 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 deramodulized spaces of irreducible connections, and they are related by the nono-billion-hodge correspondence that is the 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 stratification. And moreover, I'm going to, in the end, I'm going to make a definition of the semi-classical limit via 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 projective algebraic smooth curve. Oh, I'm sorry, 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 the 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 these two moduli space. So on one side we have the double moduli space consisting of stable Higgs pairs, and phi is a map from E to E tensor KC, canonical, that is an OSEA 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 homeomorphic by the non-Abelian hodge. Let me also briefly mention the heating vibration. So we have the dolbo moduli space, and we have a vibration over the yaw-fine space, the vector space consisting of all the sections of the polar canonical, 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. It's a homomorphic, yes. Okay, so this is this is the general framework, and next what I'm going to do is to introduce the 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 open moduli. So before, because I kind of talked about this a while ago, I this maybe I just, oh no, maybe I'm just good. So mainly I want to define the section inside the heating moduli space, inside the dolbo moduli space. So what I will consider are vector bundles, they 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 non-zero constants on one diagonal lower, and x plus is going to be the transpose, so you will have the same diagonal upper, and it's 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 live in, will be a section of the purely 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 go one diagonal, 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, 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 dolbomodular 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, it's a section on, in rank two, but otherwise it's just intersect the fiber once, but if you apply it 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 a definition of the upper, and this consists of a filtration of vector sub bundles, and the Bellison-Dreamfield definition says that nabla is a, 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, that discussions are, that, that this is a full flag, so, so the, the rank goes every step by one, but yeah, so what I'm going to, what I'm, maybe I'm just going to briefly introduce now a new definition, so let me define here, for, 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 disfiltration, this is the vector bundle, and maybe I'll just write it in words, so I, I do have a full, a full flag, and the, the, the, the last condition is fulfilled, so I do have an isomorphism between the ratios, tensor kc, 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, in, in, in r equals to 2, I have that for every lambda, a complex number, then there exists a unique filter extension like this, that is given by the short exact sequence. Sir, lambda doesn't play in your old definition, lambda plays no role at all, yeah? Uh, well, the, so, uh, it extensions, yes, it, it extension classes, so, so, so, so many this, this, this will give the first, 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 the, the, the last one to be, to be the, the last term, the, the vector model to be the last term of the extension, um, and, uh, because I want to, I want to, to consider one example, let me introduce one definition, um, so, uh, um, I'm going to recall, projective coordinate system, so many, if, if, if my ruin surface, uh, decomposes into a union of, of, 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 a, such projective coordinate system always exists, because, for example, one comes from the universal, uh, covering, that is just the upper half plane, uh, so let me, and, and this matrix, I'm, I'm asking to be in SL2 of C, this matrix, say, B, C, D, alpha, beta, SL2, so I'm going to introduce coordinates now, and I'm going to introduce Z alpha beta to be transition functions. So the last one has a beta, but there's no beta, it's your sound, the last beta, it's a B, B, alpha. No. It's, it's a coordinate change. It's called a change, I understand. The matrix, say, B, C, D, beta, so it's a beta, we change from Z beta to 0. So that makes a good sense. So this is the, the mobius code. The transformation is a little bit better for this. Yeah, fine. As, as, as, as, yeah, so that there exists, so, 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, you'll have many of them. You will. I'm very good at this. Yeah. So, so. Yeah. Okay. Okay. So this is a definition. That's a definition. So, so Z alpha beta are the transition function for half canonical, and mainly they are linear in Z beta, so in particular the first derivative will, will vanish. Okay, now I can maybe state some theorems. Maybe. Yeah. So theorem one, this, this was, was proofed with Motoki-Kombolase 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 a filtration, this is a vector space, for the vector space being, I'm going to denote V, that is, for, for, for this E0, that, that was that direct sum of line of half canonical powers of. 