Lagrangian Fibration of the de Rham Moduli Space and Gaiotto Correspondence
Dumitrescu, Olivia
There have been new developments in understanding Lagrangian fibrations of the de Rham moduli space in connection to Lagrangian stratifications of the Dolbeault moduli space through biholomorphic isomorphisms of the Lagrangian fibers. I will report recent results by different groups of authors.
 Thank you very much for the introduction. I would like to thank the organizers for the opportunity to speak. So today I'm going to present a partial holomorphic isomorphism between two modular spaces. So we have the modular space, the dolbo modular spaces, and I'm briefly going to define of stable hex bundles and the d-ROM modular spaces of irreducible connections. They are related by the non-o-billion-hatch correspondence that is a diffeomorphism. So what I'm going to do, I'm going to explain a partial holomorphic isomorphism between these, mainly between Lagrangians. So that would be a holomorphic Lagrangian correspondence by holomorphic between each of these stratifications. And moreover, I'm going to, in the end, I'm going to make a definition of the semi-classical limit via a geometrical definition. So first I'm going to start by introducing the notation and maybe some of the, I will spend some time reminding the old results. So all of this talk is going to be with the group SLN of C, maybe I will use SLR of C. And C will be a productive algebraic smooth curve. SCL? Yeah. Semi-classical limit? Oh, I'm sorry. SCL. This is the most important part, so yeah. And so I'm going to choose a spin structure and that is going to be L, a line bundle so that L squared is canonical. And because I'm working with this group, my vector bundles, well, I'm going to use both notation C and V to denote holomorphic vector bundles of rank R and degree 0. But also phi is going to be a traceless fixed field. So let me briefly mention this to modular space. So on one side we have the del bomoduli space consisting of stable Higgs pairs. And phi is a map from E to E tensor KC, canonical, that is a no-seat homomorphism. And then I want to introduce also the diromoduli space consisting of irreducible connections. And nabla is a map from E to E tensor KC. That is a C homomorphism. So these two spaces are homomorphic by the non-Abelian hodge. Let me also briefly mention the heating vibration. So we have the dol bomoduli space. And we have a vibration over the vector space consisting of all the sections of the pluricanonical. And in this case, I consider traceless Higgs fields. So it's mainly going to map these stable Higgs into the spectral data determined by the characteristic polynomial of phi. And this is an element in the heating base. Excuse me. These irreducible connections, are they flat or not? Flat. Flat. Yeah. Hologmorphic. It's a holomorphic, yes. So it's much more complicated. OK. Thank you. OK. So this is the general framework. And next, what I'm going to do is to introduce there are two Lagrangians in this space that mainly give a kind of affine parameterization of the heating base. And one will be called the heating section. And the other one will be the upper noduli. So before, because I kind of talked about this a while ago, I this. Maybe I just, oh no. Maybe I'm just going to. OK. OK. Is it. So, so, so many. I want to define the section inside the heating module space inside the dolbo module space. So what I will consider are vector bundles that are split bundles. So they split in, these are line bundles, powers of the canonical bundle, half canonical. And X minus is this matrix consisting and p is just constant. So X minus is a matrix with just nonzero constants on one diagonal lower. And X plus is going to be the transpose. So you will have the same diagonal upper. And it is going to be the commutator between these two, X plus and X minus. So that will make a diagonal matrix consisting of values, I mean, decreased by two every step. And these are the generators of the SL2 representation to SLR. And now if I consider Q to be, oh, to be a section here, I mean, to be just a vector here. This means that it has components and it consists of differential. Q2 will be a quadratic differential and so on. QR will be a section of the pluric canonical. And now the elements of the hitching section consist of E0. So they always have this split vector bundle. And phi of Q that consists of X minus. And I'm going to add, so X plus will be nilpotent. So this means that the power will cancel. So all the powers will actually will go one diagonal. And I will just multiply by Q's each of those diagonals. So that will be my phi of Q that is just an, well, has an upper triangular part. And lower diagonal consisting of this just constant. And it is not difficult to see that actually this is a stable kick bundle. And in fact, X minus is a stable kick bundle on E0. So it's a point on the del bomogelite space. And this is known as the hitching section. And it is a section of this vibration in the sense that it intersects every fiber of this once. It's a section on, in rank two, but otherwise it's just intersect the fiber once. And if you apply twice, it doesn't give you identity. So now I want to introduce another Lagrangian inside the Dirac moduli space that is the upper moduli. So I just wrote the definition of the upper. And this consists of a filtration of vector sub bundles. And the Bellisson-Dreamfield definition says that nabla is a connection is an upper. If, well, I'm using the decreasing filtration. If it satisfies the Griffith transversality condition. And this will induce an OC linear map. And this means a Higgs bundle on the quotients. And the condition is that that is isomorphism. So this being an isomorphism implies that the discussions are, that this is a full flag. So the rank goes every step by one. But yeah, so what I'm going to, maybe I'm just going to briefly introduce now a new definition. So let me define here. For any lambda in age one of CKC, so this is just a complex number. I want to define a one parameter family of filtered extension to be F lambda this filtration. This is the vector bundle. And maybe I'll just write it in words. So I do have a full flag. And the last condition is fulfilled. So I do have an isomorphism between the ratios. And finally, I want to mention a starting point. And the starting point of this I want to mention the first one here is going to be just the power of the half canonical. So it's just going to be this. So for example, for example, in R equals to 2, I have that for every lambda, a complex number. Then there exists a unique filtered extension like this that is given by the short exact sequence. lambda doesn't play a role in the diffusion. The diffusion lambda plays no role at all. Well, so. I eat extensions. Yes, it's extension classes. So many this will give the first ratio. And then by this condition, that is quite a strong condition, you can compute all the ratios. So this will be like the first. And then what I'm going to do, I'm going to sandwich like in rank two and construct another in another exam, another in another filtered extension. And right, so I want to be the last one to be the last term, the vector about the to be the last term of the extension. And because I want to I want to consider one example, let me introduce one definition. So I'm going to recall project coordinate system. So many if if if my ruin surface decomposes into a union of open sets, then I want that the transition functions. These are the coordinates on the open set are going to be given by mobius transformations. And such such project coordinate system always exists. Because for example, one comes from the universal covering that is just the upper half plane. So let me and this matrix I'm asking to be in SL2 of C, this matrix say the CD alpha beta. So I'm going to introduce coordinates now. And I'm going to do this Z alpha beta to be transition functions. The last one has a beta but there is no beta. It's your sound, the last beta. It's the beta alpha. What is the coordinate change? It's called the change in the center. The matrix say the CD of the vector plus the beta. The change from the CD to the Z. That's a good sense. So this is the mobius code. The transformation is better for this. So the universal covering is the upper half plane. And if you take Z, the local parameter here, this will induce on this Riemann surface such a coordinate system. But you'll have many of them. So, so. Okay. So this is a definition. That's a definition. So Z alpha beta are the transition function for half canonical. And maybe they are linear in Z beta beta. So in particular, the first derivative will vanish. Okay. Now I can maybe state some theorems. So theorem one. This was proved with Motoki-Kombola set 2017, I think. Is that for any lambda that is a complex number, then there exists a unique, non-trivial one parameter family of filtered extensions. For F lambda, this is the filtration. This is the vector space for the vector space being, I'm going to denote V that is for this E0. That was that direct sum of half canonical powers. So this is a statement about the filtration, and now I want to put some connections on this filtration. This is a theorem two. And maybe I'm going to mention some things. So I will give you the transition functions for this V lambda. So let me recall F alpha beta gamma to be the transition functions. This is lambda. Maybe it wasn't a good notation. To be defined as exponential of H times logarithm of Z alpha beta times exponential of lambda times d log of Z alpha beta times X plus. So these are the transition functions for V lambda. And the second part of the statement says that, well, yeah, when this is at zero, you just get back. I mean, this will make the transition function for E0. That is also my notation of E0. And finally, that d plus 1 over lambda times a Higgs times. So if I start with any element on the Hitchin section, then this is an upper on the V lambda. So this is a statement of filtration. This is a statement about actually, well, you can quite identify this filtration. So maybe I mean this first matrix, I mean, sometimes probably we denote it like Z alpha beta to the H to be to consist of exponential of H times logarithm of Z alpha beta. And this is just the diagonal matrix consisting of just, yeah, this powers on the diagonal. So this is the first part. And the second part consists of another matrix that has well identity on the diagonal. And on each of upper diagonal entries, we'll have powers of the lambda plus the derivative of Z alpha beta. As I said, because X plus it was nilpotent, so when you do write down series, you'll obtain such an object. As I said, these are linear, so that will be just a constant. So when you multiply these two, you're just going to see a matrix that has these entries on the diagonal and then upper, and then, yeah, and then upper, consisting of the derivative and powers of lambda. Okay, so now I want to talk about something that is in between these pictures, and I will come back on these theorems. So maybe I'm going to use this part here. So I'm going to introduce the notion of, well, I'm going to talk about hot modular space. So let me introduce this hot modular space of lambda connections, consisting of lambda, a vector bundle. Maybe I'm going to call it lambda, and then the connection here. So this is the modular space of the lean lambda connection. And the rule is that the light-nit rule, the lambda light-nit rule is that this is lambda times df times s plus f times ds of s, and f is just a holomorphic function, and s is just a section of the vector bundle. Usually, I like to use the notation v lambda for the connections, but it's okay. Is that a one half or one over lambda? It's one over lambda. So lambda