 So G would be a split reductive group over a local non-alchemy influence. And so O would be the ring of integers. So as usual H, the Hecker algebra is an algebra of function, compact support, which is invariant on the left, on the right, by the action of G o on G of f. So this is known to be a commuter algebra. And it is a SATA k isomorphism. I can identify it with the algebra of the dual groups. So G hat is a complex dual reductive group. So if pi is an irreducible undamified representation, that means pi has a vector, non-zero vector fixed under the action of G o. Then we know that H acts by convolution product on this vector space, which will be one-dimensionals. And this isomorphism, SATA k isomorphism. And so this pi will correspond to some sigma, some b-simple conjugacy class in G hat. And the SATA k isomorphism has been met in such a way that for every phi in H, the trace on pi of phi. So the trace makes sense because phi as a Hecker function will project the infinite dimension space on this one-dimension vector space, take the trace. So this is going to be a phi check of sigma. So now the commander break function, which can be evaluated on the sigma, on the conjugacy class. OK. So now you can try to do something a little bit different. So instead of looking at a regular function in G hat, you can look at some interesting rational functions. So let's sigma irreducible algebraic representation of the dual groups. Then one can, following Langlands, one can define the L functions, L factors. So it depends on the complex parameter s, rho, and pi. So we're really about pi, but we have auxiliary parameters s and rho. It's determinant of 1 minus sigma, rho sigma, and q to the minus s minus 1. But q is the cardinal of the residue field of the local fields. So this is, let me just denote this by the revision name. This become a function from sigma map to psi tinda s of sigma. And now psi tinda s can be regarded as a rational function on G hat, which is equivalent, which is invalid in the other action, and depending on one parameter s. So one may ask the question of whether one can construct the function. So nap psi belong to a localization of the algebra on the right. So one can ask for this psi s. One can ask, is there any psi s on the localization left-hand side that would correspond to this function on the right-hand side, right? So for doing that, you can first expand in a single-minded way. So this way, there's some from 0 to infinity of trace of sin n of rho sigma q to the minus ns. And now let me call this thing psi of n, right? Now psi n is a regular function. Then you know it has inverse attachment forms. And so we can, of course, do the left psi s to be just infinite sum, at least formally, psi tinda n to be consistent. And this becomes psi n q to the minus s, right? And we don't know how to make sense of this yet, but this presumably will have sort of a transform, this psi tinda s, which has to do with the local error factors. OK, so it is not difficult to prove that when the real property of s is large enough, this set is going to be to convert under makes sense a bit for large s. But somehow, I would like to put oneself in a situation when it just makes sense, obviously, for any s. So in particular, in some looking for a situation where the function psi, even just sum psi n makes sense by itself, don't need to make any kind of limit. So this situation actually has been devised by Brother Mann and Kashdan. This is the reason. So let me define this Brother Mann and Kashdan setting. So we are going to put ourselves in the situation as follows. So g will be an extension of gm by less than a simple group. This is not important, but we just assume this to be simple. So that is looked like gn length. So instead of doing this, you can just look at this function and twist the representation by determinant, right? So what we're looking for is the following formula. Trace of psi s on pi is the same as the trace of psi on pi tensor with determinant to the s. So for this to be true, you want to have the do on psi in this exact sequence. And actually, we want that when you ask this condition, that if you look at the rho of gn of v rho, so the Brother Mann-Kashdan condition is if you rush the rho to the central gm, it becomes the identical map from gm central in g hat to the gm central in gm v. So the condition is rho rush to gm is identity on gm. So I had to make a sign of that, but that is easy to understand what it means. So of course, if this is to be the case, then when you put c man, c man rho, then and that rushes to gm, then you get the elevation to the n's power of gm, right? And with this situation, with this condition, it implies that the psi n, psi n whose so they can transform is the trace on the n's symmetric power is supported by support of psi n is included in the matrix g in g of f, such that the valuation of determinant of g is equal to n. So obviously, we have this one support. So then we have no trouble to make the sum of them. Then it makes sense as functioned on g of f. So which can be obviously left and right invariant by the maximum compact, but which is not compact supported. OK. So that will be the main topic we try to understand this function psi geometrically. And also, you understand attempt to use the trace formula so that the main motor is going to put this function as test function of the trace formula so that I can go back to that later. So first of all, I just come back to this. I mean, historically, this function is looked very familiar because when g is g