Integrality of p-adic multiple zeta values and application to finite multiple zeta values
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34:39
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Hand fanMaxima and minimaSummierbarkeitFinitismusRule of inferenceAreaPairwise comparisonReal numberResultantLecture/Conference
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Euler anglesGroup actionPhase transitionMathematical analysisState of matterModule (mathematics)ResultantSpacetimeProjective planeDimensional analysisSolvable groupFormal power seriesPower (physics)Multiplication signLecture/Conference
Transcript: English(auto-generated)
00:25
K is introduced by both men and impenetrated by their beings, which is a real number defined by the following.
00:45
If and if some, under n-1 and n-n.
01:00
Where n-1 and n-1 are the integers satisfied with infinity, this is the real number. Where k is the sequence of finite sequence integers, which are the denominator of k and first one of n,
01:26
stretched to larger than one.
01:44
This k is called an index, even stretched to larger than one is called an even and admissible index.
02:01
Without this condition then it is called just an index. This is defined by the absolute value of k, even the sum totally weighted.
02:28
The depth is equal to the depth of the index to the n.
02:46
The version of these multiples in the values, which are defined by the initial label.
03:21
The p-n multiples in the values are denoted by p of k, where the p-n multiples in the values are introduced.
03:53
And Furosho asks that the following question belongs to the ring of integers in p.
04:14
And in his book, he says that for any n-n or n-n within the values,
04:42
belongs to the case p-n here or p-n here. By definition, the idea of the p-n generating factor divided by p, p equals pi, t equals sine model of p.
05:17
This sort of explains some applications to the theorem of finite multiple zeros values.
05:36
So, we need to introduce some more notation to explain all this up.
05:43
First, we introduce a multiples in the values, If we study by the set of all the variables in the index, it is mostly multiples in the values.
07:26
Some ring denoted by a, but its integer is a q that is just a q-algebra.
07:55
So we can consider the q-variance actually generated by these multiples in the values of weight k.
08:04
This is the section k of this space, dk minus dk is the main result in a complete relation of this inequality.
10:13
Of course, the next section, in section 2, I will give an outline of the proof of the first theorem.
10:49
According to the details of the proof, we first recall the definition of the k-algebra of multiples in the values.
11:18
We can index the series, the summation, the original summation is the same as the multiple series.
11:47
The integer is also the d-code. This is the power series, the space, the coefficient, the rational number.
12:06
And this is called the Markov group.
12:29
It is strictly marginal one, then the Markov group, which is the values.
12:42
The periodic dysfunction of the values is strictly small.
13:30
However, we can tell that the periodic integration is zero.
13:58
We can define the periodic multiples in the values, in the sense that p is strictly smaller than p is zero for something, strictly larger than one.
14:36
In my explanation, we can take this number, but anyway, for any two index in k-5, such that their weight of k is strictly larger than that of k-5,
15:17
there exists a unique algebra, k-prime, which belongs to this ring, such that multiple modes of algorithm,
15:45
the substitute there is variable p by width, p to the p over p to the p-prime of p minus one to the p. This is equal to p to the weight of k times the original particle, all the local reasons, plus some other terms.
16:19
k-prime runs over all the indices, whose weight is strictly smaller than that of k.
16:30
k-prime is this function, h, so k-prime runs exclusively by using some partial integration.
17:17
So, I have a question on your formulas, so one question is whether the empty index is allowed in some formulas, and another question is...
17:30
The empty index is allowed. So, there are conventions to define things. So, the empty index is considered as admissible.
17:45
Yes, admissible. So, and then h, k, k-prime lies in which, is what kind of thing, is it in the ring of a dagger, or...? Yes.
18:01
Ah, ok, ok, ok. Ok? Ok, ok. Ok.
19:00
And then we can define the p-
19:03
Ok, ok, ok.
