On the Resurgent WKB Analysis
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Transcript: English(auto-generated)
00:16
Thank you, Stavros. So it's my pleasure to thank the organizers for this wonderful conference, and
00:25
I'm very grateful to be invited to participate. I want to report on the joint work with Frébéric Fouvet and Ricardo Scarpa. This is work in progress on exact WKB method and parametric resurgence, a subject which was
00:45
initiated by André Boros and Jean Ricard in the late 70s. So here I mentioned just 1983, a famous paper by André, and in 84, there was this paper, Sainte application de function réjourant, and with one of the application was dedicated to the
01:08
WKB problem. And in 94, there was also very interesting preprint on weighted convolutions, weighted products. So then there was work by the Nice school,
01:23
de l'abare d'illinger famme, and there is a famous DDP paper in 93, often quoted. There was also the Kyoto school, Kawai and Takei, and I mentioned here 2005, a monograph by them, but they had many collaborators among
01:42
who, and there were among them Koike, Tatsuya Koike, who passed away sadly last year. More recently, there was a lot of activity, a renewal of interest. Here is a selection. This is a very partial bibliography.
02:01
The article by Gayoto Moore-Nietzke in 2008 was certainly very interesting about WKB and the wall-crossing formula, and the BPS spectrum, and the Conserviture at Belmonde, about the wall-crossing formula.
02:23
I quote here Pasquetti's Chiapas because of topological string theory, and Garofaledi's It's Kappa F Marignot because of the connection with Pan-Levee in Stantons. I like particularly Iwaki Nakanishi 2014, nice survey about WKB analysis and the connection with cluster algebras, and
02:49
very recently, this year, Ito Marignoshu, about TPA equations and Riemann-Hilbert problems, just to say that there is a lot of activity, a renewal of interest about this subject. So our aim is to explain,
03:05
if possible, where resurgence comes from, using a tool invented by Jean Ecole, an inductive construction of elementary non-trivial resurgence functions for which mold formalism is
03:21
particularly efficient, so I'll try to explain that at the end of the talk. So the message is that these molds help elucidate the resurgent structure, they help establish the so-called bridge equations which are behind the Stokes automorphism or the
03:40
DDP formula, and they also induce recursive constructions of, let's say, n-point correlation functions in the spirit of all the recursions we heard about. So that would be the very end of the talk. So let's start with the one-dimensional stationary
04:01
Schrodinger equation written in that form, so psi is the wave function, x is the variable, eta is 1 over h bar, so it's a large parameter, and q, here we take for q a polynomial, or it could be a polynomial plus a polynomial in 1 over eta, it could be more general.
04:25
So we are given in slightly more geometric terms, meromorphic quadratic differential, q dx square, on compact Riemann surface, namely the Riemann sphere, which is regular
04:41
outside finite set. So of course there is the pole at infinity, but the zeros of the function q count as singularities for the quadratic differential, so they are called the turning points. So we have a complex curve,
05:02
which we denote by c dot 2, so c dot is just the regular set, and c dot square is the double cover of c dot, on which we have a square root of our quadratic differential, namely p dx, because
05:22
because, so let me write this, like this, lambda zero can be written q to the 1 by 2 dx, and q to the 1 by 2 is holomorphic on c dot 2. WKB method will produce,
05:42
let's say, an h bar extension of this classical 1-form, a formal 1-form on the complex curve. Lambda is p dx, where p starts with q to the 1 by 2, and you have a formal series in one of the eta, an h bar series, and
06:04
I call it a 1-form because it will be invariant by coordinate changes, and the coefficients will be uniquely determined and they will be holomorphic on the complex curve. So the definition will be recalled in a moment. We want to study this formal series in eta with coefficients holomorphic functions in x,
06:25
and we want to study the Borel transform with respect to the parameter eta. This is parametric resurgence. So here is the definition. The Borel transform replaces eta by xi, so we get hopefully a germ, hopefully a function of x and xi holomorphic for
06:46
xi of modulus small enough. So the rule being that we divide by the factorials like this, and this should be endlessly continuable with respect to xi, in the sense that the analytic continuation exists in the large with only
07:02
isolated singularities. We repeat the Schrodinger equation. So not only will we produce this lambda, so I don't tell you yet what is this lambda, but we will also produce
07:23
solutions to the formal solutions to the Schrodinger equation in the form of the WKB ansatz. So there is this exponential i eta a, where a is the classical action function, the integral of q to the one-half, which is well defined on our
07:43
complex curve, but of course the primitive is multivalued on the complex curve. We have this q to the minus one-fourth, and then we have phi plus or phi minus if we take the opposite branch of q to the one-half. And phi plus and phi minus will be
08:02
series in one over eta, normalized by starting with the constant one. So these coefficients phi zero, phi one, like a, will be multivalued, and the series phi plus and phi minus are determined only up to a multiplicative x-independent factor, of course,
08:24
because it's a linear equation. We'll be interested in the Borel transform with respect to the parameter eta at least for certain choices of normalization, since there is this question of determining one solution that we will go and transform. Maybe certain choices are better than others.
