On Multiple zeta values and their q-analogues
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Transcript: English(auto-generated)
00:16
Thank you very much to the organizers and especially to Gerard for this nice invitation.
00:26
Oh, I see this is the last page, so I have to jump to the beginning. Sorry for that. Okay, here we go. So, I also will speak about multiple zeta. So,
00:45
thanks to Minh's talk, I will go slightly quicker on the beginning. So, as you know, here are the multiple zetas, so these iterated sums, and if you speak about multiple zeta values,
01:10
the nj's are positive integers, and even if you want the series to converge, you need n1 bigger than 2. Okay, you can go slightly beyond, you could go to nj's
01:28
positive or negative, but you must have n1 plus etc up to nj strictly bigger than j for any j from 1 to k. Okay, so as you know, the integer k is the depth, and the sum of the nj's is the weight.
01:51
Okay, so let me just make a small historical survey, but for an old one. I mean,
02:03
I start with Euler in the 18th century. He knew multiple zetas in depth 1 and 2, and he gave a lot of relations between them. Not all, we know some author relations now, but
02:25
up to, I think, 15 years ago, the only ones we knew came already from Euler. And then there's a big gap until the 1980s, but this gap is not completely empty because
02:48
for example Min indicated me that you can find some examples in higher depths in the work of Nielsen in the beginning of the last century. And then, up to my knowledge, the multiple zetas
03:07
appear first as a rule in the work of Jean Ecole in 1981. So now, what about the multiple zeta function? Now the arguments can be complex.
03:28
This is the same sum, except that the zj's are complex numbers. And then it turns out that this is a meromorphic function in k variables. And this is the work by these three Japanese
03:44
researchers, Akiyama, Egami, and Tanigawa. This series converges, provided the real part of these partial sums are strictly bigger than j for any j. And you can extend
04:02
meromorphically to the whole c to the k with a subvariety of singularities, which is described as follows. In depth k, the singular locus of depth k is defined as follows.
04:23
This is those zj's such that either z1 is equal to 1, or z1 plus z2 belongs to this set. You jump from 2 by 2 except 2, 1, 0. And or there exists j from 3 to k such that
04:48
this partial sum is smaller or equal to j. Okay, so you can see the difference between depths 1, 2, and 3. It's related to the fact that the single zeta function can be continued
05:06
analytically everywhere except at 1. And for double zeta, these holes come from the fact that Bernoulli numbers, the odd Bernoulli numbers, all vanish except the first one.
05:24
Okay, so I describe double shuffle relations because we will see some generalizations of those later. Of course, you know that the product of two single zetas is a sum of three terms, two double zetas and a diagonal term, which is again a single zeta. And of course,
05:44
the proof is here. You cut the domain into three parts, m1 bigger than m2, m1 smaller than m2, and m1 equal to m2. So the diagonal terms may appear. Okay, I can describe
06:06
most general quasi-shuffle relation. So you multiply two multiple zetas, one of depth p, one of depth q, and then you get a sum of terms of depth p plus q minus something.
06:20
This something, this small r, is the number of contractions you have in your term. And the sum is indexed over all the quasi-shuffles, the pq quasi-shuffles of type r. Okay, and these are surjections sigma from one up to p plus q onto one up to p plus q minus the type,
06:44
subject to the same relations as shuffles. But here, one sigma in the first row may be equal to some sigma of the second row, and this gives you your contraction terms.
