 Thank you very much for coming to the second lecture of this series and today we'll continue what we have started yesterday and today we'll speak about relation between the Fourier interpolation formula which I have introduced the previous time and certain functional equation which we have obtained for the generating function. So today we will not speak about energy optimization and universal optimality but rather we will concentrate on the Fourier interpolation and so here is the I'll just briefly remind the theorem which I've formulated on a previous lecture so it's a joint result with Henry Kohn, Abinav Kumar, Stephen Miller and Danilo Ratchenko and it says let d and n0 be either 8, 1 or 24, 2 so we've worked out all the details for dimensions 8 and 24 however this formula also works in more general settings and so what we have proven at least for these two dimensions that there exists an interpolation basis of Schwarz functions we number them an, bn, an tilde, bn tilde and they are radial Schwarz functions on rd and here the index n it starts from this number n0 which is either 1 or 2 and so these functions they satisfy the following property so if we now any radial Schwarz function any point x in the dimensional Euclidean space then we will have an equation like this we can reconstruct the value of this function at point x from its values at square roots of even integers for the function itself it's first derivative it's Fourier transform and the first derivative of its Fourier transform so we'll have a formula like this and we call it the interpolation formula so I think for this particular n0 then the formula is unique but for example in already in dimension 24 we could make some choices for example we could start from 1 and then there would be some ambiguity here so and so here's another result which is somehow closely related to this one which it tells that we cannot only interpolate existing functions but we can actually construct Schwarz functions with the values at these points which we like which somehow don't so the only condition is that they don't contradict the growth growth conditions which we impose on all Schwarz functions so for this we you know the following consider the following set so it will be the set of all fast decaying sequences so vector space of all sequences which we start from 1 such that the sequence decays faster than any power of the index and so now what we show that we can take any four elements in this space and construct Schwarz function which will have these values at these special points so so again let d and n0 be as above and we consider the following maps so one of them would be map we call it xi it will be the map which goes from the space of radial functions into the space of fast decaying sequences to the power four and so the function is defined in the following way so it takes function f and maps it into the collection of its values at the square roots of integers so for the function itself for its first derivative for its Fourier transform oh yes for dimension 24 it will start with two yeah but somehow still I which was just with this ambiguity somehow if we shift the numbering somehow the space will not change essentially so so here n goes from zero to infinity and so the second function which we can consider is this one we denoted by capital phi so it goes in other direction so it starts with sequences and maps them into schwarz functions and this time if you have a sequence like this alpha n beta n alpha n tilde beta n tilde and here n again runs from n0 to infinity then such a sequence would be mapped into a function like this which is given by the sum of elements in the interpolating basis who is exactly this coefficient so and so then what happens that the map xi is an isomorphism of vector spaces and if we take the composition of xi and phi it will be identity map and the same way if we compose phi with xi this would be also identity map and so also what we have done on a previous lecture so we have considered a generating series which was formed from our elements of the interpolating basis so one of them was a function we actually split it into two functions so one of them was the function which formed by functions a n and b n and the second one was by the a n tilde and b n tilde and so here tau was a variable at the upper half plane and x was a variable in the euclidean space and what we have shown last time is that now these two functions they satisfy the following functional equation if we believe in our interpolation theorem then from the interpolation theorem we could deduce the following functional equations so and the equation was like this so first f of tau and f tilde of tau they were related by the following transformation and then the both functions they are let's say linearly periodic it means that if we take a second difference of each of these functions with respect to tau here I omit variable x just for the speed of writing so then we would have equations like this and what we have seen last last lecture so we denote this by this would be like a functional equation and so what we have seen that interpolation formula it applies functional equation but now our how are you going to prove the interpolation formula so what we are going to do we are going to somehow show that the this implication it also works in different direction so what we will see today so we are going to show that if we start with a functional equation so if we if we are able to construct a function which satisfies the functional equation and the number of analytic conditions which I will specify a bit later then we can prove the interpolation formula and so now before we can do this I'll need to introduce several notations so first of what I think first thing I would like to do I would like to introduce a