So I start, oh, okay, so I see that I definitely lost Enric, so I hope I won't lose anybody

anymore. So to avoid this, maybe I will repeat what we have done yesterday.

So our basic goal was to investigate this integral as a principle, which is to determine, so to which extend the set of representation of an integer by a quadratic form in a lattice

is non-empty. And so we had a sort of natural candidate for an equivalence, for a criterion for this,

which is that, what I called, so here this is D is representable by Q in L, and here

this is this statement, so which is a bit longer, but which is elementary to check,

so the equation, so we want to solve this equation, Q to X equals D, so here we ask for having a solution to this equation in the lattice L, and here we ask to have a solution

to this equation in the real vector space, and in any of the local lattices attached to L, and here we say this, D is locally representable by Q in L, and so the goal

was to give sufficient condition for this to hold, and so I have stated a general result, and so to study this question, and especially to study this condition, we have

introduced a family of lattices which are closely related to L, so which was called the Q-genius of L, and informally this is the set of lattices, so L prime of rational

lattices, which are locally isometric to L, and so we, as I have explained, this set

can be realized as an orbit of the lattice L, under the action of some group, where,

G is an orthogonal group of this quadratic form, and so G of AF is a group of finite idyllic points of G, which is a set of sequences, GP, where GP in G of QP,

and GP in G of ZP, for almost every P. So because we have said, because any lattice

can be completely recovered by the data of its collection of local lattices, we can act on each local lattice and create a new lattice. So in a sense, why this set of

G-genius is natural to consider, because so when you are aware of the existence of

such an action of even the general linear groups on the set of rational lattices, you see that the lattice which are in the Q-genius of L are exactly the set of

local lattices which satisfy this condition, if this condition is satisfied for L. Just because if this set is non-empty for LP, then this set will be non-empty if you replace

LP by any local lattice isometric to LP. So in a sense, it's the most natural set to consider when you look at this condition, the most natural family of lattices.

And so yesterday, we finished on a consequence of the so-called regular asse principle, which is when you, so the asse principle is when you, is the equivalence of two conditions

like this where L replaced by the Q vector space Qn and LP replaced by the local vector space Qp to the n. So basically as a consequence of this definition, so it's just essentially mechanical,

the asse principle implies that if, so I should say maybe D is locally representable by Q in L

is equivalent to D is representable by Q in some L prime in the genus of L.

So a priori we don't necessarily have a representation in the lattice we want, but we have one representation in one of the, on these closely, very closely related lattices.

Okay, so this is good, but okay, so there is one thing is that, you see, so this set, the

still an infinite set of lattices, okay, because really you can make a lot of, so genus of Q is infinite, but as I explained, this problem has, I would say, additional symmetries,

but if say RQ of DL prime is non-empty, then this is the case for any of our lattices

which is globally isometric to L prime. So for any L prime prime, now in the rational G orbit,

of the lattice L. So really the question of representability, it does not, it's not about

L prime, but rather on this global isometric class. And now we have a, you have a really fundamental result, so of Ermit and Minkowski, which says the following, which says that

if you look at the set of, so of global isometric classes in the genus, so that set

is finite, okay. So, and maybe I will continue with the definition, so the cardinality of that

set, so cardinality of that set, it's, it may be noted HQ of L, and it's called the,

the, let's say, the Q class number of, and so this theorem of Ermit Minkowski, it,

of course, he generalized the Gauss theorem and the finiteness of the class number of binary quadratic form. So, and later we will see a further generalization of it in the context of adults and algebraic groups. But so, so then if we say this and add it to this

remark about symmetry, loosely you could say, so the integral as a principle holds

up to finitely many obstructions. So being that, okay, you, as a principle gives you

that you have a representation in some of these lattice classes, but maybe not the one you, you might want, but there are only many finitely many possibilities if the integral as a principle fails, okay. So, but at least in particular, so if the class number of the lattice is one,

so if, if the genius of L is reduced to the global isometric class of the lattice,

so which is obvious, then the integral as a principle holds. And so there are a number of quadratic form for which this, this class number is one. So

example is when n is two, three, or four, then you have the square quadratic form, so the Euclidean quadratic form, and in that way you recover the theorems of Fermat,

Legendre, and Lagrange, about sum of two squares, three squares, and four squares.

