 Thank you very much for the invitation to participate in this very interesting conference and to speak. It's a great pleasure. So I'd like to talk about WKB, and this is, I think, what Professor Iqal calls, Varroes Resurgence. So the plan of the talk is one to talk about how to do the endless continuation in some particular case. And then the second part, I'd like to talk about some more recent considerations to try to compute the exponent. And the second part is joint with Ludmil and Khabzarkov, Pranav Pandit. And then for the first part of this work was also with Alex Nohl. Then for kind of a subsequent part, which is not really related exactly to this question, subsequent work in the symbolic geometry with Fabian Haydn. OK, so first the basic setup. So the basic setup is we have an equation. Let's start with an equation of the form, so I'll try to write this in a kind of standard form with an h bar. So h bar d plus or minus, let's say, plus a of z times a vector equals 0. So this is our basic ODE. And so it's singular in the variable h bar. So h bar going to 0. And we'll really think of this as a variable 1 over h bar going to infinity. So then let's diagonalize the matrix. So this is going to be the spectral curve. If we try to diagonalize the matrix, of course we want to know whether the eigenvalues of the matrix, the eigenvalues of the matrix, are multivalued functions and that forms the spectral covering of z. So what assumptions would that be? A of z, yeah. Well, some assumptions. Let's not be too specific. Let me make some assumptions. Let me first diagonalize, and then we'll make the assumptions later, OK? If particular is diagonalizable, yes, one assumption. For example, yeah. We'll have a more specific problem in a minute. But kind of the problem is that part of the discussion isn't going to not really exactly applied to this case, as you'll see. But anyway. OK. So that means, well, let's choose a matrix, s of z. Also, I'm being a little bit nonspecific about what z is. So z is a one-dimensional complex, z is a Riemann surface. Come back up? No, no, I mean, if it was come back, we would really want to look at the universal cover anyway. So roughly speaking, it's the disk, OK? But there's this question, but I think which has already been evoked in previous talks, the question about the completeness of some kind of metric, you know, whether you sort of, and we're going to somehow or other assume that you're not going to sort of fall off the edge of z when you do your flows, OK? So let's choose s of z so that s inverse a, s is a diagonal matrix, which I'll for now write a1, an. So let's just sort of, I mean, a is really the matrix of one form. So maybe when I write, if I write the variable, then I'll have a dz, and if I don't write the variable, that includes the dz, OK? So let's just choose, let's just look at an example. So I think it's important, I never really did this example previously, but this is important to do this example, which is a is the matrix 0, 1, v of z, 0 times dz, and then s, so s, the best thing to choose for s actually is v to the minus one quarter. Sorry, v to the one quarter, I think it's minus one quarter. There's a plus or minus, maybe, and then s inverse, if I'm not mistaken, should be, but this might be mistaken. Something like, you can do the thing and tell me if I'm, but anyway. And now, what is s inverse times ds? So s inverse a, s, so the diagonal matrix is just what you think. So the v of z, it's really the coefficient of a quadratic differential. And it's v to the one half. The diagonal values are v to the one half and minus v to the one half. You guys all know this better than I do. Then s inverse, but maybe an interesting point here is s inverse dz, ds. So when we make the gauge transformation, we get a term which is s inverse ds, which is kind of important. That, if I'm not mistaken, is like, so the point here is that this has a logarithmic pole on the off diagonal pieces. And we chose the one quarter here so that it has zero on the diagonal pieces. Okay, so now we get a new equation, which is so let's call the new a is just the diagonal matrix. This should, sorry, this should be r. This is the rank, this is the rank. And then b is this extra gauge term. And the new equation is d plus one of h bar times the a, the new a plus b. So this is the equation I actually want to look at. And then in the example, b actually has logarithmic poles at the turning points. The turning points, well, in general, that's for the example. But the turning points in general are the points where a i minus a j vanishes. Now what's the question? So the question is we fix two points in our Riemann surface. We fix a path from one point to the other, and let's let m, so m depends on h bar, is the transport matrix from p to q, okay? So this is gonna have logarithmic growth. So this had logarithmic growth which is bounded by some constant in the size of one over h bar. The constant is basically the size of the matrix a along the path gamma, okay? So this is an entire function of h bar, let's see. It's a function of one over h bar. I mean, I'm calling this h bar for obvious reasons here, but maybe the better parameter to think of is one over h bar, which I think some people are calling eta, okay? So that's the question. So