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Geometric Langlands from a TQFT perspective February 27, 2008

Posted by Joel Kamnitzer in Algebraic Geometry, geometric Langlands, tqft.
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In my continuing attempts to clear the backlog in my brain, I would like to tell you about the talks of Nadler and Gukov from Miami workshop which I was at a month ago. (Actually I really want to tell you about Kontsevich’s talks but I don’t think that I understand them well enough to do that.)

Ever since the work of Kapustin and Witten a couple of years ago, a TQFT interpretation of geometric Langlands has been available. However, I had never “internalized” it until these talks. It gives a nice conceptual picture which makes some constructions in geometric Langlands less mysterious and hopefully makes the whole subject a bit more accessible.

In this view of things, geometric Langlands concerns the equality of two 4D TQFTs, which will denote by A and B. A and B both depend on the choice of a semisimple algebraic group G. Or more precisely, if we want A = B, then we should have A depending on G and B depending on its Langlands dual group G^\vee. They are 4D TQFTs, so they assign a number to a (closed) 4-manifold, a vector space to a 3-manifold, a category to a 2-manifold etc and related morphisms to bordisms of such objects.

I will start with a 2-manifold C. The first surprise is that A(C) and B(C) depend on more than just a topological structure for C — in particular we assume that C is actually endowed with the structure of smooth projective algebraic curve. Then we define A(C) = D-mod(Bun_G(C)) and B(C) = QCoh(Conn_{G^\vee}(C)). Here Bun_G(C) is the moduli space of algebraic principal G bundles on C and Conn_{G^\vee}(C) is the moduli space of algebraic principal G^\vee bundles with connection on C. To continue the explanation, D-mod means the category of modules for the sheaf of differential operators (equivalently the category of perverse sheaves) and QCoh means the category of quasi-coherent sheaves.

Let’s look at a particular example of the above. I will start with a slightly strange curve C, which in particular is not projective. Take C to be two copies of a formal disk D which are glued together on the compliment of the origin in each one. We denote this curve by E. It is a sort of algebraic geometry version of how a topologist would construct S^2 — instead of getting \mathbb{P}^1 you get this E. Let us think about A(E) and B(E).

Every G-bundle on D is trivial. Let us pick trivializations on each copy of D in E. Comparing the trivializations gives us an element of G((z)) and overall we see that Bun_G(E) = G[[z]] \setminus G((z)) / G[[z]] .

On the other hand, let us examine G^\vee bundles with connections on E. David stated this this space is the same as BG^\vee , though I must say that at the moment the reason is not clear to me.

Hence we see A(E) = B(E) is just the usual geometric Satake correspondence which relates G[[z]] equivariant perverse sheaves (D-modules) on the affine Grassmannian with representation of G^\vee (equivalently coherent sheaves on BG^\vee ).

Now, we are in the position to think about some 3-manifolds. Let C be again a smooth projective curve and fix a point x \in C. Consider the 3-manifold M which looks like C \times I but where we remove a ball around (x,t) for some t \in (0,1) . Our M has three boundary components (one from each end and one from the ball), so we expect a functor A(M) : A(C) times A(S^2) \rightarrow A(C) — and similarly for the B version.

Now, we take the perspective that A(S^2) = A(E) is the Satake category. This category acts on A(C) by a push-pull diagram that I won’t get in to right now.

Similarly, B(S^2) = B(E) is the category of G^\vee representations which acts on B(C) — actually I don’t quite understand this action.

Now, the statement of unramified geometric Langlands is that for any curve C, A(C) is equivalent to B(C) is a manner compatible with these actions of A(E) and B(E).

There is also a version of this for the ramified case which I had been hoping to write about as well — perhaps another time. Also Ben-Zvi and Nadler have used some intuition from this TQFT picture to give some nice applications which they hope will lead to a proof of Soergel’s conjecture for real groups.

As a final note, I should say that these TQFT are not just some mathematical fiction. According to the work of Kapustin-Witten and then later Gukov-Witten, they can actually be described in some gauge theoretic terms and the equality of A and B comes from something is physics called S-duality.

Comments»

1. anonymous - February 27, 2008

But even the statement that A(C) = B(C) is surely still unproven, right?

If you have some enthusiasm left, would you explain how to get A(S^2) to act on A(C), etc.? Is this true for any general 4d TQFT A?

2. A.J. Tolland - February 27, 2008

I should be answering questions in this thread, but instead I’m going to ask one: Is the 4d GL-twisted N=4 Yang-Mills topological field theory actually a TQFT in the mathematician’s sense of bordisms and functors? I seem to recall Peter Teichner saying that the other twisted 4d Yang-Mills theory that mathematicians care about — Donaldson theory — isn’t really a TQFT in the usual way, that you can only slice up into smaller bits with positive 1st Betti number?

3. Joel Kamnitzer - February 27, 2008

anon -
The statement A(C) = B(C) is unproved. The statement A(E) = B(E) is a theorem on the other hand. Sorry, perhaps I should have been more precise on this.

By considering that 3-manifold M I mentioned above, one can see that A(S^2) should acts on A(C) — actually one action for each point x in C. In this case, we can construct the action as follows. Consider the curve C’ obtain by taking two copies of C and gluing them together away from x. Then we consider Bun_G(C’). This space has two maps to Bun_G(C) and one map to Bun_G(E) — near the two x’s, C looks like E.

