 Okay, so thank you very much. Thank you very much for the invitation. It's a pleasure to be here. And yeah, so I changed the title a little bit, but it's mainly the same things. Talk a little bit about resurgence in matrix models and in string theory, but with some emphasis on polynomial equations. This is a work in progress with my students who are all here, Salvatore, Max and Robert, and my collaborators, Anisette and Bonk. All right, so a little bit of motivation. So if you come from the physics side, this could be something that could interest you. How can I compute non-perturbative quantum gravity? So that's a very hard question. And string theory tries to address that, and it does generically so perturbatively. So that's sort of the free energy of some string theory, which is this famous sum over a remand surface of different genera. And that series is a synthetic. So we would like to go sort of beyond it, find a non-perturbative construction of string theory. Not so much a definition. I will not have a lot to say about that, but about a construction. And in the process, it could be nice that whatever this non-perturbative construction is, that might be a trend series, could give us some hints on the semi-classics of what's going on. Just not perturbative semi-classics, which is encoded in that sum over remand surfaces, but perhaps some non-perturbative semi-classics. And the simplest example to do so is in two dimensions, where quantum gravity is described by the polynomial one equation. So if you don't care so much about physics, but math, you care about that equation, which is up there. That's the polynomial one equation. And this describes the specific heat of minimal strings with no matter, which is basically to the gravity. And the solution is this specific heat, and you can construct the free energy and the partition function or the tau function by just following some specific rules, where the string coupling relates to the z of polynomial in the way that's written there. And here's the free energy. This perturbative series is a synthetic. This coefficient grows factorially fast. We would like to go beyond perturbation theory and say something a little bit more precise. And this can be said for other equations. So another equation that could be of interest is polynomial two, because it relates to the supergravity. So that will also be interesting. The equation is there, the partition function or tau function is there, the relation between the z of polynomial and g string slightly different. But again, we have a perturbative series with a synthetic coefficients grow factorially fast. There's many other examples. We could have a higher order polynomial type equations, and so on. Mostly I will talk about polynomial one. I'll say some things about polynomial two. And I'll also say some things about discrete polynomial one equation. So that's sort of, let's say, more physical motivation. But there's also a mathematical motivation to look at this. These are the famous poles of polynomial equation. So these are plots for the polynomial one equation. The polynomial one equation has a z five symmetry. So there are sort of five pizza slices on the complex plane of z. And it has movable singularities. Excuse me. There are movable singularities, which are double poles. This has been classified, actually over 100 years ago now by Buttreau. And poles can accumulate in one pizza slice. This is so-called tritranquet solution on three adjacent pizza slices. This is the tritranquet solution or everywhere. That's the general solution. So you could ask, what is the exact location of these double poles? Well, they depend on initial data or boundary data or trans series parameters, whatever you want to say. But if I tell you what's that initial data, where exactly are they? So could we say something about that? To do so, we have to do Stokes phenomenon. And this plot, you probably all have seen it because it amazingly made it to the poster of the conference. But so now it's more famous than it was before. So I'm very happy for that. And in order to do Stokes phenomenon, I'm sorry, in order to find where all these zeros are, you have to cross these lines where there are singularities on the Borel plane. That's the Borel plane and that's a line with singularities. And so I need to know how to cross all Stokes lines in order to tell you where all these zeros are. But every time I need to cross a Stokes line, I need to know about Stokes data. So if I want to go everywhere, I need to know all Stokes data. And because the problem is nonlinear, well, I have an infinite set of numbers. There's a quotation mark for a reason. I'll tell you that in a minute. So I need to know all these numbers. So that's a part of the project that we have ongoing. Can we find all those? And we would like to have a bonus, which would be, can we also have a physical interpretation of what these poles are and what these Stokes data is? That's less clear. And this does not only hold for these Polyvue solutions, but also for classical Polyvue equations, but also here for discrete Polyvue. This comes from a matrix model. This is the discrete version of, say, the Polyvue 1 equation, where r now replaces u. It's the specific heat of a matrix model. And, well, I'm not telling you what that figure is. I'll tell you that in about half an hour, 45 minutes. But those are also poles, double poles of these discrete Polyvue r. And I'd like to know that's a density map. This was done in a computer. It was a slow calculation in a cluster. I will try to also find out where are all these guys? Can I actually pinpoint them, do something a bit better than just a density plot? Again, I'll mostly focus on Polyvue 1, say a little bit about Polyvue 2 and discrete Polyvue. Okay? So that's sort of the motivation. And so I'll tell you a little bit about old work, how we constructed Polyvue Trance series, and how we would try to approach resumming them and trying to get some information out of them. And then we will see that these Polyvue solutions, or Trance series to be more precise, they'll have a property which is associated with the fact that they're resonance. So I need to understand what's the good context to deal with resonance in the Polyvue context, and how will that influence what goes on with Stokes data? So Stokes data is, let's say, slightly different from non-resonant case. And I will show you these numerical work that we've been doing on how to try to find out what are all these numbers and how they organize themselves. Then I'll slightly change gears, assuming that, okay, so let me already give you the punchline. We don't have all the numbers, we have many numbers, but we don't have all of them, so that problem is still not solved. But let's assume that maybe I had thought for all of them, and we still hope that we will do that soon. Then I can move on and now find out where are all these poles, where are they located. Both for the Polyvue case and for the matrix model. And then in the end, I'll tell you if there's time left, I'll tell you a little bit about resummations and there's some classical decodings, plural, and I'll tell you why I'm using plural here. All right, so let's just start, only if you have any question. Again, that's the question I want to solve, and I want to try to find a trans-series solution to it. So the Protervative Solution at large Z, that was small string coupling before, looks like that, and I can grab that ansatz, plug into the equation, and I get the recursion equation for the coefficients u of g, and I just, you know, put it in the computer and