is not zero. Sorry? So we're assuming lambda is not zero. Yeah, yeah, yeah. So I'm going to, it's one over lambda, sorry. So this is right now a family of, yeah, I'm going to come back on this. So lambda is not zero. Okay, so the delin put these two definitions of the Higgs bundle and connection into this definition of the lambda, of the lean lambda connection via this. And we can notice, I mean, when lambda is zero, then you have, this becomes a Higgs bundle, while if lambda equals to one, it becomes a connection. So we should think about this as being, I mean, this, if here is the modular space, if this is the modular space of the, if this is the dolbo modular space, and this is the diram modular space, this is kind of in between. So this corresponds to lambda equals zero. This is the fiber over lambda equals to zero. So it's lambda inverse of zero. And this is the fiber over lambda equals to one. And in between there are these lambda connections. Now there exists a, this is the reaction on the Higgs field, on the dolbo modular space, and mainly it will be induced from the c star action on the hot-moduli space. So let me talk about this. So we have a c star action on the hot-moduli space, and the way it acts, I'm going to take t and non-zero parameter and look at the connection. And I'm going to send it well into this. And the fixed points of this action will be the ones that are left invariant under this c star action. So you can see that this happens if lambda is zero. So these fixed points happen if and only if lambda t is t is lambda. So this means lambda is zero. So the fixed points are Higgs bundles. And they are denoted as hot bundles. So I'm going to, so this denotes the c star fixed points on the c star action on the hot-moduli space. So now probably I want to look at one example. Maybe I'm just going to write here. So I have this exercise that I did maybe two days ago. So I'm going to save this notation above. So this is maybe a remark. So I'm going to start with phi of q on the Higgs section. And then I have the following, that lambda, v of lambda and d lambda plus phi of q is a lambda connection. So this guy that appeared here if you multiply by lambda is just a lambda connection. The second statement is that E0 comma x minus. So E0 was the split vector bundle into half canonical powers. And x minus was this constant. So one diagonal below is a fixed point for this lambda connection. And the third part is that the limit when t approaches 0 of t times lambda v lambda and lambda d plus phi of q. So these are just these operas, but now they became lambda connections. It's just, well, the same point is the same point. And finally, the fact that the grated hodge bundle associated to this lambda connection is going to be E0 and x minus. So these are some statements we can easily check. So the way to see this, what the c star action does, well, I'm just going to apply the theorems above in order to prove this statement. So z alpha beta, we said that these were the transition functions for half canonical. What's the difference between 2 and 3? Or maybe what is the meaning of 2? I mean, yeah, I mean, yeah, what do you mean? Oh, this is the fixed point. So if you add here by t and you, you, you, you, um, Is this c star action? Oh, sorry. Did I? It's okay. It's okay. No, no, it's okay. I just don't understand. So it's okay. It's okay. It's okay. It's okay. It's okay. It's okay. It's okay. Right. I mean, but I will mention in the end that this will be the same thing. But you know, I just wanted to, to give you the, the computation here. Okay. So that alpha beta are the transition functions for this. So I can take like a times that alpha beta where a is a non-zero constant. And this will be the same transient. I mean, the transition function for that. So, um, maybe I'm going to denote like F alpha beta and this is the A lambda and these are the transition function corresponding to this, to this V V lambda. And, um, so this, um, yeah, so what, what, um, what, what happens there that if you, if you, if you, if you just multiply by a non-zero constant, the second part, the second matrix will not be affected because, um, because this A is just a constant. So it will be killed by the, by the D. So you'll have the same thing, but only the, the first part will be affected in the first statement of the theorem. So for example, I'll, I'll take that, I'll have that X of H times, um, log of A times that alpha beta will just be the sum between, uh, uh, will, will be the product, sorry, the sum and then the product between this. So for example, this will be probably my notation A to the H and this. So, so, so, so I'm just, I'm just saying that this is just going to be A to the H and this is just this diagonal matrix that has just powers of A on the diagonal and then F alpha beta, the old ones. And this just comes from here. Okay, so when, when I, when I write, I mean, I'm going to take, um, so I'm, I'm going to act by T and when, when I write my, my T times phi, well, T times phi in this new transition function, probably what, what, what will happen is that I have to multiply by this new, uh, A to the H and have to conjugate by that. So this will, this will, uh, give you that, so it's the same Higgs field, T of phi, but it's written as, um, A to, uh, A to T, uh, and then phi of A to 2i minus 2 times qi. So this, this is just the vector q. I mean, q, q consists of the vector qi of the, uh, differentials and it's just going to have this, this constant. So, so if, if, if, yeah. Um, so, um, so, so for example, if q is equal to zero, um, then, then if you look at the definition of the Hitching section, this will, phi of q will be just, uh, phi of zero will be just x minus. So, um, Sorry, what happened to this? It, it, it raised to H into, what is zeta? Zeta alpha beta, there is a term in the, Zeta alpha beta? No, no, no, it was, it was the first term. For first term, it was written. The second term, you know, the second term. Yeah, yeah, so, so, so, uh, nothing. So, yeah, I mean, oh, yeah. Sorry, I didn't understand. It's good, good. Yeah. Okay, so, um, so if, uh, if, if, if this q is the, is, is, is the origin of the Hitching paste, then, then, then, then this is just x minus. And you can see here that for, if, if you choose a to be the square root of t, then this is going to be mapped in the same, in the same point. So, therefore, this x minus comma e zero is a c star, uh, fixed point for, for q equals to zero. Um, while if, if q is non-equal to zero, uh, then, um, if, if you choose again, equals to the square root of t, um, then, what, what, what will happen here is that this will be just phi of, of, um, and then this will be t i minus one q i. So, when you take the limit when t equals to zero, now the q's will be, will be killed. So, the only one surviving will be x minus. So, this proves that the limit, um, when t approaches zero of this is going to be the same point. Yeah? Because, uh, because, because i is going to be bigger equal than two in my notation. So, all the differentials will come with t times q two and then t squared, q three and so on. So, you can take the limit at zero and obtain this q minus. So, this proves, um, I mean one just is a reformulation of the first theorem and this proves two and three together. And now you can use probably theorem number one to prove four and I probably will not do that but I can explain how to do it. Uh, is it okay? Questions? So, uh, what is the graded hot bundle? Well, um, yeah, probably, um, yeah, probably, um, yeah, probably, so, um, I would have to compute the ratios of this, uh, filtration and if you remember the definition of the, this one parameter family, a filtered taxation, this came with this fixed ratios. So, these fixed ratios in the end, they were known and there were powers of the half canonical. So, um, so to compute a, um, yeah, so, so it's an easy computation to see that the associated, the graded hot bundle from, from theorem, consequence of theorem number one of, of, of this is, is, is just this because, uh, because, yeah, you, you will have that, you will have that at each, each step you'll have to take these ratios that will be like, um, K, half canonical K to R minus one or R minus three and so on and then you'll have to twist by the canonical. So, that will be an isomorphism in the end and this X minus says that you have just this constant on lower diagonals and they will give you the isomorphism between this. Okay. So, this was just an exercise in order, um, for, yeah, to, to get, um, in order to, to get some terminology. Um, okay. So, let me, let me mention some theorems now. Yes. I mean, they, yeah. They, so, so here it is just, uh, uh, yeah, they, um, exterior derivative, yeah. But did we, did we choose some special property of the, of the transition functions or any, which is any transition function? It's a canonical shift, so, it makes sense. So, tomorrow at the start we chose some transition function, some arch, the S alpha beta. Is this going to depend on the choice? When you have the F alpha beta and the S alpha beta? S alpha beta. Oh, Zeta. Yes. Uh, it, so, so it's going to, to probably depend on the spin choice, right? So, this Zeta alpha beta came from the, yeah, spin structure, but you choose just the core, projective coordinates. You don't need a solution at the end of the division to start somewhere? No, no, it's projective structure. Right. Projective structure. Right, I'm not using the solution here. You get symmetric power of the two dimension representation, you get flat connection, yeah. I mean, here, here, on, on this board I kind of defined everything I needed for this. You started with the complex projective structure? Yes. Yes, yes, yes. That was the, right. If you choose a different complex projective structure, you'll have a different base point there, is that right? Yeah, yeah, yeah. Okay. Okay, so let me now measure some theorems. Should I do that? Maybe, maybe I'll just erase this board. Okay, so, so, so, so, so theorems and this is mainly the work of Carlos Simpson, is that for every lambda v lambda, that is a hot bundle, so for any lambda-dealing connection, then there exists, or maybe I should say to fix a lambda first, fix a lambda on the base. So I'm just going to fix a lambda and look at the fiber. And I will choose a lambda-dealing connection, then there exists a unique section. This star goes to the hot module ispace, passing through lambda, the lambda, and the connection. The limit when t goes to zero always exists. So fixed, so let me just denote m double of c star consisting of this fixed points, p alpha, where alpha is a class of variation of hot structure. And for any point, p zero here, then if I look at the inverse, I mean if I look at all the points, all the points are fixed under this section in p zero and I restrict to lambda equals to one, this is a holomorphic Lagrangian. Oh, that's a different sigma, right? You use sigma for section. Yeah, yeah. Sorry? No, no, you just on the left, you say sigma is half of c star. Oh. And now it's sigma, it's something else here. Right. Number two, you meant sigma. So the notation is not sigma, you just confuse this notation. Yeah, sorry, I just switched the all notations in terms of h bar and sigma just before. Yeah, you can erase that sigma over there. Here? But then what? Okay. Sigma is number two math, limit, limit. Oh, yeah, yeah, yeah. Yeah, yeah, yeah, yeah. So maybe this one. In section, you have one exception? No, maybe c star invariant section. Yeah, yeah, it's a c star invariant section. So, okay, so what do I want to say? Sigma of lambda, whatever, is equal to that limit. Yeah. This is the sigma, yeah. Okay. So, all the points that are fixed under this, all the, so for any, so here I have the set of all the