Well, thank you, Jung, and thanks to all the organizers for bringing us together. So first, I'd like to start by saying that I'm a young user of resurgence, and I find this theory totally fascinating. So to me, it's like a piece of magic from 21st century that fell in the 20th century. And therefore, I want to first thank all the founding fathers, Professor Akal, for developing this theory, for originating it. And I hope this modest contribution will be application of resurgence. So in this sense, this talk is applied resurgence. In the course of the talk, I'll try to give you both conceptual piece of content, Food for Thought. And that will be probably the main part, because these will be questions that I don't know how to address, or I'm hoping some of you in the audience will see the pattern or the structure or deeper reasons for it. And then I'll give you also computational content. This is actually my territory. I like to compute things. And in the end, I'll try to explain how to compute expressions like this. So the end result will be a Q-series with integer powers, integer coefficients, and lots of other cool structure that's useful A in topology, and B for studying modular properties, modular forms, for example, typical Q-series development is reminding you of modular properties or modular forms. And the function on the board is one of the outcomes. Again, I'll try to explain where it comes from, how it comes about, but I'll generate this as a Q-series so I can write, I betray many terms in the Q-expansion, but I don't know what this function is. And therefore, one question to the audience, that's already your first homework, is if anybody help recognize this function, is it a Q-hyper geometric series? Is it a num sum? What kind of gadget it is? So I'll give you first 20 terms in case you already guessed what it is or as a check.  It has a name, right? But I want a kind of closed form in the sense of sum over M and then write some closed form expression. And I don't know this, I'm honest. I'm not joking, but there is a way to generate it up to any order. And as a test, so first 20 terms may or may not be enough. If you can email me or look a recent paper with Manolescu, you can find how to generate these things. Again, in the end I'll explain. And I give some later terms just as a cross check or so you can determine. And indeed, as Maxim points out, it has a name. So we would naturally call it analog of conceivage, a gear function which originally was something, again I wasn't present, in fact I wasn't in the field the other time when Maxim introduced it. And Don loved this function. So he called Maxim's identity strange identity which they explored. And that function naturally applies to trefoil knot which we may encounter later today. And this thing is supposed to be analogs for a closed manifold, namely hyperbolic manifold produced from a figure eight knot. So, and again, my goal is to teach you how to produce such conceivage that your type functions in greater generality. So this will be some topological applications, some modular applications, but again I want to put it in a broader context. And the central theme of this talk therefore, so starting with conceptual things before you lose me in details, is that we'll go from normal, non-integers to integers. And again, main question will be why this happens. So the typical setup in which as a user, I apply resurgence is where we start with some perturbative series in coupling constant that I'll call H bar. This is a formal power series with zero radius of convergence. We do a Borel transform. So that will be easy perturbative as a function on the Borel plane and that variable I'll call psi. So that's a Borel plane variable. And from there, another step is to do inverse Borel transform or produce something with better analytic properties but the same asymptotics as the original series. And this gadget in this talk will call Z hat. In topology, it has a precise meaning. So it's a very specific function but in the first part of the talk, I want to use this just as a name and then view it a little bit more broadly. So perturbative thing for us will be sum over M, A M H bar to the M where main property is that this perturbative coefficients A M will not necessarily be integer. That's actually typical case and many problems I'll consider today. So this is where rationale stands for. I mean, we start with non-integers. On the Borel plane, but let's assume they're rational. This is not important. In fact, there will be examples where they're not even the rational but in algebraic number fields. But the miracles will be happening kind of in stages. So on the Borel plane, this usual Borel transform for user is like me is basically taking this perturbative coefficients and dividing by gamma. If M is integer, I can write just factorial. If it's a little bit shifted from integer to integer plus a half, then gamma is a better way to do it. And we put psi to the M. So that's another series. And then of course on the Borel plane, we ask, okay, what are the analytic properties of this function and analyze it and then go back? But the miracle for me will be that after the dust settles and we do this procedure, go to Borel plane and back, the result here will be of the form up to some overall power of Q which may not be integer power. This will be actually power series with integer powers and integer coefficients. So the result will look kind of like that. Very good. So Q and H bar, I was going to say that next where Q is exponential of H bar. And this will be one of the central equations in the entire talk. And to those of you who are familiar with enumerative problems, I want to point out that this is very similar as in Gromov-Witton versus Donaldson-Thomas theory. If you have Gromov-Witton invariance and Donaldson-Thomas invariance of a Calabi-Yau threefold, they're actually related in a very similar way and share many similar properties that Gromov-Witton invariance are typically rational because there are stabilizers, Donaldson-Thomas invariance are integers. And this is not a joke or not an analogy. It will be true in context that we're gonna discuss today. And therefore this relation, which is also standard relation between parameter, which is producing generating function of Gromov-Witton invariance H bar and Q, which is used in Donaldson-Thomas series is exactly the same as here and for a good reason. Thank you. Any questions? I'm sorry, I'm upset because when H bar goes to zero, Q goes to one, so all terms are the same order in a series. Exactly, and that's precisely, unfortunately, unfortunately that's the kind of applications where I'll need it. So that's a great question. So indeed, whether we like it or not, at least in problems that we encounter in topology and modularity in this talk, what will happen is that natural expansion point for Q series is, of course, Q equals zero, whereas natural expansion in terms of H bar is around this point where Q equals exponential of H bar goes to one. And the reason it has something to do with conceivage is that there are analogous limits and various other rational points, whereas Maxim pointed out for that original conceivage function, some miracles happen. So indeed, the goal for me will be transferring information from this domain near Q equals zero to this domain near Q equals one and asking why certain things happen. In particular, asking why do we get integer coefficients in a very large class of problem? The integrality emerges, and that's the integrality, out of thin air, so that's pretty cool. I'm not an expert on resurgence, but I suspect that some part of explanation has to do with structure of this B of the perturbative on a Burrell plane, and then natural question is, can somebody produce conditions under which this is supposed to happen? So again, I don't know full answer to either of these questions, but I'll try to say a little bit what we believe is responsible for integrality emerging out of nothing. So that's the main point. In this error from Gromov written to Donaldson Thomas, do you have Burrell resonations from where? So at the present day, I think about going from Gromov written to Donaldson Thomas as an analogous phenomenon. It's a change of coordinates, no interval transformation can come from G to G, probably to GT. But it's the same type of phenomenon, so. And again, I wanna use this not just as analogy as a very specific tool. Okay, so this is fun. Again, integrality out of thin air, but then just to make it even more fun, I want you to think about the following problem. Let's start and go the other way. So from integers to non-integers. So this is actually not so complicated and mysterious. So we know that one plus two plus three is integer. So we start with integers, do something with them and get integers, that's not very surprising. But how about one plus two plus three plus four and so on? What does it equal to? Thank you. And that's a non-integer. That's a rational number, right? I just overshoot infinity by a little bit. Yeah, exactly. So therefore, we'll encounter both phenomena where we go from non-integers to integers, that of course is more fun. But also, what's important is that in explaining this, what will be playing a role is this kind of phenomenon where you go from integers to non-integers. And let me explain that in the next half an hour or so. Suppose you have a system with a phase space in quantum field theory, this could be field space that via some Morse function or action function you can project to the Borel plane. So Borel plane is basically plane of values of that. So that's our Xi, that's plane of values of the action. And then on a Borel plane, we pay attention to singularities or poles. So I'll give them name, say alpha and beta. And then we integrate over corresponding rays, R plus, copies of R plus, starting at a given critical point, which are images of stationary points in the phase space or field space where fields or variable satisfy equations of motion and this rays or copies of R plus represent cycles of steepest descent spent by trajectories of steepest descent, sometimes called left shift symbols in the field space or phase space. So the phenomenon that often happens in the situation or interesting phenomenon that happens is when we change some parameters or even if we don't change parameters, but we want to know how much this stationary point beta contributes to this integral when we do alpha. Either way, if you think about tilting this ray, what may happen is that it will go through the pole of through the singularity of the other critical point alpha and beta. And as a result, expression for integral will change and the typical way to describe this is to say that in the phase space or field space, integration cycle, yem alpha will change by the amount which is proportional to integration cycle beta with coefficient, which in various contexts could be loosely called either stocks coefficients or trans-series coefficient. And the basic point is that when this picture holds true and alpha beta are always integer. Remember, this talk is all about rational versus integers. And second thing that holds true is that and alpha beta in absolute value is always equal to n beta alpha because in the standard Picard left shift theory, n alpha beta is interpreted as intersection number of cycles gamma alpha gamma beta. And if gamma alpha intersects gamma beta, then of course, the opposite is also true. So as a result, up to a sign which has to do with possible choices of orientation, they have to be the same. And it's clear that intersection number counts intersection point, so it has to be integer. Important thing, and this is again, I don't know full deep reason of this, but for this phenomenon to happen to get something nice and modular and having good meaning in topology, both of these will have to be violated. And the reason it happens is precisely this. If you take a couple of integers and sum them up, the answer is integer. But if you take infinitely many integers and sum them up, you're gonna get rational. And this happens as we observe in this concrete framework in the paper with Marcus and Pavel Putrov is that in class of problems where there are infinitely many settles, where each settle point alpha is replicated infinitely many times, so alpha has its images, alpha prime, alpha double