Height of motives
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Transcript: English(auto-generated)
00:00
Now I have a great pleasure to introduce Professor Kazuya Kato, from Chicago University, and he will talk on height of motifs.
00:20
Thank you very much. Hi. So the Artemus defined the height of Arboretum varieties,
00:44
the number of Arboretum varieties, the number of motifs. He defined the height of the Arboretum variety,
01:01
and then the Arboretum variety was numbered, and he proved the state conjecture for Arboretum varieties. And then, from this result,
01:24
he obtained the Arboretum starts like a koncharevich conjecture, and then he proved the model conjecture.
01:43
That was in 1983. And I propose a definition of the height of motifs, motif over motif over number motif.
02:04
And then we can imitate the argument of the varieties, but still the state conjecture cannot be proved,
02:20
because there is some finiteness, finiteness of which we cannot prove for konuchis, but still we can imitate some part of the arboretum. So I hope to explain some of these things.
02:46
So, first I review the definition of the height. So the height is defined, so for simplicity we assume that t is a rational number with q.
03:07
Then the definition of height is the following. So let omega-a be the s of the financial forms over a,
03:27
then this is the isomorphic to g-repair, g is the dimension of a. And then, inside here,
03:41
he has a space of integral differential forms, that is, here a is a narrow model of a. So this is an integral version with,
04:03
this is over the group schema over z. And then he has this space which is equal to zg. And then take integral basis,
04:22
this is z basis, omega-1 to omega-g of integral space of differential forms. And then height of a is defined as,
04:50
you know, this is the definition of height.
05:06
And I hope to imitate this definition. So, by mutual, so I assume that that is so written, we consider that we generalize this to mutual.
05:33
But first I first interpret this by using the
05:46
first the omega, this omega-a is a part of p-1 in the graph. So first we interpret that,
06:01
we interpret the a-omega matrix,
06:20
that is, so first, omega-a appears, this omega-a is
06:43
p-1 of the Houdini filtration, this is omega-a, and f-2 is z. So this f is Houdini filtration.
07:10
Then this omega-1 to omega-g is, this is x to the power basis of,
07:29
q-base of determinant, this is x to the power of omega-a.
07:45
And then that omega-a, in fact, c-base is of omega-a. And so, omega-a is this, yeah.
08:06
And then you write this lambda, then that integration, we have here, this is the metric,
08:35
metric defined by using some,
08:40
the metric defined by Cauchy's Theorem. And so, what do we have to, yeah, this is the drum Cauchy appears, and the Cauchy theory appears, and the product, some integrality should be defined on omega-a.
09:04
So we do the similar thing, so the, to be the pure-motive one, by pure-motive one, I mean, that the H-e-m-x-q-r-t,
09:28
yeah, this is t is, this is some q-order, x is smooth projective over q.
09:42
And then, and t is, t is z-hat lattice, z-hat lattice H-e-q-r-t,
10:15
So, I understand,
10:21
this is some kind of integral, the motive over q, pure motive over q, with z-coefficient, I just thought of the stable and favorable action, stable and favorable action,
10:46
with z-coefficient. And, for that,
11:05
we usually take, in the theory of motive, pure motive, we take some direct sum and all this, but for simplicity, I just, here, I just take this as such pure motive.
11:24
In general, we can take direct sum and all this, but for simplicity, I discuss about this more, pure motive. Then, we define the height of this into this.
11:40
First, so then, we have to consider such origin. So, we have to lump the, the lump of this artist, yeah. So, we have a lot of filtration here.
12:03
And then, q is the determinant of the guru,
12:23
and then, q is the positive filtration.