fixed points that live in the dolbo. And for every point here, I look at the, at all the points that get fixed under this. And this will be a holomorphic Lagrangian. And this will give the stratification of the Dirac modular space. So this is a holomorphic Lagrangian vibration. So all the points that are c star fixed points under this, under this action. Okay. So, so there exists this, there is the, this Lagrangian vibration on the, on the Dram, and you can do the same thing and restrict mainly to, to the lambda zero. And you'll obtain a vibration for, so maybe I'm going to, to just write it that if you restrict to lambda equals to zero, you'll obtain again a Lagrangian vibration of the dolbo modular space. So these are all the Higgs bundles that have that correspond, I mean, that are in the fiber, that have this, the same Higgs point. So, okay. So, so more, moreover, this is a holomorphic Lagrangian on the dolbo modular space. So the modular space, the dolbo modular space breaks into this p alpha of, of, into, into this, this is a whole homomorphic stratification of Lagrangians. Okay. So, maybe now I can, I can mention the, the theorem. The idea of that would be like some kind of vibration, I think that's the connection of, of, of, sorry, of the Dram or of, of the Dram. Okay. Okay. It's in the subspaces are Lagrangian, but that's what she meant. It's a vibration by holomorphic Lagrangian. Yeah. It's a holomorphic vibration. Yeah, the statement that those are closed subroutings, for example, is, is just a conjecture. They're not closed here, but they're, it's just a conjecture that they're closed here. Okay. Okay. Yeah, I see. So maybe. Okay. I'll finish them by holomorphic Lagrangian. Yes, yes. Okay. Yeah, same thing if you're on the other side. Well, they're, they're not, they're not closed here. Well, I mean, I, no, I don't think it's correct to say, I mean, those subroutings are not closed in that, in the Dolbo. So I'm not sure what the vibration means, but I mean. Okay. So then I just, I just leave that holomorphic Lagrangian. Okay. So, so, so the theorem. And this is with Laura Fredrickson, Church Kidonakis, Rathematio Motohikomoulas and Antinetsky in maybe it was 2016 was that this. So, so if, if, if, if I restrict to P0 that is E0 X minus and we proved in that lemma that this is a fixed point, right, this is the fixed point. I mean, and what this is a fixed point of the hitching section on the Dolbo and also on the, on the upper module. So this is what that lemma says. Then, then, then this two, then these two Lagrangians are by holomorphic. So this is the fixed point when lambda equals to zero are by holomorphic via Galloto conformal limit. And yeah, we, we basically, we basically, yeah. So this correspondence between the, the, the, the, the, the, the Higgs bundle on the hitching section and, and, and this upper and that this is the limit of the, this is the conformal limit of Galloto. Okay. And moreover, it was generalized by Collier and Wentworth a year ago to all such fixed points. So this is the theorem Collier and Wentworth. So this is a fixed point that, for any point, that is a fixed point under this C star action. Then, and then the alpha variation of horse structure. Then there exists a bipholomorphic correspondence between, so this two fibers, these two Lagrangians now, one on the double both side and the other one on the theorem side are by holomorphic Lagrangians. Via Galloto's conformal limit. And this Galloto's conformal limit is explicit. So you can see from this, so you can see that, and, and the methods here that were used are similar to the methods that were used by us in, yeah, I have very short time. They are similar to our technique and you can see that in that case, this corresponds to the longest, the full flag. Like the, I mean the very, the all corresponds to the full flag. And you can, I can try to explain for example, the, on the opposite side, you have the shortest flag when E is a stable hex, when E is a stable vector bundle. So if for a stable vector bundle E, then you have that T phi is in the dolbomoduli space for any phi. So if E is just a stable vector bundle, then this is a stable hex field for any phi. And E comma zero is the fixed point. But you have the cotangent fiber of E over the moduli space of stable vector bundles of rank card and degree zero that is contained in the dolbo. And it's a Lagrangian. So this will be the corresponding Lagrangian. And for every stable hex, for every, every stable vector bundle, there exists a unique Hermitian Einstein metric. And once you take T to be the turn connection corresponding to that Hermitian Einstein metric, the unique turn connection. Then you get this maybe by Narasimha Shashadri, E corresponds to D. And mainly the fact that E was a vector bundle of, of degree zero, it tells you that, that this has curvature zero. So, mainly the solution of Kitchin's equation split kind of the couple into the curvature zero part and then the part where phi, the, yeah, the phi, phi dagger is zero. That can be done because, yeah, phi can be a diagonal on this, on this Hermitian metric. So this is kind of a short proof for these opposite cases in the case of stable vector bundle and in the case of oppers and in between it's, it's, it, the proof is very similar. Anyway so maybe, maybe, maybe I'll try to, I'll try to give the general picture here. So, so this is the dolbo and this is the diram. And here, well here, here is, is the corresponding agrenation of the dolbo and the diram and you have this kind of slice in the hot bundle. But this, this, this is saying that, that, that, that this is just trivial. So it's going to be a product between C and then one of these Lagrangians. So if I want to define, let's say a semi, a geometric, semi-classical limit, I can, I can start from a lambda connection, go to, to the, to the dolbo side by this and then project on the pitching base. So that is, that is a geometry definition of a semi-classical limit. There is another definition coming from physics when this is more like the limit when h bar goes to zero of, yeah. I can try to define a semi-classical limit of this and the theorem is that these two agree, these two definitions agree on modulite space of uppers and that was, so that, yeah. So maybe I can, I can define a map from the diram to the pitching base via this, well, stratification or that is trivial. And this, sorry, yeah. It's, yeah, it's holomorphic, it's probably on each slice or it's not on the whole thing. It's not on the whole thing. But it's probably holomorphic, I mean, it's between these Lagrangians but it's, yeah, so probably going to be on the slides, so. But it's not on the whole, this is not holomorphic on the whole modulite space. And here, yeah. So probably I will stop here. Thank you. I have a question, which is, so at the start you got a similar type of correspondence but it depends on the choice of base point that was the projective, how much projective structure, right? Yes. Yeah, an idea, because this one depends on the Sushman-Hijian equations, right? Right, right, I mean, it goes through, yeah. So do you have an idea of how those two compare? I mean, you could, because you're ahead, your formula with D plus, right, I mean, this is also explicit. Yeah, I mean, yeah, well, I guess this, yeah, well, I think it's a similar formula to that, D plus 1 over lambda of phi. So here you, it's not the same. Maybe it's the same if, is it the same if the complex projective structure is the one given by the uniformization? Yes, yeah, good. That's exactly the case. Yes. So if the complex projective structure is not the one given by uniformization, then do you have a comparison between them? And for example, is it going to be linear on this vector space? I don't know. Yeah. I don't know. Any other questions? I think it's figure again.
 Thank you very much for the introduction.  I would like to thank the organizers for the opportunity to speak.  So today I'm going to present a partial holomorphic isomorphism between two modular spaces.  So we have the modular space, the dolbo modular spaces, and I'm briefly going to define of  stable hex bundles and the d-ROM modular spaces of irreducible connections.  They are related by the non-o-billion-hatch correspondence that is a diffeomorphism.  So what I'm going to do, I'm going to explain a partial holomorphic isomorphism  between these, mainly between Lagrangians.  So that would be a holomorphic Lagrangian correspondence by holomorphic between each of these stratifications.  And moreover, I'm going to, in the end, I'm going to make a definition of the semi-classical limit  via a geometrical definition.  So first I'm going to start by introducing the notation and maybe some of the,  I will spend some time reminding the old results.  So all of this talk is going to be with the group SLN of C, maybe I will use SLR of C.  And C will be a productive algebraic smooth curve.  SCL?  Yeah.  Semi-classical limit?  Oh, I'm sorry.  SCL.  This is the most important part, so yeah.  And so I'm going to choose a spin structure and that is going to be L, a line bundle so that L squared is canonical.  And because I'm working with this group, my vector bundles, well, I'm going to use both notation C and V  to denote holomorphic vector bundles of rank R and degree 0.  But also phi is going to be a traceless fixed field.  So let me briefly mention this to modular space.  So on one side we have the del bomoduli space consisting of stable Higgs pairs.  And phi is a map from E to E tensor KC, canonical, that is a no-seat homomorphism.  And then I want to introduce also the diromoduli space consisting of irreducible connections.  And nabla is a map from E to E tensor KC.  That is a C homomorphism.  So these two spaces are homomorphic by the non-Abelian hodge.  Let me also briefly mention the heating vibration.  So we have the dol bomoduli space.  And we have a vibration over the vector space consisting of all the sections of the pluricanonical.  And in this case, I consider traceless Higgs fields.  So it's mainly going to map these stable Higgs into the spectral data determined by the characteristic polynomial of phi.  And this is an element in the heating base.  Excuse me.  These irreducible connections, are they flat or not?  Flat.  Flat.  Yeah.  Hologmorphic.  It's a holomorphic, yes.  So it's much more complicated.  OK.  Thank you.  OK.  So this is the general framework.  And next, what I'm going to do is to introduce there are two Lagrangians in this space that mainly give a kind of affine parameterization of the heating base.  And one will be called the heating section.  And the other one will be the upper noduli.  So before, because I kind of talked about this a while ago, I this.  Maybe I just, oh no.  Maybe I'm just going to.  OK.  OK.  Is it.  So, so, so many.  I want to define the section inside the heating module space inside the dolbo module space.  So what I will consider are vector bundles that are split bundles.  So they split in, these are line bundles, powers of the canonical bundle, half canonical.  And X minus is this matrix consisting and p is just constant.  So X minus is a matrix with just nonzero constants on one diagonal lower.  And X plus is going to be the transpose.  So you will have the same diagonal upper.  And it is going to be the commutator between these two, X plus and X minus.  So that will make a diagonal matrix consisting of values, I mean, decreased by two every step.  And these are the generators of the SL2 representation to SLR.  And now if I consider Q to be, oh, to be a section here, I mean, to be just a vector here.  This means that it has components and it consists of differential.  Q2 will be a quadratic differential and so on.  QR will be a section of the pluric canonical.  