Colmez' conjecture in average
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Lecture/ConferenceDiagram
Transcript: English(auto-generated)
00:17
Okay, so I will talk about the coldness conjecture, the average, that is the precise coldness conjecture.
00:25
I didn't write the CM type, and right now we've only improved when you check the average of all the CM types. So this is a joint work with Xi Yuan, and also Bezgan, you will see.
00:51
The proof is the work of Yuan, Wu Chen, and myself. So this is also my study.
01:10
So my lecture is trying to reduce everything from the coldness conjecture average to some popular or some kind of identity.
01:23
Why is it that I'm going to finish? Because this is probably another two lectures. Okay, so I just wanted to review the first thing, what is the automatic.
01:45
So we start with the epidemic writing over a number field. Assume that this A is a neural model. Assume it is semi-stable, semi-stable, semi-abelian.
02:16
In the state of the virus, I send some torus by the skin.
02:27
Then I invoke hips, then I fix the location. So this one is the sheep, the vector bundle. It's an invariant.
02:41
I can finish this. So this is a bundle. Okay, so this is the location of the small a. So when you say neuron model, is it the connected part of the neuron model, or you assume that the neuron model is connected fibers?
03:08
Just a connected component. So this one is the highest degree of the force.
03:29
Then, so this one is a metric, and this is based on v, k, you can define a metric.
03:49
But the following method, square equals to one over two pi of g, integration of a, b, c, alpha, alpha bar, is the method there.
04:11
So my normalization is already a normalization by forces. It's different than coordinates.
04:21
So we have, this is called permission line model. In our theory of k, so we define the part of the height of a to b.
04:43
The normalization degree of k of q, the degree of this bundle. And the one thing, when I put this in the ability condition here, to guarantee that this height doesn't depend on the charge.
05:07
I mean, it will be invariant, and the weight will change. You can tell this by part extension, it will be invariant. So this height was introduced by Paulin in 3.
05:28
It is a proof of a model conjecture, a test conjecture, a Raoulti conjecture. So it's one of the most important invariants with metric geometry.
05:42
Nevertheless, there is no real method to calculate this height, unless you check it. In this case, these bundles have certain canonical sections, if you raise the power by 12.
06:03
So this is the delta function. In general, we don't have such a section.
06:25
Now we proceed to the convex conjecture. Now roughly speaking, I will call the convex conjecture average.
06:45
Now we assume an action by a certain theory. So this E is the same field, where the 3G achieves the dimension of A.
07:22
We have a so-called tie. And if I assume that A contains all the conjugates, then the tie to the inside, or the hole,
07:55
E to A, maybe let's try to get this one to the convex number.
08:07
So fix the 100-meter place for the number. So that's a useful way to do it. Maybe I'm doing something completely stupid.
08:20
Let's try to get this one to the convex number. Since I wake up all the ties, so let's fix that. Let's see here. So this is the tie. So there's only x and y in this space. So we all get right.
08:41
A is when you actually buy, or E, this is isomorphic to. You have this tie. I have a trace given by these ties.
09:13
So this is called a CM tie. So in this case, I've been in variety.
09:22
Is that true? I've been in variety.
09:42
So you can choose. We can assume that A has a good reduction. And OK if you choose K to be. So this is enlarged.
10:02
So this is the good thing about CM. The form of the height of A does not enhance. It's only the tie, phi.
10:23
So this is proved by the convex. So we can put a node like phi here. The convex has a conjecture that this B is almost like this B is a conjecture.
10:48
So the convex is a linear combination of logarithmics that are written here. L prime of sum representation.
11:18
So this is a precise conjecture.
11:22
So this is a precise form for each phi. So far there are no results. So this is proved by the convex itself. For abelian extension E of Q.
11:43
By convex yarn, one degree of E of Q is 4. So basically there are two cases of convex improve. One is by convex for abelian CL extension. Another one is by convex yarn for quadric extension.