that's supposed to be a function. So that's the function in the non-Barrell plane. That's the function that works we're gonna transform into the Barrell plane, okay? But before we do that, let me just write the expression for this function. So this is kind of important. Let me not derive this expression, but I'll let you guys imagine. So m is gonna be a sum of mi of h bar. So the sum is over, the i are multi-endices i0, i1, in. This is n not, this is different from r. So these are the matrix, these are the indices for matrix coefficients. And mi equals, let's see. Let me write the formula and then I'll write you the definition of the terms of the formula. Mi is the integral over delta n of e to the gi of z over h bar times B. Now what are the terms in this formula? So here, z means z1 up through zn. It's a point in capital Z to the n, okay? So the terms in the formula, so gi is a function from capital Z to the n into C. So gi of z1, zn is the integral from the point P to z1 of ai0 plus the integral from z1 to z2 of ai1 and so on. On the spectral curve, you should make a nursing curve or purchase a pass. Yeah. How do you follow your pass? Yeah, okay, so sorry, thank you. So we've kind of replaced z by the, the cameral cover, actually, where the, where the, the different diagonal terms are distinct. Yeah, z new. So I should, well, right, well, I guess what I wanted to say here is really, sorry, this first part was to reduce to this case. So what I really want to look at is this equation where a, I really want to look at this equation where a is already diagonal, okay? This example is to explain how, for example, in this typical quantum mechanic case, you can reduce to that case. But of course, you have to make the covering where you take the square root of v. But it's just, I mean, that's, remember, I mean, what I'm doing here is kind of limited in scope and so on. In a sense, I'm just fixing two specific points of z and a path from one to the other. So that's going to be a little bit different from, for example, what happens if you try to use Fokken, draw of coordinates, where you need to choose, choose flat sections that have a certain asymptotic property at singular points and that kind of thing. Something like that, yeah, yeah, yeah. And, excuse me, do you assume that the AIs are these things? Well, let's assume that they, yeah, let's assume that the AIs are distinct, yeah. Assuming that, yeah, yeah. And in fact, maybe I mean, kind of have, well, okay. Let me not be too specific about the precise hypothesis here. Anyway, okay, yeah, but the AIs should be distinct, yeah. Sorry, and maybe the other hypothesis which I didn't say is that the BII are zero. We can assume that also by making a gauge transformation, that's kind of important. But again, you can see that in the example. In the example, we had to choose the one-quarter there in order to get that property, okay, anyway, so this is the function. So this is the main function which controls the asymptotic, the exponential growth of our problem, okay. And it's probably useful to notice that this is also equal to gi0 of p, maybe q, I'll say, gin of q minus gi0 of p. So if we choose primitives gi of the ai, then we have something like that, plus the sum of gi k minus 1 ik of z, okay. So if you vary one of these points, z1, for example, the variation of the function g there is given by the difference of the ai1 and the ai0. Do you want to apply it on the same path? So for now, the points line, that's the next statement, which is the delta n is the set of points of the form gamma of t1, gamma of tn, where 0 less than or equal to t1, less than or equal to t2 and so on, less than or equal to tn. So for the moment, the points, the z's are the points on the path, okay. And then what's b? So bi, so bi of z is really a form. So it's just a, it's the form that you got by taking the matrix coefficients in an obvious way with these coefficients. So it's bi, 0i1 of z1, bz1. So b should be an n form, right, because we're integrating over an n-dimensional real cycle. So this is n. So we're integrating over an n-dimensional real cycle in an n-dimensional complex space and we're integrating a complex n form. Let's see who does this most often. Okay, that's all for the formula. Now, okay, let me, before going to the Burrell plane, I think this is maybe important to say before we get to the Burrell plane because then we'll do the same thing in the Burrell plane, but before we get to the Burrell plane, the main point here is we would like to apply some kind of saddle point or steepest descent method for this integral. What does that basically mean? That basically means we want to move the cycle, because this integral is basically independent of the choice of cycle and integration because it's the integral of a holomorphic form which is closed. This is a holomorphic form, but we multiply by a holomorphic function so it's still a holomorphic form, so it's still closed. So we can still move the cycle of integration and we try to move the cycle of integration so as to minimize the real part of this function g. Okay, that's if we're trying to analytically continue in the plane, I mean in the, maybe I should draw that picture about something. H bar is real? Yeah, that's, sorry, I realized I should draw