Why we have to fix markers in SFT? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-18T07:26:12Z http://mathoverflow.net/feeds/question/21919 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/21919/why-we-have-to-fix-markers-in-sft Why we have to fix markers in SFT? Mohammad F.Tehrani 2010-04-20T01:44:43Z 2010-04-29T15:50:00Z <p>In Symplectic field theory ( Hofer-Eliashberg....) and considering moduli of J-holomorphic curves asymptotic to Reeb orbits at punctures (J-holomorphic curve into a symplectic cobordism), The authors and every body else who works on SFT, fix markers both on Reeb orbit and at the puncture. Does any body clearly understand why they do that ? What is that for ? Why is that needed? </p> <p>References</p> <p>--An introduction to symplectic field theory and also --Coherent orientations in SFT </p> http://mathoverflow.net/questions/21919/why-we-have-to-fix-markers-in-sft/21968#21968 Answer by Tim Perutz for Why we have to fix markers in SFT? Tim Perutz 2010-04-20T15:06:55Z 2010-04-20T15:06:55Z <p>I'll talk about cylindrical contact homology, which is a relatively well-established part of SFT. It's a variant of Hamiltonian Floer homology for contact manifolds. Let $\alpha$ be a contact 1-form on a closed manifold $V$, and let $A\colon LV\to \mathbb{R}$, $\gamma\mapsto \int_{S^1}{\gamma^*\alpha}$ be the action functional on its loopspace. It's invariant under rotation of loops, hence not a Morse function. The critical points are the 1-periodic orbits of the Reeb vector field. </p> <p>Here are three things we could try to construct:</p> <p>(i) The Floer cohomology of $A$. Each geometric Reeb orbit with multiplicity contributes $H^*(S^1)$ to the cochain complex.</p> <p>(ii) The $S^1$-equivariant Floer cohomology of $A$. Each geometric Reeb orbit with multiplicity contributes $H^\ast_{S^1}(S^1)$ to the complex (the $S^1$ action on itself depends on the multiplicity).</p> <p>(iii) The Floer homology of $A$ on $LV/S^1$ (over $\mathbb{Q}$). Each geometric Reeb orbit with multiplicity contributes $\mathbb{Q}$ to the complex.</p> <p>Which of these things work? None of them, without substantial modification. All of them, with modifications. They are called (i) symplectic cohomology; (ii) circle-equivariant symplectic cohomology; (iii) cylindrical contact homology. In each case, the differential essentially counts pseudo-holomorphic maps $S^1\times \mathbb{R}\to V\times \mathbb{R}$, asymptotic to periodic Reeb orbits, but with subtle differences. </p> <p>Since (iii) is a quotient construction we have to allow loop-rotation; so we mark a standard point on $S^1\times -\infty$, an arbitrary point on $S^1\times +\infty$, and insist that these markers map to chosen points on the Reeb orbits. In (i), we would use the standard marker also on $S^1\times +\infty$; allowing it to vary defines a loop-rotation (BV) operator on symplectic cohomology.</p> <p>See Bourgeois-Oancea's recent Inventiones paper for info on the relationship between these constructions.</p> http://mathoverflow.net/questions/21919/why-we-have-to-fix-markers-in-sft/22995#22995 Answer by Mohammad F.Tehrani for Why we have to fix markers in SFT? Mohammad F.Tehrani 2010-04-29T15:50:00Z 2010-04-29T15:50:00Z <p>The reason is related to orientations, and more specifically to bad orbits (see the end of paper :Coherent orientation in SFT). Assume that a holomorphic curve is asymptotic to a bad orbit ( means $n-1+CZ_{index}$ is even) of multiplicity 2m (a bad orbit always has even multiplicity). Theorientation depends on the choice of an asymptotic marker near the corresponding puncture; if it is rotated by \pi/m, then the orientation is reversed. Hence, the algebraic count of curves asympttic to a bad orbit always vanishes. Of course, as soon as we restrict to good orbits only (so that the orienttion does not depend on the marker anymore), then the asymptotic markers are not needed anymore for the orientations.</p>