08:46
So I repeat, resurgence consists in being endlessly continuable in the variable xi depending on x. Maybe this property will hold for generic x, but not all x. For instance, x should not touch the turning points.
09:04
Simplest example. So let me repeat the formula for the WK beyond that here. So we have plus or minus i eta a of x, q to the minus one-fourth.
09:21
This is very, very classical, of course. Phi plus or minus of x and eta. And of course, you notice that this a of x is just the integral of lambda zero from some reference point x0 that we fix once for all.
09:45
Let us see the simplest possible example. Only one turning point and nothing else. Turning point of multiplicity m, let's say, at the origin. So we call xj the turning points and we call x infinity the pole at infinity. In that case,
10:03
you compute the action function a of x, you compute q to the minus one-fourth and it so happens that there is a closed form for the Borel transform of these series, phi plus and phi minus.
10:21
You notice that I have inserted this factor, this monomial eta to the minus mu minus one, where mu is this number depending on the multiplicity m. So this is just convenient. In this way, we get an algebraic function of xi.
10:40
So you see that if you replace phi plus by one, the Borel transform of that monomial is just xi to the mu divided by gamma of mu plus one. Here, when you take the Borel transform of the product, you have this very elementary formula. So you observe that indeed
11:01
the thing is convergent in the xi plane and indeed it has a nice analytic continuation. It's algebraic. The Borel transform of phi plus has only one singularity at 2i a of x and phi minus has only one singularity at minus 2i a of x.
11:22
Now, what is the singularity of Borel of phi plus? Let me write this. It should be clear that it will be proportional to the Borel of phi minus. Indeed, let's have a look. Borel of
11:41
eta minus mu minus one phi plus. You can write, you can factor out this 2i a of x. Let's put it in front. 2i a of x power minus mu divided by gamma of mu plus one and we have this
12:01
xi of mu, xi to the mu and 2i a of x minus xi power mu. Now, what does it mean to consider the singularity at that point? It means that we we replace xi by 2i a plus zeta and we want that xi approach
12:26
approach 2i a, we want zeta to approach zero. So you see that this will become exponential i pi mu zeta to the mu and this will become
12:42
2i a plus zeta power mu. So in fact you exactly you find the same formula as here. So in fact, if you, it's an exercise in alien calculus, if you manipulate this alien operator defined by ECAL, which is just the
13:03
the smart way of measuring singularities in the Borel plane, you find this formula. So while it's an exercise, you have this exponential i pi mu, but then you have to to study the dictionary between the singularity and the function in the xi. I mean you have the notion of major, the singularity is defined by a major and then you take the minor. So when you compute, you find this sine function.
13:28
Incidentally, that formula was written by Jean Recal in 81, in the second book of the series on a resurgent function in quite a slightly different context. It was in the context of equational resurgence
13:43
dealing with Riccati equation. And in fact, we'll see the Riccati equation in a moment. So particular case, m is one. This is the quantum Airy equation, parametric Airy equation. You find the Stokes constant, this number here, this proportionality factor will be
14:02
minus i in that case, because mu is minus one over six. And for the quantum harmonic oscillator, you find a different number. Okay, that is really the simplest example you can dream of. So now to the associated Riccati equation.
14:25
We consider the logarithmic derivative of the wave function psi, and it's an exercise. It's very clear that you agree that the if i eta p is exactly the logarithmic derivative, then its derivative is written like this.
14:43
And so the Schrodinger equation is equivalent to that Riccati equation, p squared equal to q plus i eta inverse derivative of p. And now there are only two formal solutions to that equation. Indeed, it's clear that you can choose to start with plus or minus q to the one half.
15:04
And then once you have chosen that, the rest of the series is determined. Here is the next term, it so happens that it is a logarithmic derivative, and then plus or minus this, etc. So you find two
15:23
well-defined formal series. By the way, they are related because it's if you go, you can go from one branch to the other by the covering involution. And so you exchange the two solutions by the covering involution, which defines a complex curve C dot 2.
15:46
And from those two formal series, you can recover all WKB solutions to the Schrodinger equation. Indeed, since we are speaking of a logarithmic derivative, it's just a matter of choosing a primitive of p plus or p minus, and
16:03
then exponentiating it, and then you are free to multiply by any x independent factor. So we will recover it this way when you apply this recipe, though the first term here is responsible for exponential plus or minus i eta a.
