SPEAKER 1: So the second speaker of the afternoon is Jacob Lurie.
JACOB LURIE: Am I audible? Thank you very much and I'd like to thank the organizers for inviting me to speak at this event. So this is going to be a talk about applied topology. I'm going to start off describing a problem in number theory. I'm then going to describe how it can be translated into a problem in topology and how it can then be attacked using ideas from topology.
So I'd like to start the story by saying a little something about quadratic forms. So by a quadratic form I just mean a homogeneous polynomial of degree 2 and some number of variables. So here are some examples of quadratic forms in two variables.
And when I talk about a quadratic form, I should tell you what ring I'm working over. I'll write quadratic polynomials where the coefficients will lie in some ring. So here are the coefficients I've used are plus or minus 1. That means these are quadratic forms over any ring I like.
Now, one of the basic questions you can ask if you're given some quadratic forms is are they equivalent? So you'll say one quadratic form is equivalent to another if there's a linear change of coordinates that will convert the first into the second.
So here are some quadratic forms. Now, as quadratic forms over Z, these are not equivalent to each other.
In fact, they're not equivalent to each other even as quadratic forms over the real numbers because, for example, the first is positive definite. It always takes non-negative values. The third is negative definite. And the one in the middle is indefinite, takes both positive and negative values. But over the complex numbers, these three are all equivalent because you can multiply the variables x and y by a square root of minus 1.
So the classification of quadratic forms over the real numbers is fairly simple. You can always diagonalize them and put them in a standard form where they look like a sum of squares minus another sum of squares. And the only invariant that you have in this situation is how many plus signs appear and how many minus signs appear.
So the classification of quadratic forms over the integers is much more complicated. So if you have two quadratic forms over the integers you might want to know are they equivalent, well one way you could tell them apart is by working over the real numbers, diagonalizing them, and checking signatures. So you can tell these are not equivalent over the integers because they have different signatures.
But it's easy to give examples of quadratic forms over the integers which are not equivalent but which have the same signature. So, for example, these two quadratic forms here, they're both positive definite. So they're equivalent over r, but they're not equivalent over the integers.
And one way to see that is to see what happens when you reduce modulo 3. Modulo 3, this first form, will be non-degenerate but this 3y squared will go away. So the second form will become degenerate modulo 3. So that's another way you might tell two quadratic forms apart.
If you're given two integral quadratic forms, you might reduce mod 3 or reduce mod 4 or reduce mod 5. So that motivates the following definition. So that let q and q prime be positive definite quadratic forms and implicitly here I'm assuming they are in the same number of variables. Then we say that q and q prime are in the same genus. If q and q prime are equivalent, mod N for all N greater than 0.
So clearly if you have two quadratic forms that are equivalent, they are in the same genus. But it's not clear that the converse holds. You could ask does the converse hold and the answer is no but almost. If you fix a quadratic form q you can ask up to equivalence how many quadratic forms are there in the genus of q.
And it turns out there are only finitely many. And, in fact, you can write down a formula which tells you how many there are. And that's called the Siegal mass formula.
So before I can state the Siegal mass formula for you, I need to introduce a little bit of notation. So let's say that q is a quadratic form over some commutative ring R. And I'm going to write Oq of R for the orthogonal group of this quadratic form. This just means the set of N by N matrices with coefficients in R such that they carry this quadratic form to itself. So q composed with a is equal to q.
So, for example, if q is positive definite and R is the field of real numbers, this is the usual orthogonal group of rotations of Euclidean space. And in that case and in general, there's also a subgroup, Soq R which is going to be the collection of all A belonging to Oq R such that the determinant of A is equal to 1.
OK, so now let me state the Siegal mass formula for you. Oh, one other thing I should remark about these groups is that if you have a positive definite quadratic form over the integers then Soq Z and Oq Z are always finite groups because now we're talking about rotations in Euclidean space that are required to preserve a lattice and there will always be only finitely many of those.
So it makes sense to consider the following expression. So fix q, a positive definite quadratic form, over Z.
I'm writing in the dark. Is this visible to everyone? OK.
So the mass of q is defined to be the following. So you take the sum over all equivalence classes of quadratic forms q prime in the genus of q of the following number, which is 1 over the size of the automorphism group of q prime, the size of this orthogonal group.
So if it weren't for this denominator here, this would just be counting the number of equivalence classes of quadratic forms there are in the genus of q. But we're counting them with multiplicity where the larger the automorphism group is, the smaller the multiplicity. So the Siegal mass formula is a formula for this mass. So this is a definition and a theorem is that the mass of q is equal to something.