and d and rho is a standard representation of the douon groups, then psi is the characteristic function of g and d f cap with the matrix of d of o. So it's basically just the characteristic function of the integral matrix. And this is the single case where the function is simple. There has been another case that has been calculated classically, had to do with GSP and maybe rho with some standard representation. That was calculated by Satake. And that had been reported in the book of McDonald's. In the McDonald's book, he's calculated two examples. But I understand customer and told me that there is some mistake in the calculation anymore. I mean, had to put some factors. In this case, in the book of McDonald's, he said, cut this from something. But that is wrong. Had to multiply by another factor. OK. OK. Now, the first point I'd like to mention is, in general, you would like to understand what would play the role of this space of matrix to construct this space. So for each rho, a space that could see a matrix, that's how you see it. There is a monoid. So there is a monoid associated with rho. So this would be constructed in the Wienberg theory of monoid. But here, I should mention that the theory of algebraic monoid, it was not started by Wienberg. It was started by people like Rhen and Putsch. And so it can be generalized quite fastly in this theory of spherical varieties. I think that had to do with L function, ultimately. But I think we do not understand the whole thing yet. But let's look at this case. And the thing I would use is the theorem due to Wienberg. OK. So in this theory, somehow it's convenient to not to fix g, but to fix g prime and let g vary. So we fix something simple and for simply connected g prime. And you look at and consider all couples of g and m. So g is an extension of the torus by the fix g prime. And m is an equivalent embedding. It's a normal of phi equivalent embedding of g. And that means the action on the left, on the right of g on itself, can be extended to m. And under the assumption, this actually equivalent to m is an algebraic monoid. So the left and right action can be extended to multiplication on m. What is t? t is a torus, some torus. Oh, here is split. I think if the group, if it starts up, is not split, maybe you had to look at non-split torus. But there should not be any problem. g is not a society split. No, here g is split. I assume g is split. But this can be met. g prime is split. Here it is split, but I think that can be a non-split situation. So there is something technical but important is what Wienberg called the abelianization. So we look at this map, g, m into A. A is, so let the k be some base fin. So the regular function on m, which is inverted on the left and on the right by the semi-simple group, where this map. And the monoid is going to be very flat if this abelianization is flat with a geometric, which reduce geometric fiber. And we restrict ourselves into this setting to assume you only look at this monoid for which abelianization is very flat. And in this case, Wienberg proved there is universal. So we think that in g and dk, the universal should be the matrix. That's not true. It's more complicated. For g and 2, it is a matrix. But in general, it's more complicated. It's related to the wonderful compactification of the semi-simple group first. And I think the important point is when I recall the construction of Wienberg or the construction of universal monoid, it would be clear that it had to do with the Hecker and Jepras. So construction. Sorry? You said k is a fin. k is a fin. It can be f. k is some base fin. Base fin, when you define your group. So when you say flat, it's flat or what? No, this map is flat. This map? It's map from m to a. So actually, a is a toric varietis. When you a have action of the d, and it can be proved as a fine toric varietis. So some of the monoid, the combinatorics of this gadget is kind of mixed between the combinatorics of root system of g prime and combinatorics of toric varietis. There's some kind of tricky combinatorics. OK, so let me construct one of these monoid and some of them, the universal one. So let t prime be the maximum torus of g prime. And z prime is the center of g prime. And it can put g plus to be g prime times t prime mod z prime acting diagonally. And then we have this eta sequence to z plus to t plus to 0. Where t plus is now t prime mod z prime. And this become the torus of the adjoint groups. And it is gm to the r. And this map is given by the simple root. So you fix Boren and you have simple roots that identify the maximum torus of adjoint group with the product of gm. And so of course this, we have alpha i x 10 to g plus. Now you have the rho i. So let omega 1 omega r be the fundamental weight of z prime. And then we have this fundamental representation of z prime going to gm vi of highest weight omega i. And you want to extend this rho i over g plus. And I mean, I have to try. And the only one way to do it is to write this formula. It looked complicated, maybe because I didn't choose the right normalization. But when you try, this is the only formula you can write. So that is factor i to the center. So let go of this alpha plus and this rho plus. And then you have this alpha plus rho plus from z plus to the product of gm and product of gm vi, which can obviously embed it into some of these obvious monoid matrices. And so gm plus is defined to be the closure. This is an embedding. And this is a closure of g plus. I mean, in small characteristics, it may happen that