 Yeah, okay, so thanks to the organizers for inviting me for this so so there are three lectures So they they kind of split in the following way So so the first one will be churned Simon's theory for over for real group and I will only do as you to and So, you know in this case here. We really know what this means This is the written reciting to ray of quantum invariance of three many folks right and so in two I will do new formulation of the Teichmuller TQFT so This is the worst off. I've done with renate Kashaev and then three will be finite finite dimensional integrals and Then that will to go towards turn Simon's theory for complex Relie group so for SL2C and so the integrals that we saw this morning in Maxim's talk they will appear here and For the quantum for day of dialogue will appear here So there will be no quantum for day of dialogue today, but if you're here tomorrow, you will see it all over the place Okay, so let me try to do this Witten reciting to rave Quantum invariance of three many folds and so here there will be three many falls all over the place So so I'm going to choose a link inside as three. This is a framed Oriented link Okay, and so from this I can do surgery and I get m3. This is a closed Oriented Three many fold frame means that it has a trivialization of its normal bundle Okay, so it's like a little banded link And now I'm going to pick our which is going to be positive integer Okay, and then I'm going to look at the following labeling So I'm just going to introduce little notation L C L R This is simply just all the maps from the components of L into the following label set So I'll explain why I do this in a second And so I think that in Maxime's talk at the very end capital N is actually a little R here okay, and So now this quantum invariant introduced by reciting to rave Is a really it very given very explicitly it's e to the pi i3 r minus 2 over r Sigma of L. I'll tell you what sigma of L is in a second square root of 2 divided by r cinch pi over r All of that to m plus 1. I will tell you what m is in a second also It seems to just let the H Where we're right since I'm so I'm so I'm not so I'm not singe but sign. Did I say singe? Okay, sorry What's what's in the left hand side y r tau r? Oh, so this is tau r of ML Yeah, so I use exactly notation was taken to a few yeah Okay, yeah, well, that's the part of this talk is about reminding you of all of that So this is something over this label set. This is a finite set of course There are finitely many components m number of them here is this label set one up to r minus one You take this Quantum lambda times the Jones polynomial colored by lambda. So I'll define all of this now Okay, so I move the board a bit up Yeah, coming on no so sigma of L is the signature of The linking matrix So if you just hold all your questions for a few seconds, I'm making a list of the things I didn't define Okay, so then we So this is the signature of the linking matrix M is the size of pi zero of L so the number of components of L and Lambda here Well, lambda is really a vector if I ordered the components it would be lambda one up to lambda m So this is just a product if you like when I use this notation over the components of L And then I take The quantum integer for that integer and what is the quantum integer? Well, this is Q to the N half minus Q to the minus in half Divided by Q to the n half minus Q to the minus in half and here If you like when I'm talking about this because this here supposed to be a complex number right So then Q is E to the 2 pi I over R So this specific rotor unity here and now J L or J lambda of L This is the collared Jones polynomial of L And actually this guy here It lies in C Q Q inverse where we actually can think of Q as a formal parameter But of course we can insert this value for Q in it if we like Okay, so this is the this is the quantum invariance and of course now I will tell you what the what the what the Jones pulley color Jones polynomials are but first I just want to make a little remark that of course what you can think of here is that You know you have the boundary two torus of every component So take the link remove the link from S3 take the boundary that boundary is of course abstractly a disjoint union of M to Tori and This whole theory here associates a vector space to that right So you know what I mean by these M Tori these are just M components And so abstractly they are just copies of the two torus as the boundary of the link right of the complement of the link Okay, and so the TQFT rule simply says that this here is just the M's tensor power of VR and VR is nothing but the span of lambda running in this Set and then or say lambda zero just not to make trouble with the lambda upstairs So there's a vector space spanned over C of with with with a formal basis given by the these labels and So so that's the that's the vector space that this thing is really going on in and then I have some wave function Which is just the sum over these lambdas in 1 up to R minus 1 again and now to the M's power and