Of course that's not much of a theorem yet. It's equal to something specific. But what specific thing it is depends on the quadratic form q. So let me start by telling you what that something is in the simplest case.
AUDIENCE: Can you give us an example for this there's more than one of the [INAUDIBLE]?
JACOB LURIE: Yeah, I guess I'll give you-- well, for example, I'll give you sort of an example in a second. So a quadratic form is said to be unimodular if it's non-degenerate mod p for every p and unimodular forms comprise a genus.
So if you have-- it's easy to see that if you have two quadratic forms that are in the same genus and one is uni-modular than the other is also unimodular because they look the same mod p for every p. But the converse to that, which is not obvious, is also true. Any two unimodular forms in the same number of variables are in the same genus.
So, for example, in 24 variables there's 24 equivalence classes of unimodular quadratic forms. They're all in the same genus. OK, so unimodular forms comprise a genus and that's in some sense the simplest genus.
So let me tell you what the Siegal mass formula tells you for that genus. So let q be a unimodular quadratic form. So unimodular means that it's non-degenerate mod p for every prime number p. So let this be a unimodular quadratic in N variables and let me make a remark that this condition of unimodularity is a very strong one. In particular, none of the quadratic forms that I wrote down at first is unimodular. Those were all degenerate mod 2.
So this actually implies that N has to be divisible by 8. You don't see unimodular quadratic forms until you have at least eight variables. So in this case you can say what the mass of q is and this is equal to the following expression.
So there's a denominator of 2 to the N minus 1 and a power of pi and some factors of some gamma functions and then there's some special values of the Riemann zeta function, zeta of 2, zeta of 4, all the way up through zeta of N minus 2 and zeta of N over 2 appears twice.
OK, so this is sort of an amazing formula because if you look at this expression here, it's not at all obvious that it's even a rational number let alone that it's a particular rational number that's the solution to a counting problem. So let me just give you a couple of examples to illustrate how this works or what this tells you.
So the first example I told you would appear when N is equal to 8. So when N is equal to 8 what happens? Well, in eight variables there's a unique unimodular quadratic formula. It's the E8 lattice which came up in an earlier talk. So the left-hand side is just 1 divided this 1 over the size of the automorphism group of E8 lattice.
And that's the [INAUDIBLE] group of E8 and I believe that's 2 to the 14th, 3 to the 5th, 5 squared times 7. And, of course, the Siegal mass formula tells you that this is also equal to the right-hand side which you can check for yourself if you remember what the special values are or if you have the ability to look them up.
So lest you think that this equality appearing up there is an equality between two numbers that are very small, you could take N to be 32 and in this case you can check that the right-hand side is enormous. And this has a consequence which is note that the left-hand side is a sum of numbers and all of those numbers are less than or equal to 1. In fact, they're all less than or equal to 1/2 because every-- minus 1 is always an automorphism of any quadratic form.
So this-- if a sum of numbers less than 1 is going to be really big, you have to be adding up a lot of numbers. So this has a consequence that there exists billions and billions of pairwise in equivalent unimodular quadratic forms in 32 variables. So--
AUDIENCE: Can you describe what a unimodular form is?
JACOB LURIE: It's a quadratic form that's non-degenerate mod p for every p.
AUDIENCE: Is that equivalent of having determined a 1 [INAUDIBLE]?
JACOB LURIE: That's equivalent to having determined 1, yeah. Or plus or minus 1.
AUDIENCE: What's wrong with x squared then?
JACOB LURIE: X squared-- the associated bilinear form of x squared is x and y goes to twice xy, so mod 2 of that vanishes.
So, for example, I mentioned a little bit earlier there's a classification of unimodular quadratic forms in 24 variables. There's exactly 24 of them. But there will never be a classification of unimodular quadratic forms in 32 variables because of this.
OK so I'd like to just-- let me remind you briefly--
AUDIENCE: [INAUDIBLE] unimodular or [INAUDIBLE]?
JACOB LURIE: Sorry? Well, I'm talking about quadratic forms not symmetric bilinear forms, so there's a bijective correspondence between symmetric bilinear forms that are even and quadratic forms.
So let me remind you what the Riemann zeta function is. This can be defined as the infinite sum of 1 over N to the s. But more to the point of what I want to write in a second this has an expression, an Euler product expression as a product over all primes p of 1 minus 1, 1 over 1 minus P to the minus s.
So that expression for the Riemann zeta function gives you a product decomposition of this as a product over all primes p of a certain rational number. And I want to tell you what that rational number is in a way that will maybe demystify this expression slightly. So this expression here is equal to the product over all prime numbers p of p to the N times N minus 1 over 2 divided by the size of oq-- or Soq actually of z mod p.