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, maybe I'm going to, yeah, I'm going to, to mention some things. So I can, I can, I will give you the, the, the transition functions for this V lambda. So let, 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 this is, so, so, so these are the transition functions for V lambda. And the second part of the, the statement says that, well, yeah, when, well, this is, the, at zero, you just get back, I mean, this will make the transition function for, for E0, that is also my notation of E0. And finally, that, that d plus one over lambda times a h times, times, so, so, if I start with any element on the hitching section, then, then this is an upper on the V lambda. On the, okay, so maybe, I mean, so, 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, this, this first matrix, I mean, sometimes probably we denoted it like z alpha beta to dh to be, to consist of exponential of h times logarithm, 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, the second part consists of, of, of another matrix that has well identity on the diagonal, and, and on, on, on, on each of other, on each of, of upper diagonal entries will have powers of the lambda plus the, the derivative of z alpha beta. As I said, because x plus it was nilpotent, so when you do write down, down, down series, that you, you'll obtain such, such an object. As I said, these, these are linear, so that will be just, just a constant. So when you'll, when you'll, when you'll 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, I want to talk about something that is in between these pictures. And I will come back on these theorems. So maybe, 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 heart-modulite space. So let me introduce this heart-modulite space of lambda connections consisting of lambda, a vector bundle, maybe I'm going to call it a lambda, and then the connection here. So this is the modulite space of the lean lambda connection. The rule is that the, the, the, the light-lit rule, the lambda, like, like, light-lit 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, just a section of the, of the vector bundle. Yeah, usually, yeah, I, I like to use the notation v lambda for the connections, but it's okay. Is that a, is that a one-half or one over lambda? It's one over lambda. So lambda's not zero. Sorry? So we're saying lambda's not zero. Yeah, yeah, yeah, yeah, this is, yeah, yeah, so I'm going to, to, is one over lambda, sorry, I'm going to, so this is right now a family of, yeah, I'm going to, to come back on this. So lambda is not zero, okay, so, so, so, so, so the delin put these two definitions of the Higgs bundle and connection into this definition of the lambda, of the delin lambda connection via this. And we can notice, I mean, when lambda is zero, then you have this, 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, if, if here is the moduli space, if this is the moduli space of the, if the, this is the dolbo moduli space and this is the DRAM moduli 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, in between there are these lambda connections. Now there, there exists a, this is the reaction on the Higgs field on the, on the dolbo moduli space, but 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 section will be the ones that are left invariant under, under this c star action. So you can see that this happens if, this happens if, if, if t, if, 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, so, so, so the fixed points are Higgs bundles and they're, they're noted as hot bundles. So I'm going to, so, so this is the, this is, this denotes the c star fixed points on the c star action on the hot moduli space. So now probably I want to, I want to look at one example. Maybe I'm just going to write here. So I have this exercise that, that I did maybe two days ago. So I'm going to save this notation above. So this is a remark. So I'm going to start with phi of q on the Higgs section and then, 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, oh no, e0 was, e0 was the split vector bundle into, into half canonical powers and x minus was this constant, lower that, yeah, and diagonal below is a fixed point for, yeah, is, is the, for this lambda connection. The, the, 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 uppers, but now they became lambda connections. It's just, well, the same point is the same point. And finally, the fact that, the fact that the graded hodge bundle associated to this, to this lambda connection is, is, is going to be e0 and x minus. So these are some statements we can easily check. So what, I mean, the way, the way to see this, what the sister action does, well, I'm, I'm just, I'm, I'm just going to apply the theorems above in order to, to prove this statement. So that alpha beta we said that these were the transition functions for half canonical. And, sorry. What's the difference between two and three? Or, or maybe what is the meaning of two? Yeah, I mean, yeah, oh, this is the, the fixed point. So if you add t here by t and you, you, you, um, oh, sorry, uh, did I? It's okay. It's okay. No, no, it's okay, I just don't know what's happening. So, um, it's okay, I'll update by t equal to zero limit. Yeah, so, right, right, three, it's okay. Right, I mean, but I, I will mention in the end that, that, 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, 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 transition, 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 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, 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, 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 give you that, so it's the same Higgs field, t of phi, but it's written as, um, a, 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 be it? It, it, it reads to H into, what is zeta? Zeta