prime and so on. So there is a whole tower. And each critical point is replicated infinitely many times. That's where this happens. That's where this becomes a generic phenomenon. Are you thinking of time of dimension? So I'll tell you, so very good. Next I'll address question, when does this happen? So what is this kind of class of problems? But we feel that this is crucial, namely this way of going from integers to non-integers, where trans-serious coefficients are non-integer and they're no longer symmetric. This will be playing crucial role for phenomena of distance. Again, I think there is something much deeper going on. It's definitely above my pay grade. And because I don't fully understand this deeper reasons, I want to be crystal clear about little things that we do understand. So that's why I want to stress so much integrity versus rationality and vice versa. So what is this class of problems where this picture on the right side of the board is realized? When is it relevant? So class of problems infinitely or infinite towers, let me call them towers of saddle points, where each critical point, saddle point comes in multiple infinite copies which I'll call towers. So the list is going to be quite long and there are many ways to realize this structure and it appears in many different problems. Luckily, the context which we encountered in that paper with Marcus and Pavel actually was the context for all of the five that I'm gonna list for you, I realized at the same time. So all the descriptions apply. So first of all, such problems include problems that can be described by a spectral curve in C star cross C star. So it's important that it's punctured plain for both variables. Such a curve, let me call the polynomial equation A of X, Y, so it will be given by polynomial in X and Y, but the corresponding symplectic form with respect to which I'll want to view this or holomorphic symplectic form will be DX over X wedge DY over Y and I want to contrast this with a class of problems where we look at the spectral curve in C cross C and use DX wedge DY as a symplectic form. So here, there'll be several things going on and one of them is the fact that this space is not simply connected and there are windings. So if you start with this class of problems and perform WKB analysis, you quickly find that what we call ZP trabative as a function of H bar, we'll start at life of course as exponential of one over H bar integral of log Y where you solve Y in terms of X using this polynomial relation and integrate DX over X, that's a Louisville form for this symplectic form. But what's important is that in this class of problems because if Y is a solution to polynomial equation, log of Y is logarithmic, once you integrate it yet another time you get dialogues and this exponential of dialogues is not going to be a single valued function. It's going to be multi-valued function and therefore each critical point of this multi-valued function, we'll necessarily generate exactly these kind of towers of saddle points. Excuse me, are you thinking for instance of a potential which would be trigonometric polynomial? No. Could that be? What do you call by potential? It's not some differential equation at all. Okay, yeah. It's a good difference equation. Exactly, thank you. So in fact that's another name for this kind of class of problems that instead of differential equation which would be the case in C cross C, this leads to Q difference equation. Indeed because this symplectic form, thank you very much Maxim implies that in a quantum world if I introduce quantization of the space I'll have operators, maybe let me put hats on everything Y hat, X hat is Q times X hat, Y hat. And as a result, this equation is indeed a Q difference equation. So that's the crucial difference and then of course we can use various other tools to compute higher water terms so we can generate this Z-patributive for each bar in systematic way. So for example, topological recursion is a cool tool to apply and in this class of problems which in fact I'll describe later where A is some A polynomial. This is what we studied with Kroitor-Zuchowski. Also at that time, trying to shed light on the integrality but again we didn't understand as much as, I'll try to tell you. I want to carry integration. What's the integration contour? Over path, over path on this A polynomial curve. So it's a curve in C star cross C star which looks kind of like this. So that's a contour. There's a cartoon for the curve and you want to integrate so from some reference point which you can call zero to a critical point alpha but the point I'm trying to make is that because things are not simply connected there are all these windings and they correspond precisely to the various branches of these dialogues. So that's the phenomenon which is absolutely crucial here. It seems. The critical points are fixed. Do some of them are all critical points? Well, because... The fix is normal. No, no, like I said, I don't know what's the right kind of prescription but the problem that's being solved is this. I'm going to erase it in a second but the point is you start with one critical point. You want to perform this process of Borel resumption resurgence and usually what happens is that you get contributions from some other critical points to perturbative expansion of a given one alpha. But what I'm saying is that when you have each critical point is replicated infinitely many times even if you start with a given alpha you get contribution from infinitely many betters or their images. Not like in a cartoon pointed out at the very beginning of this work that you consider the remand surface of the function and then you are all discreet and isolated and then you can just repeat the left shift symbol thing without having these infinite numbers of critical values because they all belong to very different sheets of your... Well, here they all contribute with the same, exactly same action. So the feature is that because you expand and shade all these critical points, all the images of the beta will have exactly the same value of the Morse function. I don't see at least in physics context how to separate them and as a result they all contribute exactly the same. We have to be democratic to our children and here we have infinitely many children and the betas are all equal. So therefore I cannot just say oh, let me take contribution of these two guys because I like them. It's not fair. Okay, now the next question is how do I pull this blackboard down? Yeah, exactly. Like this? That was very tricky. Okay, so where else do we see this kind of phenomena? So that was number one. Number two is trigonometric or hyperbolic. Both names are used in the literature, integrable systems where variables are exponentiated and we use two different equations which is the Baxter equation as opposed to rational integrable systems where H bar and unexpandentiated variables are boy actual variables in the face. Another application which I already mentioned is open topological strings which is another class of problems with precisely this exponentiated behavior where each critical point is replicated infinitely many times with exactly same value of multivalued Morse function. And in this case, this perturbative series with which we start the original program before we apply resurgence is indeed generating function of Gromov-Wittanen variance. And important point is that Gromov-Wittanen variance are rational numbers which is very well known in fact, clearly explained in early works of Maxine with Urimanian due to stabilizers in theory of stable maps. Another phenomenon which is actually more interesting and that's where Marcus I and Pavel encountered it is the context of gauge theory in dimensions greater than three, three or four. And here are the point is that in such dimensions above two, gauge theory is really a theory in the sense that in two dimensions and lower, gauge fields can be always replaced by scalars or fermions, but in high dimensions, they really are quotienting the space of fields because gauge transformations are not symmetries, they're equivalences. And as a result, the space of gauge connections on any three manifold or four manifold is not simply connected because if you take space of such gauge connections, namely the phase space and then impose gauge equivalence, what you're essentially doing, you're taking your phase space and curling it up so that you're creating on trivial fundamental group. So for example, in three dimensions, this becomes Z and this fact which may sound funny is why we actually have instantons and four dimensions and that's why in three dimensions, trans-simon function is only defined module integers or two pi times integers depending on how you normalize. In other words, the cartoon is such that if you have a critical point alpha in your phase space and critical point beta because now this space is not simply connected, you have all kinds of findings and that's why your Morse function becomes multi-valued. In the context of supersymmetric gauge theory, this phenomenon shows up by solving equations which normally would be minima some super potential but in this context, it's minima of exponentiated derivative of super potential which is written in terms of logs. So same thing, logarithmic variables versus exponentiated variables. So that's another typical equation, DWD log X appearing in the exponential which has to be solved in order to find critical points. For example, in three-dimensional N equals two theory which is under 3D, 3D correspondences due to some 3D trans-simon theory. So you can see this phenomenon in multiple different ways. Here you have this W, Landau-Ginsburg type potential as for example, integer times log of X squared. It's derivative with respect to log X is gonna give you just log X, exponentiated that you're gonna get X to some power equals one but again, point is that in this class of problems, X just as above takes values in C star and therefore it gives you whole tower of solutions. I mean, whole tower of critical points. So therefore, naively, what you have to do, you have to either sum over infinitely many leftist symbols. I don't know how to do it. I don't have homological tools under my belt to do it. Or you do resurgence where you just sum infinitely many poles. That's totally okay. We can sum infinitely many numbers to regularize them to non-integers. So that's what appears here and now I'm gonna erase this magic formula which is again responsible for everything else I'm telling you today. So number five, there are many more problems where this happens. And number six is, right? And number six is it happens in the context of complex transimus theory which under various dualities can be mapped into all these descriptions I mentioned. So actually all of them. How can it survive the holomorphic? With complex gauge group. So it's transimus theory where fields formally are valued in complex groups such as this L2C. But there is no holomorphic, anti-holomorphic sector. There is no complex conjugation. So just, three dimensional complex transimus theory. Complex transimus theory. Actually, yeah, holomorphic transimus theory sometimes also refers to omega wedge transimus form on a Calabria three-fold. It's a TQFT-ish theory in dimension six.  Get algebraic manifold with closed one form but it's not differential with function. As I'm gonna get critical zeros it will be, it should go to universal cover and. Oh, cool, yes, exactly, yeah. Okay, so yeah, I would love to hear more. I think I know what you're suggesting and that's a great idea, yeah. You should tell me more, yeah. That's indeed would be a big example. So that's probably in here. Okay, so now claim that already said kind of implicitly a number of times but when you do resurgence in this context you find something interesting. So first of all you find because you're resuming each residue or each contribution in the trans series appears infinitely many times this trans series coefficients and alpha beta are no longer integer and in general they're rational numbers precisely by the phenomenon that you guys told me how to resound one plus two plus three.  