12:42
So, you shift by art? Yeah, this is, yes, in fact, the height doesn't depend on the twist, so, yeah, so, the natural way is that you shift by art,
13:02
the usual positive filtration, yeah. But actually, the, the twist doesn't change the height, yeah. And then, so the, the point is, this, we take some, since art,
13:22
art's power, so the, the eye contributes by the, by the eye, yeah. So, then, photometric can be defined, is defined on,
13:42
on LQ, by, by, by Hoge's theory. That I can, I can explain. And then, LZ, LZ inside LQ, this is, this q is asymptote to,
14:01
q is asymptote to, q and LZ is asymptote to C. This is defined by using, by, by periodic Hoge's theorem. And then we take the basis
14:21
of, of FZ, then, then the height is defined to be the inverse of this. So, what I have to,
14:41
so I hope to explain in this, and then how, how we can imitate the method of photonics, yeah. So, before that, then, some, I hope to tell you about
15:01
geometric kernel. Yes. then the number field, number field, of course, is similar to, to the
15:20
compact linear surface. Or, or function field in one module, compact linear surface, maybe, maybe the function field in one module over C.
15:44
Then, the multiple, multiple over k, maybe, and then this corresponds to, to compact linear surface.
16:06
Of course, we can consider multiple function fields, but, maybe in analytical theory, it is nice to think about the variation of the structure, maybe pure, pure motif, and the variation
16:21
of the structure, over, over, over, compact linear surface, x, then, variation of the structure. Then,
16:40
then the height of h is then the h, hm should be similar to the following thing, the, the sum of r times
17:06
r. Here, the variation of the structure, each can have similarity.
17:26
By the theory of linear, even if the Hodge filter, Hodge structure has similarity, we can, we can define a vector bundle called
17:41
bundle on X, the canonical extension of, of green, that is, H0 can be defined naturally on the open set of X, where H has no similarity, but that extends canonically to the
18:00
calling X by the theory of canonical extension of green. And this, because this is a vector bundle on the group r, you take a, this is a Hodge filtration there, and then, then you can take a group r. And then, that is still a vector bundle, and so, so you can take one degree of the vector bundle, and then you take this r, you
18:21
multiply it by r. So that is, this h like robo hm is similar to, to, to, to, to, to, to this. And this, this most, this thing was already considered by, by, by, by
18:42
by Lewis and Peters and such people are. This thing is, is in some studies.
19:00
Yeah. So that this, this one is similar to, to that one. Then I, I hope to now, now I hope to explain
19:24
integrality here, integral structure, my present definition is not so beautiful, and so I, and then too much complicated, so I just, I mean, I just explain the love story about this.
19:40
Because, ah, ah, same, same. We don't, we don't hear you. Okay, please.
20:20
Ah.
20:21
Ah, ah. Yeah, yeah, yeah. Ah. I am talking about, ah, yeah, yeah. Yeah, yeah. Yes.
20:54
Maybe, may be, but one thing is to define the height without assuming that it is semi-stable.
21:03
So in the actual argument, he always assumed that the abelian varieties have semi-stable reduction in his main argument. So I am not sure whether the definition of the height for non-semi-stable thing is nice
21:35
or not. I am not so sure about this. Probably there are good ways to define height only
21:43
for object feature, semi-stable reduction. I am not so sure about this point. So then I asked the definition of metric or LQ. So then I assume that this is in LQ.
22:21
Then it is in LQ, by Hodge theory, then C tensor, H-dramis, and this one has an integral structure
23:08
coming from the T. I will repeat the twist for you.
23:21
Maybe you have to just use your heart to see. So then if you also created quotients of this H-dramis, then this is isometric to
24:08
maybe, actually, I use the H-r in minus, and this is the, this is the T tensor, R
24:38
and N-dramis, and this is the H-r.
24:46
But this bar is… You want R or… R is in the H-r. Oh yeah, yeah, I was using the H-r. Maybe the H-r is confusing with the H-r,
25:02
I use the complex number, I think. Yeah, I remember, but maybe the tataru should be also I, isn't it? And so this bar is taken with respect to this rational structure.
25:20
So this is the usual thing for Hodge theory. And then, so this is the element of C tensor Q of the H is I, H-I, N minus I,
25:46
the determinant of H-I, C, N minus I, tensor I.
26:01
By this, there is a canonical isomorphism. The map is probably from this side to this side, but by Hodge theory then you have isomorphism as well. H-C, H-C, if you write H-C, H-C is like sum of H-I, C, N minus I.
