Arithmetic Resurgence of Quantum Invariants
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Garoufalidis, Stavros
I will explain some conjectures concerning arithmetic resurgence of quantum knot and 3-manifold invariants formulated in an earlier work of mine in 2008, as well as numerical tests of those conjectures and their relations to quantum modular forms, state integrals and their q-series. Joint work (in parts) with R. Kashaev and D. Zagier.
 Thank you very much for giving me the opportunity to talk to this wonderful conference and meet new people and reconnect with people I knew from all times. So I want to tell you about arithmetic resurgence of quantum invariance and by that I mean quantum invariance in dimension 3 and such as not invariance or three-manifold invariance. I will actually not talk about three-manifolds so much. So as I understand resurgence is a property of formal power series and their analytic continuation and their expansion of those functions near singularities where the original power series emerges or reserges. So resurgence can be formulated, if you can formulate it for formal power series you can also formulate it for its Taylor coefficients so it can be formulated for sequences themselves. So you can talk about the resurgence of the sequence and factorial, you know, for instance, which is factorial divergent but you can do things to it. So as I understand resurgence is abandoned for functions which are defined using different, differential, integral equations, linear or not. And I also understand that there are at least two aspects of the theory, one of which is proofs and the other of which is numerical experimentation, symbolic or truly numerical and I will talk about that. That part of the theory is very fun, it's suited for mathematical physics but it's only one aspect, right? So I came across resurgence in 2005 when I studied a particular question and I had an agenda in mind, namely the volume conjecture. So let me give you some introduction about it. So we're talking about a knot in three sphere and to a knot there is a sequence of low-run polynomials with integer coefficients. These are called the M-colored Jones polynomial. So for instance, N equals 2 is the original Jones. So these are supposed to be polynomials defined using the representation theory of SL2 together with the n-dimensional representation and greatly generalized and so on, so forth. So one knot has a sequence of polynomials. If you wish, you can really think that the Nth-colored Jones polynomial is basically the Jones polynomial of a parallel of a knot, the N minus 1 parallel. That's correct up to some sub-leading kind of orders which are universal and depend on the knot. But there is a little theorem in this game, joint work with Tang Le, which says that the colored Jones polynomial, namely the sequence of polynomials, is q-holonomic. So let me explain what this means. It means it satisfies a linear q-difference equation of the form CD of q, q to the n, jn plus d, nc0 q to the n, jn of q. Here I drop the dependence on k. So this is true for all n, where the cj's are Laurent polynomials in two variables. So in other words, it's a linear q-difference equations whose coefficients are polynomials in q and q to the n. If the polynomial is q to the n, also polynomials in q. Not true. What polynomial of q is q to the 100? It's a polynomial of degree 100. All right, let's make n 200. If n is a variable, q to the n is another symbol. It's not a polynomial of q. Because the degree of the polynomial grows empirically or? The degree of the polynomial here. Yes, it's also a theorem that the degree of the color Jones polynomial is a quadratic quasi polynomial. And that is actually true for all solutions of linear q-difference equations. The only proof of that theorem, actually, of myself, I was inclined to talk about that, uses Le Mahler Scholar theorem, N-Piatic analytic number theory. So yeah, the degree grows quadratically. OK, so this is a theorem and true for all nodes. So how unique are the cj's? Sorry? How unique are the cj's? First of all, cd should not be 0. And there is a unique way among all q-difference equations. There is a unique way to make a minimal order q-difference equation. And that makes the so-called A-hat polynomial, which annihilates the color Jones. And it is a not polynomial. And it's an invariant itself. OK, so let us suppose that a linear q-difference equation, that the linear q-difference equation has a basis of solutions, which are asymptotic to, so when you put q equals e to the alpha over n, they are asymptotic to e to the whatever, s minus 1 times n, ns0, ns1. ns1 over n, and so on. OK, let's suppose that a linear q-difference equation has such a basis of solution. In fact, a resurgent with respect to 1 over n. But I don't even care about the resurgence. Suppose that is true, then a version of the volume conjecture holds. And the volume conjecture just says that the nth color Jones polynomial of the knot evaluated at the nth root of unity grows exponentially, and the rate is the volume of the knot times 2 pi i. That is the volume conjecture. So the volume conjecture would follow from the previous statement, and it would make a nice theorem in a fine journal. But it is not a theorem. So if you really want to look at linear q-difference equation, so you can think of them as Schroediger except that there are difference, and really except that they are high order. Even the order of the equation for the figure eight knot, the order is 3. So already you're talking about higher order, but who cares because you can make them vector values. So OK, so that's just one idea. I mean even without mentioning resurgence, just that idea alone doesn't exist. Or if it does, please let me know, and we can finish that sort of paper. So let me now give an example. So among all knots, let's start with a trefoil, which sometimes I can and sometimes I cannot draw trefoil. So there is a formula. I didn't tell you how the color Jones polynomial is defined, but it is defined in a complicated way. So the formula is the following, minus n sum from 0 to n minus 1, q to the minus nm, q to the minus 1 minus capital N, semicolon qn, I guess just to avoid confusion, this is what I mean here. That is the formula for the call Jones polynomial. This formula is terminating, so we can replace n minus n by infinity, and then decide to set capital N equal to 0. So since you use a different use, you should define the Thank you. Yeah, thank you. So let it's, now let's set n equal to 0 and call the formula. Now it's a formula, right? It's a formula, but it isn't a formula. It is still a knot invariant. The fact that it's a knot invariant is a theorem of actually Habiro, and this formula that looks like black magic because I can rewrite the color Jones polynomial in many different ways. And if I just set n equal to 0, in other words, in the zero-dimensional representation, no, actually it's not the zero-dimensional representation. It's the infinite-dimensional representation of SL2, or formal-dimensional zero. So if I do all that, then I'm going to get a knot invariant, okay? And this so-called formula has two properties. First of all, it can be evaluated at n through 2 of unity because only finally many terms, right? And it can also be evaluated when q is e to the h bar. Oh, and why can it be evaluated when q is e to the h bar? Because e to the h bar, e to the h bar n is n factorial h to the n plus o of h to the n plus 1. So in fact, it's not so hard to see that if you evaluate, so let's make now a definition, and the definition is I use capital F is for perturbative, trivial. And of course, this is a formal power series with coefficients in h bar, rational coefficients because of e to the h bar. And Geveret 1, by the way, this series is not just a series. It is the series that we were calling concevage Zagir series. And this expression. So now let me give you a theorem. So it is a very one. So what is the theorem? The theorem is, of course, that if you look in the Borel transform, so let us look at the Borel transform. This is Greek letter xi. When I write it, it looks like 7. But what can you do, right? So this is the replacement of h to the n over n factorial by xi to the n over n minus. Topologically it is a 7. And I'm Greek on top of everything. OK. So this is the Borel transform, right? Borel transform convergent for xi near 0. So the theorem that we have with the video question, which was the entry, this was my unofficial entry. It was 15 years ago, and it still is my agenda. Of course, to prove the volume conjecture, if you can prove resurgence itself, you prove much more. You prove generalized volume conjecture and other things. So you can aim for everything. I mean, that was my agenda. So we settled for something less. And here is a theorem. The theorem is that here's a formula. OK. Some thank you, Sergei. Epsilon n are the same numbers n square pi square over 6 to the 5 half, where epsilon n are plus 1 if n is congruent to 1 or 7, or 1 or 11, 5 or 7, n 0 otherwise. OK. So this is a formula. So in particular, it says that in Borel plane, the nearest