 Thank you very much for the invitation and for being there. So I'm going to tell you about joint work with Peter Josen about e-functions. But before I, so all new results will be joined with him but before I get there, I would like to start with maybe a bit of a long introduction about e-functions, their history, some examples and then the kind of questions we are going to answer. Okay, so everything starts with one of the most beautiful results in transcendence theory of the 19th century, which is about the values of the exponential function at algebraic numbers. And the final statement is, I'm calling it Hermit Lindemann Baistras. And it says that if you take n algebraic numbers and you assume that they don't satisfy any non-trivial Q-linear relation, and then you look at their exponentials, exponential of alpha one, exponential of alpha two, exponential of alpha n, then these complex numbers are algebraically independent. So there's no non-zero polynomial with rational coefficients in n variables that annihilates these n numbers. And well, the particular case where n is equal to one, then being linearly dependent means being non-zero. And then the theorem says that the exponential of alpha, so alpha and non-zero algebraic number is transcendental. And this was first proved by Lindemann, when alpha is, sorry, by Hermit, when alpha is rational in 1873. So this gives, for example, the transternance of the number e. And then nine years later, it was proved by Lindemann for all non-zero algebraic numbers. And at the end of the paper, he said that, well, this statement was true, and then sketched the argument, and it was, well, completed and maybe made rigorous by stress in 1885. And well, also the remark is that, well, this is about values of the exponential function, but you also get the transternance of pi from this, because exponential of two pi i is equal to one. And if pi were algebraic, then two pi i would also be algebraic, and one is not a transcendental number. Okay, so this theorem gives, at the same time, the transternance of e and pi, and we can say that it completely solves the question of understanding the properties of the exponentials of algebraic numbers. So, well, soon there was the question about how to generalize this theorem to other special functions. And it's not at all clear because the proof relies on the property of the exponential that the exponential of alpha plus beta is exponential of alpha times exponential of beta. And you can imagine that this is very useful because the theorem is that there are no non-trivial algebraic relations. So, using this property, you can reduce the question of starting algebraic relation to linear relations and those are easier. And in fact, you can rephrase the theorem as saying that if you have different algebraic numbers, then the exponentials are linearly independent, even over Q bar. Yeah, so that was the situation for a while. And then comes a fantastic paper by Siegel. So, the paper is published in 1929. Here is the beginning. It is dedicated to Max Dehn. I actually was, it was a time where and the mathematicians of getting in were already feeling that there was too much pressure to publish. Publish or Paris was already quite a bad thing for them. So, they had this thing of offering papers to each other and not publishing them. So, this paper, I think was written around 26 and it was sitting in Max Dehn's office for a while. For example, Andre Veil tells in his memories that he could read it when he visited Max Dehn. And it was only published later. So, at the beginning of the paper, he says, okay, so this is when he talks about this theorem by Hermit and Lindemann. He says that it uses this property of the exponential and you can already see from the first page. So, there's a formula for the Bessel function that is going to come back again soon. So, he's in particular interested in having, well, in understanding the transcendence properties of the values of the Bessel function at algebraic numbers. And one reason is that these numbers are very much related to some physics problems. So, one of them is when you study the vibration of a circular drum. So, you're hitting somewhere in the drum and then you look at the function that gives the height of the point x, y at the time t. And it is, this function is governed by the wave equation. So, that the second partial derivative with respect to time is equal to the sum of the derivatives with respect to x and y, up to some physical constant. This is inside the circle and well, the boundary doesn't move. So, there's the boundary condition that u is equal to zero on the boundary. And you can try to look for solutions, well, you can, since it's a circle, you can use polar coordinates and then try to look for solutions in separated variables. So, this part depends on the radius, on the angle and on the time. And so, you plug this in and the first thing you see is that the time is a sum of cosines and sines for certain frequency lambda. And then you see that the part depending on the angle is also something similar as sum of cosine and sine. So, here you have another parameter m. This doesn't appear if you hit the drum in the middle. So, the first thing is m equals to zero. But then you have other modes where it really depends on the angle. And then finally, you're using these two pieces of information you plug R in and you find that it is a solution of the Bessel differential equation. Okay, so it's