A linear Lagrangian relation sends a Lagrangian subspace to a Lagrangian subspace

Let $K=\mathrm{\Lambda }\cap (L\oplus W)$. Projection $p_W:K\to W$ has image $R_{\mathrm{\Lambda }}(L)$ and kernel $T=\{x\in L:(x,0)\in \mathrm{\Lambda }\}$. For $(x,y),(x^{\mathrm{\prime }},y^{\mathrm{\prime }})\in K$, isotropy of $\mathrm{\Lambda }$ gives $-{\omega }_V(x,x^{\mathrm{\prime }})+{\omega }_W(y,y^{\mathrm{\prime }})=0$; isotropy of $L$ makes the first term zero. Hence $R_{\mathrm{\Lambda }}(L)$ is isotropic.

In $\bar{V}\oplus W$, the orthogonal complement of $L\oplus W$ is $L\oplus 0$, because $L^{\perp }=L$. Since $\mathrm{\Lambda }={\mathrm{\Lambda }}^{\perp }$, the orthogonal complement of $\mathrm{\Lambda }+(L\oplus W)$ is $\mathrm{\Lambda }\cap (L\oplus 0)$, of dimension $\dim T$. For any subspace $U$ of a finite-dimensional symplectic space, the map $v\mapsto \omega (v,-){\mathrm{\mid }}_U$ onto $U^{\ast }$ is surjective: extend a functional from a basis of $U$ to the ambient space and use the isomorphism defined by the nondegenerate form. Its kernel is $U^{\perp }$, so $\dim U^{\perp }=\dim V-\dim U$. Rank-nullity and [F2] therefore give $\dim K=(n+m)+(n+2m)-(2n+2m-\dim T)=m+\dim T\mathrm{.}$ The kernel of $p_W{\mathrm{\mid }}_K$ has dimension $\dim T$, so $\dim R_{\mathrm{\Lambda }}(L)=m$.

The isotropic subspace $R_{\mathrm{\Lambda }}(L)$ has dimension $m$. The formula proved in step 2.1 gives $\dim R_{\mathrm{\Lambda }}(L{)}^{\perp }=2m-m=m$. Since isotropy gives $R_{\mathrm{\Lambda }}(L)\subseteq R_{\mathrm{\Lambda }}(L{)}^{\perp }$, equality follows; this is the definition of Lagrangian in [F1].