this is just the product of the quantum integers like this and Then I have lambda 1 sensor up to lambda M Okay, so there's a formal vector or sorry not a formal there is a vector in this vector space given exactly like this and You see the and then there is Another vector So this is kind of a standard vector and this vector here. That's the complicated vector so that's Again R minus 1 to the M. And so this is where you just take the color Jones polynomials And you make those the coefficients of another vector and then this tau M is Is just the inner product or some bilinear pairing in fact Between these two vectors here Okay, and so you should think of this guy here as the you know the wave function associated Suspended this kind of like stupid And you forget about pre-factor Yeah, oh, sorry Yeah, pre-factor there. Okay. Mm-hmm, but essentially I essentially it's like this So maybe the pre-factors that trivial function of R that and So this is the wave function associated to L lambda granted inside The moduli space of this torus T2 here, right? And so this thing here is nothing but C star to the power 2 divided by plus minus 1 to the mth power and So this year this year is the Lagrangian of flat SL to see connections that extends over the complement of L And so you think of this as the wave function associated to that and This one here is likewise the wave function Associated To Well this I can't I just want to make it very explicit. So if I here look at This torus which is the boundaries I draw a copy of it And so here there is a little disc that hits the link in exactly one point And so I call this curve the meridian and then if I look at a another curve Which is the longitude L then I have a longitude in the meridian and they have their holonomies capital L and capital M if you like and then L here is simply just These L comma m's such that L is equal to 1 So L and M are coordinates in C star power squared 2 And so there is a wave function for that Lagrangian somehow and there's a wave function for that Lagrangian somehow And then you are taking the inner product of those two So that's the kind of TQFT rules, but notice actually that complex connections here surfaces because I talked about it this way Of course inside this you also have the SU 2 moduli space And that is just u1 square divided by plus minus 1 I mean provided by plus minus 1 means you invert both of them simultaneously to the power m And so this guy here Sits inside here, and of course you can restrict the two Lagrangians if you like so these And in some sense there it's quantization of those two that are these two wave functions, right? So this is a classical surgery picture for our trans diamonds theory But I just wanted to sort of show this picture because very reminiscent of what Yan talked about and one box talked about So, okay, that's good What actually there is a whole I mean to completely make the analogy Maybe I can also say the following since this was sort of also featuring in the previous talk And so this is really the AJ conjecture Which is due to first of all gukhoff and then also to Garofer Lidius and Tang Li and So that says the following namely that this Yeah, so one way I can somehow keep track of all of these Jones polynomials if I want to I Can introduce j series of the link L simply just assembling all of these So I some of all the lambdas now all the lambdas going into all positive integers I take these guys here, and then I take t to the lambda so this thing here lies in q inverse q and Then power series in t1 up to tm okay So it's all I just record all of them like this and then this this thing here says that the series Of L is q holonomic Basically saying that there are recursion relations that these things satisfy and that they are partially determined by those and so I can In just I mean since it's really close to some of the things the maximum was saying So let me just look at q and e that acts on This set here, and I'm going to do things for naught Because it's just a little faster to do There's only one variable t here. There are variable per the component, and so what does q do? So q on such a sum of qn Sorry of a n of q t to the n. Well, it just multiplies by q to the So this is an a it just multiplies by q to the n on the nth guy and E just multiplies by t. So we of course observe that they q commute So it's a little copy of the quantum torus that we have acting on this set here and What it says is that so the real statement here is so so so that that's a theorem So there's a theorem due to Gaupholytius and Lee and it says that if you look at the annihilator of the J series Then that's non-trivial okay, and so Well that that allows you to actually define So so Star was defined an a hat Which depends on q and q and e and so this guy here Well, you have to invert e to make this all work out very well But so I don't really want to get into the details of that, but it actually lies here and then This guy is exactly constructed in such a way that if you apply so there's depends on the k and now K is my L