 Hvala za to, da ima me vse oportunitva, da se počeš tudi vzvečne konference in izgleda nekaj nekaj, in izgleda nekaj nekaj izgleda nekaj, od vseh razgleda. Tako, vzveš, da imaš arithmeti, reserzeni, kvantom invarjeni, By to, sem zelo, da je kvantum invarjenja v zelo 3, in tako, nekaj invarjenja, nekaj invarjenja, nekaj invarjenja, nekaj invarjenja. Zelo sem zelo, da nekaj invarjenja so nekaj invarjenja. Zelo, da se početno zelo, da je prišljena všeljna zelo, in tudi je analitizovana, in bi se vse inšljati za svoje vse rečenje, na kaj je zelo zelo komunist, kaj je dobro izvizila, izvizila, nekaj je rečen, zelo rečen, kaj je za všeč, za svoje tezno, kaj je zelo kojefizijen, tako da je bilo formulatojno za sekuncije. Tako da se počešte o reserciju z sekunciju n-faktorijov. Zelo, da je zelo, kaj je faktorijalne zelo, ali da se počešte. Tako da se počešte reserciju, je obandana za funkcije, kaj je zelo, da je zelo, zpravacke, vseččešnije, intergalne, vseččešnije, vseččešnije, in vseččešnije, da si, atleto čežel constructiona teori, izvelenje iz bojsm, in izvelenje, neč jaznemanje, neč teori, neč jazne, neč teori. Kdaj imam tudi o to. To je všeč jazda. To je začala na matematikovom fiziku, ali je tukaj nekaj aspekt. Ok. Zato v 2005, kaj sem stavila nekaj zelo, da sem tukaj zelo, nekaj nekaj zelo, Zato, da me vse dobro vse zelo, da so tudi o njelti. Zato, da smo prijeljali o njelti v 3, in na njelti, nekaj je nešto počak, z njeljelj polinomijomov, z kajim inželjih počakov. Zelo, ki je tudi m, zelo, zelo, zelo, zelo, zelo, zelo. Na sve. Line n equal 2 je... ...o original zones. To je tudi polinomjela z deračenjami judpose meatballs is basically the Jones Polinomial of a parallel of a knot, the n minus one parallel. That is 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 the joint work with Tang Le which says that the word Namely, ta sekunca polinomiala je kuholonomiko. Zdaj, komendijo mi početno, kako je to priča, kako je doblika vse iz nadačenu počistku. Vkaj prejze počistku, c, d, q q n, jn plus d, c, 0, q q 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. A te vtepe nomen s q to d, 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. It calls the degree of the polynomial growth and correctly, the degree of the polynomial here. Yes, it's also a theorem that the degree of the color zones polynomial is a quadratic quasi polynomial, and that is actually true for all solutions of linear q difference equations. And the only proof of that theorem actually of myself, I wasn't going to talk about that, uses Le Mahler's Scholar theorem in the analytic number theory. So yeah, the degree grows quadratically. So this is a theorem and it's true for all nodes. And how unique are the cjakes? Sorry? How unique are the cjakes? First of all, cd should not be zero. 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 zones and it is a not polynomial and it's an invariant itself. Ok, so, let us suppose that a linear q difference equation that a 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 and they are asymptotic to e to the whatever s minus 1 times n and s0 and s1 over n and so on. Ok, let's suppose that a linear q difference equation has such a basis of solution in fact resurgent in respect to 1 over n but I don't even care about the resurgence if 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 ok, 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 ok, so if you really want to look at the linear q difference equation so you can think of them as Schroediger except that there are difference and really accept that they are high order even the order of the equation for the figure 8 knot the order is 3 so already you are talking about higher order but who cares because you can make vector values so ok, so that's just one idea I mean even without mentioning it 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 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 nn q 1 minus capital N semicolon qn I guess just to avoid confusion this is what I mean here ok, 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 different use you should define z thank you so 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 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 of formal dimension 0 so if I do all that then I'm going to get a knot invariant and this so-called formula has two properties first of all it can be evaluated at nth root of unity because only finitely many terms right? and it can also be evaluated when q is e to the h-bar and why can it be evaluated when q is e to the h-bar because 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 fees 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 1 by the way this series is not just a series it is the series that we were calling concevates Zagir series and this expression ok so now let me give you a theorem so it is everyone so what is the theorem for 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 am Greek on top of everything ok so 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 and 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 name for everything I mean that was my agenda so we settled for something less and the theorem is that here is a formula ok sum 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 1 or 11 5 or 7 n0 otherwise 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 space so this is part a and let me also give you part b so therefore b cause the singularities are in the real axis what we cannot do is Borel summation on the real axis but we can do is Borel summation right before or after 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 halves times sum epsilon n times n times q to the n square over 24 and 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 f31 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 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 the work of Don Zagir it follows from Zagir's work on the so-called strange identity 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 ok it's just a baby case sorry Maxim to call that baby but but ok so now let us grow up if we may right after that comes the figure 8 not for which case the Collor Jones polynomial I'm even gonna give you the formula I hope you appreciate the fact that these are formulas for the Collor Jones and not just for what's called the Kashaev invariant n minus 1 q q to the minus nn so therefore the function that I was talking before of q is deceptively similar it is qqn q inverse q inverse n from 0 to infinity ok and we can talk about this is also Gevreg 1 by the way we have a little theorem with Tang Le which says that this is this one should be 1 minus 7 sorry n minus 1 with the q inverse 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 Gevreg 1 so you might say well but how different is this expression from this expression 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 chain simons plus 4 pi square 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 it means the trace of the meridian and the longitude is plus minus one 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 chain simons so okay this is a conjecture so rk doesn't depend on representation no it doesn't yep it's a novel shift okay alright so now so let me give you