Thank you very much for coming. So this is the first lecture of our series, and today we'll speak more about solving these functional equations I introduced before, and today we'll see how to solve homogeneous and also inhomogeneous equations. And so there is an update, so Maxim Konsevich has found the right name for the property of the representation, which we discussed yesterday, so it's the right name for it is quasi- Important for non-rubber off-past? Yes. And so today what we are going to do, we're going to, so let's talk today, we'll speak about solving functional equations, so both homogeneous and inhomogeneous, and so we will start with the easier case, so what if we do have a vanishing left-hand side, and so here as I would like to introduce a piece of notation, so let J be an ideal in our group algebra, and then for each right ideal like this, and for each even integer, we can consider the space of holomorphic functions of moderate growth on the upper half-plane, which are annihilated by this, by all elements of this ideal, so this would be the set of all functions in our class curly p, such that the slash operator of 8k with an element of ideal G applied to this function will always give us zero, and so our first goal for today would be to compute this particular space of functions, so the space of functions which are annihilated by the ideal I, which we have introduced in our previous lecture, so I, it was the ideal generated by these two elements, and so when I mean, what has been compute for us, so what we will do, we will express all the functions in this space in terms of classical modular forms, and in order to do so, we will need a brief overview of a classical modular form, so I will need to introduce a few classical objects, so and so we'll start with a standard definition of what is a modular form, and so now let gamma be a discrete subgroup of the group SL2R, and we suppose that the quotient of upper half-plane by this group has finite volume, then we say that a homomorphic modular form of 8k, it is a function in our class curly p, so it means that it has to be homomorphic at the upper half-plane, and also has nice growth at the boundary of the upper half-plane, and the condition of modularity, it means that the action of slash operator of weight k and any element of the group gamma, it leaves our function f invariant, and this would be true for all gamma in the group capital gamma, and so now we see that in principle this definition is very, somehow it's a partial case of the definition which I have just given on the upper board, only in this case for this ideal which annihilates f, it will be the set of all elements in the group ring such that the sum of their coefficients is zero, and so those are the standard notation for the space of modular forms, it will denote by mk of gamma, this would be the space of homomorphic modular forms of weight k for gamma, and so also another, and the theory of modular forms tells us that such a space is always finite dimensional, and so we also, it will be useful for us to consider a bigger space, this time an infinite dimensional space, this will be the space of weakly homomorphic modular forms, again of weight k and k for gamma, and so these functions they also have to satisfy this transformation for property of a modular form, and functions in this space they also requested to be homomorphic at the upper half-plane, but now what we do, we relax our condition for the growth of behavior of f at the boundary, so now we allow our functions to have poles at the boundary, so they have to be homomorphic on the upper half-plane and they can have poles of finite order so they are allowed to be meromorphic near carps, and so now what will be useful for us to describe the space which we want to describe, the space of modular forms for our ADLI, so for this we'll consider these spaces of homomorphic modular forms for two particular groups, for the group SL2z and for its principal subgroup of level 2 for the group gamma2, and so for the modular subgroup we have a special functions called Eisenstein series, and Eisenstein series are defined as such a double sum, so we are summing over all pairs of integers except for pair 0, 0, and so now if, so here, so here we fix k to be an even integer which is bigger or equal to 4, and for such k we see that this sum will converge absolutely, and now because it converges absolutely it's easy to see that this property will be true for every element of SL2z, and so the Eisenstein series they also have very nice Fourier expansion around the, so the only casp of SL2z is cusp at infinity and around this casp the Eisenstein series have a nice Fourier expansion given in terms of these simple arithmetic functions, so here this number bk it is a Bernoulli number, and sigma k of n it's a sum of divisors of n taking to the power k minus 1, and choose the patient for variable z not tau, I don't know because later I will have also z4, yes because we will have this kernels which will be functions of two variables, but again it also could be a random event, and so now and we consider the, we can consider the following sum, the sum which will be actually an algebra, the sum of modular forms of weight k for all weights from 0 to infinity, and now because the matrix minus 1 belongs to SL2z we know that here if k is odd then this space automatically vanishes, so the nice fact about this algebra it is that it is actually it is generated by two Eisenstein series, Eisenstein series of weight 4 and Eisenstein series of weight 6, so it would be a free polynomial ring, and so another just piece of notation which we will use in the future, we will use the standard delta function, so it is a modular form of weight 12, and the special property of this modular form is that it does not vanish at the upper half plane, and in particular it has a product expansion like this, so here q it's e to the 2 pi i z, so it is called, it's the Ramanujan kasp form of weight 12, and so kasp form means that this function vanishes at all the kasp, and so it's the property which we will use in the future, so for a function which does not vanish on the upper half plane, and also it vanishes at the only kasp of SL2z, so it's also it decays exponentially as imaginary part of tau goes to infinity, and so from here for example