3/4 Automorphic forms for GL(2)
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17:50
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26:46
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35:41
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44:36
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53:31
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01:02:27
Diagramm
Transkript: Englisch(automatisch erzeugt)
00:14
So, I hope I have sufficiently motivated the user of Adele, so I don't know.
00:28
Just what I want to point out is that so far, regarding the Adele and what we have done about them, we have basically done a set theory, so not much structure.
00:43
So now, maybe I want to discuss the additional structure which comes to the Adele, which eventually makes it possible to do analysis. So, I will start now by looking just at the Adele.
01:08
Later, I will come back to make more precise what I discussed at the end of last talk. So now, I just start with A, so the ring of Adele.
01:24
So, this is the following ring. So, it's at V of V, the set of classes of completions of Q.
01:53
So, which is V is just the prime, the set of primes, and say, a completion which is written of an infinity,
02:04
which corresponds to a real completion of Q. And so, the ring of Adele is the following restricted product over all places of the various completions. And so, it can be written R times the restricted products over all P's of the QP's.
02:30
And so, I write it R times AF, which are the finite Adele's.
02:43
And AF, so this is the set of sequences of periodic elements for P varying. And what we require is that XP belong to the ring of periodic integer ZP for almost every P.
03:12
So, you see, this is completely analogous to what we introduced, but we define this for the general linear group.
03:23
So, the condition that we add is that instead of having just periodic numbers, then here we add the condition that we add a sequence of matrices, and we were requesting that almost all of these matrices had periodic integral coordinates.
03:42
So, A and AF, these are rings. And so, what other interesting set can we find into... OK. So, we have embeddings of...
04:05
So, we have Q, which embeds into this ring of Adele's. Simply, if you take a rational element, then you send it to the constant sequence.
04:23
And because Q has a finite denominator for almost every prime, XQ will be a periodic integer. So, we have this embedding, and before we have discussed also this embedding,
04:43
which is the same, so the diagonal embedding, and so the constant embedding. OK. So, we can also see if we want the ring of Adele's inside...
05:04
So, at least the additive group of finite Adele inside the additive group of Adele's, by sending a sequence Xp to the sequence of the Xp at every prime, and then we put zero concerning the air factor.
05:34
OK. We may also embed, so far, we may want to embed also the periodic field,
05:42
so we can write delta p to be the embedding of Qp into either A or Af, which is, you have a periodic element, and you send it to the sequence
06:09
with Xp' equals zero if p' is not p. OK. So, this ring, so one important feature is that this ring,
06:39
so it's a product, or it's a so-called restricted product of topological rings.
06:47
These are even topological fields, and even with metric. OK. One may want to give to this ring also a topology, which is somewhat related to the topologies of its factors,
07:05
and so it's not a good idea to give this ring just the product topology, because it would not have very good properties, but instead one puts on it the adelic topology, and a basis,
07:36
so I will define the topology by defining the open set,
07:40
or a basis of open sets are of the form, so an open set or a basis, so you take an open set of your shape, product of all v of omega v,
08:08
with omega v in Qv is open, and what we ask is that omega p is zp for almost every prime.
08:24
So, I remind you that in the periodic topology, the ring of periodic integers, it's both open and a closed subset of Qp, so it's even an open and compact subset of Qp,
08:42
and so this, and likewise, you give the finite adels, the same as the similar topology, and it makes AF a closed additive subgroup of A,
09:01
and each Qp becomes also closed inside the full ring of adels. Okay, so a few things that one can say now that we have this topology, so then we have this subring of AF,
09:26
which is just the full product of the zp, and then this, when you restrict the adelic topology to this full product, then you just obtain the Tikhonov topology, because all these factors are compact,
09:43
and so this is an open compact subring of AF,
10:00
and it is maximal, in fact, it's the maximal, and maximal is the only one, in fact, so it's a maximal open compact subring of AF. And maybe some things which can be interesting,
10:35
maybe just to be more concrete about this basis, so a basis of neighborhood of the identity element
10:52
for the additive group structure, so of 0 in AF is given by the open,
11:08
and in fact open and compact is given by, okay, it's a sort of abuse of notation,
11:21
so yeah, okay, no, it's correct, it's just this product NZ for N an integer, so when you vary N, you take these things, and the more and more N is divisible,
11:42
the smaller and smaller your neighborhood is. Okay, so various properties which are important for A is that, so if we look at Q,
12:04
and it's this property that Q embedded into A is a discrete subgroup, and so one reason is the following,
12:22
so if you, it's discrete, so if you intersect Q, so here my embedding is a delta that I mentioned here, so if you intersect Q with this open set,
12:41
so what are you looking at? You are looking at rational numbers, which are periodic integers at every prime, which means that a rational number, which is everywhere a periodic integer, is an integer, and then this integer
13:02
has to be between minus one half and one half, so you only have 0, so it's, you see, 0 is isolated by this neighborhood, so then you get the discreteness, and so it's discrete, but it's not too small nevertheless,
13:23
because, so it's sort of exercise, so if you look at this quotient of additive groups, so what you obtain is that this quotient is homeomorphic to R modulo Z,
13:43
so it's an exercise, so in particular, this is, okay, this is infinite, but this is compact. Okay, so, and because this thing is compact, it means that M mod Q is also compact.
