Generalized Geometric Langlands is False

June 18, 2007

Generalized Geometric Langlands is False

Posted by Urs Schreiber

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I am in Bad Honnef, at a conference titled

which I mentioned recently.

Second talk was by C. Teleman. He introduced it by saying that he had given this talk a couple of times, but always with the wrong title. Before quitting giving the talk, he said, he now wanted to give it once with the right title, which is:

The generalised geometric Langlands conjecture is false for trivial reasons.

The talk mosly reviewed various versions of the geometric Langlands conjecture, starting with the Fourier-Mukai transformation and ending somewhere in the derived world. At the end Teleman talked about the results of an explicit computation of Ext-groups in two derived categories which the “generalized geometric Langlands conjecture” conjectures to be equivalent. But the computation shows that this cannot be the case.

In choosing the title of this post, I blindly followed the above decision. I am far from being able to judge to which extent this is supposed to be a surprise for experts or just a confirmation of a general expectation that the formulation of the conjecture needs more care.

I will try to provide the full details of the talk later. With a little luck. For the moment, here just the abstract:

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C. Teleman
The generalised geometric Langlands conjecture is false for trivial reasons.

Abstract: I show that a naive formulation of the derived version of the Langlands correspondence fails for $\mathrm{GL}(n)$ and the line bundles $O$ and $O(\Theta)$ . On the more serious side, the same cohomology calculations improve the Beilinson-Drinfeld results and provide added evidence for the full faithfulness of the Langlands kernel. (This is joint work with E. Frenkel.)

Posted at June 18, 2007 11:26 AM UTC

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27 Comments & 1 Trackback

Re: Generalized Geometric Langlands is False

I assume this is pretty much the same talk of Constantin’s that David Ben-Zvi has put up notes for

here.

Posted by: Peter Woit on June 18, 2007 3:41 PM | Permalink | Reply to this

Re: Generalized Geometric Langlands is False

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I assume this is pretty much the same talk of Constantin’s that David Ben-Zvi has put up notes for here.

The talk I heard was different in character. But, yes, that counter example is the one already appearing in these notes from April.

Over dinner I talked to an expert about this. Here a virtual summary of that conversation, without any guarantees on correctness.

Q: “Is this counter example supposed to be shocking, or did the experts expect that the conjecture is wrong anyway?”

A: “For those who had really thought about this the counter example shouln’t be a surprise. But many popular talks are given which state the naive generalized Langlands conjecture without pointing out that it can’t quite work as stated.”

Q: “Is there an obvious way to fix the conjecture such that it has a chance again of being true?”

A: “It is unclear how to reformulate it in full generality, but in principle it is clear what goes wrong and how to circumvent it. Roughly, the problem is that everything is fine locally, but breaks down globally.”

“Compare this to Fourier transformation of ordinary functions: there, too, one needs a conditition on the global behaviour of the function in order for the transformation to exist. Something similar happens here. There are natural stratifications on the derived categories of coherent sheaves, and one needs to somehow restrict to those sheaves which have finite support with respect to this stratification.”

“In fact, what should be true in any case is that instead of an equivalence of the two derived categories in question, one has a fully faithful functor from one to the other. So in principle the conjecture would be fixed simply by identifying the image of that functor on objects and accordingly restricting the coherent sheaves to these in that image. But how to do this in general and in detail is at the moment not known.”

Posted by: urs on June 18, 2007 7:41 PM | Permalink | Reply to this

Re: Generalized Geometric Langlands is False

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Sadly my notes for that talk drop off just at the critical moment (when I made the mistake of listening instead of just writing!) The work of Frenkel-Teleman being discussed is very beautiful, extending Beilinson-Drinfeld’s construction of geometric Langlands from a half-dimensional slice of the space of local systems to a formal (or maybe even open?) neighborhood. The “counterexample”, as Constantin explains, is less about debunking geometric Langlands (since the problem it highlights is wellknown) and more about demonstrating some of the powereful calculations they can carry out with their technology.

The “problem” with geometric Langlands is one of “functional analysis” (according to Drinfeld). Namely there are many versions of the derived category that one can imagine for a big bad space (a stack not of finite type, like the stack of bundles on a curve), involving different amounts of boundedness, pro- vs ind- versions etc., and one has to figure out precisely which types match with which under geometric Langlands. This problem is in a precise sense the same as the problem of understanding operators with continuous spectrum in the correct analytic sense, and we have not yet learned enough of the functional analysis of derived categories to know what the right technical statements should be. I am guilty of giving talks stating the naive conjecture without pointing out its flaws, but I believe that the problems we are dealing with are on a technical level that’s not relevant to the bird’s eye view one gets from a general talk in any case.