semi-norms on the space of schwarz functions so here so let f be a radial schwarz function and so we will write d of f for the we'll denote like this the let's say the radial radial derivative of fx so if our function is radial so it means that it depends only on the absolute value of x and this d of f it just it means just taking a derivative of this function f0 the same as you apply earlier vector field yes yes yes and so now we can consider the following semi-norm on the space of schwarz functions so it will depend on two indices chi and l and it's also would work for only for radial schwarz functions so it will be a supremum over all x in rd so the cladial norm of x to the power k times the absolute value of the radial derivative of f k and l are non-negative integers and so this set of semi-norms actually defines the topology on the schwarz space so so the sequence of functions converges if and only if to some limit only if it converges with respect to each of these semi-norms and so here's one important fact which is I think it's very well known but I could not find who was the first observed something like this so the complex gaussians so the functions of this which look like this for some tau complex number in the upper half plane they yes oh yes so they they span a dense subspace in the space of radial schwarz functions and here is one a bit more technical statement but I think we'll use it later in our series of lectures so now if we fix any positive i or y so the same the same is true it means that that if we consider only gaussians with imagine with for tau with imaginary parts equal exactly to y and so now you can maybe we'll guess where I'm going with this so now what we would like to do we would like to prove our interpolation formula only for complex gaussians and then to say that this the same would work for all other schwarz functions because of the of the density of this subspace in the space space of radial schwarz functions something in terms of functions of upper half plane I'm sorry it doesn't give also some kind of identification of s right a rd with some kind of holomorphic functions on upper half plane all right may yes I'll give one problem here is that of course like it's we have this space it's spans radial schwarz functions but it's not in this functions they're not linearly independent so there is like in some ambiguities in here so probably probably if you need one to choose a function on the upper half plane we need to make some choices in here yeah so from here I see how it works like in one direction from something in the upper half plane we could get a schwarz function but maybe it's not at the moment not so clear for me how to go with the other direction and so okay so now I will formulate our other long theorem which tells us exactly what I promised you so that functional equation does imply interpolation formula but except so we need to find the solution with but what somehow is a bit difficult what makes the theorem long is that we also need to specify some analytic conditions to make sure that our formula works not only formally but also analytically and so the theorem is like this so now suppose that that there exist smooth functions f and f tilde from the upper half plane cross euclidean space into the complex numbers and suppose that these functions they satisfy the following properties so first if we look at them as a functions of tau variable then they are holomorphic and tau the second statement would be that if we look at them as a function of a second variable of variable x then they have to be radial in x and third condition so we want some balance on the semi-norms of this two functions considered as a functions again of second variable and so now for all non-negative integers k and l so we have that so now we consider this the k l semi-norm of this function viewed as a function of x and tau is considered as a parameter here so the same has to be true also for f tilde no no no it's just no it's just some it's a positive some positive number delta sorry yes not maybe not a great notation and so we have the same estimate for f tilde and this would be for some positive real numbers alpha k l beta k l delta okay gamma delta k l and so so now what comes next is a condition which somehow assures that our formula it starts either from one or from two so now we want somehow in the special case when k and l are both equal to zero what we want we want actually a stronger we want to have a stronger estimate so and this has to be true for tau in the strip from minus one to one so here for all points x and the euclidean space and we also want this beta zero zero to be a positive number strictly positive number and so the this was condition four and the condition five finally it is that f and f tilde they have to satisfy the functional equation so which is now written on the upper blackboard and so whenever these conditions hold then we claim that then f and f tilde they have expansions of this of the form which is now given at the upper blackboard refer to them maybe i'll put so they will have expansions like this so let me put like a star here and sorry a value of zero which is one of two so yes yes yes so it will come a bit later yeah so for the value of zero we need somehow we need to stress on this this condition so they will have expansions of the form like star and star tilde as it's given on the upper blackboard only with the here if this condition four holds then what we can guarantee that n zero will be equal to one and for some radial functions n b n a n tilde and b n tilde and moreover for every which is a radial schwarz function the following interpolation formula will hold it would be an interpolation formula like this only starting indeed with one and so also what we will if we are able to construct a function