So I'm taking the case where you know L is k where this is a knot So it was only one component right now. I'm formulating this form and then a k of q capital E Q and e applied to the J series is zero and so the the AJ conjecture just to state that which is Which is due to the same people here to to Gukov and to Gaupholytius and Lee It says that the following that if you take a k hat of 1 comma q comma e Then this thing here is equal to at the level of detail. I'm doing it right now q comma e to the minus 2 Where what is this well Lk the Lagrangian's corresponding to case of the flat connections that extends over the knot complement That's exactly given by the zeros of That a classical a polynomial. There is some classical a polynomial associated to a knot And it's of course in the two monodromes L and M around meridian Longitude and meridian and the zero set of this is exactly the gronction for that knot And so it says that you know you can actually recover the ak right if you have this So there is a way to make this unique in a specific way and Then it gives you this conjecturally this has been tested in a number of cases where we know that this works But it's still open this conjecture But it's very similar to the sort of recursion relations the q difference recations you were talking about I forget what you call the two variables That doesn't matter right now, but it wasn't x and y. Okay, so so here they're called q and e So here you're implicitly working with particular choice of framing That's how it's usually stated whereas previously it was framed link. Yeah, Jay was including implicitly the primary factors, so yeah Yeah, so I don't want to get into that detail right now. I mean I want to just show the principle of things Yeah, okay, so because I will not really use this part of the story But I just wanted to show that it links very well with the two stories you've been talking about Yeah, I think there are versions of this for for you know arbitrary simple simply connect the league groups But you see the so so far the whole story is paralleling very much what we've seen in Jan's talk and Maxime's talk By the way, of course, like what I should have said maybe maybe I can do this on this board here Namely that this tau r of m l well This is supposed to be equal to the integral over s u 2 connections modulo s u 2 gauge group and Then e to the minus 4 pi squared and then The r minus 2 divided by 2 pi I Well, maybe I write it like this churned Simons of a divided by h bar d a where h bar is 2 pi I Over r minus 2 so 2 is the dual coxidon number here and you adjust r down by this And then you have this and of course we don't really know how to make sense of this But that's the fantastic thing about churned Simons theory because we have an exact formula for what this is by the reciting to Ray formula Okay, so therefore what's missing now is for me to give you the John color Jones polynomials then you have the invariant exactly Okay, so now I will define the color Jones polynomials for you So maybe for now I leave that Where they It's r minus 2 Because r is the size of the rotor unity, and that is k plus the dual coxidon Yeah, yeah, this is the definition with transcendence is well defined up to one and then you get exactly this Okay The pencil how you write it down, but okay, so so the thing is that in order to To give you the color Jones polynomials What I have to do is I have to introduce certain vector spaces and these are really you know Representations of the corner group group at this root of unity, but I will give them totally explicitly for you So I'm going to take an in that's in this label set 1 up to r minus 1 and Then I will introduce Vn so this vn is not the same as the vr on the other boards that you can't see right now, so it's good This is a representation of the corner group, right? And so this is span of the following guy explicitly. So this is He n Then minus n twiddle and so for this n here and twiddle is the following it is just So to n twiddle plus one is in so it turns out as a this convenient way of organizing it just to you know minus one half and So then it goes like this it has a basis Like this all the way up to en mine minus Sorry n twiddle minus one en and twiddle So it's a complex vector space like this with this explicit basis and so of course I introduce the indexing set n Which is just these indices Plus one up to n twiddle minus minus one comma in twiddle Okay, and so then the most important guy in the story is really this following braiding isomorphism So the braiding isomorphism is just a C So this is a morphism, this is nicer morphism from v tensor v up n to sensor v up m to the thing in the opposite direction like this and So I can write it out of course in coefficients So this is just a I mean C and m of e and I Sensor C mj. I can write this as sum over v and w and then C v w i J and then the corresponding thing on this side here m v sensor e and W right so I just wrote out a matrix form for this guy now. I have to give you the matrix coefficients for this guy and