12:08
So what I want to do today We're not going to prove this conjecture. We're going to prove the average theorem.
12:28
We're going to prove the following. So the fixed CM type. CM extension.
12:41
Quadratic CM, quadratic. Then we think about G to the G of the summation.
13:02
For some abelian type. Fine, for always fine. So the exact amount of so many is equal to negative half derivative of zero eta,
13:22
zero eta. One to one quarter. Log E of f, dE f. So where eta, I use a notation, but A, there is a quadratic connection to the same E of F.
13:55
But D, E, D, F are not both discriminant.
14:07
So that is the same E of F and F itself. So you have the formula D, E will be E of F, D of square. So it's not important to distribute it.
14:26
So I have two examples. I've got F of P, F. So my error function does not have a equilibrium part. So error F is not at the bottom of their function.
14:54
And the summation of all of this, how about A?
15:03
The A is the idea. Since I find it, I don't have an equilibrium part here. A few remarks. The first remark I'd like to write there is that the formula stated in the complex paper is wrong.
15:21
Second is that many things are the references wrong. So I'm not going to say that. So this is pretty correct. There are many things like 2, 5, whatever. So anyway, now I'll write some remarks. I can write it there. The first remark that I want to make is the reason I want to work on the same.
15:52
Because Jeff, the single man, came to the IAS in April 9. He gave a pretty good lecture to reduce the number order conjecture to the format conjecture average.
16:12
First of all, we don't really have some idea to prove that. So these first things combine with the recent work of Jacobi, the single man.
16:35
But the theorem implies that we all have the same error.
16:50
The single man is first. So the argument is pretty straightforward.
17:00
I'm not going to discuss this part. I think I'll probably write some notes like that. We've just made some announcements by a bunch of people, proving a good point.
17:31
So there's an announcement. I got the lecture by 1.5 in February.
17:50
And maybe someone else did the lecture too. So I am a realtor, not a pussy.
18:21
They at least come through the conjecture. I don't know what's the state of that. So this announcement is in February. So maybe right now they can't do much. Better than that, H of phi, H of phi, you connect to half. I guess in 17th century, we have E of F, zero, and E of F, zero, are modular.
18:56
The linear combinations.
19:00
You want this agreement. E may also be a combination of both two parts. We are not really sure of that. What I'm trying to say is that the proof outlined by Howard is very different than ours.
19:31
And they somehow, using Kruger's construction to correct the bottom of the height, the average by some orthogonal group, which is a really high definition, maybe 2 to the 2t or whatever.
19:49
And as you see from, I could outline in the rest of my lecture, we only use Shumra curves. So there's a big difference.
20:01
So that's the step. Is it 1 over 2t? 1 over 2t. 1 over 2t.
20:36
The proof is on two parts. So we're going to prove two parts.
20:48
The first part is what I'm going to talk about today. It's the reduce of the calculation of CM points on the content of Shumra curve.
21:04
Second part, when you see through music, so if you haven't read my book, then I'll tell you how to do that.
21:31
Of course, I have a manuscript in the end. The manuscript should be ready in a few days. Okay, this is something I'm going to talk about today.
21:44
This is, I mean, I can't talk about it. This is not fun at all. Very technical. So this is probably the most fun part. The key idea of proof is to play with the eyes.
22:01
So it's like a magical cube. You decompose it in a small piece, then reproof them. So that you can handle this thing by Shumra curve. So the point of the eye is the bigger piece. It's a vector bond of a change in dimension g.
22:20
So I will define a small piece of dimension 1, each piece. So this usually is not direct, but actually, somehow, theoretically, you can do that.
22:44
I will tell you the first part. The first thing is that the decomposition is given by a mission line bond.
23:15
I use this notation. We know that the OE acts on this bond.
23:31
So we know that when OE acts on this bond, we have some ties. So naturally, for each tau in phi, we can define the piece.