it starts competing and it computes a bunch of guys. Here are a couple of numbers, they're rational numbers, and the expansion I get is a Synthotic, so I have to deal with it. What will the trans-series look like? Well, it turns out that it's a second order differential equation will precisely say that there are two instant interactions and in such a way that they're symmetric, so that's what gives origin to resonance, and the two perimeter trans-series that we obtain looks like that. Again, I'm just giving you the structure, I'm not going to tell you what all these coefficients are. So, sigma 1 and sigma 2 are the trans-series parameters, these are the boundary conditions of poly-v, if you want, I need two of them because it's second order. These are the things that will sort of quote-unquote suffer stokes phenomenon, so when they're stokes phenomenon, these guys jump. Then I have my exponential contribution, you see that because the instant interactions were symmetric, there's an n minus m, and you can ask what exactly does that entail, talk to you about that in a second. There's some logarithms, which actually, I mean, it's a finite sum, so they're not terrible, and then there are some phi's, the capital phi's are a synthetic series, and this is what they look like, there I'm here putting them in as expansions in Gs, which remember was z to the minus 4 fifth or whatever it was, and they have different starting orders, so there's a beta which is going to be important in a little bit. The log sectors, the reason I'm not so concerned about that is because they're not independent, again this is an effect of resonance, it's sort of the hallmark of resonance in these nonlinear problems is that there's these logs sort of distinguishing borel singularities, so they all relate to each other, so I can deal with them, in fact I can just exponentiate them when I move to the partition function that I will not worry too much about. So you can just, you know, put these ansatz into the equation, get recursion relations, and compute a lot of these coefficients, these u, g, and m, compute a bunch of them, and then you can ask, well, but do we really know for sure that it's resurgent, and one way to test it is with these resurgent large order relations, so I'm going to look, I'm sorry, now there's a capital F where there should be a u, this is a slide from some other talk, so resurgent tells me exactly how those coefficients are going to grow at large g, so they have, you know, the factorial growth, the subleading exponential growth, and so on and so forth, with the one instant on contribution giving dA and the two instant on giving the 2A and so on, so I can ask if the different elements that I have on the trend series, this is at non-perturbative level, precisely reproduce the growth of these perturbative coefficients, and just a comment which is because this is resonant, there's this a and minus a, this is actually the picture I should have in mind, but that's just a technicality, so let's not worry too much about it. So can the non-perturbative content, these u, g and ms tell me about the exact growth of this perturbative series, so can they predict these rational numbers with some irrational information, like pi, these s's are Stokes data, I'll come to that in a little bit for the moment, just want to illustrate the idea, and so here's a test where I'm looking, I'm going to different order in g here up to 30 loops if you want, and I include in these large order relations different instant on contributions here up to 90 instantons, and I'm trying to see how many digits of precision and I get, and you see that I'm above 70 digits of precision up there, so I mean this is telling you that you should have some confidence that in fact what I'm finding here is a resurgence structure, so that's sort of the data that I'm not going to show you any numbers, you can look them up in any paper if you want, I'm just going to tell you that we have all these data that we're going to work with, and the same can be done for polynomial 2, so let me not waste a lot of time in this slide because it's exactly the same as the previous one, but just for the different equation, a different perturbative solution, now there's a factor of three halves, and then a different asymptotic expansion, again it's a second order differential equation with two instanton actions which are symmetric because again it's resonant, so it means that there's a string theory inside, and this is what the trans series looks like, almost the same as the previous one, so not a lot to be said, of course it's a different equation, so the relations between log sectors is slightly different, but other than that all is the same, and you can also validate this equation, and this solution, this trans series solution as being researched, okay, so now we have these solutions, what would we like to do with it? Well we could think maybe I should not try to resum, so I have all these data, I wanted to find a non-perturbative partition function for 2D gravity, I want to find exactly where are these poles of polynomial, can I play with that? So one way to do that is in the following way, so I grab each block, phi and m was an asymptotic series, and I'm going to Borrel resum it to try to get some number out, that's just a definition, in practice we don't exactly know what that Borrel transform is because we don't have infinite data, we have a limited amount of data, so we have to do some approximation to that Borrel transform and play with it, but okay let's assume that that so far doesn't bother me a lot, and then I do this sort of a Cal Borrel resummation where I put all these guys into the trans series, and I should be able to start computing some numbers, so the first thing which I already mentioned is that I still need Stokes data, so this is sort of a quote-unquote local solution, because I still need to cross Stokes wedges and go somewhere else if I want to go global, so I need to know Stokes data, so that's the first thing that I still need to worry about, and that's what I will be talking about for quite a bit next, and the second thing is that these two parameter trans series solutions are resonant, so now I really want to point out that there's that n minus m factor there, and you know if n and m are both zero that's just the perturbative sector, the zero zero sector that's great, but when n is equal to m all these diagonal non-perturbative sectors, the one, one, two, three, three, they all have the same weight, so we can ask ourselves how exactly are they playing with the perturbative sector, and is there any change in the physics of the problem that I am looking at? Let's try to address the first problem and try to understand how can I cook up Stokes data for this, only if in one case, and so for that I want to start just one step back and think that I have different instanton actions, and let's just think a little bit about what is Stokes data for general problems where there are many instanton actions, and then I will focus on the case that well I have two and they're symmetric, so let's just assume that we have k distinct instanton actions and some trans series solution, where now the trans series parameters are just all together in that k dimensional vector sigma, there's the exponential factors, and then there's the phi n's which are a synthetic series, now I'm not worried if they have a starting order or not, just starting some a synthetic series which are labeled by n, which is a vector in this semi-positive k dimensional lattice. All right, where are the Borel singularities? So we're on