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?



model co-isotropic submanifold
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.



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> **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 are **neighborhood equivalent** to this embedding.
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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:



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> Mark J. Gotay, *[On coisotropic imbeddings of presymplectic manifolds](http://www.ams.org/journals/proc/1982-084-01/S0002-9939-1982-0633290-X/)*. Proc. Amer. Math. Soc. 84 (1982), 111-114.
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