What was the critical principle in the circle? Because suppose you do perturbative resumption at the critical of a perturbative series alpha labeled by alpha, alpha is a critical point then anything else can contribute as a trans series from what I tried to explain each other critical point just like alpha itself is repeated infinitely many times with exactly the same value of Morse function. So what you get is for example there is a trans series coefficient here originally just for one guy that is produced by pole with a residue plus one what you get is infinite sum of one, one, one, one and so on. And that's exactly what you guys told me how to regular. So that's first of all you get this but second thing is even more important even though it feels less important and more technical. So you find that an alpha beta is not always equal always and beta alpha up to a sign. And that's actually again I don't know deeper reasons for this but I think that's what's important for this modularity and topological applications that I'm gonna mention next. So in particular it happens if alpha is non degenerate or Morse critical point namely expansion of function around this point is Gaussian but beta is non Gaussian. So in other words degenerate. Are you saying this always happens or this can happen? Okay from now on it's probably fair to say that I'll stick most of the applications that I'm gonna tell you in the rest of the talk is to complex trans-simon theory but I encourage people to think about other contexts where this does happen. So Maxim is making really great suggestion and again I don't know that's part of the original question why and when I'm telling you just a little bit of why I would love to have deeper conceptual understanding of this kind of phenomena. And also I would like to know when. So in particular is there a way to characterize in all classes of problems on a Burrell plane when and how this happens. But the theorem is that indeed if, so this part of the statement that if beta is degenerate and alpha is non degenerate that's supposed to be always true. So that part has nothing to do with trans-simon. It has a flavor of mixed whole structure because we have like filtration. Exactly, exactly. Only on some. Yeah, yeah, yeah. I believe that there are exactly such several ways to look at the problem and basically what's happening here is that degenerate guys try to be in their own grading more. Yeah, yeah, so yeah. Exactly. Some more sectors of spacing. Exactly. So. To a naked eye or non-expert in this field again the first statement of this theorem probably looks much more natural. The second looks like technical and non-interesting. But again I want to emphasize that it's actually second part which has more important corollary that I want to capitalize on in the rest of this talk. And it basically says that degenerate saddles are special. I want to pause. You're going to mention the theorem in what complex right here. No, no, no. The theorem in. Exponential integral. Exponential integral. Exponential integral. So like I say this theorem is more general. So it was in the paper on complex transimmons but argument applies just as well to any exponential integral. You really mean conjecture, right? No, I mean theorem. You analyze exponential integral. You ask, how does. Let's not get too confused. The theorem is a proof. It's a statement. It's a statement. Yeah, that's a provable statement and we can prove it for you. What is the statement? So the statement is that if you start with exponential integral and. Find a dimension. In finite dimension. Let's start with finite dimensional. An application we used was to infinite dimensional. But again, finite dimensional exponential integral. And before this kind of analysis, the usual analysis of Borel reclamation where each critical point is replicated infinitely many times and analyze what happens for various types of degenerations. You indeed find the stratification that more degenerate saddles don't get attached as a transfer is to less degenerate ones. So that's it. But before stating, can you give one example of this? Showing it's not the interceptor. Yeah, I'll do it later. Like I said, I wanted to start emphasizing phenomenon and then I'll get two examples. In fact, in. I mean, for example, if you say this is, I take any exponential integral with ramifications, I will get something like this. Well, I'm saying trigonometric type of integrable system. So start with system that has to do with spectral curve described by this equation A of xy where variables x and y are c store valued. I claim that's gonna be the case. I'm happy to walk you through this in detail. The example I was going to show through Churn Simons. I apologize, I didn't know what the requests would be. Next time I'll prepare the right example. It's hard to know what which is R. But I want to again emphasize the role of this degenerate saddles because for topological applications, that's very important. And I want to, one way to say it is that, again, this degenerate saddles are special. I want to put it on more strongly. I want to say that corollary of the statement is that miracles happen. So this is surely a mathematical statement, right? Stavros, you would probably agree with this. For alpha degenerate. In the case of Churn Simons theory and let me live at the SL2 or SU2 level, this means that alpha is basically abelian. So something which normally you would think is boring or abelian is playing a special role. And I want to explain why and how in the remaining part. So first, it leads to a statement which is that there is a three-dimensional apological quantum field theory in a mathematical sense of a t-flour of a t-single. There is a functor. I'll call this functor z hat, which associates to three manifolds, a number, but so normally it would be called z hat value of the functor on a three manifold. And the point is that, as you can probably guess by now, that this number is going to depend on q. So q is a parameter, but it's not just a random number that depends on q. In other words, it's not just a function of q. It's a natural q series with integer powers, integer coefficients, which is perfectly convergent inside the unit disk, norm q less than one. And in fact, in physics, so this is conjectural part. So that's actually where conjecture is. I want to be extremely clear. What's a statement? What's a theorem? What's a conjecture? This is conjectural. That this is expressed as graded early characteristic of some vector spaces. These are called spaces of BPS states, H, I, J. And therefore it gives an explanation or breathes life into meaning of the integrality. Because once you see a q series with integer coefficients, you can naturally ask, what are these coefficients counting? And claim is that in this case, they're counting something concrete. Same kind of BPS states that you get in, say, black hole counting. Now. What is the formula? You didn't write it properly. That's not it. That's some physics thing. I suggest to. Is any group related to the story of a group dependent? It's basically for every complex group, you can label this series by either root system, such as SU2, SU3, and so on. Or by corresponding complexifications, say, SL2C is actually the only example which is considered seriously in the literature. There are some statements about SLN. But yeah, it's labeled by. You don't say that you get really finite dimensional space for a surface, nothing like this. That's right. So that's actually precisely the goal of this recent paper with Manolescu, to ask what kind of TQFT it is. So in order to say it's a TQFT, you have to be able to cut and glue three manifolds along surfaces in any way. So you have to, first of all, define it for any closed and open manifold. And also, you have to ask what kind of Hilbert space you associate to surface of genus G, for example. And so it will be some z hat of sigma G. So we only know the answer when genus is equal to 1. I want to be completely clear, completely honest, what's known, what's done. This is good enough to build any three manifold, because any three manifold can be obtained by surgery operations along genus 1 surfaces. But it would be nice to extend this analysis to higher genus. And this is indeed infinite dimensional Hilbert space. So the main technical ingredient in this work with Manolescu is to define taking the traces in this kind of TQFT. And once you're at it, I want to ask another question. Is it extended to TQFT? Can we extend it further to higher categorical levels in the sense of a TIN signal? I don't know the answer to it. So I want to be extremely clear about what's done and what's known and what's not. I mean, it's conjecture that exists from physics, but that's a conjecture. But also the invariant for closed streaming for has an index, why it has a. Oh, correct. So I didn't plan to say it explicitly, but in fact, it's exactly the same kind of degenerate abelian saddle that labels this invariant. So it's actually decorated TQFT. So that will be a precise statement. So in the world of TQFTs, there are so-called spin TQFTs defined on many folds with additional structure. So it turns out that this label or these degenerate saddles lead to so-called spin C TQFT. You have to keep track of spin C structure and how it preserves undercutting and gluing. Again, that's getting a little technical. I'll be happy to discuss this or suggest to look at this work with Manalescu just from a month ago or so, for describing the Hilbert spaces and the invariance of closed-stream manifold. It's a firing sound. It's a long row. Which, what's the finest? The independent general. Oh, this dimension? No, no. This, of course, is infinite because it's a Q-series. So it has infinitely many terms. I get confused because it's dimensionless combology space because it will increase. Yeah, so it increases, but in each gradient, they're finite. So another statement, which also is a conjecture, is that dimension of the spaces is finite for each gradient. In fact, anyway, I'll say something about it in a second. It's related to degenerate connection schedules, or not? It is, it is. So it's basically, right. So, OK, maybe I didn't plan to say too much about it. But again, since you guys asked, so what's the story here? So in entire, it's like a research program. It's not even one paper or project. In whole research program in the past 20 years, I would say, with Comron Waffa, Don Zagir, Tudin Dimovter. And they decoupled, but then more recently, younger people like Pavel Putrov, Dupé, another joined in. We're trying to look for, say, SL2-seacher and Simon's theory, which is a TQFT, so that you can A, compute things, on general three manifolds. And we used all kinds of tools from either string theory and so on. And the point is that resurgence here was one ingredient which really helped. And that's what I'm trying to emphasize here by emphasizing this role of degenerate settles. So if you perform borrel resumption around the degenerate abelian flat connections, what you get is this nice function Z hat. So that's exactly what it is. It's borrel resumption or resurgence at abelian or degenerate flat connections. It's very well behaved for some miraculous reasons, 40 QFT purposes. In other words, it allows you to cut in blue. What could be done is constructing various other functions. So for example, 20 years ago, we had vortex partition functions, which were also defined by using some other integration cycles in the borrel plane in this language. And these things would be power series in variables such as Q and X with integer coefficients on a face very similar. But this would not be a TQFT in a sense that we would not know how to take, say, partition function for not complement this word partition function and glue it in, namely, cap it off. So we didn't know how to do surgeries. So it wouldn't be TQFT. Let me just ask you a question. If manifold is homologous, integer homologous spheres, no, then it will be just perturbation series for trivial kind of things. Exactly. And actually, in that case, it reproduced some of the experimental results by Hikami, Zegir, and in special cases where it's Poincare sphere or Briskorn spheres. So it gave kind of physical meaning or Donald St. Thomas-type meaning in terms of BPS states of what was happening before. So in fact, these were pioneers of it. For example, you can do resurgence around various other flat connections which are non-degenerate. For example, I'll show you a picture which has several branches. And one of them is