(I’d like to point out that from what I understand it is not the derived categories of coherent sheaves that need to be truncated, as was suggested above, but the other side, the spaces of automorphic forms/D-modules. But there are many conceivable ways to deal with this problem on both sides, and as far as I understand no one has a definitive suggestion, so people like me keep stating the naive version and when we’re honest making caveats..)

The issue here is the same as one of the basic issues in the number fields version (eg in Arthur’s trace formula): you run into analysis once you leave the world of cusp forms (the subject of the “usual” geometric Langlands statement) and deal with the entire automorphic spectrum (ie incorporate Eisenstein series etc).

To put this more concretely: the hardest part of geometric Langlands in some sense is what to do with the trivial local system! More precisely, usually one studies irreducible local systems (or Galois representations), but once you start looking at reducible ones things get complicated (this is roughly where the sigma-model description of the 2d reduction of the 4d gauge theory in the Kapustin-Witten story breaks down).

The one example where all this is easy to see is the case of the Riemann sphere! (which is also the key case in the topological field theory POV). Here there are NO nontrivial local systems, so the space of local systems looks like a point, or a point modulo $G^L$ if you keep track of automorphisms (or a “derived enhancement” of this if you are careful about transversality issues). So coherent sheaves here look (ignoring the derived issues) like representations of $G^L$ . On the other side of geometric Langlands you are looking at constructible sheaves (or D-modules) on the stack $G$ -bundles on the Riemann sphere, which is the same as double cosets for the loop group wrt its positive and negative halves. This stack has points in bijection with reprsentations of $G^L$ , so all looks dandy, until you realize it’s not discrete after all - each point has infinitely many others in its closure.. more precisely the closure of each point (corresponding to a representation) are all the points corresponding to points lower in the dominance order. So the category of sheaves looks like a pro- or completed version of Reps of $G^L$ , and has to either truncate this category in some way to get the same as the other side. (An alternative, which Nadler and I prefer and use in the tamely ramified version, is to think of the automorphic side as “functionals on” the spectral side, i.e. as a kind of dual/space of distributions.)

In higher genus of course things get more complicated, but I think this is the main kind of issue being discussed.

Posted by: David Ben-Zvi on June 18, 2007 8:33 PM | Permalink | Reply to this

Re: Generalized Geometric Langlands is False

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David Ben-Zvi wrote:

it is not the derived categories of coherent sheaves that need to be truncated, as was suggested above, but the other side, the spaces of automorphic forms/D-modules

Right, my fault! I recounted that incorrectly.

And thanks for all this information! Here and elsewhere.

By the way, is there any chance at all to have anything like an analogue of Geometric Langlands in an entirely differential geometric context?

I don’t need the full thing. I’d be happy with toy versions. Any chance?

Posted by: urs on June 18, 2007 11:50 PM | Permalink | Reply to this

Re: Generalized Geometric Langlands is False

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Urs,

There are many possible answers but I’d say Kapustin-Witten provides an excellent one (which you know). As far as I understand their theory is not exactly what we usually call geometric Langlands, in that it is a topological field theory, i.e. independent of the complex structure of the underlying Riemann surface.

I don’t know what counts as differential geometric though — are symplectic, complex and generalized complex manifolds (and hence the Fukaya categories and categories of coherent analytic sheaves or generalized complex branes) part of differential geometry? I think of geometric Langlands as a particularly interesting case of an emerging topic of harmonic analysis of derived categories, so to find good analogues you need interesting derived categories. For a Riemannian manifold I don’t know if one can find a more interesting category than the Fukaya category of its cotangent bundle (aka the constructible derived category, after Nadler & Zaslow). If you apply this to connections on a surface up to gauge equivalence you’re pretty much in the geometric Langlands world.

Posted by: David Ben-Zvi on June 19, 2007 4:24 PM | Permalink | Reply to this

Re: Generalized Geometric Langlands is False

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I don’t know what counts as differential geometric though

For instance, I would love to see a statement which doesn’t involve the word “D-modules”, but instead (I guess) refers to “flat vector bundles with connection”.

Or: do I strictly need to conceive vector bundles in terms of locally free sheaves in order to pass to a derived category?

If so, maybe then I would try to first restrict myself to just understanding Fourier-Mukai completely “differential geometric”. No algebraic geometry, just smooth spaces, smooth bundles, etc.

Is that possible?

There are many possible answers but I’d say Kapustin-Witten provides an excellent one (which you know).

