Maslov Index in heegaard floer homology - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T08:52:58Z http://mathoverflow.net/feeds/question/78017 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/78017/maslov-index-in-heegaard-floer-homology Maslov Index in heegaard floer homology yanqing 2011-10-13T12:25:18Z 2011-10-14T01:38:27Z <p>Can anyone explain what is definition of maslov index in Heegaard Floer homology? I am puzzled> Thank you.,</p> http://mathoverflow.net/questions/78017/maslov-index-in-heegaard-floer-homology/78062#78062 Answer by Chris Gerig for Maslov Index in heegaard floer homology Chris Gerig 2011-10-13T20:47:11Z 2011-10-13T20:47:11Z <p>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 <em>Maslov class</em> $\mu\in H^1(L)$ 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".</p> <p>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).</p> <p>The <em>Maslov index</em> 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''.</p> <p>We may then give the dimension of the moduli space $\mathcal{M}_{d+1}$ of (d+1)-pointed disks as $ind(\bar{\partial})+d-2$.</p> http://mathoverflow.net/questions/78017/maslov-index-in-heegaard-floer-homology/78087#78087 Answer by Paolo Ghiggini for Maslov Index in heegaard floer homology Paolo Ghiggini 2011-10-14T01:38:27Z 2011-10-14T01:38:27Z <p>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.</p> <p>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.</p> <hr>