sometimes called conjugate. This is opposite of hyperbolic branch. In that case, you can do this Borelary summation resurgence. You start with perturbative series, your resummit. And you obtain a partition function which we suspected. And now, a couple of weeks ago, we verified very carefully in precise calculations with Don Zegir that you do get a function of Q, which is exactly partition function of Anderson-Kashaev theory. But this is much better behaved object. Again, we still don't know how to cap it off, as far as I understand, so how to do surgeries. But it's not of the form where it's a power series in Q and X with integer coefficients. It's just a function. So in many attempts that we had over the past 20 years, we had various variants of this. But until this role of degenerate settles was realized, we could never get a TQFT. And claim is that this thing is a full TQFT at least at the top level. So you can do it for three manifolds with any surgery operations you want. Now, better yet, since you- The type of a guy, it depends on X as well, right? Or is this not what you mean, Q and X? Yeah, yeah, that's what I meant. Q, it does depend- It's just for Q. I guess it's of Q and X also. But okay, you suppress X. Left on side, it's only a function of Q. Yes. Oh, sorry, sorry, sorry. Yes, sorry, sorry. Yeah, sorry. That's what I meant. Yeah, sorry. Right and left-hand sign didn't agree. Anyway. So here is another conjecture. So again, separating theorems from conjectures. So I'll call it 3D modularity conjecture. Since it appears in the paper under the same name, 3D modularity, that another miracle happens for this resurgence of degenerate side-all, side-billion connections, namely this invariant that had for any three manifold as a function of Q and possibly X if it has boundary, E is in fact a character of logarithmic vertex operator algebra. So not only this BPS dimensions, this HBPS is infinite-dimensional graded vector space, it's a module of a vertex actor H algebra. And therefore, this series has some modular properties of the type that logarithmic UA people encounter. These are not usual modular properties. That's why the logarithmic, these are spoiled modular properties of the type that Ramanujan found. So this is a mock, high-depth mock and even less modular objects. There are deeper and interesting reasons for this coming from representation theory because representation category of log VOAs is not simply simple and that's responsible for the spoiled modularity properties. But I wanna point out that for purposes of topology this degenerate side-alls are cool because they produce something that behaves nice on the cutting and gluing. And I hope you agree with me that this cutting and gluing seems to be very far from degenerate versus non-degenerate. I don't know how to relate them directly. I only see that there is a correspondence.  And here there is analogous factory of modular-like objects which are interesting for studying modular properties. For example, the one I showed you in the very beginning is an example of that. So that's precisely that kind of thing. It's a serious in Q, so it's chiral. It's a serious in Q, so it's chiral. Yeah, it's totally chiral. And that's why it's just VOA, not full CFT. Okay, so I promised you to show how to do calculations. So I want to end. So there were many questions and we started late. Can I have five minutes? Well, you have five minutes. Oh, okay, I have five minutes, thank you. So I want to show you how to compute things. And also, if I get a chance, give you another homework. So one homework has helped me recognize this Q series which I showed in the beginning. It's not a joke. I have no idea how to identify it. Yeah, before we get to the picture, I want to write a couple of formulae. So what are we gonna compute? So first of all, from what I just said, hopefully you see it. Or if not, you can ask me later, look up the details, it's all written up. So let's try to compute the starting point of the complex trans diamonds theory around some flat connection. So again, Q will be to use not complements because again, you can get any three manifold by doing surgeries on arbitrary knots. So therefore, understanding theory for not complements is obviously one important but not the most important step. And here I'll go back to something I did 20 years ago or almost 20 years ago is constructing perturbative series of H bar on a not complement as three minus K. So at that point, the conjecture was that as a function of H bar and variable X, you can either do this using WKB approximation or if you have a curve A of X, Y equals zero, which in that case is A polynomial, it may have many branches as a curve over C plane parameterized by X. So you could have all kinds of branches. So one of them is usually called geometric. There is a conjugate branch. That's the one related to Anderson-Cashier theory. There is a billion one, which is what we're gonna be interested in. And there could be lots of other ones in the middle. Suppose you fix X and you want to compute perturbative partition function of complex transiments at some branch here in the neighborhood of this value of X that you fix as a function of H bar. So claim is that it's annihilated by operator A hat, which is produced by quantizing the A polynomial curve. So it's a very simple Q difference equation, which then allows you to solve. So all these choices of branches are precisely these choices that you call alpha. So alpha is either one or two or three, depending how many of them you have. And on each of them, you can solve this Q difference equation perturbatively in H bar. That's something very concrete, something explicit. And what you get is that it's of a form that we already saw earlier today, one over H bar as zero on a branch alpha as a function of X plus S one on a branch alpha as a function of X plus H bar S two on a branch alpha as a function of X and so on. So you do it perturbatively, okay? So what we wanna do, it's already a function of X and perturbatively of H bar. So that's very computable. I'll show you this computations in a second. But what we wanna do is we wanna perform this resurgence and as we describe all along, transfer information from the main near Q equals one where H bar is equal to zero to write it as a series on Q. So what we really wanna do is, exponentiate this and resum in such a way that that's a sum over M FM of Q times X to the M. So we wanna present this as a function of X and Q or ideally series in both. So that's gonna be as a hat of a not compliment. Huh? Yeah, yeah. X belongs to C star, yeah. And that curve, sorry, this curve A of X, Y isn't C star cross C star. So that's crucial for what's happened. So there is something we did. Exactly, this was implemented in this paper with Tudor Dimovta, Don Zagir, Jonathan Reynolds. And the paper is called Exact Results in Contextual and Simons Theory. Exact because we can do it to any order in H bar. So you can solve this easily on any branch to any order in H bar. So we saw no obstacle and that's why we called it Exact Results. But we didn't know what to do with this because it's only perturbative. So that's why perturbative is part of the title because at that time we didn't know what to do with this wealth of information of computing this coefficients to any order in H bar. But now we do. So anyway, I'll scroll down where I show how we did these calculations and I embarrassed myself by pointing out that we missed something obvious and cool. So where is it? Yeah, so that's this Q difference equation. Here X and Y are called L and M. So in the case of say figure eight knot, this is one of the branches called geometric. So we compute this as one as two and so on. This is another branch called conjugate, two minutes. And this is the Abelian branch. So we have all this data for functions as zero as one as two of X. So what did we do? Again, this was 10 years ago, in fact, we just had a 10 year anniversary of this paper. But we missed something completely obvious. And I wanna again embarrass myself by showing you this. We take just that result from old days. And again, something really special happens for this degenerate or Abelian connection.  We open the exponential. We basically take that old result, no new computations. That's what we did in the paper with cheap print. Just exponentiated. And you get at each order in H bar, some function of X, which has development to positive and negative powers. Let's not worry about X dependence for a second. So focus your attention on just fixed power of X. Pick X to the one half, for example. And ask yourself, do I see anything? I'm embarrassed to say we didn't. But again, it's a calculation that a high school student should be able to do. You shouldn't wait 10 years to do it.  You see a bunch of rational numbers here. These are gram of written coefficients of something. Or you can interpret this as, in many different ways. But focus your attention on X to the half. Can you tell me what this function f of m is, in that case? It's H bar, it's X bar, it's H bar. Explanation of H bar, in other words, it's Q. So therefore you're telling me that f one half, or let me call it f one on this notation, is just Q. How about next thing? So here, notations are such that only odd values of m appear and powers of X are this odd values divided by two. So next step is going to be f three of Q. So what is the value of f three? No? Zero. There is no term after the X to the three halfs. So what about f five? That's Q squared. I trust you, but see, I'm on spot giving this talk, so therefore I wanna double check. Yes, you are much better than me, it's Q squared. What about seven? Very good. And so on. So in fact, there is a very nice closed form, which Maxim, at this point, should recognize. So this is last equation I'm gonna write and stop. So I don't want to upset the chair. This is definitely not something I wanna do, but I wanna say that what this function is, really is, so what you get is f m here is epsilon m times Q to the m square minus plus 23 divided by 24. So up to factor of Q, this is m square minus one over 24, where epsilon m are either plus one if m is equal to five or seven, mod 12, minus one if it's one or 11 mod 12, or zero otherwise. So that's how integers and Q series appear out of nothing, out of thin air. And hopefully this is nice present to Maxim because this is conceivage, the gear function and original trefoil, but now we can compute it for any node, basically. Thank you, sorry for running over time. I have a question. Yes, if you are interested in divergent Q series, of course you can suppose that Q is nearby one and you can look at resurgents in a bar, but why not look directly with Q resurgents? There is a perfect Q analog of resurgents with similar to a Q with Q, Q alien deletion and all sorts of things, Q stands for that. I would love to look that. Yeah, if anyone, so thank you for bringing it up. I definitely want to learn more from you. Like I say, I'm user and probably a very bad user of resurgents, so I'd love to explore more. And then the only thing I want to say is that what's happening here is that this thing is given to us just as a function of each bar. So, but if there is a way to produce, so that's the goal, to produce Q out of it. I'd love to learn how. At the end you have a Q series. Where you have a Q series, you can apply the Q2. Well, once I have a Q series, I'm done. I just go home at that point. I don't need anything. I mean, the whole goal is to turn each bar expansion into Q expansion, so that's at least application, but I would love to hear even if it's not directly relevant to solving this problem. Just one comment. What is called Q difference equations is not the same as in these Q difference equations, because in those Q is varying. It's not fixed. Q is into the H bar. Yes, it's similar to the. So it's not like Q. Where Q is 0.9. Q itself varies. It's a parameter, but it could be 0.9. So at least in the unit discovery. Yeah, exactly. So this is what I thought people call parametric resurgence. And I want to learn this too. So again, please come and teach me parametric resurgence, because this is very relevant. And maybe just some specific result. We're going to have this China Science defined up literature. The natural six just make a generating series of partition function of three manifolds on