 And it's my great pleasure to introduce the speaker today, Dennis Gates-Goyle from Harvard and the HOS. We will speak on the Tamagawa number formula of function fields. Thank you very much. It's a great fun to be all over. Indeed, so it's Tamagawa number formula for function fields. It's a joint work with Jacob Lurie. So I'll present a somewhat simplified context. So we'll fix a curve that's a smooth, complete curve of a function field. I mean, it's over a final field. And we'll fix g to be semi-simple and simply connected group. To simplify, I'll fix it to be over-the-ground field. So more generally, this story works for group schemes over, 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 the classical way in which one presents the Tamagawa number formula. So we consider bun g. The formula stays on the, like, step of the principle g-bundles on the curve. So this is an algebraic stack over a q. It's not quasi-compact. So yet, if you have a quasi-compact stack y over a q, you can try, well, you can calculate the number of its points over a q, but points must be taken with multiplicities. So by definition, if y is quasi-compact stack, the sum over y, which runs over the set of isomorphism class of points, and for each isomorphism class, you assign a weight. 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 number will be infinite, and one runs into a convergence problem, but it's, well, it's a classical fact that this thing converges. Each group is semi-simple. So our interest is we are counting bundles. And now I'll state the Italian number formula in this case. So you said this is holds for quasi-compact stack, but now this holds more generally now? Look, there's not the whole, this is the definition. But it converges. Yeah, it's converges for this particular one. Okay. Wasn't this a work of Kai Berent, in which he had some... Kai Berent here? Yeah, yeah, yeah. He worked out. I mean, this converges is a very classical fact. It's the finite volume of the automorphism stack. That's what it is. What is, who worked out? Kai Berent. He had a way to use four power series of convergence. Ah. But you have some implicitly the, you have the fact which, that every bundle of the type considered is rationally trivial. Isn't it true? Full-simple connected, yes. And this is somehow used to relate what you're doing to the classical form, to the classical... Yes, yes. But now when you are working with this more general situation of groups over the function field, the simply connected, then it is not necessary... It is. What is... So, this one... So, this... Well, I don't want to call it a set. Well, it's... Okay. So when G is the finite volume 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 function field 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 into that discussion. Is it still doing that more general case that the bundle is a rational retrieval? For me? The fact that principle bundles on X a rational retrieval? Yes. Is it doing the more general context? Yeah, so the method is true. But let me just stick to this case, because it's my simple... No, no, you have... I'm just going to... Hold on. Hold on. Hold on. Hold on. Excuse me, there you are. I'm going to put white. Don't leave white, because it's in the camera. Do you lower the white? I lower the... I state the formula, and it says the following thing. That... So we take this quantity, we divide it by Q to the following power, the dimension of Banji. And I can tell you what that interjection is. But hypothetically, the dimension of Banji is G minus 1 times the dimension of the group, 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 on your curve. The following things. Here, you'll take Q to the power and Qx to the power of the dimension of G. Qx is the number of elements in the residue field of X. So Qx is fx, and fx is the residue field of the curve. Residue field at that point. We fix that place. And here, I take the number of elements of G. So you see here Q was in the denominator. Here's the numerator. I don't like that. We'll invert. Let me rewrite this as follows. Instead of talking about G, I'll talk about VG, the classifying stack. And now rewrite it right inside this follows. Same product. I'll write in the denominator to the dimension of VG. The dimension of VG is negative to the dimension of G. And here I write VG, which is now a algebraic stack. Count the number of points in that sense, which is exactly one forward number of points of G. All right. Now you see the two things look kind of the same. So here we're counting the point, number of points in bun G, and it becomes a product. So by the way, there's also a convergence problem because it's an infinite product. It converges. It's an Eulerian product. And the elementary argument shows that it converges. So what you kind of deduce from this is that some idea, which is of course not at all true, is that bun G is in very emphasized quotation marks the product of these VGs. It's in no sense true, but somehow it's true at the level of counting. And the number of points becomes the same. So this is the idea that we want to realize. So it's kind of a local to global principle here that it manifests itself in the county. Okay. So we'll attack this Navagawa number formula by geometry. So we'll use the forgiveness. So let me remind you it's a very general thing. So let's again YB first, an algebraic stack, which is quasi-contact. So in this case, it's a classical formula. It's a trace of the tomatoes and growth of the negative formula when the compact is supported conomology applied. So a priori, this is an elliptic number and this is a rational number. So