we see that the inverse of delta function, it will be now a weakly holomorphic modular form, nice blackboard, and so now also nice fact about the ring of weakly holomorphic modular forms, it is that it will be generated by usual modular forms, and this one more function, oh yes, thank you, and so now when it comes to SL2z, there is another class of objects which we will need, and then it will be not free for them to launch? Yeah, of course because this one it was actually expressed in terms of these ones, and so another class of objects which we will need are quasi-modular forms, and so here I will not give a formal definition of what a quasi-modular form is, instead I will give you one example, and so this example the Eisenstein series of weight 2, so now if we take definition of Eisenstein series and formally take k equals to 2, so the problem which we have with our first definition of Eisenstein series is that the double summit will not converge absolutely anymore, so we'll need to take to do some regularization, on the other hand the second definition by even in terms of Fourier expansion it still makes sense, and so we'll use this one as a definition, and so another way to define E2 is to say that it is a logarithmic derivative of Ramanujan's delta function after some slight renormalization, and so the Eisenstein series of weight 2, they are still periodic, for example we can see this from the Fourier expansion, but now when we want to check its modularity property with respect to the element s of a cell 2z, we will get transformation property like this, so this is almost a modularity, and there is a small correction term, and so now E2 is called a quasi-modular form, and so now I still don't give definition of quasi-modular forms in terms of their transformation law, but what we can do we can just consider all polynomials in E2, E4 and E6 and say that this set it will be the set of quasi-modular forms, and so now the next group which we would like to consider would be the group gamma 2, so I would like to consider modular forms for gamma 2, and these are also very classical objects, known already for a very long time, and so one easy way to describe modular forms for the group gamma 2 is to consider Jacobi data functions, so the Jacobi data functions are defined in the following way, these are three functions obtained by summation over integer lattice, so there are functions like this, and from Poisson summation formula it's not difficult to see that they are modular forms of weight one half, so the modular forms of half integral weight we did not define it in this series of lectures and in the story I'm going to tell you they will not play a role, so instead of considering Jacobi data functions we will consider their force power, so the functions we will actually work with will be the modular forms of even weight, so we just take these functions to the force power, and so now these four functions they are not linearly independent, there is one relation between them which is called the Jacobi relation, namely it's that u equals to v plus w, and now if we want to consider all modular forms of even weight for gamma 2, then this will be again a polynomial ring generated by two of these functions, and again if we want to consider the weakly holomorphic functions, so the only thing which we have to do, we have to take this algebra and add this one more element, add delta inverse, and so now the last maybe object I will need, it would be the modular lambda functions, so lambda it's defined as a ratio of capital V divided by capital U and its function from the upper half plane into a complex numbers and it takes all values except for two, it never turns zero and it never takes value one, and so lambda it is a helped module for the modular curve x of 2 which is upper half plane modular gamma 2, and so helped module it means that it is a generator of its function field, and so if you want to imagine how this modular curve looks like, so what we can do, we can consider a fundamental domain of the action of gamma 2 on the upper half plane and standard way to choose this domain it's like this, let's call it domain D, so this would be a fundamental domain of the action of gamma 2 on the upper half plane, and so from here we see that this curve it has three cusps so to say, so one of them is cusp 1 and minus 1 because 1 and minus 1 they are identified by the action of gamma 2, another cusp is 0 and the first cusp is infinity, and so modular lambda function we can extend it to the cusp, the cusp by taking the following values, so at infinity lambda function vanishes, close to 0 it tends to 1 and close to cusp 1 it tends to infinity, and now the function which we will actually need for solving our functional equations, this will be a logarithm of lambda function, but now of course so small subtleties here is that lambda function has zero at this cusp and we will go around the cusp, we will pick up a different branch of a logarithm, so we have to choose if you want to have this to consider such a logarithm as a function of the upper half plane we have to choose the the branch which we like, and so we will do it in a following way, so we will define the curly l of z it would be defined as an integral of a logarithmic derivative of lambda, and so this will be actually the logarithm of z, if z is on the imaginary axis that will coincide with the principal branch of the logarithm, and so here we will also let lambda with subindex s it will be just lambda acted by slash operator s of weight 0, so this would be by the transformation properties of lambda function we know that it equals to 1 minus lambda function, and curly l with subindex s it would be an integral like this, and this again would coincide with a logarithm of, and so now we are we have all the necessary objects to solve the homogeneous equation we were interested in from from the beginning, so let's just briefly I'll briefly remind you so that we are searching for functions annihilated by this particular ideal, and so it will be somehow easier for us to to consider the following two ideals, it plus and e minus the one of them would be generated by these two elements, and then other one by these two elements, and so after we define ideals like this, so what we can say tell about them, so first that the