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Maslov Index property.

  1. Mar 22, 2012 #1

    MathematicalPhysicist

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

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

    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. jcsd
  3. Mar 23, 2012 #2

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  4. Mar 23, 2012 #3

    quasar987

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

    Now, if [itex]\Lambda(t)[/itex] is a loop of lagrangians, then there is a corresponding loop [itex]\Psi(t)[/itex] of unitary matrices such that [itex]\Lambda(t)=\Psi(t)\Lambda_{hor}[/itex] and the Maslov index [itex]\mu(\Lambda(t))[/itex] is defined as [itex]deg(\rho(\Lambda(t)))=deg(det(\Psi(t))^2)=2deg(det(\Psi(t)))=2\mu(\Psi(t))[/itex] 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 [itex]\Lambda(t)=\Psi(t)\Lambda_{hor}[/itex] be a loop of lagrangians and let [itex]\Phi(t)[/itex] 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 [itex]\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))[/itex].

    This is the result we want, except we want it for symplectic [itex]\Phi(t)[/itex] 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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