The Tamagawa number formula over function fields
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00:00
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Transcript: English(auto-generated)
00:25
And it's my great pleasure to introduce the speaker today, Denis Ghetschourie from Harvard and D.H. West, who will speak on the Tamagawa number formula over function fields. Thank you very much. It's great fun to be all over.
00:42
So, it's Tamagawa number formula over function fields. It's a joint work with Jacob Urey. So, I'll present a somewhat simplified context. So, we fixed a curve X, a smooth complete curve over function fields, over function fields, I mean over a finite field.
01:07
And we fixed G to be, G would be semi-simple and simply connected group.
01:23
Semi-simple. To simplify, I'll fix it to be over the ground field. So, more generally, this story works for group schemes over,
01:42
actually for groups over the field of rational functions of the curve. Okay, so, I'll present the formula in the way in which, well, in geometric terms, it's not a classical way in which one presents the Tamagawa number formula. So, we consider one G, a modulate space,
02:06
modulate step of principle G bundles on the curve.
02:23
This is an algebraic stack over a cube. It's not quasi-compact. So, yet, if you have a quasi-compact stack Y over a cube, you can try, well, you can calculate the number of its points over a cube,
02:45
but points must be taken multiplicities. So, by definition, if Y is quasi-compact stack, this is the sum over Y, which runs over the set of isomorphism class of points, and for each isomorphism class, you assign a weight.
03:01
So, you take one over, you take the group of automorphisms of this point, and you count the number of points of this group over the finite field. So, you sum this thing. And it's a finite sum if your stack is quasi-compact. Now, this guy is not quasi-compact. So, this time will be infinite, and one runs into a convergence problem,
03:23
but it's classical time that this thing converges. The bigger group is semi-simple. So, our interest is we are counting bundles. And now, I'll state the Tamagawa number formula in this case.
03:44
So, you said this is holes for quasi-compact stack, but now this is holes more generally now? Well, there's nothing to hold. This is the definition. But it converges? Yes, it converges for this particular one. Wasn't this a work of Kuybert in which he had some...
04:01
Kuybert, yeah? Yeah, yeah, yeah. He worked out. I mean, but this convergence is a very classical factor. It's the finite volume of the automorphic space. That's what it is. What is Kuybert? Kuybert. He had a way to use four power series of convergence. Ah, but you have some implicitly that you have the fact that every bundle of the type considered is rationally trivial.
04:24
Isn't it true? For simple method, yes. And this is somehow used to relate what you are doing to the classical... to the classical... Yes, yes. But now, when you are working with this more general situation of groups over the function field,
04:44
the simply connected, then it is not necessary, is it? It is. What is... So... This one, so... This... Well, I don't want to call it a set.
05:02
Oh, it's... Okay. So when G is defined only over and open, then you have to extend it to the work, I mean, because... Yeah, so, okay. In that case, what you do, you kind of sustain this thing. You first extend your group over the fraction field
05:23
to a group scheme over the entire curve and then you state it. So... The way I'm going to state it is specifically after the group has been extended. And that's why I made the simplification, because I don't want to go to that discussion. And is it still true in that more general case that the bundles are rationally trivial?
05:42
For me? The fact that the principal bundles on X are rationally trivial... Yes. Is it true in the more general context? Yes, similar method is true. But let me just stick to this case, because it's more similar.
06:04
So it's hook, yeah? Okay, hold on. Hook, no, no, no. Can you state the formula?
06:22
Put white. No, it's not blank, because it's a mirror. Can you state the formula? Oh, I state the formula, and it says the following thing.
06:55
So we take this quantity, we divide it by Q to the following power,
07:25
the dimension of G. And I can tell you what that interject is. So, parenthetically, the dimension of G is G minus one times the dimension of the group,
07:41
where G is the genus of the curve. And so, here is the formula. So it's the product. It's an infinite product over places of your curve, the following things. Here, you'll take Q to the power, QX to the power of the dimension of G.
08:05
QX is the number of elements in the residue field of X. So, QX is FX, and FX is the residue field at that point.
08:28
We fix the place. And here, I take the number of elements of G. So you see, here Q was in the denominator, here's in the numerator.
08:44
I don't like that. Well, invert. So, we write this as follows. Instead of talking about G, I'll talk about BG, the classifying stack.
09:03
And I'll rewrite the right-hand side as follows. Same product in the denominator to the dimension of BG.
09:26
The dimension of BG is negative the dimension of G. And here, right, BG, which is now an algebraic stack, count the number of points in that sense, which is exactly one forward number of points of G.
09:43
All right. And now, you see, the two things look kind of the same. So, here, we're counting the number of points in bungee, and it becomes a product. So, by the way, there's also a convergence problem, because it's an infinite product, so it converges. It's an Eulerian product, and I'm actually already going to show that it converges.
