 Итак, я буду говорить о том, что 1-поинт функции в Синч-Гордон-Моноусе, это joint paper с Стефаной Негросовым. Итак, 1-поинт функции. Я считаю, что 1-поинт функции на цилиндре. Итак, 1-поинт функции дают важную физическую информацию. Итак, это не так просто в Синч-Гордон-Моноусе, но если у вас есть рациональный, например, реакция филосиры, то есть, есть просьба, чтобы конструктировать, когда у вас есть петрубация, есть просьба, чтобы конструктировать, оперировать продукт экспенсион, после петрубации, после петрубации, после петрубания, после these calculations of short distance behavior, at least of correlation functions, is reduced to 1-point functions, which carry information about the infrared environment. All of that is very well explained in several papers. Пейпо по Алешу Замолочникову в 1991 году. И в 1991 году. И еще несколько пейпос. Очень важный пейпо по Лукьянову и Замолочникову. Один пейпос в функции премьер-филосиума и сайм-гордон-модел. И еще один пейпос по Фатееву, Лукьянову и Замолочникову. Это было несколько пейпосов по булок-дотцам, которые будут важны для меня здесь. И еще для десендов. Это был тоже пейпо. Или это был кто-то другой. Так что, для сайм-гордон-модел все не так просто, потому что модел не рациональный. Поэтому опиратель-продукт-экспансион не очень простой. Но, кстати, в многих респектах, это самый простой пейпосовый модел. Так что, как мы видим, есть один пейпос в функции. Так что, даже если пейпосы не так чистые, но есть один пейпос в функции, то лучше. Так что, история была, что мы делали с Джимбом, Мива, так что мы делали корреляционные функции. Мы изучали корреляционные функции для латых моделей. И мы хотели бы это поставить в таком виде, чтобы не перенести корреляционную функцию на два или три сайта, но найти семьи, которые depend on parameters, в которых можно сделать скейлинг-лимит. Так что это мы делали. И так, мы разработали для Сайн-Гордон модел нашу собственную процедуру, для получения этой функции. Функциями, как-то нашли, для всех десентраций, и для праймерной философии, и для цилиндров. Так что, процедура, которая была объяснена по Лукьянову Фатееву Замолоччикову и Замолоччикову, это основана на так называемой рефлексии релаций. Так что, я думал, что если, например, если вы хотите идти из Сайн-Гордона и из Синч-Горнона, то у вас не есть резонабельные латых модели. Так что, наверное, это... first of all, it makes sense, first of all, to check if our results on SIGN-Гордон agree with this reflection relations, right? And so they do, and then, so it's a kind of... Reflection relations is a pure algebraic thing, so they can be easily switched from SIGN to CINCH-Гордон. And then, so using that, we kind of considered this problem of calculation, the one-point function for CINCH-Гордон model. So this is my plan. So I first formulate the problem, then I explain this relation. Reflection relations, then I explain how we solve this reflection relations, then I give the conjecture for construction of one-point function of descendant fields. Then I check it, I mean ratio of descendants to primary. Then I check it, then I explain how to construct one-point function for primary fields and then check it for primary fields and if I have time, I'll say about mystery. So actually the main problem for me are these first two parts because so this reflection relation is not very kind of clearly proved thing especially for this CINCH-Гордон case. Yes, and so... But so in some sense I'm not responsible, there is responsible persons in the audience. And another thing that... So they are so nice that no doubt they are correct. And so Jimba told me that once Alyosha Zemolochikov came to him and so explained him about this reflection relations. So and his explanation he started with the words now I will cheat you. So and then I immediately I wanted to talk with Jimba immediately I wanted to cite Pushkin but I didn't know how to translate. And so I found on the net translations into English and French. Both of them are very awful. So I'm not even sure. So French is simple to understand but ugly. But anyway I'm not really sure that Russian-speaking people will recognize what is that. So what it means to translate. It means to translate тьмы низких, истин не дороже у нас возвышающий обман. That's what I wanted to say to Jimba. Okay, so I'll start with this. So we can see the Saint-Gordon model, so Euclidean. And so Luchianov also blazed me for this Kaoplin costume but it's convenient for me to write this way. Yes. So we are on a cylinder with radius r and there is a point here and so some operator living at this point. Yes. And so let's consider the following thing. So we take some primary field e to a phi of zero. And then we... Okay, so we normal order, order of course with respect to Euclidean time, yes. And so just consider this ratio of two of two of this descendant kind of field to primary field. So just for trivial reasons because so this action is C invariant, if you wish. So invariant under phi to minus phi. So this thing if I have even number, for example, of this phi's here, this thing is invariant under this reflection a to minus a. Yes. This is obvious. So and this even it will be for some reason enough for me. But even if it is ordinary, it's not very hard to understand. So actually the system is integrable. So there is an infinite number of integrals of motion and there are descendants created by this