m plus is not normal, in which case you want to normalize it. But this is not a very important issue. And so what we will prove is this is universal in the sense that for every flat monoid can be obtained by base change from g plus and m plus. That means you have, for example, just look at m. So we have m plus going to a plus. So this is acted on by t plus. And if you want to construct another monoid, you just look at some t, some character, some homo from t from 2 to t plus. That can be extended in a more fit, pictorial variety from a to a plus, then it base change. And that is the recipe to construct any monoid. OK. So now I want to take one minute to explain why this has anything to do with the Hecker algebra and Carton decomposition. It clearly has nothing to do, because if you want that, for example, if you want g to be in k p lambda k, maybe, I mean, it's easier to write conditions for if you want to look at the union of for on lambda prime less than lambda by the even domain order. You want to write this condition. And obviously, you have to use all these fundamental ways, representation to write down. Like the k of g and n had to write all the minus. And now, if you want to write precisely, what you really want is that rho plus of, you just put this to the t. Pi lambda ng is actually in the m plus of o. So this condition become a condition that something become integral point. So the monoid teaches basically that what is how it be used. The condition that complicated some double coset can be expressed, I mean, geometrically, as integral point. So that has great to do with Hecker and Jebreuse. Now, let's come back to the L function. So L function is basically to do with the fact that we want this to be g a and this to be g a. So just look at the more. I mean, in L function, there is this s. So it correspond to some gm. And so we want to put gm here. And so we want to put a1 here. And just to see that gm, this has to be nitrogen of co-character of t plus. And if I extend to a1, it just means the dominant. So this is some lambda dominant co-character. And it's quite obviously which one you want to choose. You want to choose this one to be the highest weight of rho. So when you start with a repetitive dual group, it has the highest weight. So you just plug it in this Bindberg construction. You get a monoid. And that is the monoid we want. And this actually is this monoid. So let's call this m rho. So then it become m rho. And this has been constructed in paper by Barbara Bergman and Clashdown. So by what I just explained a little bit before, we see that the function psi is sum of psi m. So the support of psi is very basic by this equivalence. If you take the support of psi, it's exactly the highest. At least it can contain. I mean, model is equals. It's g of f cap is m rho of rho. So the support of this function, each of the integral in point of sum model. This is a Cartesian, by construction. OK. Now, when you have this monoid, you can ask, well, we have the support. What is the function? So that is very complicated. I mean, numerically, of course, it is quite something, very complicated. Because I have to do symmetric power. And then I have to do composimetric power. So this is kind of a complicated business. Well, I mean, naively, when you have complicated functions, you want to be just a power shift. So I put some kind of conductor that let l plus of m rho to be the jet space of m rho. So it's defined by the fact that point on this is the point is valued in the formal series on m. And one can ask if you check the IC of this on l plus m. This can be defined. But I mean, there are some paper literatures by Cashdown and Greenberg, right? There's two papers, the paper by Greenberg and Cashdown. And there's also paper by Drenfeldt over the different proof of the same thing. Yes, so it proved that this jet space, I mean, locally, is a product of some different dimensional time with formal distance. So at least it's locally. So whether you're. Formally locally, so it's. Formal locally. Yes, it's true that every formal neighborhood is a product of something kind of dimensional and an infinite dimensional. But at least at the end of the standard, it doesn't prove that this actually exists, right? Well, at least on the level of stocks, I'm not sure that the stock is OK. Yes, the function can define it. And actually, this can check sure things also known. OK, OK, at least OK. So maybe this is known, but I see the function, the trace of forgiveness is his side. OK, so at least, I mean, I don't really understand this local space. But I'm. What is I see? Yes, what is I see? Well, it's not clear to me yet, but I explain you some something that they had to have I see now. It's I see of this. I see of this M L plus M. Are you happy, Takeshi? OK, so now let's look at some global more on this. OK, so now let's see the smooth projective curve of a finite field. And so now let E and E prime be a principle in ribandons on C. So what I can call what is amorphism? So phi from just notational M, each E prime M, can be nothing but a section, a global section of M, and twist on the left by E and twist on the right by E prime. Look at the thing of a vector bandon, it's actually the map, a linear map, vector bandon. So if you are given yourself a monoid, you can define a lot of amorphisms. Z bandon, not E prime, Z. Z bandons, yeah. M have two arcs on the left and the right and twist it. And you can compose it also because it's a monoid. So you can compose. And also we should look at M, look at phi such that