07:04
And here, the arguments in this contracted multiple zeta are defined by the sum of this sum here. And this sum, if you think about it, it contains only one, maybe two terms, but not more. Okay, so you probably also know the
07:27
integral representation of multiple zeta values due to Maxim Koncevich, and quoted by Donzagie in the his 1994 paper. Okay, so the multiple zeta value,
07:44
here now the nj's are positive integers, is defined by this iterated integral. So you iterate as much as the weight, and you integrate the differential forms dt over t or dt over 1 minus
08:03
t in that order. So here you have n1 terms altogether, after that you have n2 terms altogether, and then here nk terms altogether. And you have differential forms dt over t,
08:22
except the last of the packet is dt over 1 minus t. So as a consequence, you have a second way to express the product of two multiple zetas as linear combinations. These are called shuffle relations, for example zeta 2 multiplied by zeta 2. It is a quadruple integral of this differential form, and you integrate over the
08:52
product of two simplices, and this product of two simplices decomposes into six
09:01
four-dimensional simplices, and then you get six terms, four times zeta of 3 1 plus 2 times zeta of 2 2. Okay, the regularization relations are a bit more funny. The simplest one was already known by Euler, zeta of 2 1 is equal to zeta of 3. And how do you obtain it? You use zeta of 1.
09:29
Okay, you will say to me zeta of 1 doesn't exist. Okay, this is just an indeterminate. So of course in in mean stock somehow you can decide that zeta of 1 is equal to zero,
09:43
but here it's just an indeterminate. And then if you write down zeta of 1 2 or zeta of anything starting with a 1, it's also divergent. So now I write down this product
10:04
in two different ways. So the upper line is just by formally using the quasi shuffle relation, and the bottom line is just formally by using the shuffle relation. And then if you
10:21
equate the two members, the red terms disappear, like in the chemical reaction. And you get, of course, this one disappears with that one, and you get Euler's relation, zeta of 3 is equal to zeta of 2 1. Okay, so here I briefly explained the double shuffle relations,
10:45
or double shuffle plus regularization. And it is conjectured that no other relations occur. For example, you have duality relations coming from the change of variable tj goes to 1 minus
11:00
tj in the integral presentation, and these duality relations should be deducible from the double shuffle. For example, this is the first example of duality, and as we said, as we have seen, it's also double shuffle. But you have other duality relations like this, and in principle
11:22
should be expressible. But as far as I know, it's still an open question. Do these duality relations come from the double shuffle? Okay, yeah. Okay, so
11:43
there are multiple polylogarithms in just in one variable. So Min already explained us, so I go very quickly here. The multiple polylogarithms at some t between 0 and 1 is defined by this sum, or it's defined by this integral,
12:05
so it's the same for then for multiple zetas, except that instead of going from 0 to 1, you stop at t. And so even if the world, well, even if the multiple zeta value
12:22
would be divergent, if you stop at t strictly smaller than 1, it still converges here. Okay, so now I want to re-express this multiple polylogarithm in a
12:41
operator way. Okay, so this is the formula you can see in the bottom of the slide. So you start with the constant function equal to 1, you apply it the operator Y, which is multiplying by this function 1 over 1 minus t here. Okay, then you apply the operator R,
13:03
R which is the primitive operator, R of f is equal to the primitive of f vanishing at 0. This is a weight 0 rotabaxter operator. Okay, so and then you continue like this. You apply
13:24
now the operator big X, which is multiplying by small x, it is function here, 1 over t, and then operator R, et cetera, et cetera. So it's it's another way to to to rewrite
13:46
the Koncevich integral representation of multiple polylogs. Okay, so of course you know
14:01
when t is equal to 1, you recover the multiple zeta. Okay, so we have this representation by words. You introduce the infinite alphabet Y. Y star is the set of words with letters in Y. You look at the linear span
14:27
of the words, and you look at the quasi-shuffle product. So my point of view is somehow dual to to the one explained by Min. Instead of using a concatenation product and
14:43
d-shuffle co-product, I will use the quasi-shuffle product and de-concatenation co-product, if you like. Okay, so here is the quasi-shuffle product. So it's like the quasi-shuffle relations. You sum up over the same quasi-shuffles,
15:05
and the recipe looks very similar. You have contraction terms because of this R here. Okay, so for later use, once I have defined this quasi-shuffle product, I can stick to genuine shuffles. I mean, I can stick to R equal to zero, so
15:26
so I take a smaller sum, and this also defines an associative product, which is an ordinary shuffle product, but still on this infinite alphabet.
