Is there a mirror of the Rozansky-Witten theory?

Question

In http://arxiv.org/abs/hep-th/9612216 , Rozansky and Witten have associated to every holomorphic symplectic manifold $X$ a 3d TQFT (three dimensional Topological Quantum Field Theory), by topological twisting of a 3d sigma model of target $X$. The 3d sigma model is not renormalizable and so not well-defined in the UV but nevertheless, the 3d topological theory is well-defined. It is a kind of odd Chern-Simons theory.

One can show that if we take $X = T^*Y$ for $Y$ a complex manifold ( $T^*Y$ is naturally holomorphic symplectic), then the dimensional reduction over a circle of the 3d Rozansky-Witten TQFT of target $X$ is the B-model 2d TQFT of target $Y$. The question is: is there a similar story for the A-model?

In other words: Let $Z$ be a (real) symplectic manifold, the A-model of target $Z$ is a 2d TQFT. Is there a 3d TQFT constructed from $Z$ whose dimensional reduction over a circle is the A-model 2d TQFT of target $Z$ ?

This question is motivated by the fact that the existence of the Rozansky-Witten TQFT in some sense "explains" the existence of a (derived) tensor product over the derived category of coherent sheaves $D(Y)$ of a complex manifold $Y$. If one thinks about $D(Y)$ as the category of branes of the B-model, this structure seems to have no physical meaning ($D(Y)$ is what is associated to a point by the TQFT but there is no smooth cobordism between 2 points on the one and and 1 point on the other hand). But in the Rozansky-Witten theory on $X=T^*Y$, $D(Y)$ is the category of line defects, it is what is associated to a circle by the TQFT and the tensor product becomes natural because there exists a cobordism between 2 circles on the one hand and 1 circle and the other hand (a pair of pants). Mathematically, the (derived) tensor product is a rather obvious operation on $D(Y)$. If $Z$ is a symplectic manifold, the category of branes of the A-model of target $Z$ is the Fukaya category $Fuk(Z)$ of $Z$. But now, mathematically, there is no obvious tensor product on $Fuk(Z)$. Nevertheless, such a thing should exist if mirror symmetry is true, i.e. if there exists a complex manifold $Y$ such that $Fuk(Z)=D(Y)$. Maybe that if the A-model analogue of the Rozansky-Witten theory exists, it will be easier to see what is the tensor product like operation on the Fukaya category.  

Answer 1

There is the 3d A model formulated by A. Kapustin and K. Vyas in http://arxiv.org/pdf/1002.4241.pdf .

The Fukaya category of $T^*X$ occurs as the category of line operators between a Dirichlet and a Neumann boundary condition for the 3d A model on $X$.

Of course, $Fuk(T^*X)$ (actually its triangulated envelope) is equivalent to the (derived) category of constructible sheaves on $X$, by Nadler and Zaslow by http://arxiv.org/abs/math/0612399 . This has a natural tensor product... maybe they're the same?

EDIT: Actually it seems Sh(X) has two tensor products... Maybe this is because of the 4d A model?

Comments

Thanks for your answer. The paper by Kapustin and Vyas is exactly the kind of things I was looking for. I find strange that to have a 3d interpretation of $D(X)$, we have to go to a bigger space ($T^*X$) whereas to have a 3d interpretation of $Fuk(Z)$, we have to go to a smaller space (a $W$ such that $Z=T^*W$). Do you have any intuitive explanation of this fact ? Another question: you answered my original question for $Z$ being a cotangent space: do you think there is a analogue for general $Z$ (for example a compact symplectic manifold)?

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I think your two questions are related. The RW model compactified on a circle gives the B-model on the same target. I assume the same should be true of the "correct" 3d A-model. Something is strange with the 3d A-model since it can only describe the A-model of a cotangent bundle. You should think of the cotangent bundle as the local picture of a symplectic manifold _near a Lagrangian_. So it seems to me what is missing is some way of gluing the A-models of cotangent spaces together into the A-model on an arbitrary symplectic manifold. Then perhaps one can look again at the 3d A-model to figure out what to do.