integrals of motion. I mean so if you wish in Hamiltonian picture you take commutator of your local field with one of these integrals of motion. Yes. So this will be again integrable operator but so its expectation value must be zero because for example, you can expectation value, you can think of it something like that. So you take this states kind of minimal energy states in what's the bar direction, you take the vacuum expectation value. These are eigenstates for this integral. So that's why so the commutator that's why actually I don't need to consider all these operators but only factor out these integrals. So now comes more tricky thing. So let's rewrite this action like Liouville plus this kind of primary field and consider this as a perturbation. So actually, so to be honest you cannot consider this as a perturbation because this field is not small. There is no mu for issues it will be small. So Liouville has this central charge like here. So as usual but still let's try to think to consider this way, yes as a perturbation. And then let's let's take the energy momentum Liouville, I mean so if you wish for Siege Gordon modified energy momentum Liouville with second derivative, yes, with this central charge. And consider this kind of descendant. And then and then we want to say that this thing is invariant under this Liouville reflection this reflection typical for Liouville model. So this is a hat because so how do you say why it is so? But if you write the perturbation series so actually there will be severely because the dimension of this operator is negative there will be severe divergent infrared divergent. And putting it on a cylinder will not help. So then there is in this paper which I mentioned on the bullet dot so it's so it's proposed to put it just to to make an infrared cutoff to put it on a sphere or in this situation probably on something like this. C-gar so and let's let's decide that it works. So and then this this guy so there invariant under this reflection. And so I will consider here only for example only even this m and n because actually once again if I take into account this this descendant of this operator this guy so this is enough. So now one more thing is that we will assume that there is a rule how to so of course there must be identification of the space of local operators for the conformal field theory for Liouville and for this for Singed Gordon. But so we assume that there is a good way to establish one to one correspondence. This is so normally when you are in a more reasonable situation. So this you just argue by when this perturbation theory is really works. So you just argue by a dimensional reasons why it is possible. But here there are logarithms rather so it's not very clear. But I will explain later how I can shape it. So let's assume that we have we can identify with this descendant in Singed Gordon model. So we can identify them with states in Liouville. Let's argue in a similar way to the Molochikov and the Molochikov and the paper on Liouville. So let's consider it's a little bit different but similar. Let's consider this must be two. For example, matrix element of some of operator in Liouville between two states which is very negative zero mode. So then essentially in that case this field is more or less free. I mean phi. And this pleats like that and I can take I can use Heisenberg into two chiral paths. So I can use Heisenberg to parameterize them. And so these are commutation relations. So zero mode. And then it's convenient for me to work with a picture. So this field e to a phi zero I will call it free. I will denote it like that. And these guys this is a vacuum. So now of course normal ordering with this guy is the same as application of this generator here. So central charge as I said is one plus six q squared. So I define virasoro generators. And so this this is scaling dimension and application of these guys is the same as application of of this generator this generator of virasoro. So and important thing is these integrals of motion in conformal field series there is in as we know due to Zomolochikov. So I think that's contrary to yesterday so I start with the presumptions that Zomolochikov knows everything. So I repeat so. It's a bit opposite for yesterday statement. But anyway, so there is a set of integrals of motion. So this is just computing elements of the universal enveloping algebra of virasoro. So first one is trivial. So second one is reasonably simple and then it is getting more complicated. But so it's a pleasure. It's a great fun to put it on a computer and compute them. Yes. And so then we decided. So I consider here only one chirality and we decided for the descendants and we decided that we work in the quotient space. So we want to factorize the Verma module with respect to action of these integrals. Yes. So now let's take two bases in this space. So what I want to do? We have this reflection reflection sigma 1. Yes. And we have this sigma 2. So this is for descendants, for Heisenberg descendants. And this is for verasoro descendants. So now if I can relate the verasoro and Heisenberg descendants so I will have some closed system of equations for this vacuum expectation values. So I take bases on the verasoro side I take just I don't know. I generate a basis by lexicographically ordered action of even generation of verasoro because all odd spins so they are somehow