15:41
So here is an example of quasi-shuffle. Here I have five terms, including these two contraction terms, but if I discard them, I have the shuffle product. Okay, so this is traditional notation. Y star convergent is equal to the set of words which don't
16:04
start with Y1, and for any convergent word, I can define zeta quasi-shuffle of the word is equal to zeta of the corresponding row of indices. But what I have gained here is that I can extend
16:24
this linearly, and the quasi-shuffle relations are equivalent to the fact that zeta zeta quasi-shuffle is a character for the quasi-shuffle product.
16:43
It's morphism. Of course, you would like to go from convergent words to any word, that's what Min explained, how to regularize the words which are not convergent. Okay, so
17:01
this is just an example of quasi-shuffle relation rewritten in terms of this quasi-shuffle product. Okay, I think I can go quite quickly on this. Okay, so now I'll quote the first result I
17:22
obtained with Sylvie Peixal some 10 years ago about extensions to arguments of any sign. Okay, because the quasi-shuffle product, you can extend it to the words with letters in Z. Okay,
17:42
so accounting for arguments of positive or negative sign. Okay, so what we proved is the following. There's a character from this extended algebra with the quasi-shuffle product into the complex numbers, so that, first of all, phi of v is equal to zeta quasi-shuffle of v
18:07
for any convergent word. Okay, secondly, for any words with letters positive or negative, such that the multiple zeta can be defined by analytic continuation, then the character
18:24
coincides with this analytically extended value. So, in particular, phi of minus n is equal to zeta of minus n, so minus Bernoulli n plus one over n plus one. So, at depth two,
18:42
phi of minus n minus n prime is equal to zeta minus n minus n prime. So, there's a value here if n plus n prime is odd, and it's given by this expression here in terms of Bernoulli numbers. So, we got a table of values in depth two, and you have two kinds of values. You have
19:11
all the same values except the very last one, the very last one in the northeast of the table. For example, this one. Okay, these values are obtained by analytic continuation.
19:28
The other diagonals are not. And, for example, you have also a paper by Ligu and Binsang a little before us. They also have a table of value.
19:42
It is different. But, of course, the good diagonals coincide because it's analytic continuation. Okay, let me give a sketch of proof.
20:01
So, it's a con primer, regularization and renormalization, basically. Okay, so we have the quasi-shuffle product. Together with decontinuation, you get a connected filtered Hopf algebra, the filtration by the depth. And then I extend this Hopf algebra to the
20:27
letters in the complex numbers, so I extend it much more. And I look at the Hopf algebra, the same Hopf algebra with the quasi-shuffle product replaced by the shuffle product. Okay,
20:45
and as a result by Michael Hofmann, these two Hopf algebra are isomorphic and there's an explicit isomorphism. So, now this big R map is a regularization map.
21:03
So, we define it on the shuffle version of the Hopf algebra. So, it's very simple. Big R of a word is the same word but with the indices shifted by z, like this. Okay, and then we will use this Hofmann exponential with this, which is the
21:31
hopf isomorphism. And then from the regularization R, I get a regularization R tilde by twisting
21:41
by the Hofmann exponential. So, now this twisted regularization respects the quasi-shuffle product because the first regularization would respect the shuffle product. So, that's no surprise. And now we define the character big phi from this algebra with the quasi-shuffle
22:08
product, defined by this. So, again, zeta quasi-shuffle, well the multiple zeta function
22:22
is isomorphic in the complex variables. So, by defining this, it's a non-trivial result but we can prove that it takes values into the meromorphic functions in one complex variable.
22:42
And then we have the Birkhoff-Konk-Reimer decomposition. Our big phi character can be decomposed like this. So, this is the convolution product and these two pieces are characters. So, quasi-shuffle relations are still verified.
23:09
And phi plus of v is holomorphic at z equal to zero for any word v. And this is what we wanted because now... Okay, these are the formulas to get the two pieces phi minus and phi plus.