it's an equality of elliptic numbers. So compact to support conomology makes sense for algebraic stacks. So... Even non-quasi-compact ones? Well, this is for quasi-contact. So I want to apply it in the case of bungee and there is a little something to prove. In fact, this equality stays true for bungee so that this non-quasi-compact is a bungee can be controlled and indeed it can be controlled. It's basic classical reduction theory. And here there is some convergence again. So far not this is quasi-contact. For bungee there is a convergence problem. Conomology, the compact converges are dual and this is for smoothing dual converges are infinite numbers because of stabilizers. So, right, but you always get in this case for quasi-compact geometric series. Yeah, I know this from... People observe that this converges formally. Yeah, but there is a convergence problem but it's kind of automatically solved because you're dealing with geometric series all the time. Okay. So, the formula stays true in its production theory. Well, inductive limitals is parts which are finite type. Yes. Yes. Okay. I want to massage this form a little bit. So, tautologically we write this as the trace of the inverse of the covariance on the dual vector states to that and the dual vector states will be what's called the Borel-Mourou homology. It's unusual, 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. In this case, the dualizing is, well, it's a shift in the tape twist of the constant. So, we can rewrite it even further and we'll rewrite this as trace of the covariance inverse on just the usual cohomology but divided by q to the dimension of y. 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. And again, in big quotation marks I want to express the idea that the cohomology of bungee it's a well-defined graded vector space is in an emphasized quotation marks the product over all places of the cohomologies of Vg. Again, it's not at all the restricted product in the sense of Adel's, this sense of product, it's not that. So, and this is what I want to assign a meaning to. So, what does it mean by taking this product? So, that Vg index x. Well, my g is constant, but indeed, what I think of when you look at the shape. The whole concept is Vg actually. I just click maps from the curves to Vg, like continuous maps here. This is an algebraic geometry, right? So, you have access which are close points but now you're doing cohomologies over fq. It doesn't make sense, indeed. So, again, take it very, very realistically. I will now assign sense to it. So, I will make sense of the right hand side in such a way that when I take the trace of the convenience I actually arrive to the product formula as written. So, that's what I will do. What is the erasure of this? Just shift the curve. Don't use the third one. It's easier. Just shift the number to one side. You think? Yeah. Trust your experience, but that's not true. It's formula looks very similar to, this is Dilphan Fuchs's homology. It's a smooth manifold. It's for people in different routes. It's infinitely manifold. Yeah. That's the Lie algebra vector fields. It has some homology. And it has kind of formula fields with some truncation, with some sort of, some sort of isopropy type. The case of sections, the correctional homotopy type or the case of sections. It's not an analogy. I mean, you can derive this from that. Yeah, it's kind of... The only problem is that, and it's actually legal derivation. I will not use it, but I'll use another derivation. What you're saying is just, it's completely legit. You can actually do it this way. And periodically, remember theory, we don't really like the algebras. So, that's why I won't do it this way. There's something called a single, or something here that's... I take a boat. I take a boat. Oh. So, that product formula is... Well, I haven't defined it. I haven't designed it right inside. Once I do, it will be the idea of what formula. Which, of course, I would say very specifically, but in 20 minutes, 80. Okay. So, now we'll switch to very different types of mathematics. We'll be doing some major encoders. Okay, just... So, here is my X, and here is the projection from the point. I call it PX. And we'll be considering the category of the right category of elliptic sheets on X. I'll just write to you what it means that two functors are adjoint. So, here it will be sheaves on the point. Well, the same as that, well, complexes of elliptic vector spaces. And there are two adjoint functors that connect these two categories. This is the homology with compact supports, and this is this upper-shoot functor. Yeah, adjoint. So, what does it mean there? Adjoint means the following, guys. Home in the category of vector spaces from some object F to V is the same thing as home in the category of sheaves from F to PX upper-shoot could be. So, now I want to change my problem. Instead of talking just about sheaves, I'll talk about sheaves of commutative algebras. Well, when you make commutative algebras with respect to what kind of product, tensor product, and what kind of tensor product, I want... So, what is tensor product? Tensor product is... You take F and G and you pull back. That's a tensor product. 