 Okay thank you Mike. So first I want to thank you Alina and Mike for invitation. I don't give much room talk before so I hope everything would work out fine. Okay so today I'm going so so is this okay? Okay so today I'm going to talk about PISO's distrust conjecture and especially for function field and also the formulation for complex. Okay so today we will first explain, recall a little bit what is PISO's distrust conjecture and then we will talk about the function field formulation and then we will mention what's the complex analog and and along the proof we will formulate the diaventine equation on function field and also complex and then finally we will give an outline of the proof of this PISO's conjecture for function field. Okay and this is a joint work with my student and Jia Liang San at Ji Guo. Okay so let's recall a little bit what is linear recurrence. So a linear recurrence is a sequence of a complex number and then they set some linear recurrence. Okay so we will use we can write this sequence in the following form. So here are polynomial over C and you plug in N and this is some complex number and take to the N's power. Okay so this is the general form for to represent linear recurrence. Okay so in particularly we can go a little bit looser we call this as exponential polynomial. So for everything here is a polynomial here is some number to the N's power and this type of thing in general is called an exponential polynomial in the number field side and later I will explain what is exponential polynomial in the complex case. Okay and on the confersely whenever you have something represented in this form as an exponential polynomial you can check it. It also defines linear recurrence. Okay so let's start with this piece of this roots conjecture. Sorry. Okay so let's begin with the linear recurrence and let's just say over a number field and D is a positive integer bigger than two and suppose each term is this power of some element in this number field for all but finite many N and then you can find another linear recurrence A over the break closure of K such that this BN equal to AN to the D's power for all N. Okay so I want to brought to your attention here is for number field they need this assumption BN is this power for all but finite many N and for function field we can do better. Okay so what about function field? So we begin with an algebraic closed field of characteristic zero and then K is a finite extension so we can find a smooth model which we call a curve C and the function field K would be the function field of this smooth model. Okay and let's denote G to be the genus of C. Okay so let's make an observation first. So suppose we have two exponential polynomials. Okay remember it's a polynomial times a bar to the N and take sum so we have an exponential polynomial and suppose this CN everything is on the constant field. Okay and then we notice in this case because K is algebraic closed so everything of CN would be a D's power so this AN to be any exponential polynomial of a function field and in this case this is still an exponential polynomial and every N's term is this power. Okay so in other words if we want to stay a piece of the conjecture we need to modify a little bit. So the following is the thing that we think should be adjusted to the following form. So suppose we have an exponential polynomial over function field K and suppose BN is some this power and I think for infinite the many N would be enough and in this case we would find the exponential polynomial and also an exponential polynomial for constants such that BN would be CN times AN to the D. So in other words unlike the number of your cases you would need to make some adjustment of a constant term. Okay so this is the thing that we think should be should be modified. Okay and what the next is what we can do. Okay so suppose BN is an exponential polynomial over function field and let's take so let's take gamma to be the multiplicative group generated by roots of this linear recurrence. So it would be generated by B1 to BL. Okay and then we here we need to make some assumption for our method. So we will have to assume that this gamma this multiplicative subgroup when intersect with this constant field would be trivial. So later I will explain why we need this for our method. Okay so under this assumption so suppose BN is this power for infinite the many N and in this case we can find an exponential polynomial and also the adjustment would be a polynomial such that every term of B would be equal to QM times AM to the D for OM. Okay so this is what we can do. Okay and so remember the key point here is we need to make some assumption and also here we can do better than number field. The assumption only need for infinity many N. Okay so what how about the complex analog. So it was developed independently starting from Ritter in the 20s and this is a generalization. So we say an exponential polynomial in complex case we say it's order Q and which is the following form. Here also the coefficient is a polynomial over C. Here is the exponential function. So it's exponential of some Q some polynomial. Okay and so so this is in the complex case we call this exponential polynomial and in the order Q is just take a maxima of all the exponents. So it's a maxima of a degree of Q1 to QM. Okay and the following is what we can do about this complex case. So for complex case if you have an exponential polynomial order Q and suppose f equal to g to the sum this power of entire function and in this case g is also an exponential polynomial of order Q. Okay so for this type of result actually it's extend the result of Mike Green in the 70s and in there he tried to do some hyperbolic problem for complex analysis. Okay so so as you can see this seems to be a good analog right. Okay so toward the truth of this