So I mentioned earlier that over the real numbers this Soq R, that's the usual special orthogonal group in dimension N. So that's a compact Lie group whose dimension is N times N minus 1 over 2.
So given that information, if you were asked to estimate about how many elements you thought Soq had over Z mod p you might say, well, it's probably p to the dimension. So that's the numerator in this expression and the denominator is the size that that group actually has, which it's not quite equal to the numerator but it's around just about equal to the numerator in the sense that this fraction is approximately 1 and when you multiply over all p that's a convergent product. And it converges to this expression here.
So I stated the Siegal mass formula originally to you. I sold it to you as an expression for the number of quadratic forms in a genus counted with multiplicity. But another case of interest might be the case like we had when N was equal to 8.
What if you happen to have only one quadratic form in genus? Then the left-hand side is 1 over the size of the orthogonal group of that quadratic form. And the right-hand side is giving you an expression which is computing the order of that group for you.
And so it's saying roughly something like if you want to know the number of integer matrices in the orthogonal group, well, you take the number of matrices that are in the orthonormal group mod p for all p and you multiply them all together. Of course, that's not sensible unless you do some kind of renormalization and this is telling you exactly how to normalize that so that it's sensible.
So this is some kind of local to global principle. It's telling you that the size of integral automorphism groups are determined by the size of finite automorphism groups, mod p for every p.
So I should remark that's an interpretation of what appears on the right here. What appears on the left has a similar interpretation, but instead of counting points mod p, counting automorphism as mod p, you're trying to count the number of automorphisms that you have over the real numbers.
Of course, over the real numbers, the automorphism group is this compact Lie group. So you don't want to count the number of points, but there's a canonical measure that you can write down and you can ask what is the volume with respect to that measure. And that's what this expression here is.
So I've told you what the Siegal mass formula says for a unimodular quadratic form. I don't want to write down what you get for a quadratic form there's not unimodular but let me just remark that it has the same flavor. It expresses the mass of q as some kind of product over all primes of some expression that depends only on what your quadratic form looks like at a certain prime.
And those expressions actually are exactly this for primes where you're quadratic form is non-degenerate mod p. And it's something a little more complicated when your quadratic form is degenerate mod p but in all cases it roughly has this flavor. It's some kind of local to global principal.
OK, so, this is the Siegal mass formula and it's a great theorem but maybe you're not interested in quadratic forms. So then I ask you, what are you interested in? Because if you're interested in anything like quadratic forms, there's a generalization of this formula which was conjectured by Weil, which we'll apply to the sort of object that you're interested in.
So let me just state the conjecture. I'll be very vague right now and I'll come back and make it more precise in a slightly different context later. But so fix a widget q over Z, and what you want to do is count the number of widgets q prime that are locally like q.
So widget means something like quadratic form, something that it makes sense to talk about having a widget over a ring. And locally like q means q prime should be isomorphic to q mod N for every N and also over the real numbers.
So now you could try and count this number and here it's supposed to be weighted just like the definition of the mass. You're supposed to count this number with multiplicity where everything counts with multiplicity 1 over its automorphism group.
And the conjecture says that this should always be equal to, well, let me say some factor c infinity times a product over all primes of some factor cp where at most primes p, what you're suppose to see is an expression like this. Some power of p divided by the order of an automorphism group mod p. So-- sorry?
AUDIENCE: [INAUDIBLE] it converge?
JACOB LURIE: Yeah, so I should have mentioned the hypothesis first, but for a widgets q and let me say whose automorphism group over R is compact. Or maybe even better I should say compact and semisimple.
So, of course, what Weil conjectured was much more precise than this, but this idea can be made more precise in a variety of cases. So an example is maybe you're not interested in things whose automorphism group is the orthogonal group. Maybe you want to study things whose automorphism group is a unitary group.
So instead of studying quadratic forms on lattices, you'll study lattices that have an action of Z bracket I. And instead of quadratic forms you could have Hermitian forms. So that's a typical example of what you could apply this conjecture to.
So this is not maybe the usual way that the conjecture is stated. What they actually noticed was that in the case of the Siegal mass formula, you could naturally interpret the left-hand side divided by the right-hand side as the volume of a certain space with a canonical measure, and that space and its measure were defined for a large class of groups. And Weil conjectured that for any simply-connected group that measure should be equal to 1. And when you unwind what it's saying, it's giving you a statement of this flavor.
So this is-- conjecture of Weil, it was proven by Weil in a number of cases. And it's actually a theorem which was proven first for split groups by Langlands and then by Lie for quasi-split groups and by Kottwitz in general.