the level. Yeah, the six have partly many single points, which will be one of the algorithms and the infinite monogram. But actually, that's a cool thing. Some people looked at it and it does appear even as one of these versions that I mentioned in physics, but I don't know how it's related to this. So do you know by nature? It should be closely connected. The whole information to find out the many are singular values. I tried to compare details, for example, for concrete examples, and it wasn't entirely obvious. So I'd be glad to explore it. Because in some sense, yeah, anyway, I... Yeah, it's a comment that, in fact, you're taking resurgence at h bar equals 0 and you're trying to get out analytical information at h bar equals infinity. And I just would like to comment that when you do the resurgence, quantum resurgence of the Schrodinger equation, that can be done. That is, there is a definite way to... The resurgence information at h bar equals 0 is sufficient to gain very complete control at h bar equals infinity. Yeah, that's why I love resurgence. That's why when I discovered this, I think it's a magic tool from 21st century. I don't know why it works, but yeah, Marcus. Yeah, so here we're not really using resurgence. You're doing a partial resumption of your series. And actually, I have to comment that this can be done in many general contests of WPB of quantum curves. So for example, I think this is, in a sense, the analog of these refined EPS invariance for quantum mirror curves. So it's the analog of this. And actually, when we were doing calculations of quantum mirror curves, we actually had to do a similar resumption. So you get something like, of course, you get something like a Q function, functions of Q with coefficients in X. So this is probably a very general phenomenon for any WPB expansion of a quantum curve. But I think it has to be trigonometric. I think in a sense, it has to be C star cross C star. That's why I was trying to sell it like that. We're observing quantum mirror curves, which are exponential. But I guess it's more, I think this has been observed for all things. That's why we were writing this paper. I want to point out that example I gave was of course something that you can see with the naked eye, I mean this resumming to exponential of each bar. But for example, this is answer for figure eight for which I didn't have time. And if you go to other knots, then you really need resurgence. To resum a bar to the n over n factorial, you don't need resurgence. So that's clear. But in other cases, you really do. So in that sense, don't take me wrong that this is easy. So that's why I still want to learn from all of you resurgent technique. If you use the ideal tetrahedral decomposition to do you see the mediately? First of all, you can compute in many states, some models, this kind of asymptotics. But I don't know how to make it in the full TQFT. In a sense, do surgeries. So that's what I was trying to emphasize. In fact, physics makes predictions for which things should be TQFT and should not. So therefore, some of them I think have no hope. But for example, Anderson Kashyap is very special in that sense because it's lowest point in the Morse function. And that's actually also has nice properties. So I could be analogous statement that that one is special. But so anyway, I think that's the only TQFT which I know. So there is something definitely going on between degenerate versus cutting and gluing. As you know, there is a relation between complex terms, Simon's theory, and four-dimensional, maximally supersymmetric, twisted, super young Niels theory. So did you think the implications of these degenerate settles from that point of view or this general story from that point of view? We tried with Cyprin. And even at the early stages of our project, there was a paper trying to analyze those modular spaces, which claims something that even contradicts what we naively thought. So I'm happy to discuss it further. But I can't say much because these modular spaces are really complicated. So that's in a sense difference between working in a phase space and a Borel plane. In a phase space, you have to work with very singular spaces which are non-compact, and you have infinitely many homology cycles. Again, that's really what you're dealing with. On the Borel plane, you're just summing numbers. So it's more easier to deal with. And here, you have various stakey behavior, which again, I don't know how to compare it. That's a great question. I think it would be very important to develop it, but it's way above my pay grade. Let's see if I can say something.


And now, the Mole Mystery Theater, presented by M-O-L-L-E. Mole, the heavier brushless shaving cream for tender skins. Jeffrey Barnes, welcoming you to the Mole Mystery Theater, the program that presents the best in mystery and detective fiction. Tonight, as a return engagement, we bring you a hard-boiled crime story entitled, Angel Face, by Walter Wilson. And with us to play the title role is one of the loveliest stars of stage and screen, Miss June Havoc, who is captivated New York with her great performance in the Broadway hit play, Green Girl. So stand by now to meet Miss June Havoc as Dolores in Walter Wilson's story of deceit and death, Angel Face. Gosh, Mr. Barnes, sounds like there's murder in the air all right. But say, let's forget about crime for a minute and look at the brighter side of life. For instance, the way a man can cheer himself up in the morning with a clean, comfortable shave, a mole shave. Yes, sir, man, it's smooth, so smooth. It's slick, so slick. It's a smooth, smooth, slick, slick shave you get with M-O-L-L-E. Mole, the heavier brushless shaving cream for tough whiskers or a tender skin. That's right, shaving needn't be painful, not if you shave with mole. Because mole is a heavier cream, it not only softens your whiskers, it stands them up straight, ready for a smooth slick shave. So you shave faster, closer, easier, and you shave painlessly with mole. The heavier brushless shaving cream for tough whiskers or tender skins. Try it, mole. And now for tonight's mole mystery, Angel Face, starring Miss June Havoc. Our scene is a fashionable and high-priced nightclub in Midtown Manhattan. Angels are so few, but until the day that was... Sorry, I'm late. Oh, gosh, Tamara, she looks so lovely. Oh, gee, thank you. I'm getting so I can't think about anything but you. All day at the bank, I keep wishing it was time to get off so I could see you. Oh, gee, you're sweet, Joe. I think about you all of the time, too. I guess I thought there never could be anybody like you, like me, anyway.    You are, Tamara. You've got a face just like an angel. Pretty things. Sometimes I almost want to cry, you say, such a pretty thing. Hey, gosh, three weeks ago we didn't even know each other, but we were meant to meet, honey. We were meant to stay together, too, you and I, for always. And we can't let anything interfere and spoil things for us ever. Why, Joe. You think that, too, don't you, Dolores? Oh, darling, what's the matter? Well, of course nothing's ever going to spoil things for us. I couldn't live without you, Dolores. I just couldn't live without you. Joe, I love you. I love you as I've never loved anyone before. You're the first person who's ever been really kind to me. Darling. But I know something's the matter. Tell me. Maybe I couldn't help, but tell me, huh? It hurts knowing there's something wrong and me not knowing what it is. So I could help. Oh, Dolores. Is it something at the bank? Joe, it's not something at the bank. Honey, honey, please, please don't... Is it? Well, it... Yes. Joe. I was going to put it back, honey. I just needed some money quick. But before, Joe, what did you do? It was important. It was the first important thing that ever happened to me. I couldn't let it slip away. Joe. Joe, do you mean to say you did it for me? All that money you've been spending on me, taking me to expensive places like this... Don't you see? I was afraid I'd lose you. Oh, don't look. I got a little money saved up. Not much, of course, but some. You take it and put the money back. Put it back if it's enough. How much was it? You do that for me? Well, of course, honey. Oh, but it's too late for that. It was only four or five hundred, but the auditor's will catch up to me in a day or two and then it'll be all up. They came unexpectedly. No, it's no use. Well, they can't do this. Look, look, we'll run away. We'll go somewhere where they'll never find us. Oh, gee, they can't do this to us. It isn't as if we wanted so much. You and I just a chance to be together and be happy. Why does the world do this to people? Two little people who just want to love each other, that's all. I only did it because I loved you so much. Oh, Joe. Now I'm hurting you too. About me, I wouldn't care, but not you. You know, they can't do it. We won't let them. I never had anything to live for before you came along. I'm not giving you up. I don't care what we have to do. Oh, you don't know what you're letting yourself in for. I don't care. Wherever you go, I'll go with you. You won't be afraid? Not with you, I won't. I won't be afraid of anything. Oh, Joe, hold my hand all tight. Darling. It'll be all right. We'll make out, won't we? Joe, I said it'll be... It's a lot of us that's going to take money to get away. Well, you can have everything I've got. Hey, I... They're going to be after me anyway, Dolores. What do you mean? A lot of money would help. There's an old saying, you might as well hang for a sheep. Joe! Look, they don't suspect anything yet, but I'll have to work fast. If I could lay my hands on some real money, say 20,000. Oh, no, we got troubles in lunch. They're going to try to send me to jail for a few hundred anyway, Dolores, and with 20,000. Oh, won't this just make it worse? Darling, it can't be any worse. Will you stick with me, Dolores? Will you? Will you, darling? I told you I'd do anything. Whatever we have to do, I'm sticking with you, Joe. Locksmith. Yeah, you want some keys made, perhaps? No. No, I want you to do a job for me. Sort of a special job. Why not? I'll make it worth your while. See, here's the address right here in the neighborhood. Mr. Dolores Still, apartment 11E801 Bedford Street. I don't know about this. I'll have to explain it to you when we get there. Well, if it's a lock, I could fix it. I told you I'd make it worth your while. Well, look, I'll have to charge you for that special job like this, leaving my shop and... Will this do? Sure, Mr. Sure. For this I can fix. Oh, gee, I worried so every minute. At least I think it's all right. Nobody noticed a thing. They won't notice. They balance the books tomorrow night. By that time, we'll be a long way away. I won't think anything tomorrow morning when you don't show up for work. No, I told one of the boys I was going to the races. He's going to tell them I phoned in and said I was sick. That'll give me a good start on it. Sure it will. It'll work out all right, you'll see.  Twenty thousand dollars. Now, look, honey, I want you to have a duplicate key to the satchel in case anything happens to me. Oh, no, no. Nothing must happen to you. Joe, it can't. It just can't.  I don't even have just two keys. Oh, no, I don't want the key. If anything happened to you, I wouldn't want the money. What good would it do me? The only reason we got it is so that you and I can have a chance of living and belonging to each other. Oh, maybe we're doing wrong. I don't know. I guess people would say it's wrong. Only it's the only way we have a chance of being together. And that couldn't be wrong, could it? No, darling. You and me, that's the rightest thing in the world. We'll have to get going in a minute. Just let me kiss you. Darling, Joe, you're trembling. Yeah, yeah, he'll pass.  Here, no, you take a little brandy.  Thanks. Yes, I need it. You're not having it? Oh, Joe, you know I never could drink brandy. Well, it feels good. That's what I needed, all right. Sure, Joe. Yeah, that's what you needed. Oh, all of a sudden. What's the matter? Joe, the coast's clear. What's that? Sure, Al. Did he fall for it, Dolores? Of course he fell for it. The sucker. As the curtain falls on Act 1 of tonight's Mollay mystery, Joe lies on the floor unconscious, leaving Dolores and her friend Al free to carry out their secret plan. Oh, Dan, it looks as if Angel Face didn't quite live up to her name, did she? True, Mr. Barnes, as true as what I'm about to say right now, which is, men, if your morning shave feels like a blowtorch, chances are you have tough whiskers or a tender skin. So shave with Mollay. The heavier brushless shaving cream and get a smooth slick shave. Yes, Mollay is the cream that's ideal for wiry whiskers or a sensitive skin.  