23:43
The tau can be omega of A tensor. I mean, this is a real variety. So maybe we're doing something and then we write it on a fixed K. The fixed K contains all of E such that A phi has a very smooth model.
24:28
So this case, which has the base, is one part. A equals phi.
24:43
But my construction here should be more general. I've been in a variety, which is given by some ring of integers. So first I define the piece of that. It's the special product of OLK.
25:01
But the best thing is that OE tends to OLK. Not like that. Because the tau can also define the momentum from OE to OLK. Then you check the third extension. So this is the part.
25:22
So you define the piece of that. So this will be a line bond on OLK. And you have a nature of morphism from one line bond to the other.
25:40
So we have a natural morphism from omega A to the tensor product of OE to the phi of omega of A and tau. I can use this one to do some decomposition. But this one does not have a nature of emission theory, emission metric. So this is not going to work.
26:01
What I'm going to do is, well, that's the common sense in mathematics. Usually, if you want to get some numbers, you always have a view of it. So what I'm going to do is, I'm going to do the same thing for the dual eigenvariety. So I have A, the dual eigenvariety.
26:28
I can form the same morphism for the dual eigenvariety. I can form a new thing. So omega A tensor omega A dual will go to tau in the phi.
26:47
But dual eigenvariety has a time. This time is the conjugation time of omega of A of tau tensor omega of A dual of tau C.
27:02
C is the conjugation time. Amazingly, this one has a canonical method. This has a canonical method. I can describe for you. So this individual one does not have the dual of A.
27:20
So this is normal in mathematics. So that's the whole point of why every time you find objects, you always bring the dual in the future. So I can describe this dual method for you. This is probably nothing surprising. That's why we call it a polarization, I mean a variety.
27:42
A polarization of an eigenvariety, I don't remember, is also a Hermitian matrix and your eigenvariety. So if you think of it another way, polarizations bring eigenvarieties to the dual as an algebra. And the different geology brings the Hermitian matrix and your eigenvariety,
28:03
which will induce an eigenvariety in the Hamiltonian model. So there must be some way to do that. In fact, so for eigenvalues, sigma of k, we have the only way to write down the eigenvarieties.
28:24
Remember, if you just write an exact sequence of cohomologies, So we divide by h1 of a sigma of c over sigma, divide by h1, I'm not going to divide that here,
28:40
a sigma 2i i of z. I keep 2i i here to the lower weight, otherwise the weight will go to the next one. Now, this is the picture you have. So the cotangent model of a sigma, of course, is a dual, it's in.
29:14
I have in my, it is a very long, two different dual. One is a dual inside a theory, in algebra, the other dual is just like a bundle.
29:24
So I'm just trying to spot the dual of the bundles. When I say A here, I mean two of the eigenvarieties. So it's not a complete example. Someone has to raise, you have to raise this thing. So in other words, if I put this thing for this one, it's not a linear dual.
29:41
I can twist by telling the twist. So the difference is by telling the twist. And the reason why we know is by Hamiltonian decomposition is h01 a sigma. So finally, Hamiltonian has a complex conjugation, so this becomes omega a sigma.
30:06
There's anti, a form of a form, let's make it equal. So this is canonically isomorphic to this one. So what do they mean? This means there is a canonical summation theory between the differential forms of A and the differential form of the dual.
30:29
So there is a relation, a perfect pairing between omega A and omega O.
30:49
A sigma dual is a complex number. But you also can use it in a more complicated method. For example, I made two varieties of boundary bundles.
31:00
You can use a one-class boundary bundle to write down the values. It doesn't matter, this is much more direct way. And if my eigenvarieties have a complex multiplication, then we can deduce a pairing between a sigma of tau omega a sigma of tau C.
31:26
So this one is the line model I care about. So I have this line model, so I've given them for this one. So in this case, we have a different notation. And omega, this A, tau, is defined to be this bundle, tau of A U of tau C, with this method right down there.
31:57
So this is the method of the conditional line model.