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 star and shriek. I want a shriek. And I might omit writing the shriek all the time, but this is always the one I mean. Okay. And one can talk about commutative algebras inside sheaves on X. And, well... So, here the... Of course, sheaves is the derived category, but I wonder about fireplace conditions because in your previous talks you had unbounded things and so on. In eladic context, people usually consider only constructively unbounded. Correct. What are you doing exactly? So, I can tell you exactly what you do. We take the usual eladic sheaves and incomplete. So, that's the formal definition. Incomplete of the DVC. Yes, the DVC, yes. DVC bounded constructively category. So, this upper shriek defines a functor like this. So, here are your forgetful functors. There's a whole oblivion. Forget that on the algebra structure. So, the point is you're defining what it means to be an algebra, a cumulative algebra using this tensor product. Yes, I mean, commutative algebras. So, this tensor probably defines what's called the symmetric monoidal... Here, yeah. Symmetric monoidal structure in the category and then it makes sense to talk about commutative algebras. So, don't speak about commutative algebras in high sense? Of course I do. Yeah. Yeah, of course I do that. To do what? No, it's commutative algebras. Of course, not in plain derivative. Yeah. It's transportable to infinitesimally spectra lines. No, you don't need spectra. No, you don't need spectra. It's just infinity category. Infinity category. Yeah, it's infinity category. So, to define what it means to be commutative, basically... Well, just to set up... And all the sociative laws and all that stuff. Yeah. Yeah, you have to set up about the infinite tale of commutative and sociative relations. And you have to disengage yourself from trying to go to the categories and do it in the proper, high-integorative context. So, this upper sheet just induces an upper sheet punk from commutative algebras. But what about this lower sheet? Beautiful diagram here. So let me write one down underneath it. Well, you have these punkers. You can take all of their left adjoints. And the diagram will still commute to the comm-alge. I'm not even labeling these arrows yet. Well, 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 punker that actually I'm interested in. 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. If you guys know what these arrows are, what is the left adjoint to the punker of forgetting the structure commutative algebra? Free. Free. If you have a sheet, you can create a free algebra. Always commutative. Very commutative. We are 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. But we'll soon get rid of the unit. For a good reason also. But write it down. Okay. So this diagram commutes. But let me be there in the eave. Let me ask, would this diagram commute? 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 there in the eave? We forget down the algebra structure and take the commutative about support. What do you guys think? Yes or no? Could you repeat that? Yeah. I'm saying, well, the diagram with free commutative pathological reasons, because you just take left adjoints and everything. But with the following, be sure, will this diagram commute? I.e. I want to compute what this left adjoint looks like. Start with the commutative algebra. She will commutative algebras. Calculate this mysterious left adjoint and then just forget the algebra structure. Will it be the same that forget down the algebra structure and compute the left adjoint? Surely not. Surely not. There's no reason, and it's completely, completely false. So, I would like to compute this. I will be interested in this function. And the reason I'm interested in it is because to make sense of this fancy product. Okay. But before I try to define it, let's do a very simple case where x is a finite set, not the curve, just a really finite set. So let's solve it in this case. So x is a finite set. What does it mean to be a sheep commutative algebras? All it means is to have a commutative algebra for each point. Okay. 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 you call it? A, in... It's an ordinary algebra. Yeah. I'm talking about this fancy higher category, but this commutative algebra problem, this problem makes sense for just the usual commutative algebras in degrees zero. We're talking about very elementary stuff. So I'm interested in this mysterious object which is supposed to be home of commutative algebras from sheaves of x. Remember, sheaves of x is just commutative algebra each point from B, well, all the Px of a sheep, but what I do, I just take this A and kind of put it on x, just put the constant algebra in each. And by definition, this right inside is product over x of commutative algebra from what I had in each point to this A. So again, right inside for each, so you have this bunch of commutative algebras, each of them is specified at a homomorphous of A and I'm looking for commutative algebras, essentially a product. You guys are good. Okay. I'm very glad that people are awake. 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-unit. This still makes sense. What, who is this guy? x is a finite set. Let's take the simplest case. Two. Two. One and two. So we have B1 and B2 without unit. No unit. Who is this guy? B1 plus B2 plus B1 plus B2. You guys are good. You have put the blank during the presentation. Put the unit and you will move. Exactly. Well, that's fine to be in this one. Indeed. And non-unit without algebra is the same as an augmented unit without 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. Okay. If you're so good, give me the answer for general finite sets. Sum of all non-unit to finite subsets. Perfect. See? Okay. Okay, but that... Sum of the partial product over all subsets of sets. Perfect. Okay, but that... Okay, that... That narrates a special symbol. Indeed, for a finite set, it's the direct sum, maybe called column I, all finite non-unit to subsets. And for each, I'll take the tensor product over all those x's. So now, how do we want to generalize it when x is no longer a finite set? Well, it must incarnate this kind of idea, right? After all, because if we have this finite set, something like that must appear in general. What am I doing on time? Okay. 