With Mollay, you shave faster, closer, easier, and you shave painlessly. Try it. See, if you don't say, it's smooth, so smooth. It's slick, so slick. It's a smooth, smooth, slick, slick shave you get with M-O-L-L-E. The heavier brushless shaving cream for tough whiskers or tender skins. Now back to Jeffrey Barnes Act 2 of Angel Face starring Miss June Havin. Bank clerk, Joe Angerton, has stolen money so he can run away with a supposed chorus girl, Dolores. But upon his arrival with the money at her apartment, she gives him knockout drops and then telephones an apparent gangster sidekick named Al. Like a life. Nice going, baby. Ah, it was essential. You sure can handle him. Well... Come on, kid, we got work to do. Don't I even rate a kiss? Sure, baby, I was only thinking maybe we'll... Come here. Is that all? Yeah, Dolores... You liked that, didn't you?  Or you couldn't get along without her now, could you, Al? Sure, you know I love your Dolores. Sure, Dolores knows her little boy loves her. She just wants to be sure, that's all. And Al wouldn't ever think of leaving his little Dolores now, would he? That wouldn't be a wise thing to do at all, would it? Take it easy. Dolores, what's eating you? You know me. Sure, I know you. We've been around a lot together, you and me. Isn't as if it's our first job. Yeah, sure, Dolores, and it won't be the last. Not if they're all, is it easy? Oh, boy, what a cinch this guy was. He fell for it even harder than that guy in Chicago. What was his name? Tom? Oh, yeah, Tom Stephen... Don't use names like that. Someday you'll say something in the wrong place. Oh, relax. Everything's hunky-dory. Yeah, I know, but... What are you worrying about Tom Stephen for?  Tom's been six foot under for nearly a year now, and even if somebody does dig up that trunk... Dolores, lay off. You know, Tom was kind of nice. He always used to call me angel face. Remember? Angel face. What's the matter with angel face? Oh, come on now, Al. Don't look so sour. We just made 20 grand, and I feel swell. Come to think of it, I feel kind of hungry. Hungry? Sure, hungry. What's the matter with that? Come on, I'll rustle up something in the kitchen. No, I couldn't eat a thing. Come on now, that's no way to talk. We got a job to do tonight. You gotta keep up your strength now. Well, look, let's get that guy out of the way before someone walks in. Nobody's gonna walk in, and him, he's out cold for a good four hours. Oh, okay. Okay, come on, let's dump him in the trunk. I'll give you a hand. Yeah, yeah, let's get this thing over with, huh? I wonder why they always waste so much more when they're out this way. I don't know. Yeah. All right, close it up. I suppose we get going now. What's your hurry? I told you I was hungry. Come on out in the kitchen.  All right, all right, I tell you, he's out for hours. We got plenty of time. Because when he comes, too, he's gonna be six feet under.  Will you come along? We'll look in the icebox.  Not the red light. Oh, will you take it easy? We're not trying to catch a train, all right? Oh, I don't like it. Maybe someone saw us. Maybe someone will stop us or something. Sure, maybe this, maybe that, maybe anything. Maybe we wouldn't have a thin dime between us if I listened to you squawking all the time. Well, go on ahead. You got the green. Okay, okay. I just feel better when we get rid of this load. I don't like driving around with anything hot like this.  We haven't got to do a hot load. Oh, yeah? We could face some murder rapping. You say it ain't hot. Not on this trip, Al. Not on the way out. That guy's alive still, ain't he? Of course, I admitted to take some fast talking to explain why we got him in a trunk. But if somebody was to stop us this very minute and found him there, it wouldn't be no murder rap. He's alive. Oh, just relax. We got no worries. Well, that's deep enough, isn't it? Yeah, okay. That'll all make the truck all right. Well, I'll drag it over. I'll tell you, I'll be glad to see the last of this. Gee, it's heavy. I'd shove it down. No. No, wait a minute. What for? I just want to see if I can hear him breathe. Hear anything? Not a thing. You're supposing... Oh, no. You couldn't hear a thing through the trunk anyway. Just out cold from that drink I gave him. Open it up if you want to see it. No, no. All right, then. What are we waiting for? Here, help me give it a shove. Okay, now shovel the dirt back in. Al, take your shovel and get to work. Al, what are you just standing there for? Comes two times air under the grout. There isn't any air, any... Al! Or what do you think the air is going on? It's all over before it comes to. Gimme that shovel! Pops covered over, you can hardly see any of that trunk now. There, you can finish it. Take the shovel, Al. Twenty thousand. I guess we can use it all right. I've done tootin' weekend. There's more where that came from, too. What do you mean? If the guy's dead, how can you... He wasn't the last sucker alive. There's plenty more where he came from. No, no, look, Talaris, don't you figure we ought to lay low for a while. We don't want to take no chances. We've got enough to last us a while now. Who's taking any chances? No one will ever pin anything on us. But I still think it wouldn't do no harm to lay kind of low for a while. Al, you're never going to change. What do you mean I'm never going to change?  Just like when I picked you up, a small-time punk.  I don't talk that way.  Small-time punk. Small-time racket for small-time dough. Oh, yeah. Well, at least I didn't have no murder wraps hanging over me till I met you. I can thank you for that. Well, of all the ungrateful...  I make a man out of you. I cut you in on some big-time stuff, and that's a gratitude I get. Just let me tell you one thing, pretty boy, and I want you to know it. The Lord, the Lord. I'm sorry, the Lord.  I say things that don't mean. Well, that's better. Yeah, it's just my nerves. See, I get to thinking about that guy in the trunk in the cops. He's six feet underground, Al, and why should you worry about him anyway? You know, Joe wasn't exactly a friend of yours, Al. After the way he made love to me, you'd hardly call him a friend of yours. I didn't like that part of it, either. Oh, the way he used to make love to me. The Lord said, say, you got a face like an angel. That's almost what Tom Stevens used to call me, angel face. He said he thought there never could be anyone like me. He had something there. He used to say he never could live without me. He said we'd love each other all our lives. The only part of our lives had mattered. And then he'd kiss me.  Didn't it, Al? Didn't that used to make you real jealous? I thought you didn't like that part of it, didn't I?  You'll have your little Dolores all to yourself. You'll like that, won't you? Yeah. We got 20 grand, and Dolores is going to buy her pretty little boy a nice present. We'll be home soon. And we'll be all alone, just Al and his Dolores. And Al is going to kiss this Dolores, isn't he?  Now, don't turn on the light for a minute. Just kiss me first. You can turn on the light now. Good evening, Dolores. I can see you weren't expecting me. This is Jeffrey Barnes again. In just a moment, we'll bring you Act 3 of Angel Face. Now a word from George Putnam. Every day, more and more people are discovering that to get real relief from the most common kind of dandruff, they must destroy the germ called Pityrosporum ovale, which many outstanding authorities say is its cause. You see, merely washing or brushing away loose dandruff has no effect whatsoever on this germ. But one thing that does work is double dandrine. Or double dandrine actually kills this germ on contact. Even in severe cases, results with double dandrine have been amazing. And the reason for double dandrine's astonishing effectiveness is a special ingredient, an active antiseptic, so remarkably efficient, many hospitals use it. In double dandrine, we call it Alzan. So stop trying to combat this dandruff with ineffective methods that actually are no better than plain water. They can't compare with double dandrine, for double dandrine destroys the cause. If you're not completely satisfied, you'll get your money back. Get double dandrine tomorrow. Dolores, it's me, Joe. Weren't expecting me, were you? You're supposed to be in a trunk six feet under. Take it easy, Al. Can't you see? He's got a gun. You're supposed to be dead. Why don't you two sit down?  No, there, right where you are. I think we have a few things to talk over. But calmly, that's why I have a gun so we can talk things over calmly. But you're supposed to be dead. You must have got out of the trunk where we were in the kitchen. Maybe you'd like a spot of brandy. You're probably chilly from your little expedition. Grave digging is so depressing, don't you think? No, don't worry. The brandy was not drugged. I thought of that this morning. How did you get out? That trunk was locked. Yes, I thought it would be. That's why I saw a locksmith this morning when you were out and had it fixed so it could be unlocked from the inside. You see, I made a few plans, too. Like putting those weights in that trunk to take my place.  Huh, you having a thing on us? We'll let the police be the judge of that. You're not going to call no cops, Joe. You can't afford to. You've got a rap waiting for you and you know it. A little matter of the bank's 20,000 bucks. All that? Well, you see, I don't work in a bank. What? Where'd you get that cash? It belonged to me. And you and I didn't meet by chance, Dolores. I'd waited a long time to meet you. I'd looked for you for a long, long time. Well, what do you want to meet me for? Because I want to see you and this punk of yours hang for the murder of Tom Stevens in Chicago. How'd you know about that?  Talk, it doesn't make any difference now. I've got all the evidence I need. It all fits together. Okay, so we did bump off Stevens. But it's still just your word against ours. No, it's your word. Your word's verbatim. I told you I did some planning, too. Well, a dictograph behind your chair has just this moment recorded your confession. What are you doing? There, that's better. Oh, yes, you wanted to know where I got that money, the 20,000 you thought I'd taken from the bank. I told you it was mine. It is, and quite logically, too, the paid claim of a life insurance policy of which I was the beneficiary. Tom Stevens of Chicago. Tom Stevens. Yes, the man you killed. Tom. That's right, isn't it? That's right, isn't it? Angel face. Angel face? That's what Tom... Yes, that's what Tom used to call you, isn't it? But how did you know? Oh, I guess I forgot to tell you. My name isn't Joe Anglin. It's Joe Stevens. You see, Tom was my brother. And now this is Jeffrey Barnes bringing down the final curtain on tonight's Mystery Theatre presentation of Angel Face, starring Miss June Havoc. And thank you, Miss Havoc, for a thrilling performance. Miss Havoc was supported tonight by Richard Widmark as Joe and Larry Haynes as Al. Next week, we present a story entitled Squeeze Play, written by 16-year-old Leonard Thompson and starring that popular favorite of radio and screen, Walter Abel. Squeeze Play is the story of a disillusioned and near-to-well lawyer who on Christmas Eve discovers a surprising and dangerous gift awaiting him in his office. He follows this up in an effort to redeem himself and his career. And in the midst of murder and courtroom dramatics, the author, remarkable young Leonard Thompson carries us to an extraordinary climax. So be sure to join us next week for Walter Abel in Squeeze Play. The original music for the Mystery Theatre is composed and conducted by Alexander Semler. Angel Face, originally entitled Murder on Her Mind, was written by Walter Wilson and adapted for radio by Don Agger. This is Dan Seymour saying good night and inviting you to be with us next Friday when the Mystery Theatre presents Squeeze Play, starring Walter Abel. This is NBC, the national broadcasting company.


