# Maslov Index property.

1. Mar 22, 2012

### MathematicalPhysicist

How do I show that Maslov index satisfies the next property (product property).

Let $\Lambda : \mathbb{R} / \mathbb{Z} \rightarrow \mathcal{L}(n)$, and $\Psi : \mathbb{R}/\mathbb{Z} \rightarrow Sp(2n)$ be two loops, then if $\mu$ is defined as the Maslov index then $\mu(\Psi \Lambda )= \mu(\Lambda)+2\mu(\Psi)$.

I have seen the proof in Mcduff of the Homotopy axiom of this index thought I am not sure how to use it to show the above.

Any hints?

Thanks, sorry if it's indeed that much easy.

2. Mar 23, 2012

### MathematicalPhysicist

Anyone?

3. Mar 23, 2012

### quasar987

Let $\Lambda\in\mathcal{L}(n)$ be a lagrangian subspace of R2n. Then there exists $\Psi\in U(n)= Sp(2n)\cap O(n)$ such that $\Lambda=\Psi\Lambda_{hor}$, where $\Lambda_{hor} = \mathbb{R}^n\oplus 0 \subset \mathbb{R}^{2n}$. This is the content of Lemma 2.31. If $\Psi = X+iY$ when considered a complex matrix, then the map $\rho$ in the proof of Theorem 2.35 is defined as $\rho(\Lambda)=det(\Psi)^2$.

Now, if $\Lambda(t)$ is a loop of lagrangians, then there is a corresponding loop $\Psi(t)$ of unitary matrices such that $\Lambda(t)=\Psi(t)\Lambda_{hor}$ and the Maslov index $\mu(\Lambda(t))$ is defined as $deg(\rho(\Lambda(t)))=deg(det(\Psi(t))^2)=2deg(det(\Psi(t)))=2\mu(\Psi(t))$ by definition of the Maslov index of loops of unitary matrices (see the proof of Theorem 2.29).

Now, getting closer to what we want, let $\Lambda(t)=\Psi(t)\Lambda_{hor}$ be a loop of lagrangians and let $\Phi(t)$ be a loop of unitary matrices. Exploiting the computation in the above paragraph + the product property for the Maslov index of matrices, we deduce that $\mu(\Phi(t)\Lambda(t))=\mu(\Phi(t)\Psi(t)\Lambda_{hor})=2\mu(\Phi(t)\Psi(t))=2\mu(\Phi(t))+2\mu(\Psi(t))=2\mu(\Phi(t))+\mu(\Lambda(t))$.

This is the result we want, except we want it for symplectic $\Phi(t)$ and not just unitary. But then the homotopy axiom takes over since any loop of symplectic matrices is homotopic to a loop of unitary ones (this we know since U(n) is the maximal compact subgroup of Sp(2n) (Proposition 2.22) and hence Sp(2n) deformation retracts onto U(n)).