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. But connected is important, unlike the finite set, because it's connected. Until this x will assign another genetic object, it's called the Ron space of x. Ron is the last name of Zebran, it's a guy in California. So, it's a genetic object. Zifron? Yeah, it's just a term. Okay, cool. Well, how do we define object symmetry as gothic quotas? Define the puncture points. Home from an affine scheme s to run of x would be the set of all finite, non-empty subsets in home from s or x, s to x. It 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? And what do you guys think? No. Or... Yeah, it's like, if it was a scheme, what would be the dimension? It's like not a scheme. So, what is it? It's nothing. It's what we call the pre-stack. So, the definition of pre-stack is an arbitrary puncture that finds schemes off. Well, you can say to sets, but I want more general because I want also to incorporate bungee to group points of which sets are a particular case. I think that in some places a pre-stack has to satisfy a part of the restriction. Here, none. Ah, okay. Just completely general. Okay. Why do I want this guy? Well, because it already appeared to find a set. And if it already appears to find a set, I want to have it in general. And grover, I want to have sheaves in it. So, how do we make sense of sheaves on such a gadget? Of course, it is a functor in the weaker sense of to the functor or whatever, because otherwise it's... What do you mean to the functor? That is, you are not working in the strict sense of strict funtals. Yeah, I guess for me, strict funtals don't even exist. Yeah, yeah, I mean it's... It's the functor in reasonable sense. So, let's talk a little bit about sheaves. Let's first talk about sheaves on schemes. So, what are sheaves? Well, for me, sheaves is a functor for a little bit not specified what kind of categories I consider. You can consider for now, triangle it, but we'll see exactly who I have to pass to be higher world. Okay. Opposite. So, i.e., to a schemess, you assign this category of sheaves. But, for a morphism, you assign a, like, sheaves. A sheep for that functor. So, now I gave myself a pre-stack. Why? What is sheaves and why? So, here sheaves are what kind of sheaves I love the... Yeah. In the sense of in completing and so on. Yeah, this makes sense without... I could do it before in completion, but let me do it after in completion. The goal is, is it, like, chance to hide one? That's exact. I'm leading it ambiguous 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 a sheave there? That's a sheave. Up a sheave. Okay. What's sheaves and why? Not easily. What's a sheave of why? Well, you're tested by a fine scheme. To specify a sheave of why, you just have to specify all of its pullbacks. Namely, so this is for every a fine scheme and a map, little y from s to y, i.e. an element little y in form from s to y, you specify plus compatibility conditions. Namely, if you have an adaptable commutative to this triangle, because y is a groupoid, something equal to specify an esomorphism, to each such, you specify a compatibility. Effer Peschik from what you had on s2 with what you have on s1. And compatibility for two-fold compositions. s1 to s2 to s3. But such a thing has a name. So this is to say, for each blah blah blah, there's a word for it. In other words, what they're saying is that sheaves of why limit over what we call defined schemes over y, opposite of the found third, schweb. And if you define using instead of effer Peschik, just effer per star, you must get something equivalent, isn't it? Because you have a smooth atla, smooth. No, nothing is smooth. Why was completely completely arbitrary? Arbiterista. Prista. Ah, prista. Yeah. You get a different theory, they are related, but it's a different gadget. So why is an object like this? Pardon me? Why is an object like this? Why is an object like this? Luckily, we still have it. But no. So for an algebraic stack, without even finding this condition, just locally. For algebraic stack with smooth atlas, you will get an equivalent definition. Ah, okay. This is what we're going to do. But for something like the wrong space, 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 limit in triangle categories. So it's ill-behaved. You will not be able to compute anything. So for that reason, So could you use one of the columns, derivators instead? Derivators? I don't know. Derivator. So is it one of the... Someone told me there's much in which was studied in various fields. My understanding is derivators are sort of like derived categories. They allow you to do limits. But they don't have all the stuff of higher... I don't know because I don't know what these are. Ah, they're really very nice. It's one of the lessons Grottendijk sort of did, I think. Yeah, there are several series about derivators. Yeah, this is... So I don't know. I know that... But does this larger series... I know. I know. It's less explicit here. It's less explicit here. So we've got this. Now let me just choose a couple of more pieces of notation of what you can do with this. Kind of by definition, there's a map between pre-stacks, which would mean the natural transformation between these bunkers. It's built into the machine that you have the upper-street-funker pullback. You can ask, does