that picture now. So for the moment one over H, let's see, H bar is a, one over H bar is a complex parameter. So one over H bar is really any point of c. And the point is that this function, well I guess we're not quite there, but yeah, but yeah. So if we think of H bar as being real and then we want to make the function as small as possible then we're interested in the real part of that. Thank you. So H bar is real or complex? No, let's keep H bar complex, but in the example I was talking about diminishing the real part of g, that's for the case of H bar real. If H bar has some other direction then you would turn by that amount and instead of the real part of g you would diminish the real part of e to the i theta times g. So let's do, okay, let's do the Braille transform. So now let's go to the Braille plane. So the variable c is going to be dual to the one over H bar parameter, okay? So let's define m hat of c defined to be equal to the zero integral from, what I wanted to stress but I said it already is that this function has exponential growth of exponential order one of one over H bar. So we can make the following integral, integral from zero to infinity of e to the minus, I guess it's a minus, e over H bar times m of H bar times d of, so I'm sorry about this but this is d of one over H bar. So I guess if we wanted to do dH bar it'll be there's a factor of H bar in there somewhere. As I said I would really think he in terms of the parameter one over H bar and this is going to be a sum of terms, so we can also do the Braille transform of each of the terms in the sum and if I'm not mistaken the formula is the following that the mi of c are the integral over the same cycle delta n of, there might be a sign problem, of bi divided by gi of z minus c, up to a plus or minus. I'm not sure, I don't remember if that was a, if there was a c minus gi or gi minus c. Basically when you Braille transform this expression you get that thing. When you plug that into here in Braille transform you get that thing. Okay. So I guess this is probably a good place to point out but so my whole investigation of this question was actually, I think in actual fact has to be said was strongly influenced by Professor A. Call's work. Although I didn't necessarily really quite realize this at the time, but I was following J. Arnaud Moll's class as a graduate student, I was following his class on Fourier transform for eletic sheaves. And, no, but yeah, okay, so I'm not sure he's laughing, but you know, at the time, especially as graduate students we became extremely adept at passing between the eletic number theory kind of picture and the demodule complex geometry type of picture back and forth basically. So, and also Le Monde said and somewhere in his course material he's had, you know, one of the references for this course is a called theory. So of course this was a whole course about eletic, Fourier transform of eletic sheaves but then when it came to have a problem in complex geometry then it was natural to do a Braille transform and then try to look at the properties of the function in the Braille plane. Okay, so now let's see. What comments about this? So the first comment is that this path of integration depends on h-bar. Okay, so sorry, depends on c. So this is well-defined for absolute value of c bigger, let's say strictly bigger than this constant c2 that shows up here. And the path of integration the path of integration from 0 to infinity is going to depend on the argument of c. So depending on whichever, if you have c which is big enough then there's always going to exist a path of integration with the property that this is exponentially decreasing. Okay, as 1 over h-bar goes to infinity because of that estimate. So that's kind of the answer to this question about whether h-bar is real or not. We have this function which is defined on the complement. So in the Braille plane our function is defined on the complement of a disk. And then the inverse transform in this situation is that m of h-bar is going to be the integral, a circular integral of maybe e to the c over h-bar times m hat of c, dc, up to a constant. And the circular integral to define m is taken around this region. Now geometrically what is this region? Geometrically this region is obtained in the following way. It's just take the image, so let's say z dot is a sort of disjoint union of things of the form zn, zi equals z to the n, okay. And then we have just a function g from z dot into the complex plane into the c, and we should really think of this as the c plane. Then we have this delta dot is sort of a sum of delta i. So we can just write this whole integral as a formal sum, you know, it's just an integral over a single cycle. So we can write sort of m hat in a formal way. And m hat of c is just the integral over delta dot of v divided by g minus psi. Sir, I think that this m hat should be a multivariate function, yeah? That you cannot choose its variable. No, no, no, but that's, in this case, so this may be, this is probably a slightly different case than some of the other resurgence properties that we're looking at. I mean it's kind of the same, we don't have like this infinite collection of poles on the imaginary axis or something like that. In this case, this m hat of c is a