 Ok, donc, ce sera le 3ème de mes lectures. Donc, le 3ème sera ce que je vais appeler la représentation graphique de la question topologique. Donc, let me recall that I define omega gn of z1 zn as some of our all ramification points residue at ramification points of a kernel that I will call kA of z1 z times and there was this omega g minus 1 n plus 1 z sigma A of z z2 zn plus a sum g1 plus g2 equal g and I1 I2 equals z2 zn and this will be, that's what I called the sum prime but which also means no disk of omega g1 1 plus I1 z I1 omega g2 1 plus I2 sigma A of z I2 Right? So, on kA was this quantity, kA of z1 z was one half of integral from sigma A of z to z omega 02 of z1 and the dot means the variable that is integrated and omega 01 of z minus omega 01 of sigma I of z Ok, that was the definition and let me also define something which is just a short term name for omega 02 Let me just define it as a short term name Ok, the idea is that we have this recursive definition and we would like somehow to encode it in a way which is convenient for doing computations So, you see at each step we will have to take product of k times some of our omegas and at the first two steps we will start with some omega 02 So, we will have some k times B and something like that So, let me just give an example, omega 03 of z1 z2 z3 You see it's sum over A residue at A of kA of z1 z And here we have, in that sum, well the first term is absent because we start at g0, we are at g0 And in that sum, there are exactly two, this product contains two terms So, we have B z2 B sigma I of z3 plus B of z3 B sigma I of z2 Which are the two possibilities of decomposing the set z2 z3 into disjoint subsets that are not empty So, we have exactly two possibilities, so this is what we have So, we shall use a graphical convention to write this quantity And for that, I will represent this as just a line with two which forks in a trivalent vertex And here I will put the variable z and here sigma A of z And this one, I will represent it as just a line z1 z2 This one has an arrow, this one has no arrow So, just a convenient way to represent this big formula is just saying this So, somehow we have z1 z2 z3 Well, so this should be unique plus z1 z3 z2 Ok, so this is just a graphical convenient notation What this picture means is exactly the formula above So, it means each time you have an edge with an arrow, you replace it by the corresponding k Each time you have an edge with no arrow, you replace it by a B Ok, excuse me No arrow means that it's symmetric Excuse me No, it's not totally, well ok There is a subtlety which is that kA is symmetric It is symmetric, but indeed when you want to compute carefully the symmetry factors You have somehow to say that one edge, so somehow there is an orientation here So, let's put a dot on the left side And no dot on the right side So indeed, you should be careful about that But you have to keep in mind that in fact k is symmetric So, in fact all this is also equal to 2 times this one z1 z2 z3 So, either you put the dot, or you don't put the dot, but you have a symmetry factor 2 Ok, that's a choice And in fact, because of this symmetry, the symmetry factors will always be powers of 2 basically Ok Now, let's look at another one with this graphical representation Omega11 of z1 Is sum over A, residue at A kA of z1z And here in the big bracket in the right-hand side In fact, the only term is the first one Which is a B of z sigmaA of z So, which graphically you represent, there is a k There is that and here you have a B Again, there is that dot Ok, so this is just a graphical representation of that formula Ok, now let's consider another one For instance, Omega12 of z1z2 Well, according to the formula, it contains sum over A Residue at A of kA of z1z And here you have, in the right-hand side, what do you have? You have an Omega03 of z sigmaA of z2 So, plus an Omega02 B of zz2 Omega11 of sigmaA of z Plus B of sigmaA of z So, let me write it the other way Omega11 of z B of sigmaA of z2 Ok, so here this was an Omega03 So, let me represent it this way Ok, this is a kind of sphere This is a sphere, it has genie0 And it has three variables Ok, this one is what I will represent as a torus With only one leg Ok, it's just a graphical notation It just means that Omega11 I associated to it a picture of something of genus1 with one leg Whatever that means Just a graphical notation For the moment, it does not mean anything In the end, what I would like to prove Is that indeed, OmegaGN is something related to MGM So, that's what we will find in the end But for the moment, this is just a simple graphical notation So, let me continue with some examples So, Omega12 Let me continue to represent Omega12 So, Omega12, so which is a torus With two legs Z1Z2 Is according to this formula This K Z1 And with two sides And here, I glue an Omega03 An Omega03 is that sphere Ok, so here, there was Z SigmaI of ZZ Plus So, there is the second term Which is that one So, here, you glue a B And that Plus, on the third term Is Z1 And here, you glue Omega11 On the B Ok So, but now The idea is to use again That this was already We have already completed this Omega03 It's the formula above And it's already a sum of graphs So, the idea is that we replace This by a sum of graphs By the corresponding sum of graphs So, this is So, we have Z1 And here, we start on Oh, yeah, here it's Z2 It's also Z2 here in the end Ok Ok, so here Omega03 is that graph So, for instance Let's Let's write it this way So, here we had Z on SigmaI of Z And here, there will be another variable Let's call it Z' on SigmaB of Z' Ok So, here we will have a vertex A And here a vertex B So, what this is This is sum over A on B Residue at Z goes to A So, this is the same thing as here And residue And when we compute Omega03 We use another variable Z' goes to B And we have So, in this picture, we have K of Z1, Z We have a KB Of Z Z' And we have B of SigmaI of Z SigmaB of Z' And B of SigmaB Of Z' Z2 So, this big formula Is So, basically this graph Represents that formula It's the same thing Ok, so this graph is just a notation For that formula But we have many other terms We have Yes, you could indeed Indeed, you could So, here Ok, let me Do like that Ok So, somehow this is Z' Or If you want, it's like I put the dots No, let me do that Z' So, this was Z SigmaI of Z Z1 Z2 SigmaB of Z' Ok, AB Plus So, this was for that term Because omega03 contains two graphs And now we can do the same thing here So, we have this And here we have this So, again we replace omega01 By its expression And so we have this Plus Ok, so in formula This means We could put a bracket here Plus B of sigmaA of Z Z' B of sigmaB Of Z' Z2 Plus, so this was the second graph So now we have Z sigmaI of Z Z'SigmaB of Z' Z2 Z1 will be a B Of sigmaA of Z Z' And a B Of Z' SigmaB of Z' Plus, the last term is So, if you are careful B of Z Z' B of Z' SigmaB of Z' And sorry, times I forgot the B I forgot that B There should be only two B's Each time, so sorry I forgot the one with Z2 So there must be a mistake here In that graph It was Z2 here Sorry, this was that graph So there is a B of ZZ2 Which is missing And the B, yes So this was a B of ZZ2 And here we have a B Of ZZ Sorry Completely wrong We have this one And we have a B of sigmaI of Z Z2 Which is that one Ok Ok So this is just Yes Well, if you are careful, you see that this graph Has the same value of this graph Because of the symmetry Of this with a factor 2 So in the end You could also just say that this is 2 times This graph Plus 2 times This graph You could just say that It's equivalent Sorry, no, they are not planar graphs Well, you see here Well, ok, it's not subtle They are not really planar graphs Well, in that example Indeed they are planar, but it's not always the case And oriented edges Or what Yes, oriented edges But I'm going to write it now Oriented edges form a spanning tree Of a graph Always Just because each time you apply the recursion You always start by By an edge arrow So in fact Just let me mention that The way of writing the recursion So a way of writing the recursion Is something of genus G So So genus G and with N Legs So let me put the first one On the left Ok Equals So according to my recursion Is K times And here you put Either you put Something of genus G minus 1 And with the same legs And here Z1 And here you put In all possible ways So something with genus G1 And here something with genus G2 And some subsets