16:20
Indeed, remember a is just the primitive of q to the one half that we have chosen. The logarithmic derivative of q is easy to integrate, and you get this q to the minus one fourth, and then this, the remainder here, will give rise to a series phi plus or phi minus, as we had said previously.
16:42
So what is this lambda that we had promised at the beginning? There would be one form, formal in eta in this problem. What is p? It's just the h bar extended momentum. It's the half of the difference. You have two solutions,
17:00
you take half difference, and you can check that indeed it gives rise to one form. Now each series p plus or p minus is this plus or minus the half difference plus the half sum, and it so happens that the half sum is the logarithmic derivative of the half difference.
17:24
This will help us to handle this question of normalization choices. The coefficients of p plus and p minus are uniquely determined, holomorphic on our complex curve. They have Poisson expansions at the turning points and at infinity, and you check that there is no
17:42
term in the form 1 over x minus xj. As a consequence, to choose a normalization for psi plus or psi minus, to choose the factor here is equivalent to select a primitive of p plus or p minus or equivalently a primitive of p, curly p.
18:05
So here is what we do. So we must integrate this, so let's split p, curly p, there is the first term q to the 1 half, and then there is the remainder, curly y with this factor q to the 1 half, and
18:21
then correspondingly p plus or p minus is plus or minus the first term, then this we repeat, and then we have plus or minus 1 over i eta q to the 1 half, curly y, and so we can define what we will call the xj normalized solution, xj being one of the turning points or the pole at infinity, like this.
18:45
We don't touch the pre-factor, the exponential pre-factor. Here we have p, curly p to the minus 1 half, because we are starting with this, we are using this, this natural primitive of the logarithmic derivative, and here we must choose a
19:04
primitive of q to the 1 half curly y. So we choose the primitive whose Poisson expansion on xj does not contain any constant term. That's a reasonable choice. Equivalently, this is the average of the primitives corresponding to the base point a varying along a cycle on the complex curve.
19:29
So something more about these normalization, because they play a role in the theory, so psi j plus or psi j minus, you can write it this way using these slightly different notations for the selection of primitive, and so q to the 1 half
19:46
curly y, this is just lambda minus lambda zero. And a, of course, we know what it is. So here appear the so-called Voron's coefficients. They allow to connect one
20:01
normalization to the other. How do you go from j to k? You multiply by a certain integral, and the integral of q to the 1 half curly y. Now there is a remark, let me do a picture, that y
20:21
is invariant by the covering involution. So when you multiply by q to the 1 half, you can use this formula. You have xj here, and you have x somewhere. And so instead of doing this business of choosing the
20:47
the primitive by looking at the piezo expansion, you can do that. You can say, let's do this. So there is a cut here. So you start from x star, the other point for the other branch, and you turn around.
21:03
So this is the path gamma j x, which is here. So it starts from x star, it encircles positively xj, and then it goes back to x, and you compute that the integral on that path is exactly what you want.
21:21
So this means that our coefficient pi j, our integral is just, let's say, a quantum period. It's i eta over 2, the 1 half comes from here, of lambda minus lambda 0. So what you can do in that problem is to...
21:41
Maybe I should explain what is gamma jk. I mean, here we have another turning point, xj, xk, and we are supposed to compute the difference between these two primitives. So the difference can be computed by saying that we do this.
22:08
So we find a cycle appearing in the... So this is the gamma jjk. It's the cycle around j and k.
22:23
And then we can then, next we can choose, let's say, we have our reference point somewhere here, x0, so we can choose a basis for the homology. And so this would be gamma j gamma k. So we can compute once for all these
22:41
pi j of eta. So integral from x0 to xj with the rule that we have set for the turning point following the path gamma j that we have selected in advance. So this is the definition of a normalization for each
23:05
singularity. At infinity, the situation is simpler because the coefficients decay at infinity and you get an integrable function. So you can replace this funny selection of primitives by the usual integration from infinity.
23:23
So here is what you can do. Let us split p plus and p minus. Like before, we have the first term, plus or minus q to the one-half. We have the second term, and then we call y plus or minus the rest with this pre-factor. So we define this way, y plus, y minus, and
23:43
so you see that the curly y is just half the difference. And now the point is that the normalization at infinity is just defined like this. Using the integral of q to the one-half, this y plus or y minus, integrated from infinity to x.
24:02
So the solution normalized at infinity are characterized by the vanishing of at infinity of all the coefficients except the very first one. We said that this phi plus should always start with 1. And so what is y plus? It's just a logarithmic derivative essentially of phi plus or minus infinity.
24:23
So why are they resurgent? Where does resurgence come from? For all these objects, the y plus, the y minus, the phi j plus or minus, we will deal with phi infinity, plus or minus, which is easier to define.