well-defined function, perfectly well-defined function on the complement of the disk of radius c2. Just look at the definition. This m is an entire function of 1 over h bar. It's also defined at h bar at 1 over h bar equals 0, kind of trivially. Okay, m is an entire function of this growth hypothesis. So for any value of c, we can choose the path of integration. Can you define that in half-blades? But you can just, those just glue together. No. They don't. No. Why not? We just defined this in a number this way. It's not convergent. It's real patterns. Half-blades. It's only in half-blades. Yeah, but then if you choose a different half-plane, we choose a different path of integration. Yeah, it's only good cuts, yeah. But it's the same function. No, but the picture we should draw, I mean, it's not only half-blades. I mean, if you like, we can just write this formula. No. Why not? Because it's counterintuitive. Yes. No, I don't agree. I mean, let's just write down this function. This function is perfectly well-defined for any c, which is what I wanted to say now. So this is well-defined. So m hat of c is defined if c is not in the support of g-lower star of delta dot. And this g-lower star of delta dot is supported on a compact subset. All of these cycles have the property that g of the cycle is supported inside this disk. So if we take a value of c, which is not in this disk, then this integral is just well-defined. And this integral, sorry, this function m of h bar is just equal to the path integral around this thing. I think the reason, so we were discussing in the train with Jean-Pierre, the reason why this is maybe a little bit confusing is because here we're taking, we're not talking about irregular singularities. So if you introduce irregular singularities into the question, then the direction of the stokes lines are going to depend on h bar. So you can't, so the irregular monotromy is not of this form just fixed two different points. The regular monotromy is going to be an irregular monotromy in some stokes directions, but which depend on the direction of h bar. That's maybe in a professor I call this talk was the combination of equational and co-equational resurgence. In this case, we're just talking about, for example, a compact Riemann surface. We don't have this essential, this irregular singularity question. So anyway, I mean, if you're convinced that there's a mistake, please let me know, but I don't think for the present purposes I don't think that's... Okay, so now what are we going to do here? So now what's the goal here? So the goal is to, let's write this over here. So supposedly what we're going to do here should work in that type of case also, I suppose, but that's not what I'm claiming. Okay, so now what's the goal is to do an endless analytic continuation of this function, m hat of c. So what's the idea is that the one we just noticed that each mi has an endless continuation. So this is just sort of some slightly infinite version of Gauss-Mannien for finite dimensional integral. So if this map g were a map of, a proper map of algebraic varieties, then this would automatically satisfy this property. And here we have a slightly infinite version because this integral is defined on this set z, which is typically going to be like the abelian cover of a compact Riemann surface that you get by mapping it into its Jacobian. So there's some periodicity. But the singularities themselves are sort of at least either containing, I mean either finite points or maybe in the worst case, if you have to blow up something or something, are just going to be compact subsets of z to the n. So whatever happens around a given singular point is just like the algebraic geometry situation. So this kind of adds, I mean that that's not really approved exactly, but sort of in terms of general principle. And then, so the main question too is how is that they converge? The sum, the sum of continuations converges. So let me just give the definition of endless continuation, or weekendals continuation. So let's say fix a disk. So let's fix a disk where our function is defined. And then it says that for any m, there exists a finite subset, s index m inside the c plane, such that for any path, we should start at a disk because our singular set might well have points inside that disk. So we can't just fix a singular starting, a single starting point. We need to just fix a sort of infinite set of possible starting points. But we assume that our function is continued inside the disk already. So for any path starting in D0 we need to pass m of length less than or equal to m. So this is the property which I'm claiming we can prove here. I think it's probably useful to consider a weaker property. That's why I put the word weak here. I think it's probably useful to consider a weaker property because this is kind of a rather strong property actually. So the weaker property would still require that the winding number of the path also be smaller than m. Then the statement is there exists a continuation along the path. If we don't put the winding number condition then that actually tells us that you can go any number of times you want around a given singularity without sort of introducing new singularity. That doesn't seem like it would necessarily be true in general. So