Lagrangian subspace

A Lagrangian subspace of a $2n$-dimensional symplectic vector space $(V,\omega )$ is an $n$-dimensional subspace $L$ on which $\omega$ vanishes.

A Lagrangian complement to $L$ in a symplectic vector space is a Lagrangian subspace $K$ with $E=L\oplus K$. Such complements exist smoothly for Lagrangian subbundles. Identify any initial complement with $L^{\ast }$ using the symplectic cross-pairing. If its internal pairing is $b(\eta ,\zeta )$, correct its lift by a vector $a\eta \in L$ with $\zeta (a\eta )=-b(\eta ,\zeta )/2$. The corrected complement has zero internal pairing.

The Lagrangian Grassmannian is the space of all Lagrangian subspaces of a symplectic vector space. The unitary group acts transitively on $\mathrm{LGr}({\mathbb{R}}^{2n})$, and the stabilizer of ${\mathbb{R}}^n$ is the orthogonal group $O(n)$, giving

Relative to a fixed Lagrangian subspace $L_0$, the Maslov cycle consists of Lagrangian subspaces $L$ with $L\cap L_0\ne 0$. In a two-dimensional symplectic space all lines are Lagrangian, and the cycle is the single line $L_0$ in the projective line. Fibrewise versions yield sections of a projective-line bundle and record passage through a prescribed vertical direction.

For a surface thermostat, the planes containing $F$ form the projective bundle of $T(SM)/\mathbb{R}F$. The vertical Maslov cycle is the section selecting the image of the vertical vector field of a surface unit tangent bundle. It is a smooth hypersurface. In the local projective chart $\mathbb{R}F+\mathbb{R}(V+rH)$, it is $r=0$; the lifted flow satisfies $\dot{r}=1$ on the cycle, hence crosses it transversely.

The Maslov index assigns an integer to a loop of Lagrangian subspaces. Under $\mathrm{LGr}({\mathbb{R}}^{2n})\cong U(n)/O(n)$, it is the degree of the map induced by $A\mapsto \det (A{)}^2$.

The Maslov map is well defined because every matrix in $O(n)$ has determinant $\pm 1$. The loopmaps to $e^{2\pi it}$ and therefore has Maslov index one.

For a rank-$2n$ symplectic vector bundle $E\to B$, the Lagrangian Grassmannian bundle has fiber $\mathrm{LGr}(E_b)$ over $b$. A symplectic trivialization of $E$ induces a trivialization $\mathcal{L}Gr(E)\cong B\times \mathrm{LGr}({\mathbb{R}}^{2n})$.

If $E\to B$ is an oriented real plane bundle, then $\mathcal{L}Gr(E)=\mathbb{P}(E)$ is an oriented circle bundle. Its transition rotations have twice the angle of those of $E$, so