24:41
So why is it resurgent? Let us define a vector field on our complex curve, d. d is q to the one-half, d over dx, so it's due to the square root of the quadratic differential. It's due to this lambda zero. And this way, we can rephrase the Schrodinger equation
25:04
using the unknown phi plus or minus instead of psi, and using d instead of d over dx. And here is what we get. So we have this function k, which is very important. k is given by that formula.
25:22
So it comes from the initial q, which we have started with. And correspondingly, we can change unknown in the Riccati equation. And again, we can use this vector field d, this derivative, and we get this equation for y plus and y minus.
25:44
Okay. So y plus or minus in this sense is the d logarithmic derivative of phi plus or minus, and here it's not the xj normalized solution, it's the x infinity normalized solution.
26:00
So here we are supposed to invert d. So to invert d, we multiply by q to the one-half, and we choose the primitive which vanish at infinity. So it's the x infinity based raised inverse to d. So why will it be resurgent? We can compute, in a sense, phi plus and chi plus. So these are the notation, the
26:25
d derivative of phi plus. So remember that y plus will just be the ratio of these two. We can compute as well phi minus and d minus 2 i eta phi minus. Using the Neumann series, because it's a linear problem,
26:44
so it's a very particular case of parametric resurgence. So this, you recognize here the Riccati equation for the phi series, using d. So we insert this chi plus, it's d phi plus, and so the second equation is this.
27:02
And now we say we have specified the way we solve the equation. We have specified that phi plus should start with a 1, and then the other coefficients vanish, and chi plus will vanish. In fact, we know how to invert this because here we are dealing with operators in
27:23
formal series in eta with coefficients holomorphic on the complex curve C dot 2. So k, curly k is just the multiplication operator by k, and what is the inverse of d plus 2 i eta? It's given by the geometric series. This is a perfectly well-defined operator.
27:41
We have eta to the power minus k minus 1. So you differentiate all the coefficients, and you see that you differentiate the coefficients of holomorphic functions, and you do not compensate by any factorial. So you expect the growth and you expect divergence. Series in inverse eta?
28:00
Yes. Series in inverse eta? Always, always. Oh, yes, you're right. Here, minus 1 is lacking. Thank you. So the Neumann series for this linear problem gives you this. So phi plus, you see you start with 1 and you repeatedly you multiply by k,
28:21
apply this inverse operator, and then you integrate from infinity. And chi plus, there is this difference, you have one more factor. Similarly, for phi minus, when you do the computation, you find this. Minus 2 i eta times something, which starts with 1 over minus 2 i eta, and
28:42
chi minus is this. So you have four series of formal series in 1 over eta. So we are dealing with series of formal series, but each of these terms, each one individually, is a non-trivial, resurgent function.
29:01
So why is that so? It's because the counterpart of the operator that we use, d plus 2 i epsilon eta. So epsilon here is obviously, it's either plus 1 or minus 1. Here it's a
29:20
typo, it's minus 1, 2 here. So we are using plus or minus 1. So in the Borel plane, what happens? In the Borel plane, we divide by the factorials. So we get an exponential series, an exponential of d. So d is a derivation, it's a vector field, exponential of a vector field gives you the flow.
29:42
So you see here the flow, the time t flow map of our d, so remember d is just q to the minus one half of x, d over dx. So we should
30:01
inquire about the flow of that vector field. And when you know the flow, you take it at time minus xi over 2i epsilon, you substitute in g. So if you start with g not depending on eta. If you start with capital G, which is a series in eta,
30:22
then the computation is like this. Multiplication by eta to minus k minus 1 gives rise to convolution by this monomial. And then you write the the convolution as an integral, and again you see the flow of our vector field at minus xi prime over 2i epsilon. So we get an integral operator and we see
30:47
very concrete formulas that we can manipulate and that we should apply to this. So we are taking now, we know how to handle the Borel transform of all these series.
31:02
So the claim is that the Borel transform, phi hat plus or minus, chi hat plus or minus are endlessly continueable. And where are the possible singularities? Where are they located? So the recipe is that
31:21
there cannot be a singularity at omega unless when you flow along these vector fields starting from x until minus or plus omega divided by 2i, you hit a turning point. So this is the recipe to determine the singularities of the Borel transform. So you see that the singularities in the xi plane depend of course on x.
31:47
So the idea is that the vector field that we are considering is straightened by the so-called Youville transformation, which is just the action in function viewed as a change of coordinate. And in that coordinate z
32:00
the vector field is strengthened. So you can compute the flow and in fact the quadratic differential is strengthened too. So you can use the local geometry of the quadratic differential if you want to analyze the behavior near xj for instance. So in that coordinate z, everything becomes simple. But be careful that the coordinate z is defined locally.
