 Okay, I would like to thank the glaciers Professor Ahmed Abbas and professor of peace for the invitation for the nationalization of the conference Okay, I slightly changed the title so you can compare the title with that on the program It's longer, but I think it's more precise. Okay So here let me first introduce a notation So I've so the script I've is a trivial holomorphic vector bundle on a disk on that time times Laborhood of a complex manifold. Okay, I will denote the corresponding shift of a holomorphic functions also by the same Notations of script life. So I face three. So I emphasize here if he's a free LaBla is a collection. It is a unigrable collection. Okay, it is a holomorphic on the puncture disk cross x and We say LaBla has a pole of Pangora rank on along the divisor zero cross x Okay, if it satisfies this condition, okay So if you write down the if you choose a basis for I I write down the collection matrix And then the collection matrix is of this form. So the collection matrix is of form tm Some okay DXI so XI or the coordinate is a coordinate chart for the for the for x and here You have some holomorphic functions. So let me write omega I and the plus Omega dt over t. Okay, so omega and omega I are holomorphic So let's consider two special cases the rank one case and the rank Zero case so the rank zero case is often called logarithmic power We say this collection has a logarithmic pole and it's collection matrix is of this form. So I missed equal to zero Because our collection is integrable So you will have DA plus so the matrix a the collection matrix a has a property DA Plus a y to a is equal to zero. So I write down this equation Okay, so of course all the coefficients of the of the summit of zero Let's take their restriction to the divisor zero cross x see what we can get. Okay, so we got this one Okay, so the second one the second one is just the second. Okay, this term is equal to zero Okay, but we restrict everything to zero cross x if you look at the second equation It is of the same form as DA plus a y to a is equal to zero That means this this matrix also define our collection a unigrable collection. Okay So the second equation shows that The restriction of law of law to zero cross x to this divisor So I did a little by smaller blah. It is also a unigrable collection. Okay, and the first equation shows that The morph is defined by the matrix omega but also restrict to zero to the divisor is a horizontal section This omega is called the residue of this collection. Okay, so here is the first equation this one So this means the morphism defined by omega is a horizontal morphism with respect to the collection Defined by this matrix. Okay So I summarize everything like this, so small law of law is a restriction of bigger law of law to the divisor a residue Just okay, so I will denote by r infinity that is that is a residue Okay, so this collection law of law a small law of law composed by law of law is equal to zero That means it's a unigrable and the last equation means the residue is horizontal. Okay Now let's consider the rank one case. So that is I'm is equal to one now and Again, I can write down the okay. I write the collection matrix that way and Again, I write down the integral because the equate the collection is the integral both I write down the equation for you for this condition for the for the collection to be fly and again restrict this This equation to the divisor zero cross x then what you get is It's different. Of course, it should be different from the rank zero case and that's what we get what that means Okay, the first equation means that the phi Defined by this collection matrix about not by collection me by the matrix omega i zero x dx i so that's a matrix That's our endomorphism of vector bundles with differential forms has coefficients Okay, so the first equation means a phi y to phi is equal to zero that means a phi is a Higgs field Okay, now also what is r zero so r zero is the morphism defined by omega z omega x okay restrict to zero and The second equation above means that r zero the bracket of r zero and phi is equal to zero Okay, now let's put the rank zero and the rank way case together to see what we can get Now that I let the e be a triple vector bundle on x so that a triple vector bundle on x And that a pi is a projection from p1 cross x to x and that's a pullback This e on x to p1 cross x so that's a pi star e and suppose we have an integral in a global collection la-bla on this pulling back and Also, let's assume this pulling back has a logarithmic pole Along infinity divisor and has a pole of pangora rank one along the zero divisor Okay, so we combine these two cases together now Of course because I assume e is a triple vector bundle the pullback of course It's also true vector bundle on the family of projective line Okay, now let's take the restriction of this vector bundle on the the infinity divisor because I assume that the infinity is a logarithmic So I have this collection Small la-bla is a restriction of the bigger la-bla to the infinity divisor and also So this is a flat collection on e because a restriction of pi of the pulling back of e to infinity is e again So on e we have this collection smaller blah and also the residue morphism the residue morphism is flight with respect to small La-bla and also let's restrict. Let's take the restriction of this Collection to zero divisor because I also matters the zero divisor the collection has pangora rank one Okay, so this restriction will give me a hex field phi and also give me a n-morphism R zero Okay So now we get a tuple. Okay, so we have all Complex met for the x we have a true vector bundle e and We have a collection. This is a small small la-bla Okay, small la-bla is a collection on e and also we have R zero Okay, this is this one R infinity is a residue at the infinity and phi is a hex field. Okay This gives the rest to the so-called Frobenius type structure. Okay now Let me give you the formal definition of Frobenius type structure a Frobenius type structure is a tuple okay, like above and X is a germ of complex manifold e is a vector bundle, but let me emphasize It's a free vector bundle free area everywhere. Not just locally free and R zero r infinity or n-morphism of e and the phi is a fixed field La-bla small la-bla is in group is an integral of collection on e Okay, and from this data, I can construct a bigger la-bla. It's a vector. It's a It is a collection. It is a it is a collection on the Pi star e okay on the