Understanding the Product Property of Maslov Index: A Proof Using Homotopy Axiom

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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.
 
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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)).
 
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