Why would I want to know (equivariant) quantum cohomology? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T21:06:11Z http://mathoverflow.net/feeds/question/11154 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/11154/why-would-i-want-to-know-equivariant-quantum-cohomology Why would I want to know (equivariant) quantum cohomology? Ben Webster 2010-01-08T15:28:07Z 2010-11-02T13:33:18Z <p>Let's say that I have a variety I think is interesting, and based on some papers I don't fully understand, I can compute quite explicitly its equivariant quantum cohomology in terms of explicit formulae for multiplying by a degree 2 class.</p> <p>Being something of a newcomer to quantum cohomology, I'm genuinely a bit unsure of how interesting a result this is, and have doubts about writing a paper whose content is "The quantum cohomology of variety X is blah."</p> <p>Does it tell me anything particularly interesting? Might it have cool implications in integrable systems or something like that?</p> http://mathoverflow.net/questions/11154/why-would-i-want-to-know-equivariant-quantum-cohomology/11170#11170 Answer by Kevin Lin for Why would I want to know (equivariant) quantum cohomology? Kevin Lin 2010-01-08T18:26:26Z 2010-01-08T18:55:24Z <p>More generally, why would we (mathematicians, not physicists) want to know anything about any Gromov-Witten invariants at all? For me, they're just interesting invariants of varieties/manifolds. The structure of Gromov-Witten invariants is quite rich and interesting on its own.</p> <p>Perhaps your computation can tell you something interesting about another variety, via mirror symmetry. And yes, it is often the case that Gromov-Witten invariants make contact with integrable systems, but I seem to have the impression that most of the time, the applications go in the direction of integrable systems helping our understanding of Gromov-Witten theory --- e.g. Witten's conjecture, Virasoro conjecture --- rather than the other way around.</p> <p>Another thing is that, since Gromov-Witten invariants are not really enumerative invariants, it can sometimes be interesting to try to relate them to "actual" enumerative invariants. Moreover there are various questions about the relationship between Gromov-Witten invariants and other "enumerative" invariants like Donaldson-Thomas invariants. There are sometimes also connections with things like matrix models. For example, <a href="http://arxiv.org/abs/math/0101147" rel="nofollow">this paper</a> of Okounkov-Pandharipande relates Hurwitz numbers ("actual" enumerative invariants), the Gromov-Witten theory of P^1, and matrix models. Okounkov and Pandharipande have written many papers along the lines of "the quantum cohomology of X is blah" or "the Gromov-Witten theory of X is blah".</p> http://mathoverflow.net/questions/11154/why-would-i-want-to-know-equivariant-quantum-cohomology/11171#11171 Answer by Allen Knutson for Why would I want to know (equivariant) quantum cohomology? Allen Knutson 2010-01-08T18:34:58Z 2010-01-08T18:34:58Z <p>Equivariant quantum cohomology of Grassmannians (presumably not your variety) is closely related to equivariant ordinary cohomology of 2-step flag manifolds. That's some justification, for me...</p> http://mathoverflow.net/questions/11154/why-would-i-want-to-know-equivariant-quantum-cohomology/21405#21405 Answer by Catharina Stroppel for Why would I want to know (equivariant) quantum cohomology? Catharina Stroppel 2010-04-15T00:02:53Z 2010-04-15T00:02:53Z <p>Hi, I am not really an expert in this field, but I think (equivariant) quantum cohomology rings have a very nice combinatorics and show up at many places. I few month ago I was just finishing a paper (joint with Christian Korff) which connects the somehow completely understood quantum cohomology of the Grassmannian with the Verlinde algebra (which for me is the fusion algebra of certain tilting modules for a quantum group at a root of unity). </p> <p>Have a look at arXiv:0909.2347!</p> <p>There are indeed lots of connections with integrable systems. In our work we look somehow at a very easy situation: take the affine Dynkin diagram for affine sl(n) (that means a circle with n points!) Then consider the integrable system where you can place particles at this n places. Either "bosonic" that means however many you want or "fermionic" which means at most one at each place. Now there is the operation of moving a particle to the next place. This defines you linear maps on the space of all particle confugurations fixing the number of total particles. Then we define Schur functors where the variables are these (non-commuting) operators. </p> <p>Now the quantum cohomology of the Grassmannian Gr(k,n+k) has a basis which we can identify with certain partitions or with fermionic particle configurations on an (n+k)-circle using k particles. </p> <p>Whereas the fusion algebra at level k has a basis which can again be identified with certain partitions or with bosonic particle configurations on an n-circle. </p> <p>The funny thing now is that the multiplication in either of the rings is just given by a*b=Schur poly to a viewed as an operator applied to b.</p> <p>This desription of quantum cohomology was found by Postnikov a few years ago, but he did not connect it to this integrable system. The whole thing reproves and makes explicit an old result due to Witten, Gepner, Vafa and Intrilligator that the fusion ring of gl(n) at level k is isomorphic to the quantum cohomology of the Grassmannian Gr(k,n+k) when we specialise q to 1. </p> <p>So already in this really boring Baby-example something interesting shows up, so I guess one really should study all sort of equivariant cohomology rings!</p> http://mathoverflow.net/questions/11154/why-would-i-want-to-know-equivariant-quantum-cohomology/44554#44554 Answer by Alexander Braverman for Why would I want to know (equivariant) quantum cohomology? Alexander Braverman 2010-11-02T13:33:18Z 2010-11-02T13:33:18Z <p>You might have already found an answer by now but still let me add my motivation for the subject. The reason I am interested in quantum cohomology is that I am thinking about it as the correct version of semi-infinite cohomology of the loop space of the variety in question (this point of view by the way is sometimes very useful for guessing the answer). Personally I find the definition of quantum cohomology rather ugly, but in geometric representation theory this kind of phenomenon is sort of well-known: one way to do "semi-infinite geometry" which formally is not clear how to tackle, is by working with the global curve \${\mathbb P}^1\$ instead of the formal disc. For example, one can develop rather non-trivial theory of "semi-infinite flag varieties" using spaces of quasi-maps, which is defined using global \${\mathbb P}^1\$ as well. I sort of regard this as a similar phenomenon to why quantum cohomology (which is supposed to be a local object) is defined via a global curve.</p> <p>A more "physical" explanation of the above phenomenon is this: in 2-dimensional conformal field theory it happens very often that some genus 0 correlators actually coincide with some local objects (like coefficients of the OPE and such things). One example of this is the Kazhdan-Lusztig tensor product for representations of affine Lie algebras. It is actually a local thing, but the Kazhdan-Lusztig definition goes through coinvariants on global \${\mathbb P}^1\$.</p> <p>Let me conclude by saying that while the "global" approach is often useful, because that's the only way to define things regorously, the price you have to pay for this is the fact that calculations become rather difficult and mysterious. Quantum cohomology is probably a good example of this -- many answers there become much more understandable when you think in terms of loop spaces.</p>