10:04
So, what you kind of deduce from this is that some idea, which is, of course, not at all true, is that bungee is, in very emphasized quotation marks, the product of these BGs.
10:24
It's in no sense true, but somehow it's true at the level of counting, because the number of points becomes the same. So, this is the idea that we want to realize. So, it's kind of local to global principle.
10:40
Here, it manifests itself in the counting. Okay. So, we'll attack this Namadala number formula by geometry, so we'll use the Frobenius. So, let me remind you some very general thing. So, let's begin by the first algebraic stack, which is quasi-compact.
11:01
So, in this case, it's a classical formula. It's a trace of the Frobenius, a rotenig vector formula, of the compacted supported cohomology of pi. So, a priori, this is an elatic number, and this is a rational number,
11:20
so it's an equality of elatic numbers. So, compacted supported cohomology makes sense for auto-brake stacks. Even non-quasi-compact ones? Well, this is for quasi-compact. So, I want to apply it in the case of bungee,
11:40
and there, there is little something to prove. In fact, this equality stays true for bungee, so that this non-quasi-compact of bungee can be controlled, and indeed, it can be controlled. It's based in classical reduction theory. And here, there is some convergence again. Yet, so far not. This is quasi-compact. For bungee, there is a convergence problem?
12:01
No, but the cohomology, the compacted cohomology is dual, and the cosmosing is dual. But if it's an infinite number, we've got to stabilize it. So, right, but you always get, in this case, for quasi-compact, geometric series. Yeah, I know this one. You can observe that this converges formally. Yeah, but there is a convergence problem,
12:23
but it's kind of automatically solved, because you're dealing with geometric series all the time. So, the formula stays true in its reduction theory.
12:51
We inductive limit poses, parts which are finite type. Yes. Yes.
13:03
Okay. I want to massage this form a little bit. So, I'll tautologically rewrite this as the trace of the inverse of the revenues on the dual vector space to that, and the dual vector space will be what's called the Borel-Moor homology.
13:22
It's usual, it's cohomology, but with conditions in the dualizing sheet. And, so, now, let's suppose that this y is smooth, as is our bungee.
13:40
In this case, the dualizing is, well, it's a shift and a t-twist of the constant. So, we can rewrite this even further, and we'll rewrite this as trace of the Frobenius inverse on just the usual cohomology, but divided by q to the dimension of y.
14:02
And, now, you can see why I wrote this. Well, because it's exactly the kind of quantity that appears on both sides, wants to express the following idea.
14:35
Again, in big quotation marks, I want to express the idea that the cohomology of bungee
14:43
is a well-defined graded vector space. It is, again, in emphasized quotation marks, the product of all places of the cohomologies of bungee. Again, it's not at all the restricted product in the sense of Adele's, this sense of product.
15:00
It's not that. So, this is what I want to assign the meaning to. So, what do I mean by taking this product? So, that... So, that's bungee index x, actually? Well, my g is constant. But, indeed, what I think of, the fun concept is bungee x, of course. It basically maps from its curves to bungee, like, continuous maps.
15:21
This is an algebraic geometry, right? Of course. No, but you have xs, which are close points, but now you're doing cohomologies over fq bar. So, it doesn't make sense. It doesn't make sense, indeed. So, if you take it very, very realistically, I will now assign the sense to it.
15:42
So, I will make sense of the right-hand side. In such a way that, when I take the trace of different ages, I actually arrive to the product formula that is written. So, that's what I will do. What is the erasure policy?
16:02
Just shift the two ones. Don't use the third one. It's easier. Just shift the number 20. What do you think? Trust your experience. This formula looks very similar to
16:22
the Van Fook's cohomology. It has a smooth manifold. It's for people in different worlds. It's an infinite manifold. It has a layer-algebra of vector fields. It has some cohomology. And it has some formal fields and some truncation of some nice homotopy type. You can see the space of sections.
16:41
Decorational homotopy type of the space of sections. It's not an analogy. I mean, you can derive this from that. The only problem is that, and it's actually legal derivation. I will not use it. I will use another derivation. What you're saying is completely legit. You can actually do it this way.
17:07
Parenthetically, in number theory, we don't really like the algebras. So, that's why I won't do it this way. It's also something called a single, something like this. I.T.A. Bot. So, that product formula is,
17:21
well, I haven't defined it. I haven't designed the right-hand side. Once I do, it will be the I.T.A. Bot formula, which, of course, I will say very specifically, but in 20 minutes, maybe. Okay. So, now we'll switch to a very different type of mathematics. We'll be doing some adjoint functions, okay? Just.
17:42
So, here is my X, and here is the projection for the point. I call it PX. And we'll be considering the category of, derived category of the logic sheet of X. I'll just write to you what it means that two functions are adjoint. So, here, it will be sheets on the point.
