Hamiltonian circle actions and Lefschetz pencils - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T10:46:44Z http://mathoverflow.net/feeds/question/21331 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/21331/hamiltonian-circle-actions-and-lefschetz-pencils Hamiltonian circle actions and Lefschetz pencils Don Stanley 2010-04-14T12:20:47Z 2011-07-29T19:18:24Z <p>Suppose that $M$ is a symplectic manifold with a Hamiltonian circle action. Is there a topological Lefschetz pencil on $M$, $f\colon M-A \rightarrow S^2$, such that the fibers are symplectic submanifolds of $M$ and such that the circle action restricts to a Hamiltonian circle action on the fibers?</p> <p>(After reading Tim's answer.)</p> <p>Are there some cases when there do exist such things?</p> http://mathoverflow.net/questions/21331/hamiltonian-circle-actions-and-lefschetz-pencils/21339#21339 Answer by Tim Perutz for Hamiltonian circle actions and Lefschetz pencils Tim Perutz 2010-04-14T14:08:21Z 2010-04-14T15:15:35Z <p>Hm, good question. This would be potentially interesting from the point of view of classifying Hamiltonian circle-actions, "by induction" on the dimension.</p> <p>(1) I'm pretty sure that the answer is not known to be "yes". A MathSciNet search of papers citing Donaldson's paper on symplectic Lefschetz pencils turns up nothing. </p> <p>(2) It sounds plausible to me (provided that one assumes that the symplectic class is integral). Of course, someone may spot a counterexample. </p> <p>(3) Making Donaldson's construction work equivariantly seems far from trivial. One can lift the circle-action to the line bundle $L\to M$ whose curvature is the symplectic form. We now want a sequence of approximately holomorphic sections $(s_k^0,s_k^1)$ of $L^k$, satisfying estimated transversality conditions, such that $(s_k^0:s_k^1)$ is $S^1$-invariant. But we won't be able to make the sections themselves $S^1$-equivariant, because the construction involves bombarding $M$ with Gaussian sections; they live in little coordinate charts, while the circle-orbits may be big.</p> http://mathoverflow.net/questions/21331/hamiltonian-circle-actions-and-lefschetz-pencils/21629#21629 Answer by Dmitri for Hamiltonian circle actions and Lefschetz pencils Dmitri 2010-04-16T23:40:50Z 2010-04-16T23:40:50Z <p>I think, that in order to answer this question it is worth to conisder the complex algebraic analog of this question. Namely, suppose we have a $\mathbb C^*$ action on a projective manifold $V^n$. Can we find an invariant Lefshetz pencil? </p> <p>It is sufficient to conisder the case of (complex) surfaces to spot some problems. Namely, the action of $\mathbb C^*$ in a neighborhood of an isolated fixed point should be of one of the following $3$ types: $$(z,w)\to (tz,tw), \;\; (z,w)\to (tz,t^{-1}w),\;\; (z,w)\to (tz,w)$$ </p> <p>where $t\in \mathbb C^*$ and $z,w$ are local coordinates at the fixed point.</p> <p>There are not so many examples of $\mathbb C^*$ actions on surfaces that have only these three types of fixed points. But here are two examples: first is $\mathbb CP^2$ with the action that fixes one point and a separate line. Second is the action on $\mathbb CP^1\times \mathbb CP^1$ that fixes $4$ points. In both cases there is an (obvious) invariant Lefshetz pencil. You can also take the first example and blow up several distinct points on the fixed line in $\mathbb CP^2$. Maybe this is the complete list... Surelly, from all these examples one gets also the symplectic Lefshetz pencil of the kind that you wish to have.</p> <p>I don't see why there will be much more examples if you will go into symplectic cathegory. </p> http://mathoverflow.net/questions/21331/hamiltonian-circle-actions-and-lefschetz-pencils/71602#71602 Answer by Weiwei for Hamiltonian circle actions and Lefschetz pencils Weiwei 2011-07-29T19:18:24Z 2011-07-29T19:18:24Z <p>In dimension $4$ I feel this should be proved, and let me make a fuzzy attempt. In dimension $4$ Hamiltonian $S^1$-manifolds are all rational or ruled, so except for $\mathbb{C}P^2$ which needs independent check, you may just take an $S^1$-equivariant almost complex structures and consider the self-intersection zero class which is automatically transversal. This gives a fibration where the singular points come from bubblings which should be separated on the base by perturbations. This fibration should be what you want, and the plausibility comes from that if your $S^1$-action comes from transporting the fibers instead of rotating the fibers, then your base needs to rotate, but the singular values prevents you to do so except for some extreme cases where one checks by case study.</p> <p>If this is the only construction in the 4-dimensional case, I feel it very hard to extend to higher dimensions with current techniques. Because if you try to use Donaldson's construction on $4$ dimension, it clearly never give you satisfactory answers because the Donaldson hypersurfaces are usually not spheres, so this line seems not approaching a general argument; while on the other hand, if you give up Donaldson's construction, it is already extremely hard to even just to find a smooth symplectic hypersurface, at least for me personally.</p>