this is the definition I'd like to use for endless continuation. I think there might be other possible definition. So that's the thing we'd like to prove here. Now what's the technique? So to illustrate the technique, the technique is that we need so we can make a general statement about one but the point is that if we want to get the convergence then in part one we actually need a very specific continuation procedure. Rho is a path. So any path in the c-plane of length less than or equal to m then... You can draw the picture. Yeah, so let me draw the picture. Let's draw a bigger version of that picture. So our function is a priori defined on the complement of this disk of radius c. We start out with our D0 here. This says that if I choose a length like the length of this backboard there's going to be a finite set of singularities which could be outside of here also such that if I continue along a path that goes like that then I can continue my function along any path that goes around these singularities of length less than or equal to m. Then as m increases the set of singularities gets bigger and bigger. That's this resurgence phenomenon. So now what you might say is the first thing to do of course is to start from this point sort of continue around the region and see which singularities you hit first. This property implies that there's a finite set of singularities that are hit before any other ones and then the real domain... the real sort of complement of a convex domain on which m of psi is defined is going to be the convex hull of that finite set. We're claiming any rank here. So the point here is to use a specific continuation procedure in order to get a control on the convergence so that we can treat the convergence question. And what's the procedure? The procedure is to use gradient flow. In the talk I would like to discuss the case where we're just... rather than trying to continue along some complicated path we're just trying to move this domain of integration over to the left here until it hits its first singularity. That's kind of an easier question. But what I would just like to point out is that it's the same procedure that works to continue along any sort of piecewise segmented path. We just apply this moving procedure along some neighborhood of that path and then move in this direction and so on. That requires some technical things that are not really very interesting. So let me just not discuss that here. But you can do that. Let's just assume that we're just trying to decrease the real part of G. If we're trying to decrease the real part of G then we use the gradient flow of... So what we really want to use is the gradient flow of Gij of the real part Gij. Excuse me. The path of the Riemann surface is of infinite type. Is not a problem. So as I said, I wasn't saying really what hypothesis there are. You need to make a hypothesis that says that if you actually sort of flowed to the edge of the Riemann surface then you would have decreased the real part of G by an infinite amount. I mean, to do this... To do this process we only have to decrease the real part of G by a finite amount and even along our path of a fixed length we only have to sort of decrease all the different real parts of the different rotations of G by a finite amount. So the kind of completeness properties that we'd like to make sure that we don't get to the boundary of the Riemann surface before sort of an infinite amount of decrease. That's obviously a big problem in the case where B has poles because in the case where B has poles you really need to not hit the poles of B. So you need to draw a little circle around the poles of B which causes exactly a problem of this sort. That's more complicated. I tried to discuss that in a preprint I wrote for Jean-Pierre's birthday celebration. It seems pretty incomprehensible to me so I don't necessarily claim that it has to be without error and so on. The claim is you can actually even do that too so that's why the claim says that even if B has maybe let's say loving any poles this should still work but that's a little bit more delicate. My original discussion of this was just in the case where B is holomorphic but which unfortunately doesn't apply to the example. Okay so the status of the proof and the other statement on that level is that in the case where B is holomorphic you can get a better control over the type of the singularities that are met. In the case where B has poles I don't have any idea how to control what type of singularities the function will have here. I think there's sorry. No just I mean the singularities at the singular points. No the values of the singular points no but the singularities of the function the singularities of m hat at these points could have an essential singularity so I think I mean it's going to be it's definitely going to be a trance series maybe not convergent you said probably not convergent in general I have to say I don't even know that there exists even one point where it's convergent so that may be a caveat here. In the case where B is regular you can at least show that the singularities have some kind of as in product expansion and so I'm hoping that in the case where B has logarithmic pole I'm hoping that you would be able to show that it has at least