our path to try to understand Stokes data, so let's ask where are the singularities where Stokes data is going to be related, so I have this k dimensional lattice and I need to project down to the Borel plane to know exactly where these singularities are located, and the projection map is precisely done by the instanton action, and that's the reason why we'll see that resonance is going to change the way that these guys are organized, and in this business that we're at least these problems that we've been solving, the research and functions are all simple research and functions, which means that the singularities are simple, or in other words that there is only sort of two types of singularities, the pole and logarithmic branch cuts, so here I'm not showing you the pole, and I'm not showing regular stuff, so I'm just focusing on the log singularity, so what we see is that at the elf singularity of the n-instanton sector, there's a log branch cut, which is a specific projection from this ZK lattice, and then I see the resurgence of the n-splasell sector in the trans series, and there's a coefficient in front, and this coefficient in front we were calling it the Borel residue, because basically of the pole, and that's a number, so that tells us how to go from the nth to the nth-splasell sector, so that's on the way to try to understand what Stokes data in this K-dimensional series are. Now another way that Jean has taught us many years ago I guess, is that all the singularities should be encoded in these alien derivatives that were discussed this morning, and the analog of the formula that I've just shown you for Borel singularities is that bridge equation for the alien derivative of the n-sector, so the alien derivative is only non-zero if there's a singularity, and we already discussed that singularities are this l.a, and at that point I see the resurgence of the n-splasell sector, and in front of it there's this inner product of the Stokes vector times the arrival node, if you want. So you see that this Stokes, well this is a K-dimensional vector of Stokes coefficient, and these vectors are associated to specific lattice sites only, while the Borel residues need a beginning and an ending node, so there's a relation between them, here are two examples on how to relate this, let me show you the previous one, so there were these Borel residues, these numbers, and we need to know how to relate them to Stokes vectors, because if you're actually, I will tell you this in a little bit, that we are actually computing these residues, these numbers, so we need to know how to construct the vectors, which are going to control for you Stokes automorphisms and wall crossing formulas and so on, and the relation is nonlinear, so if you're doing a numerical procedure there's going to be some error propagation. Now not all lattice sites have Stokes vectors, there's a relation there, in fact most of the lattice of the ZK lattice is completely empty, they accumulate in the old negative K-Orson, let me try to motivate why, so here I am on the 3-2 sector and what I'm showing in different colors and different fonts is that the possible alien derivatives out of that sector, single alien derivatives, then I can still take multiple alien derivatives and sort of iterate these motions, but these are the single ones, and you can see that if you think of this as creating four quadrants out of the three two nodes, the first quadrant is completely empty, and it's pretty much the third quadrant which is populated, so what happens is that this is where Stokes vectors are going to be, so there's a third quadrant which is pretty much populated and there's an infinite bunch of vectors that I should compute, on the first quadrant there's nobody, and on the second on the third I just go slightly in, and this is true generically in K dimensions, so there should be a old negative K-Orson which is packed and most of the other ones are sort of empty, and we would like to know or we actually need to know all these vectors if we are to go around and go global on the complex plane to find all these Stokes transitions, let me tell you why, so again the plot that you've seen before when you walked into the auditorium, and let's just consider on the ZK lattice direction IF, the forward IF direction, okay, it happens that, so if it's the forward IF direction is a little bit like being there on the X axis, there's just a single guy going forward, it only has a component along the X direction, so the Stokes vector is just along IF, and the Stokes automorphism for that case becomes particularly simple because on the trans series it just gives you Stokes phenomena, right, the trans series parameter just jumps along that direction than nobody else, and this is the usual story that there's an exponential suppressed guy that I need to grab along and bring for the ride, so what would be the generic structure of these Stokes automorphisms in the K dimensional case, well let's just consider for a second this generalized alien operator where the only thing I'm doing there when I define this G is just forgetting that there's a Stokes coefficient and just put the generic vector V there, and it's very simple to see that they satisfy these commutation relations, so but what I want to know is what's going on with the alien derivation, so when I specify to the alien derivative you see that the nth plus m guy has that combination, so the alien sort of lattice of algebraic structure of this alien lattice only closes if this vector is of course proportional to nth plus m and this is not necessarily required, this is going to have, so this is working at the algebra level, so when we move to the group level and we write Stokes automorphisms and would be a generalized wall crossing formulae we really need to know what happened here first in order to then write the different relations that Stokes automorphisms obey or they do not obey, so we would like to know if this what's going on exactly in in order to move everywhere as I've explained, all right now comes the problem which is that this is resonance, so I still haven't mentioned resonance and the problem is associated to this projection from this lattice that I've sort of been talking about to the borough plane, so what I'm the gray, let me just go back two slides to show, I'm sort of sitting at the same point 3 comma 2 and what I'm showing now is multiple derivations which are non-zero that I can do moving out of 3 comma 2, so you know I have here this going from 3 comma 2 to 4.4 comma 2, there is a derivation which I can then iterate and I can actually iterate infinite times, but for instance I cannot iterate down infinite times because I end up at the boundary of the trans-series structure, so that's the plot that I'm doing there and you see that in the non-resonant case where the projection map has vanished in kernel, everybody ends up on a distinct singularity but if it's resonant like Polyvia is and in fact every string theory is then these guys end up on top of each other, so why is every string theoretic example needs to be resonance because in the trans-series I need to have some sort of two different expansion parameters, I need to have a g-string the coupling squared, this sort of a hallmark of the fact that there are some closed strings and I need to have other expansions in powers of gs alone which is a hallmark that there's open strings, so they all need to be together and if the trans-series is constructed with lots of gs, g-string coupling alone, in order to find some sectors which will be closed string-like expansion in gs