And now the elements of the hitching section consist of E0.  So they always have this split vector bundle.  And phi of Q that consists of X minus.  And I'm going to add, so X plus will be nilpotent.  So this means that the power will cancel.  So all the powers will actually will go one diagonal.  And I will just multiply by Q's each of those diagonals.  So that will be my phi of Q that is just an, well, has an upper triangular part.  And lower diagonal consisting of this just constant.  And it is not difficult to see that actually this is a stable kick bundle.  And in fact, X minus is a stable kick bundle on E0.  So it's a point on the del bomogelite space.  And this is known as the hitching section.  And it is a section of this vibration in the sense that it intersects every fiber of this once.  It's a section on, in rank two, but otherwise it's just intersect the fiber once.  And if you apply twice, it doesn't give you identity.  So now I want to introduce another Lagrangian inside the Dirac moduli space that is the upper moduli.  So I just wrote the definition of the upper.  And this consists of a filtration of vector sub bundles.  And the Bellisson-Dreamfield definition says that nabla is a connection is an upper.  If, well, I'm using the decreasing filtration.  If it satisfies the Griffith transversality condition.  And this will induce an OC linear map.  And this means a Higgs bundle on the quotients.  And the condition is that that is isomorphism.  So this being an isomorphism implies that the discussions are, that this is a full flag.  So the rank goes every step by one.  But yeah, so what I'm going to, maybe I'm just going to briefly introduce now a new definition.  So let me define here.  For any lambda in age one of CKC, so this is just a complex number.  I want to define a one parameter family of filtered extension  to be F lambda this filtration.  This is the vector bundle.  And maybe I'll just write it in words.  So I do have a full flag.  And the last condition is fulfilled.  So I do have an isomorphism between the ratios.  And finally, I want to mention a starting point.  And the starting point of this I want to mention the first one here is going to be just the power of the half canonical.  So it's just going to be this.  So for example, for example, in R equals to 2, I have that for every lambda, a complex number.  Then there exists a unique filtered extension like this that is given by the short exact sequence.  lambda doesn't play a role in the diffusion.  The diffusion lambda plays no role at all.  Well, so.  I eat extensions.  Yes, it's extension classes.  So many this will give the first ratio.  And then by this condition, that is quite a strong condition, you can compute all the ratios.  So this will be like the first.  And then what I'm going to do, I'm going to sandwich like in rank two and construct another in another exam, another in another filtered extension.  And right, so I want to be the last one to be the last term, the vector about the to be the last term of the extension.  And because I want to I want to consider one example, let me introduce one definition.  So I'm going to recall project coordinate system.  So many if if if my ruin surface decomposes into a union of open sets, then I want that the transition functions.  These are the coordinates on the open set are going to be given by mobius transformations.  And such such project coordinate system always exists.  Because for example, one comes from the universal covering that is just the upper half plane.  So let me and this matrix I'm asking to be in SL2 of C, this matrix say the CD alpha beta.  So I'm going to introduce coordinates now.  And I'm going to do this Z alpha beta to be transition functions.  The last one has a beta but there is no beta.  It's your sound, the last beta.  It's the beta alpha.  What is the coordinate change?  It's called the change in the center.  The matrix say the CD of the vector plus the beta.  The change from the CD to the Z.  That's a good sense.  So this is the mobius code.  The transformation is better for this.  So the universal covering is the upper half plane.  And if you take Z, the local parameter here, this will induce on this Riemann surface such a coordinate system.  But you'll have many of them.  So, so.  Okay.  So this is a definition.  That's a definition.  So Z alpha beta are the transition function for half canonical.  And maybe they are linear in Z beta beta.  So in particular, the first derivative will vanish.  Okay.  Now I can maybe state some theorems.  So theorem one.  This was proved with Motoki-Kombola set 2017, I think.  Is that for any lambda that is a complex number,  then there exists a unique, non-trivial one parameter family of filtered extensions.  For F lambda, this is the filtration.  This is the vector space for the vector space being,  I'm going to denote V that is for this E0.  That was that direct sum of half canonical powers.  So this is a statement about the filtration, and now I want to put some connections on this filtration.  This is a theorem two.  And maybe I'm going to mention some things.  So I will give you the transition functions for this V lambda.  So let me recall F alpha beta gamma to be the transition functions.  This is lambda.  Maybe it wasn't a good notation.  To be defined as exponential of H times logarithm of Z alpha beta times exponential of lambda times d log of Z alpha beta times X plus.  So these are the transition functions for V lambda.  And the second part of the statement says that, well, yeah, when this is at zero, you just get back.  I mean, this will make the transition function for E0.  That is also my notation of E0.  And finally, that d plus 1 over lambda times a Higgs times.  So if I start with any element on the Hitchin section,  then this is an upper on the V lambda.  So this is a statement of filtration.  This is a statement about actually, well, you can quite identify this filtration.  So maybe I mean this first matrix, I mean, sometimes probably we denote it like Z alpha beta to the H to be to consist of exponential of H times logarithm of Z alpha beta.  And this is just the diagonal matrix consisting of just, yeah, this powers on the diagonal.  So this is the first part.  And the second part consists of another matrix that has well identity on the diagonal.  And on each of upper diagonal entries, we'll have powers of the lambda plus the derivative of Z alpha beta.  As I said, because X plus it was nilpotent, so when you do write down series, you'll obtain such an object.  As I said, these are linear, so that will be just a constant.  So when you multiply these two, you're just going to see a matrix that has these entries on the diagonal and then upper, and then, yeah,  and then upper, consisting of the derivative and powers of lambda.  Okay, so now I want to talk about something that is in between these pictures, and I will come back on these theorems.  So maybe I'm going to use this part here.  So I'm going to introduce the notion of, well, I'm going to talk about hot modular space.  So let me introduce this hot modular space of lambda connections, consisting of lambda, a vector bundle.  Maybe I'm going to call it lambda, and then the connection here.  So this is the modular space of the lean lambda connection.  And the rule is that the light-nit rule, the lambda light-nit rule is that this is lambda times df times s plus f times ds of s,  and f is just a holomorphic function, and s is just a section of the vector bundle.  Usually, I like to use the notation v lambda for the connections, but it's okay.  Is that a one half or one over lambda?  It's one over lambda.  So lambda is not zero.  Sorry?  So we're assuming lambda is not zero.  Yeah, yeah, yeah.  So I'm going to, it's one over lambda, sorry.  So this is right now a family of, yeah, I'm going to come back on this.  So lambda is not zero.  Okay, so the delin put these two definitions of the Higgs bundle and connection into this definition of the lambda,  of the lean lambda connection via this.  And we can notice, I mean, when lambda is zero, then you have, this becomes a Higgs bundle,  while if lambda equals to one, it becomes a connection.  So we should think about this as being, I mean, this, if here is the modular space,  if this is the modular space of the, if this is the dolbo modular space,  and this is the diram modular space, this is kind of in between.  So this corresponds to lambda equals zero.  This is the fiber over lambda equals to zero.  So it's lambda inverse of zero.  And this is the fiber over lambda equals to one.  And in between there are these lambda connections.  Now there exists a, this is the reaction on the Higgs field, on the dolbo modular space,  and mainly it will be induced from the c star action on the hot-moduli space.  So let me talk about this.  So we have a c star action on the hot-moduli space, and the way it acts,  I'm going to take t and non-zero parameter and look at the connection.  And I'm going to send it well into this.  And the fixed points of this action will be the ones that are left invariant under this c star action.  So you can see that this happens if lambda is zero.  So these fixed points happen if and only if lambda t is t is lambda.  So this means lambda is zero.  So the fixed points are Higgs bundles.  And they are denoted as hot bundles.  So I'm going to, so this denotes the c star fixed points on the c star action on the hot-moduli space.  So now probably I want to look at one example.  Maybe I'm just going to write here.  So I have this exercise that I did maybe two days ago.  So I'm going to save this notation above.  So this is maybe a remark.  So I'm going to start with phi of q on the Higgs section.  And then I have the following, that lambda,  v of lambda and d lambda plus phi of q is a lambda connection.  So this guy that appeared here if you multiply by lambda is just a lambda connection.  The second statement is that E0 comma x minus.  So E0 was the split vector bundle into half canonical powers.  And x minus was this constant.  So one diagonal below is a fixed point for this lambda connection.  And the third part is that the limit when t approaches 0 of t times lambda v lambda and lambda d plus phi of q.  So these are just these operas, but now they became lambda connections.  It's just, well, the same point is the same point.  And finally, the fact that the grated hodge bundle associated to this lambda connection  is going to be E0 and x minus.  So these are some statements we can easily check.  So the way to see this, what the c star action does,  well, I'm just going to apply the theorems above in order to prove this statement.  So z alpha beta, we said that these were the transition functions for half canonical.  What's the difference between 2 and 3?  Or maybe what is the meaning of 2?  I mean, yeah, I mean, yeah, what do you mean?  Oh, this is the fixed point.  So if you add here by t and you, you, you, you, um,  Is this c star action?  Oh, sorry.  Did I?  It's okay.  It's okay.  No, no, it's okay.  I just don't understand.  So it's okay.  It's okay.  It's okay.  It's okay.  It's okay.  It's okay.  It's okay.  Right.  I mean, but I will mention in the end that this will be the same thing.  But you know, I just wanted to, to give you the, the computation here.  Okay.  So that alpha beta are the transition functions for this.  So I can take like a times that alpha beta where a is a non-zero constant.  And this will be the same transient.  I mean, the transition function for that.  So, um, maybe I'm going to denote like F alpha beta and this is the A lambda and these are  the transition function corresponding to this, to this V V lambda.  