14:03
Yeah, so, okay, on the other hand, yeah, just for the record, if you look at the embedding of Q into AF, then you can show that
14:23
this Q, embedded this way, is dense. And again, this comes, this is a consequence from the Chinese remainder theorem, and a description of this basis of open sets
14:44
of neighborhoods of 0. Okay, so depending on whether you have the adults including the real number or not including them, you see that Q has very different behavior.
15:01
Okay, so these are some of the basics. And now I, what I want to do is to construct a bit more complicated objects, and I want to come to GLN of A and to G of F or G group of an orthogonal group.
15:24
So, okay, so if you have A, then you can consider the product AN with the idyllic product topology, and, okay, so the topology,
15:42
and you can prove a lot of, so these properties, if you adapt them, they remain true. So in particular, so QN into AN is discrete,
16:02
because it's just because of the product situation and so on. I want to play with this. And so, okay, so now what I consider is V. So I consider V an affine subvariety,
16:25
an affine variety, an affine variety. So here, this is n-dimensional affine space over Q. And so, remind you, V,
16:44
it's defined by an ideal of a polynomial, and it's a set of n-apples, such that which are cancelled by all the polynomials,
17:01
which are contained in some ideal, defining the variety in an ideal of polynomials. And so what you may want to consider is
17:21
you have this thing, you have this thing which is Q algebra, and then you have this variety, so you may want to define V of A to be the set of X in now AN, such that P of X equals zero for all P in IV.
17:45
Okay, and so here, this is a ring, and this is a polynomial whose coefficients are in Q, so they are in particular in A, so it's legitimate to compute this polynomial.
18:01
And polynomial functions are continuous on the adults, because the adult is a topological ring. So this, which is a subset of A to the n, is closed, okay, it's a closed subset.
18:25
And so then, this variety, as a closed subset of Rn, inherits the adelic topology.
18:45
And you may have a concrete description, although the set may be empty, but nevertheless, you can write it as a set of Xp,
19:02
with Xp in V of Qp, and Xp in V of Zp, for almost every p. So it may well be the case that this set is empty, nevertheless. And you may look at V of Q,
19:23
and because of this discreteness, V of Q will be, if even, so, could be empty, but V of Q will be discrete as well. It will be discrete, that's it. Okay, so the most important example for us
19:46
will be the case where V is the general linear group. So I remind you that the general linear group is an affine variety, so it's a closed sub-variety
20:02
of the product of n times n matrices times the affine line, and the ideal, defined I of GLn, is the ideal generated by the polynomial
20:23
determinant of Gij times T minus 1. So T is the coordinate of A1, and Gij is the coordinate of the matrix element. Okay, so if you take a matrix,
20:41
and you take a T, and you ask that this matrix times T minus 1 equals 0, it means that T times the determinant won't be 0, and in fact T will be an inverse of the determinant.
21:00
And then the GLn that you get in that way, which you see again as a closed subset of this product, which is A isomorphic to AQ n squared plus 1,
21:24
acquires the adelic topology, and is locally compact. Yeah, maybe I did not say this. Compact topological group.
21:42
Okay, so yeah, I don't remember if I say this. So when you equip the adults with the adelic topology, the great feature that you get is that your topological ring is locally compact and separated, so it's very good properties.