squared, resonance needs to be there, so we need to find this vectorial Stokes structure in a case which is resonant, let's see how to do that, so let me tell you now about Stokes data for Polyvia, so how would we go about computing it, I mean it would be great if there was some sort of machine that would tell me these numbers but it isn't and it's hard, so we need to find some, so certainly there's no analytical procedure that we still don't know how to get them, so we're trying to solve these numerically for starts, so if I said the resurgence functions we have to deal with are simple, so I have singularities which are poles and log branch cuts and before I showed you the log branch cut, I'm showing you the pole, that's what it is, so that's the starting order so of the n-splice case sector which appears on the right hand side, so if I know because I constructed the trans-series out of the polynomial equation, if I know the n-sector and if I know the n-splice case sector, if I, you know, I can compute the residue and extract that guy out, so I could be evaluating residues at the different Borel singularities, so you know there's a bunch of them, let's say the red, the blue and the green and I compute these residues and I extract these numbers, so in practice this would be all there would be to say, except for the fact of course that I don't exactly know what these Borel transforms are because I have limited data and I cannot compute them exactly, so I need to approximate them somehow, so the numerical method to approximate them is by using paddy approximants which basically gives you a rational approximation to the Borel transform and which is a good way to do so because being rational to have poles and the poles will simulate where these log branch cuts are, of course a and minus a are real and when you play this game numerically you get a bunch of poles which in fact simulate your branch cuts but they're all on top of each other, they kind of mingle, so you know numerically you don't really know which are the reds, which are the blues and which are the greens that you wanted to separate in order to extract these numbers out, so you need to distinguish these branch cuts somehow and a natural way to do so is precisely by using the fact that when you're doing the the inverse Borel transform or the Borel resumption use a Laplace integral and the Laplace integral has an exponential factor which effectively can you can think of it as working as a damping and sort of trying to you know wash away the effect of the green and the blue poles so that you can focus on the red one by an appropriate choice of the point where I'm doing the Borel resumption, so you can do that for the first singularity you can compute the discontinuity between the left and the right Borel resumations along this this stokes line, find a good approximation for the residue at the red dot and then if you want to compute well what about the blue singularity and the green singularity they're shadowed by the red guys so I sort of remove I've computed this approximation let's say I've computed this guy which is from the red singularity I just remove it and I can focus on the guy of the blue and the green and so forth so there is you know a clear numerical path towards finding these numbers, so how many of these numbers do I have to compute is like an infinite amount I'm gonna waste the rest of my life trying to do this, let's try to see how is the trend series organized in this resonant case to try to see if there's a minimal set of numbers perhaps I don't have to compute an infinite amount of numbers perhaps I have to compute one number perhaps I have to compute two numbers and I'm done for all of Ponevitt so let's look at the structure again the trend series is resonant and the kernel is the linear span of one one so it's interesting that what's going to happen when resonance acts the structure of the alien derivatives becomes really nice because usually the forward and the backward alien derivatives are very different but in this resonant case they reorganize into a way which is very symmetric or I prefer the word it's the transpose somehow you're looking at the transpose of what was there before let me show that in the picture maybe before we discuss that equation so you see that if I'm on the 4-3 sector and I want to compute the alien derivative with a single step of action A I just have to follow the red the red arrows so I basically I move one step to the next diagonal and I sum over the directions of resonance if I want to go two away I just do the blue calculation and three away the green calculation and so on so that's what sort of that same first formula is telling you but what the second formula is telling you is that if you want to go back you're just doing the transpose of what you just did so that those are the 3 comma 4 is the transpose of the 4 comma 3 and I do those calculations so I sort of seem to be getting rid of half the infinite numbers that I need to compute you say well you still have to compute infinite numbers okay let's see if we can do any better than that so again trend series is resonant what does that imply so as I've tried to motivate a reason why this trend series is resonant is to make sure that the perturbative sector is an expansion in powers of gs squared I have a what we would call a closed string expansion in fact I have closed string expansions for all diagonal sectors and if you look at what asymptotics asks for you or for these sectors it's going to give you some general formula of which I can say yes but on this general formula which is would be a possible expansion in powers of gs I only want to keep the expansion in powers of gs squared and this is going to give me symmetries between the Borel residues again let's just look at the picture instead of looking at the formula so what it's telling us is that if I want to compute for instance the Borel residue that takes me from 3 1 to 1 comma 3 the blue one I have to do a summation over the dashed blues red I have to do summation of dash red and so on so there's still an implementation oh sorry I went in the wrong direction there's still it's making me clear making us clear that again there's many symmetries and so it's not only that I washed half of the content away but there's also even less stuff that I want to compute so let's try to see precisely what's that left stuff and the the the last stuff that I'm going to focus is one we call forward Stokes data which is sort of the lower that all right so the next thing we learn is that diagonal of all these Borel residues I actually only have to look at the diagonal Borel residues they're enough to reconstruct all Stokes data so I'm getting less and less information which is required to you know get the grip or knows these guys so this sort of backward forward diagonal symmetry told us already that we could focus only on forward Stokes data and it turns out that if you start on a diagonal point and you go somewhere else the way that the Borel residues I showed you a formula earlier with non-linear so there's a lot of guys in those three dots the way the formula organizes is that it gives me information about say quote and quote new Stokes vector out of Stokes vector that I recursively have completed before so again let's just see a picture of how that goes on so I start with my lattice and I'm telling you that I can get rid of half the information and then I say okay now let's compute some Stokes vectors and we can compute Stokes vectors from Borel residues that start at diagonal sectors here I'm looking at somebody coming from 4 4 and 