Right. Maybe I haven’t absorbed the precise content of Witten-Kapustin yet. What I did see was what looked like a chain of plausibility arguments to the naked eye, suggesting that one can make a connection between topologically twisted dimensionaly reduced Yang-Mills and some aspects of Geometric Langlands – provided one actually understands both of these formidable entities.

I don’t really feel on solid ground with respect to this.

Also, all this tends to lose itself in a plethora of more and more technical details, with no end in sight. To get started somewhere, I am vaguely hoping that there is some concrete hands-on statement which captures at least part of the interesting aspects and is as “differential geometric” as possible.

Posted by: urs on June 19, 2007 5:40 PM | Permalink | Reply to this

Re: Generalized Geometric Langlands is False

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You can get rid of the words D-modules, but I don’t think vector bundles with flat connection are a rich enough context to have an interesting Fourier theory. What you want is a “theory of functions” on spaces. By this I mean you assign to X a “space of functions” F(X), with a product on it, and with the main desideratum being: There are pullback and pushforward maps $F(X)\to F(Y)$ for maps between X and Y. In particular this means for any “kernel function” K in $F(X \times Y)$ you get a map from F(X) to F(Y) by pullback tensor and pushforward (and more generally operators for correspondences, or “spans”). If you’re picky you might also want that ALL (reasonable) maps from F(X) to F(Y) are given by such kernels.

The derived categories of D-modules or of coherent sheaves (suitably interpreted) give such theories of functions. So do things like Fukaya categories, derived categories of constructible sheaves, categories of generalized complex branes (once they’ve been defined!), etc. – but not just vector bundles of any flavor. So do ordinary cohomology or K-theory or any “motivic theory” (one should be careful then about properness for pushforwards) — but presumably this cafe would scoff at such decategorified notions!

Once you give such a theory you can start constructing Fourier transforms, and then try for their nonabelian analogues, and you’re in geometric Langlands territory. Nothing inherently “algebraic” about the story.. e.g. Fourier-Mukai can be done for any such theory, by taking X to be an abelian group object, Y the space of “characters” on X (meaning “functions” in F(X) with multiplicative behavior) and the kernel the universal character on the product…

Regarding Kapustin-Witten I think of it as much more than a chain of plausibility arguments. In particular one giant contribution from my POV (one of many) is the realization that basically all of the structures of geometric Langlands fit extremely neatly into the package of a 4d TFT - regardless of whether you even know about N=4 SYM - and this is a powerful context in which to try to understand the whole story. (I have notes on my webpage explaining this K-W in this fashion, if it helps - and video coming soon..)

Posted by: David Ben-Zvi on June 21, 2007 3:51 PM | Permalink | Reply to this

Re: Generalized Geometric Langlands is False

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What you want is a “theory of functions” on spaces.

Or rather “2-functions”, to be precise. Right?

The 2-functions I would like to see here are something like functors $\Pi_1(X) \to \mathrm{Vect}$ with suitable properties (here $\Pi_1(X)$ is the fundamental groupoid of $X$ ).

These really deserve to play the role of 2-functions, not just because an ordinary function is a 0-functor $\mathrm{Disc}(X) \to 0\mathrm{Vect}$ , but also because we also know that the right kind of theorems do have categorified versions for these 2-functions (like for instance the categorified Gelfand-Naimark theorem, which is however at the moment only understood in the baby version where everything is finite, as far as I am aware).

you get a map from $F(X)$ to $F(Y)$ by pullback tensor and pushforward (and more generally operators for correspondences, or “spans”

Good, I understand. I’d think all this exists for functors with values in $\mathrm{Vect}$ (I entertain myself with posting about exactly that here in a series of posts about “charged $n$ -particles”). But maybe I am being too naive concerning some technical conditions here.

For instance, I don’t know if

ALL (reasonable) maps from $F(X)$ to $F(Y)$ are given by such kernels

in the case of functors to $\mathrm{Vect}$ .

I have notes on my webpage explaining this K-W in this fashion

Do you mean this: Geometric Langlands and Topological Field Theory ?

Posted by: urs on June 21, 2007 4:07 PM | Permalink | Reply to this

Re: Generalized Geometric Langlands is False

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That works – Part 4 here is a more updated version of the same. any comments most welcome! (planning to write this up for a grad course I’m teaching in the fall).

Posted by: David Ben-Zvi on June 21, 2007 4:25 PM | Permalink | Reply to this

Re: Generalized Geometric Langlands is False

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Part 4 here is a more updated version of the same. any comments most welcome!

Page 3 to 8 comes from a general argument:

a $d$ -dimensional (extended or “tiered”) QFT should assign $(d-n-1)$ -categories to $n$ -dimensional parts of its parameter space.