Another example is, another example is, okay, it's a discriminant quadratic form,

so this one has class number one, two, so, okay, so there is a bit, so a bit of, but maybe I should say, but in general, the, the class number is, is greater than one,

and so then you, you have something to prove. So maybe I will just say a few words of what happens. So maybe I should say there is one case which is maybe easier than the other, so

the, if q is indefinite, so I, so the class number may not be one, but hq of l is, is a small, and small, what do I mean by small, is small in terms of, so the

natural numerical invariant which measures the complexity of the quadratic form in terms of the

discriminant of the quadratic form, so relative to the lattice l, so you take a basis of l, and then you, you form the, what's the name, the gram matrix, and you compute the determinant of it, so I will define it as a discriminant, even if maybe it's not, okay, and, and so of the size

of, of the discriminant, in fact, if it is indefinite, the class number which, which be

basically two to the power of the number of, so will be less than two to the number of maybe prime factors of the discriminant, so it's very small, but if, if q is definite, and then hq of l grows polynomially with, with the discriminant, and even, so you have even

a precise formula, so for this you need make to, to make some primitivity assumption on the lattice l, say that it has to be maximal if, if q is integral on it, but okay, so it's a vague statement, so but, yeah, probably should be n minus one, so something like this, so it's pretty quick,

quick rule, so, so, okay, so then you, you see in general, you basically cannot escape

the last sub-principle for the most general quadratic forms, so you cannot escape these, these obstructions, except in a, in a few cases.

Always even for very, okay, so, okay, okay, good, so it's, okay, so it's, good, I trust you, sure, okay, nevertheless, so the adelic treatment enables you to completely

ignore the issue of being definite or indefinite, so even, so, okay, it may simplify things, create difficulty on the other side, so, what do I want to, okay, yeah, so now I want to explain

strategy to, to, so to, to cope to under these obstructions, and, okay, just for the second,

I will not need this, yeah, we not, don't need it, so, so, okay, so I remind you what we had,

that we had a representation in some lattice L', which was given to us by the sub-principle,

and so we have at least one, and we will start from this representation to build new ones, okay, and, and the way to do this, so we will consider H, the sub-group, so we take the stabilizer of this vector in, in the orthogonal group,

so this is an algebraic sub-group of the orthogonal group, which is an orthogonal group in n minus one variable, because it's the orthogonal group of the quadratic form,

when you restrict it to the orthogonality, ortho complement of, of the vector x, okay, xq is not zero, so you, you introduce this, this sub-group, and now look at this

orbit, so you take now the adelic point of this sub-group, and you consider the lattice L', so which is of course contained into, to the full L', which is just the, the q genus of L,

so, and so consider some lattice in, in that, in that subset of, of the full genus, so what do you have, so, so what does it mean, so you have that, let's say,

right, L' to be HFL', so which means HF is a collection of local, of matrices

contained in the local group attached to H, and so you have a L' prime p is HP times L prime p for every p, but now you have the equation that xq is in L prime p, because of this,

so, and if you apply HP to each of these, and because xq is contained in,

in all the local lattices, then you have that xq is also in L prime prime, okay, and of course q, xq is, this vector has not moved at all, so it means that q of,

xq is d, so it means that xq belongs to the set of representation of d by L prime, but what is important here is that L prime prime, this new lattice, does not

necessarily belong to the global isometric class of L prime, okay,

because you have not acted by your rational matrix, you have acted by your nadelic element, okay, so potentially one has created a new representation, in the sense that