hidden here of even. And then on Heisenberg side I say I take only monomials of even degrees of because I don't want to no need to change sign. But it's too much kind of and still will be too much. So I will choose some somehow every time. And then I want to find a matrix which relates these two just by algebra. And I give two examples. For example, on level 2 there is only one L-2 and so there is this guy on Heisenberg side and so something is I think this thing is forbidden here. I'm sorry. This is for matrix. So matrix is one by one just a number. And so you see basically on level 2 you have two verasoro two vectors but so this guy we drop it because it's this is an integral of motion. And so then for example if you go to level 4 then this matrix I take A-2 squared and here this for example to vectors and then the matrix becomes complicated and it will be harder and harder. So actually we compute it up to level 10 for some reason we want to level 10. Okay, so for level 10 it's already it's a real stuff. So now what is from that point of view what is our vacuum expectation value? So this is a linear functional on this quotient space, yes? With values which depend on A on B of course also but this is not so important. Which depend on A on this parameter of the primary field and according to what has been said so if I take this Heisenberg descendants and combine them into a vector yes? So then this vector must satisfy this equation and if I take Verasora descendants then combine them into the vector they must satisfy this vector must satisfy this equation and they are related like that and and this implies they must satisfy Riemann-Hilbert problem, yes? So if I shift A by Q so I apply one then another and if I shift A by Q then so this is the vector so then there is a matrix which is given by this yes? By this U and so values two banks of the strip like here and here are related by this matrix yes? and also so it must be but this is additional relation so it must be even even with respect to Q over 2X so of course this this problem has many solutions so you can multiply by any periodic even functions so this will fix later but so the question is so for example up to this periodic functions so how to solve this problem so I return back and I show you that so okay so here this is a scalar so this we can solve this solution is given by gamma functions this thing okay of course if you try to do it so it is hard and the solution the solution to this problem came from very different as I explained so came from our works with Jim Boyd Miva and the solution is given by the following constructions so we claim that there is another basis in this space so basis created by two fermions so okay there are only creation operators here but there will be an relation later so and then these guys are such that effectively effectively I mean so under this functional ethos something like that so effectively they under this both of these reflections and also under duality B to 1 over B which is also typical for Leuville and for Siege Gordon so they just interchange they have they have odd odd spins so this corresponds to their dimensions so let me be more precise so what is the basis so I use here I use here multi index notation so let me take multi indexes composed of of odd positive integers yes so and then the basis is given by this kind of the basis is given by this kind of products yes so then more precise statement is that I will explain so this is a structure which we obtained experimentally so long ago but so I will postulate it so I postulate that if I apply this operators to my primary field yes so then the result will be as follows so first of all there will be some something here which takes care of anti-symmetricism some cache determinant cooked of this multi indexes so then the product of some special functions for every element here and for every element here yes so this special functions are just some ok so forget about this thing this is some normalization but so basically this is a two gamma functions yes ok so and then so the result is polynomials even which with coefficients which depend only on scaling dimension and central charge then so even polynomial and odd polynomial and odd polynomial has a coefficient which is like that which contains square root so scaling dimension of course under my reflection sigma 2 are invariant and this guy changes side so and then so by construction so this thing satisfies this second reflection yes because because all the all the verasaur generators I have this satisfied and everything else everything else so this this guys satisfy as well yes so only this changes sign but this odd guy changes plus to minus yes so which means changing got a bit to gamma and this two guys also change switch one to another so this is satisfied and the same as of course for duality but then if I try to with this and that actually this it's important to say that the coefficients of this guy they have some determinants they have some denominators but this matrix can be decided from from this matrix U somehow but now if we assume and try to impose this sigma one reflection then it give me overdetermined system of equations and so you check it so level 4 level 6 level 8 level 10 and it always has solutions yes so you find solutions and this that's why so this is not a mathematical theorem so I probably so I'm not I cannot prove