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A coisotropic submanifold $N$ satisfies $(T_{p}N{)}^{\omega }\subset T_{p}N$, where the superscript denotes the symplectic orthogonal complement. Every hypersurface of a symplectic manifold is coisotropic.

Let $\dim V=2n$ and let $V_{+}=\ker (S-I)$ and $V_{-}=\ker (S+I)$ be the two eigenspaces. The involution identity gives a direct sum, sinceFor $u,v$ in either one of these eigenspaces, the anti-symplectic involution identity impliesso both are isotropic subspaces of a symplectic vector space. Such a subspace has dimension at most $n$: $W\subset W^{\Omega }$ and the nondegenerate bilinear form $\Omega$ gives $\dim W^{\Omega }=2n-\dim W$, where $W^{\Omega }$ is the symplectic orthogonal complement. As their dimensions add to $2n$, both have dimension $n$. Therefore

On $N={\mu }^{-1}(-1/4)$, the diagonal fundamental vector field $X$ is nonzero, since the fixed-point set lies at moment map values $0$ and $-1/2$. The nondegenerate bilinear form ${\omega }_p$ and $d\mu ={\iota }_X\omega$ imply $d\mu \ne 0$, so $-1/4$ is a regular value. By the regular level set theorem, $N$ is a three-dimensional embedded submanifold of the four-dimensional Complex projective plane.

At $p\in N$, set $W=T_{p}N=\ker d{\mu }_p$. Then $\omega (X,v)=0$ for every $v\in W$, so $X\in W^{\omega }$, the symplectic orthogonal complement. This complement is one-dimensional and $d\mu (X)=\omega (X,X)=0$, givingThus the level set is a coisotropic submanifold, neither an isotropic submanifold nor a Lagrangian submanifold. Indeed an isotropic subspace of a symplectic vector space in dimension four has dimension at most two, whereas $\dim N=3$. Its characteristic line field is generated by the diagonal circle action.