the what happens in the case of the figure 8 let me draw a picture for you parabolic it means around not as important yes around not complement the meridian and the longitude yeah so for the figure 8 not we expect that first of all the volume of the figure 8 is about 2.02 sorry just do you understand this in real version part of complex yes yes I could just call that complex chain simons that's right mm-hmm so the complex well there are two parabolic representations the geometric one which give the hyperbolic structure and the complex conjugate so the geometric one has volume 2.02 the chain simons is zero so therefore if you plot the numbers you find v and minus i times v where this distance is about 2 units and then there is the 4 pi square you know here which is 39.7 okay and I didn't even draw it to scale so therefore we should the singularities in borel plane are suppose of this 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 they are conjecture actually for the figure 8 says that they are all gonna be true singularities and I'll give you evidence for that okay so that's one conjecture now let me give you another conjecture is the shift zero in this case zero well if I put it here it is zero I really put them on the y axis okay the renormal on shift okay so let's make a definition little the trivial of a node k is going to be the evaluation at the kth root of unity this is otherwise known as the Kashayev invariant of the node and it is a sequence of numbers yes yes it is a sequence of numbers and the conjecture is that this sequence of numbers is a research and function in borel plane with those singularities okay that's a conjecture okay so for the figure 8 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 you know some terms here but it is easy to compute these coefficients which are factorial diversion I should also say that it's another term of tongue and myself that this also this series no, let's forget okay now what about this this is e to the plus volume divided by 2 pi divided by h bar 1 over 3 4th root of 3 you know and then 1 plus 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 square 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 dn so these are easy to compute these are much harder to compute the only way to compute them is to compute the Kashaf invariant and then compute e to high precision and peel off the 1 over n the 1 over n square 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 we have done that with donza gear and we found out 100 terms in these series here just as a as a benchmark here for these series it is very easy to find many terms so we have 100 terms here and there are other methods to compute the Kashaf invariant and many terms here I should say also a little theorem a little theorem here actually it is joined with donza gear is that the Kashaf invariant of k is computable in linear time using the q difference equation so it is a linear time computation but its asymptotix is not a linear time computation so the question is what are the asymptotix of so let's put this thing geometric here and let's call these things trivial what are the asymptotix of the nth coefficient here and the nth coefficient here and if there is any justice in the world the asymptotix of these coefficients are given in terms of themselves right so let us define here ok so let g of 1 to n b1 over 2 pi i sum from 0 to infinity of a k t n minus k minus 1 factorial divided by 2iv to the n minus k and g21 is the same except I replace here by minus 2iv same and g1n which is root over pi sum of ak geometric n minus k plus 3 halfs factorial divided by sort ofiv to the n minus k plus 3 halfs with a yeah that's it can you explain the notation t first ah yes I didn't say that phi 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 they replace h by minus h so we have three coefficients the trivial geometric and the anti one which is just minus one times the geometric ok 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 g12n the anti is asymptotic to minus 3 g21n and that can be compressed into an exq. 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 ok so in particular to highest order it says for instance that the nth coefficient here is going to grow like 1 over 2 pi i n minus 1 factorial 1 over 2iv to the n times times a0 which is 1 over 4 root of 3 ok 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 term is about e to the 365 just to give you an idea of the order of growth you take all the terms that you have here you do the optimal trankation 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 e to minus 115 with these numbers here ok so they are pretty convincing yes i didn't do g2n so the g2 would be the other one 2 here and the minus 1 here ok so you may say that in Borrel plane the asymptotic expansion of these are basically expressed in terms of the series itself ok ok so that's just varestavros why can't you transform this into a theorem by using the state integral representation after all these are one dimensional integral I'm not there yet 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 520 so for the 520 let me give you the answer because the answer looks like this ok although there is a denominator here because k is less than 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 2 i root 3 alpha minus 2 and then 1 minus 33 alpha square and 2 4 2 alpha minus 2 4 5 over 4 times 23 square there is a 2 pi 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 am choosing here the complex embedding the only reason I am putting this is because Jan was asking what are the coefficients we are 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 cast hyperbolic manifold so we computed 100 coefficients here but we computed them using sums and I want to tell you the picture here the analogous picture so there are three parabolic 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 Galov conjugates and therefore we get here three numbers v1, v3, v2 and as usual there translates by 4 pi square and the numbers can be given explicitly one of them is manifestly real this is actually an SL2R representation because we are talking about 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 2 is 5.656 the only reason I am 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 3 by 4 matrix and I will just tell you what is the analogous I am 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 0,3 minus 3 4 that is the 3 by 3 bit and the other one is 1,1 minus 1 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 square over 6 that one has to put in there ok, that was my the old part which has to do about resurgence of sequences or factorially divergent series which is mainly conjectural how to convert them into Borel plane and talk about analytic continuation why do you do that because after you convert to Borel plane you take Laplace transform and then you are in the plane you want to be in the plane where you have your analytic function your solution to the difference or differential equation but we don't have a difference or differential equation but we do have a function so my second part is about those functions so part 2 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 Taich-Miller theories