So the matrix coefficients, maybe I put them over here So that goes to there so c n m So that's given as follows So it's something like Q or not something like but it's exactly like this Q to the I minus w and Then it's a n twiddle plus w quantum factorial m twiddle minus V Quantum factorial and then it's w minus I quantum factorial n twiddle plus I Quantum factorial and m twiddle minus j quantum factorial factorial times e to the pi I over 2 and then Q plus of I j V w and this quadratic expression here Q plus of I j V w that is simply 4 I j minus 2 w minus I I minus j Minus w minus I squared Sorry, the efficiency denote by series for it is it's not CNN. I am sorry. So of course, this is I j V w. Thank you and very importantly. There is also a delta function here That says I plus j is equal to V plus w Here it's C and m Sorry, this is the typo Thank you. It's like this And what do you do this till this is I remind you is Just the not to write n minus 1 divided by 2 So so it is exactly like this Okay, so this is the crossing the braiding isomorphism given explicitly in coordinates and Then there is a little requirement Namely the following n twiddle should be great greater than so w should be greater than or equal to I Should be greater than equal to minus and twiddle and a similar one for m and twiddle is greater than equal to j We're going to be greater than equal to minus and twiddle So this has to be satisfied Okay, and so I can just quickly do the other one too So the inverse can be written in a very similar form so I mean in principle I can make it an exercise here at the school find the inverse of this That's a little Not straightforward. I think but but you can maybe guess it. So Do you want to see it? No Okay So you just ask young For the inverse if you can't find it yourself And so of course, what's the main idea now? Well, the main idea is that every time I see a positive crossing What I'm going to do is I'm going to have sun in and some in Which are the labels of the strands which were called lambdas before Okay, and then I have indices on all the edges I Jv W so I'm going to some of all edges indices And then of course this guy here gets assigned this C and M Okay, this is the exact guy, okay? Okay, so Now what I want to do for reasons that will become clear is I will actually define a representation of the right group With this and then I will close off braids and get links this way So Yes, so the whole structure the quantum group gives so the representation theory of the quantum group at this brutal unity is a modular Tensor category, that's the language of to rave and Every time you come with a modular tensor category. You can build this tqft So is this formula equivalent to like new the quantum our matrix of a quantum SL2 or something? Yes, it is, you know that if you take the universal one and apply There was representations at root of unity then you get this Isn't it the So with root of unity wasn't it Carrillo fin and chicken Okay, so what I want to do now is I want to of course build the obvious thing So I'm one of I built a little category Now so it's not the category you just asked for young But it's just the the category where I look at you know the The group point actually it's the it's a great group point where everything is labeled so objects in this category It's just the finite strings lambda one up to some lambda m say Okay, and these lambda i's they are of course in my label set. So now I go back to the lambdas So objects are like this and morphisms They are simply just so a harm from lambda one up to lambda m Come a lambda sigma of one where sigma is a permutation Sigma of M So it's a home from the first object to the second object. This is simply just pi inverse of Sigma Where pi is the usual map from the braid group Bm? So this is the braid group on n strands to the permutation group on n letters So meaning all the braids that send this label set to that label set Okay, so example you have a lambda one you have lambda two you have lambda three sorry and You have maybe something like this And then this one goes down like this and then here's lambda three Here's lambda two and here's lambda one. So that's the permutation just a cycle one two Okay, so of course, this is a group point and So now what I want to do is I want to to define the Jones representation of this group point To the usual category of vector spaces thought of as a group point So I look at vector spaces and I look at isomorphisms between vector spaces Okay, and so I can actually just tell you exactly on the let's do this figure here What am I going to do? Well, so of course what I have to do is up this braid here labeled like this. It sent gives me An isomorphism from V lambda one since a V lambda two since a V lambda three to the one down here and so what I do is I You know, there are there are crossings here, right? So this is actually minus This here is minus and this here is a Minus they turned out to be all minus Sorry, that's wrong. This is a plus, of course So there are two minuses and two pluses and