type of thing actually related to some diamantine problem and so for the diamantine problem what we can say for function field is the following. So again we have a polynomial which is defined over the function field multivariable and suppose we have to draw out some possibility right. So suppose it's not like this type of form like a constant times monomial this i is some x i is some monomial product and it's some times some this power and if this is quite apparently this is something you should exclude and so if you exclude this trivial case then we have full of following conclusion. So if you start with some units this always means it's 0 and 0 like only on s. Okay so if these are s units and suppose you then and our results that you can find some positive integer m and then some numbers c1 and c2 and point is don't depend on the d we want and n degree f and the height of ui and such that so suppose this f1 l to ful is this power in k for if for some l is quite bigger than than c1 c2 like we we can find and this height f is this is a relative height of this polynomial and if l is sufficiently large and in this case for sufficiently large l and if when you plug in this f u1 l to ful is a base power in this case this this u1 to un must be multiplicative dependent over k and and moreover we know exactly how to control this mi so there is a this m is something we can find and this the absolute value of this m take some has to be smaller than the number that we can control okay so is this okay okay so so in other words in this function for your case it simply tells us so if you have some some large large exponent when you take it take it into this function this polynomial if it's a this power then it's impossible it must be multiplicative dependent over k and for some control the size okay so now let's look at the the compressed case for compressed case the related that that fending problem is the following so let's take ui to be exponential of some polynomials again this is some integer bigger or equivalent to and then because this has some geometric application so we usually take a homogeneous polynomial so suppose it's not a this power so suppose this polynomial is not a this power and suppose when you plug in 1 0 0 f 0 0 1 the failure is not 0 so this condition is geometry means because this homogeneous polynomial will defy us hyperservice and this condition just tell you the hyperservice and the coordinate hyperplanes has to be in general position okay so suppose we have this and suppose we can we find this we take a unit inside of f and we find it's a g to the this power which is some alpha to be polynomial and in this case we also can show that this is exponential this ui has to be multiplicative dependent modulo c okay so this is the compressed compressed analog okay so let me explain uh how to prove these things because for compressed case it's easier to explain so let me first give a idea how we prove this type of result so let's take a very simple polynomial let's take this example so fx to be x 0 x x 0 square plus x n to the square okay and remember ui is exponential to the qi and take u to be u 0 to un okay so what is the key point here so let's take a differentiation of ui prime and then over ui so if my calculus is correct it should be qi prime right so which is a polynomial okay which is at the most a polynomial but what is ui ui is exponential of polynomial so you you know the growth of ui would be so much bigger than qi prime so the key point is this thing is small with respect to ui okay so the next thing is uh let's try to define a polynomial which has a small coefficient so we will take this f so the idea is uh so so let me do this first so the idea is we want to define some polynomial then when you take a derivative of f u prime it's actually equal to this polynomial plugging u so how do we do it so it's very easy here you just take a derivative for x x 0 square so you would and then plug in u 0 sorry so you would you'll get this part so let's take this as a coefficient and then use x 0 square and then you can check exactly it's satisfied this condition okay so so the the point here is now we construct another multi variable polynomials with a small coefficient and then when you plug in u you get the derivative of the original f u that's the that's the main point okay so now let's see if you have so for example if f u to be g to the this power okay and so because f prime u f u prime would be would be d times g prime times g to the d minus one so in other words this g will divide both uh you will divide f u and also divide it will also divide this uh variable so in other words g would be a common zero of this polynomial and another polynomial plugging u okay so this is the main point so now you have two multi variable polynomial and one is a constant coefficient one isn't it's not constant coefficient but the coefficient is the small growth and then you know and you we know for this type of polynomial if you plug in units then the gcd is small okay so this this is this is a result of so so suppose u zero to un are multiplicative independent module c okay and then this is the result with uh levin we we have a gcd theorem and our our result also has we can has a slow growth moving target in other words exactly uh if you're the coefficient of du is a smaller than u okay and then the gcd is small so so and remember g is g is g divided both of them so this tells uh if gcd is small then f has to be small okay and but on the other hand we all know that because uh this is for if g g to the this power satisfies this type of equation and it's very easy to see the growth of g is exactly the growth is uh maximum of the growth of u u zero to un so in other words the height of g is large because of this uh in general position assumption and but it would contradict our gcd