32:24
And then you follow its analytic continuation. So there is a kind of periodicity in the problem, but it's tricky because everything is multivalued on the complex curve C dot 2. Excuse me, can you remind me what is omega x? So it's omega of x is defined in this claim.
32:47
So what are the potential singular points? They are the points such that when you flow until that time you hit a turning point. And in a moment, you will see that in fact we are determining them.
33:01
We are here reaching an alternative definition because the flow Here I'm using this gamma j, so from x0 to xj. If I start from x0, alpha j is the value of the action, one of the possible value gamma j.
33:25
But of course if you follow different paths, you get different values, you get replicas of this singular point alpha j. So alpha j is a singular point for the flow map, because we are saying that
33:42
the vector field vanishes here, and so in finite time we hit a singular point of the vector field. On the other hand, the function k that we are using in our Neumann series is meromorphic on C, but it has poles at the turning points.
34:01
Hence in coordinate z, k has endless analytic continuation, and where are the singularities in the z variable? They are located at all the possible values of the action function a. So alpha j plus
34:20
multiples of the classical periods. So this is the description in variable z, starting from x0, but now we are starting from x, so we must shift everything by z. So the recipe which is here gives us z minus or plus omega over 2i is alpha tilde, one of these values.
34:42
So omega of x, this is the answer to your question, is plus or minus 2i action of x minus alpha tilde. And so why? Yes, yes, yes, yeah, I mean it's just the action, it's just a representation of the homology group.
35:06
Excuse me, so among these x1 to xm you count x infinity also? No, not here, not here because you cannot go to infinity in finite time. Yes, of course you can in cube 4, for x4 you can.
35:22
Or maybe a mistake and let me think. No, no, no, look z is primitive of q to the 1 half, so we are integrating, we are integrating the square root of the polynomial. So it's integrable at the turning point, but it's not integrable at infinity.
35:40
This is the non-integrable part that we are always skipping when we come, when we go to infinity, we must almost always remove that part. May I cautiously disagree? I mean, I discovered that in fact you should not remove it. It's nicer, you can get cleaner things if you don't remove it. I can believe it, I can believe it. I mean, this is probably not the definitive version of...
36:06
I would be happy to see your comment in detail on that. Thank you. So where does this condition come from? So I claim that this is the condition to have possibly a singular point. Where does it come from?
36:22
So remember we were dealing with these Neumann series of that form. So in the case of phi plus, we have to invert d plus 2i eta. So the Borel transform involves the flow at times minus xi over 2i.
36:43
And so this is why we get the condition like it was written. Flow at minus omega over 2i should be on the singular locus. Similarly for phi minus. So the next question in if we are dealing with the resultant structure of phi plus and phi minus is what are the alien derivatives?
37:08
So I'll be a bit sketchy, but because in fact it's work in progress. So these formulas are written in the literature, but our understanding is still in progress. So but
37:21
the idea would be the following. If you are dealing not with the solution normalized at infinity, but with the solution normalized at xj, the alien derivative will be governed by the sj that we have computed a moment ago. So all
37:40
everything here is like as if you are dealing only with a pole and with a monomial q of x. You can substitute q of x by this, mj is the multiplicity at xj, and you get that result. But that is true only for phi j, when you are dealing with the alien derivative at a minus alpha j.
38:06
Now if we want to deal with phi infinity, we must go from phi j to phi infinity, and for that purpose we can use the voron's coefficients. And so now that's an easy computation. The alien derivative of that guy should be the exponential multiplied by the alien derivative of that guy.
38:26
So we have the sj and we get twice pi because one comes from here and the other comes when you express phi j minus in terms of phi infinity minus.
38:41
And similarly for the other for the other one. So we can go on and get formulas for the alien derivative of the Riccati solution, because it's a general fact in resurgence, I mean this part of the theory of Jean, that if you are dealing with resonant functions, their ratio will be
39:03
resonant. Their exponential, their logarithm will be resonant. So and moreover, there is an alien chain rule if you want to compute the alien derivatives of these. So once you have computed the alien derivatives of one set of solutions, you can, in principle, you can deduce the rest of the structure.
39:24
And then the next question is, but what about the alien derivative of that coefficient? So in fact here, I'm referring implicitly to the second bridge equation and the third bridge equation. So here, this is here that DDP formula would appear,
39:42
because the location of singular points for that factor, that factor does not depend on x. It's Voron's coefficient, it doesn't depend on x. We have integrated in x. So you can guess that the singular points will be located at those points, the classical periods.