L prime may be new, may not be an obvious representation, but you could have gotten from this representation just acting trivially by your global isometry, okay, so now

suppose, so then, because you have a new lattice, suppose that this lattice L prime prime belong instead to g of q of L, so then it means what? it means that xq belong to gq L for some,

for some global element, and then you have x prime q, which is gq minus one xq,

which is a vector, belong to L, okay, so what you have is that, so let's write it that way,

so we started from a lattice L prime, and out of this lattice we constructed a family of lattices, which is so contained in the genus, and the genus, I recall you, it's the union, say, of global isometric classes, so if this orbit meets the global

isometric class we are interested in, we have obtained a representation in the lattice we want,

okay, so it's a bit strange because we start from a representation, so from a vector, so we do not move the vector, we move the lattice, but of course, so you see that when you really

get new representation because this q is twisted by this global element, okay, and so that's the strategy to prove this, so the goal is to show, in fact, what we will show is

is somewhat stronger, because there is no reason to that L has a more special role than any other lattice in the genus, we'll show that, so if you look at this orbit of L prime,

so show, so give sufficient conditions so that when you look at the orbit, at this orbit, and you project it on the genus, so what you want to show is that this projection map

is surjective, so then you see this is a statement about sort of, it's a dynamical

statement, you have a space on which a big group acts, so in fact the space is an homogeneous space, the genus is an homogeneous space of under the action of this group, because it's orbit, and you take a subgroup of the bigger group, and you look at an orbit in this subgroup,

and you want to know does the orbit fill, completely fill the space, okay, is it, so maybe it may be a bit disturbing this way of seeing,

so is there a question, so okay, so and this is a strategy that can be, so the claim is that, so what one needs, so one needs, so with respect to what I have explained in the beginning,

we need to show that there exists x, so I remind you, so basically the statement was that the

integral as a principle holds whenever d is sufficiently large and satisfies a set of additional conditions at anisotropic primes and so on, so what we need to show is that

for d large enough, so plus additional condition which were spelled in,

there exists a d, there exists a representation xq such that h, f, so this orbit is growing, so becomes bigger and bigger, so here is a silly example

where you may have an orbit which is not growing, which a priori you cannot exclude, because okay, so as a principle does not tell you much about this except its existence, so example where it will not work, assume that is growing as d grows, so

example suppose that d is of the shape denote d1 square, okay, so then suppose you have a note such that you start from a representation such that q of x naught equals denote,

so then you see that x is d1 x naught, so which is a very non-primitive vector,

so is a representation of d, but the stabilizer h is a h of, so let's say h of x is just a stabilizer of x naught, so because it's a scalar multiple of this does not

as d1 goes to infinity, so you have a small problem related to imprimativity,

so this can be solved or if you are happy with additional assumptions, say I don't know if you are happy with saying that d will be a square three or will have a bounded valuation every prime, this will not occur, okay, so I will ignore this, this is a

merely a technical point, so under the condition I have given, you can always find a good representation out of the one you started from which will be an imprimative vector, so

maybe not imprimative, but at least when this group will be growing, so yeah. Could you please trace how the right hand side of this h of a rho depends on d, just a general statement? So d, h will depend on d, yes.

So could you please just show the, like briefly? Because xq depends on d, so if d moves and again and you avoid such situation, your torus will move. So is it just xq?

If xq is imprimative, yeah. So basically you set the d that determines xq and that will determine the dependency of h a rho l prime only. Yeah, but you have to choose, so you have to choose xq and if you don't want to

make a choice, assume say that d is square three, then as d varies among square three integers, you will have a good variation of h2. Okay, is it something clear?

So this covers the exceptional cases? No, because, okay, so in the case of

exceptional square classes, so what I will explain is that then this such a statement that you cover the full space will be, so we will prove something stronger that in fact

the orbit when projected there will be equidistributed, so we want to prove that it is equidistributed, meaning that we will have to integrate the orbit along a test function

on that space and so for most test functions, so the one associated to cuspidal automorphic form, there will be no problem, but you have also a test function which are characters associated to characters and in this case it may be that if the d is in

a bad square class, your character will be trivial, so your character will be globally non-trivial, but will be trivial along the orbit, so you cannot have a sort of equidistribution.

So in this representation, the exceptional square classes will come from the so-called the residual spectrum of the adelic group. So in order to know this extra obstruction,

you have to know the shape of the lattice l, because then you need to know which are characters which can occur, and so you know how to, so you can, you told me that you have

conditions to, the statement you made that if this big enough and you only look at the primitive elements, is it true even in the case when you have exceptional classes?