then what I do now is I label say this So what I do is now I label all the edges of the diagram that I have So first of all, I have to label the top one what I choose in J1 J2 and J3 now I further labeled them by indices of the corresponding Representations, right? So I'm going to give you matrix coefficients and then down here. I have I1 I2 and I3 Okay, and I'm thinking of going upwards in the composition and so therefore now I have to label the internal one V1 V2 and V3 and So of course this one here now So JB applied to of lambda one lambda two lambda three applied to the braid B So if this is B You just sum of all the internal ledges labels. So you sum V1 in I lambda two V three, sorry, maybe it in I lambda three and V2 in I lambda One I hope it's right. Yeah, and then of course C minus C minus C minus you can fill in all the indices yourself, right? And so that's the that's the morphism on braids So this gives a very nice, you know functor between the two categories very trivial Although it's not there a non trivial part. Of course is that Oops, you know all that there is, you know this braid relation, right? so I wrote the left-hand side of one of the braid relations and So this thing here you just take the strand behind and you move it across to the other side And so that actually turns into you know the Yang-Baxter equation for these C's here Okay Sorry, okay So that's that's very nice very easy and straightforward Once you got this combinatorics going right, I mean by no means is the expression Kind of trivial you have to have some machinery that allows you to compute this But that's what quantum groups does for you And so now the Jones polynomial now J lambda of L How do I get this? Well, I simply just take a Braid presentation of this thing and then of course I have lambda one up to lambda M here and down here. I have the same thing Okay in this case here because what I want to do now is I want to close it up like this and so every link Has a bright presentation like this And so now I just give you an explicit formula for what the Jones polynomial color that lambda is You just sum J one up to J n And you take J i in the lambda i's indexing set and then you do e to the two pi i J one Sorry plus up to J n if there are n ones Divided by r and then you take the braid the Jones braid representation of B and You do J one up to J n J one up to J n So that's the Jones polynomial So the nice thing about it when I do a closure of the braid is there's a very simple correction factor on it And so this here forms a mark of trace on this whole thing. That's what Jones realized And so therefore it gives you a invariant of links And you probably don't want to comment on framing No Everything is blackboard framing Okay So that's that's what I avoid by sort of doing it this way here okay super so Let me just give you an example okay, so that you Have something to think about and look at so example That will be a running example K is going to be equal to 4 1 And so let me draw 4 1 for you. So 4 1 is this not here Just want to make it completely clear that it's the figure 8 not That's the figure 8 not right Can't really argue with that So actually Tang Lee Was the first to actually really compute a very nice formula for the color Jones point for this guy Of course, what you can do is you can just you know, there is a braid presentation of this guy. I don't think I have it right Yeah, okay, so if you want this is also the following guy sigma 1 sigma 2 inverse squared closure Where sigma 1 braids strand 1 and 2 and sigma 2 braids strand 2 and 3 so it's a closer of a three braid And you just take this word sigma 1 sigma 2 inverse and squared and that you close it That's the same as this link So then you just stick it through here and you will get a formula for it And if you're very nifty with these kind of things you can reduce it to the following expression So Jack J lambda of K This is a Quantum lambda and then m equals 0 to lambda minus 1 and then q to the m lambda and Then q to the lambda plus 1 comma q m and then q to the Lambda minus 1 comma q M so these are the pockheimer symbols The standard pockheimer symbols finite ones of order m Yeah, yeah, yeah, so a q is Product m is L equals 0 to m minus 1 1 minus a times 1 minus a q and Sorry, I mean sorry Let me write it q to the L a So it's this one explicitly Okay, so You know I can write it out also very explicitly because I think it's a nice little formula This is trivial, but nevertheless sine lambda pi over r divided by sine pi over r Summing m equals 0 to lambda minus 1 and Then e to the minus 2 pi i m lambda over r and then this product L equals 1 to m. I like to write it this way 1 minus e to the 2 pi i lambda minus L times 1 minus e to the 2 pi i lambda plus L Like that. So that's the very explicit guy Uh Quantum lambda I defined very beginning the one thing I didn't define was quantum lambda factorial, but I think Everybody got that without complaints So and of course you could I could say the same thing about the tqt vector space here, right? so