singularity, all the singularities are in the real axis. The first one is in pi square over 6. The next one is in 25 pi square over 6 and keeps going. Right? They are not at all equally spaced. So this is part A. And let me also give you part B. So therefore, B, because the singularities are on the real axis, what we cannot do is Borel summation on the real axis. What we can do is Borel summation right before or after. Sorry, Laplace right before and right after. So we have a formula for that. That's a function of x now. And the formula is it is i root 2 pi x to the 3 half times sum epsilon n times n. Times q to the n square over 24 when n goes from 0 to infinity where q is e to the 2 pi ix. This is just the end of Sergei's talk except I lied here. I should have done one more thing. This is the Borel transform of f 3 1 of e to the h after I multiply by e to the minus h over 24. I really need to multiply by that. That doesn't change anything. I mean, but it is much better for modular properties. So everything is explicit. And my challenge to you is to actually give a different proof of this theorem. So the proof of the theorem basically uses work of Don Zagir. It follows from Zagir's work on the so-called strange identity. But this is a very explicit series with a very explicit formulas for everything. So if you, in fact, if you know the answer, please try to reproduce it. It's just a baby case. Sorry, Maxime, to call that baby. But OK, so now let us grow up. If we may. All right. After that comes the figure eight knot. For which case the Collard-Jones polynomial, I'm even going to give you the formula. I hope you appreciate the fact that these are formulas for the Collard-Jones and not just for what's called the Kashyaf invariant. N minus 1 cubed N, q to the minus Nn. So therefore, the function that I was talking before of q is deceptively similar. It is q, q, N, q inverse, q inverse N from 0 to infinity. OK, and we can talk about, this is also Gervais-Guant. By the way, we have a little theorem with Tang Le, which says that this is. It does look like a polynomial. This one? Should be 1 minus 1, no? Oh, yeah, yeah, sorry, N minus 1 with a q inverse. OK. Yeah, yeah, yeah, yeah, yeah, yeah, to terminate it. Thank you very much. And that's why I put q inverse here. A little theorem with Tang Le is that for all nodes, this is Gervais-Guant. OK. So you might say, well, but how different is this expression from this expression? You know, surely if you prove resurgence of one, you can prove resurgence of the other. That is my challenge to you. So in fact, let me give you here a conjecture. So here is a conjecture, stated for all nodes, right? The conjecture is that, so this thing I'm going to call it, capital Phi trivial of H bar for a node k. And the conjecture is that Phi trivial of a node k is resurgent in Borrel plane. I mean, after you do Borrel transform, with singularities at i times volume plus Chern-Simons plus 4 pi squared times integers plus a little shift. This rk is a rational number. And here rho is a parabolic SL2C representation. So that is a homomorphism of the familiar group into SL2C, which is parabolic. That means the trace of the meridian and the longitude is plus minus 1. Among all parabolic representations, there is one that everybody knows, namely the trivial representation. It is parabolic, I mean, and has zero volume and zero Chern-Simons. So okay, this is a conjecture. So rk doesn't depend on the representation? No, it doesn't. Yep, it's an overall shift. Okay. All right. So now, three, four, five. Okay. So let me give you what happens in the case of the figure eight. Let me draw a picture for you. Parabolic means around the knot is very important? Yes, yes. Around the knot complement. I mean, the meridian and the longitude, yeah. So for the figure eight knot, we expect that, first of all, the volume of the figure eight is about 2.02. So you just, do you understand this, if you draw in Chern-Simons, it's a real large part of complex? Yes, yes. I could just call that complex Chern-Simons. That's right. That's right. So the complex, well, there are two parabolic representations, the geometric one, which gives the hyperbolic structure and the complex conjugate. So the geometric one has volume 2.02, the Chern-Simons is zero. So therefore, if you plot the numbers, you find i times v and minus i times v, where this distance is about two units. And then there is the 4 pi square here, which is 39.7. And I didn't even draw it to scale. So therefore, the singularities in Borrel plane are supposed to line up in two horizontal lines and the singularities are sort of 4 pi square apart. Now you may say, and how do you know that they are all singularities and not, for instance, some of them are not? So our conjecture actually, for the figure 8, says that they are all going to be true singularities. I'll give you evidence for that. So that's one conjecture. Now let me give you another conjecture. So is the shift zero in this case, or you just didn't write it? Zero. Well, if I put it here, it is zero. I really put them on the y-axis. OK, the renormal on shift. OK, so let's make a definition. Little phi trivial of a node k is going to be the evaluation at the kth root of unity. This is, in other words, known as the Kashyp invariant of the node. And it is a sequence of numbers. Yes, yes. So it is a sequence of numbers. And the conjecture is that this sequence of numbers is a resurgent function in Borel plane with all singularities. OK, that's a conjecture. OK, so for the figure eight, I gave you two series. I gave you this sequence, and I gave you the other one, this one. So now let me give you a 2 by 3 matrix here. So first of all, just a few terms. So what is easy to compute here? Well, given that we have a formula and we substitute q equals e to the h, and by terminating sort of stuff, it is not so hard to find out that you get some terms here. But it is easy to compute these coefficients, which are factorial diversion. And I should also say that it's another theorem of Tang and myself that this also, this series, is a, no, let's forget. OK, now what about this? This is e to the minus, no, plus volume divided by 2 pi divided by h bar, 1 over third, fourth root of 3, you know, and then 1 plus, there is a 2 pi i, there is an 11, there is a root 3, some trivial constant here, and the higher order terms. The reason I put this is to show a contrast. I meant geometric here, thank you. So the coefficients of this series are, apart from the power of 2 pi i, which is homogeneous, they are supposed to be in the trace field of the knot, which for the case of the figure 8, it is q squared root of minus 3. But they are supposed to be resurgent. So in other words, if we actually write down here, e to the volume over 2 pi times 1 over h bar, and call these coefficients here an, so these are easy to compute. These are much harder to compute. The only way to compute them is to compute the Kashyap invariant, and then compute it to high precision, and peel off the 1 over n, the 1 over n squared term and so on, and do the computation numerically, accelerated using research on transform, and finally, recognize it as an element of that trace field as an algebraic number, and we've done that with Donza gear, and we found out 100 terms in this series here, just as a benchmark here. For this series, it's very easy to find many terms. So we have 100 terms here, and there are other methods to compute the Kashyap invariant, and many terms here. I should say also a little theorem. A little theorem here actually is joined with Donza gear, is that the Kashyap invariant of K is computable in linear time using the Q difference equation. So it's a linear time computation, but its asymptotics is not a linear time computation. So the question is, what are the asymptotics of, so let's put this thing geometric here, and let's call these things trivial. What are the asymptotics of the nth coefficient here, and the nth coefficient here? And if there is any justice in the world, the asymptotics of these coefficients are given in terms of themselves. All right, so let us define here. Okay, so let G of 1 to n be 1 over 2 pi i, sum from 0 to infinity of a k, until n minus k minus 1 factorial divided by 2 i v to the n minus k. And the G 2 1 is the same, except I replace here by minus 2 i v. Okay, same. And the G 1 n, which is root 2 over pi, sum of a k geometric, n minus k plus 3 halves. So factorial divided by sort of i v to the n minus k plus 3 halves with a k. Yeah, that's it. That's it. Can you explain the notation? Because I'm T first, yeah? And T what is in the denominator? Ah, yes, I didn't say that. Fee hat, the anti-geometric, so I said that there are three representations. I mean, there are three parabolic connections, the geometric, the anti-geometric. So the anti-geometric is the one that you replace h by minus h. So we have three coefficients, the trivial, the geometric, and the anti-one, which is just minus one times the geometric. Okay. And now I want to, well, tell you a theorem, except I cannot tell you a theorem if it is not a theorem, right? So what is it? It is an