f like this then we will know that for fixed k and l then the radial seminars of the all this for radial schwarz functions a n and b n we know that they will grow at most polynomially in n and so now the last condition which we have to write is the one which assures that n zero will start not from one but from the for example from two so we have that so now find last part of the formulation to be like a one and a one tilde would be equal to be one equals to be one tilde equals zero so this will happen if and only if so the functions f and f tilde so that they not only go to zero as a imaginary part of tau goes to infinity but also go to zero fast enough so they have to be small o of the e to the minus two pi imaginary part of tau as imaginary part of tau goes to infinity in the strip where the real value of tau is between minus one and one and with x fixed and so now this is the the function which we need to construct so we see that somehow functional equation plays a role also we see that our functions they have to satisfy somehow this nice growth conditions and so these growth conditions they have to work so to say uniformly when we when we change our variable x and so whenever we are able to to construct a function like that then we will also prove the interpolation formula and by computing the Fourier expansion of this function we will obtain explicit formulas for the interpolation basis so maybe you have some questions at this point so then the next step what I would like to do I would like to prove this theorem so to prove the equivalence which is so now we have all the necessary ingredients for this so now let's suppose that we have a function with all these properties so our first step would be to show that if we have a function which satisfies these properties then it will have the free expansion of the form which we are looking for and so this is a quite easy step so step one it is so prove that we have the expansions and start tilde and so now this will follow of course the expansions they will follow from the functional equation and so from the last two equations in our system from this from the linear periodicity of f and f tilde so what we do so now we consider the function like this suppose that f satisfies the functional equation then we can consider a function like this and we look at it as a function of of variable tau and so now we see that this function of course it is holomorphic also we see that it is one periodic in tau so it is in variant under taking tau into tau plus one and the last condition that it goes to zero as imaginary part of tau goes to infinity so this follows from our condition four and so now the thing which we look at very we do a very simple thing so we go from strips like this by taking tau into that it's e to the 2 pi i tau we can map such a strip into a punctured disk and so the periodicity condition for this function that tells us that our function can be written actually as a function of this expansion so it's it's a usual free analysis so from here we see that we have an expansion like this and now because our function is holomorphic and it has only one potential singularity here at the zero we know that it is removable and so that our function already on a disk it tends to zero at this point so we will go to an expansion like this and so now what we can do we of course can do the this is how we have obtained our functions bn of x and we obtain functions an in a similar way only for this we have to consider a slightly different function obtained from f so to get the coefficients a and we have to consider function like this and now the again this function again it is holomorphic it is periodic in tau and it also also goes to zero as imaginary part of tau goes to infinity and therefore we see that f of tau x it has the expansion which we expected it to have and the same works for f tilde and so analogously the same works for f tilde so this was a rather easy easy part so i'm sorry for maybe getting you bored here but so now the next thing what we have to check we have to check that the coefficients which we have constructed they are actually schwarz functions as the functions of x and so so here our second step would be to check that an bn an tilde and bn tilde are radial schwarz functions and to do this what what we are going to do we are going to give a so given an integral representation of our Fourier coefficient eighth coefficients a n and bn in terms of the function f itself and then this way we could read properties of Fourier coefficients from the properties of of the function so so here we write an an and bn in the following the following integrals so we fix some positive number y and so we integrate this function so we just use the usual formula for the computation of Fourier coefficients of a periodic function and the same we do for bn only we remember that we like bn are not just Fourier coefficients but they are normalized and here again we are integrating over some horizontal interval from i y to i y plus one and so now we see that from part three of the hypothesis of our theorem which is just here above from here we will see that the semi norms of an and bn they are all bounded so they are finite and this already tells us that the functions an and bn they are schwarz functions so but now what we also want to somehow to prove is that they are not only bounded but that they grow at most polynomially as an index n grows and so for this we take we use a usual trick for analyzing Fourier coefficients of functions of moderate growth of the upper half plane so here what we do we take this number y which was allowed to be anything but now we take it to be one over n where n is the index of the Fourier coefficients and so from here we will see that the semi norms they grow at most polynomially with