18:03
Well, same as that, well, complexes of elastic vector spaces. And there are two adjoint functions that connect these two categories. This is the copomology with compact supports,
18:21
and this is this up-shoot function. Yeah? Adjoint? So, what does it mean that they're adjoint? It means the following, guys. Home in the category of vector spaces from some object F to V
18:40
is the same thing as home in the category of sheets from F to PX up-shoot of V. So, now I want to change my problem. Instead of talking just about sheets, I'll talk about sheets of commutative algebras.
19:01
Well, what do you mean with algebras with respect to what kind of product? Tensor product. But what kind of tensor product? I want ... So, what is tensor product? Tensor product is F and G. You take F and G and you pull back. That's your tensor product.
19:20
However, I want to pull back with a shriek. I like shrieks better than stars for a good reason. So, this is my tensor product. Inner tensor product. Pardon? Inner tensor product compared to the outer tensor product. Well, this is the external tensor product, but there are two inner products. There is a star and a shriek. I want a shriek.
19:42
And I might omit writing this shriek all the time, but this is always the one I need. Okay. And one can talk about commutative algebras inside sheets on X.
20:00
And, well ... So, here, of course, sheets are the right category, but I wonder about fireplace conditions, because in your previous talks, you had unbounded sinks and so on. And here, in the elatic context, people usually consider only constructive and unbounded. Correct. So, what are you doing exactly?
20:21
So, I can tell you exactly what we do. We take the usual elatic sheets and int-complete. So, that's the formal definition. Int-complete of the DVC? Yes, the DVC, yes. DVC bounded constructible category.
20:49
So, this upper shriek defines a functor like this. So, here are your forgetful functors. Those are called oblivion.
21:02
Forget them on the algebra structure. So, the point is you define what it means to be an algebra, a cumulative algebra, using this tensor product. Yes, I mean, commutative algebras. So, this tensor product defines what's called the symmetric monoidal structure of the category, and then it makes sense to talk about commutative algebras.
21:20
So, you don't speak about commutative algebras in high sense? Of course I do. Of course I do that. You do what? Commutative algebras, of course. It's not in plain derivative, it's in sportive, but it has spectra and stuff. You don't need spectra. It's just infinity category. Infinity category.
21:41
It's infinity category. So, define what it means to be commutative, basically? Well, just to set up. And all the associative laws and all that stuff. Yes, you have to specify, infinite table of commutative and associative relations, and you have to disengage yourself from trying to get the categories into the proper high-integrated context.
22:02
So, this upper sheet just induces an upper sheet function of commutative algebras. But what about this lower sheet diagram here?
22:33
So, let me write one down underneath it. Well, you have these functions. You can take all of their left adjoints,
22:44
and the diagram will still commute the comm-alge.
23:00
I'm not even labeling these arrows yet. Please, I'm not labeling those, but... So, this is the left adjoint of what we had before. This is px lower sheet. And this is the function that actually I'm interested in.
23:22
That's what we'll be talking about. And I have a name for it. I have not a name, I have a fancy symbol for it. Integral. Okay, just for fun, do you guys know what these arrows are? What is the left adjoint to the function of forgetting the structure commutative algebra?
23:40
To be free. Free. If you have a sheet, you can create a free algebra. Always complicated. Ready? Commutative. We're in a commutative world. So, this diagram commutes. Excuse me, am I a stupid question? Is there a unit in this? A very good question. And I'm, for now, yes.
24:03
But we'll soon get rid of the unit. For a good reason also. But, right on. Okay, so this diagram commutes, but let me be very naive. Let me ask, would this diagram commute?
24:26
So, I'm asking the following question. Start with an algebra, and I want to compute the value of the left adjoint. Is it true that, and I want to compute this left adjoint, and just regard, look at the vector space I can get. Is it true that this vector space is very naively?
24:42
Forget about the algebra structure and take the commotion with the back support. What do you guys think? Yes or no? Could you repeat that for me? Yeah, I'm saying, well, the diagram with free commuted, the topological reasons, because you just take left adjoints of everything. But will the following be true? Will this diagram commute?
25:01
I'm E. I want to compute what this left adjoint looks like. Start with the commutative, sheave commutative algebras, calculate this mysterious left adjoint, and then just forget the algebra structure. Will it be the same, but forget the algebra structure and compute the left adjoint? Surely not. Surely not. There's no reason, and it's completely, completely false.
25:24
So, I would like to compute this. I'll be interested in this functor. The reason I'm interested in it is because to make sense of this fancy product. OK, but before I try to define it, let's do a very simple case
25:42
where X is a finite set, not a curve, just a really finite set. So let's solve it in this case. X is a finite set. What does it mean to be a sheave commutative algebras? All it means is you have a commutative algebra for each point.
26:03
OK, so I'm interested, let's call it B, this commutative algebra. So I'm trying to calculate this mysterious integral into an algebra, let's call it, what do they call it? A.