co-dimension of each of these ideals will be three, remember that the co-dimension of ideal i was six, but here it's half of six both times, and also these two ideals they are related to our initial ideal i in the following case, so i epsilon where epsilon is either plus or minus one, it will be exactly those elements of the group algebra error such that the s minus epsilon times r belongs to our initial ideal i, and so now we see that ideal i it's actually an intersection of i plus and i minus, and also we have that actually i plus plus i minus it will give us all the ring r, and also we we can see that the quotient space r modulo i it will be isomorphic, so we'll have an isomorphism like this, and it will be a natural isomorphism which follows from this condition, and so now which advantages this gives us, now we have our space of functions which are annihilated by i, we can split this space into two subspaces, the corollary of the lemma would be that this space it will be a direct sum of the following two spaces, and this will be useful for us because the module of the functions which live in this space and functions which live in this space they will have quite different nature, so the functions in this space we will construct them from quasi modular forms, and the functions in this space they will be constructed from the logarithm of modular lambda function, and so now what we can do we can give a description of these two spaces, so first we have a proposition that let k be an even integer, then we know that the space of functions which are annihilated by ideal i plus in the class curly p it will be the following space, and so here the functions the function phi 2 it will be the Eisenstein series of weight 2 squared, phi 0 and it will be the tau times the Eisenstein series of weight 2 minus this correction term, and phi minus 2 it is just, and so in particular from this description we can compute the dimension of this space, so from this description it's clear that the space is finite dimensional and also its dimension can be easily expressed in terms of weight k, so in particular we know that the dimension of the space it will be zero for negative k, so it would be a space like this, and the similar statement we have for the ideal i minus position, so again let k be an even integer, then we know that this space can be written in the following way, so we again have three elements which are not modular forms for sl2z, and we can express any other element in terms of them, so this time we label them xi 4, xi 2 and xi 0, and so here our functions are the following, so xi 4 is expressed in terms of the logarithm of modular lambda function and modular form for gamma 2 of weight 4, xi 2 has similar representation, only here we use a modular form for gamma 2 of weight 2 and actually xi 0 it is just one, so in particular we see that the space of usual modular forms of weight k it is contained in this space, and so now we can also compute the dimension, so we see that the dimension of the space it will be a number like this, and so now when we have found all this, so maybe now one important point, so now when we have found the space of the solutions of homogeneous functional equation, now we also know about all the relations of between values of radial Schwarz function and square roots of even integers it's derivative and the same information for on the Fourier side, so now we not only know that it's a finite dimensional space of relations, but we also have very explicit description of it, and so now the next step would be to approach solving the inhomogeneous functional equation. Can you just ask a philosophical question? Okay for me of course this is all magic and so on, but I would be curious to know did you run here some algorithm which is just like standard in your field or was it like foregone confusion that you'll be able to solve this problem with this ideal and once you knew the description of radial? So here it's something which also surprised me a lot why I could solve the functional equations, but yeah so this actually for me this was somehow a method of trial and errors and so like what somehow, so what was the, so remember in the last lecture we had the six-dimensional representation and so what somehow was a good sign for me when I realized so I initially did this step and broke the six-dimensional presentation into two three-dimensional ones because it's easier to analyze and as one part of one piece it was isomorphic to this symmetric square symmetric power representation and this was somehow a good side and I think it's more or less known that this symmetric power representation somehow is friends with quasi-modular forms that one can be expressed in terms of another and for the second representation so I was very surprised when I could actually solve it because it was somehow with only three-dimensional representation so it's possible just to play with it and to see what it does and it was because it does have this like nice triangular structure and some like big portions of this if you factor these representations through gamma two it also simplifies a lot yeah so but still I don't have like a very good explanation why do why do things have to be so nice because we could have got some ugly representation for which we would not have nice modular objects but yeah but for priority it is clear this dimensional solutions will be fine yeah so I think that this is yes yeah this this is clear but I don't know but here I think maybe because the dimension of representations is small and maybe in small dimensions like really bad things cannot happen so yes maybe this condition that the representation does not grow polynomially probably it is somehow I don't know maybe it is not just a coincidence maybe there is a deep theoretical reason for that and maybe after that it's already it has to be something not not too bad no but because if you know our priorities it's fine dimension kind of kind of approach without thinking try to solve from computer numerically and some to guess formally yeah yeah but somehow we did not really try to get it we did not get it numerically I first got it really somehow theoretically but yeah so I think in dimension I mean so we know that space would be finer dimensional and in principle we could get something like some hyper