22:02
So locally compact, separated. Oh, I don't, maybe it's, okay, it's being separated. I put it into a topological group, say. And so you can do now the same with any algebraic subgroup,
22:22
so linear algebraic group, because you may see this group as a closed subvariety of this affine variety, and you give it with, so you consider G of A, GLn of A, which is closed,
22:44
so locally compact. And just by, okay, so just an example, if G is the spatial orthogonal group,
23:03
it's ideal, so it's, so the ideal of G, it comes from the equations, so transpose of G times Q times G equals Q,
23:27
and determinant of G is one. So here, Q is the matrix of the quadratic form into, say, the canonical basis,
23:41
and then you have this equation, which is a set of polynomial equations in the coordinate of G, and this is, again, a polynomial equation, so you have something which is closed. Okay, any questions?
24:12
So for people who are not used to this, oh, okay, so everybody's used to this.
24:22
Yeah? Locally, it admits a basis of pre-compact neighborhood of the identity, say, because it's a group. So it contains very small compact subsets.
24:49
So it's a sort of minimal thing that you want if you want to do reasonable topology. So topology, I can understand at least.
25:01
Okay. Okay, and so then you have the embedding, so G of Q embeds into G of A, and it's discrete.
25:21
On the other hand, if you look at G of Q embedded into G of AF, it's not necessarily dense. It could be quite big, but maybe not dense.
25:44
So what are the interesting objects that you might want to see? So G of A, or let's say G of AF is this restricted product.
26:01
So I just repeat once again the definition. For any linear algebraic group. So yeah, I wrote this sometime.
26:22
So G of ZP, so which are the matrix which coordinate in ZP, and maybe a more canonical way to write the G of ZP is to write it as G of QP as the stabilizer of the square lattice in G of QP.
26:42
So you have this, and then you have inside this you have G of Z hat, which is just the product of the G of ZP,
27:00
which is an open compact subgroup. Not necessarily a maximal one, so you can find, if you are ready to change coordinates, you may adjust the square lattice
27:24
to have a maximal compact subgroup, but anyway, so just at least what you can say is that the G of ZP is maximal open compact
27:42
for, ah no, I need the condition, okay, I will erase this, because of the unipotence. So if the group is semi-simple, okay, so and again,
28:04
so what is interesting is that G of AF admits a basis of a neighborhood of the identity
28:24
formed of open compact subgroups, so that's very useful to have a very very small open subset which are in fact groups and which are more overcompact. These are the principal congruent subgroups
28:42
which are, so let me call it KF of N, so for N greater than one,
29:00
KF of N is the product over KP of N, with KP of N is the set of G P in, let's say, G of ZP, so integral matrices with
29:21
G P congruent to the identity mod N, okay, so if N is invertible, the condition is void, so it would be G of ZP, but if N is divisible by P, it puts it says that G P is periodically close to the identity
29:42
element, and so when N becomes more and more divisible, you get a shrinking set of open compact subgroups. And then you can just translate these neighborhoods,
30:04
translate them to any point, so to have a nice neighborhood. Okay, so this is merely local information, so now I want to come into some important finiteness theorem, so
30:22
due to Borel, and we will make the link very quickly, so for all KF into G of AF, an open compact,
30:44
so when we look at the quotient G of AF divided by G of Q, and divided by this KF, and so this quotient is finite, and its
31:05
cardinality, so which I will write H G of KF is the class number of
31:26
this open compact subgroup. So I will come to what we have done earlier, but just a small remark, if you want to prove such a statement, it is sufficient to prove this statement
31:43
for any open compact subgroup. So, if known for one, this holds
32:01
for all, because if you have two open compact subgroups, KF and K'F, so the intersection, it will be an open compact subgroup, and because it is open and
32:22
compact, it's contained to the intersection, and of finite index, in both of the big groups. So, in a sense, to prove this theorem, you have
32:42
a choice of the open compact subgroup. So it's just formal. Okay, so now of course you will have recognized
33:08
what I something I have said earlier, so if
33:24
if G is the orthogonal group and KF is say G of L hat, so it's the product of the stabilizer of the LP
33:42
for P lattice, so this in that case, the fact that this double quotient is finite, so this is
34:00
Ermit Minkowski theorem. Why? Because you have G, so the genius I recall you, the genius of the lattice L so this is the
34:22
orbit of L under this section, and so it's naturally identified with the quotient by the stabilizer the quotient of G of AF by the stabilizer of the lattice which is G of L hat
34:41
and then if you make a second quotient so here you have the genius here you have the set of genius classes and you know that this is finite. So this theorem of Borel, it's a wide generalization
35:01
of the Ermit Minkowski theorem but we will I will give a proof of this theorem now in a special case so proof
35:21
if so for G is SOQ and so I'm asking Q is Q anisotropic so meaning that Q of X
35:41
is not zero for all X in QN minus zero so the quadratic form never vanish on a non-trivial rational vector and so for this I will just
