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I have heard from several physicists that the Kapustin-Witten topological twist of $N=4$ 4-dimensional Yang-Mills theory ("the Geometric Langlands twist") is not expected to give rise to fully defined topological field theory in the sense that, for example, its partition function on a 4-manifold (without boundary) is not expected to exist (but, for example, its category of boundary conditions attached to a Riemann surface, does indeed exist). Is this really true? If yes, what is the physical argument for that (can you somehow see it from the path integral)? What makes it different from the Vafa-Witten twist, which leads to Donaldson theory and its partition function is, as far as I understand, well defined on most 4-manifolds?

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Alexander, you might want to look at Ben-Zvi and Nadler's paper (arxiv.org/abs/0904.1247). They construct the 2-1-0 part of the GL-twisted $N=4$ 4d SYM compactified on $S^1$. However, the 3d part is not well-defined precisely due to non-fully dualizability that Urs mentioned: the vector spaces you attach to 2d manifolds may be infinite-dimensional (see p. 15). – Pavel Safronov Oct 15 '11 at 17:24
My question was really whether there exist physical arguments which tell you "how defined" the theory will be. – Alexander Braverman Oct 15 '11 at 19:46

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up vote 11 down vote accepted

From the path integral point of view, one can argue why the KW theory partition function won't be well defined as follows.

At the B-model point the KW theory dimensionally reduces to the B model for the derived stack $Loc_G(\Sigma')$ of $G$-local systems on $\Sigma'$. The B-model for any target $X$ is expected to be given by the volume of a natural volume form on the derived mapping space from the de Rham stack of the source curve $\Sigma$ to $X$.

Putting this together, we see that the KW partition function on a complex surface $S$ is supposed to be the "volume" of the derived stack $Loc_G(S)$ (with respect to a volume form which comes from integrating out the massive modes).

Now we see the problem: the derived stack $Loc_G(S)$ has tangent complex at a a $G$-local system $P$ given by de Rham cohomology of $S$ with coefficients in the adjoint local system of Lie algebras, with a shift of one. This is in cohomological degrees $-1,0,1,2,3$.

In other words: fields of the theory include things like $H^3(S, \mathfrak{g}_P)$ in cohomological degree $2$. Because it's in cohomological degree $2$, we can think of it as being an even field -- and then it's some non-compact direction, so that we wouldn't expect any kind of integral to converge.

(By the way, I discuss this interpretation of the KW theory in my paper http://www.math.northwestern.edu/~costello/sullivan.pdf)

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Of course, this is an example of the phenomenon Andy mentioned (non-compactness in field space). If you dimensionally reduce, you can see this from the categorical point of view Urs mentioned: the B-model for $Loc_G(\Sigma)$ is built from coherent sheaves on $Loc_G(\Sigma)$, but that's not a fully-dualizable category. It seems (?) to be smooth but not compact. This means that the B model for $Loc_G(\Sigma)$ is only partially defined: operations are defined for surfaces with at least one outgoing boundary. – Kevin Oct 22 '11 at 14:54
Thanks a lot - this is very interesting. I have two questions though: first, can you make a similar argument for Donaldson theory (to see why its partition function is well defined for most 4-manifolds). Second, I didn't understand the last point of your comment - it seems more or less possible to attach a category to $\Sigma$ without boundary (roughly speaking derived category of quasi-coherent sheaves on $Loc_G(\Sigma)$ - this has to be carefully defined, but in the context of geometric Langlands it is nowadays known what the right category is). why doesn't it contradict your argument? – Alexander Braverman Oct 22 '11 at 18:17
For the second point: I'm very far from an expert on geometric Langlands, so I hope what I say is not wrong. But, I would guess that the category you attach to Loc_G(Sigma) is not "proper". Proper means $\operatorname{RHom}(E,F)$ is of finite total dimension, where $E,F$ are perfect complexes (i.e. compact objects of the category). Proper is a necessary condition in order for the category to give a full TFT. There are more details on smooth + proper categories in Lurie's paper and Kontsevich-Soibelman's work. – Kevin Oct 23 '11 at 1:11
The only thing I really understand about Donaldson theory is how to twist the $N=2$ gauge theory to get a holomorphic version of Donaldson theory, which should count holomorphic bundles. (Of course this should be equivalent.) There, you expect the partition function to be a kind of volume of $T[-1] Bun_G(S)$, where $S$ is a complex surface. Since for stable bundles $Bun_G(S)$ has tangent complex in degrees $0,1$ (deformations + obstructions, no automorphisms) $T[-1] Bun_G(S)$ has tangent complex in degrees $-1,0,1$, which is fine. – Kevin Oct 23 '11 at 1:26
up vote 10 down vote