empirical observation. And the empirical observation says that this is asymptotic to 3, g1 to n. The anti is asymptotic to minus 3, g2 to 1n. And that can be compressed into an eschew symmetric 2 by 2 matrix. And finally, the trivial one is asymptotic to g1n and g2n. So therefore, you get a matrix of this kind. Okay. So in particular, to highest order, it says, for instance, that the nth coefficient here, it's going to grow like 1 over 2 pi i n minus 1 factorial, 1 over 2 i v to the n times a 0, which is 1 over 4th root of 3. Okay. So, by the way, how do we do this? So first of all, we use everything we have, namely 100 terms here. So the 100 terms is about e to the 365, just to give you an idea of the order of growth. Then you take all the terms that you have here, you do the optimal truncation. So e to the 365 divided by this are two numbers of 365 digits. You take the ratio, you subtract 1, and you find a number which is e to the minus 115. So order dropped from 365 to minus 115. With these numbers here, okay? So they're pretty convincing. Sorry, but what is this? Have g1n and g2n? Yes. I didn't do g2n. So the g2 would be the other one, 2 here, and the minus 1 here. Mm-hmm. Okay. So you may say that in Borrel plane, the asymptotic expansion of these Aeons are basically expressed in terms of the series Aeons itself. Okay. Okay. So that's just the... Barista, why can't you transform this into a theorem by using the state-interval representation? It's a one-dimensional integral. I'm just doing my first part of my talk. Right? I am still into a formal power series. I haven't done my analytic functions yet. So just a third example, but very quickly, which is the 5, 2, 0. So for the 5, 2, 0, let me give you the answer now not of the Kashyap because we... I mean just of the Kashyap invariant, sorry. The answer looks like this. Okay. And although there is a denominator here because k is less than or equal to m, it always cancels with a little bit of the numerator and you have left both to expand when q is exponential of h bar and when q is the root of unity. So in this case, the trace field is actually cubic, the only cubic field with discriminant plus, I mean minus 23, the only one. And in this case, if we did the trivial, just that would be easy. But if you did the geometric, you find 1 over 2i root 3alpha minus 2 and then 1 minus 33 alpha square and 242 alpha minus 245 over 4 times 23 square. There is a 2pi i times an h bar and higher order terms. So this field has one complex, I mean one pair of complex embeddings and one real embedding and I'm choosing here the complex embedding. The only reason I'm putting this is because Jan was asking what are the coefficients we're talking about and the coefficients are in the trace field. And conjecturally, every number field with at least one complex embedding is the trace field of some casped hyperbolic manifold. OK, so we computed 100 coefficients here, but we computed them using actually a kind of num, I mean sums and I wanted to tell you the picture here, the analogous picture. So there are three parabolic, I mean other than the trivial representation because this is a cubic field, it has three embeddings in the complex numbers and the geometric one is always parabolic, therefore, so are the other two Galois conjugates. And therefore, we get here three numbers v1, v3, v2. And as usual, there translates by 4pi square. OK, and the numbers can be given explicitly, one of them is manifestly real. This is actually an SL2R representation because we're talking about the real embedding here and the other two are 1.379 plus or minus 2.828, that is the volume of the 520. And you can actually check that this is nearly equilateral, namely the distance between v1 and v2 is 5.011 and the distance between these two is 5.656. The only reason I'm writing these numbers down is because they are given to us by the gods, they are not hyperbolic geometry, normalizes the volume, normalizes geodesics, normalizes everything, it is not up to us to mess with these numbers. And in this case, we get such a thing, so therefore, we're going to have a 3 by 4 matrix. And I will just tell you what is the analogous, I'm just going to cut the conversation down and we computed the analogous matrix with don. I should mention all these are joined with don and the matrix now is 033 minus 34, that's a 3 by 3 bit and the other one is 11 minus 1. Okay, this is sort of the matrices that I believe Sergei wanted. Oh, and there is an additional shift, I should say for the 5.2, there is a nasty kind of shift which is a pi square over 6, that one has to put in there. Okay, that was my, the old part, which has to do about resurgence of sequences or factorially divergent series, which is mainly conjecture how to convert them into Borrel plane and talk about analytic continuation. Why do you do that? Because after you convert to Borrel plane, you take Laplace transform and then you're in the plane you want to be, in the plane where you have your analytic function, your solution to the differential or differential equation, but we don't have a differential differential equation. But we do have a function. So my second part is about those functions. So part two, analytic functions of a variable tau. So tau is going to be mostly a variable in the cut plane. So it turns out that there are theories, perhaps not full TQFTs, and they only work for Casp manifolds. These theories are called quantum Teichmiller theories or quantum hyperbolic geometry, and they always work for punctured Riemann surfaces, and therefore Casp three manifolds, and they may be the wrong theories and so on, but they do give invariance of knots in particular. These theories were developed by Kashaev, Andersen, and several sort of other collaborators. And I want to focus that the building block for this state sums, oh, I should say that for a general note you might wonder how horrible does the formula look like, and the answer is not too bad. You're always doing a finite dimensional sum of ratios of quantum factorials. That's all there is. And likewise in these new theory, you're going to do not a finite dimensional sum, but a finite dimensional integral, and the building block is not going to be the quantum and factorial, but it's going to be a special function for Davis quantum dialogarism. So let me give you an example. What is the answer for the figure eight knot? So for every knot, oh, I should say here first of all, theorem for every knot, there is a function which is analytic. Well, OK, you may say, oh, well, you can take the constant function, I guess. But I mean, there is a function which is analytic, which is computed via ideal triangulations and state integrals. Did I stop? You still have one block for left. No, I know. But my numbering got almost up. Six is here. So seven, eight, nine. I learned this method from a young postdoc at MPI. Yeah, but you have to do it in advance. Otherwise, you get carried away. OK, so what is the invariant for the figure eight? It is the integral of phi b x square e to the minus pi ix square dx, where phi b is a very special function. I can write down the formula, but the formula will not really talk to you unless you have played with it. Bz, hyperbolic sine, b inverse z, dz over z. OK, so this is for devs. It's a double quasi-periodic, and it has all kinds of properties. So this is the invariant. Now you may say, OK, so this is an invariant. It is an analytic function. So what can I get out of this? Here is a theorem with Renat Kashyaf. You can think of this function as follows. So if tau is in the upper half plane. What is the impedance on tau? Ah, thank you. Thank you very much. So if tau is in the upper half plane, then up to a silly factor which I will just skip, although it's recall. You know what? I don't even need to skip. 24, and then 1 over root tau little g times big g minus root tau big g little g. Where little g is sum minus 1. It's a specific kind of num sum, and big g is the same except you multiply by 6m plus 1. 