geometric functions or but yeah and I think maybe maybe like the fact that it's only dimension three it also played role that we have nice objects which we are familiar with and not something completely strange and unusual and so yes now when we have solved the homogeneous equation now we could approach an inhomogeneous one and so so here actually equations like this the solving equations like this it's not a completely new problem and I think it actually appears also in different areas for example it's very similar to solving a Riemann-Hilbert problem or even one can think of it as it's all somehow it reminds a little bit of a classical Dirichlet problem when we know that something happens for example in a boundary and we want to extend it inside of some domain of in a complex plane but so maybe before we come to the method of solution I will the first thing I would like to show to do is to show that to solve this in homogeneous functional equation it will be sufficient for us to solve it only a neighborhood of this of the domain D and then we could extend it to the whole plane by using functional equation so so here's again maybe I'll this time I will define what D is or precisely so D is this open set so D we see that D it lies on this strip this strip where the real value of tau is between minus one and one or real value of z as written here and also we exclude the two circles of radius one half about what plus one half and minus one half and so now we have the following proposition so let K be again an even integer this time a non-negative even integer and suppose that h1 and h2 are continuous functions so then the following holds so the first property which we call it as the analytic continuation so here we suppose that functions h1 and h2 are halomorphic and let O be in a subset of the upper half plane be an open neighborhood of the fundamental domain of the closure of the fundamental domain D and so now suppose that we have a function f from complex numbers be a halomorphic function which satisfies the following transformation property properties so the first one is that if we slash it with t minus one squared we will get h1 and if we slash it with s times t minus one squared we will get h2 and so because now f is a function which is defined only in a and not in the whole upper half plane but only in this open set O so now we require that these two properties they hold whenever the left-hand side of these two equations whenever it makes sense so both sides are defined so it means that equation one is defined for one it is is defined on the O intersected with the translation of O by minus one and intersected with the translation of O by minus two so it means that this should be satisfied for all arguments which belong to this set yes yes yes if always looks something like this maybe yes it has to be for example on this on this side for two it has to be true on the on the intersection of these three sets which should be in in this area and then f extends to a halomorphic function on the upper half plane and this halomorphic function will satisfy both equations one and two so you don't need any compatibility conditions between h1 and h2 no no that we we don't need them and so this comes from the fact that the ideal i it was just really generated by these two elements and the second important property it is that because we wanted to find not only a halomorphic solution but we also wanted to assure that our function f has moderate growth and so the second property it would be the propagation of the moderate growth bound so now we suppose that f is a continuous function and it satisfies conditions one and two and also we suppose that f grows moderately on d so it means that f of tau is bounded by alpha times only s tau is not in the all upper half plane but only in the fundamental domain d then we claim that f will also grow moderately on on the whole upper half plane okay so that's okay there should be just right there should be a condition okay so I think we also know that okay yes we also have to assure that h1 and h2 also right so we we write it here so suppose that h1 and h2 they also have moderate growth right so this is an important condition yes so that and if we know that h1 and h2 have moderate growth and also if f satisfies this moderate growth conditions but only inside of domain d then we can make sure that the same holds also on the all upper half plane here you use that f exists for some small domain but is it possible to say for each h1 h2 such f exists okay so as we will come to it that actually it will exist for almost like all h1 and h2 so maybe yeah so maybe there is some maybe there is a condition of the cusps which somehow because we designed all this to be satisfied for our particular for functions but yes but somehow modular this condition and the cusps I think it will exist for all h1 and h2 and so maybe we make a small break so now we can okay continue so so on this lecture I will not give a proof of these propositions because I think the proof which we have now in our paper it's quite lengthy and technical maybe I have to think a little bit how to make it less lengthy and technical and hope I could present it to you in a second part of this series of lectures so so now maybe for today we will just take this proposition as a formulation and then I will also today I will explain how to how to get explicit formula for the solution of such a functional equation and so here we will what we will do we will assume that our solution f has a nice integral representation so for example so this is a homogeneous one and so some of what so particularly were here here we had any h1 and h2 but there is only one pair we are actually interested in and so from phenomena probably maybe we will concentrate on this one which we really want to solve however the method is more or less general so so there we will make the following ansatz that our function can be presented in the following form and here we have some new object k which is a meromorphic kernel of two variables in the upper half plane and I will present it to you in a moment so we want to search for our function in the following form and so some advantages from this presentation is that for example from this formula we can read that our function f it has to coincide with exponential function