such theorems I don't know how to do it but anyway so there is a very special basis in the verma module which which kind of respects both of these relations so why it gives me overdetermined system of equations these multipliers here are very important because so okay so if I switch A to minus A then they will produce some multipliers and I have so the result of and then I go from to Heisenberg generators here so there will be polynomial must be divisible by some huge multipliers anyway so once again so this let me formulate it exactly so now if I take once again my functional eta which corresponded to this vacuum expectation value yes and to this expectation values and I combine I evaluated on all these vectors fermionic vectors and I combine it into one vector which is called f so then this f satisfies the following system f of minus A equals J J is an operator which switches I plus and I minus so kind of trivial thing so it satisfies that yes so my originally complicated Riemann-Hilbert problem turned out into a trivial one so this is this is why there so when did I start something at three yes so this is why in this basis if I want to compute my one point functions on a cylinder or this ratio so one point function on a cylinder in this basis yes then they must be just symmetric periodic function of Q but for the moment I did not say anything about them yes so they must depend on this geometry on the radius so I don't know this is a conjecture how they they vacuum expectation values in this fermionic basis so this is a conjecture which is based first of all on analogy with sign Gordon and then on some other ideas for example you can easily compute the vacuum expectation values for the components of the energy momentum and so on so you just think about it and you come with the conjecture so I will probably if I have time I will say how one can prove it so now I will explain how I construct now this this one point functions in this fermionic basis so I start with a function called Q yes so this is a function from which all the thermodynamics all the thermodynamic quantities or in other words everything about spectrum about vacuum so vacuum vector in this matzubara in this matzubara space so which contains all the information yes about them and this function due to so I think this in this way it was explained first in the paper by Alyosha but I am not sure so probably it can be also read from lokyanov paper on one point function so it must satisfy the following equation okay and so which I called quantum run scan and so if you supplement this equation with conditions that this asymptotic that when theta goes to infinity it behaves this way I am not sure that it is correctly written here I am sorry so probably I have to define by something this is rather full I probably have to divide anyway so cosine and r are correct but so coefficient here coefficient here is wrong and this is a mass this is a mass of physical particle and this is related to my to this constant by this famous formula Alyosha Zmolochikov and so so we supplement with these asymptotics and then we have to say something about zeros in the strip from minus pi i over 2 and if we assume that there are no zeros then this is a vacuum state so then solution is unique so it is reduced to some non-linear integral equation but I will not explain that so now let me introduce let me introduce a following a following function which we always introduce e to minus y of theta is product of this 2q and I take a linear operator let's take function of theta and so then this is a linear operator which contains 4 shifts 1, 2, 3 and 4 so here there is a coefficient 1 plus e to epsilon and here are coefficients depending on this a so a is hidden only here so let me take also delta function it is not a delta function but it is similar so then I take a function I take kind of green function so I take a function which satisfies this linear equation so I apply this operator to this omega and I get this delta function kind of green function or from the right other argument it depends on two arguments and then so what I want to say is that all this vacuum expectation values of descendants not only descendants are given just by this function so actually it is convenient for me to take a following a following to consider Fourier image of this function with respect to both variables so actually this function here I am not sure so important that it is entire function of L and M which means that when L goes to infinity or M goes to infinity so I can just compute the asymptotic by taking these two things of these two things so which means that the asymptotic at infinity the asymptotic at infinity goes in this odd powers e to minus 2k minus 1 theta and asymptotic at plus infinity at minus infinity goes in this odd powers of this function and then theta evaluated at integer at imaginary integers here and here are just coefficients of these asymptotics and then the main conjecture which we don't doubt that if I take this ratio of descendants it is given by determinant which is cooked this function of these values here of these integer values here and from time to time with some addition when they coincide but essentially the asymptotics of this function omega define me all these one point functions let me mention one funny thing here that according to my according to what I explained from the very beginning