20:58
So, the sum of these three points as a normal crossing divisor has
21:00
begun, which has a local resonance to it, associated with the normal crossing divisor. And we have an obvious code p, and x is a union of x minus p sub-skin, where p runs over the subset of three points,
21:34
which means that the subset of p, the non-empty subset of the set of three elements of which, and here we develop this as a closed sub-skin of p,
21:51
corresponding to p. Anyway, this defines the final covering of the underlying scheme of this.
22:19
For the k, which is non-negative, we obtain the satellite and the direct product of the covering.
22:56
So, each and every time of the covering, the tensor product and the final product of the final set.
23:39
The component 1 has a canonical meet, combining with the block structure.
24:15
At the best symmetry, for any other i, we have seen a canonical meet of problems. And for any non-negative product, the subset of this response is defined by the element.
24:58
If i is different from i, then by removing some straight transform of some divisors,
25:08
then it will change from the exact closer range from x minus p.
25:21
Similarly, we have an exact closer range from u.
25:45
Here, a grow-up is something like this. This grow-up is necessary only when the continuity of p is equal to 2. Here, a grow-up is necessary only when the continuity of p is equal to 2.
26:15
And here, a grow-up is necessary only when the continuity of p is equal to 2.
26:24
If we obtain a canonical exact close-inversion, then this scheme has a unit-related V-U-L denoser close-inversion of u.
27:49
The grow-up is defined as, we note that these schemes are affine schemes, so we can regard these complexes just as a complex of z key modules.
28:42
And then by passing the new values, p1, p2, p3, p4, p5, p6, p7.
29:37
Also, the canonical is the approach used to find the natural action of from these action and complex machines.
29:48
So we can recover the complex and throw this into the system. Excuse me, this is complex. Mic? Yes. Why is it a special project?
30:22
The thing we can define is a clean and dry. It's defined to the training sound, like another map.
30:42
We multiply this sub i, where this sub i is defined to the product.
31:02
R-S component. I subtracted all of them. Next to the i, I don't want to divide them. Except for the h. Where i runs over to s0 isn't.
31:22
Rather, the meaning of s0 is a product of one type. k is a random thing in the program.
31:42
This embedding does not induce a dark or complex.
32:04
We have to define the monism, the complex, basically by all things. For example, this mistake is divided in one form. We don't paint a dark or complex.
33:09
Part of the singularity is placed at a grade zero. Then we denote it.
33:21
One can show by using the project frame as a crystal and coconut in this space. This is an awesome bridge. It represents a free city. Which means a canonical bridge.
33:47
We denote which building on the back. In the W. Where W is the set of walls. And the additional structure is the concrete bridge.
34:31
And it satisfies this mixed state. So this is infinite dimension, is it?
34:45
Yes. And each of the c dot is a five-dimensional cohomology? Yes. So when you take the total complex, is it a total complex where it's infinite?
35:03
Is it that you have to take infinite direct sum or infinite direct product? Yes, direct sum. Infinite direct sum? Yes. Okay, I understand what you are doing.
35:26
Also, the gradient is with respect to weight filtration. It's equal to zero when the index is odd.
35:41
And when it is even, the problem is half is very explicit. So this implies that there exists unique. Also, the dimension of one dimension. This is the group series one dimension.
36:02
Group series. Generated by the image of the eta of empty. So this implies that there is this unique theorem of H2QP.
36:27
Such that it satisfies all the properties. The first one is now is likely to be not by theta. So the theta is the problem in theta. The second one is that the image of theta of the element eta sub empty is equal to one.
37:09
The observation k equal to k, one minus one would be equal to one.
37:55
The image theta of the element eta sub empty is equal to one.
38:14
The second observation is that the image of phi of qk.
38:27
We can define the hot filtration of the compress level from standard way. I should be able to see that the case hot filtration is contained in the case divided power times.
39:14
At your times. For using these two observations,
39:47
I would like to give an outline of this problem.
40:03
We need to introduce another version of p and multiple see the colors. This is by giving a speech to form a completion, natural structure, multiplication in the future,
41:01
and how much multiplication is given by the thousandth, the five th, the seven th, the eight th.