pulling back of e to the family of projective line Okay, so the capital la-bla is given by this formator So the first term is a pulling back of la-bla. The rest is just a n-morphism with differential formats as Okay with differential form values So now our condition is that this capital la-bla must be integrable Okay So note that this capital is a bigger la-bla is a has logarithmic pole along infinity It has a pole of punk ray rank one along zero divider Okay So we get we put this run a zero case and rank one case together now and it is holomorphic elsewhere and although the condition for integrability it can be written Okay, okay, it's equivalent to the following three equal six equalities so this first one means small la-bla is Ingradable and the rest view at the infinity is horizontal Phi is a hex field and the bracket if of r zero and phi or equal to zero and then the The rest two equations just you know It's a relationship between the data at infinity and at zero. So the phi la-bla phi is equal to zero and then the last one Equation, okay So that's a definition of Frobenius to have structure. So it looks Well, they look so very complicated But let me point out that only my ro-morphic integrable collection la-bla on the triple vector bundle Pi star e on the family of projective line with logarithmic pole at infinity and a pole of punk ray rank one along zero divider and No other post holomorphic elsewhere is above the is above the form So it gives the rise to a Frobenius type structure So you can just define a Frobenius type structure as something on the triple vector bundle on P1 cross x Okay, he's a integrable collection On a free on a triple vector bundle on the family of projective line and you needed to require you need to you require that as a The collection has an algorithmic pole at infinity and a pole of punk ray rank one at zero So that give you probably more geometric and more conceptual definition for Frobenius type structure Okay. Now, of course, now we are We want to come how to construct a Frobenius structure. So you need to to start with something. Okay, so So we are going to start with okay, suppose we have a family of Disc, okay, so it's a D D means disc disc across e Suppose that we have a free vector bundle if on D cross x on the family of disc Suppose we have a collection la la on if so if it's free vector bundle. So triple vector bundle Okay, and also let's assume it has a punk ray rank one Just do you really need three will only log P1 directions and not Exits well, so I should say that I do the big way I define the Frobenius type structure X is a germ of a complex manifold. So you can just always shrink hacks So I should point at that. I'm just a station's not part of the data. Yeah, right. Yes. Okay, right. Yes So you can always a shrink hack so that is a trivial everywhere. Okay, so rank one Okay, so it's a punk ray rank one. That means the shape of rank one. So let me read Pungray rank is equal to one Okay, so now I have something on the disc I want to extend it to a Frobenius type structure That it means I want to extend it to something on P1 cross x so that okay So that's the load of the extension by F2 the Blur to okay, so I want to extend it to something like this so that is a restriction to D cross x It's a given data is a restriction and also ID infinity. It has logarithmic pole. Okay So if you can do this or not so you'll need to extend F to a trivial vector among though also on P1 cross x Okay So that's our problem So if we want to solve such a kind of problem so that we can construct Frobenius type structure Now let's first consider the special case where x is just a point. So that is We work with D and P1. So that is a classical Bocco Bocco Boccohoff type problem now Boccohoff type of problem is like follows that the D be a disc F labella is a trivial holomorphic vector bundle Okay, F is a trivial holomorphic vector bundle on D on the disc and the labella is a integral of a collection It is a holomorphic on the punctury disc But it has a part it has a pole of punk ray rank one along zero Our problem is it's a falling we wanted to find a pair F to the labella to the so that F to the is a trivial vector Bundle on P1 Labella to the is a whole integral of a myomorphic collection with not gripping a part infinity It is holomorphic outside zero and infinity and it's a restriction is a given data So that is a classical Boccoff problem. Okay It's not always solvable. So to solve this problem You need some conditions on the on the data that you start with F labella. Okay So this is the problem when you have no parameter, so maybe the parameter space is trivial. It's just a point Okay, and of course as I said you're not to construct Frobenius type structure You need to to work a family with families. Okay, so I state this one as a theorem Boccohoff problem for a family now the D be a disc Okay, X is a germ of complex manifold F labella is a trivial holomorphic vector bundle on the disk cross X Equipped with a integral myomorphic collection of punk ray rank one along the zero deviser Okay Suppose we can solve the Boccoff problem for F labella restricted to a Fiber okay D cross zero so that means if I can solve this problem just for P1 Okay Okay, so that is that is a classical Boccoff problem If you can solve the classical Boccoff problem on one fiber Then you have a positive answer for the Boccoff problem for the family then so the the the result is then there exists a unique one a unique spare after the labella to the Okay after that is a trivial holomorphic vector bundle on P1 cross X So X you have to shrink X. Okay, as I said a moment ago and Labella Tilda is Myomorphic collection. It has a logarithmic pole at the infinity holomorphic outside zero infinity and it's a restriction Okay, it's a restriction to this fiber is the Boccoff problem is a solution of the Boccoff problem The classical one. Okay, so you have so basically and also It's a restriction to D cross X is the data that you start with Okay, so that let me draw a picture for this one. So this is my X This is a P1 and This is a zero and this is a infinity and so now on a disk around a zero, okay Around a disk of zero you have the data that you start with So this is F Tilda or F labella. It is defined on D cross X D means the disk around a zero Okay And also let us see suppose. Okay, so this is the point X zero Suppose you can solve the Boccoff problem Okay for P1 for the along this fiber. Okay, so you solve this problem Okay, so this is D infinity. So you have