1 minus 7q minus 14q square, and so on. So in other words, our invariant is a combination of q series and q tilde series. This series, much like the series in Sergei's talk, are only defined inside the unit disk. And they, of course, have divergent expansions at 1 or at any cusp, any rational, any root of unity. And we have proven the existence of asymptotic expansions around every cusp with dawn. This is not totally trivial. And of course, the conjecture is that these asymptotic expansions at every cusp are resurgent, but we don't touch that. But it is a conjecture. But the amazing thing is that although both little g and big g are divergent as tau approaches a specific number, let's say in the positive reals, their combination extends in the category of plane. So now I need to connect the asymptotics of the little g and the big g with the invariants that I was telling you about. Maybe you can say how q is related to tau. Thank you. I didn't say. q is e to the 2 pi i tau. And q tilde is e to the minus 2 pi i over tau, just as in modular forms. So asymptotic expansions of little g and big g, let's say at 1. And that is a theorem. So let me call little w. So notice that my functions are not just power series in q. They are also in q tilde. And not only in q and q tilde, but they have polynomial axial square root behavior in tau. So they are not just pure q series. Sorry about that. But that's life. So we talk about this big g little q tilde here and capital the i over 2, same expression, but with a minus sign. Then when tau is e to the i theta over n and n goes to infinity, w is asymptotic to the geometric series, and big w is asymptotic to the anti-geometric, except that these series are multiplied by g of q tilde and big g of q tilde. And keep in mind that these are instanton corrections. And that is actually a theorem with Don Zagir. And something different. So this happens, by the way, when theta is between 0 and pi over 2. And something different happens when theta is between pi over 2 and pi. To actually explain what happens, I want to draw for you the walls and the crossing of the walls. So we have this. And we have this. And so on. This is a tiny angle, but how tiny is it? And the answer is 0.032 times pi. So in this sector, you see that this series is asymptotic to this formal power series in tau with no corrections. When you jump in across, you see the first power of q tilde. When you jump in again, you see the second power. And again, and again, and again. So how far can you keep jumping to see the powers to eventually see the first coefficient? Here you have to be very near. I mean, your tau has to be near the real axis. So I want to give you some numerical data now. And then I will end by connecting the formal power series to these two functions. So numerics. Let's take tau to be 150 plus i. So that's very near, but scaled down to 2,500,000. So the imaginary part is 1.3e to the minus 6. So according to that, we should actually see about four coefficients of these q tilde series. So in other words, these tau constants, or however you want to call these symbols, maybe these vorous symbols have integer coefficients. So the trivial. So we're going to take all our 100 terms that we have. And here is the function. The function. And here is the size. So the size is 234. If you take little w of tau, you find 234, just size-wise, digit-wise. If you take this, you get minus 232. The q tilde itself is e to the minus 29. If you take the trivial one, sorry, the geometric. And divide by little w and subtract 1, you are dividing these two large numbers minus 1, and you find 0 to 192 digits. And if you do it for the other one, here, the anti-geometric capital W, with the corrections, you find minus 192. And why can't you go further than 192? Because you only used 100 terms of these series, and I used them to the optimal truncation. So every one of these factorial different series has an internal error, which is only exponentially small, but you can't go beyond. And here we reach the bottom of the error, and we found all the nine terms here. And then we repeated the experiment and so on and so forth. So, OK, so what this is saying is that the little g and so these analytic functions, this is the state integral, and this is the anti-state because it replaces tau by minus tau. These functions can reproduce the trans series. How about using the trans series to reproduce the functions? So if you're actually plugging the asymptotics of these W's here, and you go back to the state integral, you're going to find something funny. And that's going to be the last thing that I want to talk about. And the funny thing is that, yeah. OK. So converse how to reconstruct the state integral from this series, factorially divergent, right? So take this expression into the minus volume over 4 pi. This series here, I remove the exponentially large term because I put it here anyhow. So 2x1 minus x, phi geometric of minus 2x over 1 plus x, minus e to the volume over 2 pi over 2 pi. Actually, pi. The geometric one, 2x over 1 plus x, and the geometric of minus 2x over 1 minus x. OK. So this is a series, right? This is a power series in x with constant coefficient that and that and whatever the constant term is here. Yep. Yep. So this series is not factorially divergent. It has a radius of convergence equals 1. And in fact, it is equal to the state integral of tau when tau is 1 plus 1 minus x, well, 1 plus 2x over 1 minus x. OK. Convergence of products, bilinear products of factorially divergent series is convergent. And to explain that, let me say that the coefficient here of this series, the coefficient of x to the 140 is e to the 260. But after you multiply them out, the coefficient in this product of x to the 140 is 0.0141. OK. I'll stop here. So question. Yeah. Does the originals change your variables just to make them soccer-abolic singularities? Yeah. So the state integral has an essential singularity at i equal to 0, which are these series. What I do is I expand it at tau equal 1. So how do I expand tau equal 1? I can set 1 plus epsilon. But just to center things better, I send it like that where x is near 0. And then if I do that, then my radius of convergence is going to be 1. So full reconstruction of the Laplace transform by passing the Borel plane. I still don't know why these are singularities. Any more questions? Yeah. But pure curiosity, do you have any control numerical or analytical of the growth at infinity in the Borel plane of the functions? How tame are they? Yeah. They are exponentially growing. Yeah. They're exponentially growing. Well, actually, as you well know, there's been a PhD of the case of Pricotovet. That elucidates in my new detail all the resurgence aspects. I think you have to ask. Yes. Now, whether 4 pi squared, for instance, is a singularity. Oh, I should tell you that. So Marcos Marinho had 20 terms of this series. And then he did a Pade approximation in the complex plane to locate the singularities. And he found one singularity here. And then kind of a cloud that wasn't conclusive. And then we tried with 100 terms, and we found another one here with some sigma. I don't know how many sigmas. But whether 4 pi squared is a singularity and whether this statement is true, that goes beyond the thesis of Schroff-Fettas-Scharma. That did not exist 15 years ago when I visited you. This is new stuff. Yeah, but the PhD came after that. Yeah. And it's not a question of finding the synthetic equivalent, actually. Given that you know the exact syntactic structure of the Taylor coefficients, you can get not only the location of the next singularity, but actually the exact Taylor expansion. It's not immediate. The nearest one, yes. No, no, but all the others can be, there is a trick which is called the ping-pong. The ping-pong that I know involves only the two nearest singularities, or in this case... No, no, it doesn't. There is a trick that can take you anywhere. If you think you have the ping-pong game at hand, I suggest to try this into the conservative series that is actually missing a lot of singularities. Sir, sir, sir. In the case of figure eight, and actually it is mentioned at the introduction of the PhD, that the method in theory extends to all notes. And we are not interested in that, but there is an extension in another direction. For all power series, it is exactly the same syntax, let us say. So in a sense, it is less and more general than the volume structure. But actually there is a full picture. And to come back to Voro's question, indeed there is exponential at most. And it is mentioned that it grows at infinity on the normal race. So there is no...it is not a matter of conjecture. It is a dumb thing. I will write since you challenged me. I will write this. Okay. Remove these numerics. So the state integral, this I41 of tau, can be written because of this combination of series in the following way. And this, by the way, if Carlos asks, they are absolutely convergent, and in fact with analytic expansion. So here is the formula for the hundredth instanton term. 7, 7, 5, 6, 9, 7, I over root tau g, minus root tau 8, 7, 3, 0, 8, 2, 2, 0, 8, 2, 5, 3, 6, 3, big g of q. That's the 500 instanton term. If you can reproduce these numbers, whatever black magic or box you use, please let me know. But I don't think that you can reproduce these numbers because it's one thing to talk about, I think, a general theory versus specific values of Stokes constants and specific functions. But maybe I'm wrong. So this is my challenge. Well, actually the arithmetic nature of the Taylor coefficients is totally irrelevant in that case. Okay. You might be obsessed by this, but actually it's not, as far as resurgence is concerned, it is totally, it has nothing to do with it. Yeah, but parabolic flat connections have geometric origin. And even those constants, like the volume, have geometric origin. I mean, yeah. No, but our argument was only about the status of the state. Is it a conjecture or is it not? I was saying this is established. Okay. Well, you're welcome to your opinion. Any more questions? No, let's say thanks. Thank you.