respect to the index n for k and l fixed and so now of course if we consider an tilde and bn tilde everything works analogously and so from here we see that the consequences of this is that the interpolation formula it converges absolutely so sorry not the interpolation formula but so the right hand side of the interpolation formula it will converges converges absolutely for every function f which is a radial schwarz function and so and now we also see that this the right hand side of the interpolation formula it will define a continuous functional in the space of schwarz functions it defines a continuous linear functional on the space of radial schwarz functions and so now finally we are ready to prove the interpolation formula but now the proof of the interpolation formula it will actually follow from the density of Gaussians in the space in the space of schwarz functions so now let's fix some x0 in the dimensional Euclidean space and we consider the following linear functional which goes from the space of radial functions radial schwarz functions for example into r and it is given by the following this may be we did not define the lambda of f to be the following functions here we take the right hand side of the interpolation formula at point x0 and we subtract the left hand side and so now the interpolation point at the interpolation formula at point x0 it is equivalent to the fact that this linear functional lambda vanishes and so now because we know that this linear functional it is continuous so it will be sufficient for us to check vanishing only on the set of complex Gaussians that lambda is continuous and therefore we know that now it suffices to prove that lambda of each complex Gaussian for example for all tau in the upper half plane but now we see that that this condition it is just equivalent to to the functional equation and so this finishes the proof and so here one more remarks and now we see that the solutions of our functional equation they will give rise to interpolation formulas on the other hand what we could also consider we could consider the this the homogeneous version of this functional equation where on the right hand side we have only zeros where we don't have this complex Gaussian but we have only only zeros and then such functions they will correspond to relations between values of Schwarz functions on the square roots of even integers as derivative and the same information from the Fourier side so yes so lambda it's a functional on the Schwarz space so it's continuous linear functional on the Schwarz space and the topology on the Schwarz space is given by the by this set of semi-norms so that gives you zero yes i want to prove the function zero check on the dense subset yes yes yes so suppose that that h is a solution h and h tilde are our solutions of the homogeneous equation and this homogeneous equations probably I will write them on the next blackboard and so the equations would be like this so h of tau so then it's we also know that from this two last equations that h will have a Fourier expansion like this so and okay so here again somehow but we want them to be homogeneous solutions but we also want h to satisfy this gross conditions which are sure that n starts from one okay so let's do it like homogeneous was this a special piece of the function equation no you so i just like the first condition i replaced it so there i had some function on the left hand side but here i replaced it by by zero and you forget about dependence on max yes yes homogeneous equation yeah no no no it's okay i know maybe it's not a great terminology but i think again in many areas we do have it like we have like differential equations and whatever equations if we have something on the left hand side we say that it is a homogeneous sorry yeah so yeah so this is kind of a functional equation but cn plus i tau square root of 2n maybe some numbers dn so e to the 2 pi i and tau so we know that we will have Fourier expansions like this so here i i did not spell out the conditions which ensure that we have Fourier expansions starting from one not from zero but i hope you will excuse me for this so we'll have Fourier expansions like this and so now we have that for just for the as a consequence of the proof of the corollary of the proof of the previous theorem it will be now then that for any radial Schwarz function in a d-dimensional space we have the following relation we will know that sum from we know that such a sum always will be zero so in a some sense we have if you remember in the beginning of our lecture we have introduced the maps xi and phi and so this tells us uh this formula describes for us a subspace of uh so we have this map xi which went from the radial functions into the space of fast decaying sequences and now we see that the so to say orthogonal complement of this space it will be somehow isomorphic to the space of so space of solutions of this homogeneous functional equation but here again of course we have to consider such as solutions and we expect these functional solutions also to have this nice uh moderate growth property as a function of tau and then we can assure that the Fourier coefficients grow at most polynomially and then at least yeah and such a pairing it will always make sense it will be always convergent and so what you said before it means that this equation homogeneous question does not have solutions no it has it has solutions again depending on d so for example for if we restrict it like this like in dimension eight it will have only trivial solution but for example if we start from n equals one in dimension 24 it will have probably one non-trivial solution up to linearity and somehow in principle everything what we are saying here it would work in uh any dimension uh and so in a somehow usually well as