And, um, so this, um, yeah, so what, what, um, what, what happens there that if you,  if you, if you, if you just multiply by a non-zero constant, the second part, the second  matrix will not be affected because, um, because this A is just a constant.  So it will be killed by the, by the D.  So you'll have the same thing, but only the, the first part will be affected in the first  statement of the theorem.  So for example, I'll, I'll take that, I'll have that X of H times, um, log of A times  that alpha beta will just be the sum between, uh, uh, will, will be the product, sorry,  the sum and then the product between this.  So for example, this will be probably my notation A to the H and this.  So, so, so, so I'm just, I'm just saying that this is just going to be A to the H and this  is just this diagonal matrix that has just powers of A on the diagonal and then F alpha  beta, the old ones.  And this just comes from here.  Okay, so when, when I, when I write, I mean, I'm going to take, um, so I'm, I'm going to  act by T and when, when I write my, my T times phi, well, T times phi in this new transition  function, probably what, what, what will happen is that I have to multiply by this new, uh,  A to the H and have to conjugate by that.  So this will, this will, uh, give you that, so it's the same Higgs field, T of phi, but  it's written as, um, A to, uh, A to T, uh, and then phi of A to 2i minus 2 times qi.  So this, this is just the vector q.  I mean, q, q consists of the vector qi of the, uh, differentials and it's just going  to have this, this constant.  So, so if, if, if, yeah.  Um, so, um, so, so for example, if q is equal to zero, um, then, then if you look at the  definition of the Hitching section, this will, phi of q will be just, uh, phi of zero will  be just x minus.  So, um,  Sorry, what happened to this?  It, it, it raised to H into, what is zeta?  Zeta alpha beta, there is a term in the,  Zeta alpha beta?  No, no, no, it was, it was the first term.  For first term, it was written.  The second term, you know, the second term.  Yeah, yeah, so, so, so, uh, nothing.  So, yeah, I mean, oh, yeah.  Sorry, I didn't understand.  It's good, good.  Yeah.  Okay, so, um, so if, uh, if, if, if this q is the, is, is, is the origin of the Hitching  paste, then, then, then, then this is just x minus.  And you can see here that for, if, if you choose a to be the square root of t, then this is  going to be mapped in the same, in the same point.  So, therefore, this x minus comma e zero is a c star, uh, fixed point for, for q equals  to zero.  Um, while if, if q is non-equal to zero, uh, then, um, if, if you choose again, equals  to the square root of t, um, then, what, what, what will happen here is that this will be  just phi of, of, um, and then this will be t i minus one q i.  So, when you take the limit when t equals to zero, now the q's will be, will be killed.  So, the only one surviving will be x minus.  So, this proves that the limit, um, when t approaches zero of this is going to be the  same point.  Yeah?  Because, uh, because, because i is going to be bigger equal than two in my notation.  So, all the differentials will come with t times q two and then t squared, q three and  so on.  So, you can take the limit at zero and obtain this q minus.  So, this proves, um, I mean one just is a reformulation of the first theorem and this  proves two and three together.  And now you can use probably theorem number one to prove four and I probably will not  do that but I can explain how to do it.  Uh, is it okay?  Questions?  So, uh, what is the graded hot bundle?  Well, um, yeah, probably, um, yeah, probably, um,  yeah, probably, so, um, I would have to compute the ratios of this, uh, filtration and if  you remember the definition of the, this one parameter family, a filtered taxation, this  came with this fixed ratios.  So, these fixed ratios in the end, they were known and there were powers of the half canonical.  So, um, so to compute a, um, yeah, so, so it's an easy computation to see that the  associated, the graded hot bundle from, from theorem, consequence of theorem number one  of, of, of this is, is, is just this because, uh, because, yeah, you, you will have that,  you will have that at each, each step you'll have to take these ratios that will be like,  um, K, half canonical K to R minus one or R minus three and so on and then you'll have  to twist by the canonical.  So, that will be an isomorphism in the end and this X minus says that you have just this  constant on lower diagonals and they will give you the isomorphism between this.  Okay.  So, this was just an exercise in order, um, for, yeah, to, to get, um, in order to, to  get some terminology.  Um, okay.  So, let me, let me mention some theorems now.  Yes.  I mean, they, yeah.  They, so, so here it is just, uh, uh, yeah, they, um, exterior derivative, yeah.  But did we, did we choose some special property of the, of the transition functions or any,  which is any transition function?  It's a canonical shift, so, it makes sense.  So, tomorrow at the start we chose some transition function, some arch, the S alpha beta.  Is this going to depend on the choice?  When you have the F alpha beta and the S alpha beta?  S alpha beta.  Oh, Zeta.  Yes.  Uh, it, so, so it's going to, to probably depend on the spin choice, right?  So, this Zeta alpha beta came from the, yeah, spin structure, but you choose just the core,  projective coordinates.  You don't need a solution at the end of the division to start somewhere?  No, no, it's projective structure.  Right.  Projective structure.  Right, I'm not using the solution here.  You get symmetric power of the two dimension representation, you get flat connection, yeah.  I mean, here, here, on, on this board I kind of defined everything I needed for this.  You started with the complex projective structure?  Yes.  Yes, yes, yes.  That was the, right.  If you choose a different complex projective structure, you'll have a different base point  there, is that right?  Yeah, yeah, yeah.  Okay.  Okay, so let me now measure some theorems.  Should I do that?  Maybe, maybe I'll just erase this board.  Okay, so, so, so, so, so theorems and this is mainly the work of Carlos Simpson, is that  for every lambda v lambda, that is a hot bundle, so for any lambda-dealing connection, then  there exists, or maybe I should say to fix a lambda first, fix a lambda on the base.  So I'm just going to fix a lambda and look at the fiber.  And I will choose a lambda-dealing connection, then there exists a unique section.  This star goes to the hot module ispace, passing through lambda, the lambda, and the connection.  The limit when t goes to zero always exists.  So fixed, so let me just denote m double of c star consisting of this fixed points, p  alpha, where alpha is a class of variation of hot structure.  And for any point, p zero here, then if I look at the inverse, I mean if I look at  all the points, all the points are fixed under this section in p zero and I restrict  to lambda equals to one, this is a holomorphic Lagrangian.  Oh, that's a different sigma, right?  You use sigma for section.  Yeah, yeah.  Sorry?  No, no, you just on the left, you say sigma is half of c star.  Oh.  And now it's sigma, it's something else here.  Right.  Number two, you meant sigma.  So the notation is not sigma, you just confuse this notation.  Yeah, sorry, I just switched the all notations in terms of h bar and sigma just before.  Yeah, you can erase that sigma over there.  Here?  But then what?  Okay.  Sigma is number two math, limit, limit.  Oh, yeah, yeah, yeah.  Yeah, yeah, yeah, yeah.  So maybe this one.  In section, you have one exception?  No, maybe c star invariant section.  Yeah, yeah, it's a c star invariant section.  So, okay, so what do I want to say?  Sigma of lambda, whatever, is equal to that limit.  Yeah.  This is the sigma, yeah.  Okay.  So, all the points that are fixed under this, all the, so for any, so here I have the set  of all the fixed points that live in the dolbo.  And for every point here, I look at the, at all the points that get fixed under this.  And this will be a holomorphic Lagrangian.  And this will give the stratification of the Dirac modular space.  So this is a holomorphic Lagrangian vibration.  So all the points that are c star fixed points under this, under this action.  Okay.  So, so there exists this, there is the, this Lagrangian vibration on the, on the Dram,  and you can do the same thing and restrict mainly to, to the lambda zero.  And you'll obtain a vibration for, so maybe I'm going to, to just write it that if you  restrict to lambda equals to zero, you'll obtain again a Lagrangian vibration of the  dolbo modular space.  So these are all the Higgs bundles that have that correspond, I mean, that are in the fiber,  that have this, the same Higgs point.  So, okay.  So, so more, moreover, this is a holomorphic Lagrangian on the dolbo modular space.  So the modular space, the dolbo modular space breaks into this p alpha of, of, into, into  this, this is a whole homomorphic stratification of Lagrangians.  Okay.  So, maybe now I can, I can mention the, the theorem.  The idea of that would be like some kind of vibration, I think that's the connection  of, of, of, sorry, of the Dram or of, of the Dram.  Okay.  Okay.  It's in the subspaces are Lagrangian, but that's what she meant.  It's a vibration by holomorphic Lagrangian.  Yeah.  It's a holomorphic vibration.  Yeah, the statement that those are closed subroutings, for example, is, is just a conjecture.  They're not closed here, but they're, it's just a conjecture that they're closed here.  Okay.  Okay.  Yeah, I see.  So maybe.  Okay.  I'll finish them by holomorphic Lagrangian.  Yes, yes.  Okay.  Yeah, same thing if you're on the other side.  Well, they're, they're not, they're not closed here.  Well, I mean, I, no, I don't think it's correct to say, I mean, those subroutings are not  closed in that, in the Dolbo.  So I'm not sure what the vibration means, but I mean.  Okay.  So then I just, I just leave that holomorphic Lagrangian.  Okay.  So, so, so the theorem.  And this is with Laura Fredrickson, Church Kidonakis, Rathematio Motohikomoulas and  Antinetsky in maybe it was 2016 was that this.  So, so if, if, if, if I restrict to P0 that is E0 X minus and we proved in that lemma  that this is a fixed point, right, this is the fixed point.  I mean, and what this is a fixed point of the hitching section on the Dolbo and also  on the, on the upper module.  So this is what that lemma says.  Then, then, then this two, then these two Lagrangians are by holomorphic.  So this is the fixed point when lambda equals to zero are by holomorphic via Galloto conformal  limit.  And yeah, we, we basically, we basically, yeah.  So this correspondence between the, the, the, the, the, the, the Higgs bundle on the hitching  section and, and, and this upper and that this is the limit of the, this is the conformal  limit of Galloto.  Okay.  And moreover, it was generalized by Collier and Wentworth a year ago to all such fixed  points.  So this is the theorem Collier and Wentworth.  So this is a fixed point that, for any point, that is a fixed point under this C star action.  Then, and then the alpha variation of horse structure.  Then there exists a bipholomorphic correspondence between, so this two fibers, these two Lagrangians  now, one on the double both side and the other one on the theorem side are by holomorphic  Lagrangians.  Via Galloto's conformal limit.  And this Galloto's conformal limit is explicit.  So you can see from this, so you can see that, and, and the methods here that were used are  similar to the methods that were used by us in, yeah, I have very short time.  They are similar to our technique and you can see that in that case, this corresponds  to the longest, the full flag.  Like the, I mean the very, the all corresponds to the full flag.  