 Thank you very much for giving me the opportunity to talk to this wonderful conference and meet  new people and reconnect with people I knew from all times.  So I want to tell you about arithmetic resurgence of quantum invariance and by that I mean quantum  invariance in dimension 3 and such as not invariance or three-manifold invariance.  I will actually not talk about three-manifolds so much.  So as I understand resurgence is a property of formal power series and their analytic  continuation and their expansion of those functions near singularities where the original  power series emerges or reserges.  So resurgence can be formulated, if you can formulate it for formal power series you can  also formulate it for its Taylor coefficients so it can be formulated for sequences themselves.  So you can talk about the resurgence of the sequence and factorial, you know, for instance,  which is factorial divergent but you can do things to it.  So as I understand resurgence is abandoned for functions which are defined using different,  differential, integral equations, linear or not.  And I also understand that there are at least two aspects of the theory, one of which is  proofs and the other of which is numerical experimentation, symbolic or truly numerical  and I will talk about that.  That part of the theory is very fun, it's suited for mathematical physics but it's only one  aspect, right?  So I came across resurgence in 2005 when I studied a particular question and I had an  agenda in mind, namely the volume conjecture.  So let me give you some introduction about it.  So we're talking about a knot in three sphere and to a knot there is a sequence of low-run  polynomials with integer coefficients.  These are called the M-colored Jones polynomial.  So for instance, N equals 2 is the original Jones.  So these are supposed to be polynomials defined using the representation theory of SL2 together  with the n-dimensional representation and greatly generalized and so on, so forth.  So one knot has a sequence of polynomials.  If you wish, you can really think that the Nth-colored Jones polynomial is basically  the Jones polynomial of a parallel of a knot, the N minus 1 parallel.  That's correct up to some sub-leading kind of orders which are universal and depend on  the knot.  But there is a little theorem in this game, joint work with Tang Le, which says that the  colored Jones polynomial, namely the sequence of polynomials, is q-holonomic.  So let me explain what this means.  It means it satisfies a linear q-difference equation of the form CD of q, q to the n,  jn plus d, nc0 q to the n, jn of q.  Here I drop the dependence on k.  So this is true for all n, where the cj's are Laurent polynomials in two variables.  So in other words, it's a linear q-difference equations whose coefficients are polynomials  in q and q to the n.  If the polynomial is q to the n, also polynomials in q.  Not true.  What polynomial of q is q to the 100?  It's a polynomial of degree 100.  All right, let's make n 200.  If n is a variable, q to the n is another symbol.  It's not a polynomial of q.  Because the degree of the polynomial grows empirically or?  The degree of the polynomial here.  Yes, it's also a theorem that the degree of the color Jones polynomial is a quadratic  quasi polynomial.  And that is actually true for all solutions of linear q-difference equations.  The only proof of that theorem, actually, of myself, I was inclined to talk about that,  uses Le Mahler Scholar theorem, N-Piatic analytic number theory.  So yeah, the degree grows quadratically.  OK, so this is a theorem and true for all nodes.  So how unique are the cj's?  Sorry?  How unique are the cj's?  First of all, cd should not be 0.  And there is a unique way among all q-difference equations.  There is a unique way to make a minimal order q-difference equation.  And that makes the so-called A-hat polynomial, which annihilates the color Jones.  And it is a not polynomial.  And it's an invariant itself.  OK, so let us suppose that a linear q-difference equation, that the linear q-difference equation  has a basis of solutions, which are asymptotic to, so when you put q equals e to the alpha  over n, they are asymptotic to e to the whatever, s minus 1 times n, ns0, ns1.  ns1 over n, and so on.  OK, let's suppose that a linear q-difference equation has such a basis of solution.  In fact, a resurgent with respect to 1 over n.  But I don't even care about the resurgence.  Suppose that is true, then a version of the volume conjecture holds.  And the volume conjecture just says that the nth color Jones polynomial of the knot evaluated  at the nth root of unity grows exponentially, and the rate is the volume of the knot times  2 pi i.  That is the volume conjecture.  So the volume conjecture would follow from the previous statement, and it would make  a nice theorem in a fine journal.  But it is not a theorem.  So if you really want to look at linear q-difference equation, so you can think of them as Schroediger  except that there are difference, and really except that they are high order.  Even the order of the equation for the figure eight knot, the order is 3.  So already you're talking about higher order, but who cares because you can make them vector  values.  So OK, so that's just one idea.  I mean even without mentioning resurgence, just that idea alone doesn't exist.  Or if it does, please let me know, and we can finish that sort of paper.  So let me now give an example.  So among all knots, let's start with a trefoil, which sometimes I can and sometimes I cannot  draw trefoil.  So there is a formula.  I didn't tell you how the color Jones polynomial is defined, but it is defined in a complicated  way.  So the formula is the following, minus n sum from 0 to n minus 1, q to the minus nm, q  to the minus 1 minus capital N, semicolon qn, I guess just to avoid confusion, this is  what I mean here.  That is the formula for the call Jones polynomial.  This formula is terminating, so we can replace n minus n by infinity, and then decide to  set capital N equal to 0.  So since you use a different use, you should define the  Thank you.  Yeah, thank you.  So let it's, now let's set n equal to 0 and call the formula.  Now it's a formula, right?  It's a formula, but it isn't a formula.  It is still a knot invariant.  The fact that it's a knot invariant is a theorem of actually Habiro, and this formula that  looks like black magic because I can rewrite the color Jones polynomial in many different  ways.  And if I just set n equal to 0, in other words, in the zero-dimensional representation, no,  actually it's not the zero-dimensional representation.  It's the infinite-dimensional representation of SL2, or formal-dimensional zero.  So if I do all that, then I'm going to get a knot invariant, okay?  And this so-called formula has two properties.  First of all, it can be evaluated at n through 2 of unity because only finally many terms,  right?  And it can also be evaluated when q is e to the h bar.  Oh, and why can it be evaluated when q is e to the h bar?  Because e to the h bar, e to the h bar n is n factorial h to the n plus o of h to the  n plus 1.  So in fact, it's not so hard to see that if you evaluate, so let's make now a definition,  and the definition is I use capital F is for perturbative, trivial.  And of course, this is a formal power series with coefficients in h bar, rational coefficients  because of e to the h bar.  And Geveret 1, by the way, this series is not just a series.  It is the series that we were calling concevage Zagir series.  And this expression.  So now let me give you a theorem.  So it is a very one.  So what is the theorem?  The theorem is, of course, that if you look in the Borel transform, so let us look at  the Borel transform.  This is Greek letter xi.  When I write it, it looks like 7.  But what can you do, right?  So this is the replacement of h to the n over n factorial by xi to the n over n minus.  Topologically it is a 7.  And I'm Greek on top of everything.  OK.  So this is the Borel transform, right?  Borel transform convergent for xi near 0.  So the theorem that we have with the video question, which was the entry, this was my  unofficial entry.  It was 15 years ago, and it still is my agenda.  Of course, to prove the volume conjecture, if you can prove resurgence itself, you prove  much more.  You prove generalized volume conjecture and other things.  So you can aim for everything.  I mean, that was my agenda.  So we settled for something less.  And here is a theorem.  The theorem is that here's a formula.  OK.  Some thank you, Sergei.  Epsilon n are the same numbers n square pi square over 6 to the 5 half, where epsilon  n are plus 1 if n is congruent to 1 or 7, or 1 or 11, 5 or 7, n 0 otherwise.  OK.  So this is a formula.  So in particular, it says that in Borel plane, the nearest singularity, all the singularities  are in the real axis.  The first one is in pi square over 6.  The next one is in 25 pi square over 6 and keeps going.  Right?  They are not at all equally spaced.  So this is part A. And let me also give you part B. So therefore, B, because the singularities  are on the real axis, what we cannot do is Borel summation on the real axis.  What we can do is Borel summation right before or after.  