36:00
recall something so I introduce so the adults it's tightly linked with the notion of lattice so I will translate this statement into some concrete statement about lattice for which we will prove the theorem and even something
36:24
stronger so let me introduce L of Rn to be the set of lattices in Rn so not necessarily rational
36:42
basis the Z-module generated by real basis of Rn and so let me take so as I said to prove the Borel theorem it is sufficient to do it for any open compact subgroup so I will take
37:02
an open compact subgroup of that shape so I take L rational lattice and I consider the following map so from the adults into the real lattices
37:23
okay which let's say I have a needle element gr times gf and I map it to the following so I take gfl so then I get a rational lattice and now I
37:43
act by my real element let's say by minus one something like this okay and what do I get in under this map so
38:05
this map okay so what do I get here so the image is
38:22
the union of lattices of real lattices which are in the
38:42
gr orbit for the natural action so ln r acts on the real lattices and so in the gr orbit which are in the gr orbit
39:01
of some lattice in of some lattice in the genius
39:25
and so really you have identified okay so you have this map and this map of course it's not
39:41
injective because you can see that okay you can see that the the group of rational elements if you define this map that way the gq acts so maps
40:02
does not do anything so what means this map gives you it's an identification between the quotient ga modulo gq modulo g of l hat and it gets identified
40:23
with the set of lattices and this thing okay or you could do it directly also it
40:42
will be identified with the disjoint union of gr modulo gamma i
41:02
okay where okay let's say that so let me select a set of representatives of the genius classes so I take a set of representatives of these
41:26
genius classes and yeah even I don't even know that the genius classes is finite it's just a set of representatives and so what
41:42
is gamma i so I don't want to make a mistake so it's gq intersected with gfi kf gfi minus one
42:01
yeah and where gfi is such that it sends l to the lattice li so okay and so you can check
42:23
it's a set theory you can check that you have this identification and this identification it's a nomiomorphism of so this this is a discrete subgroup of gr so this has a topology and you take this
42:42
disjoint union inside so to speak inside the set of full lattices so and you get this so okay so now I'm going so this is completely general but now I am going to use my
43:02
hypothesis so up to scaling q we may assume that q of l the lattice so the quadratic form is integral
43:21
on l and so now so for any x in l not zero this implies that q of x in absolute value is greater than one because it's a non-zero integer but so
43:42
you can see that if this property is true it remains true for any so then it implies that this is true so q of l prime is integral for all l prime in the genus of l
44:01
because if you want to check that something is integral you need only to check it for every periodic element so for any p for any prime number and then the lattice l
44:21
prime you cannot distinguish much from the lattice l because they are they differ from each other by an orthogonal transformation so you have this property and so for any x in l prime and not zero you have that q of x
44:42
is greater than one but then if you just apply also a real orthogonal transformation you have that for all x in l prime in
45:01
g a of l where g a of l is just this map that I have written there you have x not zero so you have that q of x is greater than one
45:21
because all these lattice are all isometric to each other and this is good because what it says is because q is a quadratic form so this implies that for all x
45:42
in l prime prime different from zero then the Euclidean norm of x if this is big the Euclidean norm of x has to be big by some constant but which depend only on the lattice
46:00
l and not on l prime so the second thing that that you should know is that you may look at the volume so the Euclidean volume of the lattice l prime so which is which is rather the
46:22
volume of l prime and what you can prove or check is that this volume is in fact equal to the volume of the original lattice so the volume is fixed so now what you have is that
46:41
you have a set of lattices all of the same volume which have no small vectors so now you have Mahler compactness criterion
47:01
implies so you have to show that this set is closed but the Mahler compactness criterion implies that so which is a standard result in the geometry of numbers implies that the set is finite
47:23
g r of l is compact so not only it shows that this quotient is compact but also that this union is finite so you get both in one strike
47:40
any some questions so this is very basic the simplest case of proof of Borel theorem but it's so on general the use of Mahler compactness so for which groups will it apply? so any anisotropic group you do exactly the same thing
48:03
so the torus so you get the finiteness of the class number of a number field no I'm not sure but ok
48:21
ok I don't know the proof of Borel so you can if you want to have the proof of Borel finiteness theorem it suffice to ask because it's published at IHES so so you can even
48:40
bring it back with you in the plane I don't remember the volume but I ordered it for myself so then you see that
49:17
so at least you see that if q is anisotropic
49:27