a D infinity. So that's a disc around the infinity So I will use that later. Suppose you can solve the Boccoff problem along Along along a fiber then our result is that you can solve the problem Boccoff problem for this family So that is you get a solution F Tilda labella Tilda is a restriction to the disk around a zero is That is a solution is that is a data you start with is a restriction for the along this fiber is a solution is a solution for the For the for the classical Boccoff problem. Okay Okay, so that's that give you a positive answer, but moreover one can prove the following your solution Okay, it's unique your solution if you're restricted to disk times To the disk at the infinity Okay, it is equal to what it is equal to the pulling back of your solution Okay, so it's a pulling back of your solution at the special at that special fiber So that means what that it means your solution actually does not deform the data at infinity does not deform your solution at The infinity so here at infinity and let me just write I'm deformed at infinity Okay So that is that is due to the some kind of rigidity property of logarithmic post logarithmic collections Okay. Also, we have an algebraic version of the Berghoff problem let G labella be a C tau tau inverse module equipping with a collection with post at zero and infinity and with regular Significate at zero. So this is just an algebraic version of a vector bundle on q1 Minus zero and infinity. Okay, so Let's suppose G zero is a free C tau sub module of G with pangray of rank one pangray rank one such that G zero tensor this polar lower on polynomial is equal to G So that it means G zero. So let me also draw a picture. Okay here. So basically you have P1 that is a P1 Deleted zero did it infinity on this on the delete zero and infinity you get this vector bundle G It's a free vector bound. It's a trivial vector bundle and suppose at zero you can extend G to G zero Okay Around so G zero is a free is it's also free C tau sub module. So it's a free vector bundle Pangray rank one and it's a restriction to the to P1 minus zero infinity is this G Okay Then the birth of problem is to find Free sub module. So free module C tau inverse module G infinity also with logarithmic So you want to extend a G to G infinity? Okay, so that the G infinity is a free vector bundle near your neighborhood of infinity and also we want G zero G infinity restricted to P1 minus zero infinity all equal to G And also we want to the vector bundle that you glue G zero and you can glue G zero G infinity along G So you want to the the the glue the vector bundle defined on P1. This one is a trivial vector bundle. It's a tree It's free. Okay. So that's the algebraic version of the birth of problem. Okay So now we want to give a criteria for for the answer for this algebraic version of birth of problem to be to be solvable Okay Okay, let we be the delin extension of this Maromorphic collection G lab law because the G lab law has Nogorithmic collection so you can define the so-called delin extension the 90s and actually infinity is just the fiber of this delin 90s at infinity and Molo Dormey, so you have a Molo Dormey around infinity X on it Let's consider an increase in fact filtration on this actually infinity. Okay So how this filtration how we got to this filtration? So you say we start with something G zero Okay, so G zero is a data you start with. Okay, so it's something on zero. So it's a Pungray of rank one. It has Pungray rank one Okay, so we consider this one G zero model divided by tau K is of course equal to tau G zero divided by tau K plus one Because the G zero say G zero is a C tau module So tau times G zero is contained in G zero. So we get G zero model. Okay. It's containing G zero model Okay, plus one so power minus K G zero give you an increase in filtration on G Okay, so that give you a G increasing filtration on G now this increases filtration on G will induce a filtration on actually infinity Okay, so it's so that G in G K actually infinity is this increasing filtration defined from this light is G zero Okay, now I can start state as a Marihiko settles criteria that N be of the new potent part of the more order me T Suppose there exists a mixed whole just structure on the nearby cycle of G at infinity So that is a whole to filtration is given by the filtration that the increase in filtration that I just defined and the weight Filtration is a more order me filtration Okay, and then we can solve the Berghoff problem. Okay So let's recall a fact. Okay, so it's a theorem proved by the lean Okay So it's if you know F is a punctually pure least ql ball shift on P1 minus zero or maybe deleted some other points Then the more order me filtration on F 80 infinity So F 80 infinity you consider as a representation of the local Galois group Add infinity. So on this on this F 80 infinity basically F infinity is also the one inch in cycle Libre cycle of F 80 infinity So on this galore representation, you have the weight filtration and also you have also the more order me filtration Now Dereen prove that the more order me filtration is the same as a weight filtration probably up to a shift Okay, so this condition. So let's compare this condition this this theorem with the condition of more he will settle Okay, so his condition say site You want to the okay So you have a mix horses structure the weight filtration should be the more order me filtration Okay, so at what time so in the edit case when weight filtration is the same as more order me filtration Well, if you start if you local Galois representation comes up from a global one Representation of the global Galois representation so that the global one is a pure then you have that condition Okay, so for this reason, let me propose another aortic version of the Berghoff problem Okay, so that 80 be the generic point of the consolidation of of the projective line over the finite field and it takes a strict localization at zero and the light 80 be the Generic point of the strict localization and suppose you have a row Representation of the local field. So it's something like you have something here. Okay, or maybe let me draw pictures here It's P1. Oh, yeah, I already have the picture here. So but here you have a representation. Okay Let's find a condition of course, so it's not always solve but let's let's try to find some conditions on row under this condition There exists the least puncture a pure shift ql bar shift I found P1 minus zero infinity It should be a 10 D run finite