Sorry, Laplace right before and right after.  So we have a formula for that.  That's a function of x now.  And the formula is it is i root 2 pi x to the 3 half times sum epsilon n times n.  Times q to the n square over 24 when n goes from 0 to infinity where q is e to the 2 pi  ix.  This is just the end of Sergei's talk except I lied here.  I should have done one more thing.  This is the Borel transform of f 3 1 of e to the h after I multiply by e to the minus  h over 24.  I really need to multiply by that.  That doesn't change anything.  I mean, but it is much better for modular properties.  So everything is explicit.  And my challenge to you is to actually give a different proof of this theorem.  So the proof of the theorem basically uses work of Don Zagir.  It follows from Zagir's work on the so-called strange identity.  But this is a very explicit series with a very explicit formulas for everything.  So if you, in fact, if you know the answer, please try to reproduce it.  It's just a baby case.  Sorry, Maxime, to call that baby.  But OK, so now let us grow up.  If we may.  All right.  After that comes the figure eight knot.  For which case the Collard-Jones polynomial, I'm even going to give you the formula.  I hope you appreciate the fact that these are formulas for the Collard-Jones and not  just for what's called the Kashyaf invariant.  N minus 1 cubed N, q to the minus Nn.  So therefore, the function that I was talking before of q is deceptively similar.  It is q, q, N, q inverse, q inverse N from 0 to infinity.  OK, and we can talk about, this is also Gervais-Guant.  By the way, we have a little theorem with Tang Le, which says that this is.  It does look like a polynomial.  This one?  Should be 1 minus 1, no?  Oh, yeah, yeah, sorry, N minus 1 with a q inverse.  OK.  Yeah, yeah, yeah, yeah, yeah, yeah, to terminate it.  Thank you very much.  And that's why I put q inverse here.  A little theorem with Tang Le is that for all nodes, this is Gervais-Guant.  OK.  So you might say, well, but how different is this expression from this expression?  You know, surely if you prove resurgence of one, you can prove resurgence of the other.  That is my challenge to you.  So in fact, let me give you here a conjecture.  So here is a conjecture, stated for all nodes, right?  The conjecture is that, so this thing I'm going to call it, capital Phi trivial of H  bar for a node k.  And the conjecture is that Phi trivial of a node k is resurgent in Borrel plane.  I mean, after you do Borrel transform, with singularities at i times volume plus  Chern-Simons plus 4 pi squared times integers plus a little shift.  This rk is a rational number.  And here rho is a parabolic SL2C representation.  So that is a homomorphism of the familiar group into SL2C, which is parabolic.  That means the trace of the meridian and the longitude is plus minus 1.  Among all parabolic representations, there is one that everybody knows, namely the trivial  representation.  It is parabolic, I mean, and has zero volume and zero Chern-Simons.  So okay, this is a conjecture.  So rk doesn't depend on the representation?  No, it doesn't.  Yep, it's an overall shift.  Okay.  All right.  So now, three, four, five.  Okay.  So let me give you what happens in the case of the figure eight.  Let me draw a picture for you.  Parabolic means around the knot is very important?  Yes, yes.  Around the knot complement.  I mean, the meridian and the longitude, yeah.  So for the figure eight knot, we expect that, first of all, the volume of the figure eight  is about 2.02.  So you just, do you understand this, if you draw in Chern-Simons, it's a real large  part of complex?  Yes, yes.  I could just call that complex Chern-Simons.  That's right.  That's right.  So the complex, well, there are two parabolic representations, the geometric one, which  gives the hyperbolic structure and the complex conjugate.  So the geometric one has volume 2.02, the Chern-Simons is zero.  So therefore, if you plot the numbers, you find i times v and minus i times v, where  this distance is about two units.  And then there is the 4 pi square here, which is 39.7.  And I didn't even draw it to scale.  So therefore, the singularities in Borrel plane are supposed to line up in two horizontal  lines and the singularities are sort of 4 pi square apart.  Now you may say, and how do you know that they are all singularities and not, for instance,  some of them are not?  So our conjecture actually, for the figure 8, says that they are all going to be true  singularities.  I'll give you evidence for that.  So that's one conjecture.  Now let me give you another conjecture.  So is the shift zero in this case, or you just didn't write it?  Zero.  Well, if I put it here, it is zero.  I really put them on the y-axis.  OK, the renormal on shift.  OK, so let's make a definition.  Little phi trivial of a node k is going to be the evaluation at the kth root of unity.  This is, in other words, known as the Kashyp invariant of the node.  And it is a sequence of numbers.  Yes, yes.  So it is a sequence of numbers.  And the conjecture is that this sequence of numbers is a resurgent function in Borel  plane with all singularities.  OK, that's a conjecture.  OK, so for the figure eight, I gave you two series.  I gave you this sequence, and I gave you the other one, this one.  So now let me give you a 2 by 3 matrix here.  So first of all, just a few terms.  So what is easy to compute here?  Well, given that we have a formula and we substitute q equals e to the h, and by terminating  sort of stuff, it is not so hard to find out that you get some terms here.  But it is easy to compute these coefficients, which are factorial diversion.  And I should also say that it's another theorem of Tang and myself that this also, this series,  is a, no, let's forget.  OK, now what about this?  This is e to the minus, no, plus volume divided by 2 pi divided by h bar, 1 over third, fourth  root of 3, you know, and then 1 plus, there is a 2 pi i, there is an 11, there is a root  3, some trivial constant here, and the higher order terms.  The reason I put this is to show a contrast.  I meant geometric here, thank you.  So the coefficients of this series are, apart from the power of 2 pi i, which is homogeneous,  they are supposed to be in the trace field of the knot, which for the case of the figure  8, it is q squared root of minus 3.  But they are supposed to be resurgent.  So in other words, if we actually write down here, e to the volume over 2 pi times 1 over  h bar, and call these coefficients here an, so these are easy to compute.  These are much harder to compute.  The only way to compute them is to compute the Kashyap invariant, and then compute it  to high precision, and peel off the 1 over n, the 1 over n squared term and so on, and  do the computation numerically, accelerated using research on transform, and finally,  recognize it as an element of that trace field as an algebraic number, and we've done that  with Donza gear, and we found out 100 terms in this series here, just as a benchmark here.  For this series, it's very easy to find many terms.  So we have 100 terms here, and there are other methods to compute the Kashyap invariant,  and many terms here.  I should say also a little theorem.  A little theorem here actually is joined with Donza gear, is that the Kashyap invariant  of K is computable in linear time using the Q difference equation.  So it's a linear time computation, but its asymptotics is not a linear time computation.  So the question is, what are the asymptotics of, so let's put this thing geometric here,  and let's call these things trivial.  What are the asymptotics of the nth coefficient here, and the nth coefficient here?  And if there is any justice in the world, the asymptotics of these coefficients are given  in terms of themselves.  All right, so let us define here.  Okay, so let G of 1 to n be 1 over 2 pi i, sum from 0 to infinity of a k, until n minus  k minus 1 factorial divided by 2 i v to the n minus k.  And the G 2 1 is the same, except I replace here by minus 2 i v.  Okay, same.  And the G 1 n, which is root 2 over pi, sum of a k geometric, n minus k plus 3 halves.  So factorial divided by sort of i v to the n minus k plus 3 halves with a k.  Yeah, that's it.  That's it.  Can you explain the notation?  Because I'm T first, yeah?  And T what is in the denominator?  Ah, yes, I didn't say that.  Fee hat, the anti-geometric, so I said that there are three representations.  I mean, there are three parabolic connections, the geometric, the anti-geometric.  So the anti-geometric is the one that you replace h by minus h.  So we have three coefficients, the trivial, the geometric, and the anti-one, which is  just minus one times the geometric.  Okay.  And now I want to, well, tell you a theorem, except I cannot tell you a theorem if it is  not a theorem, right?  So what is it?  It is an empirical observation.  And the empirical observation says that this is asymptotic to 3, g1 to n.  The anti is asymptotic to minus 3, g2 to 1n.  And that can be compressed into an eschew symmetric 2 by 2 matrix.  And finally, the trivial one is asymptotic to g1n and g2n.  