23:32
And maybe I recall quickly how it works. Your meromorphic function algebra is split into two
23:42
parts, the pull part a minus and the formal series part a plus. Okay, and this is so-called minimal subtraction scheme in physics. And then you can define recursively phi minus and phi plus
24:02
with these formulas here. The red term is called a Bogoliubov preparation map. And then if you take the opposite of projecting on the pull part, you get the counter terms. And projecting
24:21
on the formal series part gives you the renormalized character. And then the character we look for, you just evaluate the plus part at z equal to zero. This you can do. And it still verifies quasi-shuffle relations. Okay, so we were quite happy with this result.
24:49
And then, a few years after that, we wanted, with Kurocheh Braimifard and Johanna Zinger, we wanted to describe all solution of the problem. Describe the set of characters which
25:06
extend multiple zeta functions in that sense, I mean by respecting the quasi-shuffle relations. And there's a renormalization group behind this. Okay, and
25:29
yeah, let me start with a very general result. I start with a commutative connected filtered Hopf algebra. And a will be any commutative unit of K algebra,
25:45
and ga will be the group of characters with values in a. The product is given by the convolution, and we have this conilpotence property, which it is crucial to make some
26:02
recursive arguments. Okay, so the proposition goes as follows. You suppose that big N is a right co-ideal with respect to the reduced co-product. The reduced co-product is this delta tilde, when you start from delta and extract the two extreme terms. Okay, and
26:28
and then if you look at this set TA, you look at those characters alpha which vanish on the right co-ideal. And this is a subgroup of GA.
26:43
And the proof is quite simple, it's a five-line proof. Essentially, if you take two elements alpha and beta in the TA, you want to prove that alpha beta minus one is still in TA, to prove it is a subgroup. And
27:07
essentially, you say that the inverse of a character is given by composition by the antipode. You replace the antipode by its recursive expression, and then you end up with alpha
27:21
minus beta of w prime here. And now you use the right co-ideal property to say that this vanishes. So we call it the renormalization group associated to the co-ideal
27:44
n. And then comes this trick. You look at a partially defined character, that means it's defined only on the co-ideal n, and you suppose that it respects the unit, and
28:08
it's multiplicative whenever the three terms are defined. I mean, whenever the three terms belong to n. And now I look at x zeta a is equal to those phi's in the group GA, so that
28:25
phi restricted to n is equal to zeta. Okay, and it turns out that this is a principal homogeneous space, so this we proved by with the Kurushe Brahimi-Fahrt, Johan Zinger, and Yanqiang Cao. This is a principal homogeneous space, that means that
28:47
this left action here is free and transitive. And the proof is quite similar to the proof that TA was a group. Okay, I'll skip this. Okay, now we apply this general framework to
29:03
our multiple zetas. So the base field is the rational numbers, our Hopf algebra is this one, with the arguments of any sign, I remember you. And the target algebra will be just
29:20
the complex numbers. The right co-ideal is the linear span of non-singular words, that means that the words which write down like this and which fulfill these three conditions, which are the contrary of the singularity conditions of the beginning of the talk,
29:40
right? So n1 has to be different from 1, n1 plus n2 has to be outside this set, and this partial sum has to be outside this set. Okay, and this right, okay, this big N is
30:04
obviously a right co-ideal for the concatenation, if you look at these three conditions here. And there's a supplementary property, it's also stable by contractions like this. Okay, you gather some consecutive letters and you sum them up like this.
30:26
And n is stable by these properties. Okay, so sigma will be the complementary of n, okay, and so sigma k will be those words so that the corresponding row belongs to the
30:42
singular locus at depth k. And the partially defined character zeta will be zeta, okay, will be our multiple zeta values, either by convergence of the iterated sum or by
31:02
analytic continuation. It can also happen. Okay, so now the set of all solutions to our initial problem is given by the renormalization group acting on one solution. Fortunately, we had proved with Sylvie Péchat that one solution exists, okay.