one-dimensional torres and there is a There's one thing I think I do want to say Well Just by by giving you just at least one example of all the things I've talked about let me just give you the a A polynomial of this guy So if you know you do the same things as I talked about before you have these two wave functions that corresponds to the Lagrangians and blah blah blah I don't want to write all of that again, but if I look at Lk, then this is actually the following This is just a k of lm equals to zero and ak of lm This guy is l minus one times minus l plus lm squared plus m to the four plus Two lm to the four plus lm squared and to the four plus lm to the six Minus lm to the eight So that guy Okay, so that's the Lagrangian. We're quantizing and somehow we are getting this wave function out of it So the beautiful thing about your assignments theory is that it gives you explicit formula for all these wave functions So there's no, you know, I mean ambiguity in any things here. It's it's it's really nice Yeah Yeah All right, I just want to do one more thing namely So so far I've just told you about surgery on frame blinks But I can also do rational surgery And so I just want to give you a stack of three many folds and they're corresponding quantum invariance. So Where am I so Yeah, so so in there is of course a quantum representation representation Namely look at the mapping class group of the two tourists. Well, that is else SL to see matrices, right? This is mapping the tourists to itself. And so this guy here goes into you are minus one so unitary R by R R minus one by our by own matrices and If I now define psi zero as the one that is one inside this vector space VR T2 so maybe I should just recall for you that this is span of Lambda in one up to R minus one and then it's like this and so I just take the first vector like this Okay, then Sorry, that is right. There was a little Thing missing there. And so if I define psi a over b to be row R of the matrix a b c d Apply to this vector psi c here Then I get a vector You know a new vector in this vector space and the thing is that if I want the quantum invariant of the manifold which is obtained by You know, let me take a knot By doing a over b surgery on this guy Well, then that is simply psi for the knot and then you take the pairing with a over b So that's the way it works With this TQFT rules and I can actually write down This row R of a b cd explicitly for you and So now I do this because you will see that that gives you some of the quadratic terms in the formula tomorrow that Maxime was talking about so This guy here Explicitly is well some factor pre-factor that depends on our which I don't want to do Then some over mu is equal to plus or minus one. I write it this way sum over n equals to zero up to the size of b Minus one Mu and then exp of a long expression not too bad, but it's like this to be our a j squared minus 2 mu j K plus 2 are in mu plus K plus 2 are in mu squared like this So there is a Like that and there is a one like that. So there is explicit expression for this representation like that Okay, it's a unitary representation. So that's why the CR is there to make sure they're all operators are unitary and it's really a representation and Then you see that you just get this very simple formula For this guy here by combining these Okay, so What time is it? Yeah, good good So what I want to do now is I want to formulate so the the main conjectures in this field about all of these invariants as I see it Okay In your story there was no There was in the W in the WRT and there was this formula here, right? yeah, but it was absolutely clear that Reciteekin and Tureev took ideas from Witten's paper on this to write down what they had if you doubt about that ask Kulian Yeah So so now I want to make a couple of conjectures So the first one is due to Witten Okay So so one and that's Witten if you like But in some sense, it's just what you will expect if you look at a path integral like this Okay, so what we expect is that tau r of some close three-mini-fold m Maybe I write ml since we've been doing this all the time is Has an asymptotic expansion as r goes to infinity Which is of the following type? you sum through the finite set of s u2 trans simons values then you take X of Minus 4 pi squared divided by 2 pi over r minus 2 and Then r to the d theta. I'll say that what that is in a second B theta and then 1 plus and then set theta of 2 pi i over r minus 2 I'm sorry Yeah, so so there should be a Cta here. Okay. Okay. Yeah, and so This guy here set theta of each bar Well, that's a formal series. It's most likely divergent in most cases some cases not but some of it are and All the examples we know this exponent is inside half integers as far as I know and B theta is a real number So there is an asymptotic expansion in the Poincare sense Meaning that you truncate all these series at some large in Take that finite sum and take it on the other side take the absolute value and then that grows like one power less in our In yeah in our minus 2 as our my as