In geometric Langlands we are dealing with 1-categories associated with 2-dimensional spaces, so if this is related to any $d$ -dimensional QFT, then for $d - 2 - 1 = 1 \;\; \Leftrightarrow \;\; d = 4 \,.$

(On the top of p. 6 it seems you mention the construction that is used in what is sometimes called “topological conformal field theory”, where Hom moduli spaces of surfaces are replaced by their cohomologies. Is that just a side remark or does it actually play a role in relating geometric Langlands to 4d QFT?)

I guess on these general grounds one can also already see that if Langlands comes from any QFT, then from a gauge theory:

generally, the $(d-n-1)$ -category which a QFT assigns to an $n$ -dimensional part $X$ of its parameter space is the $(d-n-1)$ -category of sections of a $(d-n)$ -vector bundle over the space of fields $\Phi : X \to \mathrm{target}\; \mathrm{space}$ .

For geometric Langlands these sections live over the moduli space $\mathrm{Bun}_G(X)$ which is roughly $[X, B G]$ . So if there is any QFT at work in the background, it is one whose “field configurations” are $G$ -bundles (similarly for the other side of the correspondence).

So this way, from general properties of (extended) QFTs, we can see that if geometric Langlands is at all related to QFT, then to 4-dimensional gauge theory.

That’s nice.

By the way, is there any evidence for or against the idea that

Knot invariants are to 3-dimensional Chern-Simons theory as Khovanov homology is to 4-dimensional topologically twisted Yang-Mills theory?

The latter would be the natural candidate for “categorified Chern-Simons”, wouldn’t it?

On the last pages, starting with page 11, you mention the main keywords of the Kapustin-Witten construction.

I think I do understand the general idea, but I don’t know how much this is really understood in detail.

For instance these t’Hooft line operators that are supposed to give the Hecke operations: in all discussions that I have seen they are described in terms of singularities in the path integral. What am I to make of this, mathematically?

It seems clear that these t’Hooft operators really ought to be defect lines as they appear, rigorously (though clearly not in the required generality), in the FRS framework, e.g. Duality and defects in rational conformal field theory. More recently, there has been progress on explicitly identifying these defect lines with target space structures that essentially amount to the kind of pull-push correspondence we expect to see here. (I mentioned this here.)

Posted by: urs on June 21, 2007 5:57 PM | Permalink | Reply to this

Re: Generalized Geometric Langlands is False

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This might be ignorance of TFT, but I don’t see why the ‘t Hooft construction is any different than the usual boundary conditions one considers in TFT or inserting vertex operators in CFT — you integrate over a space of gauge fields on the complement of the tubular neighborhood of a loop, and impose a given boundary condition (given by a version of the Dirac monopole, pushed to a singularity of G-gauge fields using a cocharacter $U(1)\to G$ ). That’s all these disorder operators are I think.

The boundary of this tubular neighborhood is $S^2\times S^1$ , so it is not surprising that what you get is identified as the K-group (or $HH^*$ ) of the category you assign to the two-sphere, namely representations of the Langlands dual group (whosse highest weights are given by such cocharacters). Now this category corresponds to something in codimension 2, so maybe the right words for it are not boundary condition but defect or something, I’m not familiar with this.

K-W’s paper explains very beautifully how precisely these monopole singularities correspond to Hecke correspondences - i.e. the Hecke correspondences are given by moduli of solutions to a monopole equation (Bogomolony eq) in 3-d with given boundary conditions and prescribed singularities. Of course the paper is not a math paper, and one would have to work quite hard to make it into one, but it’s excellently written and quite detailed.

As for avoiding D-modules, I’d recommend the setting of constructible sheaves – it’s a purely topological construction, very flexible, with a good pull/push formalism, and is in some sense I believe the minimal such containing flat vector bundles (secretly it’s the same as (rh) D-modules in an analytic setting but we don’t have to mention that..) The one caveat is to have a reasonable formalism you have to pass to the derived category – otherwise it’s as if one is trying to do cohomology with only 0-cochains..

Posted by: David Ben-Zvi on June 21, 2007 7:00 PM | Permalink | Reply to this

Re: Generalized Geometric Langlands is False

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Silly question: how do you pushforward a functor from $\Pi_1(X)\to Vect$ along a closed embedding? I’m probably missing something basic here, but I don’t know of such a pushforward for flat vector bundles. Constructible or perverse sheaves are defined essentially as a completion of flat vector bundles under pushforwards, and coherent sheaves are defined as a completion of (algebraic or analytic) vector bundles in the same way.. I guess in algebraic topology one looks at modules over the suspension spectrum of the based loops on X as an analog of this — is that what comes out of your construction?