or quantum hyperbolic geometry and they always work for punctured Riemann surfaces and therefore casp and they may be the wrong theories and so on but they do give invariance of notes 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 comes 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 factorians that's all there is and likewise in this 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 so let me give you an example what is the answer for the figure eight note so for every note oh I should say here first of all theorem for every note there is a function which is analytic well okay 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 left no I know but my number in God almost stopped six is here so seven, eight nine I learned this method from a a young postdoc at MPI yeah but you have to do it in advance otherwise you get carried away okay so what is the invariant for the figure eight it is the integral of phi b x square e to the minus pi i x 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 okay so this is for devs quantum dialogarithm and it is a very it's doubly quasi periodic in so properties so this is the invariant now you may say okay so this is an invariant is an analytic function so what can I get out of this here is a theorem with Renat Kashaev you can think of this function as follows so if tau is in the upper half plane what is the difference on tau 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 well you know what I don't even need to skip 24 and then one over root tau little g times big g 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 6n plus 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 these series much like the series in Sergei's talk are only defined inside the unit disk and they of course have divergent expansions at one or at any kasp, any rational any root of unity and we have proven the existence of asymptotic expansions around every kasp we've done this is not totally trivial and of course the conjecture is that these asymptotic expansions at every kasp 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 cut plane so now I need to connect the asymptotics of the little g and I need the invariance that I was telling you about ah 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 ok, 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 actual square root behavior in tau so they are not just pure q series 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 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 antigeometric 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 donza gear 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 question this is a tiny angle but how tiny is it and the answer is 0.032 times pi ok so in this sector you see that this series is asymptotic to these 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 you keep jumping to see the powers so you know to actually 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 scale down to 2500 so the imaginary part is 1.3 e 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 voros symbols have integer coefficients so the trivial so we're going to take all our 100 terms and here is 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 if you take this 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 trankation so every one of these factorial divergence series 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, so forth 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 actually plug in the asymptotic of these w's here and you go back to the state integral you're gonna find something funny and that's gonna be the last thing that I want to talk about and the funny thing is that so converse how to reconstruct the state integral from this series factorial divergence so take this expression e to 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 pi 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 so this is a series this is a power series in x with constant coefficient that and that and whatever the constant term is here so this series is not factorially divergent it has 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 1 plus 2x over 1 minus x combinations of products by linear products of factorial 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 Is the origin of this change of variables just to make some similarities ah, yeah so so so this the state integral is ah has an essential singularity at i equal 0 which are these series what I do is I expand it at tau equal 1 so how do I expand i 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 if I do that then my radius of convergence is going to be 1 so full reconstruction of the Laplace transform bypassing the Borel plane I still don't know why these are singularities yeah 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 ah yeah they are exponentially growing yeah, they are exponentially growing actually as you well know there has been a phd in the case of figure 8 that inocidates in my new detail all the resurgence aspects I think you yes now whether 4 pi square for instance is a singularity oh, I should tell you that so using Marcos Marino had 20 terms of these series and then we he did a pade approximation in the complex plane to locate the singularities and he found one singularity here kind of 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 square is a singularity and whether this statement is true that goes beyond the thesis of Shreveda's Sharma that did not exist 15 years ago when I visited you this is new stuff the phd came after that it's not a creation 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 expansion the nearest one, yes all the others there is a trick the ping-pong that I know involves only the two nearest singularities or in this case 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 in the case of figure 8 and actually it is mentioned at the introduction of the phd that the method in theory extends to all notes but we are not interested in that but there is an extension in another direction for all power cities in the context so in a sense it is less and more general than the volume conjecture but actually there is a full picture and to come back to Vorel's question indeed there is an exponential at most a growth of infinity on the normal rays so it is not a matter of conjecture it is a don't thing I will write since you challenged me I will write OK remove these numerics so the state integral this I for one 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 hundred instanton term 7 7 5, 6, 9, 7 I over root tau G minus root tau 8, 7, 3, 0 X3 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 general theory versus specific values of stokes constants and specific functions but maybe I am wrong so this is my challenge actually the arithmetic nature of the data coefficients is totally irrelevant in that case you might be obsessed by this but actually it's not as far as resurgence is concerned because it has nothing to do with it but parabolic flat connections have geometric origin and even those constants like the volume have geometric origin then I mean argument was only about the status of the state is it a conjecture or is it not I was saying this is established ok ok ok ok