generically phi lands in G. That means generically it's an isomorphism of bandons. Then what I mean is obvious global analog of this loop space is the space of a pair of E and phi, where E is ribandon on C and phi is a map from E0 to E. So it's generically an isomorphism. Sorry? E0 is the trivial bandon. E0 is the trivial bandon, yes. And these have an amorphism to biobilization. This has obvious map to the same thing for E, EGM, and for A. So what we get here is the space of a pair of, so this map to the space of a pair of L and of L and maybe anpha, where L is a ribandon. And anpha is a global section of L over the curve, yes. So this is basically just union not zero of CMN, CM effective divisors on the curve. And so let me call this CN. And so X also be the component to the union of XN. It's actually a map to CN. I mean, this is very standard, I think, in the case of G-contrude, GN, LEN, and phi, the trivial representation. It is on very first paper, Geometric LEN. See, this is the space it considers. And so with this, it's really not difficult to prove that the IC of this space is a global space. So if D belong to CN, so D is a sum of DI, CI, the CI our point in DI is a multiplicity. And then you want to restrict the IC of X to the fibers over D. So I can call this map to a system name, AP. So AP minus 1 of D is to be a CMDI of some perverse shift a lambda. So a lambda is a perverse shift corresponding to representations. And that would be corresponding to DI 1 to 1. And then when you collect, it gets symmetric powers. So this map, to this fact, I mean, the proof can be proven by the same proof in slow-mo paper. So give this function. But somehow, I mean, maybe it's totally stupid. But I think the interesting point is even if you cannot define, I mean, I don't know how to define the IC on the jet locally. This looks like a local. That means you can obviously map X to, X can be mapped locally to M of O, V, mode GOV. And it's stuck if you like to. So they let go of this C. And so you see that the trace of Frobenius on the IC of X is the pullback of the product function of psi v locally. So that is kind of, at least for me, an indication that there should be the IC as mixed in. And this psi v local, this map is not very smooth. But it's coyotes. Yeah, but if this formula is smooth, if you consider the divisor, which is concentrated at one point, in the neighborhood of that divisor, this map is formula smooth. I see. Which is easy to prove. And so this is actually. So this is why I said that this conjecture is known, because this global statement is really easy. And that kind of implies that local conjecture immediately, because it's easy to prove this map is formula smooth. It's not formula smooth as a map from all effects to this thing. But in the neighborhood of devices which are concentrated at one point, it is formula smooth. OK. OK, so well, so this function seems to have a nice geometric interpretations. So there is what you really want to do with these functions. OK. So let me do it. So there is an endoscopy proposal of a plan lens. So let now an automorphic representation pi to be product of pi v in a discrete representation is the automorphic space. So here you can have the two choice. Either you can fix the centron character, or fix an edan. I think it's going to fix an edan somewhere. So just to make this space have finite volumes. So assume this representation and let me find everywhere. Then you can write that L of s rho pi is a trace of product of psi v on pi tensor determinant to the s. So it's very tempting to put the trace, to look at the trace of product of all the psi v on the whole L2 of this with some. And that would be a kind of sum of L function. It is convergent when it can be proven. It can be coming for an s. Well, this is s. I'm not going to mean that anything is proven. But as far as the spectrum side, you understood the right-hand side by the spectrum side by look of lab bit and finish. I understand that it can be proven that it converts up to the vertical line of the treatment representation, the first form of the treatment representation. So it converts. So for example, if you just sum over all cospital representations, this will be convergent? Probably, but I don't know. I expected this. On the geometric side, it is in mysterious. But we do expect that have a geometric side. Actually, by the same kind of very similar construction, you can explain it a little bit later. You have nice and modular space that one can work with. So here, what I want to do with this? So using a trace formula, you're not working with just one function. You would like to have a whole space of test function to put the trace formula. So I think we would like to have a space of Schwarz space, depending on role, on every place. And this Schwarz space have a very specific element that is the function psi v. So that would replace the case, the usual case, when you have this space of function with compact support and smooth on g of fv. And so this kind of role equals 0. And then you have the characteristic function of k. And for every role, I think this, at least I find the standard of Bradford-Mann and Kashlan problem, there should be this Schwarz space depending on role. With this basic function, sorry, so this is role equal to 0. This is role equal to 0. No, the space is, as this is some kind of Schwarz space that's depending on role. At role, what do you mean by role equal to 0? If role is representation, if role is a representation, what does it mean? Oh, the trivial representation. Thank