40:00
So I stop here my comments on the resurgence, because I want to move on to something different. We have said that we would reach the resurgence structure using these representations of the solutions normalized at infinity by means of these Neumann series, series of elementary
40:24
non-trivial resurgence series. A way of motivating what follows is to ask, can we do the same for y plus and y minus? So we have said, we are confident that it's a ratio of resurgence series, it will be resurgence. But can we analyze y plus and y minus in the same
40:44
manner that we analyzed phi infinity plus and phi infinity minus? So here is what happens. We have our four series of series. Here is a compact way of rewriting them.
41:02
I introduce those series m indexed by plus or minus epsilon 1, epsilon 2, epsilon r. These are strings, these are words, the letters of which are plus or minus. So when the word is empty, you have 1. And then how do you define m associated with a word of length r?
41:27
So it's an inductive definition, so you invert this operator d plus 2i eta, the sum of the letters. You invert it using primitive at infinity, vanishing at infinity, if the sum happens to be zero.
41:44
And then you apply that to b with the first letter. b with the first letter is either k or 1. And then this is the function, the series associated with the truncated word, in which you have removed the first letter. So is that clear for you? For instance, start
42:03
with phi plus. So you have, what is this nth term? So we have 1, so you multiply by k, you apply this operator, multiplication by k, and applying the operator, this is multiplying by b and applying d plus 2i eta when there is only one letter, which is epsilon 1 equal to plus.
42:24
But then immediately after that, you apply the inverse of d infinity, which is that case when epsilon 1 plus epsilon 2 is zero. So this is why I wrote here m minus plus. I started with plus, so that was the, and then I have a minus, etc. So these are elementary
42:48
series. This is a family of series indexed by words on the alphabet plus or minus one. This is the definition of a mold, a family of objects indexed by words.
43:02
But here we see only a slice of the mold. We are using only alternate words of length, of even length or odd length, but only alternate words. What about the other words? What about the rest of the mold? What will it do for us? So I repeat here the definition of the mold,
43:23
the inductive definition. I repeat what we have obtained so far, and here is the claim. We can represent the solutions to the Riccati equation in the same way. So here we have series with coefficient one and only peculiar words, but here we will use
43:43
a different subset of words, and we have different coefficients. These coefficients beta plus and beta minus are integers. They are very often zero. To be non-zero, beta plus requires the sum of the letters to be one, and the value is computed by iterating these operators. So you
44:05
have two elementary operators, d over dy minus y squared d over dy. You see that there is homogeneity going on here, because b plus is homogeneous of degree minus one, b minus of degree plus one. So when you do the composition, you see that the product applied to y will have degree
44:29
one minus the sum, but if the sum is one, it's a constant, and that constant is an integer number. So this is the recipe to compute beta plus, and the recipe for beta minus is here.
44:42
So now we are interested in words whose sum is minus one. And so I claim this, and I also claim that if you forget the factor minus one to the r plus one, instead of finding y minus, you will find one over minus two i eta plus y minus. So why is
45:03
that true? I would like to spend five minutes to explain that. It's a very elementary part of mold calculus, as defined by Ekal. So let's rephrase it here in our context. The key point is a certain family of quadratic relations. So we are dealing with a mold on a certain alphabet,
45:26
which was plus or minus one, with value in the ring r. It was in our case the ring of formal series in eta whose coefficients are holomorphic on the complex curve C.2. So we said a mold is just a family of objects indexed by words, so we can say this is the
45:44
collection of values of a sequence of functions. So giving a mold v is like giving a sequence of functions, and the nth function depends on n arguments. These quadratic relations that we will
46:02
write are called symmetric relations. So here it goes, v0 should be a constant, it's a function of nothing, and that constant should be the unit of our ring, and then for all pq, for all p-tuple b, for all q-tuple c, the product should coincide with this sum. So it's a sum of
46:28
values of vn, n is p plus q. So you take the sum of all subsets of the set of indices 1 to n, so i is the subset of cardinality p, j is the complement, cardinality q, and what is this?
46:48
This is the n-tuple a, the unique n-tuple such that when you extract the i part, you recover b, and when you extract the j part, you recover c. So I wrote the formula in this way because it's
47:04
reminiscent of something which is well known in topological... previous example? Yes, here v is general mold, and we will apply this to m. So this is just a more general definition. A1 am was plus and minus? Exactly.