3 3 down to get information on that Stokes vectors minus 2 minus 3 I need to look at two relations to construct a two-dimensional vector and this will give me immediately those vectors just thought of guys that came from the diagonal and then I just keep iterating because next I want to compute the minus 2 minus 4 and the dots is what I just computed before and then next I might want to compute the minus 2 minus 5 and the dots is what I computed before and so on and so forth so I'm computing all these guys so now I know okay there's it seemed like there was a lot of stuff that I needed to compute I've sort of went down let's just apply this numerical method of computing residues let's use all these symmetry properties that reduce the amount of data I have to compute to some sort of minimal data so what exactly can you compute of these infinite lattice so back in 2011 this was the first time we looked at this problem we use the synthetics to try to compute these numbers and well it was okay we computed a couple of num many of those numbers are not known so it was nice but it's not the efficient way so it you cannot get enough precision to get all these numbers so with with with the method of residues we have managed to go on quite a bit so we already have a lot of information about the first two diagonals and which is in blue is numbers that we already know and in gray are numbers which are being computed so we're now doing the third diagonal and you can ask why are you computing them right now why didn't you compute them yesterday and show this all of them because this is actually a very computer intensive calculation so what I'm telling you here is the precision that I can get on these different diagonals and x is sort of the x position right so here you can see that there's the first the second and we're beginning to have the third and the x is just yes I know can you just hold all the two slides and I will show you what the numbers are they're not integers I can tell you immediately I will show you in two slides I mean again I don't want to show too many numbers but I will show two numbers all right so as I was saying the x is telling you where are you on this lattice and the s is telling me on the x axis and the s is telling me the diagonal and you see and that n I will tell you what n is in the second is the stokes data that I'm computing and you see that when I want the first diagonal I can get a lot of precision in some relatively quick way I mean okay when you're going down to minus four we're already looking at five days of computational time and if we keep going down down it's going to be harder but the problem of going to higher diagonals is that precision drops very quickly so I ready to get 10 digits of significance it's a four-day computational time and because there's all these non-linear relations between stokes data and boreal residues errors propagate really fast and basically they tell me nothing about relations that I would like to uncover so that's why the gray dots are being computed now because it's probably going to take on the matter of weeks to have those numbers with the precision that we need all right so now to us try to start answering what Maxime was asking well okay great so ideal this numerical work but I would like to see some numbers out so what exactly are these numbers that you're computing so here unfortunately the colors are not great but what was blue and it's like greenish here is stuff we know and the I'm not sure what the other color is but it's stuff which is being computed but we can guess a lot of stuff analytically guess a lot of stuff of what we have found okay so let me tell you about what we managed to guess the first thing is we can go back to this this thing I told you earlier whether the algebraic structure of the ellen lattice closes or does not close which will have of course influence on what type of stokes automorphism relations and wall crossing formulae can get so at the first two diagonals it does numerically within our errors we verify that there's a closure of the algebra and if we assume that that's going to be true for any action and whenever the third diagonal numbers are out we will find out if we are going in the right direction or not this tells us that the stokes vectors are all of this form so this is a vector two entries and this is not a binomial number so r plus one s minus r minus one up to a number a proportionality constant so all I have to do now is know what these numbers and are these proportionality constants are and there can be reconstructed as I said before just from looking at the diagonal Borrell residues so this is what they look like walking down the first diagonal we can infer relations between all these guys so we can sort of starting from let's say boundary stokes data boundary of the grid I can walk down the diagonals and construct stokes data inside the grid and those are the first relations for the first diagonal so the one is telling me that I'm looking at information from the first diagonal there's a bunch of data functions coming out and there's a couple of ratios of the guy which is at position x equals one and x equals zero but if I know the guy at x equal one and x equals zero I can predict the other guys similar relations can be done for the second diagonal I'm not going to write them they look pretty much the same and we know how to change diagonals at fixed x equal one so we sort of know how to change diagonals if we're on the boundary of the grid so our hope is that if we know how to change on the boundary of the grid and as we know how to populate the grid and we're done with those two numbers so what are those two numbers oh sorry a plot before they have very suggestive patterns so they have all these Poissonian it's not but look alike structure on the diagonals and the numbers are those so it looks like by it's not known yet that this infinite set of stokes data reduces to two numbers so one which is known for actually a long time was computed by Frasad David in the 90s is that number and the other one is that number there 2.439262690 which analytically we don't know what it is so let me report on a failure we tried to guess it and we're still trying so this is so far it's only a partial failure we still haven't given up so we came up with this code my students actually they're very good where you grab a function with a couple of parameters which are integers in a given range so here's an example I take ABCD integers I compute the ratio so I have a rational number I have a zeta function evaluated at C I raise it to the power D and then I can try other combinations I can put in pi square root of pi I can put logs I can put zeta functions I can put polygamma functions and so on and this code tries many millions of possibilities per minute over the course of almost four weeks and we failed nothing came out so I mean of course there's reasons why we might have failed which is that we have these bases where we try to guess this number is not big enough or it might be say the result of the submission of a non-trivial series so I mean it need not be answered in such a way we don't know so we still have so this is ongoing we still hope to pinpoint that number and finally close this story but that's what we have so far that's for how many digits of this number do you have this is on the rough of 100 yes sorry but sometimes we fail Polyv2 not going to tell you about it just give you this is computationally more intensive because remember Polyv2 has a cubic term instead of a square term so we have less guys we have those those first