I don't know Some subsets which I called I1 And some subsets which I called I2 So this is the way Of writing the recursion Ok So let me now state the theorem So if you recursively Apply this Exactly in the way I did it For the example of omega12 If you apply This recursively What do you have What do you have So the theorem Which is kind of obvious Which is that omegaGN Of Z1 ZN We have our graphs G Which belongs to a set That I will call GGN Of Z1 ZN That will be a set of graphs And we have a product So So I will describe this set But basically this set of graphs So G Has 3N vertices 3 valentes vertices It has 3G-3 Plus N Edges So some of them are oriented 2G-2 Plus N Oriented Edges Forming a spanning tree Of the graph It's a tree with root At Z1 So it's always So basically you will always have Something like a tree You have a tree The root is always Z1 OK And you're going to have some OK let me take another OK And you're going to have Also some bees Some non oriented Ones So you want the graph to be OK Let me Do that That OK You want the graph to be 3 valentes So every vertex is 3 valentes And It has So N-1 External legs Are Non oriented And At Z2 ZN And you have G Non oriented Edges That form G loops So in the end you have a graph With G loops OK But there is a non local constraint Is that Those Internal edges Non oriented edges Can only go from a vertex To one of its descendants Or ancestors You are not allowed to go from One branch to another So these are not Feynman graphs There are less such graphs Than Feynman graphs Non Going From a vertex To Its Descendant OK And there are some additional things Each vertex carries A I will say color Which is A ramification point An element of R So each vertex carries a color A B, C, B E, F, G Et quelque chose And each And it carries Also a variable So each vertex V Carries a color AV And the variable ZV And the theorem is that This Residue At ZV Goes to AV So product for all vertices You have a product For arrowed edges Which are the form V1 Going to V2 Or let's say V going to V prime K AV prime Of ZV And a product of non arrowed edges B Sorry V prime Of Some B of ZV ZV prime So you see that's exactly what we have been doing And the proof of that theorem is completely straight forward What is not Totally obvious is that constraint That you want to go Only from one vertex to its descendants But it's basically the way You construct those graphs recursively That implies that And it's not difficult at all So for instance if you want to see What are all the graphs that contribute To this omega 1, 2 You see that it's all the graphs That I've represented Also Yes sorry Again At each vertex there is a left side On the right side It could be You will have some powers of 2 Ok So vertices That are There is a left vertex Left side on the right side So let's say that a vertex Always looks like that Ok Yes Yes Yes I teach vertex There is one in going oriented edge But the outgoing edges Can be both oriented Or both non oriented Or one oriented, one non oriented All possibilities exist So this is very convenient Because it allows to represent Very easily all those quantities So instead of remembering Formulas You just have to remember this procedure And you can It's easy And also this graphical representation Is really the key to To proving many other theorems For instance That's how I use this representation To prove that the omega g n's are symmetric And I use this representation To prove in fact many other properties Of a topological recursion This graphical representation Is very convenient because it makes things Recombinatorial to prove So The next Another remark Another representation It's instead of Remember that my k a of z 1 z I represented so far As z 1 z sigma i of z But it's also Convenient to represent it By a thickening So instead of having just legs You thicken the legs And somehow Three dimensional representation But just notice that here Outgoing there is only In fact there is only one variable z On both legs The two variables are somehow They are the same One is just the copy Another copy of the first one So you would like to think That there are two outgoing things So it looks like a pair of pants But in fact Which splits into two In fact so somehow it's like a pair of pants But With a boundary Which is a disc that has been pinched At one point, that's the good way of representing things Because there is only one modulus Associated to the boundary There is only one boundary But somehow it will You can glue two things So when you're going to glue it You will have that possibility So z 1 Of z But splits into And same thing for b Of z 1, z 2 Well, which was omega 0 2 of z 1, z 2 Which is Which I represented So far like that But I will now represent it as just the cylinder z 1, z 2 So With this representation For instance I have at omega 0 z 1, z 2, z 3 Now I represented As this So now this is truly a pair of pants z 1, z 2, z 3 Is According to my representation So something like that And here you glue Two cylinders z 2 z 3 Plus the other possibility So which is z 2 And so on And for instance omega 11 Which would be This quantity Is going to be just That And here you glue a cylinder So it's very inspiring picture Unfortunately We are not truly able to give it Meaning in geometry for the moment But I would say it's a beautiful picture The theorem above Says that if you want to compute omega g n So corresponding to So basically it says So the topological recursion first it says That if you have something Of genus g With n boundaries Is Well Basically it's all the possibilities To Remove It looks like all the possibilities to remove A pair of points So you see here that There is one more hole here You are indeed creating the extra hole Plus Ok All the possibilities to do that So this is just a graphical representation Of a recursion For the moment it does not really mean something For instance in the Mirzarani's case For hyperbolic geometry It is not known What would be the good line to cut Really give a meaning to that representation It is not known Does it exist, it is not known A simple ramification point For the moment we have simple ramification points In fact that's my next section In fact This graphical representation Allows to give generalization To higher order ramification points That's exactly what I was going to say No Which is my number So it's just my 32 Ok So far indeed I have given the formula for topological recursion Only assuming that we had Simple ramification points So this graphical representation Makes it easy To define also the case of higher order Ramification points So that was my Higher order ramifications So remember I have The spectral curve is something We have a certain remand surface sigma And we have a projection To the base So sigma and sigma 0 And we have that projection which I call x And typically Near a simple ramification point In the vicinity of a simple ramification point The map Is 2 to 1 And there are exactly 2 branches That can be exchanged By an evolution Now assume that we have A higher order ramification point Somewhere Where we have several branches Coming to So now in a vicinity We have not a 2 to 1 map But a N to 1 map So imagine a A R ramification point Of order Da And da Larger than 2 So da is the number of branches That meet at this point Which means that locally There is a group There is a local Let me say local Galois group Da That permutes the branching The branches If sigma Sigma belongs to GA Is more or less equivalent To say that x Of sigma of Z Equals x of Z In a neighborhood Of A Typically we are a cyclic group But when you go to I think When you go to Cameral curves Well there is a possibility To take something more But yes indeed for curves This is a cyclic group So typically this would be Zda So indeed for a simple ramification Point this is Z2 GA contains exactly 2 elements The identity and an evolution But for higher order points It can be something more And so let me define What I would call So for every K Such that K is between 2 And the Order of GA Let me define The equivalent Of a kernel K But now it will carry A small K So somehow this one was the K2 KA Of Ok Let me call That one P And let me call All V over Z1 up to ZK Ok For the moment they are like Independent variables Excuse me Ok Ok Let me change the notation Z1 yes P1 Pk Ok, let me change My notation a little bit So it is not confusing And so by definition It will be slightly different