41:32
Because who keeps this delivery?
41:47
We define phi of kz. This is the sum of the parenthesis associated with the p and the d.
42:07
The image is theta of theta of theta sub del mu. This is the fourth one.
42:52
I describe phi and phi of z.
43:27
This is the first one. No commutative form of theta b. Or another completion form. Yes, no commutative form of theta b.
43:47
We define the cosmogonality of theta.
44:22
We substitute the variable in a and b with a over p and a over p and take the consciousness function.
45:07
The domain is theta b. That is the number of theta b and theta b.
45:58
The difference is less than that.
46:49
Less than that.
47:18
The composition times the usual version, k times one over kl,
48:24
equal to the k, which is the finite sum, is the same expression as the mod p times theta b, but we regard it not as a component of theta,
48:44
as a function of theta b. This is called the finite hull multiple-armed sum. This is the conjecture second two, might be called conjecture two, just a year ago,
49:01
but it is already proved by David, and because he won't present in the special case,
49:22
instead of containing his conclusion, or one of his children.
49:54
Just the convention of the indices, so you had in the beginning the condition that k1 is at least two,
50:05
called admissible, but then here we don't have this condition. In this case, this assumption is unnecessary. So all the definitions of the p-adic things are without this assumption, or only this? All the definitions are without this, in the p-adic.
50:28
In the whole p-adic story, you don't assume any admissibility of the index? Or just here? Yes, yes, yes. No, over. We don't assume. Okay. Here, in blue.
50:41
Yes. It's a funny thing. I'm going to give a sketch. Are you familiar with this proposition? No. We need to do something more interesting.
51:06
The question is, does z0 have a solution?
51:23
I guess the first thing is to show, and this is some form of power series, also known as power series, coefficient in some ring, and Coleman, and this is called the Coleman function,
51:40
is to use a power series expression, so we only have a subring of power series times,
52:03
and all the power series, or we know they are open, but the whole part, coming to the index,
52:27
is equal to the power to the whole over thing. So move up a little bit here,
52:43
and multiply by these properties and the coefficient of and some in some way conditions. And also,
53:02
here is some form of programming barrier notation,
53:30
and this is also doing fine. Also, so we can evaluate it as t is equal to one,
53:40
but it is equal to, in some cases, larger than the values, but the important is the following properties we need there,
54:23
and we need to write what is the position of t. So this is a product of a composition of
54:46
model, model, model, model. Then this type of product is double, which is equal to or the one which ends with e,
55:02
so this corresponds only to the one corresponding to each other. Obviously, the term in some coefficient,
55:20
term in some term, which is not, belongs to the type of product, and this is equal to the relationship of the single, the relationship of the product. Take this combination
55:40
and then you take the coefficient, and then we need the first type becomes the form of this combination. It doesn't appear in the terminology of long, polynomial long. taking the,
56:00
consider the coefficient indicator a in the last one's type
57:17
is f of a w, the one a w,
57:21
the other one's a inverse
57:45
to the equation,
58:51
the second part is equal to
59:00
in turn,
59:21
this polynomial is something like is a ring
59:40
something like in some integrals, right? In some circles, I might fit to the round course of this thing, and calculate the next table.
01:00:02
This is complex integration. This integration is the realization of the box effect around the orientation of the box.
01:00:26
Also the realization of the box is the close substrate of the wave, which is the metallic energy of the waves. When the substrate is constructed in the internet,
01:00:42
it is filtered around the form of the substrate of the wave.
01:01:10
This is the idea generated by a technique.
01:01:24
The natural structure of the wave network is the nature of the input.
01:01:44
We have a few added equations for it. This does the work of doing a cheetah.
01:02:01
If we fix something, then we have a map. This is wrong. The inner structures will be in the diode zone.
01:02:33
The intersection of the programming will be increased and P0 will be off.
01:02:43
And then taking a positive input. So we put in the map. This is not a gradient. You can see the map generated by the element usually given by T.