infinity and it's restriction to F zero Is restrict a restriction to 80 is a given representation. Okay Why I call this one as a Berghoff problem because you see one in the Marigold saddle Criteria when he saw when he want to solve the Berghoff problem He just well he put some conditions like the weight of filtration is a monotony filtration But we know the weight of filtration is a monotony filtration if you study ways Study way something pure. So in my version of the Berghoff problem You know, I just put the I wanted the solution to be pure So you just try to extend your local representation to a pure representation And of course you want a representation to be permanently ramified at infinity because that's also the solution for the Berghoff problem You also want to the something to be tame at infinity But I put a quote on the Berghoff problem because this AI to work And does not capture all the conditions in the status criteria in status criteria You also needed some conditions on the whole destruction whole defoultration. Of course, it's impossible, you know to You know to to to have the AI to counterpart of the pure whole defoultration. Okay. Anyway Okay, now let's Go to the construction of Sabah and a to a they construct some kind of Frobeno's type of structure starting from a long degenerate a convenient to lower on polynomial Okay, so let me first introduce some in some notations WJ are some vectors with integer Components and if it's just a lower on polynomial. Okay, so it's exponent just comes from this WJ and The data is Newton-Portigian of f at infinity the defoultration of Newton-Portigian at infinity is the convex hole of those Exponent W1 to Wn and zero Okay, and we suppose we our condition It was so we make the assumption that f is convenient and long degenerate convenient means what convenient means zero lies in the interior of data and Long degenerate means a following if you for any phase sigma some sigma of data not contained in the origin Okay, you consider the part f sigma. So that is you take those those lower on polynomials those terms in lower on polynomials But you only take those those terms with exponent with exponents lies in this phase of sigma and Then you you wanted it. Okay, you solve this equation the partial derivatives all equal to zero if this If there is no such critical point then waste for any sigma then we say f is long degenerate. Okay, and Also like the G1 GM be another family be a family of lower on polynomials. So you have I'm Law on polynomials and we can consider this deformation. This is a deformation of the lower on polynomial f Okay so you started with a small f and then you deform this lower on polynomials by Introducing those acts as a parameter x1 xn Okay We say f is a sub diagram deformation of f if all the exponents of this lower on polynomial G1 GM Lies in the interior of data. So do data is It's a convex hole of the exponent of the lower on polynomial life okay, and You notice that if I feel if I feel a sub diagram deformation then you can also calculate you can also calculate the The pollute the Newton political party infinity of this deformation Okay, because we if a G1 GM if this is a sub diagram deformation Then you say because you assume G1 GM their exponents less in the interior of data. So this won't change the Newton polyhedral at the infinity. Okay We say f is a universal unfolding if the images of G1 GM is Jocobin coefficient and formal basis And one can show that if G1 GM form a basis then this deformation has some kind of universal property But I don't want to state that Now it's a construction of Dwight and Sabah and we consider the twisted the wrong complex. Okay, so this is the wrong complex so you see Tn means the end dimension of the horrors am is all fine am is I'm dimension of fine space Am I mean also say am is a parameter space for this deformation You see the coefficients as the coordinates of X1 and Xm. Okay, and you consider the reality Okay, so you consider the torus the relative torus Okay, the torus parameterized parameterized by am and consider this the drum complex But now I do not consider the Euro drum complex. I consider the drum complex with D twisted by e to power tf Okay, and Then I can and then you can show that the drum co-homology actually is non-trivial except for the n ns co-homology So you can so G is the ns co-homology of this one What is a G0 G0 just means okay? You you take a You just replace this capital F the deformation by small f So that is you do not deform the f you can't you just consider the trivial deformation of f You get it we will get this is G0. So you can one hand show that a G is a free C X1 Xm module taught how your small user that is geometrically means it's a free vector bundle is trivial vector bundle on P1 minus 0 infinity Family, okay, you can also define some collections on it on G and G0 Of course, the G0 is actually a special case of G because G0 it correspond to the trivial deformation of the Doron polynomial and One can show that a G lab law has a regular similarity at a T0 and also one can define your light is G0, okay? you just replace that that parameter T by 1 or 1 over tau so and also you just move this t But originally it's a t times df you move t to the front of D. You're going to 1 over t Okay, so this is a definition of G0 So G0 is basically almost the same as G But you know it you want to make it become a light it so you change the parameter and then you can show that a G0 is Of is a trivial vector bundle on P1 minus 0. So that is my G0, okay? Okay, so this is it is a light is of G. So we get G also So G is something on 0 minus P1 minus 0 infinity And then lab law define myomorphic collection on G0 with primary rank 1, okay? G0 is called the briskon light is associated to this sub diagram deformation So this is works only for sub of the diagram deformation You think that those criteria do I end up about prove that a book of problem is solvable for this pair G G 0 and of course G This one is one correspond to the trivial deformation. That's also special case of the sub diagram deformation So so you of course, they also solve this one for this pair. They also solve that both of proof Problem and you know, let's consider the universal unfolding now the problem becomes more complicated There is no algebraic way to define Briskon light is so not me. Well, I will skip this part. Let me just mention that Well, when you define G G 0 like this You need you have to rely on some transcendental procedure roughly speaking G