From the previous one It will be integral From A To P1 Of omega 0 2 of Z1 And so That means the variable which we integrate And the integration path Is taken within the neighborhood Of A And product from J equals 2 to K Of omega 0 1 Of P1 Minus omega 0 2 Of Pk Of PJ So this is just the product And I will represent that Instead of representing it As before as something with So it has Z1 And here instead of having 2 Things on the right hand side It will have K Of them So P1 up to Pk So it will be represented This way Sorry No, they are not ordered Sorry They are ordered But in the end It's exactly like before There is the symmetry Which is that the value in fact Will be the value given by the particular regression Will be independent of that order But for the moment Z1 Yes Z1 plays a totally different role Z1 But they are symmetric with respect to all variables No K n'est pas symétrique Sorry Yes But not now Before I was taking I was not starting from A I was starting from Sigma of P1 And I had a one half in front So it's different But it will give the same result For the topological regression So it's different Before what I had defined Was the symmetrization of that one But since everything is symmetric In the end You can indeed symmetrize That gives the same result Now the definition The definition is that Omega g n Of Z1 Zn equals So I will need some little space Some other all branch points Some other All K Equals To the order Of the group Residue at P Let's call it P Goes to A And here I will put Let's put that before Some other all subsets So this is a notation A is a subset Included in Ga Minus Identity So I will take all subsets Of my group Except identity And I will take all subsets Of cardinal K minus 1 So that's just a notation To say subset of cardinal K minus 1 Ok So now I will have A Residue At Z goes to A K A K Of Z1 Which Was my first variable here And The next variables are Z And All the Set of sigma Z For all sigma belonging to A So I have K minus 1 Of them so that makes K variable Here Times And now I have Something That's going to be Like the product of omegas That appear in the right hand side A sum of product Of omegas Just before writing it In letters Let me write it graphically So graphically what I want to say Is just the following thing Graphically what I want to say Is basically nearly the same picture As below Except that now I have Not only 2 but I can add 3 4 and so on So what I want to say is that Is the following Ok 1, 2, up to N Is the sum Over all possibilities So you have Here K of them So you have that sum of K Ok, this was Z1 Z2 ZN Ok, let me Here you have G And here So K equals 2 2 GA And here you can glue things Some of them Can get Disconnected So from a source like that So here you can have Something like that Ok, that could be a possibility Ok, you have to split Into all possible ways So either you could have 1 connected piece Or 2 disconnected I mean 2 connected components Or 3 connected components The only thing you want Well, one thing you want Is that every of the Z2 Up to ZN variables Here, in the right hand side Must be connected to something there I mean, for instance You are not allowed To do something like that And here is something Like that Well I mean Just found something like that Ok, but it's just That when we are going to write the sum of products We have to be careful that we don't have such things Ok So we have all possibilities To do that Just remark one thing If you have L components If you have L components If you add all the Euler characteristics So some from I equals 1 To L of Ki I Plus Les characteristics Caractérismes Plus Non Nope Sort pas sur le genus que vous pouvez avoir ici. Et le constrain sur le genus, c'est juste que tous ces genus, g1, g2, etc., vous avez ce sume de i equal à 1 à l de gi, ce qui doit être equal à g minus k plus l. En fin, c'est tout ce que ça dit. C'est ce qu'on va avoir dans ce... Donc, ici, nous avons ce sume de gi equal à g minus k plus l. Donc, si nous voulons écrire la définition précise de la braquette, qui est ici, nous devons prendre toutes les possibilités pour expliquer ces choses k dans deux parts. Donc, c'est le sume de nos partitions. Donc, pour moi, c'est le sume de nos partitions. Le sume de la braquette est equal à partitions de k, de ces éléments, de z et le set, le set de sigma i et de z, quand sigma est à l'a. Donc, ce sont les partitions. Donc, partitions, c'est à dire que je n'ai pas l'ordre des sub-sets. C'est important pour les facteurs symmétriques. Donc, c'est le sume de nos partitions. Et maintenant, quand j'ai choisi des partitions, donc, par exemple, ce sera la première partie et ce sera la seconde partie. Quand j'ai choisi ces parts, j'ai dû décider de quoi je les collais. Donc, maintenant, pour chaque choix de partitions, j'ai deux... Let's say mu is mu1 muL. These are my parts. So, now I have to take sum over g1 plus, as I say, gL equals g minus k plus L. And sum over i1 iL must be equal to all my external variables. Mu equals to the vector mu1 muL. Mu is equal to those of partitions. So, it means that these are the parts. It's just notation to say that it's the parts. But how do you know that there will be L parts? Well, L is by definition the number of parts. That should be equal to this. Since sigma, the partition depends on the sigma. Sigma is coming from this cardinality k minus 1. In fact, no, the partition really depends on the cardinal of sigma. So, how it is equal to L? The L is the number of parts. L equals the number of parts of mu. So, L depends on mu. It's an L of mu. L of mu. So, I should write it this way, L of mu. L is a function of mu. Some partitions have one part. Some partitions have two parts. Some partitions have three parts. On that most, they can have k parts. So, now, we have a product from 1 equals to the number on all parts of omega gi. Now, we have the cardinal of the part mui plus the cardinal of the part of the set ii. Well, the main difference is that parts cannot be empty. Whereas sets cannot be empty. Parts cannot be empty by definition of a partition. But sets, here in the union of sets, the sets can be empty. You have mui and ii. Ok. So, that's it. With a restriction, and it's the same as before, no disk. You want to have never an omega zero one appearing in the right-hand side. May I turn to this case with many variables? P1 is distinguished. Yes. But at the end of the day, it's symmetric. Yes, in the end of the day, it's symmetric. So, that works. Ok, that's the definition. This is the definition of the topological recursion when you have higher order branch points. And the other one was just a special case of that one, when all branch points are for order one. Oh, ok. Yeah, it was a different place. P1 is distinguished. P1 is distinguished. Yes, indeed. P1 is distinguished. But in the end, the result is independent of how you distinguish it somehow. So, that's the definition for higher order branch points. And there is a very beautiful thing. Well, there is one easy way to obtain higher order branch points. So, an easy way of obtaining a higher order branch point is taking a limit of several simple branch points coming together. And so, we have a definition of the omega gn when we have only simple branch points. And now we take a limit where several simple branch points coalesce together. And then we have another definition of the topological recursion. Is the limit of the first one equal to the other one? And the answer is yes. So, that's a theorem. Limits of… Well, I'm not going to write it in full details, but simple branch points to higher order branch points. Well, in that limit, well, basically, what the theorem says is that the limit of omega gn equals the omega gn of the limit. So, omega gn is continuous. I will later say how omega gn's are related to integrals over mgn. And in that case, so, it's all related to the… To give formalism, I'm not going to enter the details, but basically, this theorem says that the ancestor potential is continuous. Ancestor potential. The total ancestor potential, that's what… So, basically, Milanoff used this theorem to prove that the total ancestor is continuous and that has lots of consequences. For instance, it allows to prove Faber-Pendari-Pendet for conjectures about our spin intersection numbers related to usual intersection numbers. So, because simple branch points will be related to intersection numbers and higher order branch points can be related, for