this upper-street-funker admit a left adjoint? But if you're working with electric sheets from the way the theory is set up, it does always admit a left adjoint. It's a map of arbitrary pre-stacks. So on the pre-stack, but still you're working with the category of affine scales, is it a finite type, or... Yes, yes. But they write the hind schemes and that's a finite type. Over your brown fields. In this case, it's a cube. Happens to be over the top. And just as a metro notation, let me introduce this. So we go back to the projection to the point, or PY, instead of writing PY below a sheet, I'll write co-homology to compact supports. That's notation. And when I pull back from the point, my QL, I'll just denote it by the dualizing notation. But now let's combine these two pieces of notation and let's take the co-homology of compact supports of the dualizing. Do you guys have a name for it? Bradley. No, no, no. It has a name. All right, move. No, no. It starts with H. It's called homology. Compact supports of homology. But now I will not... I don't like the letter H. I'll write the letter C. I want to call it co-chains because we are really... Chains. I put the lower start. We are in a higher world, but the individual homologists it's really kind of chains. All right. I'm sorry, F over shrink is between Y and Y2 or between the shift category. Aversive. Particle result to all people. Thank you. All right. So this letter joint just exists by abstract kind of argument? No, it follows formally from the fact that from there it's formal. So... There's really some geometry down. You use... You use the entire theory of electric chains. Just look. In this theory we don't have upper stars and lower stars. We have just the shrieks. If you want lower start you have to say pay extra. So why am I talking about this? Because we won't do it on space. We'll start going back. So we are... Like we are in the highest point of the trajectory now we start. We start with B which is a commutative algebra in sheaves. If you remember what we did with finite sets we actually, from B we produced another object let me call it factor B which was actually a sheave. So these alters were 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. And I'll get... It's a... Unital is non-unital 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. But I mean it's unital but you're forgetting the factors unital. And we did the same procedure as in the case of finite sets. Namely the fiber factor B let's say just few valued points of the wrong space corresponding just multiple points of the initial space X is the tensor product of the eye of the sheave fibers of the initial and these points. Like we did in the finite set remember it was BX1 times BX2 Unital In the non-unital case. Because you're taking wrong. Yeah. You might have the unit, just forget it. Okay. And now, so... So if it has a unit the sheave fibers have a unit also? They do. Okay. Finally, so it happens to the homology with compact supports over the wrong space. So I went to the older discussion to define, to construct this left adjoint. Here's what I'm saying. It's really kind of fancy, right? You have to go to the space X 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 to explicit formula you construct the wrong space from your B construct this sheave from the wrong space vector B and then take its homology. You say homology or co-homology? It's co-homology with compact supports. It's the only thing that exists in our formula. Because you want homology at all. Ah, homology is co-homology with compact supports homologizing specifically. So you need enough machinery to actually prove that it makes sense, otherwise you need enough machinery because all of those signals are with higher things. So you need some nice machinery that allows you not to We don't have it. So we have the machinery now. Okay. So Louie's higher algebra is that machinery? Alright. So now let's go back to Samagawa. One step closer to Samagawa. So, let me just say that now a certain symbol will appear on the backward which hasn't been introduced. I'm wondering if you'll catch it. I'll take a very particular b. And I'll take a very particular c. Both are algebras in a curve and I claim that I have a map from b to c. Is she lower star? I haven't introduced... I've only introduced homology. I haven't introduced co-homology. What is that? So I claim that homology of anything is naturally co-algebra. Therefore, it's dual is an algebra. But in which sense the dual? Revenue. We are in Vect. Take the linear dual. Vect, according to what you said, includes direction. So it's a symmetric monoidal category There's operation of duality. There's operation of duality. It makes sense. No, but you said that sometimes you take the int completion. So you take... It's an infinite dimension. But the dual of an infinite dimensional guy is an inverse limit of the dual of the finite dimension. But it's fine. The dual of a co-algebra is nonetheless an algebra. But not the converse. Not the converse. The dual of a co-algebra. Of course, it's an algebra with a huge uncountable... However, it just so happens that it finds the national each-chromological degree. So this duality is not as horrible as it would be in general. So each-chromological degree is fine? Yeah, but I don't care. It formally gives the definition. It makes sense. It's huge, but it's okay. Now I claim there's a map from B to C. But let's say, there and easily, where it comes from. As algebras. It means that for every point of the curve from the co-homology of BG to the co-homology of Bungie, where does it come from? So for each point of the