So here I have but additional condition, oh yeah, so no no, but the group still, the group h will grow and it will fill a good portion of that space, but it will avoid some finite, so some

finite places, yeah. Okay, so maybe, so there is a, so this strategy, okay,

so besides the Duke theorem, there is a very nice place where this strategy is implemented and which goes at higher level, but in which really it's, so I say it's a dynamical statement,

but there is a place where it gets really dynamical, it's in the paper of Ellenberg and Venkatesh, so when they study not representation of an integer by a quadratic form, but when they study representation of quadratic lattice in a higher rank quadratic lattice,

so you can make just exactly the same reasoning, so except that you will not consider a stabilizer of a point, but stabilizer of several independent vectors, and so it's an invention as paper,

I don't remember the date exactly, actually it's not there, someone has Google, so if you want, you can check, so there it's really, so the dynamical treatment is somewhat different, okay,

so I wanted to, so what I, maybe I want, okay, so but what I want to explain maybe is

how this strategy allows you to prove even, so even more precise statement, so

okay, so suppose now we have proven that for D large enough, and we have had sufficiently many reasonable conditions on D to ensure that we have the

integral as a principle for every lattice in the genus, every one, so because, okay, so then one can look for more refined problem such as the position, so we know that RQ of D

L, okay, is non-empty, so what are the positions of the vectors, the x,

and then so there is a various way to look at the position, so the most obvious one is then, so we scale this set of representation by D to the one-half, and this set

is contained into Rn, so plus or minus one is the sine of D, and what is this thing, so

minus one of real vectors taking value plus or minus one, so this thing is an ellipsoid

or an hyperboloid depending on whether the quadratic form is definite or indefinite, and so this is a fixed space, and so what you may want to look at is for the

values L that you are interested in, how does that set fill this space, or how are these vectors distributed into that hyperboloid, okay, and so this by which theorem, this hyperboloid is

an homogeneous space, so wait for E of R, so this thing can be identified with, so it can be written as an orbit

of a single point, let's say x naught, so you choose your favorite point here, and you take the orbit, and you obtain the full space, and so as such it's homogeneous,

so x naught does not need to be a rational, it's a real vector, and so because this is a homogeneous space, it admits G R invariant measure,

so which is uniquely defined up to scalar, which comes from taking the quotient of the R measure

on these two groups, and okay, so then one has the following equidistribution statement, so I will, okay, let me fix some notation, so I will write gen q of L here is L1 is L,

so just to fix IDs, L2, a set of representative of the of the genius classes, so I write, okay, so

what does wit say? wit, wit's theorem, wit's theorem, that an orthogonal group acts transitively

on a sphere, on a, yeah, on the level set of a quadratic form, okay, so the set of,

you fix a set of lattices which represent all the various genius classes, so you know that this set of representations are non-empty when D is large enough plus additional conditions, and so the theorem, which is that for all I, so the set is equidistributed on

Li, sorry, yeah, is equidistributed on my hyperbolic as D goes to infinity plus

conditions, so the condition to guarantee that integral as a principle holds,

and specifically what do I mean by equidistributed, so what, okay, so suppose you take phi one, phi two, two compactly supported function on that hyperboloid,

continuous compactly supported, so no problem with, so my definition of equidistribution will be set proof, and so then we assume that, so this integral is not zero,

okay, then the ratio, so you take, you average over all representations

of D by Q in the lattice Li of psi of xi over D to the one half, and so you make the ratio with the same, but with phi two, so then this ratio, so for D large enough,

it will be well defined, and it will converge to the ratio of the measures,

and this is valid for every, so every choice of representative, so, okay, so in a sense,

your integral representations are well distributed. So as I understand the D big enough, it depends on the lattice too, right, yes, yes, yes, it depends on the lattice, yes,

is there a kind of uniform state with respect to the lattice, so, so when n equals three, you may face the z equal zero, so maybe, but yeah, you can make

so modulo z equal zero issue, you can make something really, so the constant D will depend polynomially on the discriminant of the Q discriminant of the lattice.