31:28
So we investigated a little bit this group, just to say that it's very big. It's infinite dimensional. If you look already at depth three, you see there's a lot of room
31:45
inside it. Okay, so now I jump to Q-Analogs. Okay, so let me remind what is a Jackson integral. It's given by this ugly formula,
32:02
but maybe it's more easy to grasp it. I like this. So here is an example with Q equal to one roughly. Okay, and it's quite obvious that
32:21
when Q goes to one, the classical limit gives you the ordinary Riemann integral. Okay, but here our Q will be considered as an indeterminate. Now I can define a weight
32:40
minus one rotor-backster operator by summing up like this. PQ of f of variable t will be f of t plus f of Q t plus f of Q square t, etc. Operator Q t is invertible because
33:00
this is due to our algebra A here. A is this algebra of formal series in two variables, but starting with t. Okay, and PQ is invertible, and the inverse is given by the
33:22
Q difference operator f of t minus f of Q t. You have a modified Leibniz rule for the Q. So this is Leibniz identity, but you have a diagonal term here. And we end up with three equivalent identities, one for the P's, one for the D's, and one mixed. Okay, so now I can
33:51
define multiple Q polylogs the same way. You replace the Riemann integrals operator by the Jackson integral operators. Okay, but this is exactly the same recipe. So for later use,
34:10
I will need the operator y bar of multiplication by this function t over one minus t. But for the moment, I just mimic the formula for the polylogarithm, replacing the Riemann integral
34:23
operator by the Jackson integral operator. Okay, so now I will speak about the Ono-Okuda-Zudilin Q multiple zeta values model. So we call that multiple zeta is given by
34:42
evaluating the polylog at one. Here we evaluate the Q polylog at Q. And in terms of nested sum, you have this formula. So in the numerator you have Q to the m1, and in the denominator you have powers of Q integers. Okay, so for convergent words,
35:17
when Q goes to one you recover the ordinary multiple zeta. When Q goes to one in a reasonable
35:26
way. Okay, we have an alternative description in terms of the operator PQ. Okay, so
35:44
ZQ bar is the modified multiple zeta. This is ZQ multiplied by one minus Q to the minus the weight. Okay, so this is this iterated sum, and it is given by this row of operators.
36:02
But here I use the operator big Y bar, which is the operator of multiplication by this function. We have other models for QMZVs. I just mentioned a few. The Schlesinger model is the same,
36:22
but you evaluate the Q polylog at one, not at Q. You have the Zhao Bradley model, which is the denominator. The numerator is more complicated, this Q to this power here. And you have the Bachmann-Kuhn multiple divisor functions model, and maybe a few other ones.
36:51
Okay, but we like this model by Ono, Okuda, and Zudilin because it's defined for arguments of
37:00
any sign. If you take nj's in Z, it still makes sense, and you even have this estimate when Q is a complex of modulus strictly smaller than one. All the sums converge, and you have this. And you still have this operator formula for computing the Z bar Q, okay?
37:32
Of course, with negative powers of P or positive powers of D. Okay, here are some examples.
37:44
Zeta Q bar of zero. Bar, without bar is the same. It's Q over one minus Q. A row of zero is just elevated to the power K. Zeta bar Q of minus one is just this,
38:02
so it's Q over one minus Q minus Q square about over one minus Q square. And, interestingly enough, Z bar Q of one is this Lambert series. Sum of Q to the m
38:21
over one minus Q to the m. Okay, and here, not only we have meaning for any signs, but we have a complete double shuffle picture. Actually, Takeyama had also a double shuffle picture, but in the Bradley model, so
38:46
the picture is more complicated, and it doesn't extend to arguments of any sign. Okay, so first of all, I'll describe the Q shuffle relations.
39:06
Okay, maybe I have to hurry a bit. So you look at the alphabet here with three letters, but you suppose that DP is equal to PD is equal to one, okay, and we have a third letter Y.