at all square roots of even integers because here we have this sine squared which will be responsible for double zeros and also this particular shape it will be useful for us when we will prove the positivity of f on the imaginary axis and so now what is this kernel k so the kernel k is the following function function with the following properties and so theorem which we prove is that for dimensions d which are equal to 8 and 24 there exist a unique meromorphic functions k which is k of t on the product of two upper half planes and this function satisfies the following properties so first property is that we want all the poles of this function to be concentrated only at those points where two arguments of this function are sl2z equivalent so we say that for a fixed z in the upper half plane the poles of the k of tau z in tau they are all first they are also only simple zeros and the second condition is that they are all contained in sl2z orbit of the point z been married to two points maybe maybe maybe not somehow we don't use any kind of product expansion yeah it was a nice formula for the difference of of jc actually like something in which you multiply get the collomorphic function yes but but somehow we don't we don't use the product formula in this it's not that we did not try it's just yeah somehow at some point really hope that it might be useful for example for proving positivity but it did not we could not make it work like this and so now what we also want this function k we wanted to satisfy the homogeneous functional equation so so here we take slash operator with respect to variable tau and z is considered as a fixed parameter and third for for z which is at the upper half plane and any element of group algebra so now we have the following properties so now if we could take the we apply slash operator to invariable tau to our kernel k and then this residue it will be some number but now we know if we apply an element or if a would be an element in our ideal we know that this function would be zero so the residue also will be zero so what we we can see from here is that the residue it will actually depend only on the class of a modular right ideal i so and of course clearly by linearity it's also a linear map so here phi somehow we see that it has to be some linear map and now what we do we just specify a concrete linear map we are interested in so it is the following linear map which is defined by the following conditions so now how can we define this map we need to choose some nice set of nice basis for this vector space and define the value of phi on the elements of the basis so it is defined in the following way so now maybe just a small remark that now the elements the following elements of group algebra they will spend will spend this quotient space and so now what we can do we can just define and also yes yes yes so somehow yeah so now because it has pulse so not what we will do we will for example define our f by this integral in for example in the on this domain d so it's actually in domain d because because of this lot so many of this residues they vanish so actually we can do define our function by this integral in the domain d and then what we will do we will extend it analytically from the domain d to the whole upper half plane and so now what we also need the first condition in our theorem it will assure that k also has because we are all the time concerned about the growth of our functions so we also need to put some growth conditions on k near the boundary of upper half planes so what we we require that the following functions or the functions so if we take our for example k8 of tau z then we multiply it by a difference of two j invariance to remove all the poles at the diagonal and then also to make sure that our function does not grow too fast at the boundary we multiply it by the cusp form delta and so this will be enough to do it for dimension eight and in dimension 24 actually our function it will grow far faster we have to allow faster growth of k at the cusps so we have to multiply by higher power of delta of z so we want these functions to be in the class curly p with respect to both variables with respect to variable tau and with respect to variable z and also so thus now this will be guarantee for us that our interpolation formula starts at the right place so we also want our function to vanish at the cusp as tau goes to infinity it has to have order of vanishing which is for the kernel which corresponds to dimension 24 actually the order of vanishing has to be higher and this should be true as the imaginary part of tau goes to infinity and so now let me briefly explain you the idea of the proof how we prove the existence of such a kernel k so we prove it just by constructing this function explicitly so what we do is just give an explicit formula for the kernel but now how do we find this formula from such a conditions so here the idea which we use is that all all these four conditions they will also imply that our kernel k it has nice model modulator properties not only with respect to variable tau but it also has certain nice modular properties in variable z and then we can find the for first the right modularity conditions in variable z then what we do we also describe these conditions in terms of a certain ideal and then we find the space of functions annihilated by this ideal and this already gives us a finite dimensional space especially if we use this the condition four which I've written up there then we have a finite dimensional space where we can search for our kernel k and then for example condition three of the theorem it gives us several linear constraints for for the function and of course also together with condition four it also gives linear constraints but so let me explain you what kind of ideal this will be and so here is another small piece of notation I actually have a question about the condition three what is the meaning of this condition where did it yes okay so the constraint that you yeah so unfortunately I really wanted to have like a very clean and nice way of like seeing which which which linear functional we wanted to use but essentially like our choice of phi it is it was predefined by our ansatz by our particular shape of f so you take this ansatz and you plug it to the functional equation