instance, to our spin intersection numbers. And so, this theorem says that you can obtain our spin intersection numbers as limits of usual intersection numbers. Let me mention that this theorem is not trivial at all. Because, for instance, if you call epsilon the distance between two branch points, two simple branch points, each term in the formula seems to have poles in powers of epsilon. So, it seems that each term could diverge as epsilon goes to zero. But when you take the sum of all terms, in fact, all the poles disappear and there is a limit and the limit is equal to that. So, it's not trivial at all, but it's true. Non, non. Ok. It works if the coalescing branch points lead to a smooth higher order branch points. So, the spectral curve is still smooth. So, I did not write all the details. In fact, there is another theorem when it is not smooth that I'm not going to talk about now. But again, in fact, the topological recursion is always somehow well behaved under those limits. It always, in some sense, commute with the limits, except that when it's not smooth, it's after a rescaling that it commutes. But still it's, I mean, basically the limit of the topological recursion is always the topological recursion of the limit up to some rescaling. So, it's well behaved and it can also be compared to the Crippant Resolution Conjecture. Well, ok. But so, let me now go to something else. Yeah, I understand that you may be explicitly say, but can we explain this limit thing? Ok. Let's take the example of x of z equals z to v r z to v r plus epsilon z. Well, let's put epsilon to v r minus 1. Ok. dx, ok. Let me put minus r. Ok. Ok. When you take the, you see that when epsilon equals 0, you have 0 is the branch point of order r. Ok. And when epsilon is not equal to 0, dx of z is basically it's r r z to v r minus 1 minus epsilon r minus 1. Ok. So, which means that the 0's of dx are epsilon times the roots of unity. 2 pi i j over r minus 1. Am I right? Yes. So, so basically you have several, so you have here you have several roots of unity. Ok. Each of them is a simple branch point. So let's call them a j. Ok. Each of them is a simple branch point. When epsilon equals 0, you have only one branch point at 0, which is of higher order. Ok. So the question is, if you compute the topological recursion with that curve for epsilon non-zero, you use the usual formula for the topological recursion. You find some omega g n's. Do they have a limit when epsilon goes to 0? And the answer is yes, and the limit is precisely the one I computed there. And it's not trivial because each term can have negative powers of epsilon. For instance k, the kernel k has negative powers of epsilon. But it turns out that when you sum all the graphs and everything, which one is easier to compute? It really depends. No, in fact, well, the last one is probably easier to compute. Well, no, it really depends on your problem. Well, I mean, in the end you know that there is no pole in epsilon. But it's not so obvious from the definition. When you take the first definition, so when epsilon non-zero, each term has poles in epsilon. But when you take the sum of all points and so on, all the poles disappear. And there is a limit when epsilon goes to 0. It's not a trivial theorem. So let me now say something else. So now let me go to the third part. And I'm going to go into your business. So, this will be my third part, which I called, I called it A, B, C, D, T, N, S, which is really what Maxime and Yann introduced. So let me just make first a remark, which is that since the topological recursion is computing always residues at branch points, everything will be just in the end, combinations of the Taylor expansion coefficients of residues at branch points. And remember that sometimes we had some residues with some kA. OK, let me write for instance this way. And here we have a B of Z, Z2. We are going here to take a residue at Z, going to Z, very close to A. So we need to Taylor expanse, sorry, this was omega02. We need to do the Taylor expansion of this omega02 when Z is close to a branch point, but with Z2 arbitrary. So let's do this Taylor expansion. So let me take, so let's come back to the simple branch points. Everything could be done for higher order branch points, but in that case for simplicity. So the local variable near A, instead of, so Z was an abstract point on the curve, but so let's me give real zeta A of Z, which is just the square root of X of Z, maximum X of A. So this is a good local variable. I'm just a bit confused. You're taking A as a ramification point, not the branch point. Yes, ramification point, sorry. Ok. I should say ramification, so excuse me, I often make the confusion between the two. And just some image of the ramification. Yes, yes, yes. The branch point is the image of the ramification point on the base curve. You're right. But it's kind of abuse of language I make from time to time. So ok. So this is a good local coordinate in the neighborhood of a branch point. Just changing zeta to minus zeta. But so the idea is now we want to expand omega 02 in that coordinate. Omega 02 of Z, Z2. We would like to expand it. So in two powers of zeta, let me leave some space. Zeta A of Z to some power. D zeta A of Z because it's to some power. Ok, let me take directly the odd powers. No sorry, even powers. 2k. Times a coefficient. And the coefficient will be a one form in the variable Z2. And let me give the name to it xa Ak of Z2. So it's just the name of that coefficient. Up to a detail I like to put a minus sign. I like to put a 2 to the power k on the 2k minus 1 double factorial. Ok. It's just a... It will be more convenient for practical applications. Plus terms which you ignore. Yes, plus the odd parts. And which will play no role because each time because of a symmetry of k so somehow all the odd terms will always disappear from the end computation. Ok. So another way of saying that is that xa Ak of Z2 let's call it is the residue is simply minus 2k minus 1 double factorial over 2 to the k. Residue when Z goes to A of 1 over zeta A of Z to the 2k omega02 of ZZ2. So it's simply that. Yes, it must be 2k plus 1 or 2k minus 1. Yes, 2k plus 1. Yes, yes, of course. Yes. Ok. So it's just the definition. So by just the definition these are the coefficients in the Taylor expansion of omega02. And so it's obvious from the definition of a topological recursion that all what you will get in the end will be combinations of those coefficients just because that's the only thing residues can do. Yes, they have poles at ramification points. So in fact it's easy to see that xa Ak of Z behaves as 1 over zeta A Z to the power 2k times while there is a power Did I write it? So while there is a 2k 2k minus 1 factorial plus analytic at A. So basically there is a polar part times something like 2k minus 1 double factorial over 2k something like that plus analytic at A. So it means that subtracting that to that is holomorphic at A. Yes, and analytic everywhere. So that's the only pole. So it's a pole of order 2k and there is no other pole. 2k plus 2 I think it's 2k plus 2. So is it an assumption that there is no other pole? No, it's by definition of this the only pole of this can be at A. There can be no other pole. It's a consequence of that definition. So now let me say something which is kind of obvious but very deep so that let's call it a proposition. Omega gn of z1 zn is a certain combination of let's say, so we have pairs a1, d1 a2, d2 an, dn of so basically we will have a coefficient let me leave some space i equals 1 to n of xi ai di of zi so basically in each variable zi, we can have only those things. And the coefficient let me give the name fgn of so I take your notation a1, d1 a2, d2 so there exist such coefficients it's just because every raise you formula is just going to do that it cannot do something else now the question is what are those coefficients and if you write the topological recursion it gives a recursion among the coefficients it gives a recursion among the coefficients well now first let me say what are so for examples and so sorry another property is that the coefficients fgn must be well first this sum is finite it's easy to see that the sum is finite again because by taking raise views you always extract only a finite number of terms in the Taylor expansions and the sum is finite and the coefficients fgn are polynomials of the Taylor expansion coefficient of omega 01 