four guys in orange was the first time we looked at this back in 2013 we computed them and now we're having a second run at this problem see if we can be better off I mean certainly we have more data we have that blue data the if you look at these n coefficients so the properties we find are very similar the n coefficients they look the same but we still don't have enough data for me to be telling you that there's only two numbers and full stop so let me leave it at that all right so now let's imagine that we had not failed in finding that other number well we do have it numerically so you could say well I'm happy with just numerical results and let's now start going everywhere across Stokes line trying to find where are these poles of Polyv2 can can we do something about it so remember this is the picture I showed you already there's this z5 symmetry of Polyv1 the movable singularity is organized into these five pizza slices there's a three tronca solution where there's only one guy the tronca solution where there's three pizza slices next to each other and the general solution to permit a trans series and I'm going to tell you results for the tronca solution all right so there's these double poles I had already mentioned Polyv solution generically have double poles they depend on initial data and there's a relation between the solution of Polyv and the partition function or the tau function and so the double poles become simple zeros of this partition function so you could think that these these different fields of poles are describing for you different phases of the system we have and let's just I'm let's just make the problem as simple as possible just look at the one parameter trans series so let's forget resonance for a second and let's try to to see what we can we can do and the trans series is this double sum where I have an asymptotic series there and then I have to sum over instanton numbers so you could ask a very naive question which is well why don't I really don't like to do perturbation theory I hate doing all this Borel-Paderi summation so let's not do this summation first and let's look at summing over instantons instead maybe just using the first leading guy and as I told you way back there is a starting order for each of these trans series that beta coefficient it turns out for these polymer solutions the growth is linear so this gives you a hint of what you can do so here's a picture of what's going on so what I'm showing you here is the different and the different instanton sectors of the trans series and as it goes up are the entries of the asymptotic series and in green if it's there are some entries and in gray if there is no body so the fact that the starting order is linear is telling you that you know there's that linear direction so what I don't want to do is perturbation theory so I don't I don't want to be summing up and then going around so I can ask no let's just forget about provision theory let's just look at the first guy and do the direct summation over that diagonal so basically what we do is we introduce a composite variable tau which has a trans series parameter in the exponential and then what I'm going to do is just sum overall the the leading order guys so the u0 here would be just the first coefficients of that first red arrow on the picture on the left and I just sum all of those and then I would do so iteratively for the next guys so if you want is the picture here on the right I'm just doing iteratively for the next guys and surprisingly all these series are convergent so this is a result which is known as trans asymptotics in the mathematical literature so here here's what it is at leading order for panlevere so in appropriate I mean without surprise of course in appropriate variables this u0 is the wire stress elliptic function it's degenerate because it degenerates down to just a rational function and you know you say okay that's nice because that's a rational function which has definitely a a a second order pole that I was expecting to find at tau equals one you're saying well but wasn't there an array of poles well yes there was an array of poles but remember that tau is this exponential of z which I still need to invert that's the Lambert w function and when I invert the Lambert w function I basically get the first array of poles directly and then as I put in more subleading contributions so the as I put on the next red arrows in the diagonal I get the second array corrections of the first the third array corrections of the second and first and so on so this is looks like it's a more efficient way to try to find where are the locations of all these panlevere poles but you'd say well but I still need to go along all these diagonals something order by order to get the next guy and the next guy and I really don't want to do partition theory I don't like it so can we do any better well turns out that you can look at the partition function and the partition function has a structure this is the same structure is the starting order of the different guys in the partition function it's sort of different so I can still do this linear summation which here it's just basically a finite guy so convergence is trivial because I'm just something to guess and what I get is one minus tau so I get the simple zero that I was expecting to find and again it's telling me that the zeros are at tau equals one and then I could do the middle plot and just do everybody else but I could do instead say no there's a quadratic growth of the leading order of the partition function of the tau function so why don't I sum directly over all these guys at once and if I do so I actually get all arrays of poles all arrays of zeros so I get the position of all those guys of course it's still not the exact position because I still need to have small adjustments to their position in powers of one over z but it's a beginning on trying to find the exact location of all these guys so what does these equations look like so here's this quadratic resumption at the level quadratic because the starting order of the partition function goes like n squared and that's what the partition function looks like and at leading order it actually gives a sort of familiar result that had been seen well by marcos and Bertrand previously in the literature which is very much fair function like except with a with a barn's contribution and then as you go to a higher order I'm just going to flash some formulas so as you go to higher order these get some extra terms with some polynomials with that look like that I mean it's not very a lot of things to do and if you put in the two parameter trend series things get slightly more complicated because now we have these harmonic numbers also coming in but you can construct these partition functions this is a construction which is very much associated to this sort of sigma one sigma two parameter trend series might not be the most efficient way to look at it so an alternative is to just think well but that was resonant to start off with so the trend series that we should be looking at is the one which is modded out by the generator of resonance and so an explicit formula for the partition function of two parameters can be written in a way which is first of all very symmetric or very transposed if you remember this transposed symmetry between different sectors of a resonant trend series so basically here of the second formula the second and third lines you can think of them as the transpose of each other and you can do these trend series summation to which one of these blocks and you can then go in the direction