yes in homogeneous yes but then then by shifting contours I will see which which residues I will need at each of the points and but also I kind of tried to find a nice or at least nicer algebraic description of this process but till now I could not yeah so we do it essentially by by hand just by and so so let's consider the following the star operator which acts from our group algebra to the group algebra so it is an involution and this or anti-involution and it works like this so if we have element of a group algebra here a i are some coefficients and gamma i are elements of the group x numbers and gamma i elements in psltc so just take an inverse of each group element and so now an important fact which we will use the following simple lemma it tells us that suppose that for we have two elements m and n in sl2z then we have the following condition if we want to compute a residue at tau equal to nz of k which is slashed but now in variable z with weight 2 minus d over 2 so the weight which is complementary to d over 2 then what will we get we will get the following entities and this in this case of course star is just we can replace it by minus one and so now what we do now we define a new ideal i tilde which will be defined in terms of our ideal i it will consist of all elements m in the group algebra such that this functional equation of m star it is zero for all n in r and then i tilde will also be a right ideal and so now so now what we want to do we want to see what happens if we apply an element of this ideal i tilde to to the function to the kernel k so suppose that m isn't this ideal then what we know we know that for all elements n in sl2z we know that the residue of point tau equals to nz of this function kernel k slash twiz so we know that this residue it will be the same as a residue of at point tau equals to z of this time k slash twiz tau and by our condition three we know that this will be the same as phi at point n m star and so but now here we use the definition of ideal i tilde so we know that n m star at here there is yes we know that this has to be zero by the definition of this happens by the definition of i tilde and this happens by condition three of the theorem okay so only and this was by the lemma above so and so now what we see that again from from the way we have defined our kernel actually from condition one we know that the only poles of such function they can be at those points where tau and z are sl2z equivalent to each other but from here we see that all the residues at all these points are zero and because we had only first order poles from here we see that this function purely k which is slashed by slash operator of weight two minus d over two now it has to be holomorphic in tau variable and also now we see what kind of gross conditions we had for for k so now from the gross conditions now we will see that this function it would belong to the class to the class p and from the condition two we know that it also annihilated by the ideal i so okay for it d over two so again as a function of tau and except for this we know that we also have from the last condition of the theorem we have that we know that we have certain vanishing at the cusps and now because we have an explicit description of this space we can see that the vanishing conditions at the cusps they can be satisfied if and only if this function vanishes and so here from here we see that the kernel this kernel function slashed by an element of ideal i tilde has to be zero for all and so now from here what we can do now after we know this we can compute the space of functions which are annihilated by this ideal i tilde and it turns out that i tilde it's also a nice ideal for example the co-dimension of i tilde is also six and their so to say the representation which is attached to i tilde it will be the representation which is dual to the one which was associated with i and so this description of this space so now we can so now we can describe and so now we use the fact that now we know that this such a product z into this power which depended on d now such a function it has to belong to the to this space so the other product of this to find a dimensional spaces okay so here only here we have to add 12 and here we have to add it also becomes positive so now we have to search for our kernel essentially in a space like this and this will give us a explicit formula for for the kernel kd as you have just mentioned one of the products to the model curve yes yes and so and also this kernel it can be somehow this kernel after multiplication with the difference of j invariance and after multiplication with delta functions so it becomes like a polynomial which can be expressed in terms of a modular functions v and u which we have introduced before also in terms of eisenstein series and in terms of logarithms of lambda function and so somehow so it is a finite expression but it is not very small so maybe at some point I would like to discuss the computational aspects of our proof and maybe then I will show you how this formula looks like and unfortunately it's not as small as we hoped for and so now maybe the last statement for today so maybe I'll outline a bit our now after we have constructed kernels so how do we construct the function f so maybe just to be more precise about the undoes which I've made there so again for tau in the upper half plane and x in rd what we do we will set that f of tau okay so and here's but here it's important think it's also a bit yeah we have like it is one additional technical complication it is that our kernels curly k they actually they don't decay as in the second as imaginary part of z goes to infinity so in for for d equals 8 they will grow linearly in z and for each lattice we will even have a pole so we'll have a pole in z as z goes to infinity and so the integral which I gave which I've written there it will converge only if the absolute value of x is big enough and so we also assume that the absolute value of x it has to be bigger than this like square root of two and zero minus two so it means that in dimension eight we have only a slight problem around zero but in dimension 24 it means that we have to avoid the ball of radius two around zero and so now for the such tau and x we set our f to be the square root of z thank you and so