dimension grows so what we will see later that the space of such solutions is always finite dimensional and actually we can we have a very good description of it very explicit and the dimension of space it grows with dimension of the space of solutions grows with dimension of our Euclidean space we are considering so should we make a small break maybe for five minutes now after we've shown that interpolation formula is equivalent to existence of a certain nice solution of a functional equation so now the equation which we have how to solve this functional equation so both first how homogeneous and then of course also inhomogeneous so so the next our next step which probably will take us sometime probably more than just second half of this lecture so there's probably something I will continue in May so how to construct and f tilde and so for this what will be useful for us to look at f and f tilde as a function of of a of tau variable in the upper half plane and to study the modularity properties of these two functions so so now let me introduce some notations so let's consider the following group so we consider the group ps l to z so it's a group of all uh two by two matrices with integer coefficients uh module all the subgroup which consists of identity and minus identity matrix and so now this group it acts on on the upper half plane by the linear fractional transformations and because it acts on a upper half plane it also acts on functions from the upper half plane and so there is a one particular action which is given by the slash operator so what we do we fix the function f from the upper half plane into complex numbers let's fix an integer k then and let's fix one particular element of of the group so and probably it probably suggests to be even integer so now we define the slash operator it acts on functions in the following way so we have a action like this and so this action it it is actually an action of a group so if we have two elements of in sl to z then the product of two elements acts on function in the same way as we first act by the first element and then act by the second element and so now because we want to consider this a bit more general functional equations so another object which we would like to introduce would be the group algebra of the group psl to z so so let r be the group algebra and now of course the action of slash operator it also extends to this group algebra by linearity so slash operator of weight k now it is extended can be extended to the group algebra by linearity yes so it is a right action and so now also as we have written it so the slash operator it acts on any functions without any restrictions on the upper half plane but we would like to consider one particularly nice class of functions so namely the functions which have a moderate growth so now we say that a function f on the upper half plane it has moderate growth if there exist positive constant alpha beta and gamma such that the absolute value of this function f at point tau is bounded by the following quantity so what we insist that our function it grows at most polynomially as tau tends to the real line or as tau goes to infinity and it's somehow this definition that actually agrees with the definition of moderate growth which is given in the theory of symmetric spaces so so this should be true for all tau in the upper half plane okay so at some point we will actually need some functions which are not continuous and have this growth so I maybe I'll just here I will just save this notation for all functions maybe or maybe also it's not very good analytic properties but for holomorphic functions we will have this class curly space curly p so this would be the space of all functions from the upper half plane into complex numbers such that f is holomorphic and has moderate growth and so now an important property it is that if we apply a slash operator to a function then if our initial function had moderate growth then the new function also will have moderate growth so if f has moderate growth this will imply that f slashed by gamma also has moderate growth and so now with this after we have introduced the slash operator so now what we can do we can rewrite our functional equations so so here again so rather standard notation it is that we consider two particular elements of sl2z one of them it will be a matrix like this t it is a matrix like this so these two elements they act particularly nicely on the upper half plane so s is an involution and it sends tau into minus one divided by tau and t it acts on the upper half plane just by translations by one and so now the group psl2z it is generated by s and t these two elements and so for between these two elements there are two relations only two the first they already said that s is an involution so s squared is one in psl2z and the product of s times t it's also an elliptic element but of order three and these are the only two relations between the generators and so now what we can do now we can rewrite the functional equations for f and f tilde in terms of a slash operator so the f tilde can be written as here this last one is always from the group algebra okay so first it's true for any element of the group but also for the element of yeah because of this thing it's of course this set of functions of moderate growth or class of functions of moderate growth it is a linear space if we add two functions with this growth we get a function which cannot grow too fast either so and so now let's write this in terms of the action of a group algebra this would be an equation like this so this first equation that again expresses the periodicity properties of f the second one translates periodicity properties of f tilde and the last is the functional equation which we have obtained from the interpolation formula for