So this is the statement that I'd like to talk about but I'd like to talk about it in a slightly different context. So we're doing number theory now and right now this is sort of number theory over q. We're thinking about the fraction field of the integers as q. There's an analogous statement if you replace q by any finite extension of q.
But another thing that you can do when you're doing number theory is to replace fields like q in number fields by function fields. Instead of working with fields that are finite extensions of the rational numbers, you could take a field like the integers mod p and the join of variable t.
So a field like this or any finite extension of a field like this is called a function field and these are analogous to finite extensions of q in many ways. And many problems in number theory make sense in both contexts.
However, the typical behavior is when you have a problem that makes sense for number fields and makes sense for function fields, the function field problem might be a lot easier because in the case of function fields you have a lot more tools, mainly tools coming from algebraic geometry, that are available that are not available in the case of number fields.
So this statement was sort of an exception to that rule in that this was known in the case of a number field but not known in the case of function fields. So what I'd like to do is first tell you what the statement is over a function field and then tell you how it translates into a problem of topology and then describe an attack on that topological problem.
So let me fix some notation for the rest of this talk. So Fq is going to be a finite field with q elements and sigma is going to be an algebraic curve over Fq. So that means something that's described as an algebraic variety, a solution to polynomial equations with coefficients in Fq.
So let me draw you a picture. You should think of sigma like it's some kind of a Riemann surface because if instead of working over the field Fq I was working over the field of complex numbers, a one-dimensional algebraic variety would look like a Riemann surface. And this gives you good intuition in general.
So an algebraic curve has a well-defined genus, so let me call that genus g. And for each point belonging to sigma-- so by that I mean what in algebraic geometry would be called a closed point of sigma, there's an associated residue field at x which is a finite extension of Fq. So this data is roughly going to be our replacement for the integers and the rational numbers.
So the other data that I want to fix. I'll let G with a subscript complex numbers be a complex semisimple Lie group. Now, a complex semisimple Lie group always admits the structure of an algebraic group in a conical way. So what that means is it can be described as an algebraic variety inside the set of N by N matrices for some N. It can be described as solutions to some polynomial equations.
So a main example if you're not familiar with the theory of algebraic groups, just imagine for the rest of this talk that what I'm talking about is SLN C. So this is the set of N by N matrices with determinant 1 and the condition of having determined 1 is just a polynomial equation in the matrix entries.
So a feature that you see in this situation and is in fact a general feature is that the polynomial equations that describe this algebraic group, you can write them down over the integers in a conical way. So we can talk about SLN not just of the complex numbers but over any ring we like. We can talk about SLN Z mod p, for example.
So this means-- well, so let G be the associated algebraic group which makes sense over the integers. So that means that it makes sense to talk about G of R for every ring R. And in particular it makes sense to talk about G-bundles in algebraic geometry. So the meaning that I'm going to give to widget in this context is G-bundles.
So consider the following sum. This is the sum over G-bundles on sigma-- let me call my G-bundle script p of 1-- over the size of the automorphism group of p. So if you have a G-bundle over this algebraic curve, over Fq, it will always have a finite automorphism group just coming from the finiteness of this field Fq. So each of these individual numbers is well-defined.
Now, let me warn you that this sum is in this context is not finite anymore. I've lost the analog of the compactness assumption that I had on my group earlier. This is typically an infinite sum. So, for example, if G is the group SLN we're talking about studying vector bundles on our curve sigma, and typically there are not just finitely many isomorphism classes of vector bundles. There are infinitely many.
Nevertheless this sum turns out to be well-defined. It converges. And Weil's conjecture tells you what this sum converges to. In other words, it roughly tells you in some weighted sense how many G-bundles are there on the curve sigma.
So what this turns out to converge to is q to the 1 minus g times the dimension-- little g is the genus. Multiply by the dimension of this group. So some factor of q times a product over all x in sigma of, well, an expression like what we had before.
So we write down a rough estimate. How many automorphisms do you think a G-bundle has or-- I'm sorry. How many elements do you think G has over this field kappa x? Well, a rough estimate is the size of kappa x raised to the power of the dimension of G. And you should divide by the actual order.
So really I should not label this Weil's conjecture but I should label this the split case. This is what I've stated here is the analog of the version of Weil's conjecture that was proved by Langlands.
AUDIENCE: Do you propose [INAUDIBLE] the size of kx to the [INAUDIBLE]?
JACOB LURIE: The size of kx to the [INAUDIBLE], yeah.
AUDIENCE: Sir, can you say that all the [? four ?] [? limits ?] curve [INAUDIBLE]?