So therefore, you get a matrix of this kind.  Okay.  So in particular, to highest order, it says, for instance, that the nth coefficient here,  it's going to grow like 1 over 2 pi i n minus 1 factorial, 1 over 2 i v to the n times a  0, which is 1 over 4th root of 3.  Okay.  So, by the way, how do we do this?  So first of all, we use everything we have, namely 100 terms here.  So the 100 terms is about e to the 365, just to give you an idea of the order of growth.  Then you take all the terms that you have here, you do the optimal truncation.  So e to the 365 divided by this are two numbers of 365 digits.  You take the ratio, you subtract 1, and you find a number which is e to the minus 115.  So order dropped from 365 to minus 115.  With these numbers here, okay?  So they're pretty convincing.  Sorry, but what is this?  Have g1n and g2n?  Yes.  I didn't do g2n.  So the g2 would be the other one, 2 here, and the minus 1 here.  Mm-hmm.  Okay.  So you may say that in Borrel plane, the asymptotic expansion of these Aeons are basically  expressed in terms of the series Aeons itself.  Okay.  Okay.  So that's just the...  Barista, why can't you transform this into a theorem by using the state-interval representation?  It's a one-dimensional integral.  I'm just doing my first part of my talk.  Right?  I am still into a formal power series.  I haven't done my analytic functions yet.  So just a third example, but very quickly, which is the 5, 2, 0.  So for the 5, 2, 0, let me give you the answer now not of the Kashyap because we...  I mean just of the Kashyap invariant, sorry.  The answer looks like this.  Okay.  And although there is a denominator here because k is less than or equal to m, it always  cancels with a little bit of the numerator and you have left both to expand when q is  exponential of h bar and when q is the root of unity.  So in this case, the trace field is actually cubic, the only cubic field with discriminant  plus, I mean minus 23, the only one.  And in this case, if we did the trivial, just that would be easy.  But if you did the geometric, you find 1 over 2i root 3alpha minus 2 and then 1 minus 33  alpha square and 242 alpha minus 245 over 4 times 23 square.  There is a 2pi i times an h bar and higher order terms.  So this field has one complex, I mean one pair of complex embeddings and one real embedding  and I'm choosing here the complex embedding.  The only reason I'm putting this is because Jan was asking what are the coefficients we're  talking about and the coefficients are in the trace field.  And conjecturally, every number field with at least one complex embedding is the trace  field of some casped hyperbolic manifold.  OK, so we computed 100 coefficients here, but we computed them using actually a kind  of num, I mean sums and I wanted to tell you the picture here, the analogous picture.  So there are three parabolic, I mean other than the trivial representation because this  is a cubic field, it has three embeddings in the complex numbers and the geometric one  is always parabolic, therefore, so are the other two Galois conjugates.  And therefore, we get here three numbers v1, v3, v2.  And as usual, there translates by 4pi square.  OK, and the numbers can be given explicitly, one of them is manifestly real.  This is actually an SL2R representation because we're talking about the real embedding here  and the other two are 1.379 plus or minus 2.828, that is the volume of the 520.  And you can actually check that this is nearly equilateral, namely the distance between v1  and v2 is 5.011 and the distance between these two is 5.656.  The only reason I'm writing these numbers down is because they are given to us by the  gods, they are not hyperbolic geometry, normalizes the volume, normalizes geodesics, normalizes  everything, it is not up to us to mess with these numbers.  And in this case, we get such a thing, so therefore, we're going to have a 3 by 4 matrix.  And I will just tell you what is the analogous, I'm just going to cut the conversation down  and we computed the analogous matrix with don.  I should mention all these are joined with don and the matrix now is 033 minus 34, that's  a 3 by 3 bit and the other one is 11 minus 1.  Okay, this is sort of the matrices that I believe Sergei wanted.  Oh, and there is an additional shift, I should say for the 5.2, there is a nasty kind of  shift which is a pi square over 6, that one has to put in there.  Okay, that was my, the old part, which has to do about resurgence of sequences or factorially  divergent series, which is mainly conjecture how to convert them into Borrel plane and talk  about analytic continuation.  Why do you do that?  Because after you convert to Borrel plane, you take Laplace transform and then you're  in the plane you want to be, in the plane where you have your analytic function, your  solution to the differential or differential equation, but we don't have a differential  differential equation.  But we do have a function.  So my second part is about those functions.  So part two, analytic functions of a variable tau.  So tau is going to be mostly a variable in the cut plane.  So it turns out that there are theories, perhaps not full TQFTs, and they only work for Casp  manifolds.  These theories are called quantum Teichmiller theories or quantum hyperbolic geometry, and  they always work for punctured Riemann surfaces, and therefore Casp three manifolds, and they  may be the wrong theories and so on, but they do give invariance of knots in particular.  These theories were developed by Kashaev, Andersen, and several sort of other collaborators.  And I want to focus that the building block for this state sums, oh, I should say that  for a general note you might wonder how horrible does the formula look like, and the answer  is not too bad.  You're always doing a finite dimensional sum of ratios of quantum factorials.  That's all there is.  And likewise in these new theory, you're going to do not a finite dimensional sum, but a  finite dimensional integral, and the building block is not going to be the quantum and factorial,  but it's going to be a special function for Davis quantum dialogarism.  So let me give you an example.  What is the answer for the figure eight knot?  So for every knot, oh, I should say here first of all, theorem for every knot, there is a  function which is analytic.  Well, OK, you may say, oh, well, you can take the constant function, I guess.  But I mean, there is a function which is analytic, which is computed via ideal triangulations  and state integrals.  Did I stop?  You still have one block for left.  No, I know.  But my numbering got almost up.  Six is here.  So seven, eight, nine.  I learned this method from a young postdoc at MPI.  Yeah, but you have to do it in advance.  Otherwise, you get carried away.  OK, so what is the invariant for the figure eight?  It is the integral of phi b x square e to the minus pi ix square dx, where phi b is a very  special function.  I can write down the formula, but the formula will not really talk to you unless you have  played with it.  Bz, hyperbolic sine, b inverse z, dz over z.  OK, so this is for devs.  It's a double quasi-periodic, and it has all kinds of properties.  So this is the invariant.  Now you may say, OK, so this is an invariant.  It is an analytic function.  So what can I get out of this?  Here is a theorem with Renat Kashyaf.  You can think of this function as follows.  So if tau is in the upper half plane.  What is the impedance on tau?  Ah, thank you.  Thank you very much.  So if tau is in the upper half plane, then up to a silly factor which I will just skip,  although it's recall.  You know what?  I don't even need to skip.  24, and then 1 over root tau little g times big g minus root tau big g little g.  Where little g is sum minus 1.  It's a specific kind of num sum, and big g is the same except you multiply by 6m plus  1.  1 minus 7q minus 14q square, and so on.  So in other words, our invariant is a combination of q series and q tilde series.  This series, much like the series in Sergei's talk, are only defined inside the unit disk.  And they, of course, have divergent expansions at 1 or at any cusp, any rational, any root  of unity.  And we have proven the existence of asymptotic expansions around every cusp with dawn.  This is not totally trivial.  And of course, the conjecture is that these asymptotic expansions at every cusp are resurgent,  but we don't touch that.  But it is a conjecture.  But the amazing thing is that although both little g and big g are divergent as tau approaches  a specific number, let's say in the positive reals, their combination extends in the category  of plane.  So now I need to connect the asymptotics of the little g and the big g with the invariants  that I was telling you about.  Maybe you can say how q is related to tau.  Thank you.  I didn't say.  q is e to the 2 pi i tau.  And q tilde is e to the minus 2 pi i over tau, just as in modular forms.  So asymptotic expansions of little g and big g, let's say at 1.  And that is a theorem.  So let me call little w.  So notice that my functions are not just power series in q.  