so if I want to have descendants I have to have the same number of beta star here and gamma star and beta bar star and gamma bar star but as it is written here so it is obvious only number of the right-hand side makes sense if the number of beta stars bar is equal to number of gamma stars plus gamma by star so this is important and this is used to to go from descendants to the primary fields so later I will construct also the one point function for primary fields but talking about descendants what can we check with descendants actually not too much because not much is known about one point function for descendants but one thing can be checked so when R goes to zero then there is a following procedure which was explained in Valois lecture so you have you argue that you have a long plateau so this so you can move this R into the coupling constant because everything depends on dimensionless combination and then there is a huge plateau reflection from two boundaries so you basically it's a very good approximation you have very good approximation if you just consider the dynamics of zero mode the dynamics of zero mode is described by this equation and so this is a reflection amplitude and so if we are talking we are interested in the ground state so this turn is zero so we take this equation and what does it do it allows me to compute the momentum here in terms of radius of the cylinder and then of course this procedure implies that if I take if I take yes, so that if I take my one point functions for example for the operator L-2 yes L-2 then in the limit it must be given by obvious formula for descendance of L-2 where P is replaced by this function of R on the other hand it is given by my function theta and there are these two multipliers yes so now this function theta is computed from my I wanted to write formula more beautiful than they are so this function defined by this omega which is defined from this difference equation but you can redo it in terms of integral equation and this integration can be easily iterated and so then you can indeed compare it and so this is what my student is doing now and here so if for example this dimension less quantity is like that or something I don't remember exactly for varieties of A and B you have six digits agreements and this explains actually that my choice I said from the very beginning that it is not quite clear for this non-rational model how should I identify the field non-ambiguously identify the field from massive from Louisville to massive model so this explains that I did correctly yes so and now there is one more funny thing so this is for this I don't know so this explanation in of the same type as as let me see of the same type of this reflection equation but look let's see so here for example you see I consider descendants yes I took the normal ordering with this zz and zbar zbar components of the energy momentum tensor but I did not take the trace so because if I would take the trace I of course would change the primary field but in fact so this fermions they are very well prepared for changing the primary fields as well namely so this is a formula which is kind of completely this formula is just taken from our computation for Saint Gordon model without much without big changes so if I consider descendants of this shifted field yes it can be expressed in terms of original field to which also these beta stars gamma stars are applied yes but now the number of beta stars is bigger than the number of what of gamma stars so I have additional thing here and what does it mean so this plus minus so these are multi indices and it means that every element that I add this to M to every element of these multi indices and then you can ask me so what happens if this guy becomes negative and if it becomes negative it becomes annihilation operator and so annihilation operator so I mean like that for example yes and so they are canonical with this function t which I had before and so I had there is this function cM of A the last thing which I have to explain here and this function cM of A is given by some mass cooked of gamma functions and this is of course a ratio actually a ratio of one point functions in infinite volume but so here I just put it by hands but so inside Gordon this was a result of some honest computation and we were very happy when it happened to coincided anyway so now I can by the same formula of course I claim the same formula by the same formula I can express the function of any shifted field by unshifted one through the same function actually this procedure obviously requires some consistency because so for example the same thing here I can compute using my function cM of A-MB and the result must coincide so actually they coincide from only one equation so this is the same as we proved with Miwai Jimba some time ago and so this equation can be proved so now we start for primary fields we are much more we are much happier because more is known so there are several possibilities to check things so let's see for the field for the simplest shift we have this kind of formula so because this thing is what when I applied it means one beta star beta one beta one star gamma one bar star yes so this is given by this formula and so when theta goes to infinity so this when r goes to infinity this theta goes to zero then this is just this is just a ratio of this location of the logic of one