of finding these general zeros okay I have about 10 minutes so I'm just going to be slightly faster on what happens for the partition function but basically I just want to show you plots I'm not going to show you a single formula I don't okay so what happens when we go discrete polynomial so those are solutions to matrix model so m here is an n by n matrix which I want to compute this integral this partition function and I want to do so starting by going to large n this is the famous a tough limit where that combination gs times n is kept fixed the gs is the coupling which appears there in the exponential and v is some polynomial potential for the moment and if we do this then the free energy has again an asymptotic expansion like the ones we had before so don't worry about what's I don't know why that in brackets that survives from some other talk it doesn't matter for this so here's one example that you can solve exactly which is the Gaussian matrix model the potential is just m squared and then I just have to do n by n Gaussian integrals and you can do so exactly and find a solution at fixed n fixed number of eigenvalues uh of what we call the canonical partition function and it's a Barnes function times some powers of n squared times some powers of n it's not too hard to write down the exact grand canonical partition function so some overall these n's and what you see is a formula that you know you've seen already this is the leading term that comes from polynomial one so you see that well maybe a similar structure is at play here so let me show you some just plots for the quartic matrix model and there's a the quartic potential in rad and these n eigenvalues they usually accumulate next to critical points of the potential you know either the one two or third critical points and there's famous solutions out in the literature where they all accumulate around one critical point of the potential that usually go by the name of the one cut or another one which they accumulate around symmetric configurations that goes by the name of two-cut solution and you can look at these solutions and try to see if where are these the analog of the poles of polynomial in this matrix model case in this discrete case you have to solve discrete polynomial so again i'm just flashing you the equation again i've shown you this before r is the analog of u and i'm expecting that it's going to have double poles somewhere which should translate to simple zeros of the partition function and it turns out that there's a long story i don't want to tell that you can also construct a two-permeter trans series which is resonant now everybody has a p dependence so instead of looking just at sequence of numbers i'm looking at sequence of functions but the ideas are pretty much the same and so we've done this construction and did the resurgence checks actually before let's not go into that so let's just assume that this was computed what can we play with it well oh the colors are not great but i'll try to explain so this is the phase diagram at complex t at complex tough coupling and we find some specific region so the the either solid or dotted lines are anti-stokes lines so these are phase boundaries this is remember the stokes line is when an exponential guy must come along for the ride an anti-stokes line is when it becomes of the order of the perturbative so there's a phase transition if you want it takes the instant ons can take dominance so the one and the two are the one cut or the two cut configurations those should be bluish not sure what color you're seeing then the outer lobes with the dashed curves this is where eigenvalues accumulate on the three possible critical points of the potential and they're all at the same level of energy they cost no energy to go moving around so this is a very oscillatory configuration in fact Bertrand told us many years ago that there are some theta functions taking dominance there and out in i think it was supposed to be rad there's a trivalent tree phase where eigenvalues accumulate in some trivalent configurations but i'll have nothing to say about that so let's just focus on region one where there's a one cut solution and the green region where there's a three cut solution so how are these things behaving so i go directly to the matrix model which is discrete it's n by n so i'm not looking at the large n expansion and i compute these specific heat the r's which from the point of view if you want a orthogonal polynomials i can think of constructing orthogonal polynomials for the matrix model based where the measure is basically the one that's giving me by the potential and those are n's will be the recursion coefficients for these orthogonal polynomials and i'm plotting here their absolute value and their phase and you see that in the blue region where there's the one cut they're sort of well behaved and once i move out there into the green region where there's these three cuts and they should be oscillating they start oscillating so you can ask how much can you tell about this eventually can i tell exactly everything about these oscillations by using the trans series and the methods i talked to you about so here is trying to go with the perturbative trans series alone just you see that the perturbation theory is pretty good on here on the one cut region but then once i go into the antistokes it just doesn't see what's going on so you can ask well let's try to put instantons and i put the first instanton correction and it already sees that something is going very different as i go inside but it doesn't see a lot inside i can just go slightly inside and by putting more instantons it doesn't get terribly better i get probably a better handle of the first guys here but i still cannot see the leading oscillations it's a little bit like trying to come up in the prolific case with grips with seeing the first array of poles so then we can do this linear analytic trans series summation to leading order in gs and i see the first hump so that's like seeing the first array of zeros but again i don't like to do perturbation theory i don't want to keep going well we can't keep going and do next to leading i see the second array of zeros i see the second oscillation and on and so forth but of course the punch line as you probably all have guessed is that if i'm going to do the quadratic case i already get immediately the picture of all these oscillations in one go they're not all perfect you see that for instance that black dot over there is slightly lower i mean it does see that there's an oscillation but it doesn't quite reach it but it's already pretty close and now g string corrections should allow me to be on spot of them and so the real zeros that we would want to compute are or poles i mean again all these things translate to each other are these liang zeros so this is the plot that now you should recognize that i showed you at the beginning of the talk there was a density plot that run for a couple of weeks at the closest at cern and we can now reproduce that plot with basically specific positions so the question is whether you know this final trend series can still be resumed and get exact positions of everybody or not all right so i have five minutes left so let me just be very quick in fact i don't have a lot to say about these semi-classical decodings so non-perturbative definitions are great but they're not always available in string theory and perturbation theories is pretty intuitive so i kind of know what's going on because i look at this sum over Riemann surface of different genera but they're certainly complete