JACOB LURIE: The closed points. Closed points. So the points over Fq are the ones for which this field has exactly q elements.
OK, so this more precisely is really the split case version of Weil's conjecture. The general case of Weil's conjecture would give you an analogous formula where the group G was not constant but instead it was a family of groups parametrized by this curve. So that's the case where this was not known over function fields.
But let me just stick to the simpler case. I want to describe an attack on this-- so I want to describe a strategy for proving that these two numbers are equal and it will generalize to the case where G is not constant.
OK, so the first thing we want to do is take advantage of the fact that we're working over a function field and not over a number field. Function fields are supposed to be easy.
So what's easy about this case? Well, when we were talking about the Siegal mass formula we were trying to count the number of quadratic forms in a genus and that counting problem didn't have a lot of structure. We just had a set of equivalence classes of quadratic forms. But G-bundles have an algebra geometric parametrization. Meaning I can talk about G-bundles over this curve sigma which is a curve defined over a finite field, Fq.
But I could also talk about G-bundles on the curve that I get by taking some extension of Fq. I could, for example, replace Fq by its algebraic closure, in which case I'll see a whole lot more G-bundles.
So we can encode this structure by introducing an object that I'll call BunG sigma. So I'm going to call this the moduli space although the real word I should use is stack. So this is a moduli space G-bundles and what does that mean? It means that giving a map from y into BunG sigma should be the same data as giving a G-bundle on sigma cross y.
So the word stack refers to the fact that when I say these pieces of data are equivalent. I'm not talking about isomorphism classes of bundles but I want to keep track, want to remember that G-bundles have automorphisms.
So this object is a stack but it's what's called an algebraic stack, meaning it's something that you can, for many purposes, think of as an algebraic variety. You can think of it as the set of solutions to some polynomial equations. In other words, you can write down some equations such that solving those equations gives you a G-bundle.
So in those terms, what are we trying to do? Well, we're trying to look at solutions to those equations inside the field Fq. So what we want is to know the number of Fq points of this moduli space.
So here a number is, again, we're counting these with multiplicity. So this by definition really means the left-hand side of that expression here. It's a sum over G-bundles p of isomorphism classes of 1 over the size of the automorphism group. All right, so if this wasn't a stack, if it was just an algebraic variety, then all of these numbers here would be 1 and we would just be counting the number of points in a more traditional sense.
So let me ignore the fact that this is a stack and just think of it as an algebraic variety for a moment. The question that we are then given, we have some algebraic variety over Fq and we want to know how many points it has over Fq. Well, this is the subject of another very famous idea of Andre Weil.
So to explain this idea, so let's suppose we have some algebraic variety y. So let me describe this algebraic variety. It's the set of two [INAUDIBLE] x1 through xk that satisfies some polynomial equation.
So let's just say for simplicity it's one polynomial equation-- f of x1 through xk is equal to zero. And here I mean that I want this to be an algebraic variety over Fq meaning the coefficients of this polynomial live in Fq.
But when I think of this as an algebraic variety maybe I want to think of but I'm looking for solutions that lie in the algebraic closure of Fq. So let's think of this as a subset of the algebraic closure of Fq to the power k.
So now inside this subset you have a finite set consisting of solutions where each of the xi's belongs to the finite field Fq. So that's what I'll denote by y of Fq. Well, it's a subset of y and it also lives inside Fq to the k.
So now how can you describe this subset? Well, the easiest case might be to think about when there's just one variable and there are no equations.
So how do you describe Fq inside Fq bar? Well, Fq has q elements and it's the set of solutions to a polynomial equation of degree to q. It's the set of all x in Fq bar such that x to the q is equal to x.
Now, we're working in over a finite field which means that-- so q is a power of some prime number p and we're in characteristic p which means raising to the pth power has magical properties. So taking a number and raising it to the pth power is additive. It's always multiplicative but in characteristic p it's also additive. So raising to the qth power is actually a ring homomorphism from Fq bar to itself.
And what this buys you is the following equation. If you take f of x1 through xk, if you raise all of these variables to the qth power, that's the same thing-- I'm allowed to move the q to the outside. That's the same thing if f of x1 through xk raised to the qth power. So the operation of taking all of the coordinates and raising them to the qth power carries this algebraic variety y to itself.
So there's a map from y to itself which, roughly speaking, just raises all of these variables to the qth power. So this is called the Frobenius map. And if you're interested in the set y of Fq, the set of solutions to these equations with coefficients in Fq, this is the set of fixed points of the Frobenius map.