alpha beta, there is something. Is it alpha beta? Ah, no, no, it was, for first term, it was written. No, no, no, no, no, no, no, no, no, no, no. Yeah, yeah, yeah, so, so, so, uh, nothing. So, yeah, I mean, oh yeah. It's good. Okay, so, um, so if, uh, if, if, if, if this q is the, is, is, is the origin of the Hitching base, 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 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, uh, 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, you call them 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. Um, 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 for questions? So, uh, what is the graded hot bundle? Well, 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 extension, 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, it's an easy computation to see that, uh, that the associated, the graded hot bundle from, from theorem, consequence of theorem number one, uh, 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, um, k, uh, 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, a 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, uh, to get, uh, um, in order to, to get some terminology. Um, okay. So, let me, let me mention some theorems now. Yes. Yeah, uh, they, so, so here it is just, uh, uh, yeah, then, 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, the excess. 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? Uh, S alpha beta. Or is it the L, I don't know. Oh, zeta. Yes. Uh, it, so, so it's going to, to probably depend on the spin choice, right? Or, so this zeta alpha beta came from the, yeah, spin structure, but you choose just the core, the projective coordinates. Do you need a solution of hidden divisions to start somewhere? No, no, it's projective structure. Right. Projective structure. Right, I'm not using the solutions here. You get symmetric power of S2, two dimensional presentation, you get flat connection, yeah. I mean, here, here, on, on this board, I kind of defined everything I needed for this. Ah, you started with a complex progressive structure? Yes. Yes, yes, yes. That was the, right. So, if you choose a different complex, a regression structure, you'll have a different base point there, is that it? Yeah, yeah, yeah. Okay. Okay, so let me now measure some theorems. Um, maybe, maybe I'll just theorize this board. Um, okay, so, so, so, so, so theorems, and this is mainly the work of Carlos Simpson, um, is that for every lambda v lambda, that is a hot bundle, um, so for any lambda-dealing connection, then there exists, um, 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, I will choose, uh, uh, lambda-dealing connection, then there exists a unique section. C star goes to the hot module I space, passing through lambda v lambda and the connection. The, the limit when t goes to zero always exists, fixed. Um, so for, um, let me, let me just denote, uh, m dolbo of c star consisting of these fixed points, p alpha, where alpha is a class of variation of hot structure. Then for any, for any point, p zero here, then um, if, if I look at the inverse, I mean, if I look at all the points that are, uh, all the lambda, yeah, all, all the points that are fixed under, 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 a sigma for section. Yeah, yeah. Uh, sorry. No, no, so you just on the left, you say sigma of c star. Oh, and now it's sigma, it's something else. Right. Number two, you meant sigma. Section is not sigma. You just confuse the notation. Yeah, sorry. I, I, I just switched the, the, all notations in terms of h bar and sigma, uh, just, just, just before. Yeah, yeah, you can erase that sigma over there. Here. Oh, here. But then what? Sigma is a number two math. Limit, limit. Oh, yeah, yeah, yeah, yeah, um, yeah, yeah, yeah. So maybe this one. Section in harmonic section. No, maybe c star invariant section. Yeah, yeah, it's a c star invariant section. So, so, okay, so, um, what, what do I want to say? Sigma of lambda whatever is equal to that limit. Yeah, this is the sigma, yeah. Okay. So all, all the points that are fixed under this, um, all the, so for, for any, so, so, so, so, 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, um, at all the points that get fixed under this. And this, this will be a holomorphic Lagrangian. And this will, will, will give the stratification of the Dirac modular space. So this is a holomorphic Lagrangian vibration. Um, so all the points that are c star fixed points under this, under this action. Okay. So, so, so there exists, uh, there is the, this, uh, 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, uh, vibration for, um, um, so maybe I'm going to, to just write it that if you restrict to lambda equals to zero, you'll obtain again, uh, 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, the same Higgs, uh, point. So, um, okay, so, so more, more over, this is a holomorphic Lagrangian on the dolbo modular space. So the modular space, the dolbo modular space breaks into this p alpha of, uh, uh, uh, into, into this, this is a sort of, a holomorphic stratification of Lagrangians. Okay. So maybe now I can, I can mention the, the theorem. So the theorem, um, Can I point out one thing which is the, the idea that that would be like some kind of vibration, I think that's a connection. Of, of, of, uh, sorry, of the DRAM or of, of the DRAM, okay. Okay. I mean, it's in, the subspaces are Lagrangian, but that's what this should mean. It's a vibration by holomorphic Lagrangian. No, it's not holomorphic, it's not holomorphic, it's not holomorphic. The idea, the, the statement that those are closed