one over you know one over C minus A0 I'm hoping you can show that the function has at least growth less than that but that's that's conjectural for the moment it's a sense of curve actually finite dimension is not one dimension it's almost a problem yeah these are finite each term is a finite dimension integral so each term is pretty much controlled it's a sense of curve in product of corpus of curves yeah that's easy so I mean that's easy so one in principle is easy two I won't be able to explain really why it's true anyway but what I would like to just say is that in order to do two in a good way you need to treat one in a systematic way rather than just by making a general statement like that which is as follows but this leads to something interesting leads to an interesting geometrical thing so let's see how to do that but so for now what we would like to know is that rather than treating a general path let's just treat analytic continuation in the plane like that which corresponds to the case h-bar real let's just try to push our cycle as much to the left as possible and to push our cycle as much to the left as possible we're basically just going to flow by the gradient flow of this function so now let me just explain what the basic problem here is that this is kind of the main phenomenon that's happening I think I'm not going to get to the second part of the top unfortunately ok so the main phenomenon is that this integral is that delta n you know delta i which is delta n is a relative homology class it's a cycle of boundary so remember this is a real it's a real triangle in a real n-dimensional triangle in an n-dimensional complex space this is z to the n so what are the singular points of delta delta should be on the board here or I just erased it delta is here what are the boundary points of delta well it's when t1 equals 0 or t2 equals t1 and so on or tn equals 1 or you can just see in this integral the singular points are when z1 equals p z1 equals z2 and so on up to the zn equals q so the singular divisor is z1 equals p union z1 equals z2 so it's a union of I guess n plus 1 divisors inside z to the n so it's just a complex simplex and this integral is going to be independent of the choice of representative in here in this relative thing however the gradient does not preserve the boundary so what do we need to do we need to descend the cycle so we need to inductively we need to inductively descend the cycle the boundary first and then descend the whole cycle so let me just draw the pictures I think this should become relatively clear with the drawing but the point being that we need to we want to do this in a systematic way so we want to apply some actual flows to descend the cycle because that's the only way we'll have a control over the size of the cycle as a function of n so what happens if we do this for delta 2 so delta 2 is a cycle like that so if we draw this inside if I'm not mistaken one way direction here so this is z1 equals z2 this is z2 equals q this is z1 equals p so these are complex planes these are complex remount surfaces so this is z2 and this is z each of these things here each of these guys is zn-1 and they intersect in zn-2 and so on so what we need to do to descend so if we apply the gradient flow on this triangle only then it's no longer going to have the boundary on the boundary of course that gradient flow is a part of the picture but we should first apply the gradient flow to the path on the boundary like that then do that on each boundary piece so one thing to notice is that boundary pieces have their own indexation if this is a zi this is going to be like a zi prime the functions g are compatible so the function g on the z with its indexation on the boundary is compatible with the other one so we apply the gradient flow on the boundary then we take the triangle plus the homotopes on each of the boundary pieces I mean this is the flowed path that sort of might have hit a saddle point here for example so this is a flowed down path here but we have to take the full homotopy of this path and then add that to the triangle and now we can flow this entire piece down now the new cycle so this part here is on the boundary so we can use this as a new boundary okay so the new cycle is going to be everything obtained by applying the flow plus what was obtained by applying the flow to the boundary pieces okay so it's going to be sort of the bottom of this picture plus the side okay excuse me I don't understand why the vertices are separated by the gradient flows because the g function on this thing is different from the but here we're flowing by a gradient flow which might does something to the points p and q also but of course it means that the flowed path of this boundary is this plus this plus this I'm doing the two dimensional case this picture is the one dimensional case I want to try to say here in the two dimensional case there's the flowed down part from here we flow down this segment that gives us this segment that we need to add this piece and this piece which are given by the flows of the boundary so the same principle in two dimensions is going to tell us that the full cycle we want is the flowed down cycle of this stuff plus the boundary pieces okay now the point is that the resulting cycle is