So proposal of Andre Weil is that we should then be able to count the number of points on the left-hand side using the Lefshetz fixed-point formula. So the number of points of y over Fq should be given by the trace of the action of Frobenius-- well, let me say on compactly-supported cohomology. It should be given by the following sum, an alternating sum of the trace of the Frobenius on the i-th compactly-supported cohomology group of y.
So if we were working in ordinary topology and all of these fixed points were simple fixed points, then something like this would be true. But the obstacle to proving something like this is making sense of the right-hand side. After all, y is not an algebraic variety over the complex numbers. It's an algebraic variety over a finite field and there isn't a good topological space to take the cohomology of in this context. But making sense of this idea and proving this theorem is one of the great successes of the Grothendieck school of algebraic geometry.
So in the context of algebraic varieties over Fq, this statement is true provided that by cohomology here you mean a tall cohomology. So Grothendieck introduced a cohomology theory for which this is a sensible and true statement.
OK, so I actually want to state that slightly differently. Let me assume that y is smooth of dimension d. So in that case, if we were an ordinary topology, we'd have an oriented manifold and compactly-supported cohomology would be the same as homology.
Well, it's kind of the same in algebraic geometry. The only difference is that you have an orientation but it's not preserved by this Frobenius map. So if you switch from compactly-supported cohomology to homology, it introduces a little bit of a factor. And what you actually learn is that the number of points-- if you state this in terms of homology-- the number of points of y over Fq is equal to a rough estimate for how many points you expect, q to the dimension, times a similar sum where you put trace of Frobenius now acting on the homology of y.
So this is true for smooth algebraic varieties and now let me mention this is also true when y-- if you replace y by this thing I call BunG sigma, which remember is not quite an algebraic variety. So this is not a formal consequence of the usual Grothendieck Lefschetz trace formula, but in this case this was proven by Kai Behrend in his thesis.
So and what's this dimension? Well, that's that same power of q that appears up there just by chance. So if you use this formula for the number of points, then you've converted this Weil's conjecture into a statement in topology.
JACOB LURIE: Oh, yeah. Yeah. I should emphasize-- oh sorry, this equation for ordinary algebraic varieties? Or--
AUDIENCE: No. I guess for the sides.
JACOB LURIE: Yeah, yeah. So well one thing to be careful of is that for ordinary algebraic varieties, this side is an integer, this is a finite sum, and they're equal to each other. In this situation we're interested in, as I mentioned before, this is defined by that sum, which is an infinite sum. And this thing actually ends up also being an infinite sum because BunG has homology in infinitely many degrees.
So the statement is that both of these expressions, which are potentially infinite sums, are convergent sums and they converge to the same thing. And dealing with that convergence is exactly what makes this more difficult than the usual Lefschetz trace formula.
But let's take it. So if you believe this statement then what is the Weil conjecture saying? The Weil conjecture is now saying that if you look at summation minus 1 to the I times the trace of Frobenius on Hi of this moduli space of bundles, that this is given by-- has an Euler product expansion. It's given by a product of local factors at the points of your curve.
So let me just write lambda x. Lambda x is that number up there but I'm not going to convince you of that in this lecture.
So this is what we want. We want a statement like this. And what it comes down to do is understanding the homology of BunG. So I want to say this is a problem of topology, although maybe that's a little bit misleading. It's a problem of a tall topology. It's a problem of topology in the world of algebraic geometry.
AUDIENCE: Yeah, actually, what do you mean by homology?
JACOB LURIE: I mean a compactly supported--
AUDIENCE: I mean cohomology in a palpable homology. But I don't know how to define a tall homology.
JACOB LURIE: Well, you could define it, for example, as the duel of a tall cohomology or as compactly-supported cohomology with coefficients in the dualizing complex.
So we want to understand the topology of this moduli space of bundles. So now I want to-- let's step back for a minute and pretend that we're not working over Fq. Pretend that we're working over the complex numbers. So then sigma is a Riemann surface and G is a complex semisimple Lie group.
Oh, I should have mentioned there's a missing hypothesis up there. G should be simply connected. If it's not simply connected, there's a correction you have to introduce.
And what we're interested in is studying, well, again G-bundles on sigma. And here we have our choice. We could study a whole amorphic G-bundles on sigma or we could just study arbitrary G-bundles on sigma.
And let's attack the problem first just from a purely topological standpoint. Let's try studying just arbitrary topological G-bundles on sigma. So those are classified by maps from sigma into the classifying space BG. Maybe I should write BGC.
So let me just state over here roughly the problem could be stated as follows. So let M be a manifold let's say of dimension n. So I have in mind the case where M is our Riemann surface sigma, so n is equal to 2. And let x be a pointed space. And the space I have in mind is the classifying space for this group G.