01:03:21
This doesn't depend on the choice of the beginnings. We can take the potentials to be just P0. So we have P0, P0, P0, P0, P0, P0, P0, P0, P0, P0, P0, P0.
01:04:03
And collection of image under C, P0, P0 divided by some power of P0. Then the consequence of C1 is that of the diode. In brown, this image contains the sub-regional finite event.
01:04:41
Consequence of A, P0, P0, P0.
01:05:42
This is straight to the root of P0.
01:06:33
And this is green in P0 value. This is green in P0 value.
01:06:42
Motific extension. But we can approximate the range data by the user. So that's the end of this talk.
01:07:09
I hope you enjoyed it. It is surprisingly large for this test. The thing is that the release must be detailed.
01:07:20
It is nearly equal to the release balance. Antipode inversion over here. Also data divided by it. The P is greater than P0. Finally, I would like to address this. It is important that we can show that this equality is important in this equation.
01:07:51
So we obtained the lab from the experiments of the entire exam. We obtained the two subjections.
01:08:07
It is a non-reflection event. And also it gets conjectured by, in some sense, by an egg or a egg. The question is that you have disconnected these two subjections.
01:08:28
Thank you very much. Thank you very much. Are there any questions from Paris?
01:08:41
Okay, so are there any questions from Paris? So the relation, so can you explain the relation of the, you define certain expressions using truncations in the definition of the, you define p-adic multiples of the virus, but also you define those in the k-necos again from,
01:09:06
so there were finite sums in F. I don't find it in my, ah, this one.
01:09:24
Yes, so can you explain the relation between the finite sum and the multiples that are? This one? Can you explain? This is the finite sum, so we can consider it as p in the version of this.
01:09:42
Then we obtain something like this. This is the what? This is the finite sum. This is the finite sum, and if the larger p is equal to the both p, what will be the value? And it is called, and this is the name of the finite multiples of the virus.
01:10:05
And, I'm sorry, this is the least p-adic multiples of the virus, but it is related, it goes from definition to the unlike of this version of the p-adic multiples of the virus.
01:10:28
So, by using this formula, we can relate the finite sum and some reduction of the names of the p-adics. So, as a colony of, as a colony of p-adics, if all the structures are smaller than p-adics,
01:11:17
then that appearing in this finite sum is a p-adic integer, in this case,
01:11:55
and it won't be equal to the finite sum.
01:12:05
By using this, we can relate the finite sum of the p-adics. Not really speaking, but really speaking. And also, I have not explained that.
01:12:22
Also, the lineage of the p-adics, multiples of the p-adic multiples of the p-adic multiples of the p-adic multiples of the p-adic.
01:12:43
So, you also said that some of the map was not a graded map in the blackboard there, and H was graded, I suppose. H? So, the map from H to V is not graded. So, we are not graded.
01:13:00
So, what does it, what does it do? Can I have this one? Yes. Yes, this is not graded. Does it send an estimate? Yes, this has a gradient form, which follows the definition,
01:13:25
and this also has a natural gradient. Yeah, I don't, so every element goes to a finite combination of things,
01:13:42
or every element goes to a finite sum, or? Yes, finite sum. Okay, so, are there other questions?
01:14:05
Okay, so I think that's all for the rest. Okay, is there any questions or comments from the team? No questions. So, I wonder if there's any relationship between the real zeta,
01:14:23
body zeta value with the modulated zeta value. Oh yes, there is another comparison, a usual comparison map from H, there's a map from H to R, by using the whole Bayesian drum,
01:14:42
by using the Bayesian drum from this point, and this map to the modulated, or to the modulated zeta value. Do your results have implications for real modulated values?
01:15:00
Oh yes, the first one, by using this whole group sum, it is proved that the space of modulated values of this space is bounded by the dimension d. Okay, that's not something that you're talking about here.
01:15:23
Oh, no, no, no, no, no, then my result is only merely for a finite project. Oh. No. There are no more questions from Beijing.
01:16:12
Okay, thank you.
01:16:24
Okay, let's have some.
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