26:28
This is a Hodge decomposition. So then this bar is inside the determinant of H-I, N minus I,
26:51
and then the bar changes the I and G, and this is the indices.
27:02
So then you have the bar is inside the determinant of H-C, H-C, N minus I, I, C, N. And if, yeah, tensor I, tensor I.
27:25
So to replace this I by N minus I, then this is equal to the determinant of H-I, N minus I.
27:42
Just I is replaced by N minus I. Then you take the tensor product of this then you have I and determinant of H-I, C, N minus I.
28:06
And then you have I plus N minus I, so you have N. And then, but then this is the tensor product of determinant is determinant of the direct sum.
28:27
So you have this. And so then you have the positive decomposition tensor called the direct sum of the tensor called H-C. So, yeah.
28:45
Then H-C, you have an integral structure. So H-C is a, H-C is a C tensor with an integral structure H-C. And so you have, this is C tensor determinant of Z, H-C.
29:09
And then you have tensor N. Now here you have the basis, Z basis here. You have Z basis here, Z basis.
29:20
And then, then, then now you have, then now the, the, the, the, this, I mean to decrease it. If I need to read up, just this tensor is divided by N.
29:43
You can write it. Here is a complex number, so the, the, the metric is divided to be, to be, to be the square root of this.
30:05
So this defines the metric form in L2. Now, now to define the integral structure. And my, my definition is too much complicated at present.
30:25
So, so I am, maybe it is not so good to, to explain what it is because I, if I start with us, it is then that we will finish my, my, my talk. So, so, so, yes, so, so now the integral sort are co-affirming with Z, then,
30:50
then, then we have to define, it is this, the, the such piecing intersection with Z piecing.
31:09
So it is enough to define this P for each P, P prime. And then, now I will use the periodic So, so by periodic theory, the, the B-dramed tensor, periodic theory tells that
31:28
B-dramed tensor H-dramed in X is isomorphic to, to B-dramed tensor, tensor, tensor H in Q.
32:03
The integral structure here, the integral structure here, so I take P here, T is, T is, and so, so, and that is, then if you take
32:27
the, the d-dramed vector, that is d-dramed vector is, is over Galois stable path, right? So you take a Galois stable QP path, next path, next path.
32:50
And then now, now you have a QP tensor H-dramed in X is, is equal to this, by definition it is B-dramed of this V-P, V-P is QP tensor.
33:14
Now, the, the d-dramed just, just defined to be the fixed part of the whole series.
33:24
The, the, the d-dramed is, the d-dramed is the Galois variable.
33:47
It is not clear how, how we can define our integral structure by using the integral structure T-P. And so, in my opinion, the integrality in the periodic Ho Chi theory is still not in the best situation.
34:06
And so, we wonder how to transport the continuous objectivity to the continuous objectivity. In a good case, the Compton-Raphile theory, if the Ho Chi filtration satisfies the so-called axis of good structure,
34:42
reaction at P, and the Ho Chi filtration has been chartered, if A is Ho Chi long, and if A plus P minus Y is zero, then for such a crystalline representation will be, then you have a function D is defined.
35:25
And such a thing is exact for such things. And then the Ho Chi, this is the Ho Chi filtration, the Ho Chi filtration on this side defines the Ho Chi filtration on this diagram Vp.
36:03
The integral thing is defined. And then you can use this integral structure to define the LZP. LZP is just the LQ.
36:23
In fact, you can define the integral structure in this way.
37:20
But in general, if X has a bad structure P, then it is a very delicate problem to define an integral structure on the diagram side. My general definition is not so good, so I probably it is better to move
37:41
to the LZP, how to imitate that, to explain how to imitate for things like this. I have some pre-printing in archival in the QP of Gangaraku-Mochi and Vp.
38:35
And then because you have an integral structure in Q1, Qp, Tp, all by using this integral structure,
38:44
then you can define some integral structure by using this integral structure here. So I defined the integral structure by such method. And in fact, if Vp is E of the abelian variety, then such case,
39:10
then the vertex definition, the result coincides with the definition of the integral structure.
