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Donaldson-Thomas invariants are the (virtual) Euler characteristics of moduli spaces of elements of the derived category of coherent sheaves (with some fixed Chern class, satisfying some stability condition, etc.) which bear some relation to "holomorphic Chern Simons theory", whatever that is.

Should I think of these as "A-model" or "B-model" invariants?

On the one hand, DT invariants come from the bounded derived category of coherent sheaves, which is what features in the B-model. On the other, there is the MNOP conjecture which tells me that the DT invariants of a CY 3-fold are ``the same'' as the Gromov-Witten invariants of the SAME 3-fold, which are A-model things.

As I understand it, according to Costello, if I take the cyclic A-infinity category built out of d-b-coh and run it through his machinery, I should get the topological string corresponding to the B-model. But (modulo my confusion on the matter) according to Kontsevich and Soibelman, if I take the cyclic A-infinity category built out of d-b-coh and run it through their machinery, I should get DT invariants. So what is going on?

edited Jul 4 at 15:47

Yosemite Sam
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asked Jul 4 at 15:29

Vivek Shende
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This is a good question and I hope someone can provide a sensible answer. My vote is that DT theory is B-model --- it seems to me that most predominant way that physicists talk about DT theory is in terms of counting D-brane states and there they are definitely thinking of D-branes in the B-model. It shifts the question to what exactly is this A-model/B-model duality (the MNOP/Gopakumar-Vafa conjecture) which is not mirror symmetry (since as you point out, the manifold doesn't change). – Jim Bryan Jul 5 at 22:48

Jim Bryan · Accepted Answer · 2012-07-09 17:05:21Z UTC

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I got a (very) short answer to this question from Nikita Nekrasov who emailed me with "[DT theory is the] B model per se. The GW/DT correspondence is the duality between the A model and the B model (S-duality)".

I had not been aware that the GW/DT correspondence is an instance of S-duality. Recall that S-duality occurs in various contexts and is characterized by taking the coupling constant from one theory $g_{s}$ to the inverse $1/g_s$ in the other. So in the GW/DT correspondence, the GW expansion is when $g_s$ is small and (this point was clarified for me by Ooguri) the DT expansion in the variable $q=exp(-g_s)$ is valid when $1/g_s$ is small.

edited Jul 9 at 17:05

answered Jul 6 at 14:50

Jim Bryan
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	I guess that the difference between Ooguri's change of variables $q=exp(-g_s)$ and the usual one in math $q=-exp(ig_s)$ is minor -- it amounts to redefining the GW and DT invariants by signs depending on the genus and the euler characteristic respectively. – Jim Bryan Jul 6 at 14:50
	I believe this paper is the relevant reference: arxiv.org/pdf/hep-th/0403167.pdf – Jim Bryan Jul 6 at 15:17
	Hi Jim! Thanks for the answer and comments. Does that mean there's supposed to be a formulation of (homological) mirror symmetry entirely in terms of coherent sheaves? – Vivek Shende Jul 8 at 13:38
	In principle, yes, but you would have to have a good handle on higher genus mirror symmetry. The basic question of where do the (higher genus) GW invariants go under homological symmetry is something that I don't think is very well understood. They are quantum corrections and I don't think people know much about the quantum corrections on the coherent sheaf side. – Jim Bryan Jul 9 at 17:04
	A related comment for these dualities (in string theory): T-duality is the underpinning of Mirror Symmetry, and S-duality is the underpinning of the relation between Donaldson and Seiberg-Witten invariants. – Chris Gerig Jul 9 at 22:59

Richard Thomas · Answer 2 · 2012-08-01 16:12:29Z UTC

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Ha, I think I'm going to disagree.

DT invariants are (more-or-less) independent of the complex structure on the CY: they are invariant under deformations of the CY.

However they depend on the (stringy/complexified) Kähler structure, or stability condition, or whatever. Hence wall crossing etc.

So even though they appear to be defined using D(coh), they're rather insensitive to that. What they are sensitive to is the small piece of data that's often forgotten -- a stability condition. So really one should think of them as defined in terms of a point in the space of stability conditions, or the stringy Kähler moduli space.

Ideally they'd be invariants of only the symplectic structure, just like GW invariants, but until MNOP is proved in full generality that's not known even in the projective case.

This makes them sound like "A-model invariants" to me, but then I'm not very sure about the physics language.

answered Aug 1 at 16:12

Richard Thomas
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I think the wall-crossing aspect in physics comes directly from the B-model interpretation. These invariants are counts of BPS states which (unlike states which don't saturate the bound between mass and central charge) tend to be invariant under deformations of the parameters of the theory except when they meet walls of "marginal stability". Isn't this is the context of Douglas's Pi-stability which is the physical precursor of Bridgeland stability. Continued below: – Jim Bryan Aug 1 at 16:38

Because of duality, the partition function can be viewed as both A-model and B-model, but since we are talking about the invariants --- the coefficients of the expansion of the partition function at $q=-exp(i\lambda)=0$ --- I think we are forced to call them B-model. If you do the expansion at $\lambda=0$ you of course get A-model. Have I shouted loud enough to change your mind? :-) – Jim Bryan Aug 1 at 16:42

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Isn't mathoverflow supposed to be for well defined math questions with well defined answers? I think we can blame Vivek for causing all this trouble. – Jim Bryan Aug 1 at 17:28

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I'd be very happy to just blame Vivek. That seems much the best solution. – Richard Thomas Aug 1 at 18:22

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Jim, I'm not sure I fully understand your comment about expanding about different points, but (as you say there) the answer is probably that you want to call DT invariants "B-model", but for the mirror CY. The way I interpret Vivek's question, that would make them "A-model" for the CY he started with. Anyway I think our previous two solutions are preferable. – Richard Thomas Aug 1 at 18:31

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Are Donaldson-Thomas invariants “A-model” or “B-model” ?

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