an equation of degree two, differential equation of degree two. And there's this term where you see lambda and m. So, this equation has a unique holomorphic solution around zero, which is the Bessel function of order m. And it is, but you can look for the power series and the differential equation gives you a recurrence relation for the coefficients and then you find that the coefficients are like this. So, this half minus one to the n and divided by m factorial, m plus m factorial. So, m is the order again. And then set over two to this power. So, this is the power series representation and something that is going to be important later on is that it also has an integral representation. So, actually this is the integral of exponential of set over two t minus one over t against this differential form where you see the order again. And if you, well, if you use Cauchy's theorem, then you need to compute the residue and then there's a two pi i appearing. So, you divide by two pi i. Yep. So, this is the Bessel function, which is the holomorphic solution to that equation. And then the conclusion of the story is that the radial part is this Bessel function evaluated at lambda times r. Lambda was the frequency that we got at the beginning. But then you plug in the boundary condition. So, this is r equals to one. And then this tells you that the gm of lambda has to be zero. So, you find that the frequency is a zero of the Bessel function. And this was actually known much, really much earlier than Siegel's paper. And it appears, for example, in a paper by Bourget in 1866, where he makes the conjecture. So, you can plot the first Bessel function, g0, g1, g2, and so on. And, well, you see that they seem to have no zero other than zero. So, when the other is strictly positive, here you get a positive power. So, it vanishes at zero. But this seems to be the only one. So, and I went to see Bourget's paper and I was surprised to find that immediately after the paper in the volume where it is published, you get the referee report. So, this is the ones who presented the paper and then they explain what he does and why it is good and should be published and so on. Okay, so, that was one of the motivations for Siegel, this question about zeros of the Bessel function and something that it's not too hard to see, for example, by integration by parts or there will be a comological interpretation later on is that there's a recurrence relation between the Bessel functions of different order. So, you look at J M plus K, then you can write it as J of M and J of M minus one. And what appears is a polynomial with rational coefficients in one of our set, which is a explicit polynomial. It's called Lommel polynomial in the literature. And, okay, so now let's assume that there's a common zero to J M and J M plus K. Okay, so this means that these two terms here vanish. So, the other one has to vanish as well. But J M is a solution of a differential equation for the two and the derivative of J M is essentially up to sign J M plus one. So, if these two vanish at the same time, it means that the function of the derivative vanishes and then you can use the equation of all the two to deduce that the function will be identically zero. Okay, so what it's easy to see is that J M and J M minus one does not vanish at the same place. So, the conclusion is that this has to vanish at this potential common zero. And then this says that this zero, if it exists, is an algebraic number. Is there a question in the chat? Yeah, there's a question by Christopher Lloyd. No, sorry, they have absolutely nothing to do it's just a clash of notation that I didn't realize. Okay, so now let's park this for a while. We will go back to it later. So, that's Seagal's motivation and to try to generalize as he can, the mid-Lindemann biostatium he introduces the notion of an E function. Okay, so what is an E function? It is by definition a power series with algebraic coefficients. This is the field of algebraic numbers. It is convenient to write it as A n divided by m factorial set to the n. No mystery about the name. If you put all A n equals to one, then you find the exponential and E is for exponential. And besides having algebraic coefficients, you ask for two conditions. The first one is that it is a solution of linear differential equation with polynomial coefficients. So there exists some differential operator for the n, pn derivative to the nth power and so on such that annihilates the function. You want this to be non-zero, non-zero differential operator. And this amounts to saying that actually the coefficients A n over n factorial satisfy some recurrence relation. They satisfy a recurrence relation with polynomial coefficients. And then if it is true for A n over n factorial, is it also true for A n? So for example, even if I, here I ask that the coefficients are in Q bar, you immediately get from this recurrence relation that they are in a number field. So coefficients will lie in a number field. And then the second condition is the one that allows one to do a diophant and approximation of transcendence theory. And it is that we want the coefficients, okay, so we want the coefficients to be bounded by C to the n too, so to grow as most geometrically. But the coefficients are algebraic numbers on the one hand and we want to do diophant and approximation on the other hand. So you need to do this in the right way. And the right way that means that you must bound