47:22
Here is a typical example of mold satisfying this quadratic relation, and our m is very much of the same nature. So here is the standard way we rephrase this property in mold calculus. This is absolutely equivalent to using the shuffling of the two
47:45
words b and c. So our mold m satisfies this. So our mold m is defined on the two-letter alphabet. It is symmetrical and you can check it by induction on the length of the sum of the
48:02
lengths on the sum p plus q, and it's not difficult because you have an inductive definition and you have a characterization. It's defined through a differential equation with a boundary condition at infinity, so it's not difficult to test this property. So our mold is symmetrical. What good does it do to us? So remember that the Schrodinger equation
48:24
was written for phi plus in the form of a system. So this is the system for phi plus chi plus. Equivalently, we can say this is a vector field or this is a derivation
48:41
acting on that algebra, the algebra of formal series in y1, y2, whose coefficients are in r, and again the same typo, this should be eta inverse. So our coefficients take value in r, and let's consider these derivations. So d acts on the coefficients, d is the differential d over
49:08
d, essentially, and so we have d over d y2, d over d y1. You remember that b plus is k, so b plus y1 d over d y2, this is what comes from this k phi, because phi is y1,
49:23
and b minus is 1, so this chi here is responsible for this y2 d over d y1. Let us call L0 the first part of the operator, let's call b plus bar this operator, b minus bar
49:40
that operator, so we are dealing with this derivation, and this derivation encodes the Schrodinger equation in a certain sense. So these are derivations. Now let's consider this operator, so this huge sum, sum over all possible words of m, epsilon, and this product of
50:02
derivations. So this is formally its convergence, so it defines an operator, and the inductive definition of our rule precisely says that it conjugates L and L0, so it's like a normal form problem, we are conjugating what is given to us to something
50:25
simpler, we have gotten rid of these two terms. But because the mold is symmetric, theta is an automorphism, that's a very general principle. Why is that true? It's because you see how theta
50:42
is constructed, we are using the mold as coefficient, and here we are using product of derivations. Each derivation satisfies Leibniz's rule, a product of derivation satisfies a generalized Leibniz's rule, and when you write the generalized Leibniz's rule, because they do
51:00
not commute, you see the shuffling coefficients appearing, and by duality, if the mold is symmetric, then this guy will satisfy that the product is sent to the product, instead of Leibniz's rule. Some of all possible words, including the empty word, all possible lengths,
51:25
it's a huge sum. And you don't put a factorial? No, no factorial here. So it's an automorphism. What are the automorphisms of the space of formal series? They are all substitution
51:41
automorphisms. So we know in advance that theta f will be f composed with something, and that something is obtained by evaluating theta on y1 or y2. In fact, in our problem, when you do this, you find that this little theta, this formal change of coordinates in the y1, y2 space,
52:05
is this. So everything is very simple in that case, because the problem is linear. But wait a minute. So we obtain, I don't do the computation that it's not difficult, you obtain that. So you can believe it, I mean, after all, what will happen when we evaluate this
52:22
product of operators on y1 with so elementary vector field, it will be very easy to compute, you will see on the alternate word, it's clear. So you get that. Okay, but remember, that operator is an automorphism. So I repeat the formulas here, definition of b plus bar, b minus bar, definition of theta, and this is what we have obtained so far.
52:47
Now, what is the general solution to the Riccati equation written for the y function? It's the logarithmic derivative of the general solution to Schrodinger. So you see the general solution to Schrodinger is sigma plus and sigma minus are just parameters.
53:07
So since I focus on the equation with a plus, I see the multiples of phi plus and the multiples of phi minus appear with this correction exponential minus two i eta a. a is lacking here.
53:23
This exponential minus two i eta a, because I'm switching from phi minus to phi plus. So we can introduce sigma, which would be sigma minus over sigma plus, and you see that this takes the form of a linear fractional evaluated on this
53:44
monomial. So let's introduce theta for Riccati to be that. So this is a formal diffeomorphism in the y space with coefficients depending on x and eta.
54:00
So morally this y is just y2 over y1. Because theta is an automorphism, because of the symmetricity, when you compute theta of y2 over theta over y1, it's the same as theta of y2 over y1. So you can compute that thing, theta Riccati of y, as
54:24
the evaluation of that big operator capital theta on y2 over y1. So previously we were just evaluating on y1 or on y2, but now we are evaluating on y2 over y1. And what do we find? Well, the image by just one operator b plus is one, and the image by one operator b minus is
54:47
minus the square. So maybe you remember the formulas for b plus and b minus, they were just as simple as this. So this is how we prove the claim. So we have obtained that formula,
55:02
that y plus, the solution to Riccati, is given by this mold expansion with coefficients beta plus, which are produced by these operators, elementary operators acting on y. This is a very very elementary instance of mold calculus, but what does it
55:21
get? Where does it get us? We obtain this, we have a recursive definition of this m, and when we solve the induction, this is what m epsilon is. So epsilon is an arbitrary word, and so we are inverting that operator, multiplying by b epsilon one, inverting that one,
55:42
multiplying etc. And whenever a suffix of the word has zero sum, the prescription is that we should use the primitive vanishing at infinity. Okay, let us define the resonance level of the word as the number of zero-sum suffixes.