our goes to infinity It's strange the small little jobs. Yeah, yeah I have a quick question. So yeah Here you sum over the value of transimons for flat connections Connected components of that connect in some other words Can there be two flat connections with the same value of transimons and they're in by different values of d theta or other things for Yeah, they could possibly be but then they just come in the same thing here Because it's just that you take the maximum of the d theta's for that and then you clump it all together So the whole idea is that you see when you write it this way The set here is determined by the right-hand side and the d theta is determined and the b theta is determined So it's not the normal sum over flat connections, which actually don't make sense in general right because well the set But do you have we've summed away that and just take the sums of this of the transimons for you I think it's true I think there's nothing fishy about this It's completely true Okay, so I mean what could happen okay is that you have our arithmetic progressions on these instead And then maybe look over them it's just we don't know but add them if you like But I mean we do see these real numbers here leader these are atomized the torsional square root or master my distortion, right Those we see No, no, no, no, no their examples where they don't cancel for cypher fiber many folks we know they occur no All right, so then there's sorry a cashier Volume conjecture So it says the following when you take the limit as r goes to infinity of 1 over r and then lock of You take j you take j r So notice that r is not allowed label, right? It's just one outside So it has quantum to mention zero, but what you do is you do this and Then you divide by the guy for the odd not in the way I have normalized things Over r and you so you evaluated the same rotor unity as the label Then this here is conjecturally is equal to 2 pi 1 over 2 pi times the volume of s3 minus K Again, what is jr of? I'm not on not you was the honor. Sorry. Yeah, you was the honor And you probably mean log of absolute value Yeah, yeah, yeah, absolutely like that Okay, so that's the volume conjecture it's been tested in many many different cases And then let me just Do one more conjecture Which are kind of things that I would like to have set up Before I write the main research and conjecture and then I'll stop so These are the GPP be invariance gukhoff pay Protroph and Wafa invariance Which I think also took its origin in some work of gukhoff and Marino and the third guy was put rough I think and so these are said had invariance labeled by spin C structures of ML and so Sege, please allow me to say that they are normalized so they lie in here for now And so a is a spin is a spin C structure of ML so the index by spin C structures ML is the manifold we obtained by doing surgery on L Yeah, yeah, and so this guy here is supposed to be the sum Where we sum over i and j and then minus one to the i q to the j and then the dimension of H ij of ML comma a So they're supposed to be some vector spaces or modules and so this H ij in the sort of physics way of talking about this These are the vector spaces of B PS states and And so but these are sort of covenants type homologies one would like to construct. This is a j supposed to be down here So they're not really constructed yet mathematically as far as I know. Here's ML as well But these series here can be actually be constructed some specific examples, okay? All right, so with the arm with these sort of Conjectures here. I want to formulate another conjecture which is Yeah, let me leave the example out there and I hope I can fit it on these these boards here Yeah, yeah, yeah So so what I would actually I want to put it in the main middle here. So let me put this up So it's what I would like to call sort of the the research and conjecture So it it doesn't really need exactly those conjectures and something that independent of it But in it the formulation I give it now it it will depend at least on witness asymptotic expansion conjecture But so the research is conjecture and so there are lots of of names attached to this suddenly Star Wars should go there Marcus should go there Gukov should go there Putrov and so on number of people so But let me just write it out. So it won These series here are resurgent So the series from up there from the witness asymptotic expansion conjecture are resurgent We can to Borel Do the Borel some of these whoops Can't really bend in this way And so these guys here are defined On the complement and Of some discrete set and which set do we expect it to be? Well, we expect that if you take minus 2 pi I in normalization I have a chance I'm so the complex chance I'm in values Then this is this is exactly the set modulo Set Sorry, what? Yeah, sorry, it should be shifted by theta This guy here shifted by theta Okay Then 3 of course these B. Theta's Satisfy a wall crossing structure may be analytic, okay? and if we take So so for Suppose I take theta C this stands for the conjugate