And approaching field goal range already are the bobcats again to give the boys boys right up the middle once again The mustangs the ball out the ball is out Stevenson says they have it waiting for the official signal And they do mustangs ball And just as you said they've been stout against the run coach they Extend a really impressive streak. Yes now that that's 56 straight games in which they've had a turnover and Couldn't have come at a better time for them You know knock back on their heels a little bit by that return, but you know a nice job by their defense And they're getting some people of the football in the middle there And that's what that's what happens when you get people to the ball you generate some turnovers and You know, maybe a big momentum shift here for the Mustangs 56 straight games with at least one take away as coach pointed out the last time the Mustangs did not force a turnover was November 3rd 2012 against like homing and so six years Almost since the last time they didn't have a turnover


This is Chris Pike. My friends call me Big C. I'm back in action today, I'm back on Freedom Central and I got a new series for you guys. That's right, for those of you that are new to YouTube and video editing, or even if you haven't done that and you're thinking about becoming a video editor and a YouTuber and you're not sure where to start, this is the place to start. This is the series for you. What am I talking about? I am going to teach you for free using only free software, so no cost to you guys at all, free everything, how to become the best video editor you know. I'm going to teach you step by step how to do it. Again, using free software. Now that sounds like a big promise, but here's the thing. I am an expert in Adobe products. I'm an expert in Photoshop, Premiere Pro, and After Effects. I've got 2.8 million views on my channel and I've got all those Adobe certifications. Here's the thing, I'm going to take that expertise and that experience, I'm going to take it and apply it to this software, which is HitFilm Express for video editing and Audacity for audio recording. Again, both free pieces of software and I'm going to show you and hold your hand to becoming the best video editor you know in the steps you need to know them guys. This is going to be an amazing series. If you're just, if seriously, if you're thinking about becoming a YouTuber and you want to do it and you know you can do it, but you know you're a little bit afraid of the software or it looks kind of, you know, difficult or imposing, please just follow this series. I'm going to break it down step by step for you and it is not nearly as hard as you think it is. You're going to learn everything you need to know guys. This is going to be an incredible series. Stay tuned for your first episode. I'll be releasing it shortly and again, join me on this journey. We're going for it. Alright guys, thanks for watching this. Stay tuned for the next episode. I'll be back soon.


I've officially cracked the code. I've been sitting here for like an hour trying to figure out how I can get an offensive lineman at wide receiver.  I wanna beat someone online with my chubby boys at wide receiver catching passes. That's my goal and I finally figured it out. Now I remember Wheel of Mud last year a bunch of times I would log into a game and then all of a sudden like my right tackle was running a route that I thought my tight end was supposed to be running. Now that is because you can actually sub in left tackles and right tackles at tight end. Now I didn't do that in Wheel of Mud but for some reason if you don't have a backup tight end the game will automatically fill it with a right tackle and this is the even funnier part. If you don't have a tight end and you don't have a right tackle to fill the position it will fill the tight end position with your middle linebacker. I'm not kidding, I literally went out and got this ridiculous Ryan Shazir for this video. So every single person that we're gonna throw to in this video is gonna be an offensive lineman in at the tight end position but we will also have the option of putting Ryan Shazir on offense at the tight end slot. I don't ask me why it does this, I couldn't tell you. So my first starting tight end is gonna be David Bakhtiari. I want Joe Thomas out there running routes too so I'm gonna put in Lindstrom, perfect. And now I can put Joe Thomas as my backup. And if I need another tight end to throw two it's gonna be Ryan Shazir. Is this a realistic way to win the football game? Probably not. But if I play good defense you can win doing literally anything. So I'm gonna try it out. Not only that, but I'm gonna play weekend league. We're gonna be going up against some super sweaters with Joe Thomas and David Bakhtiari. Wish me luck. Also I'm excited to partner with Bessie again to show off their incredible weatherproof shoes. Top five worst things all time is getting wet socks in your shoe. And this right here is your solution to that problem. This is one of my all time favorites from Bessie, the men's everyday pearl whites. Obviously they look really good. But they're also waterproof, slush proof, sand proof, slick proof. So they keep my feet completely dry no matter the condition. They're also very flexible, super lightweight and very, very comfortable. And I wanna show off this waterproof stuff right now, okay? New camera in three, two, one. Hello. Here's my totally dry socks. Here's my pearl whites. Ready? Look at how easy this is. Boop. Boop. Crispy. And here's the coolest part. Just chilling underwater. Totally dry. Dry, look at that. Click the link in the top of the description to check out Bessie and use my code MMG for $25 off. Bessie, thanks for sponsoring guys. Enjoy the rest of the video. Justin Simmons, Todd Gurley, Tyrone Matthews. That's not necessarily a God squad. I like to see that in weekend league. All right, starting out on defense. This is gonna be standard. So, you know, we just gotta lock up. Oh, start out with a run. Oh, that's not what I meant, bro. I thought it was a play action. Got me pretty good. Probably another run. I'm gonna QB Spy Durand James. Yup, yup, yup. Oh, we're all over that. Let's go. Second and seven. He's got Mario to flashback at QB. I like it. Let's go on this 88 speed shaz. Oh, and underneath, you're so late. Inside zone. Inside zone. Get me onto my ladders. He's running inside zone. He's not, he's not.   If you're underneath, he's gonna juke and then I rock him. No! I'm hitting really hard. I'm just not playing very good defense. This is a run. I'm so ready for this. If it's not a run, it's a touchdown. Okay. Okay, we're good. We're good. Beyond Sanders getting in there. Come on. Throw it. Where you gonna throw? No way! What are you doing?  I'm gonna win this game. I'm gonna win this game and David Bakhtiari's gonna have four touchdowns. All right, so hear me out, okay? If we go with this formation here, we can sub Ryan Shazir there. So now I got 88 speed Ryan Shazir, middle linebacker Bakhtiari and Joe Thomas all on the field. Do I get the snap off? Yes, I do. And away, eight, eight. Got him. You know what, he's alignment. We're gonna give him the benefit of the doubt. He just wasn't looking. I wish I could tell you, I knew what he just did there, but I don't. I have no idea what he's trying to do. Joe Thomas is open! Oh! Yes sir, yes sir. I'm a sense Shaz deep. I'm a block, Joe Thomas. I'll go bow on a wheel route and I'll get Julio out of there. Let's see what this looks like. Bakhtiari underneath, big, whoa, whoa! I just ran an unintentional hook and ladder. Did I just find a game breaking glitch? Second and three Bakhtiari. Oh, that was like a bad pass. All right, we're gonna go to Shaz on this one. Ah! I keep hacking, but it's not working.  Look at this. It puts Deon Sanders at fullback. Deon Sanders, why? I don't know. I wish I knew. This whole video is just devolved into absolute anarchy. Deon Sanders is the read option on Peek's Own Bubble. We're gonna hand it off to him. Let's see how this goes. Deon, Deon, Deon! Just got a seven yard rush for prime time.  There's no way you have. All right, I'm running play action, double pose. I'm literally just gonna run straight over to Bakhtiari and hopefully I can hit him on this corner route. First and 10. Oh, it is play action, okay? He knows I'm throwing it anyway. We should have, oh! Okay, Bakhtiari, he broke it up. Let's see who's in on verticals. Bakhtiari, the man, and the man alone is in on this. Does he have it? Does he have the space? Shit, he keeps running main coverage. I don't think I can convert. We just need a really big play here. Fourth and 10. Thomas got him again! The level of frustration I would be at if I was this man right now is indescribable. Let's try Shaz underneath, block the rest of them, clear out Julio Jones. We're gonna have to hit Ryan Shazir here. Release! Is it there? Dude, this is so frustrating. I just want this to work.  He's usuring my Bakhtiari.  This is probably main coverage. I'm sad to say it boys, but if there's somebody who's open, I'm gonna throw it. You what?  I'm sorry. I'm sorry, I have to. I feel bad doing that. I wanted to only go to the boys, but I gotta play for the win, you know? I'm gonna block Tony G. Let's see if Bakhtiari's there if his user goes to it. His user's not there? Oh, Tyran Matthew computer comes down, makes an amazing plan, is gonna get a pick six. He didn't even do that. It's his computer free safety. It ain't over till it's over, baby. I ain't giving up. I hope that comes back to bite him. I hope that that decision right there gets this man. I love this play against main coverage. I don't know if Bakhtiari's gonna be fast enough for this play, but we gotta try it, right? Ooh, ooh. I don't even know who I just threw to. Oh, that's a pick.  