Let $(V,\omega)$ be a $2g$-dimensional symplectic vector space. I'm trying to understand the Maslov triple product. I know that it can be defined in a variety of ways, but for the applications I'm interested in the right definition is as follows. Let $\lambda_1,\lambda_2,\lambda_3 \subset V$ be Lagrangian subspaces. On the subspace $W = (\lambda_1 + \lambda_2) \cap \lambda_3$ of $V$, we can define a bilinear map $\ast$ via the formula $$(a_1+a_2) \ast (b_1+b_2) = \omega(a_2,b_1) \quad \quad (a_1,b_1 \in \lambda_1, a_2,b_2 \in \lambda_2).$$ It is not hard to show that this this is well-defined and symmetric. Then $\mu(\lambda_1,\lambda_2,\lambda_3)$ is the signature of the resulting inner product. It is easy to show that $\mu(\lambda_1,\lambda_2,\lambda_3)=-\mu(\lambda_2,\lambda_1,\lambda_3)$. What I'm having difficulty showing is $$\mu(\lambda_1,\lambda_2,\lambda_3) = -\mu(\lambda_1,\lambda_3,\lambda_2).$$ This seems tricky; the vector spaces that the relevant inner products are defined on don't even seem to have the same dimension. Can anyone help me? The paper I'm reading is here; the fact I'm trying to prove is Lemma 2.1, which is stated without proof.

ps : I'm not sure how to tag this, so I just chose some tags that are related to things that use the Maslov index.


you can find a proof in Turaev's book Quantum Invariants of Knots and 3-manifolds p183: The bilinear form vanishes on $\lambda_1 \cap \lambda_3$ so it is equivalent to a form on $((\lambda_1 + \lambda_2) \cap \lambda_3)/(\lambda_1 \cap \lambda_3)\simeq ((\lambda_1 + \lambda_3) \cap \lambda_2)/(\lambda_1 \cap \lambda_2)$

$a_3=a_1+a_2\mapsto a_2=a_3-a_1$

and under this isomorphism, the quadratic forms correspond.

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    $\begingroup$ Welcome to Math Overflow. When I click the link I don't get to see a preview of any pages, perhaps because I'm in the US rather than France. It's generally good practice to summarize what you're linking to so that a reader can see how the question is answered even if they can't click the link. $\endgroup$ – David White Nov 12 '13 at 17:43
  • $\begingroup$ sorry, I did not realized. $\endgroup$ – B. Patureau Nov 12 '13 at 19:35

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