39:22
But I think the precise thing is too much complicated around here. So the method is roughly like this. My present method is roughly like this. But I imagine that there is a better method to define the integral structure there.
39:45
My definition is too much complicated, too strange. So this is not so good to define, to study the family of motifs.
40:02
So this works, my method works only for individual motifs. So if you consider that you like to study the family of motifs, then my definition method is not so good. So I hope that there should be some good method which works for a family of motifs, or a family of variations of peer recovery presentations.
40:30
But I'm sorry, I stopped this. And then I explain how to imitate things in his proof of the data conjecture for homomorphisms of abelian varieties.
40:57
And then I explain, we can imitate some part.
41:02
So he had two things, that is, to prove that, one thing is that he has some AM, isogeny to AM, then the AM is bounded.
41:38
And another thing is that the abelian variety's height is finite.
41:56
Here, in fact, he is dimension G, and G is fixed, and he assumes that his semi-stable texture will be there.
42:12
And what I mean is that he has such a thing as this here. And the reason for, this reason is that his height is, his height is essentially equal to the
42:29
height of the, this is because, yes, I think, yeah, yeah, so I hope to explain a little more.
42:51
That is, yeah, that is, yes, it is because it is essentially equal to the height of the, yeah, yeah, it is the point of the,
43:28
You think some polarization? Oh, yeah, polarization should be fixed, yeah, yeah, yeah, so everything should be fixed, yeah.
43:41
Therefore, let alone that if you have a large variety, then on that space, then the rational point, if you pick bound, you have, if you pick some C, then there are only finitely many rational points on the large variety.
44:08
Of the height smaller than C. So, this is not equal, but similar to this. So, Hultings proved that his height is almost nearly equal to the height of the point A, which is a point of the marginal space of the varieties.
44:30
Yeah, so by this then, he proved this. And Hultings, again, is that he needed to prove that, he needed to prove that if you have an exact sequence like this, and then this is Galois cube.
45:00
I am now assuming K is cube for simplicity. W are stable, stable subspace of AP. The key point of the proof is, key point of the proof of Hund.
45:24
For example, the T-conjecture of the Abidiam varieties, the T-conjecture is that the Hund is over AP, Galois cube,
46:12
And this is adjacent to the case A is B. So then the key point is to show that for such things there are some Fs in QP and A such that A is double.
46:42
And then from this then consider the integral version TPA, some DCP version.
47:05
Then the TPAN is an average variety which is isogeneous to A such that TPAN is B to ZN TP A plus WZP.
47:22
So then you have a modulism of LAMO. Prove these two things.
47:41
Then by this then there are only finitely many isomorphism classes in this N. So he has many many isomorphism things. And then the isomorphism, he found many many because there are only finitely many isomorphism classes.
48:12
So assume that for simplicity this is by replacing A by AEN1 by A then assume that all those are isomorphic to A.
48:35
Then you have a isogeny from A to A and you have AEN.
48:46
And then by replacing the set of AEN1 by some infinite subset of AEN1 then you have a limit.
49:15
This is the endomorphism. By those you have many many isomorphism and then you can find many many non-trivial isogeny.
49:24
And then you have many many non-trivial elements in N to A. So here the problem of BP is assumed to be A here. And then the problem is that you should find many elements of N to A but the first thing is in this way.
49:44
This is a compact space so if you replace the set of AEN1 by some infinite subset then you can have convergence.
50:01
And so then because this A is such a thing then N goes to infinity then this becomes zero so the DPA becomes WZP.
50:21
Of course the limit of this is WZP. So that was his result. And so then I cannot imitate this part but I can imitate this part.
50:44
This part is because here we need to use the fact that there is a mutual space of an abelian variety. But for motifs there is no mutual space in general motifs.
51:03
So this part cannot be imitated. I cannot imitate this part but this is okay. So then I hope to imitate that part. So the consequence is that if we have such finiteness for motifs then we can prove the data conjecture.
51:24
The data conjecture is that, the data conjecture is that the ZP is equal to GQ.