submarines, for example, is, is just a conjecture. They're not closed here, but they're, uh, it's just a conjecture that they're closed here. Okay. Yeah. I see. So maybe buy for the morphic Lagrangian. Yes. Yes. That's everything in here on the other side. Well, they're, they're not, they're not closed. Well, I mean, I don't think it's correct to say, I mean, those subriders are not closed in that, on the dolbo. So I'm not sure what vibration means, but I mean. Okay. Then, then, then I just, I just leave that holomorphic Lagrangian. Yeah. Okay. So, so, so the theorem, um, and this is with Laura Fredrickson, Josh Kidonakis, uh, um, Rafa Maceu, Motohikom Ulasen, and Tinecki in, maybe it was 2016. Was that, uh, this, uh, so, so if, if, if, if I restrict to P zero, that is E zero X minus, and we proved in that lemma that this is a fixed point, right? This is the fixed point. I mean, um, and what the, this is a fixed point of the hitching section on the dolbo and also on the, on the upper moduli. So this is what that lemma says. Then, then, then these two, that these two Lagrangians are by holomorphic. So this is this X four. When lambda equals to zero are by holomorphic via Gaioto conformal limit, Gaioto's conformal limit. Yeah, we, we basically, we basically, yeah, so this correspondence between 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 Gaioto. Okay. And moreover, it was generalized by Colliar and Wentworth a year ago to all such fixed points. So this is the theorem Colliar Wentworth, 2018, that for any point, um, that is a fixed point under this C star action, um, then, and then the alpha variation of host structure, then there exists a by holomorphic correspondence between, so these two fibers, these two Lagrangians now, uh, one on the dolbo side and the other one on the Uram side are by holomorphic Lagrangians via Gaioto's conformal limit. And this Gaioto's conformal limit is explicit. So you can see from this, um, so you can see that, and, and the methods here that were used are similar to the methods that were used, uh, uh, by us in, uh, yeah, I have very short time, um, are similar to, is similar to our technique. And you can see that in, in, in that case, this corresponds to the longest, to the full flag, like the, uh, I mean the very, the, the upper corresponds to the full flag. And, um, um, you can, you, I, I, I, I can try to explain, for example, uh, the, on, on the opposite side, you have the shortest flag when E is a stable hex, uh, when, when E is a stable vector bundle. So if for a stable vector bundle E, then you have that E phi, um, is in the Dolbo modular space, uh, for any phi. So if it's 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 modular 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, um, for every stable hex, uh, for, for every stable, uh, 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, uh, the unique turn connection, uh, then you get this maybe by Narasimha Shashadri, E corresponds to D. And mainly the fact that E was a vector bundle of, um, 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 five, uh, the, yeah, the phi and phi dagger is zero and that can be done, uh, uh, because yeah, uh, phi can be, uh, diagonal alone on this, uh, Einstein Hermitian metric. So this is kind of a short proof for these opposite cases, uh, in the case of stable vector bundle and in the case of uppers. And in between it's, it's, uh, it, the proof is very similar. Um, anyway, so maybe, um, maybe, maybe I'll try to, I'll try to, to give the general picture here. So, so this is the double and this is the DRAM. Here, um, well, here, here is, uh, is the corresponding Lagrangian of the, uh, Dolbo and the DRAM. And you have this kind of slice in the hot bundle. Um, 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, uh, one of these, uh, Lagrangians. So, um, if I want to define, um, uh, let's say a semi, a geometric, semi-classical limit, I can, I can start from a lambda connection, go to, uh, to the, to the Dolbo side by this and then project on the pitching base. So that is, uh, that is a geometry definition of a semi-classical limit. There is another definition coming from physics. When, uh, this is more like the limit when each part 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 modular space of uppers. And that was so that, yeah. So maybe I can, I can define a map from the DRAM to the pitching base via this, um, uh, well, stratification or that is trivial and, uh, this, sorry. Yeah. Uh, it's, it's, it's, yeah, it's a holomorphic pro, it's probably on, on each slice or, um, it's, it's, it's not on, 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 slice or. But it's not on the whole, this is not holomorphic on the whole modular space. And here, yeah. So probably we'll stop here. Thank you. I have a question, which is, so at the, 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? You have an idea, because this one depends on the solution and vision equations, right? Uh, 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, you're ahead of your formula with D plus, uh, right? I mean, this is also explicit. Yeah. I mean, yeah. Well, I guess this, yeah, I, well, I, I think it's a similar formula to that D plus one over lambda of five. 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, by the uniformization? Yes. Yeah, good. That's exactly the case. Yes. But so if the complex projective structure is not the one given by the uniformization, then do you have a comparison, uh, between them? And for example, is it going to be linear on this, on this vector space? I don't know. Yeah. I don't know.