32:09
We created some conditional line model, morphism of two conditional line bundles.
32:30
So there is a construction, there is a morphism of two line bundles, omega A conditional line bundle,
32:56
U of the tensor product of the tau omega of A of tau.
33:08
So this is the conditional line bundle. We have two conditional line bundles, so we can do something. So the first one, we proved the theorem. So the theorem is proved in the paper.
33:21
It's not really hard to do that. So in fact this is basically, the height of this thing is actually 5. The height of this thing is also actually 5. Because the ball in the height depends on the time. Not only depends on the time, because when you compare the ball and the height, you are really more than a number. So it's really conjugated class all the time.
33:41
So I mean, the dual ability variety is definitely the type of IC. It's the same kind of IC class. So if we compute the height of A and the height of A, we really get the same thing. So it has two measures, right?
34:01
Well, let's find the number with the metric. All the other metrics. Yeah, but there are different metrics. Different metrics. So first you give a number with the metric. Yes. And then you use this error to get the metric. Right. So I have the same kind of morphism is isometric.
34:24
So let's define isometric. Let me define some notion out there. Let's give a definition. It is quite old.
34:45
A is going to be the one and a half of the degree. Of course, I need to normalize it the same. Degree, maybe. A cubed. Degree of A cubed.
35:04
Right, because I write the twice. Yeah. One proposition I proved. I mean, our purpose is this definition. We are having weak decomposition. We pick up a different variety of decomposition.
35:20
But in fact, this decomposition depends only on the type. It actually depends only on the type of two types. I have a 5 and a 12 inside here.
35:43
So I can write down this one. I write 5 to be the sign drawn with 12. So I can define this one. I can rewrite this in. Rewrite this in. It depends on H, cosine, and so on.
36:04
So cosine is the same type, but only has t minus one element. 12 is the one for them. So again, we have the first popular decomposition formula for the size of a small piece.
36:23
So there is a theorem here. For the horizontal height, H of phi minus summation.
36:40
You can multiply the two pieces by 12. And the cosine, you get H of cosine of 12. This one actually is pretty easy to calculate. One over E phi over Q and log.
37:02
Discriminant of phi, and discriminant of phi C. So this is a review. Discriminant of phi. Discriminant of phi is a trace map.
37:20
So somehow you can define whether it's like linear map. So you can define discriminant of the trace map. But you cannot define it globally. You have to define it discriminant of a time, a monopoly. So we are going to learn that.
37:42
I said that to prove the theorem, I first need to decompose the horizontal height of a small piece. Now I want to regroup this piece. Regroup the piece I defined. So for the horizontal height, cosine of E of degree G minus 1.
38:08
As we do here, I define the height of cosine of H cosine of tau plus H cosine of 12 R.
38:24
So this one is the two CM types extending the horizontal height. So again, what we really proved in the paper is about the partial type of the theorem.
38:50
The partial type is a very funny thing. I have a CM field. I check only the partial bit. Then we can show that the partial type is intact.
39:05
That's not the kind of type at all. So it's given by one of the two teams of L prime of F, L prime of 0 eta, of L of F, 0 eta, plus minus 1 of 4G.
39:25
This is minus 2. 4G log E of F. So this is actually pretty surprising. Because, I mean, the common adjective of average, all the things.
39:47
We can use all types. We just take one type, which is composed of small pieces. Then we take the partial type, add a little bit, we get the same. So this formula is very similar, of course, with your price, common adjective.
40:02
But 4G does not imply this one automatically. Because we said use all types. We said use four types only. This is the same, but I realized in fact of the proof.
40:23
Okay, so this is the first step. I manipulate the bottom ice. I cut a small piece. I go to them. Nothing different. I will use seven things I can compute.
40:47
So now I come to unitary. So somehow, remember, this H plus sign is almost like H.
41:07
H plus five, I mean two types. H by one plus H by two. Let me see.