They are also in q tilde.  And not only in q and q tilde, but they have polynomial axial square root behavior in tau.  So they are not just pure q series.  Sorry about that.  But that's life.  So we talk about this big g little q tilde here and capital the i over 2, same expression,  but with a minus sign.  Then when tau is e to the i theta over n and n goes to infinity, w is asymptotic to the  geometric series, and big w is asymptotic to the anti-geometric, except that these series  are multiplied by g of q tilde and big g of q tilde.  And keep in mind that these are instanton corrections.  And that is actually a theorem with Don Zagir.  And something different.  So this happens, by the way, when theta is between 0 and pi over 2.  And something different happens when theta is between pi over 2 and pi.  To actually explain what happens, I want to draw for you the walls and the crossing of  the walls.  So we have this.  And we have this.  And so on.  This is a tiny angle, but how tiny is it?  And the answer is 0.032 times pi.  So in this sector, you see that this series is asymptotic to this formal power series in  tau with no corrections.  When you jump in across, you see the first power of q tilde.  When you jump in again, you see the second power.  And again, and again, and again.  So how far can you keep jumping to see the powers to eventually see the first coefficient?  Here you have to be very near.  I mean, your tau has to be near the real axis.  So I want to give you some numerical data now.  And then I will end by connecting the formal power series to these two functions.  So numerics.  Let's take tau to be 150 plus i.  So that's very near, but scaled down to 2,500,000.  So the imaginary part is 1.3e to the minus 6.  So according to that, we should actually see about four coefficients of these q tilde series.  So in other words, these tau constants, or however you want to call these symbols, maybe  these vorous symbols have integer coefficients.  So the trivial.  So we're going to take all our 100 terms that we have.  And here is the function.  The function.  And here is the size.  So the size is 234.  If you take little w of tau, you find 234, just size-wise, digit-wise.  If you take this, you get minus 232.  The q tilde itself is e to the minus 29.  If you take the trivial one, sorry, the geometric.  And divide by little w and subtract 1, you are dividing these two large numbers minus  1, and you find 0 to 192 digits.  And if you do it for the other one, here, the anti-geometric capital W, with the corrections,  you find minus 192.  And why can't you go further than 192?  Because you only used 100 terms of these series, and I used them to the optimal truncation.  So every one of these factorial different series has an internal error, which is only  exponentially small, but you can't go beyond.  And here we reach the bottom of the error, and we found all the nine terms here.  And then we repeated the experiment and so on and so forth.  So, OK, so what this is saying is that the little g and so these analytic functions,  this is the state integral, and this is the anti-state because it replaces tau by minus  tau.  These functions can reproduce the trans series.  How about using the trans series to reproduce the functions?  So if you're actually plugging the asymptotics of these W's here, and you go back to the  state integral, you're going to find something funny.  And that's going to be the last thing that I want to talk about.  And the funny thing is that, yeah.  OK.  So converse how to reconstruct the state integral from this series, factorially divergent, right?  So take this expression into the minus volume over 4 pi.  This series here, I remove the exponentially large term because I put it here anyhow.  So 2x1 minus x, phi geometric of minus 2x over 1 plus x, minus e to the volume over 2  pi over 2 pi.  Actually, pi.  The geometric one, 2x over 1 plus x, and the geometric of minus 2x over 1 minus x.  OK.  So this is a series, right?  This is a power series in x with constant coefficient that and that and whatever the constant term  is here.  Yep.  Yep.  So this series is not factorially divergent.  It has a radius of convergence equals 1.  And in fact, it is equal to the state integral of tau when tau is 1 plus 1 minus x, well,  1 plus 2x over 1 minus x.  OK.  Convergence of products, bilinear products of factorially divergent series is convergent.  And to explain that, let me say that the coefficient here of this series, the coefficient of x to  the 140 is e to the 260.  But after you multiply them out, the coefficient in this product of x to the 140 is 0.0141.  OK.  I'll stop here.  So question.  Yeah.  Does the originals change your variables just to make them soccer-abolic singularities?  Yeah.  So the state integral has an essential singularity at i equal to 0, which are these series.  What I do is I expand it at tau equal 1.  So how do I expand tau equal 1?  I can set 1 plus epsilon.  But just to center things better, I send it like that where x is near 0.  And then if I do that, then my radius of convergence is going to be 1.  So full reconstruction of the Laplace transform by passing the Borel plane.  I still don't know why these are singularities.  Any more questions?  Yeah.  But pure curiosity, do you have any control numerical or analytical of the growth at infinity  in the Borel plane of the functions?  How tame are they?  Yeah.  They are exponentially growing.  Yeah.  They're exponentially growing.  Well, actually, as you well know, there's been a PhD of the case of Pricotovet.  That elucidates in my new detail all the resurgence aspects.  I think you have to ask.  Yes.  Now, whether 4 pi squared, for instance, is a singularity.  Oh, I should tell you that.  So Marcos Marinho had 20 terms of this series.  And then he did a Pade approximation in the complex plane to locate the singularities.  And he found one singularity here.  And then kind of a cloud that wasn't conclusive.  And then we tried with 100 terms, and we found another one here with some sigma.  I don't know how many sigmas.  But whether 4 pi squared is a singularity and whether this statement is true,  that goes beyond the thesis of Schroff-Fettas-Scharma.  That did not exist 15 years ago when I visited you.  This is new stuff.  Yeah, but the PhD came after that.  Yeah.  And it's not a question of finding the synthetic equivalent, actually.  Given that you know the exact syntactic structure of the Taylor coefficients,  you can get not only the location of the next singularity,  but actually the exact Taylor expansion.  It's not immediate.  The nearest one, yes.  No, no, but all the others can be, there is a trick which is called the ping-pong.  The ping-pong that I know involves only the two nearest singularities, or in this case...  No, no, it doesn't.  There is a trick that can take you anywhere.  If you think you have the ping-pong game at hand,  I suggest to try this into the conservative series that is actually missing a lot of singularities.  Sir, sir, sir.  In the case of figure eight, and actually it is mentioned at the introduction of the PhD,  that the method in theory extends to all notes.  And we are not interested in that, but there is an extension in another direction.  For all power series, it is exactly the same syntax, let us say.  So in a sense, it is less and more general than the volume structure.  But actually there is a full picture.  And to come back to Voro's question, indeed there is exponential at most.  And it is mentioned that it grows at infinity on the normal race.  So there is no...it is not a matter of conjecture.  It is a dumb thing.  I will write since you challenged me.  I will write this.  Okay.  Remove these numerics.  So the state integral, this I41 of tau,  can be written because of this combination of series in the following way.  And this, by the way, if Carlos asks, they are absolutely convergent,  and in fact with analytic expansion.  So here is the formula for the hundredth instanton term.  7, 7, 5, 6, 9, 7, I over root tau g,  minus root tau 8, 7, 3, 0, 8, 2, 2, 0, 8, 2, 5, 3, 6, 3, big g of q.  That's the 500 instanton term.  If you can reproduce these numbers, whatever black magic or box you use,  please let me know.  But I don't think that you can reproduce these numbers  because it's one thing to talk about, I think, a general theory  versus specific values of Stokes constants and specific functions.  But maybe I'm wrong.  So this is my challenge.  Well, actually the arithmetic nature of the Taylor coefficients  is totally irrelevant in that case.  Okay.  You might be obsessed by this, but actually it's not,  as far as resurgence is concerned, it is totally, it has nothing to do with it.  Yeah, but parabolic flat connections have geometric origin.  And even those constants, like the volume, have geometric origin.  I mean, yeah.  No, but our argument was only about the status of the state.  Is it a conjecture or is it not?  I was saying this is established.  Okay.  Well, you're welcome to your opinion.  Any more questions?  No, let's say thanks.  Thank you.
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