theta theta value So the conjugate trans diamonds value. So this is a trans diamonds of conjugate Representation so think about the manifold is hyperbolic it has a hyperbolic representation the conjugate of that is another representation And so it has a trans diamonds value. That's the one. I'm calling theta C then that guy is If I take the Laplace transform of B theta C Well, this is supposed to give the set a k the Taikman or T curve T okay Laplace transform and You know that finally five sorry, this is kind of a little funny order here, but you know somehow the set had a G p p v invariant is obtained by a finite Fourier transform Of the B. Theta's let me put it little wakily like this But so these set a hat should be determined by these B's as well these B. Theta's Yeah, yeah, we're faces and stuff like this like Sergei The four I mean only the value that corresponds to the conjugate representation So I take B theta corresponding that take B corresponding to the conjugate representation and then I do Laplace transform of that guy That should give you It's conjugate for the hyperbolic one So you take the hyperbolic one and you conjugate it and then that has a Transcendence value its imaginary part is the smallest possible. There are among all the complex trans diamonds And you have file a fine ray going down stokes, and so that's the one we found with which we're not And so this is the sort of general research and conjecture So and what I will try to do tomorrow is Then to show you the new formulation of this guy here So to actually give you at least one of the bees we are the this inverse Laplace transform And then I will talk about in the second like the tomorrow talk about finite Intervals which gives you exactly the real Transcendence and then I will look at other left-shits symbols and give you in particular this one from there in particular in this example And the other ones will be other left-shits symbols So all right Thank you To say what you mean by resurgence in one Yeah, it means that You know when you do the borrel transform to this Then that's convergent in the small disc and it is endlessly Anaglytically continuable in the complement of some discrete set and the set is in that set as William pointed out shifted by theta Just minus the two is just a refinement of one Yeah, I tell you where the singularities are That's right, and then the b-theta satisfies the wall crossing system That would be an analytic wall crossing system. That will be a further information And then the question would be how unique are they if given that system? I don't quite know and then finally there is a way that we actually know one of them already Explicitly by by my work with Renat. So we have a formula for that guy that corresponds to the conjugate representation But this this last statement we know that is not always true Yeah, no, that's right, this is this is for very specific It's in some specific direction that that works I agree with that Yeah, I agree Yeah, so if I remember right these z hat a's I'm supposed to have some kind of a Nicest in thought is when it goes to a root of unity with some kind of expected behavior. Oh, that's what it says Oh, I'm sorry. Oh, I think that the radial limit conjecture. Yeah Do you see some second yeah, and it's true in some cases, but I'm not sure whether it's true always We don't talk about it and I think You can see on the d-status or we'll have some examples Where it doesn't work as far as I understand from him. That's not that's what he says to me So I didn't put it there Details like it's a more of a comment for example Thought to Somebody should look at zero surgery on 5 to not because they're the set of yours to be a little bit shifted So it's set of terms time and values where we see the polls, but I cannot explain the shift. I hope that somebody Smarter can can explain the ship. So it's here in surgery on five to two That's the first simplest example or Again, it looks like it fails. That probably they're missing something. I'm sure it's in man. That will be true In my story of the symmetric measures is rational coefficients Well, no, no, I mean sure right, I mean, you know if you this is very reminiscent of the Neumann-Sagear data, right? And the Neumann-Sagear data determines the triangulated three many fold and so some of them are three many fold others are not and That's why I said maybe the example the baby example you looked at is maybe not a three meaningful I don't know. I don't think it's a close three minute. It's actually we know It's not a close three meaningful because of its asymptotics Okay, because it should have gone like it went exponentially fast But it should have But it's it's really doesn't for the for the trivial flat connection, right? So so yeah, I mean, of course you can map three many folds to certain matrices. I Agree and then you can study the general one. That's fine. I mean, I don't see any problem with that Some people like three many folds, right? By the way, it could be fun to formulate. What's the volume of these matrices?