Posted by: David Ben-Zvi on June 21, 2007 4:41 PM | Permalink | Reply to this

Re: Generalized Geometric Langlands is False

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Silly question:

Thanks for asking this. So let’s see if what I have in mind works, or else, understand what it is that makes it fail.

how do you pushforward a functor

By using the adjoint of the pullback.

Let $f : X \to Y$ be a map. Let’s write $f : \Pi_1(X) \to \Pi_1(Y)$ for the corresponding functor on paths.

Precomposition with $f$ is the pullback functor on functors:

$f^* : \mathrm{Func}(\Pi_1(Y),\mathrm{Vect}) \to \mathrm{Func}(\Pi_1(X),\mathrm{Vect}) \,.$

I am imagining that this has an adjoint $\mathrm{Hom}_{\mathrm{Func}(\Pi_1(X),\mathrm{Vect})}( V_1 \,, f^* V_2 ) \simeq \mathrm{Hom}_{\mathrm{Func}(\Pi_1(Y),\mathrm{Vect})}( f_* V_1 ,\, V_2 ) \,.$

If so, $f_*$ is my pushforward.

An elementary simple baby toy example of this I talked about here.

I discussed how to understand fusion products this way (in the finite case) here. (Details of the computation appear here).

So, what could happen is that as we impose more structure on our functors (smoothness, holomorphicity maybe, or the like) the pushforward simply ceases to exist. This I am not sure about.

Posted by: urs on June 21, 2007 4:59 PM | Permalink | Reply to this

Re: Generalized Geometric Langlands is False

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Rereading the discussion, I would like to emphasise again that I am not claiming that I understand generally how to obtain the pushforward of functors with values in vector spaces in a way that makes good sense beyond the finite case.

Rather, I notice that these look like the right thing to look at in finite examples, and I would like to understand if there is a way to formulate something like Fourier-Mukai or geometric Langlands using similar structures (instead of D-modules, in particular).

Posted by: urs on June 21, 2007 6:23 PM | Permalink | Reply to this

Re: Generalized Geometric Langlands is False

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Let’s see if I can understand the pull-tensor-push construction mentioned above , in the context of representations of groups (equivariant K-theory over a point).

We fix a group $G$ , and the only space we will consider is the point $pt$ , which we think of as being acted on by $G$ . (After graduating from this example, we can try $G$ -sets, like in the Tale of Groupoidification.) We define our space of functions $F(pt)$ as the space of equivariant vector bundles over $pt$ , i.e.

(1)

F(pt) = Rep(G).

As I see it, $pt \times pt = pt$ , and the pullback operation

(2)

\pi_1^* : F(\cdot) \rightarrow F(\cdot)

should just be the identity map. Is the push-forward map “take space of sections”? Mind you, not flat sections, otherwise we’d have shot ourself in the foot, surely. Anyhow, “take space of sections” is probably just the identity map too.

Ok, so let’s see what happens. Given a kernel $\rho \in F(pt \times pt)$ , in other words, given a rep $\rho \in Rep(G)$ , we want to understand the resulting map

(3)

\hat{\rho} : F(pt) \rightarrow F(pt),

that is,

(4)

\hat{\rho} : Rep(G) \rightarrow Rep(G)

induced by pull-back, tensor with $\rho$ , and push-forward.

That just sends

(5)

\sigma \mapsto \rho \otimes \sigma.

Is that right? If so, here are some questions:

(a) What can we say about the extent to which all maps are given by such “kernels”? In other words, how general are the maps $\rho \otimes - : Rep(G) \rightarrow Rep(G)$ ? You can’t get all linear functors $Rep(G) \rightarrow Rep(G)$ this way; for instance, you can’t get one which sends an irrep into an irrep of a lower dimension.

(b) Can one see a Fourier-Makai transform in this context? Is it just “take the character”… somehow?

(b) How about derived categories? Do we need to put them into the mix?

Posted by: Bruce Bartlett on June 21, 2007 8:06 PM | Permalink | Reply to this

Re: Generalized Geometric Langlands is False

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Bruce,

I think the push-forward in the case you describe is just the identity. Push-forward is defined to be the adjoint functor to the pullback functor. The adjoint of the identity is the identity.

Can one see a Fourier-Makai transform in this context?

What I would very much like to do is to go through a kind of Dijkgraaf-Witten version of geometric Langlands, if possible at all, where we let the group $G$ be finite and do everything over finite groupoids and think of vector bundles as functors from groupoids to vector spaces.