and omega 02 and remember that I had written that omega 01 near a branch point a was something like was something like again I put some powers I think there was a 2 to the k let's I define it this way plus some k t a k 2 to the k over 2k plus 1 I think it was double factorial zeta a of k zeta a of z to the power 2k plus 1 times 2 zeta a of z d zeta a of z and I think there was a 2 here so basically it says that the fgn will be polynomials of those coefficients t a k's and polynomials of omega 02 of z1 z2 if I expand that in a vicinity so let me leave some space if I expand that in a vicinity when z1 is close to a and z2 is close to b let me subtract the pole delta a b d zeta a of z1 d zeta b of z2 over zeta a of z1 minus zeta b of z2 to a square ok, there must be some Taylor expansion coefficients some k and l sorry, and here there was plus of course even part and here also I'm going to write only the part that has a good symmetry k l and let's call the coefficients 2 to the k plus l 2k minus 1 double factorial 2l minus 1 double factorial b a k b l zeta a to the 2k zeta b of z2 to the 2l d zeta a of z1 d zeta b of z2 ok, so the fgn's so what we get is that the fgn's fgn equals a polynomial of the t a k's and the b a k b l must be a polynomial of all those variables in fact if you think about it of how the recursion worked remember that b b so which was omega02 appeared there was an omega02 for each non-arrowed edge in the graphs and the k was in fact it contained an integral of omega02 so somehow the coefficients of omega02 appear for each edge of a graph so in the end in those variables this is a polynomial of degree 3g minus 3 plus n which was the number of times they can appear and for the t a k's it's more complicated it can be a polynomial of well the degree of a polynomial in the t a k's is more subtle but let me write some examples well when you compute omega03 of z1 z2 z3 you can compute it using the recursion formula and the result c'est très simple it's sum over a t a0 x i a0 of z1 x i a0 of z2 x i a0 of z3 that's the final answer for omega03 for any spectral curve it's very very simple so what does it mean is that the coefficients f03 basically the coefficients are zero except so the only coefficients that are non zero are this one a0 so with the same a equals t a0 that's the only non vanishing coefficient in this polynomial ok another one that's interesting is omega11 so let me write the coefficients of omega11 and let me write just the coefficients that are non zero ad sorry a1 equals I think it's 1 over 12 or 1 over 24 sorry it's 1 over 12 t a0 and f11 of a0 is 1 over 12 t a0 t a1 plus t a0 p a0 a0 so that's an example you can compute it so which means that you can see what omega11 is but it's just that of course this 1 over 24 sorry 1 over 12 you're going to see it's an intersection number we're going to see that in a moment but just now let me say that the recursion so the topological recursion implies a recursion on the coefficients fgn and let me collectively note those coefficients alpha1, alphan where alpha is a pair ad ok but now let me say that alpha1 that the alphas are in a set whatever it is so it's the set they belong to a set of indices ok they belong to a set of indices and what is the recursion the recursion can be written in the following way so so there exists so basically there exists so there exists so the theorem there exists 4 tensors that I will call abcd these 3 will be rank 3 tensors and this one will have rank 1 ok and the definition is just so the definition of d is just d alpha will just be the coefficient f11 of alpha by definition of alpha1, alpha2 alpha3 will just be by definition the coefficient of f03 so that's the definition so for the moment the theorem says nothing and the subtlety now the interesting thing is that fgn now for 2g-2 plus n larger than 1 fgn of alpha1 alphan equals sum over let's call them beta and gamma of c of alpha1 beta gamma times on here we have our usual combination fgn of n plus 1 gamma delta alpha2 plus sum g1 plus g2 equal g on i1 i2 equals alpha2 alphan fg1 1 plus i1 beta i1 fg2 1 plus cardinal of i2 gamma i2 except that now I take in the sum I don't want f01 but I also don't want f02 so that's what I will call stable which means no disk and no cylinder so it's a stronger thing plus so this was this is the end of that parenthesis plus sum over sum from j equals 2 to n sum over beta of the tensor b alpha1 alphaj beta times fgn minus 1 of beta and you take alpha2 to alphan but you remove alphaj so this theorem is totally obvious from the definition and writing what c's on the b's are is is really I can give you the explicit formula c's on the b's basically they are computed as ratios involving k and b so they are just very simple combinations but the interesting thing is that it takes this very general form but now you, well now the question you could ask is 4 tensor abcd et apply this recursion, what does it give did you say why they are tensors well it just means they depend on 3 indices but that's a good question ok ok ok ok to make even tensor you need to say that they are the coefficients of some tensor corresponding to some choice of vector space so you have to build a vector space and all that I'm not going to enter the details but for the moment it just means that it's a function of 3 discrete indices just the coefficients of something when you choose and to say they are tensors you should say how they transform when you change the basis in that vector space and all that I'm not going to do that the experts are here yes some of them are up indices but I don't want to enter the details but I'm just going to cite your theorem I'm just going to cite your theorem which is that sorry so it's a tensor in a classical sense not in algebraic sense of extending the scalars I thought that no no no even to tensor product of some spaces ok but let me just say so of conservation soil ballman which is probably from last year no 2017 or 1617 or 16 I don't remember but it's very recent but so your theorem is that now if you define let's define the following function so let's so define a vector an infinite dimension vector let me call it x which will be which will be which has coordinates x alpha alpha belongs to the ok no sorry a is not a good set a good name ok so v indices alpha just belong to a certain set whatever it is and x is just a vector with coordinates x alpha I'm going to define a function of x which is exponential sum over g so it will be dependent on 2 things x on the h bar h bar to the g minus 1 sum over n so everything is in b exponential sum over n 1 over n factorial sum over alpha 1 alpha n x alpha 1 x alpha n fgn alpha n so somehow I'm just saying that I take the fgn as the Taylor expansion coefficients of some function so I just define a function from the Taylor expansion from its Taylor expansion I also introduce an h bar corresponding to the expansion in powers of g and I define that function ok that's the definition it really makes sense if the fgn are symmetric sorry it's formal it's a formal series ok for the moment it's just a formal series but the main theorem is that it is there are some operators that annihilate it and those operators are defined as so define the operators l alpha which is h bar d over dx alpha minus sum over beta on gamma so let me call a alpha beta gamma x beta x gamma plus 2 times b of alpha beta gamma x beta d over dx gamma so h bar d over dx gamma plus c alpha beta gamma h bar d over dx beta h bar d over dx gamma ok minus h bar d of alpha ok that's the definition of some infinite family of operators you have an infinite family of operators on the CRM then for every alpha l alpha applied to psi equals 0 well there is a kind of you can go somehow backwards so if you have a psi that is annihilated by those operators then then the coefficients fgn have to satisfy this recursion somehow but there are some subtleties which is that first is there a psi so and for that you have to say that the l alphas should satisfy some commutation reactions should be close on the lubrication sorry so basically one requirement is that the l alphas should satisfy a certain reaction but some over gamma and let me call the question f alpha beta gamma l gamma should formally algebra and in fact so now you can ask the question given four tensors a b c d if I take arbitrary four tensors a b c d and I define recursively fgn by this formula will I get something interesting