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So any non-empty word writes uniquely like this, P to the n1 Y up to P to the nk Y with nj's positive or negative. Now define the bar Q shuffle of
39:42
a word by the Q shuffle of the corresponding row of nj's and extend linearly, okay. And you have a Q shuffle product on this space of words, and it's defined recursively by these
40:02
four relations, and these relations are just abstractions of these relations we have with respect to these three operators, okay. And we were able to prove that the product
40:23
is commutative and associative, and the Q shuffle relations are just written like this. Z bar Q shuffle is a character for this shuffle product. So we have also Q quasi-shuffle.
40:43
So now Y tilde is the infinite alphabet but in both directions with nz. The internal product is the same, Zi contracted with Zj will give you Zi plus j. Y tilde star is the set of words
41:07
with letters in Y tilde. Star will be the ordinary quasi-shuffle, and our quasi-shuffle product is the ordinary quasi-shuffle moved by this operation here. Just this shift on the first
41:29
letter of the word, okay. And I define the quasi-shuffle Z bar function value like this,
41:41
and I extend linearly. And the Q quasi-shuffle relations write like this, okay. It's a character for this quasi-shuffle product. So I give you an example of Q quasi-shuffle relations. For two letters, the product of single Q zeta functions is given by
42:03
the ordinary quasi-shuffle term plus a weight drop term, okay. So the weight is not conserved, and if you go to the ordinary, well, non-modified Q zetas,
42:21
you have this one minus Q in front appearing. So you can, it's coherent with the fact that when the limit Q goes to one, the weight drop term disappears, and you get the ordinary quasi-shuffle relation. Okay, but here you have no regularization relations. You have this swap or this bridge,
42:50
I don't know how to, well, this change of coding between the two alphabets, big R, Gothic R, and here is the recipe. And the change of coding writes like this.
43:10
You jump from quasi-shuffle zetas to shuffle zetas with this. But that's all. This is defined for arguments of any signs, so no problem. So we have quasi-shuffle relations,
43:28
shuffle relations, change of coding, and that's all. So I give you an example of computation using double Q shuffle. I look at this product, okay, and there's one appearing here. So when Q
43:47
goes to one, maybe something strange will happen. So let's see. So first of all I use the quasi-shuffle relation, so I have not three terms but six terms, the ordinary quasi-shuffle
44:01
term and the weight drop term. So using Q shuffle, so I translate into the shuffle words, this product, okay, and so I have to shuffle the word py with the word ppy.
44:21
And then I use the recipe given by the Q shuffle formulas here. No, sorry, here we go. Yeah. And, okay, in a few lines, at the end I have the Q shuffle bar of this row,
44:44
this combination of words. And I translate back to z bars, okay. So now this term above is equal to this term below. There are quite a lot of nice cancellations,
45:01
and at the end you end up with this, okay. If you remove the bars, you have this formula here, and when Q goes to one, you go back to the Euler relation, which was obtained by regularization. But here our regulator is the parameter Q, so to speak, okay. And interestingly
45:29
enough, this identity can be found in a work by Bailey in 1936. So this was pointed out
45:42
to us by Vadim Zudilin, and Bailey attributes this identity to Bell, and it was learned by Hardy to him. So it's quite interesting. Okay, so it's time to conclude.
46:03
Maybe there are other relations than the double shuffles. It seems that there are also duality relations, but I have no time to to develop this. We would like a more combinatorial description of the Q shuffle product, and we would like to find a compatible co-product. Okay, this is
46:26
related to this naive question. I spoke about how to to renormalize quasi-shuffle relations. We succeeded in that in some way. What would mean extend the shuffle relations to the negative
46:45
side? That means that you will have to shuffle two packets of cards containing a negative number of cards, so it doesn't make sense, naively. But with this Q parameter, it does make sense
47:01
in some way. So we would like to find a Hopf-Algebras structure here. This is work in not progress for the moment, but we have still the project to go back to this.