26:22
It's a null denied algebra. Yeah, you, I'm talking about this fancy higher category, but this commutative algebra, this problem makes sense for just the usual commutative algebras in degree zero. We're talking about very elementary stuff.
26:40
So I'm interested in this mysterious object, which is supposed to be home of commutative algebras. Well, sheaves of X, and remember, sheaves of X is just commutative algebra each point from B, well, probably PX of a sheik, but what I do, I just take this A and kind of put it on X,
27:03
just put a constant algebra in each. And by definition, this right-hand side is product over X home of compulge from what I have at each point in this A.
27:20
So, again, right-hand side, for each, so you have this bunch of commutative algebras, and just each of them is SFI homomorphs of A, and I'm looking for commutative algebra to match that product. You guys are good. Okay, I'm very glad that people are awake.
27:40
All right, and now we're going back to Luke's question. Take the unit out. What do you get then? Now my algebras are non-unital. This still makes sense? Who is this guy? X is a finite set. Let's take the simplest case.
28:03
Two. One and two. So I have P1 and P2. Without unit. No unit. Who is this guy? P1 plus P2 plus P1 plus P2. You guys are good? Yes, unit for the blank, unit for the unit. Put the unit and you will move.
28:22
Exactly. Well, that's fine to leave this one in, indeed. And non-unital algebra is the same as an augmented unit algebra. And so you know what to do. Put the unit in, take this as a product, take the unit out. That's what you get. Perfect.
28:41
If you're so good, give me the answer for general finite sets. Sum of all non-empty finite sets. Perfect. Okay. Okay, but that... Sum of the partial product of the whole subset of sets. Perfect. Okay, but that... Okay, that narrates a special symbol for a finite set.
29:26
It's the direct sum. When we call it column i, all finite non-empty subsets. And for each, I'll take the tensor product over all those x's...
29:44
...dx. So now, how do we want to generalize it when x is no longer a finite set? Well, you must concatenate this kind of thing here, right? After all, because if we have this finite set, something like that must appear in general.
30:02
How long am I doing on time? So now x will be a curve, but in general, what I'll say now to define this integral, x will be arbitrary connected scheme.
30:23
But connected is important. Unlike the finite set. Because it's connected. And to this x, we'll assign another geometric object. It's called the Wron space of x.
30:41
Wron is the last name of Zilang. It's a guy. It's a geometric object. Is it Wron? Yeah, it's his student. Nice student. Okay, cool. Well, how do we define objects in geometry?
31:01
As both of you taught us, define the function points. Hom from an affine scheme S to Wron of X will be the set of all non-empty subsets in Hom from S to X.
31:32
Looks pretty horrible. So you can ask, is it representable by... See, no multiplicities here. Really subsets. You can ask, is it representable by a scheme?
31:42
And what do you guys think? No. Yeah, it's like, if it was a scheme, what would be the dimension? It's not a scheme. So what is it? It's nothing. It's what we call a pre-stack. So the definition, a pre-stack is an arbitrary functor
32:18
affine schemes of. Well, you can say two sets, but I want more general
32:24
because I want also to incorporate 1G to group points of which sets are particular sets, particular hits. I think that in some places a pre-stack has to satisfy a part of the restriction. Here, none.
32:42
Just completely general. OK, why do I want this guy? Well, because it already appears to affine a set. If it already appears to affine a set, I won't have it in general. And moreover, I want to have sheaves in it. So how do we make sense of sheaves in such a gadget? Of course it is a functor in the wicked sense of pseudo-functor
33:01
or whatever, because otherwise it's... What do you mean pseudo-functor? That is, you are not working in the strict sense of strict functors. Yeah, I guess for me, strict functors don't even exist. Yeah, I mean, it's the functor in the reasonable sense.
33:32
So let's talk a little bit about sheaves. Let's first talk about sheave and schemes. So what are sheaves? Well, for me, sheaves is a functor.
33:58
For a little bit, not specify what kind of categories I consider.
34:02
You can consider, for now, triangulated, but we'll see exactly who I have to pass to the higher world. OK. Opposite, so, i.e., to a skewness, you assign this category of sheaves, but, for amorphism, you assign a like-sheaves, the sheaf of that functor.
34:37
Now I give myself a pre-stack why.
34:41
What is sheaves and why? So the sheaves are what kind of sheaves? Yeah. In the sense of incompleting and so on. This makes sense without, I could do it, I could do it before incompletion, but let me do it after incompletion. And categorize it by just the higher ones.
35:02
That's exactly, I'm needing the biggest for the next two minutes, so you can say, you can do it either triangulated or higher, but in a few minutes, I'll explain why I have to stick to higher. Not in a few minutes, in one minute. I'm sorry, that's an F over shriek there? That's a shriek, upper shriek.