And you can, I can try to explain for example, the, on the opposite side, you have the shortest  flag when E is a stable hex, when E is a stable vector bundle.  So if for a stable vector bundle E, then you have that T phi is in the dolbomoduli space  for any phi.  So if E is just a stable vector bundle, then this is a stable hex field for any phi.  And E comma zero is the fixed point.  But you have the cotangent fiber of E over the moduli space of stable vector bundles  of rank card and degree zero that is contained in the dolbo.  And it's a Lagrangian.  So this will be the corresponding Lagrangian.  And for every stable hex, for every, every stable vector bundle, there exists a unique  Hermitian Einstein metric.  And once you take T to be the turn connection corresponding to that Hermitian Einstein metric,  the unique turn connection.  Then you get this maybe by Narasimha Shashadri, E corresponds to D. And mainly the fact that  E was a vector bundle of, of degree zero, it tells you that, that this has curvature  zero.  So, mainly the solution of Kitchin's equation split kind of the couple into the curvature  zero part and then the part where phi, the, yeah, the phi, phi dagger is zero.  That can be done because, yeah, phi can be a diagonal on this, on this Hermitian metric.  So this is kind of a short proof for these opposite cases in the case of stable vector  bundle and in the case of oppers and in between it's, it's, it, the proof is very similar.  Anyway so maybe, maybe, maybe I'll try to, I'll try to give the general picture here.  So, so this is the dolbo and this is the diram.  And here, well here, here is, is the corresponding agrenation of the dolbo and the diram and  you have this kind of slice in the hot bundle.  But this, this, this is saying that, that, that, that this is just trivial.  So it's going to be a product between C and then one of these Lagrangians.  So if I want to define, let's say a semi, a geometric, semi-classical limit, I can,  I can start from a lambda connection, go to, to the, to the dolbo side by this and then  project on the pitching base.  So that is, that is a geometry definition of a semi-classical limit.  There is another definition coming from physics when this is more like the limit when h bar  goes to zero of, yeah.  I can try to define a semi-classical limit of this and the theorem is that these two  agree, these two definitions agree on modulite space of uppers and that was, so that, yeah.  So maybe I can, I can define a map from the diram to the pitching base via this, well,  stratification or that is trivial.  And this, sorry, yeah.  It's, yeah, it's holomorphic, it's probably on each slice or it's not on the whole thing.  It's not on the whole thing.  But it's probably holomorphic, I mean, it's between these Lagrangians but it's, yeah,  so probably going to be on the slides, so.  But it's not on the whole, this is not holomorphic on the whole modulite space.  And here, yeah.  So probably I will stop here.  Thank you.  I have a question, which is, so at the start you got a similar type of correspondence but  it depends on the choice of base point that was the projective, how much projective structure,  right?  Yes.  Yeah, an idea, because this one depends on the Sushman-Hijian equations, right?  Right, right, I mean, it goes through, yeah.  So do you have an idea of how those two compare?  I mean, you could, because you're ahead, your formula with D plus, right, I mean, this is  also explicit.  Yeah, I mean, yeah, well, I guess this, yeah, well, I think it's a similar formula to that,  D plus 1 over lambda of phi.  So here you, it's not the same.  Maybe it's the same if, is it the same if the complex projective structure is the one  given by the uniformization?  Yes, yeah, good.  That's exactly the case.  Yes.  So if the complex projective structure is not the one given by uniformization, then do  you have a comparison between them?  And for example, is it going to be linear on this vector space?  I don't know.  Yeah.  I don't know.  Any other questions?  I think it's figure again.
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Thank you very much for the introduction. I would like to thank the organizers for the opportunity to speak. So today I'm going to present a partial holomorphic isomorphism between two modular spaces. So we have the modular space, the dolbo modular spaces, and I'm briefly going to define of stable hex bundles and the d-ROM modular spaces of irreducible connections. They are related by the non-o-billion-hatch correspondence that is a diffeomorphism. So what I'm going to do, I'm going to explain a partial holomorphic isomorphism between these, mainly between Lagrangians. So that would be a holomorphic Lagrangian correspondence by holomorphic between each of these stratifications. And moreover, I'm going to, in the end, I'm going to make a definition of the semi-classical limit via a geometrical definition. So first I'm going to start by introducing the notation and maybe some of the, I will spend some time reminding the old results. So all of this talk is going to be with the group SLN of C, maybe I will use SLR of C. And C will be a productive algebraic smooth curve. SCL? Yeah. Semi-classical limit? Oh, I'm sorry. SCL. This is the most important part, so yeah. And so I'm going to choose a spin structure and that is going to be L, a line bundle so that L squared is canonical. And because I'm working with this group, my vector bundles, well, I'm going to use both notation C and V to denote holomorphic vector bundles of rank R and degree 0. But also phi is going to be a traceless fixed field. So let me briefly mention this to modular space. So on one side we have the del bomoduli space consisting of stable Higgs pairs. And phi is a map from E to E tensor KC, canonical, that is a no-seat homomorphism. And then I want to introduce also the diromoduli space consisting of irreducible connections. And nabla is a map from E to E tensor KC. That is a C homomorphism. So these two spaces are homomorphic by the non-Abelian hodge. Let me also briefly mention the heating vibration. So we have the dol bomoduli space. And we have a vibration over the vector space consisting of all the sections of the pluricanonical. And in this case, I consider traceless Higgs fields. So it's mainly going to map these stable Higgs into the spectral data determined by the characteristic polynomial of phi. And this is an element in the heating base. Excuse me. These irreducible connections, are they flat or not? Flat. Flat. Yeah. Hologmorphic. It's a holomorphic, yes. So it's much more complicated. OK. Thank you. OK. So this is the general framework. And next, what I'm going to do is to introduce there are two Lagrangians in this space that mainly give a kind of affine parameterization of the heating base. And one will be called the heating section. And the other one will be the upper noduli. So before, because I kind of talked about this a while ago, I this. Maybe I just, oh no. Maybe I'm just going to. OK. OK. Is it. So, so, so many. I want to define the section inside the heating module space inside the dolbo module space. So what I will consider are vector bundles that are split bundles. So they split in, these are line bundles, powers of the canonical bundle, half canonical. And X minus is this matrix consisting and p is just constant. So X minus is a matrix with just nonzero constants on one diagonal lower. And X plus is going to be the transpose. So you will have the same diagonal upper. And it is going to be the commutator between these two, X plus and X minus. So that will make a diagonal matrix consisting of values, I mean, decreased by two every step. And these are the generators of the SL2 representation to SLR. And now if I consider Q to be, oh, to be a section here, I mean, to be just a vector here. This means that it has components and it consists of differential. Q2 will be a quadratic differential and so on. QR will be a section of the pluric canonical. And now the elements of the hitching section consist of E0. So they always have this split vector bundle. And phi of Q that consists of X minus. And I'm going to add, so X plus will be nilpotent. So this means that the power will cancel. So all the powers will actually will go one diagonal. And I will just multiply by Q's each of those diagonals. So that will be my phi of Q that is just an, well, has an upper triangular part. And lower diagonal consisting of this just constant. And it is not difficult to see that actually this is a stable kick bundle. And in fact, X minus is a stable kick bundle on E0. So it's a point on the del bomogelite space. And this is known as the hitching section. And it is a section of this vibration in the sense that it intersects every fiber of this once. It's a section on, in rank two, but otherwise it's just intersect the fiber once. And if you apply twice, it doesn't give you identity. So now I want to introduce another Lagrangian inside the Dirac moduli space that is the upper moduli. So I just wrote the definition of the upper. And this consists of a filtration of vector sub bundles. And the Bellisson-Dreamfield definition says that nabla is a connection is an upper. If, well, I'm using the decreasing filtration. If it satisfies the Griffith transversality condition. And this will induce an OC linear map. And this means a Higgs bundle on the quotients. And the condition is that that is isomorphism. So this being an isomorphism implies that the discussions are, that this is a full flag. So the rank goes every step by one. But yeah, so what I'm going to, maybe I'm just going to briefly introduce now a new definition. So let me define here. For any lambda in age one of CKC, so this is just a complex number. I want to define a one parameter family of filtered extension to be F lambda this filtration. This is the vector bundle. And maybe I'll just write it in words. So I do have a full flag. And the last condition is fulfilled. So I do have an isomorphism between the ratios. And finally, I want to mention a starting point. And the starting point of this I want to mention the first one here is going to be just the power of the half canonical. So it's just going to be this. So for example, for example, in R equals to 2, I have that for every lambda, a complex number. Then there exists a unique filtered extension like this that is given by the short exact sequence. lambda doesn't play a role in the diffusion. The diffusion lambda plays no role at all. Well, so. I eat extensions. Yes, it's extension classes. So many this will give the first ratio. And then by this condition, that is quite a strong condition, you can compute all the ratios. So this will be like the first. And then what I'm going to do, I'm going to sandwich like in rank two and construct another in another exam, another in another filtered extension. And right, so I want to be the last one to be the last term, the vector about the to be the last term of the extension. And because I want to I want to consider one example, let me introduce one definition. So I'm going to recall project coordinate system. So many if if if my ruin surface decomposes into a union of open sets, then I want that the transition functions. These are the coordinates on the open set are going to be given by mobius transformations. And such such project coordinate system always exists. Because for example, one comes from the universal covering that is just the upper half plane. So let me and this matrix I'm asking to be in SL2 of C, this matrix say the CD alpha beta. So I'm going to introduce coordinates now. And I'm going to do this Z alpha beta to be transition functions. The last one has a beta but there is no beta. It's your sound, the last beta. It's the beta alpha. What