that my gradient flow of the given function is always I should take into account the period of boundary which I should start with now yeah that's basically the idea so the new boundary of the cycle is this so it still goes through the vertices okay but when you write the new ones on the boundary no the vertices of the new triangle are still the same as the vertices of the old triangle but we had to flow I mean so what happens if you think about this maybe it's hard in two minutes but if you think about this for a little while if you've got a bunch of different flows and I think Misha even wrote a paper about this if you have a bunch of different flows you actually get the following structure which is you have a sum over labeled trees let's call that delta for the tree T or something like that what's delta for the tree let me just draw this like this so this is our original path gamma and then so in this picture there's a piece which corresponds to the original triangle that's going to be the tree these two trees that's going to be trees like that okay so with a one so this is the segment zero one okay the top thing of the picture is the segment zero one where these attach those are you might say the places in the original cycle delta which are at the origin of the point okay so if we take our original triangle and flow down we get the picture T1 T2 1 1 okay but we have some other pictures which are for example just T1 now I'm probably going to have a mistake here but there's several different possibilities we need a label here so there's going to be like S1 1 or there's going to be a S1 S for example there's several different parts of the picture which correspond to this to a point here right the point here we got flowed I guess so here we flowed down by the gradient flow of we flowed down all the way by this gradient flow I guess the point is that so kind of a detail here is we don't really want to do the gradient flow inside the Z2 we really want to do first the gradient flow in one direction then the gradient flow in the other direction so those are going to be either the 1S or the S1 cases so there's maybe two two segment maybe probably S0 there's going to be really two segments so this picture okay and the the picture on labeled trees is we have a tree labeled with 0, 1 or some variable S and we have the top edges are labeled T1 up through TK okay so there's going to be variables S1 up through S and minus K the variable is T1 up through TK okay and the image of this so this diagram is labeled by the now in the regions of the diagram we have the indices and so on and the image point corresponding to this diagram is obtained as follows which is you flow along the gradient flow for the function G you follow along the gradient flow for the function real part of G I K I L for the time designated by the variable okay now what are the critical points I should probably finish here what are the critical points the critical points are the case where you end on all the different critical points so the critical points of this whole picture are the case where the bottom of the tree these are all the different critical points okay so if you think about it this is exactly the picture of spectral networks so these gradient flows so to be more precise there's really spectral networks but not for an imaginary direction but for a real direction and the real direction you could actually choose because in here you could either flow this way or that way it doesn't really make a difference if you're trying to move to the left okay I'm trying to follow the structure that you say that each leaf corresponds to a critical point right so the what are the places where you can't flow any further the places where you can't flow any further when you've hit a critical point in all the different variables if you hit a critical point in some of the variables but not the other ones then you can keep flowing in the other variable so the places where you can't flow any further so these are gonna generate the fimbles the places where you have critical points on all the bottom vertices of the tree basically so it says that the locations of these critical points are things obtained by by doing the gradient flow down to a critical point along each of these edges of the tree basically that's essentially some kind of spectral network type of picture so let me just finish by a caveat which is that in the case where that we were discussing before the more advanced case where the B has maybe first-order poles at the turning points then what I don't know how to control with this procedure is the fact that any variable is supposed to stop at a turning point so this point here might be a turning point but for a different variable not necessarily the I0, I1 variable so that's different from Gaia Tomorowski so a priori from this procedure in any case the locations of the poles you have to add extra possible locations of the poles basically and that's why I wanted to say in part 2 which is that maybe there's some hope using our work with Ludmila and Pranov on harmonic maps to buildings maybe there's some hope at least in the case of SL3 of getting a more precise information on the location of at least the first pole basically namely I think the conjecture is that the location is