51:51
So if M is H, M, X, Q, R and T then this P is just T.
52:09
So the consequence is that, I hope to explain how I can imitate this part. So the consequence is that if such motifs have high finiteness,
52:36
such motifs with fixed huge numbers, G over Q, in that sense, such motifs with integer structure is finite,
53:00
then this data conjecture is true.
53:22
So I hope to explain how we can imitate that part. So assume that M is like that and we are given a projection U. And then you prime a curve stable, curve Q bar Q stable.
53:53
And then M is just equal point of P.
54:12
And then the T, M, N is H, M, X, Q, R, T, N.
54:29
Here T, M is a kernel of U, P. And then T, P, N is equal to Z, M, N in the same way as us.
54:57
So then, assume that T is of good reduction at P.
55:24
And then we add some A, such that if A is the total in each drum and T minus 1 is Z. Assume this, then what we can prove is that this is just similar to the proof of Hulkin.
56:08
The proof is the proof. So we have exact sequence. So you have an exact sequence by the theory of Hulkin.
56:47
You have a gradient quotient. So then, this is a minjum.
57:39
This is a weight.
57:44
And then T is sum of R, R, gradient, R, T, drum.
58:05
We have this. This is true for any such B. B is a global representation.
58:20
And then for global representation, which is weight M and which is drum at B. So weight M representation, then you have basically T.
58:41
Reduce it to that case. Because this is V is equal to delta V. T of T is equal to M for sum M.
59:08
Because this is a one dimensional representation, so isomorphism. Over, over, sorry, over some number.
59:29
You can then, in this case, then S is determinant, V is determinant.
59:41
This is a Hasundo Newtonian. Then, confusing, so die, die, die, die. Then this is minus M and T. T is also.
01:00:00
Yeah, yeah, that is, so then this is Sachi, you have, so this is Guru Ai, multiplied by Ai, works very well here. And so then, by this then, now if you compare issue with height,
01:00:31
then it is equal to N T V, divided by T N S V, and this is N1.
01:00:45
The reason is that if you compare these two, then on the Hoch side, then you define the metric by using the integral structure here.
01:01:01
But that changes a little. And for the definition then, you define the difference. So the index is equal to, sorry, the index here is equal to, index is equal to this 2 times, 2 times, 2 times T V.
01:01:55
So this is, this is, oh yes, yes, because the difference comes from V,
01:02:04
so the difference comes, the difference in N and N comes from the, yeah. So H Z divided by H Z, N is equal to U divided by T Z.
01:02:21
And so from this then, the length of this U, and the tensor comes from this way. And so the metric, the metric, and on the other hand, the metric difference of the integral structure comes from this U.
01:02:52
And then you are with, then the difference of the integral structure is like this.
01:03:01
So the, in the definition of the height then, you define the integral, you use the integral structure and the metric, and the integral structure changes by this, and the metric changes by this, so that they cancel, so that there is no difference. So this is the proof of that independence.
01:03:24
So then, if you miss this minus then, by the same argument, I have to delete that. Ah, sorry, I have to stop here. I did not explain the height of mixed motifs, and I could not explain how this story is related to the story of the destination, which structures.
01:03:58
Ah, the, the, the, the, yeah, the, if you consider as a family, and
01:04:09
then, then, then, just to finish this, the motif over, over curve over longer field,
01:04:29
then, then, then the formula is that the, the, the, the, each is the log of the height.
01:04:42
So at the, at each point, then, then you have a specialization of the motif, and then you consider the height. Then this is something like, like the, the geometric height times the, this is the average formula.
01:05:09
And, and the, yeah, the, the, such a thing can be proved in, in certain cases, ah, for mixed motifs. And, and the, the, such a thing is, and then, again, the, the, and the, and the, and the internal structure, I don't know.
01:05:33
The geometric contributes to the, the, yeah, the, contributes to the, the Archimedean part of the height.
01:05:43
And, and then, ah, ah, to prove such a formula, then, then we have to study the, the designation of the, the structure at the point of the singularity. Ah, I'm sorry, I have no time to, ah, I'm so sorry, sorry, I have no time, I have to finish, yes, yes, yes.