curve I can map co-chains on BG to co-chains on Bungie. Evaluation map. Evaluation map. So you have a map from X times Bungie to BG. So for each point of the curve you have a map from Bungie to BG you have pulled back from co-homology. So there are four. Adjunction. From here we obtain the map from the... Where this product or all the BGs come from. Yeah. I should have said maybe sorry. The definition of what I mean by this is so that symbol that made no sense by definition means that. Maybe you should also mention that those and it was developed how to calculate what the space of maps from some C-double completely small dimension to high connected space. I think it should be sensitive to what you do. Except in algebraic geometry. Of course this is what we do. Yeah. But in algebraic geometry it doesn't work except when x is a curve and the target is BG. I am because it starts with what I mean. It always works. No, no, no. Correct. But in algebraic geometry you have dimension 2 of curve. Yeah. But if you start with the surface it is not convergent. So. Adjunction. So you get the map from here and now I state the real theorem. By the way, this map makes sense when x is an arbitrary algebraic variety and now x is a curve and G is simply connected. Sorry. Then the map is an isomorphism. So if this theorem has a name this is the idea about them. Of course they didn't write it like this because at the time there was no round space or anything like that. However, the left hand side of the book is very tense. It can be computed very explicitly in terms of the cohomology of BG. So is this the paper? Yes, it's that paper. But they work overseas. Yes. They include it over complex numbers and they include it by the method of Maximus talking about. It was the game of differential geometric proof. It's analysis. Yes. So it was analytic proof basically along the lines of Maximus explaining but the statement, if you write it in this form it makes sense in general and it holds an arbitrary ground. It holds algebraically. So this is the kind of real assertion and as a corollary so once you have that you can formally take it's a work but numerical product form. So if you take the trace of the Frobenius on both sides you actually arrive to the product form of the road before. That's the product of traces of inverse Frobenius on the cohomologies of this thing will be the trace of the inverse Frobenius on the cohomology of this. So the real work is here. So the question is how much time would I have? Five minutes? Five minutes. So why you get exactly the product instead of those finite before we said that we get something we had the product of subsets You can see the product of things which is one plus something if you have infinite product you can write it as some of the infinite subsets taking the product of one plus some numbers. Yes, but this means it is one plus something and the one is out but here the one is the little bit of work required here and I'm not saying it's immediately evident But you are still taking things which are a unit or not the same capital star BG as a unit? Yes, so the way we use it is as follows. So we actually rewrite it the left-hand side, the way that a TN bot is actually a symmetric algebra of something and we calculate and see. So maybe if you want in question time I can provide the idea of a formula to say well that thing is explicitly and from there show how to deduce this. So in the remaining now three minutes I can try to say how we group this. So it also requires group it's not like that. If you wish this is a local to global principle we are saying that so the left-hand side is essentially a local from finite subsets of the curve and the right-hand side is global it really has to be kind of an automorphic object, it's a bungee. So something needs to be done. So it doesn't come for free So let me say what the key geometric ingredient is we deduce this gadget called a finder's money but it's the wrong version I don't have time now because we have more details in question time so it's the modulated space of a bundle equipped with a rational trivialization and there is this map that forgets the rational trivialization and it also projects the mod space away from which sets we have the rational trivialization So the key geometric result is that the homology so this map is homology is a homology equivalence So this map is locally a product so it takes every bundle that admits a rational trivialization what it says is that for a given bundle the space of rational trivialization is homologically trivial that is to say that the space of rational functions from a curve 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 but actually calculate this homology and prove that it's trivial what's the whole homological contractable is it contractable? it is it's actually space of maps from a point to G is not contractable not at all but rational maps is not point to G but rational maps includes constants maps but it's okay so I can tell you the reason I mean there are a lot of comments on how that comes about and this is specific to dimension 1 so none of that would be true it was either lower dimension as you say point or the surface it's really kind of the thing so what is the definition of the Rangresvanyan over a base S? it's bundles on S cross X I'll say that in a second so maybe I have to officially finish because it's 11.05 answer this and everything else shall we do that? thank you okay so we will take few questions so maybe we start as usual Tokyo and then Beijing and then Paris okay so Tokyo, do you have some questions? thank you thank you thank you thank you yeah