But right now I have to prepare for teaching an exercise group…

Posted by: urs on June 22, 2007 6:59 AM | Permalink | Reply to this

Re: Generalized Geometric Langlands is False

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Back from my duties, I realize I should have read Bruce’s message more carefully. He did of course say that the pushforward in his example is the identity!

Here is the program I would like to know more about:

Fix some finite group $G$ . For any 2-dimensional space $X$ , get $\mathrm{Bun}_G(X)$ be the (finite) groupoid of $G$ -bundles over $X$ .

This is supposed to be our configuration space fields on surfaces for a finite group gauge theory.

Then I think we want to be looking at functors $\mathrm{Bun}_G \to \mathrm{Vect}$ to be thought of as vector bundles on the configuration space.

I don’t know if we will have to pass to the derived category of such functors. But if we have to, is there anything that could stop us?

What is known about the derived category of complexes of functors from a finite groupoid to vector spaces?

Posted by: urs on June 22, 2007 9:43 AM | Permalink | Reply to this

Re: Generalized Geometric Langlands is False

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Bruce,

Let me try your problem with different notation. A point with G symmetry is to me the stack pt/G, or if you prefer the classifying space BG - in either case I’d like to keep G in the notation. There’s a map from pt/G to pt, and pushforward for vector bundles is taking the invariants of a representation of G.

In any case let’s try your exercise: we look at pt/G times pt/G, on which vector bundles are $Rep G\otimes Rep G$ . We’ll use these as kernels for functors from Rep G to itself. If I want to construct a functor which takes $\rho_1$ to $\rho_2$ for any two reps I use something like $K=\rho_1^*\otimes \rho_2$ : pulling back $\rho_1$ and tensoring with K we should find the space of invariants for the first G is $\rho_2$ as a rep of the second G.. so I think the idea works here.

To see a version of Fourier-Mukai let G be a finite abelian group and $G^\vee$ its dual group (group of characters), and consider the bundle on $(pt/G)\times G^\vee$ giving the universal character of $G$ (i.e. on the fiber over a point of $G^\vee$ , i.e. a rep of $G$ , put the corresponding rep of $G$ as a vector bundle on $pt/G$ ..)

Posted by: David Ben-Zvi on June 22, 2007 2:53 PM | Permalink | Reply to this

Re: Generalized Geometric Langlands is False

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in either case I’d like to keep $G$ in the notation.

Yes. I think where Bruce talks about the point, we really want to see the category with one point and $G$ worth of morphisms. That’s nothing but the groupoid presenting the stack you mention.

I usually call this guy $\Sigma G = \left\lbrace \mathrm{pt} \stackrel{g}{\to} \mathrm{pt} \; | g \in G \right\rbrace \,.$

Then $\mathrm{Rep}(G) = \mathrm{Func}(\Sigma G, \mathrm{Vect}) \,.$

Next we want some correspondence space. I don’t think the mere cartesian product will do what we are aiming at, but as a warmup it is clearly a good choice.

So there is a span $\array{ && \Sigma G \times \Sigma G && \\ &{}^{\mathrm{out}} \swarrow && \searrow^{\mathrm{in}} \\ \Sigma G &&&& \Sigma G }$ where the arrows are the obvious projections on the two copies of $\Sigma G$ .

Hitting this with $\mathrm{Func}(-,\mathrm{Vect})$ we get the cospan

$\array{ && \mathrm{Func}(\Sigma G \times \Sigma G, \mathrm{Vect}) && \\ &{}^{\mathrm{out}^*} \nearrow && \nwarrow^{\mathrm{in}^*} \\ \mathrm{Func}(\Sigma G,\mathrm{Vect}) &&&& \mathrm{Func}(\Sigma G,\mathrm{Vect}) }$

which you want to think of as $\array{ && \mathrm{Rep}(G\times G) && \\ &{}^{\mathrm{out}^*} \nearrow && \nwarrow^{\mathrm{in}^*} \\ \mathrm{Rep}(G) &&&& \mathrm{Rep}(G) } \,.$

For a given rep $\rho \in \mathrm{Rep}(G \times G)$ the pull-push is

$\mathrm{out}_* \circ (\cdot \otimes \sigma) \circ \mathrm{in}^* \,.$

By the way, this tensoring in between also has a nice diagrammatic formulation. We should realize that, at least as soon as we want to make a connection to QFT, the vector bundles appearing here ought to be sections of 2-vector bundles. (The trivial 2-vector bundle, in our case, since there are no twists in sight.) Then what looks like a functor $\Sigma G \to \mathrm{Vect}$ really becomes the component map of a pseudonatural transformation. And the operation

first pull

then tensor

then push

is simply pull-push of transformations of 2-functors (as described here).