across perimeter space so i already showed you that if i just take perturbation theory i'll never be able to see any of those zeros there so i need to do better and so resurgence tells me that i should complete perturbation theory with non-perturbative trend series and marcos likes this idea of semi-classical decoding where i have this black box and i want to understand this black box with semi-classics so the trend series is sort of translating for me what are the different semi-classicals which are important in different regions of perimeter space so can we play this game here for this we can think of you know uh pollivier solution the zeros of pollivier and i have this um this decoding by the trend series that i've shown you and you can ask can i understand what are the different semi-classics that are at play in this trend series okay so again let me just flesh you things that you've already seen so that was the pollivier one equation which had this perturbative intuitive expansion and that was the two perimeter trend series solution and i want to now not only understand the perturbative guys but also have some intuitive grasp on the non-perturbative guys and so there's one way to try to approach it is just to stay within the realm of 2d gravity and then i can say i don't know the zero zero sector is the string expansion that's perturbative 2d gravity the n slash zero sectors these are some objects non-perturbative objects we know in string theory which are known as z z brains so i can see them there as well but if i start looking at the generalized instanton so the n slash m sectors maybe the interpretation is not so obvious and it might not be so obvious because the semi-classics might be clear with a different reorganization of the trend series this is not necessarily that when you write the trend series you immediately wrote it in the way that the semi-classical decoding is obvious and that's of course associated again the resonance i already told you this when n is equal to m all the diagonal non-perturbative sectors have the same weight so what should we be doing well we should be modding out the trend series by the generator of the resonance so this is what the free energy i showed you a similar formula for the partition function this is what the free energy should look like and now let's just focus let's forget about the instanton corrections of this reorganization of the trend series and just look at the first line so what is this first line exactly so if you use explicit data again i'm not showing you a lot of numbers but please trust me if you use explicit data that we have for the trend series it turns out that you can rewrite the first guy so the sum over all the diagonal sectors in this way and you know maybe the mathematician half of you will say this doesn't mean me anything but maybe the physics half of you will say well i've seen this formula before because this look very much like so the product of sigma one and sigma two looks like something which is usually goes by the name of a chen panted factor and these guys are genus gh puncture contributions to an open string um a partition function with the correct powers of gs so then it you know it's quite fast actually to think that what you should do is to introduce in a tuft like parameter s which is the product of gs times the chen pattern factor sigma one times sigma two and then a tuft like limit where gs goes to zero and sigma one sigma two goes to infinity at fixed s and you find that that guy is actually reorganizing into a closed string theory and but it involves a modulus dependence now it depends on s this is a tuft like parameter so you can ask what exactly is going on there seems to be some topological strings inside these minimal polynomial strings and yes i mean this relates this is not exactly their answer but it relates to the answer of bonelli at all based on earlier work by gamma union yorgov and lisavi that there is in fact a topological string inside this polynomial strings which is given by uh basically the nekrozava kunkav dual partition function at that specific gauge theory conformal point so here's the question that guy depends on a modulus so again it has to be one of the cases that john was telling us earlier today that this is not a case of equational resurgence like the whole polynomial story so far has been uh because in order to compute those guys i need to use something which is known as the holomorphic norm equations which are co-equational resurgence kind of things and this is work in progress and the little introduction to what we're trying to do you can see in the talk by sozane so that will continue there but i'm afraid that for me the time is up so let me just tell you that i hope i've convinced you that these resurgence-run theory solutions work well for polynomial equations and that we actually we don't know the full set of polynomial stokes data but we know a lot already uh well i'm okay we can always know more of it um we find these movable singularities in the discrete case in the continuous case we have a pretty good handle of where they are and we're approaching hopefully exact expressions doing some sort of uh exactly the kv uh and semi-classical decoding is should be thought of semi-classical decoding because depending on how you look at how you write the train series you get different semi-classicals out of course there's still many open things we don't know the whole set of stokes data not yet we can ask whether there's more semi-classical decodings than the ones that i showed you maybe and whether these you know whatever stokes automorphisms we're going to write in the end is what crossing like formulae we're going to have module properties say for the partition function for the tau function and get a you know closed-form expression for the full non-perturbative partition function there is a proposal by Bertrand Marcos for the matrix model and it should be somewhat very similar to that with the another the novelty coming from resonance let's say and um that's about it i think my time is up thank you very much sure sure yeah at some point very early in your talk you said you have reduced to small and small non-parameters and then before you get just two numbers i think you get like genetic series in two variables yeah is it true let's go back to see what exactly so the two numbers were those so that you want before yeah because before i think you get something reduced to functions on two variables kind of is doing this yes sometimes we get vectors was that the vectors that you're asking so we get so the stokes stokes vectors are vectors of which we know the say the x and the y entry uh you should shoot this is and no we need those guys so those guys there those are labels they tell me basically which diagonal am i looking at and where on x i am so it's just a way to label all these numbers but all these numbers then satisfy these sorts of relations so if i know the first sort if i know how to start the recursion then i can go walk down the diagonal i was not busy people for part of this this is kind of pretty crazy some kind of change of variables into variables i don't think i know maybe the champion has something interesting we we have not we've put alpha to zero i'm sorry for zero yes yes for zero is similar to one yes but you shall find zero you have a two or zero instant i agree you we've just an alpha equal zero the the the non-results that i've showed you which was this is when alpha is equal to zero and that's what's ongoing so maybe at some stage we will do alpha non-zero but not yet thank you very much