you. And so with this basic function, that makes sense to make the S-Roll not a little bit trivial. Because there was a condition that there was this multiplexer group that you wanted to have there. Well, I mean, I just think that it I think it just should have some kind of trace formula that apply depending on role. So here I think this kind of diverging approaches. There's approach by Barbara-Mann-Kashlan, and later, I think, pursued by Luang LaFoq, who would like to prove the functional equation of this AOS-Rope by using the trace formula on geometric size, trace formula on this side. And so there need to be the geometric size of some function in this place, phi v, where phi is equal to psi almost on place. It is a system of some L function. And one would like to go to the L1 minus S-Roll of p-dual. And here you would like to go to the space of S for the contraband derivative of V-check. So there should be some things that are not fully transformed and between these four space. And the equality of geometric size would look like a Poisson summation. So I think Luang has even candidate for this space, if I understand correctly. So that is one proposition. So there's another proposition. What is the test product? So you have a test product of Schwarz space. So what is this test product? Sorry? Can you say again? You wrote the test product of Schwarz space. So what is this test product? So the test of Poisson, we call it had this function, the basic function. Next one. So the test of what of V? Yes, so product over V. And phi v is equal to psi v for almost on V. So I mean, obviously, one does a full return from the psi v to the d-dual psi to make it consistent. OK, so now, beyond those copies, I mean, in Leng's approach, he's not, originally, he was not looking at the functional equation. He was more interested in looking at the bones that appears in this right-hand side. And we hope that the proving factoriality is more or less equivalent to partitioned automorphic representation into different backs. Back corresponds to subgroup of the d-dual group that receive the Leng's parameters. And which back you want to put your representation depend on all of the bones you get in LS row. So control the bone that appears on the right-hand side would tell you what to put in different backs. So that is, I mean, naively, what would I to do? And as always, you want to try to use the geometric side with this complicated function to prove whatever you want to prove on the spectral side. But of course, it's rather intimidating because, I mean, for the American convergence. I mean, you cannot really know how to define this space to start with at the moment. So we can say some kind of long way to go to this. But at least I think that it's reasonable, proper to list. Because at least it seems like you have an isometric interpretation of the left-hand side. So now let me explain something that I've been trying to do for a couple of years without much success. OK, so now we have this infinity point in C. And we also choose a map from gm to the center of g so that you can twist g-bandon by divisor. So every e-bandon, you can do e of infinity, just for a vector bandon. You can twist it by infinity. OK. So now the quotient of ga mod gf infinity to the z, go, now corresponding to the category of g-bandon. But then we decide that v to bandon, we just give an isomorphism to this two-bandon. So just think about the vector bandon. You can decide that v, you give an isomorphism with v twisted with o-infinities. Now, when you look at the rheumatic cell orbital integral, so it's quite easy to guess that you want to look at the isomorphism of vector bandon or of bandons. So we look at our space of m. It is a space of, so let e to be an object in this. Then the endomorphism of v, the m endomorphisms. The m is a map from v to v and infinity. And n goes to infinity. E is the same as v. Yeah, it's the same as v. Then you look at the space m is a space of e and phi. And phi is just endomorphism of e. So this is to be, you can write this to the limit which now becomes e and phi with now phi map e to e and infinity. So this looks exactly like a hitching vibration. And you can define it, so let me call this mn. So this is very similar to the hitching vibration. This is the m map. OK, so we want to use, so hitching vibration of this f to mn to an, an using some kind of adjoint quotient on the monoid and built from the adjoint quotient. mn is this modular space of endomorphism of e bandons. But what is the index n? n is because I have to add the. Oh, it's the same n as over there. It's this n, adjoint quotient of a monoid. And you want to start to study the geometry of trace formula by using this hitching vibration. Now the first point, the first issue you can meet is if you put I, I mean, you have to do with IC because you cannot work with other shift. You don't do it very, very well. But then what would be the IC of this mn, right? You just put the IC of this mn, what would be the right function? Because it's not the same as the space of xn I defy earlier. But this question seems to be resolved by, by buti. I guess. OK, so, so now once we get this, well, well, and then, of course, there's very much speculation because even the hitching vibration, when you, when you start to work with the important cone or hyperbolic space, it's a lot of problem. It's not done yet. I mean, the hyperbolic part has been studied extensively and I mean solved by work of Le Mans and Chaudoir. But for an important cone, it's fine. No, it's not, it's