41:20
Let me try it. If I remember, this is the definition here. You take the sign, two types. Somehow, you can compute. You can compute the sign by a single linear variety.
41:43
A0 is equal to A by one plus F by two.
42:03
So the phi one, phi one, phi two, the final type is expanded. Also on this one, you have a morphism by tau of E.
42:25
So in fact, this will be true. So the final distance, H0, is exactly equal to H of A by one. I'm sorry, tau phi A plus H2.
42:41
A by two, tau phi A. So this gives us a lot of flexibility. It's not only H plus sign. H plus sign can be independent of the time you choose. Also, you can compute it by any linear variety, which is actually the same.
43:02
So the problem, what I'm going to do, is to have a very special linear variety in the workarounds. So I fix, I fix the, I fix the base, fix the tau at the real number.
43:26
And the B of Poisson's algebra, the moment you communicate, the B is the tau, and so on and so forth.
43:49
So again, the tau will be taken out of all the places. Then I assume that the E has invented the B, right?
44:02
So in this kind of B, there will be E plus E times J, T of X, equals X bar over J, J squared, see that? We have this model. So then I pick up an old B, a maximum, content of old B.
44:34
So in this way, I can form a linear variety by the forming method,
44:46
like A zero to B, some such structure I'm going to find later on, some such structure, divided by the old B.
45:00
So the first structure I put on this thing, the B, there's a Q of real number, and two keys for the E. Sorry, I guess it has B cross. Yes. Yeah, no, I agree. So I have my polynomial algebra on the two keys.
45:23
One piece is in this part, another piece E over J, and the real number Q. I'm going to put a factor structure on this polynomial, such as that this one has a type, type one, this one has a type, type two.
45:43
But recall that both phi one and phi two are CM-type, extended by partial CM-type assignment. So I get this linear variety. So this is the one we're going to use. But this linear variety we'll define.
46:01
There will be a special point. So this A zero is corresponding to a point P inside. I write X nature, because I have something else for nature.
46:21
So this is a Schrodinger curve, universal Schrodinger curve. To define a Schrodinger curve, I have a picture here. I already found that there's a Hermitian method here. Remember, B would be B with plus sign E.
46:43
This is the Hermitian norm. The Hermitian norm is defined by the following method, by E, A plus B J, C plus D J. Because I made a Hermitian, right? So you can take AC bar, E minus B, D bar, A squared.
47:01
So this is E, E. So I define the Hermitian norm. Once the Hermitian norm, you calculate the unitary group. I actually now calculate the similar unitary group. V, I only care about the kinetic component. So this actually is given by delta F cross,
47:21
modulus, P cross, and E cross. And this one acts on E is given by the following. B, E, and X is given by R. I mean, this, my B is a left, E minor,
47:40
so I get E, X, B inverse. So that's the action. You check this action. That's the unitary, the Hermitian space of E of dimension two. So I use this one to write down the Schrodinger curve. You said that B was unramified at all points but one.
48:10
I think you made that. You said that B at all Archimedean places but one. One Archimedean place. It's unramified. One Archimedean place, a town, a place. That's not what you said.
48:22
I find a place I don't care. So that's not what you said. Is that what I said? You said it was ramified at one place. It's unramified. Yeah, it's pretty hard to find a one.
48:41
I hope I create a one like that. So I'm not going to... That's what I said? Okay. I found all the clears. These are these nodes. I'm not really sure what I was going to say. Okay. So the structure H, the convex structure,
49:05
T of R is the Schrodinger data that H, T of R. So in this case, we have found a Schrodinger curve. It's basically as I said before.
49:22
Maybe it's a K. Sorry. And this is C. Let me write down. Give a name. Call it by G. Okay, by the G there. G of Q. Actually, in the convex conjugating, class of H is actually as a model to the congruent of a bank,
49:43
the G of Q bar. So you get this thing. So this is actually a pair of ties. But the T-R ties are in varieties. You have an identical variety. You have an amorphism.