35:21
OK, what's sheaves and why? Now, what's a sheave and why? Well, you tested by a fine scheme. To specify a sheave and why, you just have to specify all of its pullbacks. So this is for every affine scheme and a map little y from s to y,
35:42
i.e. an element little y in pull from s to y, you specify plus compatibility conditions, namely if you have an adaptable commutativity of this triangle,
36:09
because y is a groupoid, it's not that they're equal, it's specified in an isomorphism. To each such, you specify a compatibility,
36:23
F over shriek from what you had on s2 with what you have on s1, and compatibility for twofold compositions, s1 to s2 to s3.
36:51
But such a thing has a name, so let's just say for each blah blah blah, there's a word for it.
37:09
In other words, what they're saying is that sheaves and why is the limit over what we call affine schemes
37:21
over y, opposite of the upper shriek. And if you define using instead of F after shriek, just F after star, you must get something equivalent, isn't it? Because you have a smooth atlas.
37:40
No, nothing is smooth. y was completely, completely arbitrary. Arbitrary stack? Pre-stack. Ah, pre-stack. You get a different theory, they're not related, but it's a different gadget. So why is an object like that? Pardon me? Why is an object like that? Why is an object? Luckily, we still have it.
38:01
But no, so for an algebraic stack, besides having a fairness condition, just locally... For algebraic stack with a smooth atlas? Yes. You will get an equivalent definition. Ah, okay, this is what I thought. But for something like the one-space,
38:20
you will not get an equivalent definition. Not at all, actually. All right. But now comes the moment to explain why I actually need higher categories because, well, it's a well-known fact, it's a very bad idea to take the limits in triangular categories. So it's ill-behaved. You will not be able to compute it. So for that reason...
38:41
So could you use, what do they call them? Derivators instead? Derivators? I don't know. Derivator. So is it... Some intermediate machine which was studied in Paris, yeah. My understanding is, derivatives are sort of like derived categories. They allow you to do limits, but they don't have all the stuff of higher...
39:03
I don't know, because I don't know what these are. Ah, they're really very nice. It's one of the lessons Groton-Dietz sort of did, I think. Yeah. There are several series about derivatives. Yeah, Max Groth, who's... So I don't know. I know that for... Yeah, but is this a little larger series?
39:23
We're doing a little... I know, I know. Yeah, it's less exclusive here. So, we've got this. Now let me just introduce a couple of more pieces of notation of what you can do with this. Kind of by definition, if you have a map between three stacks,
39:45
by which we mean the natural transformation between these functions, it's built into the machine that you have an upper-street function, a pull-back.
40:00
You can ask, does this upper-street function admit a left-a-joint? Well, if you're working with analytic sheaves, from the way the theory sets up, it does always admit a left-a-joint. This is a map of arbitrary... Arbitrary three stacks.
40:23
So, on the three stack... Let's say you're working with the category of affine schemes, let's say a finite type, or... Yes, yes. Whenever I have a finite scheme, then I have a finite type. Finite type over...
40:41
Over... Over the brown fields. In this case, it's a cube. It happens to be over the brown. And just as a matter of notation, let me introduce this. So if you go back to the projection to the point, py,
41:04
instead of writing py below a sheet, I'll write cohomology with compact supports. As notation. And when I pull back from the point, my QL, I'll just denote it by the dualizing notation.
41:25
But now, let's combine these two pieces of notation, and let's take the cohomology of compact supports of the dualizing. Do you guys have a name for it? There are G, I think. No, no, no, it has a name.
41:40
Alright, who? No, no. It started with H. It's called cohomology. Compact supports cohomology. But now, I don't like the letter H, I'll write the letter C. I want to call it co-chains, because we are really in the weeds.
42:02
Chains. I put the lower star. We are in the higher world, so we're not taking individual cohomologies, it's really kind of chains. Alright. Now, go...
42:29
So this letter just exists by abstract limit kind of argument? No, it follows formally from the fact that lower should exist for affine schemes.
42:41
Ah. Just from there. I see. From there, it's formal. So... There's really some geometry there, that's not... Well, yeah. You use the entire theory of elastic machines. Just note, in this theory, we don't have upper stars or lower stars, we have just the peaks.
43:01
If you want lower star, you have to, say, pay extra. So why am I talking about this? Well, because we won't be around space.
43:28
We'll start going back. So we are in the highest point of the trajectory, now we start.
43:40
So, we start with B, which is a commutative algebra, in sheafs. Remember what we did with finite sets? We actually, from B, we produced another object, let me call it factor V,
44:06
which was actually a sheaf. So these algebras are no longer unital? They don't need to be unital. In our case, they happen to be unital, but they don't need to be.
44:24
And there's a... And I'll get... It's a... The unitality, non-unitality is the core of our work with Jacob. And I'll get to that discussion at the very end. So let's hold on to this.