here's the statement which I have to prove is that now the function tau goes to f of now it will extend holomorphically so it extends to a holomorphic function on an open subset of the upper half plane which contains containing the fundamental domain the closure of the fundamental domain d and this extension and satisfies the functional equation we desired for so whenever the left hand sides are defined and so now this is a situation which is similar to the to the one we discussed in a proposition previously so again what we have to do first we have we show that we can extend now f only to a small neighborhood of a fundamental domain d and then we could use the and also for this extension we will have to prove the all the estimates we are interested in and then using the proposition which I formulated before then we will be able to come to see that the extension to the all upper half plane also exists and that it also satisfies the necessary growth bounds so thank you very much yes so the result of so similarity here but


Sam here with latent sports cards. We are ripping 20 22 a panini elements football 12 box full case break number eight Random division style here is a look again at your random results As always you can find these on the website under the results tab guys We are down to six spots on Flawless six to go where you can almost taste it almost taste it But before we do that we got to do this in our next break So let's uh, let's get ripping on this good luck everyone I'm gonna update our night sheet. We're down to six to go on flawless. Heck. Yeah. Thank you everybody Let's do it. Good luck here Pull something big man. The one-on-ones out of here. It's probably a nuclear the super short prints. Those are really sick Got some NT collegiate on after this man.  All right You got the AFC North, right racer You're chilling man on that one. I would say when I see slabs And zap a time. Oh, yeah, always in zap a time empty sick. I haven't seen that yet We've got Matt Stafford that is out of 199 and we've got Lamar Jackson 88 out of 95 Very nice as well FC North. Ha, thank you ice man. That thing was so crazy I was like shaking after it dude the entire store came like running to the window to see what it was four out of ten Tyree kill Going to the AFC East nice Jersey there Tyree kill patch that's sick We've got 110 out of 125 that is Zemir White Going to the AFC West There you go, Nathan G picking up a Zemir White. This is so cool going to the AFC South Chris M heavy metal 35 of 49 Darius Leonard That is so cool man. Heavy metal auto Darius Shaquille Leonard six to go on flawless That is so awesome. Thank you guys so much Cannot wait to rip that. Let me tweet this out really quick. It's too. Good luck here everyone See what we got Chase young To 199 and we've got Russell Wilson on the tungsten Broncos country. Let's ride.  Are you here? Nathan? Are you here? Nathan tell me you're here, dude Nathan G Nathan G. Where are you at? Dude? This is nasty Nathan Nathan Oh, where is he? That's an H my head Nathan Dean's here. All right. Thank God Dean's here. We're good. We're good There he is How about going to the AFC East? one of one Damian Harris Jones Taekwondo Dude, that is so nasty. Oh My god, dude Dude, that is so cool one of one Nathan let's go buddy 101 Mac Damian Harris Taekwondo and those kind of nasty dude kind of nasty You let's go. Congrats man swoosh. I don't go right into that right into the mag We've got how about a sauce gardener? Going back to the AFC East Cleaning up. What's up Papa Ken coming in here now? What's up everybody? Also Caleb is live over on late in gaming. I don't know if we went back into his room. Oh Just ended never mind. He's not live Friday met a zoo release he will be live steel signatures Cordell Stewart 74 out of 79 Going to the AFC North. There you go Richard. Nice one there All right, we're only two boxes into this guys. Let's go Here we go. Good luck Broncos country. Let's ride a mericium 55 out of 95 and Zenon Matt Stafford 38 out of 54 We've got NFC East Devonta Smith that is one out of 199 Nice one there. AFC there NFC East. We've got going to Kyle P That's NFC West Tyler Higby 78 out of 125 Geez Nathan is killing it right now The G is cleaned up better for sure. How about a breeze hall rookie steel signatures? 59 out of 99 go into the AFC East and grads on that one Nathan I'll see some fire guys. There's more fire to be had in here for sure as cases going kind of hard AJ Brown to 199 and Drake London 41 out of 54 going to the NFC South God's AFC North back to Richard G That is Deontay Johnson 25 out of 54. Oh, that's so sick, dude Wow, dude, this case is nuts NFC East go into Josh each How about a Brian Dawkins on the atomic signatures? 6 out of 7 on the weapon X Dude, that's crazy. Let's go dude Yep, six out of seven on the atomic auto that's nasty dude look at the background on that thing That's so crazy Then we've got sky more RPS rookie steel signatures 62 out of 79 going to the AFC West That's Nathan G again We're literally a third of the way through this case dude That was Josh each good hit there Josh. Oh, yeah, boy. Oh, yeah Don our resident Eagles fan. We have two Eagles fans in the building right now. Actually Let's go nuclear breeze hall dude Nathan This is insane, dude Super super super short print Yeah, dude, that is nuts Not numbered on the breeze hall dude the G the Nathan G Literally is cleaning up man. That's unbelievable, dude. I've hit I've opened like eight cases of this and hit two We've got a mercy on that's going to the AFC North Kyle Hamilton there you go Richard 47 out of 95 We've got AFC South Brandon cooks I don't know what that's numbered to me one second here. That is three out of 70 and then we've got 33 out of 75 Proton RPA Christian Watson NFC North there you go drew on the board and NFC South this time going to Derek you Jonathan Vilma 41 out of 99 Those new peers are so sick Alex hit one of those actually Get the Garrett Wilson those until I pulled to the two jet rookies out of those basically Just keeps going dude Got Eden Hutchinson to 199 and a mercy on Deandre Hopkins 29 out of 95 there you go Kyle pick up a D hop Got 16 out of 47 Derek Henry on the two-colored critical mass patch He is got Reverend Jordan 11 out of 50 going to the AFC South Chris M and Chris M also picking up