50:02
You have the polarization of level structure. So this H is available variety. Of course, this is the one over the reflexive field. Over there. So, I mean this.
50:21
So some skin over E reflexive field. So this is a variety of dimension. Let me try to calculate. This is 2G.
50:42
You have, you have your form over E with an amorphism of A over S. So this is going to be a tie. My one plus my two, I really think that will be a popular case.
51:00
So I'm not going to write it out there. Because this line is best in the case of amorphism. Do you make sense? The longer will be some correlation from A to A dual.
51:21
I'm not going to describe. I mean, I can do everything probably in the kinetic. So this will bring the velocity from motion to the grain. We will give the reflex and E cross. We will give the conjugation. Then you have the level structure.
51:42
So this will be compatible with Psi E. But Psi E is a summation of then you have to define Psi Q of x and y to be the face field Q x squared to lambda x squared to y
52:00
to put the face field Q of lambda x, y, y. So then E will be f squared to lambda x, y, and b. So you've got a synthetic form there. That will give you the polarization of this thing.
52:23
Great! So there's not so much time to talk about this unit version of writing. And the reason I do that is because I want to make a connection
52:41
between these omegas. So that's A. A. A. But A over S. We have A is equal to M. A X is the universal of even variety.
53:03
So I have to find the line for M to be omega A X tau times omega A X U or C. So now the question is
53:20
you find the extender. One extender. This end. And write this end again. Two integral models over O of E
53:41
of this end. But unfortunately, I don't know what I'm going to do. You will have some trouble to study the integral models with the integral curve. And you'll probably get the integral model and it'll be smooth. But fortunately, the integral curve is
54:02
much better to do that. What can you do is I will do it. I don't do this end. I will do it. I don't want to work correctly by a modular problem. So I will not work on
54:22
the Poissonian integral curve instead. The reason is this integral model may not be regular. And this case is regular
54:41
if you choose the level very well. By the work of Twinter it's not really the case for the old Twinter case. So I will not be able to talk a lot
55:02
So now I will just write down how to construct this Schrod curve. So I have a Schrod curve X defined by Poissonian method of u.
55:20
So this is a Schrod curve. And this Schrod curve is parameterized as the height structure and b is the real number. But it's a type that has a height structure. You have a direct sum of b i, b i is equal to b times the real number. The f of tau i.
55:42
So this has a height structure of type of negative one zero plus one zero and b one of zero zero So presumably this is not
56:03
there. This is not a height structure for a linear variety. So what I'm going to do is your tensor to rewrite this your tensor e to the real number
56:21
e to the real number We have a height type Poisson so you get e one, e two e g Here you can use Poisson Here you have to type
56:41
zero zero So that's amazing. That's a height. That's a Poisson type. But e to the real number this is a cell that has a space of e So you can rewrite this as a tensor product of e to the real number
57:02
e to the real number e to the real number So this is what you're going to type How do you type y? y is the same So it's h zero This one is h one This one is the Poisson type
57:22
So in this case you get a morphism from two products, two Schmula curves y to x nature So this one is a Schmula curve a Schmula variety dimension zero defined by Poisson
57:41
So in this way you'll find the error by the plus sign So then the integral model of s will give you the integral model of s this one So finally we'll reduce the computation
58:02
to the following texture then rewrite it by a theorem Then the combination
58:24
Assume final place modified to both
58:40
e and b So now what do you have? You have many ways to choose b Then there's h plus sign h plus sign is equal to hl of p
59:01
So this l is a Poisson model r is an integral model of the Poissonian Schmula curve p is a special point fixed by the integral
59:20
So this is what I can say is y, z, z the Groziger type formula No, this is not Groziger So we do this part Then, what's left? What's left is the Groziger type formula This is, no sorry I'm too busy, okay
59:43
From here, already get h plus sign of h of p and here we rewrite the formula by negative l f of 0 beta
01:00:02
LM0H, minus plus one-fourth log EV over D of E of M. So this is what I've done in the YCC formula in the Roza K-type.