is a free transform of the Gauss-Mannin system for the for the for this universal deformation And G is a trivial holomorphic vector bound on p1 cross P1 minus 0 minus infinity cross x now now you have a parameter x. What is that x? x is of is a labor is a labor hood Of zero in the alpine space am because when I define the universal unfolding you say The universal unfolding has a parameter space its coordinates s1 to xm So it's parameterizing am but now when you construct the briskon light is Uh, okay. So you can only define the over a small labor hood of zero. Okay Okay, so that is my x and G uh, okay So you also have a collection of light has a regular signality at infinity G zero is the lightest Of G. Uh, punkery rank is equal to one. Okay When restricted to the to the parameter x equal to zero This briskon light is for the universal unfolding coincides with one algebraically that I just defined for the trivial deformation Okay Because do I understand about what if I use in subtle criteria they prove that the birth of problem Uh, it's solvable along this special point that is for the trivial deformation So now because by the theorem I stated a moment ago you can solve the birth of problem for the family. Okay for this With us guys a guide of frobinus type structure on the universal unfolding But to get a frobinus manifold structure you have to do more. Okay, uh, you needed to find us a primitive form Primitive for to transplant is a frobinus type structure to the tangent space of the universal unfolding parameter space Okay, we'll also need to put a metric because by definition A frobinus, uh, manifold is something that you have a frobinus type structure on the tangent space. Okay Also, you have more you also need a metric. Okay Another approach is to start with a solution of the birth of problem for sub-diagram deformation A moment ago, I also said that do I and sub-bar they prove the Using subtle's criteria They also prove the birth of problem is solvable for family for which family for the family defined by sub-diagram deformation If the sub-diagram satisfies certain Generating conditions, then there is a theorem of hardening and mining to show this this theorem shows that the The solution, okay The the frobinus type structure on the sub-diagram deformation has a universal has a universal deformation This is the universal deformation will give you the frobinus manifold structure. Okay Okay, in summary, we start with a bridge coordinate for if Okay, so this is a trivial deformation and we Okay, what is a bridge coordinate is bridge coordinate is obtained as a free transformer of the Gauss-Mannin System associated to this lower on polynomial. This lower on polynomial is a morphism from torus to a1 And then you solve the birth of problem using subtle's criteria Okay, so you get the solution of the birth of problem along one fiber Okay This bridge coordinate is for this a trivial deformation has a deformation Which is a bridge coordinate is for the universal unfolding. So that is g0. So g0 is a bridge coordinate is for the universal unfolding of of that f Then you you extended. Okay. So then you extend this birth of a problem solution of the birth of problem Okay for this special fiber and for this data you start with then you get a solution for the family Okay, you get a solution of the birth of problem for the briskon 90s of the universal unfolding with us guide the Frobinus type structure now orismatically So i'm interested in the aod or maybe characteristic p part. Okay, so what i'm going to do Okay, so originally you started with briskon 90s So it's related to the free transform of the gauss-mannin system Okay, now gauss-mannin system just correspond to rf strike ql bar. So that's gauss-mannin System and then you take it's a free transform. Okay, but then dwine subar proved that The birth of problem is solvable for this one. Okay, they use a set of criteria So of course, we should verify that this free transform of rf strike ql bar should satisfy The conditions of set of criteria now for me the aod criteria just means Okay, it must be a purely safe on p1 minus zero and infinity Also, it should be a time at infinity or maybe a zero depends on the coordinate that you choose and also hope also you hope that It has a pangori rank one. Okay, so orismatically it means that It's slopes at a zero at infinity should be less than or equal to one Okay, actually we're going to prove something similar result for the free transform of rf strike Capital f capital f is a sub diagram deformation. So basically I'm going to do the same the same thing for the sub diagram deformation Okay, then we have to deform your solution Okay, you get a solution for this one point Along one fiber and then you need to deform And then after that you need to study the after deformation local deformation Then you try to extend this one to a solution on the on the whole space Okay, so then we study the deformation of this free transform with prescribed local monodromatic zero and infinity Okay, so Here are some notations fq is a finite field Psi is a long true additive character The way that I'm going to study the free transform Of rf strike ql bar is just is to use hyper geometric functions Hyper geometric shapes or may I even call it gkz hyper geometric shapes uh An analog of the classical hyper geometric integral. Okay, I write that one. Okay, that's an exponential integral is this Is this um exponential sound? Actually, this exponential sound has been studied by denerff and noether And also by odolfson and sperber In their papers, they only consider they consider the special case where those yj is constant Okay, but not uh, it's the idea of guy found or maybe philosophy of guy found is that if you want to study Okay, maybe study just a special exponential sum is not you may not capture Uh, it's a whole property. You should start a family of them. You write the parameter You write the you just introduce some parameters and study some families. Then you will get some more information Okay, so this one. So we just tonight is a coefficient yj to change Okay, then you then you get the hyper geometric functions or refined field. Okay Now how to define a uh, what is a corresponding? Okay, use if you have a family of exponential sums, then you should have a galawar representation or maybe an aortic shape Okay, what is that aortic shape an is n dimensional fine space t is n dimensional torus Okay, so I can't define a morphism like this So this morphism is defined by this Laurent