Okay, so I want to explain this, I want to explain how it goes, this equidistribution

statement, because I want then to bridge to the full Adélie group, but so I would, okay, so,

so what I explain, okay, let me, okay, let me call gamma i to be g of li, so I mean the stabilizer of the lattice in gq, so then embedded into gr, so this is a discrete subgroup of gr,

and so this is what we said in the beginning, this group acts on the set of representations, so then which are split into a Dijon's union of orbits, okay, so there is a finiteness theorem,

but this union is finite, and it's a variation of the on the Ermit Minkowski theorem, so here

this is a finite, so the set of representation splits into a finite union of gamma i orbits, so when we will analyze our sum, so let's look at the numerator, so we will split it into

according to these orbits, and so this is what we want to evaluate, so,

okay, so this phi is phi one, say phi is phi one or phi two, okay, so now what is phi? Phi, it's a function on this hyperboloid, but because of this identification of phi, you can

view phi as, okay, you can view phi as a function on gr, which is, on gr, which is write h x naught of r invariant, yeah, I want to explain to,

yeah, I want to explain duality principle, so this is a function which is right, invariant on the right, and okay, so, and the compactly supported function on that quotient, you can, you may assume that they are of a specific shape, so by density,

that psi is of the following shape, so psi of, let me say, x equals,

so it's something of that shape for, yeah, and phi, a compactly supported function on gr.

So the space, the compactly supported function on the quotient are generated by this function which are obtained, which are made invariant by average g. So then what we are looking at, so it's a sum over these orbits of a sum over gammas in gamma

i of integral, an h integral of this phi of gamma i g i with a few notation, h naught, d h naught,

and okay, and this, you can just switch summation and see this as a sum over these orbits

of integral of x naught of, okay, what is my, okay, we'll call it little phi of gamma i

g i h naught, d h naught, yeah, where phi of gamma i g is just the function which is obtained by averaging over gamma

of big phi of gamma g. So, and this is a function which is continuous, compactly supported on the quotient gr modulo gamma i.

Okay, so the, so this, by what I've made that I explained, so the equidistribution statement

for the representations, rq of d, rq of d l i, it's equivalent to an equidistribution statement

for, so for actual i on this ellipsoid, it's equivalent to an equidistribution statement

for a collection of h x naught of r orbits in the space, in the quotient gr modulo gamma i.

Okay, so it's a standard duality principle. So, and so here then you see we are looking for

an equidistribution problem in the quotient of a group, a discrete subgroup, so now we start to smell automorphic forms. So, and but, okay, but here, so it's a statement in a sense

which was made on purpose, it was made for individual gamma i's. However, it's much better,

if you want to try to solve this problem, it's much better to look at it as an equidistribution statement for a collection of h x naught r orbits in the following space, in this disjoint union.

Okay, and I will explain, so I will explain next time that this thing is naturally identified with the following quotient. So, you take gr times gaf divided by gq modulo this group, so g of l

OK? And it's a bit false, but plus the collection of orbits.

So what is it? The collection of orbits can be realized as a projection on that quotient of the orbit.

So HR times HAF times, I would say,

yeah, I want to say, yeah, let me say just the L prime that we started with. So HAF is Adélie group, which was introduced here

from our primitive representation. So then, in a sense, when we want to prove the more precise version, so this equidistribution theorem, we have to look not only at how this adelic orbit equidistributes

in the space of classes of rational lattices, but we will have to look at how this big sort of complete orbit equidistributes in this big adelic quotient.

And so to study this equidistribution problem, you have to analyze this quotient and things like this. OK, thank you. Questions? Yes?

Can we think of RQDL as the equi-operator acting on the symmetric space demotic? No, no, it's what would be a equi-operator? RQDL, the axon on the left of RQDLi. Where? Top board, topmost board. This one? The topmost board, yeah, second line of the topmost board.

Topmost? No, no, it's not a equi-operator. So when n is 4, this thing can be interpreted in terms of equi-operators.

But not, so this orbit, you can relate it to equi-operators and the resolution comes from bounds for eigenvalues of equi-operators. But for n is 3, this orbit is related to L functions.