01:00:22
So finally, that's the formula I've reduced to. So we reduced to some calculation. And we should continue to mark that. Once we continue to mark that, I will not have time to describe that. Maybe, okay, I'll make it in two minutes. So I'll describe that.
01:00:40
What we proved is here, it's the same method that we've proved. You should go zagging one way where you have one side, and then you have kernel, other side which are left to your kernel. Then you can get a difference. And when we proved zagging, what we do, we're trying to show this difference completely to all custom forms.
01:01:00
What we proved is that the code is connected. Of course, in that case, we don't care. We proved the code is connected. In fact, if you only have one kind of custom form, you throw a lot of jumps to the track. You don't care about that. One more time, the code is connected.
01:01:23
You have to bring all the tracks back. So we brought another 50 pages to deal with the tracks. We started recycling all the tracks. And the paper, all these are YCC papers, all these are mass studies. So when you get the paper,
01:01:40
a lot of the tracks you can bring back, then give your code is connected. Okay, so that's it. Let's start with Terez. Do you have any questions for Terez?
01:02:12
Okay, do I have any questions for you? So you use an integral model.
01:02:22
So any integral model you're doing, or you have some special model? So I need an integral model to handle the transformation. You remember, I mean, the one I'm doing is the differential form is naturally 50 inches
01:02:44
of the Spenser map. So you need to have a fully integral model to handle the Spenser map. And unfortunately, I usually remember, unless your key is a good friend, you will not get a good Spenser. So there's some error there.
01:03:01
So that's the kind of estimate. I don't believe ever that you'll get a good Spenser. For arbitrary ones, you have such a problem. But at the convenient Schumacher, what do you do? You don't have, I think, a variety, but you'll have all the TDPs for groups. So you still tend to find that the hydrogen combination
01:03:23
is just one connection to study the Caler Spenser map. So that's one other reason I moved from So in other words, somehow, I constructed an integral model for the use of Schumacher,
01:03:41
but there's already a problem. Even if you have a regular and a total F, you can change the TDP nature to be singular. So I really don't know. I mean, anyway, it's much better to work on the container Schumacher to study the Caler Spenser map. Also, another weird thing is the Caler Spenser map,
01:04:00
in the literature, most of them are wrong, because they, I mean, some literature, they messed up between the cotangent model and the cotangent model, because there's a problem of the thing. So we really do very carefully to study the Caler Spenser. So that's the reason I needed that.
01:04:23
Thank you. Okay, so finally, any questions from Beijing? So you proved the average case,
01:04:42
so any recovered Caler Spenser, are there any examples of any recovered Comer Scrupper by using this group structure? So you are asking what is my next project? Okay, because you have an actual group structure. Okay, so I believe you need, I mean, I didn't find that. If you work on U21,
01:05:01
instead of U11, you should get everything. But that needs a lot of work, because first thing, you needed to, I mean, you needed to work really hard on our theory, and it should not be kind of bogus, and our theory says we needed to solve some,
01:05:23
one question, which is, I still don't know how to solve it, is construct a harmonic form, I mean, a k harmonic form, on a line bundle. So if your line bundle is a
01:05:41
square of curvature, it's like a club, the new thing is I'm using the U and I theory, and the existence is not very clear. So that is done, then we have some hope. You see, what I do is,
01:06:00
do you remember the combination of Psi, To, To Ba together? If I have the unitary time U21, what I have to do is H twice To. So if I can do this one and the unitary varieties on the unitary of U21,
01:06:22
I should be able to compute the formula H To Psi twice plus H To Ba Psi. Right? So I get this is the U21. And this is the unit, the unit one gives me formula H To Psi
01:06:42
plus H To Ba Psi. So you see, I can get a different blanket. So wait for next year. Any more questions?