polynomial, but this Laurent polynomial the coefficient yj now is a parameter Okay, it's indeterminate. Okay And okay, so we have these projections To to to tn and to an Now the gkz hyper geometric shape is defined to be this object. So you use edge This morphism edge edge is that Laurent polynomial with variable coefficients Use this morphism to pull back the aortic shape and then push forward using pi two Then you will get something on a n right get something on a n Now by gross only trace formator you can calculate the trace of Frobenius on this shape And this one this trace is exactly that exponential sum that family of exponential sum Okay Now we have the formula theorem one can show that hyper geometric That gkz hyper geometric shape is a mixed perverse shape on the affine space an It's way with weight less than or equals to n plus n And its rank is minus It's related to the volume of that polytope at infinity. You didn't polytope at infinity and also we can do more We have combined we can write down the formula for the rank of the weight w sub quotient So that is you You can use the weight to define something like something like pangray Type, you know polynomial and you can you have a comb combinatorial formula for the pangray polynomial defined by weight Okay, now another another result Suppose v is as a risky open subset of the affine space so that this v parameterized So that is a parameterize those Laurent polynomials, which is non-degenerate Okay, or moment I'll get it. I define what is the meaning of non-degenerate Laurent polynomial, okay, then one can show that on this open subset v The hyper geometric shape is smooth And it vanishes for i not equal to minus n. So you only have i equal to minus n to worry about Now suppose zero less in the interior of la la but actually this is my condition because I assume my Laurent polynomial Is long is convenient convenient just means zero less in the interior of the polytope Of the newton polytope at infinity If zero less in the interior then the hyper geometric shape is a smooth shape pure of weight n And also its rank is equal to n factorial times the volume of the newton polyhedron Okay, so let's consider this a special non non-degenerate Laurent polynomial, okay And let's suppose we have a sub diagram deformation Uh, you can enlarge w one w n by adding those exponent of molomials, okay with non-zero coefficients of g i In this way, okay You won't change the the polytope because the exponent of g i they all lie in the interior of this data, okay And also because of this because now the exponent of g i also It belongs to w one w n so you can write a g i as also the Laurent polynomial of this type And f xt f xt is that sub diagram deformation, okay So you write down the formula for f xt But I put a t prime here because later I needed to introduce another parameter, okay So this one, okay, it's again a Laurent polynomial and again it's non-degenerate and convenient Okay, so x is equal to am so that is my parameter space for the for the sub diagram deformation What is capital f capital f is is the morphism defined by this uh by this sub diagram deformation, okay Because a sub a sub diagram deformation is just a family of Laurent polynomials, right So this f family is this I've just defined a family of morphism from torres to a one And let's denote by five this morphism. So I just my p one x one x m to that coefficient So what is this coefficient is this one, okay So remember that what I'm interested in, okay f is Is a is a sub diagram deformation RF Shrek is a gospel mining system And I take the free transform of the gospel mining system. So that is the analog of g, okay That I introduced before Okay, so now we we have the following proposition This free transform of the gospel mining collection of the gospel mining system is what is a pudding mic by this five Of the hyper geometric shape. So that is if you want to study The gospel mining say the free transform of the gospel mining system You can you only need to work you only need to study the hyper geometric shape. Everything is determined by the hyper geometric shape Okay Moreover, we have the following the image of Five, okay that capital five if you would delete zero and infinity So capital five is defined on a fine line, okay If you did it zero from it then its image is contained in the is contained in the open sub side v parametrizing non-degenerate low-round polynomials a moment ago I said that This hyper geometric shape has a nice property when we restrict to this open side v Parametrizing non-degenerate low-round polynomials, right? Okay. So putting these two facts together I get the following corollary when restricted to p1 minus zero minus infinity Then this gospel mining system free transform of the gospel mining system one ish is for i not equal to m minus one Okay, for h m minus one. It is a purely this shape of weight n also I should say it's weight or it's rank. I forget to say it's rank is n factorial times volume data and Also, it is it is pure it is okay So it's a pure shape that is you know, it satisfies the condition of settle criteria um And also we know that it is tamedy ramified at zero it has slopes less than or equal to one the infinity So that means, you know, it's a punk array that correspond to the fact that the punk array rank is less than it's equal to one, okay Okay, so so far everything is good. Okay. It's uh completely parallel to what happens in the You know in the work of dua and sabah Now i'm going to okay. I will skip this part. Okay Now i'm going to study the deformations. Okay, because you see that when sabah Do I they construct a Frobeno's type structure? They first solve the bulk of problem for for special fiber and then they study So then you have to the deformation. Okay the deformation and then That is a brisco knight is and then you try to extend this one to to something on p1 cross x. Okay So so arismatically We started with this one This is a special fiber the one that I just solved using hyper geometric shape and then that's uh, okay So later I will introduce some deformations at zero and also adding infinity And then now the problem becomes a deformation of aortic shapes. Okay, so I want to uh to be on You know more general context f is one of the following type field. It may be a fine field f a l A fine extension or maybe a fine or maybe ql ball or maybe they're finite extensions Or maybe they're algebraic quarter. Okay, so the theory will work for all these types of field C is a category of artinian