point functions so now let's consider the following function yes let's normalize by this infinite radius this is as we said this is already a q-periodic function of A because so all the non-periodicity all these reflection relations are already taken care of so for this thing there are several results so there is a formula which were found by Leclerc and Mossard by some procedure which is which looks a little bit strange they did thermodynamics they worked in this channel and used usual form factor which looks crazy because basically you have to to make some cutoff and so on but they introduced cutoff in rather tricky way so they have some prescriptions how to regularize infrared infrared divergences which manifests themselves because you have to take farfax on the diagonal of course have poles so they divergent but they regularize somehow and so they arrive at the following formula yes so there are series like that so which you can compute term by term and so this measure is like that right and so there are some functions some canals phi here and some functions here which you can compute from known for singed Gordon primary field for factors yes and then what do we do? you just shift it and take ratio and rearrange the result in the same way from one integral, two integrals three integrals and so on and and then I don't know the complete proof and I never care but so you can check three first terms as I said my function theta or omega or theta this is more or the same is given by some linear integral equation yes so and you can check that first three terms just coincide with three iterations of this yes so this is we put check here so now classical case so finally what we are doing we are computing the functional integral right and so if this suppose that this a here is big it is some finite no 2a for some reason it is some finite alpha by b yes so as usual so as you do and consider limit b goes to zero and then so this this guy is this functional integral is sitting on a classical solution which on this kind of classical solution which so what are the boundary conditions so the boundary conditions of course that it goes to zero then there and there is a singularity like that at the origin so then you have to evaluate the action on such solution and so e to this action of a b squared is or minus action of b squared is this one point function in that kind of classical approximation action you have to again like and do you have to regularize a little bit so Lukyanov write his paper formula which is which is first to Lukyanov and the Molochik fund published so if you think about it then more or less you can derive it yes but it takes some time but anyway so the formula is there is some mess here so this mess we already actually we don't care so this is this infinite infinite volume contribution and then there is this nice function here this integral so if you wish dialogarism yes so this function depends on alpha ok so then the ratio if I take the ratio so I remove this integral and the ratio my favorite ratio the finite is already finite quantity like that then I take my function omega and I and so my equation now there was this shift and the shift contained 1-b squared 1-b squared over 1 plus b squared yes and b now is equal to zero yes it's classical case so that's why this is this is one so this there were four shifts but now there are only two of them coincide pairwise coincide so you have this kind of equation so this equation you can solve so first of all factorize this guys and then solve ok so and you introduce such a function and then this requirement that my omega answer coincides with Lukianov results in the following identity yes so first you see and this is kind of you have a heart attack yes but then eventually so it is not so bad because consider this this Riemann-Hilbert problem yes so with this boundary condition so this asymptotics yes so what is this thing so obviously this is its behavior at infinity yes at plus infinity so this is solved by kind of by trivial taking logarithm and so on but let's solve it in a stupid way let's write a singular integral equation yes so and let's iterate it once and then you will hit the right hand side I don't know but what is funny here is that look I have this exact quantum answer and for example if you think about quasi classical correction so this it is given by essentially by determinant of this Laplacian plus this cosine this phi classical phi classical is not a simple is not a simple function I must say which means that I know how to compute this determinant so I have no idea how do I compute this determinant so actually this nobody knows even Lukianov I don't know about Sasha okay so now Sasha knows everything so now now about mystery so there is an interesting observation by Alyosha which he did long ago so you see this is a funny equation because there are four different shifts as I said and so what does mean for different shifts essentially they are independent so Alyosha says okay so let's consider this kind of equations this is the latest Liouville equation so it is indeed a Liouville equation in a sense that in the continuous limit like that it is Liouville and it is as good as Liouville equation in a sense that you can write you can again introduce two functions depending on light con variables and so general solution is given by through these two functions so that's very