Posted by: urs on June 22, 2007 3:58 PM | Permalink | Reply to this

Re: Generalized Geometric Langlands is False

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By the way: I worked through toy examples of Fourier-Mukai in this language in QFT of Charged n-Particle: T-Duality.

For the finite case, this is precisely the formalism needed here, I think (and for the non-finite case it should provide the right structure but needs amendment by various technicalities).

What I still don’t know is what for instance the right correspondence space (or rather: category) would be to use for a finite version of geometric Langlands. And if we do have to pass to derived categories.

Posted by: urs on June 22, 2007 4:13 PM | Permalink | Reply to this

Read the post Waldorf on Parallel Transport Functors
Weblog: The n-Category Café
Excerpt: Review of parallel transport functors by Konrad Waldorf.
Tracked: June 18, 2007 10:54 PM

Re: Generalized Geometric Langlands is False

Forgive me for the stupid question, but can someone formulate the geometric Laglands conjecture in a clear manner? Thx!

Posted by: Squark on June 22, 2007 9:52 AM | Permalink | Reply to this

Re: Generalized Geometric Langlands is False

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can someone formulate the geometric Laglands conjecture in a clear manner? Thx!

This is copied from notes taken in a lecture by T. Pantev:

Geometric Langlands Conjecture:

Let $G$ be a complex reductive group and let ${}^L G$ be its Langlands dual group.

Let $C$ be a compact smooth curve of genus $g \geq 2$ .

There exists an equivalence of the (bounded) derived category of coherent sheaves on the moduli stack of $G$ -local systems over $C$ and the (bounded) derived category of modules for derivations on the moduli space of principal bundles with structure group the Langlands dual group ${}^L G$ : $c : D(\mathrm{Coh}(\mathrm{Lo}(G))) \simeq D(\mathbf{D}_{\mathrm{Bun}_{{}^L G}}-\mathrm{mod})$

2) such that $c$ maps structure sheaves of points to Hecke eigensheaves.

Luckily, David Ben-Zvi is arround to correct whatever needs to be corrected here.

Posted by: urs on June 22, 2007 10:00 AM | Permalink | Reply to this

Re: Generalized Geometric Langlands is False

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Ok, let me try to understand what’s going on for $G=C^*$ . Hopefully, David Ben-Zvi will keep me from leading you too far astray. (I tried G a finite abelian group, but I got very confused.) In this case G is canonically isomorphic to LG. (In general, if M is a free abelian group of finite rank, then the Langlands dual to $Hom(M, C^*)$ is $Hom(Hom(M,Z),C^*)$ . But, when M has rank one, there is a canonical isomorphism between M and Hom(M,Z) – the one so that the resulting innner product on M is positive definite.) To avoid confusion with the complex numbers, I am renaming our curve to be X.

For any group G, G local systems are classified by $Hom(\pi_1(X),G)$ , modulo conjugation by G. Here G is abelian, so “the moduli stack of G-local systems over X” is just $Hom(pi_1(X),C^*)=Hom(H_1(X), C^*)$ . This is an algebraic torus of (complex) dimension 2g; it might be best written as $H^1(X,C)/H^1(X,Z)$ . I’ll abbreviate this space by S(X) in the future.

A C^* principle bundle is another name for a line bundle, they are classified by Pic(X). Pic(X) has countably many components, I assume that there is some technical condition telling me to just look at $Pic^0(X)$ . This is an abelian variety of (complex) dimension g, it might be best written (by the Abel-Jacobi theorem) as $H^{0,1}(X)/H^1(X,Z)$ .

I’m going to go on a crazy tangent here. It seems to me that there is a canonical projection (of complex vector spaces $H^1(X,C) \to H^{0,1}(X)$ and this should induce a map of complex varieties $S(X) \to Pic^0(X)$ . The fibers of this map are affine spaces of (complex) dimension g. I don’t see any canonical choice of section, so I’m going to figure this is an affine bundle but not a vector bundle. However, if we took the vector bundle corresponding to the affine bundle $S(X)$ , it would be a trivial bundle whose fiber over each point of $Pic^0(X)$ is canonically $H^{1,0}(X)$ . Since $H^{1,0}(X)$ is the cotangent space at the origin to $Pic^0(X)$ ; I’m going to propose identifying this vector bundle with the cotangent bundle $T^*(Pic^0 X)$ to $Pic_0 X$ .

