Quantum cohomology rings as invariants - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T17:10:53Z http://mathoverflow.net/feeds/question/47322 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/47322/quantum-cohomology-rings-as-invariants Quantum cohomology rings as invariants Cat 2010-11-25T13:04:29Z 2010-11-25T18:12:05Z <p>Let $X$ and $Y$ manifolds. What kind of relations between them (like homeomorphism, diffeomorphism, homotopy equivalence) gives an isomorphic quantum cohomology rings?</p> http://mathoverflow.net/questions/47322/quantum-cohomology-rings-as-invariants/47331#47331 Answer by Barbara for Quantum cohomology rings as invariants Barbara 2010-11-25T14:55:42Z 2010-11-25T14:55:42Z <p>Quantum cohomology, and more generally Gromov Witten invariants, are invariants under deformation. Given a family $X$ of compact symplectic manifolds over a base $B$, to a path in $B$ one can associate an isomorphism between the quantum cohomology of the fiber $X_0$ over the starting pooint of the path and the qc of the fiber $X_1$ over the end point of the path. Homotopic paths (homotopic wrt end points) induce the same isomorphism. In particular $\pi_1(B,0)$ acts on the quantum cohomology of $X_0$.</p> http://mathoverflow.net/questions/47322/quantum-cohomology-rings-as-invariants/47352#47352 Answer by Jim Bryan for Quantum cohomology rings as invariants Jim Bryan 2010-11-25T18:03:47Z 2010-11-25T18:03:47Z <p>If $X$ and $Y$ are Calabi-Yau threefolds which are obtained from each other from a simple flop, then their quantum cohomologies are isomorphic (Li and Ruan proved this). Lee-Lin-Wang prove the analog of this in higher dimensions. One of the things that makes this interesting is that the <em>classical</em> cohomology rings are not (in general) isomorphic, but the quantum cohomology rings are.</p> <p>More generally, if $X$ and $Y$ are birational and $K$-equivalent (i.e. there is a resolution of the birational map $X\leftarrow W \to Y$ such that $K_X$ and $K_Y$ are isomorphic when pulled back to $W$), one expects their Gromov-Witten theories to be "equivalent". In some situations, "equivalent" will imply isomorphic quantum cohomologies, but not in general. To formulate the general equivalence properly requires Coates and Givental's Lagrangian cone formalism. Crepant resolutions are a special case of this and so the "crepant resolution conjecture" is part of this set of equivalences. Here are some papers on the subject (please forgive my laziness at only putting arXiv numbers and not journal references):</p> <p>Li-Ruan arXiv:math/9803036</p> <p>Bryan-Graber arXiv:math/0610129 </p> <p>Iritani arXiv:0809.2749</p> <p>Lee-Lin-Wang math.AG/0608370</p> <p>Coates-Ruan arXiv:0710.5901</p> http://mathoverflow.net/questions/47322/quantum-cohomology-rings-as-invariants/47355#47355 Answer by Bruno Martelli for Quantum cohomology rings as invariants Bruno Martelli 2010-11-25T18:12:05Z 2010-11-25T18:12:05Z <p>I am not an expert in the field, but I guess that there are examples of diffeomorphic symplectic manifolds with non-isomorphic quantum cohomologies (hence homotopy equivalence, homeomorphism, or even diffeomorphism alone is not enough to give an isomorphism on quantum cohomology).</p> <p>Two simply connected smooth 4-manifolds having the same <a href="http://en.wikipedia.org/wiki/Intersection_form" rel="nofollow">intersection form</a> are homotopy equivalent by a theorem of Milnor. More than that, they are homeomorphic by a famous theorem of Freedman. However, very often these manifolds are not diffeomorphic. An example of homeomorphic but non-diffeomorphic pair is the <a href="http://en.wikipedia.org/wiki/Barlow_surface" rel="nofollow">Barlow surface</a> and the blow-up of the complex plane in 8 points. Note that both manifolds admit a Kahler structure.</p> <p>In the book of McDuff - Salamon <i>J-holomorphic curves and quantum cohomology</i>, they describe in Example 7.3.6 a construction due to Ruan of two diffeomorphic non-deformation equivalent 6-manifolds. The manifolds are the two 4-manifolds just described, both multiplied by $\mathbb{CP}^1$ (by a result of Wall, these two homeomorphic 4-manifolds become diffeomorphic after such a stabilization). </p> <p>Ruan uses quantum cohmology to prove that these two diffeomorphic symplectic 6-manifolds are not deformation-equivalent, so my guess is that he uses that their quantum cohomologies are not isomorphic as a tool.</p>