local f i gibberish with rescue field also be f morphism on c just f i gibberish homomorphism k is algebraic algebraically closed field of characteristic p Um x is a smooth projective curve over k. Okay Is is a finite subset of x but actually in my application I only needed to take x to be p1 and i suggest zero and infinity But I want to be in more general context Eight is is a generic point of the streak localization It is a generic point of p1 And then for each is we fix our embedding our embedding of the local Galois group to the global fundamental group Uh, let's suppose i f is a least shift. Okay on x minus i so that is i f is a shift on p1 Okay on this special fiber Row zero is a corresponding representation Now my problem is deforming how to deform this iodic shift or maybe how to deform this row zero Okay, here is a theorem Let r be a complete Lothering local f algebra with residue field also being f That's supposed to have a row is okay row is a system Galois representations of the local field. Okay, so let me draw a picture here So I just draw a picture for p1 And I say just zero and infinity Okay, now suppose I have a row is so that is uh row is here is we have a row zero That is a representation of the local Galois group Okay to r And also we have a row infinity. It's another represent It's a Galois representation at infinity Okay, so that is at zero and infinity. I already introduced some deformation Okay, but along this line along this line. We start with the representation It is representation corresponding to that shift row. Okay Corresponding to that shift row here. I do the best row zero. Okay Row zero is just a representation of the pi one p one minus zero infinity Okay to uh gl Now this is just i fit just a residue field of our mod n. Okay So I have something on the fiber and I have some deformations along zero and infinity Now when you can solve this problem, okay When you can extend this one to a global deformation Now the criteria is this suppose x equal to p1 so just like this Suppose anamorphism of f is equal to f. So this condition actually is not serious because Usually we just start with a shift which is irreducible. Okay, absolutely irreducible then by shul armor And the f must be f. Okay, and suppose there exists a representation lambda It is of rank one so that a row restricted to this local Galois group is equal to determinant row i And no lambda is also a deformation of determinant Row zero. Okay, so that means Suppose you can solve this problem solve this problem. You can extend all this data After taking determinant. Okay, suppose after taking determinant you can extend this You can extend this data to something globally defined. Okay, so that's my condition Then the result is that there exists a deformation row Okay, row is a more is a fundamental is a representation representation of the fundamental group and Okay locally, it's just a row i's and also is determinant must be row So that is if you want to solve this aortic birth of problem for a family Basically, all you need to do is to solve the problem for the rank one determinant. Okay So, let me give you some examples I will take this example Lorentz polynomial is t1 plus tn minus one and then plus their rest protocol of their product Now why take this example this example? Okay, so it's proved by baronico of that Frobenius manifold structure on the universal and folding of this Morphism is isomorphic to the one attached to the quantum cohomology of p p i minus one and also automatically it's also very interesting because The free okay the the Gauss-Mannis offer is a free transform of the Gauss-Mannis system Attached to this i is closely related to the crux to my son Crux to my son is also very interesting example, you know in arithmetic, okay And okay, so let's take i f to be the free transform of this Gauss-Mannis collection and one can Calculate its local you know locally you can find calculate its monotomy So I've restrict to eight infinity is equal to by that but that's those audience show shift It's a direct sum of audience show shift And I've restrict to eight zero is ten million ramified with unipotent monotomy and single Jordan block Okay, I will denote this one by u. So that is u is a Representation of the local Galois group. It is ten million ramified. It is unipotent with a single monotomy. Okay And I want to okay. So now I want to study its deformation First I will just introduce some deformations on infinity at infinity. Okay now this is a Galois group local Galois group at infinity has a time Prime a opod, okay prime a opod that might okay map subjectively to the o one by this standard amorphous And let's take a character of the z o one to kill To this formal power series You just map the generator to one plus s i okay take their take their composition You will get a deformation. Okay, you will get a representation. Okay This representation. So this representation I will denote by chi i this representation One can show is a universal deformation of the trivial character Okay Now by our serials that is I just mentioned a moment ago. Okay, uh, there exists a representation this representation. Okay, it's uh It lands into the coefficient ring. It's now it's q l The formal power series with n variables, but now you need to model the idea Okay, so that a row mod out the maximal idea is the representation row zero Row restrict to infinity is that u now why why I wanted the restriction to 80 to be u That is because when I mentioned the classical Solution to a family Well, okay when I stated the theorem of do I know subar about the Construction of the Frobenius type structure at the infinity is undeformed. Okay At the infinity is undeformed because it's it is due to the rigidity of logarithmic Collection, okay So I just keep the infinity to be undeformed but not because I change the coordinate system now becomes row restrict to 80 Is undeformed is the u but a row restrict to infinity. Okay, so you have some kind of deformation Okay, and then one can prove that is such a data so your local data can be extended globally. Okay, so this is an example but what is How to do how to prove such kind of result? Suppose we are given a row zero. So that is representation that we start with for convenience. Let's assume the determinant is equal to one and uh, let's also suppose we are given some Matrix non-singular matrix P zero i's for each i's, okay Now I'm going to define some Functors deformation functions for an a belongs to this arting algebra This category