4
$\begingroup$

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.

$\endgroup$
5
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.