Why do I do this? Because we are supposed to relate coherent sheaves on $T(X)$ to D-modules on $Pic^0(X)$ . Now, for any smooth space $Y$ , there is a close relationship between D-modules on Y and coherent sheaves on $T^* Y$ . For simplicity, I’ll sketch this when Y is affine (even though it isn’t in our application). Let Y=Spec A, and let DA be the ring of differential operators on Y. There is a filtration on DA by the order of the operator. The associated graded of this filtration is a commutative ring, call it gr(DA), and Spec gr(DA)=T^*Y. This means that there is a functor from D-modules on Y equipped with a filtration to coherent sheaves on $T^*Y$ equipped with a grading. Exploiting this functor is one of the main ways that people study D-modules, even when the D-modules don’t come with natural filtrations. I don’t know enough here to say anymore.

So, if I understand this correctly, in the special case where $Y=Pic^0(X)$ , geometric Langlands says that we can make this correspondence very precise by two tricks (1) passing to the derived category and (2) using $S(X)$ instead of $T^*(Pic^0 X)$ .

At this point, I have a nagging feeling that I made a mistake in identifying C^* with its dual, even though there was a canonical way to do it! I suspect that this had the effect of identifying $Pic^0 X$ with its dual. Jacobians of curves are always isomorphic to their duals, but for a general abelian variety, $A$ and $A^{\vee}$ is not isomorphic. I suspect that the right statement here is that, for any abelian variety $A$ , there is some canonical affine bundle $E(A)$ over $A$ , whose associated vector bundle is $T^*(A)$ . And the right statement is probably that the derived category of D-modules on $A^{\vee}$ is equivalent to the derived category of coherent sheaves on $E(A)$ .

Anyways, that’s as far as I get. Anyone want to fill in the next steps?

Posted by: David Speyer on June 22, 2007 5:24 PM | Permalink | Reply to this

Re: Generalized Geometric Langlands is False

Typo correction. In the paragraph beginning “Why do we do this?”, the symbol T(X) in the first line should be S(X).

Posted by: David Speyer on June 22, 2007 5:37 PM | Permalink | Reply to this

Re: Generalized Geometric Langlands is False

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A little plug might be in order: I have various lecture notes here, of which the first MSRI talk and the second Oxford talk are particularly relevant - ie cover essentially the same picture David explained clearly above (plus there are videos).

In the abelian case ( $G=C^\times$ as in David’s post) the geometric Langlands theorem essentially boils down to a Fourier transform for flat connections on abelian varieties, due to Laumon and Rothstein: Any sheaf with flat connection on an abelian variety A can be written as a direct integral of geometric characters of A.

Definitions: 1. a sheaf with flat connection means quasicoherent sheaf, + a flat connection – an operator $\nabla$ differentiating sections, with zero curvature. These are also known as D-modules.

a geometric character means one of two (equivalent) things: a. just a line bundle with flat connection on A. b. a flat connection (or D-module) with the character property: its fibers are given canonical identifications $L_{x+y}=L_x\otimes L_y$ , varying well in families.

Line bundles of degree zero on A are labelled by the dual abelian variety $A^\vee$ , while line bundles with flat connection (or geometric characters) are labelled by the space David called S (at least complex analytically – algebraically, as David mentioned, we should think of S as a nontrivial affine bundle over the dual abelian variety $A^\vee$ , modelld on the cotangent bundle.)

Direct integral means equivalence of derived categories, between D-modules on A and coherent sheaves on S, s.t. skyscrapers on S go to geometric characters on A. Since any coherent sheaf is put together in some sense from skyscrapers, this gives a precise meaning to saying any D-module on A is put together from geometric characters.

Well there’s more details in the notes (and more in the book based on those notes that’s theoretically in preparation now), so maybe I should refer to them and save some energy for actually writing up the notes..

Note to David: This story is only about $Pic^0$ but really in geometric Langlands we don’t want to restrict to $Pic^0$ —- all D-modules on the stack Pic are equivalent to coherent sheaves on the stack S of line bundles with connection (up to some derived subtleties). The cool thing is that the Z of components of Pic corresponds under this duality to the $C^\times$ stabilizers of points in Pic or S: the stack of line bundles on a curve is what we usually call Pic times $BC^\times$ , and reps of $C^\times$ are labelled by integers..

Posted by: David Ben-Zvi on June 22, 2007 6:57 PM | Permalink | Reply to this

Re: Generalized Geometric Langlands is False

The link given for David Ben-Zvi’s notes may be wrong or outdated.

Posted by: Todd Trimble on June 22, 2007 7:17 PM | Permalink | Reply to this

Re: Generalized Geometric Langlands is False

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Oops, thanks, it’s http://www.math.utexas.edu/~benzvi/Langlands.html (a mysterious extra character appeared on the previous post).

Posted by: David Ben-Zvi on June 22, 2007 7:23 PM | Permalink | Reply to this

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June 18, 2007