MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Can anyone explain what is definition of maslov index in Heegaard Floer homology? I am puzzled> Thank you.,

share|cite|improve this question
It's the index of the linearisation of a certain differential operator (between two Banach manifolds), that gives the expected dimension of the moduli space of pseudo-holomorphic discs in the symmetric product ${\rm Sym}^g(\Sigma)$. Take a look at the original papers, especially at Ozsváth, Szabó: "Holomorphic disks and invariants for closed three-manifolds", published in the Annals of Mathematics. These aspects are discussed in section 3. What puzzles you about this definition? – Marco Golla Oct 13 '11 at 13:00
Please say more about what puzzles you. There is no tax on words here. – S. Carnahan Oct 13 '11 at 17:40
up vote 2 down vote accepted

Working in the Fukaya category (objects are Lagrangian submanifolds $L_i$ of the symplectic 2n-manifold $M$ with some extra hypotheses), assume $2\cdot c_1(M)=0$. Then we can trivialize $(\bigwedge^n_\mathbb{C}TM)^2$ and choose a nowhere-vanishing section $\Omega\in\Gamma((\bigwedge^n_\mathbb{C}TM)^2)$. This induces a "phase-squared map" $\phi:LGr\rightarrow S^1$, where $LGr$ is the Lagrangian Grassmannian bundle over $M$, which on Lagrangians is $\phi(L):(d\;vol_L)^2/\Omega\rightarrow S^1$. Now $H^1(LGr)\cong\mathbb{Z}[\phi]$ so there exists a bundle $\tilde{LGr}\rightarrow LGr$ (fiber-wise universal cover). The Maslov class $\mu\in H^1(L)$ is the map induced by $\phi$, and $\mu=0$ iff $L$ lifts to $\tilde{LGr}$ (lift $l:L\rightarrow \tilde{LGr}$). Such a pair $(L,l)$ is a "graded Lagrangian".

For $x\in L_0\pitchfork L_1$ there exists a unique path $\gamma:[0,1]\rightarrow \tilde{LGr}$ (up to homotopy) connecting the lift of $x$ in $\tilde{L}_0$ to the lift of $x$ in $\tilde{L}_1$. Now let us work locally with $L_i=\prod_j\mathbb{R}$, and $\phi$ has components $\phi_j$, and $\tilde{\phi}_j$ is the lift of $\phi_j$ to $\tilde{LGr}\rightarrow\mathbb{R}$. Then we have an index $i(x)=\sum_j\ulcorner \tilde{\phi}_j(x\in \tilde{L}_1)-\tilde{\phi}_j(x\in\tilde{L}_0)\urcorner$ (ceiling notation).

The Maslov index of the operator $\bar{\partial}$ (defining a moduli space of J-holomorphic strips), is then $ind(\bar{\partial})=i(x)-\sum_j i(x_j)$, where $x$ is your ''incoming point'' on the $J$-holomorphic strip and the $x_j$ are your ''outgoing points''.

We may then give the dimension of the moduli space $\mathcal{M}_{d+1}$ of (d+1)-pointed disks as $ind(\bar{\partial})+d-2$.

share|cite|improve this answer
See Paolo's remark below. – Chris Gerig Nov 16 '11 at 20:12

The symmetric product doesn't satisfy $2 c_1 =0$, therefore the machinery about graded lagrangians explained in Seidels's book cannot be applied. In Heegaard Floer homology the Maslov index can, and in fact does, depend on the homotopy class of the disc. As Marco said, you should explain us what puzzels you, so that we can give you a better answer.

You might also want to look at Lipshitz's paper "A cylindrical reformulation of Heegaard Floer homology", where he works out an explicit formula for the index.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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