the answer is in general no for just some simple reason if the tensors a b c d are completely arbitrary fgn will not be symmetric so so there is a constraint on a b c d and yes this is the constraint yes exactly this is the constraint on a b c d but but indeed you discovered that so the constraint on a b c d is that there is such reactions but again those reactions are not so trivial to solve and basically what are the tensors a b c d that satisfy those reactions well that's what structure but basically what I want to say is that it's not easy to find examples of a b c d s that satisfy this so for which the fgn are symmetric it's not easy to find some of them but you see that if we started the other way around from a spectral curve we always get something that satisfies that the question is somehow is there really something else that satisfies that and it's not totally clear at the moment what else there could be and I'm not sure yes but in fact we have many generalizations of topological regression which I did not talk about which go into the non-commutative realm and I think this could in fact be in it so and also I said that so far I was considering only cases where there is a finite number of ramifications points so the number of ramifications points was finite but if you allow it to be infinite with some grading something to give meaning to a sum I believe that then it could be this generalization what you are talking about in fact so it's not clear to me how different those structures are really and I think the answer is not known at the moment excuse me G starts from 0 so there is a G equals 0 to infinity so there is a first the first power is negative so exactly like you have usually in WKB expansions it starts so it starts typically with exponential 1 over h bar so which means that to really give a meaning for that as a power series applied by h bar to really give a meaning and then when you want to apply the operator somehow exponentials commute well with operators so you could rewrite everything in terms of what is in the exponential h bar is complex it's a formal variable so since h bar has a negative power in the leading term so this means that this equation is true as an expansion in power of h bar all the coefficients of powers of h bar give 0 so you cannot take h bar goes to 0 this limit doesn't matter you can't because you have h bar in the coefficients well there is a kind of limit h bar no what psi is divergent as h bar going to 0 psi is divergent but many quantities are somehow having a meaning as h bar going to 0 to go into deeper details about that but so basically this is the way this is an algebraic way of rephrasing the topological recursion so the idea is you can have some vector spaces transfer over it some of those quantum error structures it also leads to the same topological recursion but now I want to go to intersection numbers what is the rectangle of this? sorry well you should ask them no no, this is a wave function wait, you can if you introduce a little dagger you can write the number operation what is it? yes it's somehow via 0 oh yeah ok no I think it was here ok but now let me go to modular spaces sorry well I suggest that you pursue this discussion later I don't want I want now to move to something else so in fact the idea is that now we would like so the idea is omega gn we saw it in the case of Mimir Zarani was something like an integral over some mgn of something could it be the case in general and in fact our graphical representation will allow to say that yes it is and I want to say exactly I also show you those coefficients fgn in some examples for instance this f11 as a 1 over 24 so it's very suggestive that there are intersection numbers and in fact I also say that all these coefficients fgn are universal polynomials of the t's and the b's ok they are universal polynomials so you can so one way to compute them is take some special examples of spectral curves where you know what they are so you take your favourite examples and you compute them and it was so using that it was possible there is a theorem that I am going to write in a minute saying exactly what are those polynomials in terms of intersection numbers basically the coefficients are combinations with intersection numbers so but for before I want to introduce intersection numbers so I some of you know very well what they are but I want to be introductive so I'm so let's define mgn is the moduli space of Riemann surfaces of genus G with n marked points on modulo isomorphisms basically two surfaces with marked points are isomorphic if there is a map of the other which is holomorphic so basically this is a set of Riemann surfaces of some genus G and you have n marked points p1 up to pn ok of some genus and the marked points are labeled so it means that you in the isomorphisms I don't allow to exchange p1 with p2 for instance p1 is 1 as label 1 attached to it there is another space which is very useful so these are just smooth Riemann surfaces well it is known that this is a finite dimensional manifold but non compact of dimension and it has a complex structure and it's of complex dimension 3G-3 plus n ok sorry I forgot to say that assume that 2G-2 plus n is positive ok ok and the compactification has been defined by Deline and Mumford the compactification of this is nearly the same thing but now we shall allow surfaces to be non smooth p1 pn and they are what I call nodal and stable and so it means that now sigma gn can be something like that it can be made of several components even some of them can do that with some genus some of them can disappear and the the marked points ok let me put them like that ok some components can have several marked points some can have none sorry this is not stable ok now it's stable ok ok and the constraint so nodal means that there are nodal points ok you can always distinguish nodal points by saying that a nodal point somehow will become a pair of marked points I will use that and just so now these are the points p1 p2 up to pn and pour vous dire que c'est mgn vous voulez que la caractéristique totale soit 2g-2 plus n donc des summes de toutes les caractéristiques donc 2-2gi minus nombre de points spécifiques donc pour chaque i il doit être equal à 2-2g-n et stable signifie que chaque chose doit être négative donc chaque component doit avoir une caractéristique négative ce qui signifie stable donc c'est la stabilité signifie que chaque caractéristique de chaque component doit être négative et de toute façon c'est la complexion de cet espace c'est la fermeture de cet espace et il y a une topologie que je ne vais pas décrire et parce que c'est un espace très compliqué parce que ce n'est pas un manifold ce n'est pas même un orbifold c'est un stack mais je ne vais pas l'entraîner c'est un manifold mais il y a des compagnons de différentes dimensions non non non ce n'est pas le cas comme mgn ok mais il y a des compagnons de différentes dimensions c'est un mgnon de 35 oui c'est un mgnon de 35 ok ok c'est juste que je ne suis pas familiar avec le nom mais je veux juste définir le bandel cotangent le bandel cotangent donc sur chaque je vais définir le bandel li sur mgn ou mgn bar je vais définir ce avec i equals 1 à n c'est le bandel qui est fiber c'est le espace cotangent donc fiber sur un point sigma gn p1 pn c'est le espace cotangent sigma gn pi donc sur chaque point pi à chaque point pi il y a un espace cotangent qui est un espace dimensionnel donc localement c'est isomorphique donc c'est un bandel li c'est un bandel li c'est un bandel li c'est un bandel li ok c'est un bandel li et puisque c'est un bandel li vous pouvez compter une classe charme et il y a seulement une classe charme c'est une classe charme la classe charme c'est psi i c'est c1fli et c'est la 2 formes sur mgn bar c'est la 2 formes et vous pouvez l'intégrer sur les cycles de mgn bar mais depuis que c'est la 2 formes vous pouvez l'intégrer sur les 2 cycles si vous voulez l'intégrer sur les mgn bar vous avez une bonne dimension donc vous allez définir la définition de la nombres intersection donc il y a une notation mais si si c'est d1 plus dn c'est 3g-3n et maintenant je vais toujours écrire dgn c'est 3g-3n c'est plus fort alors vous définissez vous définissez donc c'est juste une notation tau d1 tau dn index g c'est une notation c'est l'intégration de la mgn bar de psi 1 d1 psi 2 d2 psi n dn c'est maintenant une forme de bonne dimension c'est une forme qui a