the complex quotients and so here for the notation so here I use this upper index tau because f it's a function of two variables and so this is just to stress that I apply slash operator with respect to the variable tau the first variable and so now what we can do somehow it's a bit difficult to work with two functions simultaneously one thing which we can do we can remove for example function f tilde from this functional equation and transform it into functional equation which will contain only f and so we do it in a following way so we see that from the last equation we can see that f tilde it's actually equals to e to to a function like this and so now we can rewrite the first two equations in terms of f and so from here we get that f the first equation does not change and the second equation can be written in the following form so now this would be the functional equation we want to solve for so we and now we want to solve this functional equation for f as a function in the class p and so again if we just require f to b and p we could have some extras maybe the solution might not be unique but as we will see later somehow the solution of such a homogeneous equation it will always contain only five it will be always finite dimensional but we also remember that we have this extra gross conditions on f close to the cusps and this will give us the uniqueness of the solution and so now we what we would like to do we would like to consider the following ideal inside of the group algebra so we consider the right ideal and so this will be ideal which is generated by these two elements of the group algebra so it is generated by and so so we see that in some sense among the properties of this ideal they will determine for us the properties of the solutions of the functional equation and it turns out that this ideal is quite nice maybe better than we could expect from a random ideal in this ring so it has the following properties which will be later important for us so first is that if we consider the quotient of the group algebra by this ideal and look at it just as a vector space then this vector space will have dimension vector space over complex numbers then it will have dimension six and so here like one can we can observe immediately that six is smaller than infinity which is a good news for us so this gives us some hope about so for solving the functional equation a second rather nice property is that so the ideal i it is actually freely generated by these two elements so it is i is a free r module generated by these two elements and so for example because it's a free module what it gives us it gives us that now if we know the action of these two elements on f then we actually will know the action of any element of the ideal on f at least theoretically can you remind for the sake of physicists what a free r module means okay so it means that somehow if we take any combination here like this element times some element of r plus this element times another element of r and we ask whether it equals to zero and it will be equal to zero if and only if these two coefficients which we applied to the generators they both have to be zero so it means that we don't have any hidden relations between these two generators and so the last property which is probably a bit difficult for me to state but yeah so then so the if we because we have a we have this quotient which is a vector space and now our group psl2z and also the group algebra they will act on the on the right on this finite dimensional space so it means that we will have some for example we will have a six dimensional representation of psl2z associated with this ideal and now it turns out that this representation is quite nice so the so the representation of psl2z of on the module it has so I'm not sure if it's right terminology or not but it has what I call a polynomial growth and so let me explain this to you in a moment so it means that if we for example if we have a representation on six dimensional space it will give us this after some right choice of basis and dealing with the fact that this is actually a right representation so but we can for example transpose it and get a representation of psl2z in gl6 and for example so let's also assume that we have a standard basis chosen here so for each element in psl2z we can associate a six by six matrix and also have to transpose to make it left representation and not right representation and so then the lemma which we proved is that there exist some constants so which don't depend on anything so absolute constants c and n which are positive real numbers such that so each entry of this matrix sigma of element abcd is bounded by so it's polynomially it grows at most polynomially depending on this uh entries abcd and d of a matrix in psl2z here again so here abd is any matrix in sl2z and this property it will be important for us when we will try to to prove that the solution of the functional equation which we have found it does belong to the class currently p it should be this property something standard in geometric group theory so it's certainly known in geometric group theory but i for example i do i could not find like whatever for example very good definition for this very good name for this property so i don't know maybe if you know what is like a more correct name for this property yeah but it certainly is very natural and property and it's quite useful so and also maybe i should stress that somehow it's not a trivial property so if we take some random representation of psl2z we don't hope for property like this for example it could happen that we map our matrix at t into some element t into some matrix which has an for example eigen value which is not of absolute value one and then of course we can compute representation of t to the power n and the