I was just smart. That was not my smartest play of all time. First in 10, he just throws a dinky underneath one. God, he's breaking every tackle. This is annoying. I don't think he's running this. Ooh, he is. Let's go. Fourth to one. I got it. Oh, I really do have it. I didn't think I got that.  Let's see if Bakhtiari gets helpin' in. Uh, I'm sorry, boys. This is dramatic. I'm sorry. I have to throw that pass. It's a touchdown. I have to throw it. I really can win this game. I will get redemption for Bakhtiari. Game is winnable. Dude, we just scored 17 seconds on the clock. All right, seven to 13. Okay, man, I'm startin' with the ball. We really can pull this off. All right, let's see if Bakhtiari gets open underneath. I have not gotten to throw much to him. Makes me sad. He dropped. Ooh, let's give him an out-route. I think an out-route will be open for him. Yup, yup, yup. Think if I throw that to Joe Thomas early enough, it's good. So there's the in-route. Yup, early, early. Ah! They do not make it easy out here, boys. I'm telling you what, they do not make it easy out here. I'm trying, boys. I'm trying. Fourth and 10. So Bakhtiari's in this weird press coverage. I actually think that Bakhtiari is gonna be my route here. I'm serious. Nope, he got press covered way too hard. I'm sorry. I gotta go to Odell for my fourth down conversion. I feel bad about it though. All right, I got, I got, ooh Bakhtiari. A little lay down three yards, don't mind if I do. I gotta hit him on the out-route again. The out-route was too crispy not to throw twice. Look at that. Look at that separation! Ah! God, he's just stuck. This guy's getting embarrassed. All right, Bakhtiari underneath. Oh, his first drop in a while, man. There's a very finite amount of things that I'm capable of. Ooh! Look at the separation on that slate route. Dude, he had the separation. I should have thrown it earlier. Let me see if I can get that again. Bakhtiari is gonna convert this. I'm telling you. Let's see what his user goes on. There it is. Nope. Oh no, I sent Leo and it took his man too close. Do you guys smell a run? I would love a run right here. Oh, play action. I got the corner out. I got him both. Where are you gonna throw? Where are you gonna throw? I'm really gonna do this. That was a nasty return bar prime time too. And this man, disrespecting me, went for two. And now he's gonna lose the game because of it. Shenan, done it, bro. You really about to lose to David Bakhtiari and Joe Thomas as my primary wide receivers. He did not rage quit. You did not just rage quit against me. Maybe he truly was experiencing network issues. All right, the man is in five wide. Let's see if this is a vertical or it is a vertical. There's nothing he can throw here. Not an open at all. Oh, Jamal Adams was right there for his trash pass. So close to icing the game right there. Ooh, I don't have Aaron down in the right spot either. Read option, nice play. Nice play call. And he does not go down, that's sketchy. Yep, there it is. Read option again.  Give me that fumble. I just clocked your QB.  Are we gonna play back off? He's gonna run or read it. That has to be a joke, right? I mean, what I just watched is a joke, right? I like a little bit of Wildcat. Let's see if I can get it on the jump. Dude, that's there. Ah, he keeps getting in. I guess I'll send him straight up the middle and I'll just throw an L. Ooh, barely got off of that. Can I throw this? Oh, no I do not. Oh, darn it. I'm gonna actually lose. Oh, no, no, no, it's not over yet. It's not over yet. I can still get this ball back. All right, we're run committing. He's gonna pitch right side. This is an oppressing one. We were so closely securing the victory. It just didn't work. It just straight up didn't work. I tried my hardest. Three receptions, 21 yards. Two receptions, 34 yards. Joe Thomas had two drop passes. David Bacchiori had seven drop passes. Hey, you can't say I didn't try, boys. You can't say that. All right, I love you guys. I'll see you in the next video. See you.


This systematic review evaluates the potential utility of lab-on-a-chip, LOC, technology for antimicrobial susceptibility testing, AST, of Staphylococcus aureus, highlighting its ability to provide rapid access to minimum inhibitory concentration data and aid in selecting appropriate antibiotics and dosages for treating challenging infections caused by this microorganism. This article was authored by Carlos M. Ardila, Mateo Zuluaga Gomez, and Annie Marcella, V-Bars Biles.


Deep in the vast and unforgiving territory of Wind River, a murder has taken place. One that can only be solved by an expert hunter and an FBI agent. Loosely inspired by true events, Wind River will keep you on the hook and not out in the cold, even though it is frigid up in these mountains. Let's talk about Wind River for a couple minutes on Adam Does Movies. The reason I'm talking about this movie today is because Patreon supporter Isaac over on patreon.com slash adamdoesmovies at the mithril level demanded I talk about it. Unlike most of the movie requests I get, I had actually seen Wind River. Granted, it came out in 2017 and I didn't remember a lot of it. So I rewatched it with my wife, thinking this was okay. I remember it being all right. It's actually better than all right. It's a solid flick. I wouldn't go out and recommend it willy-nilly to people, but certain individuals will really dig it. Now I will say, it is a depressing film all around since it is inspired by true events that take place on this Indian reservation. A lot of stuff happens to these people and it's not good, specifically younger women. They go missing often. It's the only region in America where they don't keep stats on what happens to these young women, mainly because they don't have the resources. Writer and director Taylor Sheridan does a good job presenting this to the audience, showing them just how dire the situation is for the people up there. They often turn to drug use, selling, and other sketchy things you wouldn't necessarily put on the job resume. Our two leads here are a couple of Avengers, Hawkeye and Scarlet Witch. You have every man hero, Jeremy Renner playing a no-nonsense hunter. He takes down wildlife. That's what he does for a living. He's also been dealing with the divorce and a major loss in his life that we'll get revealed later on in the film. Joining him is Elizabeth Olson as Jane Banner. She flew in from Vegas. She's the only FBI agent that was even remotely close and clearly the FBI doesn't prioritize killing that happens on an Indian reserve. This is a who-done-it murder mystery. I'm not gonna give away the game because that's half the fun in watching these types of movies is how it plays out who's behind the dirty deeds done dirt cheap. I will just give you the basic rundown. The movie shot very well. It's under two hours, which I love, which I think is absolutely great. And even with the hour 45 or whatever, it feels a little padded. There's a couple scenes, mainly around the snowmobile porn, as I'm calling it, where I don't need to see Jeremy Renner driving around, turning once in a while. It's just, it's like, we don't have enough action in this movie. Let's shoot some snowmobile driving around for a while. That'll get him hooked. I didn't need it. The movie has a decent pace, even though it is a slower moving affair. You have likable characters. They have great chemistry together. You have some nice supporting actors coming and going.  That's the big takeaway. This woman's raped and left for dead out in the cold. It's sad.  Joining the tight script and solid visuals is a nice compelling score. It's somber as hell, like the movie, but it keeps you engaged. I'm a sucker for a good investigatory film. This one nails it. And what makes it a little bit different than its competition is the Indian aspect of it. We get to see some of the culture there. We get to see the impacts of hundreds of years of poor decision-making and how it's played out for these people and their culture. I don't consider this an amazing movie by any means, but it is a really solid, easy one-time watch. I've seen it twice. It was actually pretty pleasant the second time as well. There was a few years in between which helps, but good cinematography, good characters, good story. I mean, you can't ask for much more, right? All right, that's my thoughts on Wind River. Isaac, thank you for the recommendation. It was good to watch this one again, honestly. Now, if you're not a patron supporter at patreon.com slash adam does movies, or right here can absolutely become one. It helps me and the channel out immensely. You can just hit that join button on YouTube and pick your tier. If you wanna recommend a movie, become a Mithril member. You can do it for one month even. It's month to month if you want, or you can become a member for a full year and get up to four movie review requests. I would really appreciate it. It helps a ton. Thanks again, Isaac and all supporters. Please like the video if you got something out of it. Subscribe to the channel if you're new here and somehow stumbled upon a Wind River review in 2023. And hopefully I'll see you next time. Stay warm.