results because g has to be small okay so so this is uh very uh simple minded uh uh thinking of how to do this i think this idea has been there since green and koba jalan year uh but uh but at that time they don't have a gcd but now we have a gcd so this can actually put into work okay okay so how about function field okay so let's take the same example and uh we we like to write our finding for function field okay so we have some s unit for this un to and we have some large error so the key point is again for key point then remember what we want to plug in now we want to plug in ui to the error's power so let's take a derivative and so this this looks bad every time we need to take a different error right but it's okay so the derivative doesn't change so much you just have some constant multiple of ui prime over ui but what is the height of ui prime over ui so so for if you think about ui is a polynomial and when you take a derivative and you divide by ui then we know the the only zero can occur would be when ui has zero and the only pole comes from the zero of ui and the pole of ui moreover it's only this only simple zero and simple pole in other words the height the the pole can every pole here is simple when you take about polynomial all the pole are simple and it only comes from zero and pole in other words the the height will be bound by this the distinct zero and pole so in other words here it's at the most twice of the height ui and because it's a function field you need to make some adjustment on differentiation so here the correcting term is 3g okay so so what so for function field this is what i say so you you need to take a different local parameter and then for this here the zero and the pole comes from the zero of ui and the zero pole of ui and they are simple and for here is the thing you need to adjust them okay so that's why you get 3g okay and here t is in general we take when you take u prime is we take du over dt okay sorry that's a lot of notation okay so now so we also want to define what is dlf so ui are fixed so i write the arrow here so it's a similar idea we do the same thing okay and so when you plug in u1 arrow to u1 arrow then you get the dlf okay so again then you are considering the gcd of this polynomial and the f okay and so in other words we need to estimate the gcd of f and this dfl and what about this dfl so it depends on l but if you look at our formulation here so the derivative actually only the l only comes from only on the coefficient so it doesn't affect the height of the coefficient okay so in other words for the function field case we need to deal with some gcd and moreover the gcd of f so we need to deal with gcd of many couples many couples and then for this part of gcd the invariant is the height of the coefficient it's fixed okay okay so but the bad news is we don't have a gcd in high in a multivariable of this form yet so we only have resulted for n equal to 2 by kovac and zanier so in other words we need to work out another gcd theorem okay so so the gcd theorem uh is uh the following so let's so for the function field case so we have two polynomials and they are co-prime not constant okay so again similarly we fix if show greater than zero and uh there's always we can find an integer m and then do your ci everything is effectively and depend only on if show such that when you give all the n top holes so suppose the height is so it's supposed the height is very big then in this case we will have uh some of the top holes would have so remember in the complex case we would go directly say they are multiplicative dependent over a constant but in function field case we cannot do that but we can say the height is bounded okay so this is something different and but also uh this this this m this exponent can be bounded and m is something we can find it explicitly and suppose it's not not this case suppose not this case then in then we have gcd bound okay so actually i don't like this for new vision too much because it kind of uh anyway you don't feel it's complete but um but but let's look at the the thing we want to apply so that when we want to apply this theorem we want to apply to u1 g1 error gn error at all to be a large power so in other words when error is large enough this equation is always correct so we don't have this one and so but and then when error is large if this is not not not constant then the height would would also go very big greater than error and so in other words this one will not will fail so in that case we will really gcd theorem only this hold okay so but that would rely when error is very large so in other words we we will have the following a syntactic formulation so again f and g are polynomials over function field and uh so we have some so here we don't need to because it's a fix so we just say we have some something non zero and in this case uh again we find the m and c1 c2 so the point is you if you have sufficient large n large l okay and uh in this case for sufficiently large l either either you have g1 to gn are multiplicatively dependent over k or you have a gcd bound so the gcd is small okay so so i don't like the previous version but this first one is okay it's fine it's the thing we can get okay so so for gcd theorem is for function field i think is kind of fascinating uh like for function field case sometimes for gcd you you can find the uniform bound and so for example uh so alina also have some results to to have a uniform bound for gcd of function field but here it's not uniform here it's depend on g i so but i think to find the uniform bound should be very hard okay okay so okay so finally uh let's talk about uh how how this gcd theorem uh sorry how the this is a diameter equation formulation for function field can help us to find the the function field pixels okay so let's remember what