this idly quotient so g a modulo g q modulo g l at it's a very nice space because it's a compact space
49:42
even you can remove this quotient because then this thing will be compact because you quotient by your compact but this set it has a very concrete interpretation it's a finite union of quotient of a real group by discrete subgroup
50:01
and I remind you this is basically the space we wanted to study we found when we were looking for the distribution property of representation of an integer by
50:22
a quadratic form so I remind you the drawing ok maybe I will yeah ok it's not exactly so what did we add on what side if we think classically
50:41
we add here a collection so we had made some gymnastics and what we have reached it's a collection of
51:03
h x naught orbit so and what is
51:22
ok we had this orbit since it was g x so for some representation g x i is a real matrix here defined by g x i of x naught equals x i over square root of d
51:41
so our problem about equidistributing the representation was about looking at a set of h x naught orbit into this this union and on the other side we had to look so
52:03
this thing was essentially so it's not exactly true because maybe it's ok you have this map and it's not exactly an identification so it's more an inclusion and here on that side you have
52:21
and I want to be correct g ok g r of g f prime and we have this
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and what are the g r and g r prime ok so g r is a real matrix defined that g r of x naught equals
53:02
x q over d to the one half and g f prime it was a finite adelic element such that g prime f of l
53:20
equals l prime ok so it's not always the case that this quotient which you send which you project into this quotient corresponds to doing this but it's a subset
53:42
of this quotient in general and under some condition it can even be the full set of representation that you want to have so just I remind you the notation so you have the g equals s o q then you had produced out of
54:02
as a principle representation of d by q in some lattice l prime and we have this subgroup
54:21
of g which was a stabilizer of x q into g and so you can do exactly what we did here we have this quotient and we look at this an orbit of this subgroup into this big quotient
54:43
ok and so what do we want to do ok x naught in the very beginning we had this ellipsoid I remind you we have r q d of r
55:02
plus or minus one of r n so it was a unit sphere or unit hyperboloid and this and if you pick x naught into this hyperboloid then you get an identification
55:20
so this is a nomogeneous space under g r so you may identify it with g r with h x naught with h x naught of r which is I wrote where
55:43
x naught was this maybe I should write x naught just to be consistent so maybe I switch a bit the notation from previous talk but
56:00
ok so you make a choice just to identify this hyperboloid with a nomogeneous with a quotient so it's just an arbitrary choice and it's a reference point so to speak ok so and ok what are we going to
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prove in the end so to prove the equidistribution statement ok so g a and h a are locally compact
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topological groups so they admit left invariant
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half measure so let's call it mu g and because g is g q is a discrete subgroup then you obtain by taking the quotient with the
57:20
counting measure then you get a measure which I will call again mu g between brackets on that quotient and in fact in our case and so and then you have mu h
57:44
on g a mod h a mod h q ok so let's again assume that q is anisotropic
58:00
so what I will say is basically true but it becomes a bit simpler in that case so you have these two measures and so there is ok because these plates are compact these measures so mu g and mu h are finite again because you
58:20
have a finite in fact the measure mu g is also right invariant from general things so it's not true for a general linear algebraic group but left invariant half measure is also right invariant but in that case the groups are it's true so
58:43
so so right g a invariant and the finite measure ok and so you can normalize
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to be of volume one and so we normalize to have mass one
59:21
so the measure of the total space is one which is always possible and so what do we want to do we want to prove that for any phi which is a continuous function on say g a mod g q and what
59:42
do we want we want this continuous function in fact to be a continuous function on the quotient by this so we will ask it to be g of l at invariant because we see the compact
01:00:00
group, this subspace is a closed subspace of the full space. OK, I just write the statement I want to have. So I have this.
01:00:22
And what I want to prove is that I consider the integral of phi of h d mu of h.
01:00:42
And so what I need to prove at the end is that this integral converge to the integral.
01:01:03
So as d goes to infinity, and I remind you, there are some additional possible additional conditions. And so what happens is that it's possible here to take our functions phi to be of a very special shape.
01:01:27
So this function here that I am considering, so you can see these are continuous function on this union of quotient.
01:01:41
So these are perfectly non-exotic objects, I would say. And these are automorphic forms. And so what I will explain or describe next time is the necessary background to understand this statement.
01:02:23
Thank you again.