56:06
How many terms here do vanish? How many times do we need to use the integration from infinity? The rest of the times, we use only the geometric series with d, you remember, but because of those zero-sum suffixes, our m epsilon will involve an n-fold integration from
56:28
infinity. And this is slightly reminiscent of something we've seen in the talk about topological recursion. So we are tempted to group together all the words with the same resonance level n. So let us introduce wn plus or minus as the sum of the beta. So
56:47
in this big sum, we extract the words with resonance level n, and we can call that n-point component, or n-point correlation function, and then our y plus is the sum of that.
57:04
So what we obtain here is that we have a recursive definition which produces automatically a sequence of objects whose sum is really the solution to the Riccati equation. So afterwards, you just need to integrate from infinity to exponentiate, and you get a WKB solution.
57:26
And you know that it was mentioned in the previous talk that in every case topological recursion does produce the WKB solution, but here we have a different angle, a different way of constructing recursions, and so let's say this is food for thought. We plan to
57:47
investigate that further. Thank you for your attention. So to prove the claim, you need to prove that the
58:03
sum of terms converges somehow or other. You are referring to the last? Claim, or? I mean, anywhere. The claim and then the... Oh, well, there are claims of different nature. For instance, that one, where is it?
58:21
That one... Okay, do you mean on the resurgence? Yes. Yes, I skipped that part, but indeed, to prove resurgence of phi infinity plus or phi infinity minus, we should establish that this series of formal series gives rise in the Borel plane to a series of functions, and that's a fact,
58:45
no problem, but so then we need normal convergence on the compact subsets to on a certain Riemann surface. Yes, indeed, this is the strategy, the reasonable strategy, which was indicated by Jean Recal in as early as 84.
59:01
So my question is, at the end, are you planning to say that basically the wn's have some kind of... there's like not too many wn's, and each one sort of gets smaller and smaller, and that that's why it converges? I mean, the series for y plus is...
59:23
we can apply the same strategy directly instead of invoking a general theorem which has the ratio between two resurgence series must be resultant, we have access directly to the to a decomposition of y plus in elementary pieces in the Riemann surface. So that's the strategy, yes, it would be to
59:43
and then after regrouping we would like to see what happens, but well it's work in progress as I mentioned at the beginning and... May I ask another question? So you had wn's, maybe in the next slide?
01:00:00
Here, just topological recursion. Yes, here they are. Right there, right there. So are these Wn's supposed to be factorial divergences? Y plus is, but these are Wn's themselves.
01:00:27
Yes, everything here is resurgent. I mean, this is a decomposition. I mean, in n you mean factorial, with respect to n it's convergent. With respect to n, we have convergence here.
01:00:41
The divergent takes place only with respect to one over eta. So this is a series of, a resurgent series, which is convergent for the topology of resurgent functions. Because in topological recursion, those sums are factorial divergent. So in fact, it's a different angle. So we are doing something else.
01:01:01
Yes? I think the point is, in fact, that those should be factorial conversion, because when you do an n-fold integration, you get a one over n factorial. And then maybe there's only an exponential number of terms. That is indeed the case. Yeah, but in topological recursion, it doesn't happen, apparently.
01:01:20
So here, the computations are organized in a different way. So this is what we wanted to propose to that audience, to see a different way of organizing the computations. Andrei? I would like to ask you, I'm very excited by this normalization at infinity.
01:01:42
Because in my work, I was guilty. I think I didn't realize that it could be done. So later, I made one normalization. And the advantage of normalizing at infinity, including the divergent part, is that you get something which is translation covariant.
01:02:02
Now, is your normalization at infinity explicitly also covariant by translation? It's not mine. I mean, it's in the literature. It's, I mean, in FAM, it's already in FAM. Yeah.
01:02:21
And in fact, in Ekal's paper in 84 already. So is it covariant? I mean, the trick is that lambda 0 is not integrable at infinity. So you have to remove that part. You can't define it. You can't define the integral to infinity by a way, a classical version of theta regularization.
01:02:44
So in a way, it, but perhaps it's, finally, it might be allowed to what you are doing. So I'm very excited. So I didn't think too much about it. Because for us, the idea is that we fix once for all this x0. And it appears only in this pre-facto.
01:03:02
It's a totally different business when we are referring to the integration of that guy. Because here we have a whole series. So theoretically, we might change the selection of primitive at each term of the series. And then we would get crazy things. But here, it's just once at the beginning. So we didn't pay much attention to that.
01:03:21
But I would be interested in your comment about zeta regularization. So I haven't quite understood. To write a full solution, you would have still to write some bridge equation for the Boris coefficients, right?
01:03:42
And the second part was not some method to get around. Like, you would still have to? Yes, it's, yes. These are two avenues of research, yeah. I mean, I switched at some point from one direction to another. Yeah, yeah, yeah.
01:04:06
Thank you.