is the coordinate change? It's called the change in the center. The matrix say the CD of the vector plus the beta. The change from the CD to the Z. That's a good sense. So this is the mobius code. The transformation is better for this. So the universal covering is the upper half plane. And if you take Z, the local parameter here, this will induce on this Riemann surface such a coordinate system. But you'll have many of them. So, so. Okay. So this is a definition. That's a definition. So Z alpha beta are the transition function for half canonical. And maybe they are linear in Z beta beta. So in particular, the first derivative will vanish. Okay. Now I can maybe state some theorems. So theorem one. This was proved with Motoki-Kombola set 2017, I think. Is that for any lambda that is a complex number, then there exists a unique, non-trivial one parameter family of filtered extensions. For F lambda, this is the filtration. This is the vector space for the vector space being, I'm going to denote V that is for this E0. That was that direct sum of half canonical powers. So this is a statement about the filtration, and now I want to put some connections on this filtration. This is a theorem two. And maybe I'm going to mention some things. So I will give you the transition functions for this V lambda. So let me recall F alpha beta gamma to be the transition functions. This is lambda. Maybe it wasn't a good notation. To be defined as exponential of H times logarithm of Z alpha beta times exponential of lambda times d log of Z alpha beta times X plus. So these are the transition functions for V lambda. And the second part of the statement says that, well, yeah, when this is at zero, you just get back. I mean, this will make the transition function for E0. That is also my notation of E0. And finally, that d plus 1 over lambda times a Higgs times. So if I start with any element on the Hitchin section, then this is an upper on the V lambda. So this is a statement of filtration. This is a statement about actually, well, you can quite identify this filtration. So maybe I mean this first matrix, I mean, sometimes probably we denote it like Z alpha beta to the H to be to consist of exponential of H times logarithm of Z alpha beta. And this is just the diagonal matrix consisting of just, yeah, this powers on the diagonal. So this is the first part. And the second part consists of another matrix that has well identity on the diagonal. And on each of upper diagonal entries, we'll have powers of the lambda plus the derivative of Z alpha beta. As I said, because X plus it was nilpotent, so when you do write down series, you'll obtain such an object. As I said, these are linear, so that will be just a constant. So when you multiply these two, you're just going to see a matrix that has these entries on the diagonal and then upper, and then, yeah, and then upper, consisting of the derivative and powers of lambda. Okay, so now I want to talk about something that is in between these pictures, and I will come back on these theorems. So maybe I'm going to use this part here. So I'm going to introduce the notion of, well, I'm going to talk about hot modular space. So let me introduce this hot modular space of lambda connections, consisting of lambda, a vector bundle. Maybe I'm going to call it lambda, and then the connection here. So this is the modular space of the lean lambda connection. And the rule is that the light-nit rule, the lambda light-nit rule is that this is lambda times df times s plus f times ds of s, and f is just a holomorphic function, and s is just a section of the vector bundle. Usually, I like to use the notation v lambda for the connections, but it's okay. Is that a one half or one over lambda? It's one over lambda. So lambda is not zero. Sorry? So we're assuming lambda is not zero. Yeah, yeah, yeah. So I'm going to, it's one over lambda, sorry. So this is right now a family of, yeah, I'm going to come back on this. So lambda is not zero. Okay, so the delin put these two definitions of the Higgs bundle and connection into this definition of the lambda, of the lean lambda connection via this. And we can notice, I mean, when lambda is zero, then you have, this becomes a Higgs bundle, while if lambda equals to one, it becomes a connection. So we should think about this as being, I mean, this, if here is the modular space, if this is the modular space of the, if this is the dolbo modular space, and this is the diram modular space, this is kind of in between. So this corresponds to lambda equals zero. This is the fiber over lambda equals to zero. So it's lambda inverse of zero. And this is the fiber over lambda equals to one. And in between there are these lambda connections. Now there exists a, this is the reaction on the Higgs field, on the dolbo modular space, and mainly it will be induced from the c star action on the hot-moduli space. So let me talk about this. So we have a c star action on the hot-moduli space, and the way it acts, I'm going to take t and non-zero parameter and look at the connection. And I'm going to send it well into this. And the fixed points of this action will be the ones that are left invariant under this c star action. So you can see that this happens if lambda is zero. So these fixed points happen if and only if lambda t is t is lambda. So this means lambda is zero. So the fixed points are Higgs bundles. And they are denoted as hot bundles. So I'm going to, so this denotes the c star fixed points on the c star action on the hot-moduli space. So now probably I want to look at one example. Maybe I'm just going to write here. So I have this exercise that I did maybe two days ago. So I'm going to save this notation above. So this is maybe a remark. So I'm going to start with phi of q on the Higgs section. And then I have the following, that lambda, v of lambda and d lambda plus phi of q is a lambda connection. So this guy that appeared here if you multiply by lambda is just a lambda connection. The second statement is that E0 comma x minus. So E0 was the split vector bundle into half canonical powers. And x minus was this constant. So one diagonal below is a fixed point for this lambda connection. And the third part is that the limit when t approaches 0 of t times lambda v lambda and lambda d plus phi of q. So these are just these operas, but now they became lambda connections. It's just, well, the same point is the same point. And finally, the fact that the grated hodge bundle associated to this lambda connection is going to be E0 and x minus. So these are some statements we can easily check. So the way to see this, what the c star action does, well, I'm just going to apply the theorems above in order to prove this statement. So z alpha beta, we said that these were the transition functions for half canonical. What's the difference between 2 and 3? Or maybe what is the meaning of 2? I mean, yeah, I mean, yeah, what do you mean? Oh, this is the fixed point. So if you add here by t and you, you, you, you, um, Is this c star action? Oh, sorry. Did I? It's okay. It's okay. No, no, it's okay. I just don't understand. So it's okay. It's okay. It's okay. It's okay. It's okay. It's okay. It's okay. Right. I mean, but I will mention in the end that this will be the same thing. But you know, I just wanted to, to give you the, the computation here. Okay. So that alpha beta are the transition functions for this. So I can take like a times that alpha beta where a is a non-zero constant. And this will be the same transient. I mean, the transition function for that. So, um, maybe I'm going to denote like F alpha beta and this is the A lambda and these are the transition function corresponding to this, to this V V lambda. And, um, so this, um, yeah, so what, what, um, what, what happens there that if you, if you, if you, if you just multiply by a non-zero constant, the second part, the second matrix will not be affected because, um, because this A is just a constant. So it will be killed by the, by the D. So you'll have the same thing, but only the, the first part will be affected in the first statement of the theorem. So for example, I'll, I'll take that, I'll have that X of H times, um, log of A times that alpha beta will just be the sum between, uh, uh, will, will be the product, sorry, the sum and then the product between this. So for example, this will be probably my notation A to the H and this. So, so, so, so I'm just, I'm just saying that this is just going to be A to the H and this is just this diagonal matrix that has just powers of A on the diagonal and then F alpha beta, the old ones. And this just comes from here. Okay, so when, when I, when I write, I mean, I'm going to take, um, so I'm, I'm going to act by T and when, when I write my, my T times phi, well, T times phi in this new transition function, probably what, what, what will happen is that I have to multiply by this new, uh, A to the H and have to conjugate by that. So this will, this will, uh, give you that, so it's the same Higgs field, T of phi, but it's written as, um, A to, uh, A to T, uh, and then phi of A to 2i minus 2 times qi. So this, this is just the vector q. I mean, q, q consists of the vector qi of the, uh, differentials and it's just going to have this, this constant. So, so if, if, if, yeah. Um, so, um, so, so for example, if q is equal to zero, um, then, then if you look at the definition of the Hitching section, this will, phi of q will be just, uh, phi of zero will be just x minus. So, um, Sorry, what happened to this? It, it, it raised to H into, what is zeta? Zeta alpha beta, there is a term in the, Zeta alpha beta? No, no, no, it was, it was the first term. For first term, it was written. The second term, you know, the second term. Yeah, yeah, so, so, so, uh, nothing. So, yeah, I mean, oh, yeah. Sorry, I didn't understand. It's good, good. Yeah. Okay, so, um, so if, uh, if, if, if this q is the, is, is, is the origin of the Hitching paste, then, then, then, then this is just x minus. And you can see here that for, if, if you choose a to be the square root of t, then this is going to be mapped in the same, in the same point. So, therefore, this x minus comma e zero is a c star, uh, fixed point for, for q equals to zero. Um, while if, if q is non-equal to zero, uh, then, um, if, if you choose again, equals to the square root of t, um, then, what, what, what will happen here is that this will be just phi of, of, um, and then this will be t i minus one q i. So, when you take the limit when t equals to zero, now the q's will be, will be killed. So, the only one surviving will be x minus. So, this proves that the limit, um, when t approaches zero of this is going to be the same point. Yeah? Because, uh, because, because i is going to be bigger equal than two in my notation. So, all the differentials will come with t times q two and then t squared, q three and so on. So, you can take the limit at zero and obtain this q minus. So, this proves, um, I mean one just is a reformulation of the first theorem and this proves two and three together. And now you can use probably theorem number one to prove four and I probably will not do that but I can explain how to do it. Uh, is it okay? Questions? So, uh, what is the graded hot bundle? Well, um, yeah, probably, um, yeah, probably, um, yeah, probably, so, um, I would have to compute the ratios of this, uh, filtration and if you remember the definition of the, this one parameter family, a filtered taxation, this came with this fixed ratios. So, these fixed ratios in the end, they were known and there were powers of the half canonical. So, um, so to compute a, um, yeah, so, so it's an easy computation to see that the associated, the graded hot bundle from, from theorem, consequence of theorem number one of, of, of this is, is, is just this