of arting algebra's r a is what r a is just a deformation of row zero Okay, so row is a representation of the fundamental group of the projective line of x minus those x Is coefficient nice in a and if you mod out as a maximal idea of a you'll get a row zero And also I want to determine a row to be one and also ps is just some kind of coordinate chart at the local at the local point. Okay um Okay, also I want to r a is equal to the isomorphic classes of it Okay, so there is a way to define when to search pair when to when to search data Equivalent okay, but you know, it's essentially say is a row and a row to a equivalent then You know, you put you put this you consider the equivalent class and also locally We also have a functor r s a r s a is this is just a representation of the local Galois group Okay, I wanted to be a deformation of row zero Okay Now one can show that r and r s they satisfy the Schlesinger's conditions and they are pro-representable And also we have canonical morphisms. So if you have a global deformation, then you restrict to the local one Then you'll get a local deformation. So you have a morphine performance global one to the local one Now the fibers of this morphism are just the representations of the fundamental Groups with pre prescribed a local monotony. Actually, these functors also appear in quiescent's walk. Okay Now we have the following theorem all this functor That's a suppose n dot f is equal to capital is equal to f that means you basically assume f is Absolutely irreducible. Okay Then these funters are smooth r s this local funters are smooth these morphism are funters Okay, so you you put you take a global deformation restrict to each local Galois group Okay, so you get such a funter to the product this funter is also smooth So basically the universal deformation rings are just a power series Okay, and also we have a formula for the tangent spaces Okay Now in the special case when iF is given by that special Lorentz polynomial Well, we have such formulas for the for the tangent spaces and also for the dimension of the fiber Now in the transcendental case, we deform briskon 90s associated to iF by the briskon 90s associated to the universal unfolding That is your deformation spaces of dimension n. Okay And also we keep the logarithmic lattice at the infinity and deformed. So that's what we did in For the when we construct a Frobenius type structure Now this suggests what this suggests that we should choose a subspace of our infinity Okay, it should be of dimension n corresponding to the briskon 90s Of the universal unfolding. Okay, and also at infinity you just choose a trivial deformation Okay, because at infinity we do not to deform the the data. So we just map Every object a to u. So it's a constant deformation And then we try to find the element in the fiber. Okay, that corresponds to this Frobenius type structure But I must confess that I haven't find Such candidates that is or at infinity, of course, you do not deform anything Okay, but at zero what is The correspond what is the the deformation corresponding to the universal To corresponding to the briskon 90s of the of the universal deformation. Okay, so But also you can see that the theory Of deformations for aortic shapes is a bit different from the deformation theory for collection So probably, you know, when I work on this part of the problem Maybe I'm thinking well, maybe probably I'm working on the wrong context Maybe the more general context should be Instead of studying aortic shapes, you should study paedic theory. Okay, study the deformation of crystals. Okay So that seems to be more Natural, you know for this type of problems. Okay, but I haven't think seriously about that type of problem Okay, we'll stop here. Thank you Any question The dimension don't matter because you consider deformation of Local system Right, yes Right, yes, probably so maybe I should consider something dimension less than But even for that, okay If you do not deform the local monodromy the fiber is also, you know You can see the fiber is of dimension n times n minus one plus one We should consider something choose a subspace of dimension n from the fiber. That's also a problem Which I don't know Yes, but you know, he studies The the the Lambert field case. So you also some have some alba. Yes Yeah, of course the paedic part is very important in his work. Yeah, but you know, you see the definition almost the same Okay, he he just used a different terminology. He called a framed deformation He should fix some basis locally But what I did is fix another basis, but you know some matrix, but they are basically the same thing Is a question In the in the Birkhoff problem classical, they I mean you have this notion of trivial bond around P1 Right And in the eladic Okay No, I do not have the corresponding eladic counterpart for something for shape to be trivial for vector bundle being trivial So the condition I replace is just, you know Pure this condition, but I really don't know some other conditions that correspond to this triviality Of free shift. I don't know actually Because I mean they seem to be different For instance, if you take the free transform of a Pure technically ramified Eladic shift on a one. Okay Is it a solution to your problem? I mean, do you know do you know it? Well, it depends. Some if you got something pure Okay, if it is a pure then of course you got it, but some you don't know whether it is pure Okay, because it seems that the right The right analog in the complex case, I mean should not take into account the triviality I mean if you I mean if you don't have the condition in the Right, yes You don't do not need the condition in the In the complex case. So, I mean it's just a little simpler Oh, you mean in the complex case, you really don't need the the vector bundle not trivial? Yes, I mean you don't I mean Since you don't have the analog in the eladic case, so I need to To refer to it or to use it in the complex case, I mean, okay It's a little Simpler than the true work of problem. Okay. Yes. Okay. You mean my okay So this one is weaker in some sense. Yes But I think even for this one you you do not have a positive answer always, right because you see At least the the the the Galois representation you start with the local one showed, you know It has monodermal filtration has weight for transition coincides. So that's all right one condition My restriction, but I don't know whether that's that restriction. It's it's it's a sufficient condition actually