nice this he wrote one of he wrote equations then he says that let's just consider solutions which satisfy this additional condition yes kind of I don't know of the morphic solutions and so then and then he says ceta is equal to pi i u yes and so obviously you will obtain this equation right this is this makes sense because it really tells you something interesting about this equation so what I want to say is that about my function omega so now if I take this equation this discrete Liouville equation and I just consider this variation yes I take a linear equation for the variation and I obtain a following a following operator linear operator so now let's assume that for some reason so now I want to say that so I am looking for solutions which satisfy this kind of equation yes so which under this shift so it does the same thing but also this multiplier primary field comes somehow I don't know why so this gives me this kind of green function for this gives my function omega so why is this so I don't know probably in future some clever people will tell me that you just take your functional integral and you immediately observe yes yes okay yes you can ask it later so and since I have two more minutes I will say that the way how you can prove things so actually for example for one point function there is a representation in terms of separated variables for sign Gordon I wrote some time ago but Lukanov correctly remarked that Sing Gordon is the best possible application here so it is you can write some infinite dimension also integral so of course you can take from minus n to n and then regularize somehow of this form yes and so this gives me this one point function and this new are related to b squared like that this new and new tilde are interchanged look this is a kind of generalized matrix integral because in matrix integrals you have just van der moin squares normally yes you have van der moin squares so here you have one van der moin on one variables and another van der moin on other variables so I claim that I can compute exactly this integral yes so ratio of this integral but then you can of course you can reproduce it so the question is so this must be a way to prove my results I mean to to explain starting from here that the solution is given by my formula thank you very much this expression in some sense with some function q which was q of course must satisfy the equation as I said yes but Sergio proposed something like this so this formula as I said so this formula is just it follows from screen and separation of variables I am saying that I can compute it exactly yes but the problem is how to calculate q q q is given by tba equation tba equation is not linear equation but it is better than any linear equation so it converges iterations converge very rapidly yes and how to regularize this product no this I think you can find in lookian of papers so in the same way more or less from minus n to n there is of course some divergent multiplier but he computes the power of n many I am almost confused why you don't see any sign of freezing transition in your computation because in in your literature to the gravity we know sometimes transition will change parameter that seems to be dominated by localized peak and also in the semi-classical freezing transition I mean in the semi-classical equation you would if you write it there is no solution for when alpha is big enough so there should be a transition when you increase by looker if alpha is bigger than that's an effect which is important actually so I don't know but for some reason the analytical properties with respect to alpha for sign Gordon are complicated they are not as simple as given by this reflection equation but for change Gordon honestly I don't see any problems it's periodic it cannot be because semi-classical semi-classical equation has no solution when alpha is bigger look it's further to value alpha equal to what what is the problem no, semi-classical equations change Gordon is a better source you are talking about you are talking about Penlevi no but for example about Penlevi problem there is a short short equation semi-classical period the semi-classical equation okay I see what you mean so so you are talking about this thing yes there is a problem with this multiplier but there is no problem with this multiplier so the previous slide previous slide before slide yes so that the two equations at the bottom of the slide has no solution if alpha is bigger is it true? this is called mining you are talking about you are talking about casps okay yes this is okay this is understand but this is rather yes, this is rather a problem about this thing yes here you will have presumably both of this 1.5 1.5 function in infinite but this thing is alright so the thing is that by anand you impose a periodicity in alpha yes yes yes is it really correct to do that by extracting you are freezing transition which is the same thing and you cannot it's not true that you can impose a periodicity when you go in some regime you have you can I don't think the same I am not I don't think I have to go too far so it's enough for me for example so I probably I don't need to go to you are talking about q about something like q so I don't need to go from minus q to q I can go from somewhere inside if I take a plus a plus q yes maybe this is because we are getting there any one more question thank you