because, uh, because, yeah, you, you will have that, you will have that at each, each step you'll have to take these ratios that will be like, um, K, half canonical K to R minus one or R minus three and so on and then you'll have to twist by the canonical. So, that will be an isomorphism in the end and this X minus says that you have just this constant on lower diagonals and they will give you the isomorphism between this. Okay. So, this was just an exercise in order, um, for, yeah, to, to get, um, in order to, to get some terminology. Um, okay. So, let me, let me mention some theorems now. Yes. I mean, they, yeah. They, so, so here it is just, uh, uh, yeah, they, um, exterior derivative, yeah. But did we, did we choose some special property of the, of the transition functions or any, which is any transition function? It's a canonical shift, so, it makes sense. So, tomorrow at the start we chose some transition function, some arch, the S alpha beta. Is this going to depend on the choice? When you have the F alpha beta and the S alpha beta? S alpha beta. Oh, Zeta. Yes. Uh, it, so, so it's going to, to probably depend on the spin choice, right? So, this Zeta alpha beta came from the, yeah, spin structure, but you choose just the core, projective coordinates. You don't need a solution at the end of the division to start somewhere? No, no, it's projective structure. Right. Projective structure. Right, I'm not using the solution here. You get symmetric power of the two dimension representation, you get flat connection, yeah. I mean, here, here, on, on this board I kind of defined everything I needed for this. You started with the complex projective structure? Yes. Yes, yes, yes. That was the, right. If you choose a different complex projective structure, you'll have a different base point there, is that right? Yeah, yeah, yeah. Okay. Okay, so let me now measure some theorems. Should I do that? Maybe, maybe I'll just erase this board. Okay, so, so, so, so, so theorems and this is mainly the work of Carlos Simpson, is that for every lambda v lambda, that is a hot bundle, so for any lambda-dealing connection, then there exists, or maybe I should say to fix a lambda first, fix a lambda on the base. So I'm just going to fix a lambda and look at the fiber. And I will choose a lambda-dealing connection, then there exists a unique section. This star goes to the hot module ispace, passing through lambda, the lambda, and the connection. The limit when t goes to zero always exists. So fixed, so let me just denote m double of c star consisting of this fixed points, p alpha, where alpha is a class of variation of hot structure. And for any point, p zero here, then if I look at the inverse, I mean if I look at all the points, all the points are fixed under this section in p zero and I restrict to lambda equals to one, this is a holomorphic Lagrangian. Oh, that's a different sigma, right? You use sigma for section. Yeah, yeah. Sorry? No, no, you just on the left, you say sigma is half of c star. Oh. And now it's sigma, it's something else here. Right. Number two, you meant sigma. So the notation is not sigma, you just confuse this notation. Yeah, sorry, I just switched the all notations in terms of h bar and sigma just before. Yeah, you can erase that sigma over there. Here? But then what? Okay. Sigma is number two math, limit, limit. Oh, yeah, yeah, yeah. Yeah, yeah, yeah, yeah. So maybe this one. In section, you have one exception? No, maybe c star invariant section. Yeah, yeah, it's a c star invariant section. So, okay, so what do I want to say? Sigma of lambda, whatever, is equal to that limit. Yeah. This is the sigma, yeah. Okay. So, all the points that are fixed under this, all the, so for any, so here I have the set of all the fixed points that live in the dolbo. And for every point here, I look at the, at all the points that get fixed under this. And this will be a holomorphic Lagrangian. And this will give the stratification of the Dirac modular space. So this is a holomorphic Lagrangian vibration. So all the points that are c star fixed points under this, under this action. Okay. So, so there exists this, there is the, this Lagrangian vibration on the, on the Dram, and you can do the same thing and restrict mainly to, to the lambda zero. And you'll obtain a vibration for, so maybe I'm going to, to just write it that if you restrict to lambda equals to zero, you'll obtain again a Lagrangian vibration of the dolbo modular space. So these are all the Higgs bundles that have that correspond, I mean, that are in the fiber, that have this, the same Higgs point. So, okay. So, so more, moreover, this is a holomorphic Lagrangian on the dolbo modular space. So the modular space, the dolbo modular space breaks into this p alpha of, of, into, into this, this is a whole homomorphic stratification of Lagrangians. Okay. So, maybe now I can, I can mention the, the theorem. The idea of that would be like some kind of vibration, I think that's the connection of, of, of, sorry, of the Dram or of, of the Dram. Okay. Okay. It's in the subspaces are Lagrangian, but that's what she meant. It's a vibration by holomorphic Lagrangian. Yeah. It's a holomorphic vibration. Yeah, the statement that those are closed subroutings, for example, is, is just a conjecture. They're not closed here, but they're, it's just a conjecture that they're closed here. Okay. Okay. Yeah, I see. So maybe. Okay. I'll finish them by holomorphic Lagrangian. Yes, yes. Okay. Yeah, same thing if you're on the other side. Well, they're, they're not, they're not closed here. Well, I mean, I, no, I don't think it's correct to say, I mean, those subroutings are not closed in that, in the Dolbo. So I'm not sure what the vibration means, but I mean. Okay. So then I just, I just leave that holomorphic Lagrangian. Okay. So, so, so the theorem. And this is with Laura Fredrickson, Church Kidonakis, Rathematio Motohikomoulas and Antinetsky in maybe it was 2016 was that this. So, so if, if, if, if I restrict to P0 that is E0 X minus and we proved in that lemma that this is a fixed point, right, this is the fixed point. I mean, and what this is a fixed point of the hitching section on the Dolbo and also on the, on the upper module. So this is what that lemma says. Then, then, then this two, then these two Lagrangians are by holomorphic. So this is the fixed point when lambda equals to zero are by holomorphic via Galloto conformal limit. And yeah, we, we basically, we basically, yeah. So this correspondence between the, the, the, the, the, the, the Higgs bundle on the hitching section and, and, and this upper and that this is the limit of the, this is the conformal limit of Galloto. Okay. And moreover, it was generalized by Collier and Wentworth a year ago to all such fixed points. So this is the theorem Collier and Wentworth. So this is a fixed point that, for any point, that is a fixed point under this C star action. Then, and then the alpha variation of horse structure. Then there exists a bipholomorphic correspondence between, so this two fibers, these two Lagrangians now, one on the double both side and the other one on the theorem side are by holomorphic Lagrangians. Via Galloto's conformal limit. And this Galloto's conformal limit is explicit. So you can see from this, so you can see that, and, and the methods here that were used are similar to the methods that were used by us in, yeah, I have very short time. They are similar to our technique and you can see that in that case, this corresponds to the longest, the full flag. Like the, I mean the very, the all corresponds to the full flag. And you can, I can try to explain for example, the, on the opposite side, you have the shortest flag when E is a stable hex, when E is a stable vector bundle. So if for a stable vector bundle E, then you have that T phi is in the dolbomoduli space for any phi. So if E is just a stable vector bundle, then this is a stable hex field for any phi. And E comma zero is the fixed point. But you have the cotangent fiber of E over the moduli space of stable vector bundles of rank card and degree zero that is contained in the dolbo. And it's a Lagrangian. So this will be the corresponding Lagrangian. And for every stable hex, for every, every stable vector bundle, there exists a unique Hermitian Einstein metric. And once you take T to be the turn connection corresponding to that Hermitian Einstein metric, the unique turn connection. Then you get this maybe by Narasimha Shashadri, E corresponds to D. And mainly the fact that E was a vector bundle of, of degree zero, it tells you that, that this has curvature zero. So, mainly the solution of Kitchin's equation split kind of the couple into the curvature zero part and then the part where phi, the, yeah, the phi, phi dagger is zero. That can be done because, yeah, phi can be a diagonal on this, on this Hermitian metric. So this is kind of a short proof for these opposite cases in the case of stable vector bundle and in the case of oppers and in between it's, it's, it, the proof is very similar. Anyway so maybe, maybe, maybe I'll try to, I'll try to give the general picture here. So, so this is the dolbo and this is the diram. And here, well here, here is, is the corresponding agrenation of the dolbo and the diram and you have this kind of slice in the hot bundle. But this, this, this is saying that, that, that, that this is just trivial. So it's going to be a product between C and then one of these Lagrangians. So if I want to define, let's say a semi, a geometric, semi-classical limit, I can, I can start from a lambda connection, go to, to the, to the dolbo side by this and then project on the pitching base. So that is, that is a geometry definition of a semi-classical limit. There is another definition coming from physics when this is more like the limit when h bar goes to zero of, yeah. I can try to define a semi-classical limit of this and the theorem is that these two agree, these two definitions agree on modulite space of uppers and that was, so that, yeah. So maybe I can, I can define a map from the diram to the pitching base via this, well, stratification or that is trivial. And this, sorry, yeah. It's, yeah, it's holomorphic, it's probably on each slice or it's not on the whole thing. It's not on the whole thing. But it's probably holomorphic, I mean, it's between these Lagrangians but it's, yeah, so probably going to be on the slides, so. But it's not on the whole, this is not holomorphic on the whole modulite space. And here, yeah. So probably I will stop here. Thank you. I have a question, which is, so at the start you got a similar type of correspondence but it depends on the choice of base point that was the projective, how much projective structure, right? Yes. Yeah, an idea, because this one depends on the Sushman-Hijian equations, right? Right, right, I mean, it goes through, yeah. So do you have an idea of how those two compare? I mean, you could, because you're ahead, your formula with D plus, right, I mean, this is also explicit. Yeah, I mean, yeah, well, I guess this, yeah, well, I think it's a similar formula to that, D plus 1 over lambda of phi. So here you, it's not the same. Maybe it's the same if, is it the same if the complex projective structure is the one given by the uniformization? Yes, yeah, good. That's exactly the case. Yes. So if the complex projective structure is not the one given by uniformization, then do you have a comparison between them? And for example, is it going to be linear on this vector space? I don't know. Yeah. I don't know. Any other questions? I think it's figure again.
Lagrangian inside the Dirac moduli space that is Right. I mean, but I will mention in the end that is that I have to multiply by this new, uh, A to the connection of, of, of, sorry, of the Dram or