In a "fully defined" TQFT the spaces of states are necessarily finite dimensional. This follows simply from the fact that the correlators assigned to the cap and the cup cobordism (the "2-point functions") equip the space of states with the structure of a dualizable object in the corresponding monoidal category of vector spaces, which are precisely the finite-dimensional objects.

Similarly, in a "fully defied" extended n-dimensional TQFT (a "fully local one") the "n-space of states" assigned to the point is a fully dualizable object.

But there are TQFTs with non-finite state spaces, and extended TQFT with not-fully dualizable $n$-space of states. In the case of d=2, these are (somewhat misleadingly) known as TCFTs . Famous examples are the A-model and the B-model. And the Kapustin-Witten 4d TQFT reduces to these in certain compactifications (see for instance Kapustin's review pages 17-18).

So how can this be? The answer is that a "TCFT" is a TQFT that, as a cobordism representation, is defined only on the subcategory of cobordisms that are called "non-compact" or "with positive boundary". Roughly speaking, this is simply the subcategory obtained by discarding the cup (or the cap) cobordism. This removes from the TQFT the requirement to have dualizable state spaces, but otherwise retails all the structure of a TQFT.

For an extended such TQFT (a "fully local one") the state 2-spaces (those assigned to the point) still have lots of nice structure, even without being fully dualizable. One says that they are Calabi-Yau objects.

A detailed discussion of all this is in is section 4.2 of Lurie's On the classification of TFTs

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Thank you. I more or less know Lurie's paper. My question was to what extent this applies to the Kapustin-Witten theory and specifically why does it happen that for some other twist the theory is "more defined" (or am I mistaken here?) – Alexander Braverman Oct 14 '11 at 13:58
Concerning to what extent this applies to KW theory: as I tried to indicate, at least we know that it has compactifications to 2d that in certain parts of the parameter space reproduce the A-model and the B-model. For these 2d TCFTs we know excactly what's going on (via Lurie's section 4.2). Since these are simple special cases induced from KW theory, it seems to follow that KW theory is "at least as non-fully defined" as this. Not sure if this helps, but this is the statement that I can see so far. – Urs Schreiber Oct 14 '11 at 15:29
Thank you. In fact, orginally I wanted to know whether there is a way to predict how well defined given TQFT will be by looking at the path integral. But what you wrote is also very helpful! – Alexander Braverman Oct 15 '11 at 17:50
When you say that the A and B model have non-finite (dimensional) space of states, do you mean in the case of a non-compact target? I would naively think that the A model with a compact target $X$ has a finite-dimensional space of states (namely the quantum cohomology of $X$). – Andrew Neitzke Oct 21 '11 at 17:01
No, I mean the "2-space" of states, the A-oo algebra of string states assigned to the point. This is not a fully dualizable object for the A- and B-model. – Urs Schreiber Oct 22 '11 at 18:41
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up vote 8 down vote

I would think that Donaldson theory is also not a 4d TFT strictly speaking -- after all, there are some 4-manifolds for which it has metric dependence. Isn't that enough to violate the letter of the law?

In that case it's usually said that the reason for the failure is some non-compactness in field space (one finds this claim e.g. near the bottom of page 5 of hep-th/9709193). I suppose a similar problem could afflict the Kapustin-Witten twist of N=4 super Yang-Mills.

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Thanks. I would actually be glad to understand the pattern: is there a way to predict how well defined given TQFT will be by looking at "the space of fields" and the path integral? – Alexander Braverman Oct 20 '11 at 3:32
I am not fully competent to answer (and I am not at all sure that the issue I am mentioning is the same one your original interlocutors had in mind), but here is a partial comment. Looking at hep-th/9709193 I get the sense that the non-compactness that is relevant is that of the space of supersymmetric (or "BRST invariant") field configurations. If that's right, then there would be a potential issue whenever the equations that determine the supersymmetric field configurations define a non-compact moduli space. – Andrew Neitzke Oct 20 '11 at 4:56

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