une dimension correcte pour être une forme maximale vous pouvez l'intégrer et c'est typiquement c'est un numéro et c'est typiquement un numéro rationnel il s'agit de q et c'est même un numéro positif je ne sais pas q plus ok et donc ces sont les numéros et ils sont appelés les numéros de l'intersection et c'est seulement une définition et juste un remarque si d1 plus dn n'est pas equal à 3g-3n vous devrez les définir comme 0 donc c'est très convenu c'est-à-dire que chaque somme où la condition de dégrise n'est pas satisfaite va disparaître de la somme ok, juste pour vous donner quelques exemples qui sont connus tau 0, tau 0, tau 0 0 est 1 tau 1 en genus 1 est 1 à 24 et ainsi ok, il y a ces numéros et ils sont reminiscent de la coefficient f03 11 que l'on avait et je veux dire donc juste un remarque que ce sera la classe capa memford donc définition de memford classe capa donc vous voyez que il y a un map forgetful de mgn plus 1 à mgn bar pi ce qui est le map forgetful ce qui signifie que vous n'oubliez pas la last mark point ok si vous portez la classe psi n plus 1 à mgn plus 1 sorry donc à mgn plus 1 vous pouvez le pousser vous pouvez le pousser à cette classe et c'est pas ce que je vais dire c'est pas correct mais c'est la définition de la classe capa à moins dans chaque intégral c'est ce qui va arriver donc l'intégration de la position de la last mark points sera équivalent à l'intégration de la classe capa k et c'est une forme 2k donc capa 1 est une forme 2 capa 2 est une forme 4 et capa 0 est le numéro et en fait capa 0 est la caractéristique donc c'est utile pour nos propositions une autre chose qui est utile c'est c'est tout de même ce que cette picture c'est ici mon but c'est d'écrire une même forme mais j'ai besoin d'ingrédients vous voyez pour chaque surface de nodules il existe une sorte de représentation graphique de surface de nodules donc on s'appelle le graph dual le graph dual sera un graph pour lequel les compagnons deviennent vertices et les points nodules deviennent agis donc ici j'ai 1 component, 1 component, 1 component j'ai p1 p2 ici j'ai 1 point nodule 1 point nodule ici j'ai 2 points nodules par rapport à ces compagnons ici j'ai 1 leg extérieur et 1 leg extérieur et je vais récolter la genus de chaque component donc ici c'était 0 et il y a 4 des points spécifiques donc on peut dire que c'est 0,4 ici c'est 2 et 1 point spécifique c'est 2,1 c'était 1 avec 4 points spécifiques 1,4 et ici c'était 0,3 mais c'est tout de même les compagnons de MGN donc les compagnons de MGN les compagnons de MGN sont dans la barre de MGN et les compagnons de non, ici je parle de la boundary de MGN dans la barre de MGN mais MGN est ouvert dans la barre de MGN ok ok, je suis un petit peu je suis très introducteur ici je veux juste dire que je vais dire comment vous allez dans la boundary vous pincez un cycle vous créez 1 point nodule donc ce sont les 1 dimension 1 boundary ok même les boundaries ne peuvent pas être connectées c'est une boundary c'est complexe que vous avez mentionné c'est une boundary ok je veux juste être très sketchy à ce niveau je veux juste dire que je veux write this thing that delta MGN whatever that means it's basically MG minus 1 N plus 2 plus sum of G1 plus G2 equal G sum of N1 plus N2 equals N of MG1 N1 plus 1 MG2 N2 plus 1 ok ok, on stable sorry ok ok, je veux juste dire that there is a map in fact in bad space sorry sorry, sorry, not that sorry, it's just a product ok this is union ok on sum here means union also ok, I just want to say you have certain classes of boundaries that are identified with either basically with creating a nodal port and let me call that so here I'm not going to really to talk about boundaries but classes of boundaries so corresponding to those different strata so it's just so I'm not so familiar with that notation but I know how to compute the intersection numbers corresponding to that so I will call that map L basically and in fact I will go from the classes of that into classes of that ok ok what I just want to say is that in practical computations I just, basically I just want to explain the notations and let me state the CRM because it's time very soon let me state the CRM and I will and I will first say what is the CRM for when there is only one ramification point on our spectral curve and then I will write the CRM with an arbitrary number of ramification points so k is s so which was my sigma x omega 0 1 omega 0 2 has one ramification point and that is simple and let's call it a ok let's take this simpler example which was the case for MIRZARANI we had only one ramification point so the CRM which is by me in 2010 and soon after I did the general case of arbitrary number of ramification points but the CRM is the following that omega gn or if you want the coefficients fgn sorry fgn a d1 a dn is an integral over mgn is 2 to the dgn times integral over mgn bar ok, sorry let me write it as intersection numbers so there will be an intersection number product from i equals 1 to n of psi di those are this di's times a certain class and what is that class that class I write it sum over k of t hat ak kappa k exponential 1 of those boundary classes sum over k and l of ba k al of the image by this l of the psi corresponding to another point so ok, let me write it this way k psi p minus l ok I will explain the notation so basically it's a kind of integral over mgn bar of some class but we just need some explanation of how you compute this and if you want to recover omega gn remember that omega gn of z1 zn is just the sum of our d1 dn of fgn dn sorry, a d1 adn of kad1 of z1 kadn of zn it was just that and remember that this were the coefficients it was some other letter sorry it's psi i to the di yeah but but k was some other letter the coefficients of omega 02 the Taylor coefficients of omega 02 no, zeta was the local variable and I think psi was the so it was omega 02 of z1 z2 was the sum of zeta a of z1 to the 2k d zeta a of z1 and the coefficient was not psi a k of z2 it was psi, oh sorry sorry you're right so just one thing for the moment I had defined sometimes ta ks which were the coefficients of zeta to the 2k plus 1 so my omega 01 was basically 1 over ta0 zeta plus sum over k 2 zeta d zeta ok but you see this is different names here I have ta k and here I have the hat a k these are not the same but they are nearly the same I would say it's just transform but the idea is that if you take the Laplace transform of omega 01 of z e to the minus ux of z ok basically if you expand this integral in 2 powers of u you will recover the ta ks there are some 2 to the k over 2k plus 1 double factorial or something like that but you see the Laplace transform because somehow it will kill this denominator 2k plus 1 double factorial but so if you do this Laplace transform so just in order to make sure that it's well convergent that it has a so let's kill the so 2u square root of u over square root of pi I think if you take that now so it's nearly Laplace transform and here you will integrate over something which is the x minus 1 of x of a plus r plus ok so a certain control basically you extract the control on which x of z minus x of a will be positive so then the integral will be convergent if you expand that in 2 powers of u basically you recover the ta ks but if you expand the log of that in 2 powers of u you recover those coefficients so let's expand so sum over k minus sum over k of t hat k u minus k so that's the definition of those coefficients t hat k and it's very easy to compute and those coefficients I had already defined and remember that it was basically omega 0 2 of z1 z2 was basically d zeta this way d zeta 1 d zeta 2 over zeta 1 minus zeta 2 to the square plus sum over k on l of 2 to the k plus l over 2k minus 1 double factorial 2l minus 1 double factorial those coefficients b a k a l zeta 1 to the 2k zeta 2 to the 2l d zeta 1 d zeta 2 ok ok, let's l delta c'est d'actualité quand on le fait ok, je veux mais c'est exactement... je vais donner des exemples mais je vois que c'est terminé je veux juste pour appeler ça dans le cas de Mirzharani pour montrer que ça implique Mirzharani's recursion qu'est-ce que 1 par i i est square root of minus 1 où est i ? i e to the power 1 by i sigma over delta e to the intersection psi i to the i n'est pas square root of minus 1 c'est 1 by i, qu'est-ce que c'est ? n'est pas square root of minus 1 n'est pas square root of minus 1 n'est pas square root of minus 1 donc ici c'est 