47:21
And so I said to you the parameter Q is a regulator. What about renormalizing when Q goes to one? Okay, interestingly enough, when I gave this talk two weeks ago, David Broadhurst
47:46
asked me what about the other routes of unit. Well, good question. But there's a very nice pre-print by Dorrigoni and Kleinschmidt about the behavior of the Lambert series for Q goes to
48:02
one or Q goes to the routes of units. So at least for zeta Q of one, we have a complete picture here, and maybe the task would be to do the same for the other Q multiple zeta values.
48:22
And this could be a good research program for the future. Thank you very much. I'll take a minute. Yeah, yeah, yeah, maybe.
48:43
But you have to say if you want to add something. No, no, no, I have finished. Thank you very much. I just wanted to have the gallery back. Okay, questions.
49:05
Yeah, okay, let's go. So the negative materialities, zeta at negative analysis, you say that you obtain by an electric conjugation. So do you compare the result obtained
49:27
by Rho in its PhD digits, the values obtained by finite part of asymptotic expansion?
49:46
Yeah, they are quite different of the rational number. But on the way, this is coherent with the process obtained by finite part of asymptotic expansion.
50:05
Okay, but you say you obtain by this, you say analytic continuation is coherent with process of asymptotic expansion? Yeah, because the value gamma comes from
50:28
asymptotic expansion does not come from analytic expansion, except for the kind of analytic extension. Okay, but outside this point, where analytic extension is possible,
50:46
you have a kind of freedom of choice, which is encoded in this renormalization group. You see clearly what that means exactly, but
51:04
if we don't have a coheren between shuffle and shuffle, you are already saying so that with analytic continuation, I don't see why.
51:22
Okay. And so before that, you say to me that the rational number you obtain just satisfies the different product and not the different objects, do you?
51:40
Yeah, yeah, yeah. Do you confirm that? Yes, I do confirm, and more than that, the shuffle product, at least naively, has no meaning in the negative side. But no, I already say that the meaning of the shuffle product and quadratic shuffle product
52:07
give two process, analytic process, to obtain rational number. Okay, yeah. Okay, always the process of singular expansion takes a finite part,
52:26
analytic expansion takes a finite part, and we get, again, happen like diagram. Okay, but we have also arguments with mixed signs, and then there are no rational numbers anymore. Yes, yes, it is just polynomial shuffle with rational series,
52:45
start of the plane, because they satisfy all of shuffle products. The start of the plane as exposed in the top of the graph today, the start of the plane satisfies shuffle products.
53:03
And take the start of the plane, shuffle, no complete polynomial, they all satisfy shuffle products. Okay. Do you confirm that? I have, yes, I don't see, I don't follow the link, but maybe there is
53:24
some. Let us study thoroughly. I had just a small question. You spoke about Q-shuffle product.
53:42
Yeah. Sorry. So I was thinking about the Q-shuffle product we did in a non-commutative symmetric fraction three, but it is not the same, at least presented like that.
54:02
Our shuffle, Q-shuffle does degenerate at square, at roots of unity also. Okay. My question is, you said that it was commutative and associative. Yes. Which is not trivial when you see the relations.
54:25
Do you have, is your argument direct computation or by, do you have some For quasi-shuffle, you say?
54:41
For quasi-shuffle, no, it's quite easy, because it's a kind of transport of the ordinary quasi-shuffle. You can see this expression here. Okay. So that from this, it's not so difficult to show the
55:02
Operate, yes. Okay, you have a reason, and for Q quasi-shuffle. No, this is Q quasi-shuffle. This product. Sorry, Q quasi-shuffle, okay. For the Q shuffle. This for the Q shuffle with all these relations that
55:22
Yeah, okay. Okay, yeah, the Q shuffle, ah yes, it was before. Actually, it's approved by induction on the number of letters in the expression.
55:44
I understand, yes, it is a sort of word, direct by direct computation. For commutativity, it's easy, for associativity, it's tedious. It's not difficult, but it's a bit lengthy, but okay.
56:01
Chapouton told us, but why don't you say that the representation you have is faithful, and then it will be obvious, but we didn't succeed in proving that the representation we have is faithful, so we had to redo abstractly the proof.
56:22
Okay, I understand.