it's similar to nobody has a question in Tokyo sorry Beijing, do you have some questions? I have a question just go ahead go ahead so it's a question so it says a paper by Franco and go where they propose a geometric version of the traceformer and Artyapod's form occurs at the spectral side of the traceformer so do you see there's a possible connection between this proof of Tamagawa number and the proposal of geometric traceformer? good question I don't know, I haven't studied that paper it's an official detail I should do it, just thanks for the suggestion yeah because in the number few cases Tamagawa number can take off course as proof as a consequence of the traceformer consideration so it seems that there should be a connection maybe in the formula few cases the traceformer as they say is interpreted as a consequence of Artyapod's fixed form of the broken data that's a traceformer okay I'll go and look other questions from the gene can you say a word on the proof of the left-hand traceformer for Banji? yeah so how do I deal with these infinities, I mean right, the convergence yeah so it's a reduction theory so let's deal with SL2 for example so how does Banji for SL2 look like? well, it looks as follows you can take an open piece which are bundles which are not very unstable yeah yeah, you take bundles which are not very unstable well in general Banji is stratified by the degree of instability there are several stable bundles and then for each integer m there are bundles that look as follows so they are extensions of where L has degree-specific m m is a positive integer so this looks very very simple it's a stack of extensions of this like this and you explicitly see and then you sum up over all the terms m and you get a geometric series so on each stratum Banji looks as a finite union of geometric series and that's how you control this so in other words you prove for stacks of for algebraic stacks of finite type and in this case this guy it's not a finite type but it's again a finite union of things that look something like finite type time something that looks like a geometric series I'm going to say geometric series I mean it's 1 over point over a n this is a finite translation group but it's how important you can really get to that series you may depressify each piece of a n yeah these as you vary the degree other questions from Eugene? no more questions so questions inters so concerning this one 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 total topology they are rationally trivial on the total space there is some result who? Dreamfield Simpson and now the rationally trivial in this context is trivial outside the scheme which is finite over the base so in this case so when the group is simply connected it's much much stronger than that it says that if you take you take any non-empty divisor yes on the curve on the curve then after and it's how they change of your scheme of parameters it becomes trivial for simply connected sorry for semi-simple for semi-simple this is true for semi-simple bundles for constant groups you mean 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 a simply connected case and again outside the given yes you specify a section where a total base change of the scheme of parameters the thing trivialized outside the given section so now concerning the definition of the Tasmanian run yes can you say what it is? so first of all I said that it maps to here into here so I'd better if I want to s point to this I have to specify point of here and point of here so from s to the Tasmanian run first of all it's a pg it's a g-bundle on s-convex it's a point s point of bun g also it's a finite subset in home and now I specify my crucial Tasmanian side pattern I specify a trivialization p0g is a trivial bundle on x-convex without when I remove a certain divisor which divisor I take the union of the graphs so I run through this finite set I for each element I take the map from s to x I take this graph and throw it out okay so this roughly explains this and then there is another very question but still concerning this integral of those px upper shriek of c star bg so we work there with kind of tensor product of thing which ever you need but so I want to understand roughly why it is like what conserves stated about the product of plus 1 plus alpha i is 1 plus finite product it's a great question so no literally when you take the finite set you get exactly that if x is a finite set yes the way this integral is defined remember integral is comodable compact support of what we call factor f that is the comodology of the guy which is so I wonder if we should maybe take another question another question then should I answer one more question so again the integral is by definition the comodology of the compact support on the 1 so if x is a finite set you literally get what Max Tim 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 well in this case is this joint union of this 1 2 and 1 2 and what's factor b so you get b1 2 b1 tends to be 2 and comodable compact support is just direct sum ok so you get the ok and then how this is the product of b1 times 1 plus b2 how this is the product so you're asking why do I call this this quotation mark so then you you want to get something which is the numerical product formula which was actually a product ok so maybe I should maybe we should terminate now except if you can say just we'll take that pardon yeah it is ok so maybe we take another question there is no other questions ok then we would say goodbye since it should be now 7 p.m. in Tokyo exactly we need to go ok so goodbye and see you in one month bye bye bye