a Rookie steel signatures Trey von Walker 31 out of 79 dude just keeps going His case is nuts and we just got to the halfway point literally just got to the halfway point of this This is insane dude insane. What a case this is it bond needs that car to 199 And cobalt that'll have been cooked 12 out of 27 going to the NFC North. Oh, baby Let's go on flawless breaking at the end of the night. We are almost there almost there We've got David Montgomery going to the NFC North That is seven out of 27 nice one there We've got 42 out of 99 that is gorgeous going to that is Kyle with the NFC West Tyrion Davis price very nice and how we long 20 out of 35 AFC West going back to who else but Nathan Caleb actually said that this scared me a little bit. He said if you guys don't watch Meta zoo release on Friday over on Leighton gaming He's gonna take two empty Pringles cans and super glue them and then when I'm not looking He's gonna stick them both on my hand, so I can't break anymore very weird very like specific I don't know why he's gonna do that, but he said he's gonna do that So I would watch met a zoo on Friday Not clearly you're not gonna want to miss it if that's what's at stake He would be full very foolish to miss it for sure All right, here we go AFC South we got Alec Pierce to 199 and Aiden Hutchinson 77 out of 99 a mercy him got nice AFC North. How about a Lamar Jackson? 190 out of 199 on the fusion there you go Richard Right, isn't that a very strange thing to threaten to do to someone? very strange Got going to the NFC East and the Kobe Dean with the Eagles That is 154 out of 199 seriously jealous of that nuclear As a cool that is a cool hit man, and then we've got Seven of ten NFC East again Carson Strong There you go NFC East Josh H picking up some steam here So they're killed have a go on We've got say Quan Barkley to 199 and Matt Stafford 67 out of 74 on the tungsten there you go again Kyle nice one there got Justin Herbert That is going to the AFC West 115 out of 199. We've got one Thornhill that is 29 out of 125 Again, AFC West and go into the AFC South 100 out of 199 John Mechie the third on the RPS rookie steel signatures doubly appealing true Rory true All right next one up here That's what's nice to is like we have so many so many sports that just keep going and going Right, that's what I'm saying. Yeah, they're always televised to for the most part We've got Austin Echler AFC West and Zenon is 50 15 out of 54 Jonathan Taylor AFC South We've got Zach Wilson 71 out of 199 go into the AFC East Got Tony Pollard nine out of 25 AFs or NFC East Who there we go There we go Richard Richard coming alive now We've got rookie steel signatures George Pickens 15 out of 27 There you go Richard. Congrats, buddy. That is a nice guard Love the silver ink on these things. All right guys two to go Two to go Robinson, Jr. 37 out of 74 Deebo Samuel 39 out of 54 going to the Niners yeah, I just want to check because this this redemption on the back. I have not seen yet I just want to see where he goes To this one right here. Yeah, that Pickens book. Let's nasty Nathan. I saw that I saw a picture of that things crazy Did you hit that here I'm assuming if that was you all right perfect we've got That is 34 out of 75 Jonathan Taylor Matt Ryan Alec Pierce AFC South We've got I Got too excited dude. I just got too excited Rory rookie neon signs Romeo Dobbs go into the NFC north with the Packers and Dude this one is so sick How about a heavy metal auto redemption Charles Haley?  Oh Must be a different one you pulled it on your own Gotcha, gotcha gotcha Charles Haley. There you go Kyle really cool Hall of Famer to add NFC West he is a checklist as a Niner so that'll be going to the NFC West Last box of the bridge. Let's see something crazy to end us. Why not dude? Why not in Pierce? AFC South to 199 Cortland Sutton 15 out of 95 AFC West We've got Chris Olave going to the NFC South Followed by a Damon Clark for the Cowboys 69 out of 99. That's going to be NFC East on the Damon Clark pretty nice way to end it here NFC East. We've got Josh H Johan Dotson 98 out of 99 on the rookie steel signatures. Thanks for the sub by the way Billy Appreciate you buddy That'll do it for the breakouts. Let's do a recap man. This is so awesome. This is so much fun to rip Chris Olave Sates Quad Colts Zach Wilson Justin Herbert AFC West Lamar AFC North David Montgomery NFC North Derek Henry AFC South Brandon Cooks AFC South John T. Johnson AFC North Devonta Smith NFC East Tyree Kill AFC East We've got Johan Dotson NFC East Damon Clark NFC East Charles Haley NFC West. That's gonna be so cool Romeo Dobbs NFC North George Pickens AFC North Tony Pollard NFC East John Mechie AFC South Juan Thornhill going to the AFC West Carson Strong NFC East Nicoby Dean NFC East Howie Long AFC West Tyrion Davis Price NFC West Trayvon Walker AFC South Revan Jordan AFC South Christian Watson NFC North Jonathan Vilma NFC South Skymore AFC West Brian Dawkins NFC East. That's so sick Breece Hall AFC East Tyler Higby NFC West Cordell Stewart AFC North sauce gardener AFC East Darius Leonard AFC South and Zemir White AFC West our Maggard. We had a nuclear super super short print Breece Hall going to the AFC East and Hit of the break one of one Damian Harris Mack Jones Taekwun Thorin Go into the AFC East with Nathan That will do it for the break guys. Thanks again for the fill next up. We've got NT collegiate


Welcome to All Hands Update, I'm Petty Officer Jen Blake. The MQ-4C Triton unmanned aircraft system landed at Naval Air Station Pentux River, Maryland September 18th after completing its first cross-country flight. This is an exciting day. I personally witnessed history being made again. Here at Triton coming from the west coast to the east coast, it's longest flight ever. Not just geographically, but time in the air. It was just goosebumps. It was a phenomenal sight to see and to experience with the team. Triton completed 15 flight tests showing its ability to operate at various speeds and altitudes. So this is really the future of naval aviation. I'm a P3 pilot by trade, I've flown my entire career and Triton is part of the P3 replacement. And so when I think about that and think about where this is going to take naval aviation is just incredible, the technology, the capability, it just redefines how we do things.