do we want to do we have some linear recurrence and when you're plugging in every turn is an n's power okay and then we want to find the we want to say oh actually you can find the another linear recurrence and so b n would be some a n to the d and times some polynomial okay so so that's what we will try to do so let's begin with b the this linear recurrence so this is the the thing we are given okay and uh and here so we will take uh let's choose u1 to un to be basis of this gamma generated by the roots of b okay so um so then in other words uh in this formulation it's standard that we can find the Lorentz polynomial and uh so here is the part the polynomial type okay so such that b n would be equal to this Lorentz polynomial and to the f a m to the u1 m to the un m okay and then you can cure out the you can take out all the denominator and it doesn't affect our results so let me just say let's show f is in is a polynomial okay so now we have b m equal to f to the m u1 m to the un m and now let's look at our assumption our assumption says uh the intersection is trivial so in other words this tells us uh u1 to the un are multiplicatively independent module k so this is uh I don't know whether it's a technical condition but anyway now the method we use here we try to strongly rely on this condition so we make we there's no way we can take this out okay okay so now we have the setup so we have b m and it's a polynomial uh this term is a polynomial and here is u1 m to un m okay so by our our assumption this is a this power for infinitely many okay so for from the asymptotic from the diagram team results we just mentioned so for for each f m so when m is fixed okay and we we know we we plug in u to the m and uh if it's big enough okay then our diagram team equation tells us that uh f m would be those trivial case so it would be some constant monomial depend on m and then some polynomial to this power as long as m is sufficiently large okay okay so now we have a bunch of term uh we know it's in this type of form so how do we conclude that f itself is this type of form so uh this is I think this must be known somewhere but I but uh when but we give some uh reproof so suppose you have a polynomial which is not not kx zero they are called prime and suppose you plug in m they are called prime what am I talking about so if you have two called prime and when you specialize and so for for all but finite many they have to remain called prime okay so in other words uh if if we know for for infinitely many they are in this type of form and because you know uh if if originally I write an equation for f and support and so suppose uh I write down all the irreducible term and so for infinitely m they have to be uh called prime and if so in other words once you know for infinitely m they are in this type of form then there's only one possibility that uh this f is in some polynomial times xi times g to this power okay so uh so now we know the f itself is in this type form and we are actually quite close so so uh so this monomial is always something you just take an extension you can hear it so let's just assume that we don't have that monomial term let's look at this term okay so suppose f uh can be written in some polynomial times g to this power and uh let's assume even further this is also something can taking care of so assume case q is monic and so and also q has no this power okay okay suppose now let's deal with this type of f so what does this tell us so you try we have infinitely many m and you plug in m and take u1 to the n un to the n and it's a simply qm and g to the d so what does this tell us so remember bm is a this power okay and here is a this power so this tells us qm is also a this power for infinity many n okay so so so this is really a question so suppose you have a a polynomial monica this power three and suppose you you take an infinite many n constant such that it's a this power so is it q to be a this power or it's q q has to be constant coefficient because it's not it's this power three so in that case q must be constant coefficient so this is uh which is a problem and the so we uh so the result just tells us in this case q must have a constant coefficient okay so sorry it seems to go very fast i will save a lot of time for questions so finally just let me recall uh what is uh generalized uh ends power problem so this is a joint work of past and me so uh our result is the following so this is a more complicated form so if you have m is bigger than this number and now you start with some some polynomial which is not constant coefficient monica degree n so you can factor out polynomial with the constant coefficient part and this part is no constant coefficient okay and then uh that's a further further uh so actually is a polynomial with no constant coefficient term and then let's take out their monica irreducible vectors and suppose that they are written in this type of form okay now suppose we have some mu which is larger than every e j okay and let's buy the some distinct element from constant and so the question is the following so if you are looking at this polynomial you take a i and at every point they have a large multiplicity say the multiplicity is divisible by uh this this large number so our theorem says in this case you don't have much choice in this case h would be simply equal to product of j equal to l g j to the mu's power so in other words uh in in our previous case because we assume g to be this power three so it cannot be in this form so so therefore this actually must be constant and so in other words you have only this constant part so uh this is how to uh apply this theorem to this uh this power to this uh piece of this roots conjecture so uh wow i think uh this is the last slide last page of my slide so maybe i can uh uh spend some time for question