44:41
But, I mean, it's useful that you're forgetting the fact that you know. And we did the same procedure as in the case of finite sets, namely, the fiber factor B at, let's say, just few valued points of the run space,
45:20
corresponding M-tuple of points of the initial space X, is the tensor product I of the chic fibers of the initial B and these points,
45:47
like we did in the finite set. Remember, it was BX1 times BX2. In the non-unital case. In the non-unital case. If you might have a unit, just forget it.
46:01
Okay, and now, so... So if it has a unit, the chic fibers have unit also? They do. Okay. Finally, so the answer for the integral happens to be homology with compact supports
46:23
over the run space. So I went to all of the discussion to define, to construct the sub-adjoint.
46:45
Here's what I'm saying. It's really kind of fancy, right? You have to go to the squeeze stacks and all this stuff, and you can't do otherwise. It's really kind of... You have no choice. You must do it. If you want explicit formula, you construct the run space from your B, construct this chic from the run space factor B, and then take its cohomology.
47:02
You say, homology or cohomology? It's cohomology with compact supports. It's the only thing that exists in our formula. Because you want cohomology on top. Ah, cohomology is cohomology with compact supports, co-visualizing specifically. So you need enough machinery to actually prove that it makes sense.
47:20
Otherwise, if you... Pardon? You need enough machinery, because all of those things are with higher... Yes. So you need some nice machinery that allows you not to... And we do have it. So we have the machinery now. Okay. So Louis' higher algebra is that machinery.
47:41
All right. So now let's go back to semaglobons. So one step closer to semaglobons. So let me just say that now a certain symbol will adhere backward, which hasn't been introduced. I'm wondering if you'll catch it.
48:12
Take a very particular B. So take a very particular C.
48:34
Both are algebras.
48:54
I claim that I have a map from B to C. C lower star?
49:04
I haven't introduced... I've only introduced homology. I haven't introduced cohomology. What is that? I wonder what you mean, because that's also... All right. So I claim that homology of anything is naturally co-algebra. Therefore, its dual is an algebra.
49:20
But what makes sense to dual? Linear. We are in vect. Take the linear dual. No, but the vect, according to what you said, includes... Yeah, yeah. So it's a symmetric monoidal category. In there it makes... There's operation of geology. You just... There's operation of geology. It makes sense.
49:43
No, but you said that sometimes you take the int completion. So you take... Yeah, yeah, yeah. It's infinite dimensional. Okay. But the dual of an infinite dimensional guy is an inverse limit of the dual of the five dimensional pieces. But it's fine. And so... The dual of co-algebra is nonetheless an algebra.
50:04
But not the converse. Not the converse. Ah, the dual of the converse. Of course, it's an algebra just with a huge uncountable... Okay. Perfect. However, it just so happens that each of these has finitely many... is finitely natural each comological degree.
50:21
So that... So this duality is not as horrible as it would be in general. So each comological degree is finite. Yeah, but I don't care. Formally given the definition, it makes sense. It's huge. That's okay. Now I claim there's a map from B to C.
50:50
So let's say there in the eely where it comes from as algebras. It means that for every point of the curve, I have a map from the cohomology of BG to the cohomology of bungee. Where does it come from?
51:02
So for each point of the curve, I can map co-chains on BG to co-chains on bungee. So for variation... A variation map. So you have a map from X times bungee to BG.
51:20
So for each point of the curve, you have a map from bungee to BG. You have pullback and cohomology. So that's... So therefore, a junction. From here we obtain a map. From there... That's where this product of all the BG's come from.
51:43
Yeah. I should have said maybe... Sorry. When she would create an algebras,
52:00
the definition of what I mean by this is the symbol that made no sense. The definition means that. Maybe we should also mention that those rational homotopy theory and it was developed how to calculate homotopy space of maps from some C double copy of a small dimension to high connected space here.
52:23
I think she was sensing what you do. Except in algebraic geometry. Of course this is what we do. But algebraic geometry doesn't work except when X is a curve and the target is bungee. Of course it stops with four dimensions.
52:42
It's infinity. It always works. No, no, no. Dimension three. Right, right, right. Correct. But in algebraic geometry, you have... Dimension two of curve. Yeah. But if you start with surface, it's nice to guess it's not convergent. So, a junction.
53:00
So you get a map from here and now I state the real theorem. By the way, this map makes sense when X is an arbitrary algebraic variety, G is an arbitrary group whatsoever, and now X is a curve and G is simply connected, then...
53:27
And semi-simple. Sorry. Semi-simple, simply connected. Otherwise, no. Then, the map is an isomorphism.