not known. But you can at least expect that it's the best of the case. This is completely determined by the comology of the genetic fiber. This kind of monorhobe cell. So the fiber of this is a bit of a variety. And we have to say that the answer is very complicated. But it can be some kind of a recipe to write this answer, form the representation of monodromy on the H1 of the billion varieties. And so we would think that then the comology of Mn, that is somehow what we are interested in Finne, that is trace formula, is somehow a kind of, should be a comology of pred group. I mean, in many cases, it looks like this kind of answer, that is what happens. And of course, hopefully, beyond endoscopic underture of Le Mans, can be a chance to lead the question of stability in the comogy of pred group. That had been studied a lot by topologists. So I mean, the bridge is not built, but I mean, there seem to be one. Thank you. Thank you. Maybe we start with some questions from Tokyo. So you are a great group. What is this? OK. So the map Mn to An, so how can I look at Takeshi? Should look at this or this? OK. So the map from Mn to An is basically a destination of billion varieties. So the basic thing you can think about is the family of true covering a projective line. So that is determined basically by the ramification divisor. Then something bad can happen when the divisor become hemant multiplicity. But before that, the divisor become multiplicity free. You have a nice family of covering, so you have a nice family of billion varieties in Jacobians. And you have representation of the fundamental group on the H1 of that a billion varieties. So that is a very well-known representation of the bread group, that is the bread group on the H1 of a billion varieties. And in that case, I think I want to improve some stability's result and go to infinity's. If optimistic enough, there should be some similar in more general situation. Thank you. And then I'll ask a question from Tokyo. Thank you very much. Yes, then I'd like to ask about the beyond the endoscopy. So you consider the sum, geometric side. So if we want to compare two groups, then with which kind of relations expected for your dysfunctional side? Can you ask, I don't understand what kind of something I couldn't hear well. What kind of relations? Relations, relations. So in the endoscopic case, we expect the magic of several of them. Yes. In general case, which kind of relation is expected for? I mean, it's not really a belief that there is a relation between operandegons itself. I mean, in the endoscopic case, I mean, this guy is a very wonderful case when you have a map between this shock space, and then they just have some operandegons, but in general, it seems to be rather a transform, the transforms that if you know all operandegons, then you know other operandegons. Just one not determined by one, but by the whole family, because they're all transforms. But in this beyond the endoscopy approach, as I understand, at least it's not about operandegons, it's about the whole sum of operandegons, and more precisely, some kind of asymptotic behaviors. So it's not really reasonable to try to have any precise answer about this homology, but how when n go to infinities, then there should be comparison between two different groups. This is as precise as I can give an answer for a time being. No, thank you. That was from Tokyo. OK, so we go to Beijing. So any questions from Beijing? Let's give her. I think there's no questions from Beijing. OK, the question is from there. So, Halgasson, do you expect a closed form answer for this homology? So for example, if you really can see in the case of three-bill row, then basically you're talking about homology of bun g for which there is a simple answer. So do you believe in kind of similar type answer in general? No, it's L-function. You cannot say nothing about L-function, right? The point is much easier, things. So the point is kind of asymptotic of the coefficient. You get the sum of the mn in the simplest case, like you just truncated directly the series. You took the sum of a and qn up to some big n. And how it grows saves you something about the points. And that's what we want to compare different groups. And that should be much easier to say about precise value of this. Halgasson, do you know anything about what I do? Would you like to do anything about it? I mean, you're only talking about, so to say, the spherical case here. So do you know anything about smaller complex groups? I mean, if you can see there, for example, you mentioned the Schwarz space, which should, and so for example, we can prove that it has the same characterization by means of the row sheaves, if you can see that's the what-word part. Conjecturally, it should be the same for the Schwarz space, although there I'm not sure how to formulate this precisely. But at least in the lower case, you can do similar analysis. But here, you were only talking about computing the trace of one particular function, which is invariant. But in principle, you had to put other functions. Yeah, so my question is whether you have any conjecture about what, for example, if you put some invariant functions. I mean, here you formulate some kind of league conjecture about what this can mean. I mean, at least about as far as conserving the pawns, it's not really important to the local information. I mean, I cannot formulate anything. But I think for the pawn question, it's not local, it's not very relevant. Other questions? Let's go to the speaker again.