So to construct the functor D-mod(A(C)) x D-mod(A(E)) -> D-mod(A(C)), we pullback the two D-modules to Bun_G(C’), tensor them together and then push forward to Bun_G(C).

Usually the action is described slightly differently, but I think that this way of thinking about it is more in line with Nadler’s philosophy.

AJ - I guess it is not a TQFT is the strict sense, since the category you assign to a 2-manifold C depends on its structure as an algebraic curve. But I’m really not sure.

4. Noah Snyder - February 27, 2008

What about at the 3 and 4 dimensional levels, does it only depend on topology there? Or is there still an algebraic structure playing a crucial role?

5. Joel Kamnitzer - February 27, 2008

I think that no one knows how to give a mathematically precise description to what is assigned to arbitrary 3 and 4 manifolds. I suppose that it will depend a priori on a metric and then hopefully on just an algebraic structure (at least on the 4 manifolds). I’m not certain.

6. David Ben-Zvi - February 28, 2008

I think one confusing thing is the relation between
geometric Langlands and the 4d TFT (N=4 SYM in the GL twist) studied by Kapustin-Witten. The latter is a topological field theory in
as strong a sense say as Donaldson theory - in particular
it does not depend on a complex structure on the surface C.
The former depends on the complex structure,
though there’s a flat [but nonintegrable!]
connection over the moduli of complex structures on C.

The honest TFT has B(C)= coherent sheaves on
the character variety of pi_1(C) (representations of pi_1 into
the dual group up to conjugation), which is independent
of the complex structure on C. The A side is trickier - it is
roughly the category of A-branes — an enlarged version
of the Fukaya category of Lagrangians — on the moduli
space of solutions to Hitchin’s equations (Higgs bundles),
except without any stability assumptions.. algebraically
this looks like the cotangent to the moduli of bundles
and A-branes look like D-modules, but presumably these
are ANALYTIC not algebraic D-modules.. in any case the category
of A-branes depends only on the symplectic structure on this
Hitchin space, which is independent of the complex structure on C.
This is explained nicely in a recent review by Witten on the arXiv.

(AJ - I don’t know how exactly one is supposed to handle
Donaldson theory as a TFT, but often with TFTs there
is some extra structure implicit on the manifolds — orientation,
spin structure, etc — so presumably when one formulates
Donaldson theory axiomatically one of these choices will be - or
rather imply - the choice of a chamber in H^2 in the case of a 4 manifold with b2+ =1 or something like that..)

David’s picture with the funny curve E is adapted to make the
“honest” TFT match geometric Langlands more closely.
The way it would be formulated in conventional TFT just uses S^2 -
in any TFT Z, what you assign to a sphere always has a multiplication and acts on Z(C) for any C of the same dimension.
The difference between A(S^2) and the Satake category A(E)
is roughly one of completion while B(S^2)
(as an abelian category) is exactly Rep of the dual group.

7. Joel Kamnitzer - February 28, 2008

Thanks for your explanations David. I knew it wouldn’t be long before you jumped in and corrected my mistakes. Thanks and I apologize to D.N. for mangling his talk.

One simple question: why does Rep(G) act on QCoh(Conn_G(C)) for any curve C?

Judging from the usual statement of geometric Langlands, the one thing that we can see about the action is that a representation V acting on the structure sheaf of a point, O_E, should give O_E \otimes V(E)_x. That V(E)_x means form the associated vector bundle and then take the fibre at x.

8. David Ben-Zvi - February 28, 2008

Hi Joel - thanks for the report, which I think is a great service
(I don’t know what DN did but I certainly
intentionally and unintentionally
obfuscate some of those points when I talk..)

The action of Rep_G is exactly what you write — for any point
x in C and any rep V there is a vector bundle W_{x,V} on
Conn_G(C) and the operator (”Wilson line”) is just tensoring with
this vector bundle. The bundle has fiber V(E)_x at a connection
E, as you explain. Since we’re just tensoring with a vector
bundle it is automatic that any skyscraper just gets tensored
with a vector space.

In fact a nice way to think of things is that Conn_G(C)
is “Spec” of the commutative ring (really tensor category)
which is the tensor product of a copy of Rep G for every
point x in C (ie the global unramified Hecke algebra) — more
precisely we also need to build in the fact that the Rep G actions
are identified for infinitesimally close x’s (ie there’s a flat
connection on them).
Since the Hecke operators make A(C) a module for this
commutative algebra, it corresponds to a “quasicoherent sheaf”
on “Spec of the Hecke algebra” — more precisely a sheaf
of categories.. and geometric Langlands says roughly
that this sheaf is nothing other than the “structure sheaf”,
ie Coh(Conn_G(C)) itself..

9. Scott Carnahan - February 28, 2008

E seems to be quite a strange beast. It looks like E’s detection of separatedness in targets is analogous to S^2’s detection of asphericity in 3-manifolds.

From what you wrote, it looks like you’re defining G-connections on E by gluing. Does this agree with the crystalline definition? I tried to complete the diagonal in E x E and got really confused. Nonseparatedness means the diagonal embedding is not a closed immersion, but maybe that’s not a problem, since descent data are defined by identifying pullbacks.

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12. Vishal - March 18, 2008

Ok, I just found out WordPress has deleted quite a number of backslashes from this post.

13. Joel Kamnitzer - March 18, 2008

I restored them. I wonder why that happened.