53:47
So, and this theory has a name. This is the idea of what we want. Of course, they didn't write it like this because at the time, there was no run space
54:00
or anything like that. However, the left-hand side, although it looks very tense, it can be computed very explicitly and you can do it very explicitly in terms of the cohomology of BG. So this is the paper. Yes, it's that paper. Yes. They proved it over complex numbers
54:21
and they proved it by the method that Maximus was talking about. We gave it differential geometric proof. It's analysis. Yes, it was analytic proof, basically along the lines of what Maximus was explaining, but the statement, if you write it in this form,
54:43
it makes sense in general, and it holds an arbitrary geometry. It holds algebra dramatically. So this is the real assertion and as a corollary, once you have that, you can formally take, well, it's a word,
55:04
but numerical product formula. So if you take the trace of the Frobenius on both sides, you actually arrive to the product formula we wrote before.
55:20
The product of traces of inverse Frobenius is on the cohomologies of this thing will be the trace of inverse Frobenius on the cohomology of this.
55:40
So the real work is here. So the question is how much time do I have? Five minutes? Five minutes. So the way you get exactly the product instead of those finite, before we said that we get something,
56:00
we add the product of finite sets, subsets of sets. Yes, but you can see the product of things which is one plus something. If you have infinite product, you can write some of the infinite or finite subsets.
56:20
Taking from the product of one plus some numbers. Yes, but this means it is one plus something, and the one is out, but still the one... There's a little bit of work required here. I'm not saying it's immediately evident.
56:42
But you are still taking things which are a unit or not, the C hat of star Vg as a unit? Yes, so the way we use it is as follows. So we actually rewrite it, the left hand side,
57:00
the way that Etienne Bott rewrote it. It's actually a symmetric algebra of something, and we calculate and see. So maybe, if you want, in question time, I can write the Etienne Bott formula, saying what that thing is explicitly, and from there show how to deduce this.
57:24
So in the remaining three minutes, I can try to say how we prove this. So it also, well, it requires proof, it's not like... If you wish, this is a local to global principle, we are saying that...
57:41
So the left hand side is essentially local, we assemble from finite subsets of the curve. And the right hand side is global, it really has to do with the kind of kind of automorphic objects, bungee. So, something needs to be done.
58:06
So it doesn't count for free. So let me say, what the key geometric ingredient is, we introduce this gadget called the finders' monument. But it's the wrong version.
58:22
Again, I don't have time now, but we have to add more details in question time. So it's the modulated space of a bundle equipped with a rational trivialization. And there is a snap that forgets the rational trivialization. And it also projects the mod space. Away from which sets you have rational trivialization.
58:43
So the key geometric result is that the homology... So this map is homology equivalence.
59:06
So this map is locally a product. So basically every bundle admits a rational trivialization. What it says is that for a given bundle, the space of rational trivializations is homologically trivial. That is to say that the space of rational functions from a curve,
59:21
which is a horrible, horrible, but manageable pre-stack. Imagine, we are talking about the space of rational functions. You can make sense of it as a pre-stack. Rational maps from the curve to the group G. You can make sense of it. You can actually calculate its homology and prove that it's trivial.
59:40
What we call homological contractable. Is it contractable? It is.
01:00:01
But the space of maps from a point to G is not contractible. Not at all. But rational maps is not contractible. Yeah. But rational maps includes constant maps. But that's okay. We've got more room. So I can tell you the reason. Again, I'll be happy to comment how that comes about.
01:00:21
And this is specific to dimension one. So none of that would be true if it was either lower dimension, as you say, point, or the surface. So what is the definition of the Rand-Rasmanyan over a base S? It's Anderson on S cross X.
01:00:41
I'll say that in a second. So maybe I think it has officially finished because it's 11.05. And then I'll answer this and everything else. Shall we do that? Yeah. Thank you. Okay. So we will take a few questions. So maybe we start as usual.
01:01:01
Tokyo, then Beijing, and then Paris. Okay. So, Tokyo, do you have some questions? Any questions?
01:01:28
Yeah, it's similar to nobody has a question. Beijing, do you have some questions? There are no questions.
01:01:43
Just go ahead. Go ahead. Sorry, I have a question. So there's a paper by Frankel and Go where they propose a geometric version of the trace formula. And the Arteopod formula, it occurs at the spectral side of the trace formula.
01:02:02
So do you see there's a possible connection between this proof of Tamagawa number and the proposal of geometric trace form? Good question. The answer is, I don't know. I haven't studied that paper. It's in efficient detail. I should do it. Just thanks for the suggestion.
01:02:21
Yeah, because in the number few case, Tamagawa number can take our course as proof as a consequence of the trace form consideration. So it seems that there should be a connection there. Maybe in the formula few case, the trace form, as they say, is interpreted as a consequence of the Arteopod trace form.
01:02:50
OK, I'll go and look. I just, I haven't. I'll do it.
01:03:01
Other questions from Beijing? Can you say a word on the proof of the Snapchat trace formula for Bungie? Yeah. I mean, so how do I deal with these infinitive, I mean, right? With convergence? Yeah. So, it's a reduction theory.
01:03:25
So, let's deal with SL2. OK, just for an example. So how does Bungie for SL2 look like? Well, it looks as follows. You can take an open piece which are bundles which are not very unstable.