1 by i ok ok, je vais je vais le dire en un moment c'est 1 by i c'est l'image donc c'est tout de même c'est juste que vous vous poussez oui, c'est le boulot et c'est le boulot correspond à un point nodal un point nodal qui est identifié à deux points sur le boulot de boulot et vous pouvez prendre les deux classes de psi correspondant aux deux points donc ça juste signifie que vous poussez au boulot de boulot à l'image de l'image de ce boulot de boulot donc c'est juste l'image de ces classes de psi associées à les deux deux halves du point nodal pour faire une expansion et en fait cette expansion signifie que vous devez faire les summes de tous les geographes possibles de toutes les surfaces nodales donc quand vous expandez l'exponential au premier ordre il y a un au prochain ordre il y a basically un point nodal au prochain ordre il y a deux points nodales puis trois points nodales donc vous ne pouvez pas en avoir plus que 3G minus 3 plus N non deux points et oui, un autre remarque que ce que vous avez dit c'est que la fgn d'A1 D1 dA1 Dn est 0 si D1 plus Dn est plus grand donc c'est ce que cela implique cela implique que seulement finitiquement beaucoup de ces coefficients sont non-zero c'est ce que j'ai dit c'est un polynomial seulement finitiellement beaucoup d'eux sont non-zero記得 que beaucoup de ces intersection des numéros quand vous ne respectez pas la condition tous ces intersection des numéros donc en fait presque tous les termes dans cette somme sont acceptés mais je vais faire l'exemple de Nier-Zarani si la condition de dimension est satisfaite la condition de dimension est 0 alors vous avez des points oui alors il n'y a pas non la condition que vous avez dit c'est la condition de dimension des vis des vis des degrés de toutes les formes qui apparaissent dans la somme doit être la dimension de la dimension de la sphère de Mgn doit être equal à Dgn donc quand la dimension de la forme n'est pas la dimension de la sphère sur laquelle vous voulez l'intégrer vous avez 0 c'est ce que je veux dire ok alors, let's take an example donc c'était donc ma courbe ma courbe spectrale c'était x of z c'était z square et y of z c'était minus 1 over 4pi sorry 4pi sin 2pi z et omega02 de z1z2 c'était juste dz1dz2 1 over z1 minus z2 c'était square alors, en cet exemple le coordinate local zeta est vraiment equal à z en cet exemple zeta, parce que vous vous souvenez que zeta z c'est square de x z minus x a et ici en cet exemple vous avez a equals 0 c'est juste z donc ça veut dire que cet exemple c'est précisément ce que vous vous subtraque quand vous faites cette expulsion donc ça veut dire que dans cet exemple vous avez b a k b a l c'est equal à 0 pour tout k l donc ça va de toute façon, vous ne serez pas concernés par ce terme en cet exemple mais maintenant, ce qui est intéressant c'est de compter ces questions t hat k donc, quand vous voulez compter ces questions t hat k vous devez prendre une ligue de e minus ux y dx et vous devez intégrer donc, basiquement x z y z dx z et vous devez intégrer z juste sur r ok juste un remarque c'est pardon r, r dans la variable z c'est notre plus en vx variable donc juste un remarque c'est que e minus ux y dx c'est juste equal à 1 over u donc c'est juste intégrer par part ok, c'est convenient donc en fait ce que je veux dire maintenant c'est d y z et square root of u e to the c'est juste absent, parce que x of 0 x of the branch point is 0 donc, vous devez compter ça ce qui est 2 square root of u over square root of pi intégrer e to the minus u z square et ici c'est minus donc, vous avez minus 1 over 4pi mais quand vous avez la délaration de ça vous avez un 2pi qui arrive donc minus 1 over 2 et donc vous avez un cos 2pi z ce que je vais faire d'expansion 2pi z plus d'expansion minus 2pi z d z et il y a 1 over 2, donc 1 over 4 donc je vais simplifier minus over 4pi si je suis correct ok et et bien, c'est très facile pour compter cette intégration c'est très facile et à la fin, ce que vous trouverez c'est qu'au 1er, c'est 2 fois la même intégration donc c'est minus square root of u over 2 square root of pi fois et basiquement, vous avez minus euh euh 1 over sorry, minus 2pi square over u c'est ce que vous trouverez euh, sorry minus pi square over u non, sorry, il va être minus pi square root of u et basiquement, c'est times square root of pi over square root of u donc ce qui sera minus 1 over 2 minus pi square over u donc ce qui signifie que ma t hat a1 est pi square et ma t hat a0 est la log de minus log 2 basiquement tout ce que je veux considérer c'est minus 2 mais donc, qu'est-ce que le théram est un petit peu tard qu'est-ce que le théram dit, c'est que omega gn z1 zn est summe d1 dn product de xi ai ai di mais vous pouvez vérifier mais c'est la même chose que 2di minus 1 euh, sorry, plus 1 double factorial d zeta ai over 2 to di zeta ai to do di plus 2 times et ici vous avez psi 1 to d1 psi n to dn times exponential pi square capa1 et vous avez, sorry et vous avez, vous avez 2g minus 2 plus n et on a minus 2 euh, sorry 3g minus 3 plus n et minus 2 pour la puissance donc je dois avoir fait le erreur 2g minus 2 plus n euh ok, si vous êtes attention avec les powers de 2, en la fin vous recouvrez donc c'est la classe de vile peterson et vous voyez, en la fin ce sont les volumes hyperbolas donc ce sont les volumes hyperbolas donc, c'est-à-dire que ce CRM prouve que les volumes hyperbolas sont satisfaits par la récution topologique et en fait, nous avons procédé de l'autre côté nous avons dit que la solution de la récution topologique est les volumes hyperbolas en utilisant le CRM donc ça prouve le Mirzarami je n'aime pas cette prouve parce que la partie la plus difficile de la prouve c'est d'occuper la transforme de la function sin et si vous appliquez la même chose pour la function Lambert vous trouverez donc, de nouveau, vous devez compter la transforme de la function Lambert et la transforme de la function Lambert est la fonction gamma la exponentielle donc la expansion Taylor donc la expansion asymptotique c'est la fonction de la log de la fonction gamma c'est la formulae sterling elle intervient les nombres Bernoulli et avec presque aucun effort vous trouverez la formulae ESG pour avoir ces nombres et la partie la plus difficile de la computation c'est d'occuper la transforme de la function Lambert et vous trouverez que c'est la fonction gamma donc j'aime cette prouve et aussi, si vous appliquez le Mirzarami du Calabia Manifold C3 vous appliquez le Mirzarami C3 de nouveau, vous compliquez la transforme Lambert et ce sera la fonction beta Euler la fonction Euler beta c'est un produit de 3 fonctions gamma le Mirzarami C3 c'est l'équation e2vx plus e2y plus 1 c'est 0 alors, vous devez prendre un framing plus fy un framing f et si vous compliquez la transforme Laplace donc maintenant compliquez le minus ux y dx avec y et x par cette formule c'est très facile de compliquez et c'est la fonction Euler beta donc c'est quelque chose de gamma u gamma f u par gamma de 1 plus f u quelque chose comme ça qu'est ce que vous parlez de c3 c3 et c'est un mirar c'est un mirar c'est ça oui, c'est quelque chose de bien con je ne vais pas entrer dans les détails mais le plus difficile de la compétition c'est de compliquez la transforme Laplace c'est 3 fonctions gamma c'est 3 fonctions gamma et si vous expliquez ça par la force de l'U vous avez des nombres b2k par 2k 2k-1 x u-1-2k et depuis qu'il y a 3 vous avez 1 plus f 2k plus 1 minus 1 plus f 2k plus 1 et ça se termine maintenant si vous je pense que c'est un minus si vous êtes attention mais maintenant ça se termine que vous devez prendre l'exponential de summe de b2k 2k 2k-1 et c'est suffisant vous répliquez la u par la classe kappa ok alors il y a plus d'un demi de ces déviseurs en fait il y a quelques autres termes je n'ai pas besoin d'interpréter les détails mais c'est tout de même ce n'est pas le cas et ce qu'il dit c'est que si vous regardez le mirroir de c3 vous appliquez le CRM ce que vous avez c'est les intégrations c'est le formula donc on a le CRM mais dans le même CRM vous avez la recursion l'esg formula et vous avez beaucoup d'autres choses et la plus difficile part de la compétition que vous avez c'est de la forme de la plateforme de la fonction je trouve que c'est magnifique donc la prochaine fois je vais vous montrer ce qu'il donne quand vous avez plusieurs points de branche je vais vous expliquer ce que ça veut dire mais je n'ai pas besoin d'interpréter le temps est terminé merci