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Which part of physical B model is not rigorous? the physical theory of B model,if it is not mathematical rigorous just because the Feynman integral,but it looks like for me the space is finite dimensional, what the problem is causing that it still not rigorous?

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I see the articles constructing the B model mathematically,but I don't know why they do this,or do that ,because there is no relating to original physics. – HYYY Mar 6 '10 at 3:55
What do you mean by the physical B-model? There's the IIB string and the topological B-model, but I don't know of a physical B-model. – Aaron Bergman Mar 6 '10 at 4:53
O,I mean topological B model, so, for example, why is it not rigorous constructed by physicist,the problem is in path integral(but it seems it is finite dimensional) or anything else?Thanks! – HYYY Mar 6 '10 at 13:37
By the way, what do we really mean by mathematical defintion of B model, and what is the relation to the definition by physicists?(I said no problem in feynman integral part) – HYYY Mar 7 '10 at 1:51

To define (as Kevin Lin does above) the B-model purely as the derived category of coherent sheaves is fine and rigorous, but it ignores the higher-genus aspects of mirror symmetry -- which was the original question. As I wrote above, Kevin Costello gives a rigorous description of the higher-genus amplitudes, but it is still conjectural whether this agrees with the physics. The issue is that higher-genus string amplitudes depend on an integration over the moduli space of Riemann surfaces (or a space of maps from them, depending on the model), and this demands compactification. The full, non-topological theory is of course an ordinary two-dimensional quantum field theory, with all the usual difficulties in making the path integral rigorous.

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Hi Eric, maybe this is a distinction between the B TFT and B topological strings -- the derived category I think knows everything about the former in all genus via Costello/Kontsevich-Soibelman/Lurie reconstructions of closed strings from open, right? but of course as you say it knows nothing about integrating over moduli of curves/extension to nodal curves etc. – David Ben-Zvi Mar 9 '10 at 12:56
Hi- Costello's construction of the GW partition function (0509264) does involve the moduli space of Riemann surfaces. He works with topological conformal field theories in that paper. If his result differs from the physical B-model, it's due to a difference in the compactifications of the moduli spaces used. But the reconstruction results of Lurie et al (of course I don't myself understand how the cobordism hypothesis is proven in this case) do produce invariants at all genera that may have nothing to do with GW theory, as you state, due to use of TFT's vs. TCFT's. -z – Eric Zaslow Mar 12 '10 at 2:58
Hi-- I was thinking of his 0412149, a precursor of the cobordism hypothesis where all genus topological B-model (eg) is recovered from the derived category. I didn't think there was any difference between TCFT and (differential graded) TFT, only a question of whether you pass from TFT to topological strings, calculating a topological string partition function by integrating over moduli spaces (as 0509264 does)? – David Ben-Zvi Mar 17 '10 at 19:19

Kevin Costello's mathematical definition of the B-model (math/0509264) is rigorous. It's an open problem whether this definition agrees with the BCOV construction, as far as I know.

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BCOV is closed-string B-model, correct? The Costello paper you reference does not give all of the closed-string B-model, only a part of it. – Kevin H. Lin Mar 6 '10 at 0:11

Open-string B-model seems to be ok: derived category of coherent sheaves in the Calabi-Yau case, matrix factorizations in the Landau-Ginzburg case.

The following are closed-string B-model, but genus 0 only:

  • See Barannikov-Kontsevich for the Calabi-Yau case:

  • For the isolated singularity Landau-Ginzburg case, see the work of Kyoji Saito (for example, take a look at the book of Hertling on Frobenius manifolds).

For the higher genus closed-string B-model, see this question.

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Thank you for you two the answers. but I mean, the physical theory of B model,if it is not mathematical rigorous just because the Feynmann integral,but it looks like for me the space is finite dimensional, what the problem is causing it is still not rigorous? (Perhaps I still need to look at the physical theory of B model to see what is going on) – HYYY Mar 6 '10 at 1:05
I assume the space you are referring to here is, in the Calabi-Yau case, the cohomology (of polyvector fields) of the manifold, or the Jacobian ring in the isolated singularity Landau-Ginzburg case. Indeed these are finite dimensional. I do not know any physics and I do not understand the BCOV paper, but I think that they consider not just cohomology of polyvector fields $H^\ast(\Lambda^\ast T_X)$ but also $\Gamma(\Lambda^\ast T_X \otimes \Omega^{0,\ast}_X)$, which is infinite dimensional. Perhaps this is a source of problems, but I don't know --- this is just a naive guess. – Kevin H. Lin Mar 6 '10 at 1:59
Thanks!Here space refers to the space of field configuration that the Feynman integral performs on. – HYYY Mar 6 '10 at 3:11

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