01:03:44
Yeah, yeah, yeah, yeah, yeah. You take bundles which are not very unstable. Well, in general Bungie is stratified by the degree of instability. There are semi-stable bundles. And then for each integer n there are bundles that look as follows.
01:04:02
So they are extensions of, like this, where l has degree specifically n, n is a positive integer. And so this type looks very, very simple.
01:04:22
It's a stack of extensions of this and this. And, well, you explicitly see and then you sum up over all integers and you get geometric series. So on each stratum, so basically what happens is Bungie looks as a finite union of geometric series.
01:04:40
And that's how you control it. So in other words, you prove for algebraic stacks of finite types. And in this case, this guy is not a finite type, but it's, again, a finite union of things that look something like finite type times something that looks like a geometric series.
01:05:01
When I say geometric series, I mean it's 1 over a n, point over a n. Well, this is a fine translation, but you can't point to it. You literally get geometric series. You made the classifying space of a n.
01:05:22
Yeah, so these as you vary the degree. Okay, other questions from Eugene? No more questions. Okay, thank you. So, questions in Paris? So concerning this...
01:05:42
Run space? So first of all, there is a result that I think I heard in some seminar, but I don't remember exactly. So the fact that the bundles are rationally trivial, but also over a base that is locally for the topology, they are rationally trivial on the total space. There is some result that is...
01:06:00
Drinfeld-Simpson. Drinfeld-Simpson. Drinfeld-Simpson? Yeah. And now the rationally trivial in this context means trivial outside the scheme which is finite over the base? So in this case, so when the group is simply dependent, it's much, much stronger than that. You said that if you take any momentum divisor on the curve, then after an entire base change of your scheme of parameters,
01:06:35
it becomes trivial outside of the divisor.
01:06:41
Ah, okay. So simply connected... Sorry, for semi-simple it means... For semi-simple, this is true. For semi-simple bundles... Okay, for constant groups you mean. Okay, so this is proved for constant groups for group schemes. Well, it's proved only in our paper with Jacob and it's proved in the simply connected case.
01:07:03
And again, outside the given... Yes, you specify a section and after a total base change of the scheme of parameters, the thing trivializes outside the given section. Okay, so now concerning the definition of Drasmanyan run. Yes. Can you say what it is?
01:07:21
Yes, absolutely. So first of all, I said that it maps to here and to here. So I'd better, if I want to ask points of this, I have to specify points of here and points of here. So, home from S to Grasmanyan run is the bundle. First of all, it's a PG, it's a G bundle on S times X.
01:07:48
It's a point, S point upon G. Also, it's a finite subset in home and now it specifies my crucial Grasmanyan time pattern.
01:08:01
I specify a trivialization, P0G is a trivial bundle, on X times X without, when I remove a certain divisor. Which divisor? I take the union of the graphs.
01:08:21
So, I run to this finite set I, for each element I take the map from S to X, I take this graph, I throw it out. Okay, so this roughly explains this and then there is another very question that I, because, excuse me for this,
01:08:41
but still, the work concerning this integral of those PX, upper roof, upper shriek of C star BG on there. So, we work there with tens of, kind of tens of product of seeing whichever you need. But, yeah, so I want to understand roughly why it is like what conservationists
01:09:04
stated about the product of plus one plus alpha I is one plus finite product. It's a great question. So, first of all, literally, when you take the finite set, you get exactly that.
01:09:21
If X is a finite set, then the way this integral is defined, remember integral is cohomology with the back support of what is called factor F. That is the cohomology of the guy which is…
01:09:45
So, I'm wondering if we should… We should maybe take another question and… Should I answer the first question or should I… Yeah, yeah, yeah. Okay, so, again, the integral is by definition cohomology with the back support of one.
01:10:07
So, if X is a finite set, you literally get what Maxime said, because factor B, what is it? So, let's say, suppose X is a finite, it's a two element set, what's factor B?
01:10:21
Well, Iran, in this case, is this joint union of three elements. It's one, two, and one, two. And what's factor B? So, you get B1 with two. B1 tends to B2.
01:10:41
And cohomology with the back support is just direct sum. Okay, so you get the… Okay, and then how this is the product of… One plus B1 times one plus B2.
01:11:00
How this is the product of… So, you're asking, why do I call this, this quotation product, tensor product, quotation marks? So, then you want to get something which is the numerical product formula, which was actually a product formula… Okay, so maybe I should, maybe you should terminate that, I'll answer this. Okay, so if you can say it just in a few words.
01:11:23
Pardon? You cannot answer just in a few words. Yeah, it is. Okay, so maybe we take another question. There is no other questions. Okay, then we will say goodbye, since it should be now seven years in Tokyo. Yeah, exactly, we need to go to…
01:11:43
Okay, so goodbye and see you in one month. Bye-bye.