entries of such matrix would grow polynomially yeah but also i just can have a question from the people get local system on more the space of electric curves yeah does it get a variation of what structure is it kind of geometric yeah so now we'll see so so this representation we get that it is quite nice so i will yeah so so it is for for example the solutions of homogeneous equations they can be expressed in terms of classical modular forms so it is a quite nice representation so maybe i will spend last 10 minutes on giving you an idea how so how we proved this statement we actually proved it just by analyzing our representation and recognizing some pieces of it so our representation it consists of some very very nice pieces and for those pieces it's more or less obvious that this property holds so so now let's give an easy description of sigma and so for this we consider the following representations the first let row three be the symmetric power representation of dimension three and so this representation on the level of entries of the matrix it can be given in the following form so with one possible choice of bases so it's a symmetric power representation and so this actually is the unique representation of psl2z which can be extended to psl2r to say up to after an isomorphism this is the only one and so one way to understand it it's also the way psl2z acts on homogeneous polynomials of degree two and so another representation which we consider it would be so here we take a representation of sl2z into representation of dimension two and it's given by it's we will give yeah so it is this time it has to be sl2z no not okay so we define it by action on generators and so now this will be a two-dimensional representation and this representation it has actually a finite image so the kernel of this representation it will be just the group gamma two which is the principal congruent subgroup of sl2z so the subgroup of all matrices which are equal to the identity matrix modulo two and so the index of gamma two in sl2z is six and the image of p6 it is isomorphic to the dihedral group of a triangle so it will be just a group of symmetries of a triangle and so now we need another another object which would be a map from sl2z2z2 and so here with z2 so we have we also give it action on generators so the s is mapped into zero and t is mapped into this so we consider this as a row vectors and so now what we require we require that this map v it satisfies the cycle formula it is so it is related to this representation row two and so now from these pieces we can finally construct our six dimensional representation so our six dimensional representation that actually splits into a direct sum of two three-dimensional ones so one of them it will be the representation row three and the second one it will be a three-dimensional representation which is constructed from this finite representation row two and this co-cycle so now we define the representation so we define row from I think a priori it's from sl2z gl6 and it looks like this so it is matrix which looks like this so this is a three by three part and this is another three by three part so and so now we have the our representation sigma it is equivalent to this one so after some change of coordinates one minus one in the representation row two x not really okay so maybe I have to yeah maybe I have to double double check this what happens to okay so I remember that at some point it did work but maybe I have to double double check what what I does is this so of course this representation it should factor through psab is to be well defined for psl2z as well so and so for the but now the fact that we have such an easy presentation of our representation sigma it actually helps us solving for example the homogeneous equation and from this we can deduce that the solution of a homogeneous equation it has to be some combination of classical modular forms also quasi modular forms and logarithms of modular forms but so the somehow like this part the symmetric power representation it will be responsible for the quasi modular forms side and so this the cacycle which we will have here it will bring for us logarithms of modular forms and now for the next time so of course we want what we want to do the next time so next time we would like to so next in the next lecture we would like to solve the the homogeneous functional equation and probably before we do that we will also have a small introduction into the theory of classical modular forms and also our first our big goal would be to just to solve the functional equation with an trivial left hand side and so here we will solve our equation in the following way so we will construct our function f as in the following form it would be an integral from zero to infinity integral of this form and so here where k would be a function meromorphic function on two copies of the upper half plane and so this function it will be a solution of a homogeneous equation with respect to first variable and this will be true for all elements of the ideal i in the group algebra and also it will have the singularities of k will be only at the points where tau and z are psl to z invariant to each other and there will be a special condition on the residue at this point and so probably then it's all for today thank you of which uh you guys it is some six by six man i think it it is given in our paper we have to say what was the explicit form of g we also have to say what is explicit form for sigma because there have been some choices which we haven't made explicit on a blackboard we do it in our paper but they are rather arbitrary i would say so it's just so i'm gonna say that like for for a particular choice of basis in in this in the quotient space so this would be the the row would be just equal to to sigma