01:06:14
Questions, ah, first, ah, from Paris.
01:06:26
Can you compute the height of the motif attached to an eigenform? Sorry, sorry. Can you hear me? Ah, I'm sorry, yeah. Yeah. Can you, could you repeat the question? Can you compute the height of the motif attached to an eigenform?
01:06:45
Ah, motif, you mean motif. That is a hard question, yeah. And, yes, in the language there is a big progress in Lang Lang's correspondences.
01:07:05
So, ah, the such thing should be studied by, from the side of modular forms, I think, yeah. Yeah, but I never studied such thing, yeah. Yeah, yeah, it really is very nice to, to think about the, the automobile form side of this, of this talk, yeah, yeah.
01:07:29
I, I have no idea about this, yeah, yeah, yeah, yeah. So, I have a question that I think is mainly about recalling all the facts. So, in the case of paintings, I think it was known before that the Galois representation is semi-simple, if I recall correctly.
01:07:49
Oh, yeah, yeah, yeah, yeah. Is it? Yeah, I, oh, yeah, I assume that the motif here, the pre-emotives are colonizable. So, ah, I think the semi-stabil, the stability follows from that thing, yeah.
01:08:04
I, but, but, semi-simple, simplicity is, I think the, is the result of the, I, yeah, I, I once thought about this and then,
01:08:23
simplicity was okay, I, because we can assume that the, the motif has, we, we assume that the motif has, has polarization, yes. Then, yeah, I, I think that is okay, yeah. Sorry, I, I cannot tell you the reason for where we are.
01:08:44
So, another question related to, to this Tate conjecture. So, of course, there is the question of semi-simplicity on the side of motifs. So, Janssen proved it even more logical in numerical equivalence, I think. Does it follow from the Tate conjecture that you, that the homological and numerical equivalence are the same?
01:09:09
Maybe it's well-known, but can you recall what is known about… I am not social. I, I am not social, but, yeah, yeah, yeah, sorry, I, I don't, I cannot answer really about this question, yeah.
01:09:34
Yeah, semi-simplicity is a question, some problem, yes.
01:09:42
And I, yeah, so, I am sorry, I, I, my brain is not working so well. So, I hope to think about it later, but I am sorry. At present, I cannot answer, yeah, yeah, yeah.
01:10:10
So, any questions or comments from Beijing? So, are, are there special cases where the finalist assumption is satisfied?
01:10:22
Ah, that is so. If the homogeneous feature have some modular space, then, then that may be okay. But for other, other cases, I am not social. Yeah, yeah, I don't have any, any, any, anything about the finalist, yeah, yeah.
01:10:44
I, I, I have no idea, yeah. Yeah, sorry, I, I have no, no, no result about the finalist, yeah, yeah, yeah. Questions?
01:11:03
So, there are no more questions from Beijing. Any questions or comments from Tokyo? So, you can formulate analog of Voita's conjecture for, for what use? Ah, yeah, some Voita, the, the, yes, the, the, the, that is the log p, log p here, p,
01:11:36
m has a bad transaction at p. Ah, then you have to compare, compare the, the height of n.
01:11:47
And then, then the Voita's conjecture, for multi-version, the, the multi-version of the Voita's conjecture is something like this. Ah, but, ah, we have to put some coefficient and it is not clear what, which, what
01:12:02
is a good coefficient here. Yeah, so, so, ah, the questions are here, this here. What is this here? Yeah, the, roughly speaking, the, the, I don't know, p, p, p, p, p, p, p, yeah. The, the Voita's conjecture is, is, should have, for motifs should have such shape,
01:12:25
ah, yeah, roughly speaking, yeah. And, and what should be e, e or what you think should be discussed, yeah. And, and, yeah, some, and also a geometrical analog of this should be considered to,
01:12:40
to, to decide e, yeah. Ah, I have some, I, I, I made some about, about this, but, but yeah, some, some study can be done here, yeah. Yeah, yeah, yeah.