going to be this type of picture but where this is the I0, I1 turning point and so on I guess I should stop here let me just comment just how would you want to prove that this converges well basically when we do this flow we're always decreasing the real part of the function okay and then we're going to cut off our we're going to cut off our cycles when we decrease that by a finite amount so the size of the cycles is going to have some kind of condition like the sum of Si less than or equal to some constant and this generates an exponentially small region in fact together with the fact of course obviously the original delta is exponentially small sorry and factorially small this generates a factorially small size and the original cycle has a factorially small size so rough and then the calculation is that sort of the number you have to calculate the number of cells you get here is some constant to the n so constant at the n divided by n factorial and that's kind of the very rough so I'll stop so comment in the spectrum nature this is still open it's an exponential bound it doesn't appear on the number of spectral networks yeah it's a little complicated because it has trees no but so okay and this gives a slightly more precise information which is that it's only the spectral networks that you get by this procedure so it means that they have to go up and then cut the path so I think there's only finitely many we don't have closed cycles yeah it's good well for a given path and we're fixing the p and the q and so on then the number of spectral networks that are going to cut through the path in that way is fine yeah it seems to me that this aspect of the theory which you have covered should go over to higher dimension and I especially think of the banyan block formalism which I briefly flapped on which is a system of integral recursion with integrals exactly like the one on the top there with the denominators and so this would be very nice if you can prove because banyan block is formal if you can use to prove resurgence in higher dimension yeah maybe that was certainly being fun to do let me just comment maybe that as far as I know I mean but maybe I mean you know better but I mean from just kind of a very a priori perspective we're really using the fact here that you have a closed form formula for the function we're trying to compute so we're not so this is kind of not going in the same direction as what you guys are proposing which would be fantastic which is to take the perturbative theory and sort of create the resurgence function here we sort of already know what the function is we want to maybe prove some additional properties like continuation is very similar to what banyan block wrote in terms of iterated integrations but of course there's a lot of rigorous work to do but are they finite dimensional integrals? yes okay okay special networks in higher dimensions well maybe I mean I mean the initial the initial idea is just to use some kind of gradient flow to find the critical point so in the rank 2 case the quantum resurgence the ridiculous case you recovered the location of the singularities as found by a car found on a bar and in the location of singularities okay but what about the nature of the singularities how come that in your formalism you don't reach the nature of the singularities because they have very precise predictions for that well so in the general case this is kind of related to the each term m i c I mean if you look at this integral what are the singularities if bi has logarithmic singularities then we have we're gonna have some kind of integral like maybe Marcin can do this on top of his head something like d z 1 up to d z n divided by something like z 1 up to z n but then times some kind of summation of z i minus c I think the integral maybe something like that so I think you can show that this integral has one over you know norm of c type of a behavior but I was just doing that this week actually I was trying to think about but that's gonna be the same that we're gonna have this type of behavior for any value of n so what that really means is that each m i is gonna have a series expansion perhaps at the singularity points but the series expansions all have the same order of initial term so the when you make the sum the series expansion if it existed the coefficients of the series expansion in the sum will themselves be series in the coefficients of each of the coefficients here which is different from and that's like one of the main simplifications in the original case where b is regular in the case where b is regular the powers of c that happen in the series expansion go up so any given coefficient is just a finite sum as far as I can tell you also have powers of logs in those series that also go up so the sum is not gonna be a regular thing it's gonna be a trans series with higher powers it's gonna be like sum of c to the n times log c to the n something like that but where m the value m goes up as a function of n at least that's in the regular case in the irregular sorry that's in the case I mean don't mean regular I mean the case where b is holomorphic in the case where b has poles I don't even see why the powers of log could all happen at the first term as far as I can tell maybe I'm missing something but