And what we want to do is describe the space of all maps from x into BG-- or, sorry, the space of all maps from M into x. So if, for example, G was a discrete group, a discrete abelian group then a map from M into BG-- well, homotopic classes are just classified by H1 of M with coefficients in G.
So this you can think of as a kind of cohomology/ although if G is non-abelian, it's a sort of non-abelian kind of cohomology of M. So we would like to describe this space and we would like to describe it in a way that has a sort of local to global flavor. We would like to describe it as something that has to do with contributions from individual points of M.
So if we were talking about ordinary cohomology, there's sort of a local to global principle that governs ordinary cohomology which is poincare duality. And what I want to describe now is sort of a non-abelian analog of that.
So let me state for you a theorem. So versions of this go back to Graeme Segak, Dusa McDuff, and more recently [INAUDIBLE] but a statement which is addressed at understanding this.
So I want to take that mapping space which I want to think of as a kind of cohomology of M and express it as some kind of homology of M. So the first thing that I want to do is make it look more covariant.
So if U is an open subset of M, I can consider maps from U into x that are compactly supported. That means that outside a compact set, they carry everything to the base point of x. Now, compactly-supported maps, that's a covariant [? founter ?] of U because if U is contained in a larger open set then I can extend any compactly-supported map just by carrying everything outside of U to the base point.
And, of course, I should have mentioned before, M is compact. Let me write that now. If M is compact then compactly-supported maps from M into x is the same as all maps from into x.
So we get a lot of things that contribute to this mapping space. So every time we have an open subset of x, we have a map from the left-hand side to the right-hand side. So now you can take the colimit of this over, say, all open sets U and that's obviously going to be the same as the right-hand side because there's a final object. Just take U equals M.
The interesting statement is that you don't need all open sets. You only need to think about very simple open sets. So here I should really write homotopy colimit. And if you take the homotopy colimit over all U which are homeomorphic to a union of disks.
AUDIENCE: [INAUDIBLE] x.
--this left-hand side. Oh sorry, that M is an X. Absolutely.
OK, so theorem. This map is a homotopy equivalence if x is n minus 1 connected. So I want to call this non-abelian poincare duality. So let me just tell you first one reason you might call this non-abelian poincare duality is that from this you can recover ordinary poincare duality.
So an example, if you want ordinary poincare duality, you should take x to be an Eilenberg-MacLane space. Let's say if you want coefficients in Z, take x to be KZm for m large.
And then what happens? Well, if you take the ith homology group of the left-hand side this ends up being isomorphic to the homology of M in degree-- this should be plus n minus m. And if you compute pi I of the right-hand side, what you get is the cohomology Hn minus I of M with coefficients in Z. This is in the case where M is oriented, what really goes here is the orientation local system.
And that looks like that number is wrong. That should be an I minus n plus m. And this map is just a cap with the fundamental class of n.
And so the statement that this map is a homotopy equivalence is the statement that this map is a isomorphism for each I and that's the usual statement of poincare duality.
So when x is in Eilenberg-MacLane space, this recovers poincare duality. That's one reason I'd like to call it non-abelian poincare duality.
But the other case I'd like to make is that this is a local to global principle. This so let me tell you first why I want a theorem like this. I want a theorem because maybe I want to say something about the topology of maps from m into x. Maybe I want to compute its cohomology groups. For example in that problem over there.
Well, that's hard for two reasons. It's hard because M might be complicated and it's hard because x might be complicated. And what this is is a mechanism that allows you to separate those two complications.
So the right-hand side, M and x are all tangled up with each other. But in the left-hand side there are sort of separate because if you're interested in compactly-supported maps from U into x where U is a disjoint union of disks, well if U is just one disk then that looks like the n-fold loop space of x.
In particular, it doesn't depend on M. And if U is a disjoint union of disks, that just looks like a product of finitely many copies of the n-fold loop space of x.
So these individual terms here that are appearing here, they're really independent of M. Where M appears is in the diagram that you're taking the colimit over. The common [INAUDIBLE] of this diagram depends on M but it doesn't depend at all on x. You're just thinking about subsets of M that are homeomorphic to a disjoint union of disks. That poset doesn't see x.
So that's why this might be useful. It gives you a way of separating the complications of M at x.
So now I want to take this idea and bring it to algebraic geometry. So this was a theorem of topology but it has an analog in the setting that I'm really interested in for studying Weil's conjecture. So let me tell you what that analog is.
So let's look at the space we want to understand, which is this moduli space for G-bundles on sigma. And I'm going to make another space which in some sense-- instantiates this idea. So I'm going to call this RanG of sigma and this is going to be a moduli space for the following data.
