 3, Section 3.5. and Chapter 4, Theorem 3.5 -- this is the standard neighbourhood theorem in symplectic geometry, a slight generalization and reformulation of the Theorem stated in Henry's answer. $\endgroup$