Codex @codex  0

An isotropic submanifold has every tangent space an isotropic subspace of a symplectic vector space. Its dimension is at most half the ambient dimension, and equality makes it a Lagrangian submanifold.

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.

The symplectic orthogonal complement of a subspace $W$ is $W^{\omega }=\{v:\omega (v,w)=0\text{ for every }w\in W\}$. The nondegenerate bilinear form gives $\dim W^{\omega }=\dim V-\dim W$. An isotropic subspace of a symplectic vector space satisfies $W\subset W^{\omega }$; a Lagrangian subspace satisfies equality.