So a point of RanG sigma means a finite subset of sigma, a G-bundle p on sigma, and a trivialization of p on sigma minus S.
So what is this saying over here? What is this non-abelian poincare duality theorem say? Says if you want to understand the topology of all maps from M into x, you might as well only think about maps from M into x that are supported on a disjoint union of disks.
Of course, you might as well make shrink those disks and make them very small. You might as well think about maps that are supported in a small neighborhood of finitely many points.
So in this algebra geometric setting, the maps are G-bundles and we don't have things like open disks but it does make sense to talk about G-bundles that are supported near finitely many points. Just demand that your G-bundles come with a trivialization away from that finite set of points.
So this space here, this is an analog of the left-hand side of this statement here. But it's a sort of moduli stack that lives in algebraic geometry, and of course there's a map from this RanG sigma to BunG sigma which just forgets the finite set S and forgets the trivialization.
So a theorem, this map is [? iso ?] on homology, an algebraic geometric analogy of that statement up there.
So why does this help you?
JACOB LURIE: Well, this space is like the-- it's like you built that space out as the colimit of all of those spaces.
OK, so why does this help you? Well, now you can implement this idea of sort of treating the complications coming from sigma and the complications coming from G separately. So to do that, let me introduce one more object that I'm going to call Ran sigma which is going to be the same definition up here except without the G.
So this is just the moduli space of finite subsets of sigma. And there's a map from RanG sigma to Ran sigma that forgets the G-bundle and forgets the trivialization. So this Ran sigma doesn't depend on G. It doesn't have anything to do with G-bundles and, well, I see that I'm just about out of time so maybe I don't have time to explain this in anymore detail.
But what you want to do in order to compute the homology here is to use some kind of Leray spectral sequence for this map. You want to express the homology of this space on top as something like the homology of the space on the bottom with coefficients in the homology of the fibers.
So in this case the bottom doesn't depend on G and the fibers don't depend on the global structure of sigma. So that ultimately gives you a local to global formula which looks exactly like this. But I think I will stop there.
SPEAKER 1: Any questions?
AUDIENCE: Can you explain the left-hand side of the board up there? What I know about KZm is it computes the [INAUDIBLE] topology in the space.
JACOB LURIE: Yeah, so--
AUDIENCE: And seems like the [INAUDIBLE] itself.
JACOB LURIE: Well, which isomorphism? This?
AUDIENCE: The one on the left. The one on the left.
JACOB LURIE: The one on the left?
AUDIENCE: How do you know that pi I is the Hi minus the n plus the [INAUDIBLE]?
JACOB LURIE: Well, you have-- it is a version of poincare duality I guess you could say. Well, that statement together with this statement. Poincare duality is the sum of those two statements, so which one you want to say the poincare duality is happening in is up to you. But, well, let me say a few words about how you can--
AUDIENCE: Your statement
JACOB LURIE: What's that?
AUDIENCE: You've stated the poincare duality when you wrote the left-hand equality I suppose.
JACOB LURIE: Well, what this really is, this is computing-- without poincare duality this is computing the homology of M with coefficients in a local system that has to do with the compactly-supported cohomology of Euclidean space. So poincare duality for Euclidean space allows you to make this identification. And then this local to global principal is taking you from poincare duality for Euclidean space to poincare duality for an arbitrary manifold.
AUDIENCE: Why do you need little m to be enlarged?
JACOB LURIE: Why do I need--
AUDIENCE: For the connected points.
JACOB LURIE: Yeah. So this-- well, of course, there's always [? capping ?] with the fundamental class there's always an isomorphism from F. Oh, that's the usual poincare duality statement. What happens is there's really a map from here to here and I need M to be large in order for this map to be an isomorphism.
Maybe you might ask why do I need-- in the statement of the theorem, I need x to be n minus 1 connecting point. You might ask why do I need that assumption?
And, well, let's think about what would happen, for example, if x was disconnected. So what we want to say is that any map from M into x comes from a map that's supported in a finite union of disks and if x had another component that was not the base point component, there's no way to map a positive dimensional manifold into that other component that will have compact support.
So if M is going to have dimension of at least 1, x had better be connected. And similarly if M has dimension of at least q, x had better be simply connected and then so forth.
SPEAKER 1: Anymore questions? Let's thank Jacob Lurie.
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Jacob Lurie of Harvard University gives a presentation at the 50th annual Cornell Topology Festival, May 6, 2012.
The festival was organized by the Cornell University Department of Mathematics with support from the National Science Foundation.