not only A n, but also all the Galois conjugates of A n. So that you want the max of all Galois conjugates of A n to be at most C to the n. And then you also want that the common denominator of A zero up to A n. So this is the smallest integer such that when you multiply by it, you get an algebraic integer. So you want also these to be bounded by C to the n. And if these two conditions hold, then the power is called an E function. It is not going to appear too much, but there's also a, this is quite the previous argument is that's a common zero of the vessel function must be algebraic. Yes, exactly. So in maybe 10 minutes, we are going to use the theory of E functions to rule out this possibility. Yeah. Okay. Yeah, I was going to say, if you delete the n factorial here and you keep the rest of the definition, you get something and this is called a G function. And the reason is again, that if you do A n equal to one, then you get the geometric series. So G is for geometric. Yeah, I can speak, but can I ask a stupid question? So D n is an algebraic integer. So what does that mean? It is at most C n? No, no, it's just an integer. It's a usual integer. Oh. So, yeah. So it's the, for each algebraic number, I can pick a integer. So it's not a minimal, it's not a minimal way to, okay, thank you. Okay. So, yeah, let me give you some examples. So, well, the exponential function I already gave it to you. And then the motivating example of the Bessel function. So let me just do for simplicity n equals to zero. So then this is how we defined it using the power series. And from here, you maybe don't see it. Well, that we already have the differential equation. So we need to see this growth conditions here. And maybe from this way of writing is not so clear because you need the n factorial to match the exponent, okay? So instead, you can write it as set to the two n. You divide by two n factorial and then what remains is the binomial coefficient. And then, well, it has rational coefficient. So you just need to bound the size. This coefficient, you can see that it is at most one. For example, writing one plus one to the two n and using the binomial formula. And the common denominator will be four to the n. So you can take a c equals to four in the theorem. Okay. And it's not too hard to prove from the definition that this class of functions is stable and there's sums, products, derivatives. And you can also take the primitive that vanishes at zero. So out of one example of a funny function, you get many of them by doing these operations. Okay, so I told you that the motivation was to generalize this kind of transcendence results. And here is how Siegel did it. So Siegel did it for the Bessel function. And with a technique that in principle could apply to all E functions, but there was some technical condition called normality. And at the end of the day, it was maybe harder to prove normality than to prove the theorem. So Schiedlowski in 55 found a way to bypass this normality condition. And then the very precise statement I'm going to give here is due to Andre and Boyker's in, so the final statement is due to Boyker's in 2006 and relies on the structure of the differential equation of E functions that was understood by Andre. So the statement is the following. We are going to look at the N E functions. And we assume that the solution of the vector F1 to Fn is a solution of a linear system. This should be a cube bar, sorry, and actually everywhere in this slide. So they are solution of a linear system with rational coefficients. So this is not really a restriction on a given E function because we know that it is a solution of a differential equation. So you can always take its derivatives and consider the companion metrics of the system. And then the theorem is about the special values. So we are going to evaluate these functions at an algebraic number alpha. And there are two cases that should be excluded. So first one is alpha equals to zero. Well, this is because these are power Cs with algebraic coefficients, so at zero they take algebraic values and there's just nothing to say about it. And more interesting is you need to exclude the situation where alpha is a pole. So this is an ASA metrics of rational functions and you don't want these rational functions to have a pole at alpha. And if this is the case, the theorem says that every algebraic relation between F1 of alpha up to Fn of alpha, polynomial with algebraic coefficients that vanishes at these complex numbers actually comes from an algebraic relation between the functions. So you can lift these to a polynomial Q such that now it's in one more variable and it vanishes identically on the functions. And when you specialize set to alpha, you find the polynomial you started with. And well, something I didn't say that it's implicit here is that because of this growth condition, actually this power Cs has infinite radius of convergence so you can evaluate that any algebraic number. So the question about the pole is to exclude some examples. So you can do the following thing, for example, you take exponential of set which is a transcendental function. So according to this, it should take a transcendental values but then you multiply by set minus one, okay? And then you get a function which is still transcendental but that vanishes at one now. And when you write the differential equation, you will see a set divided by set minus one in the matrix. So this pole explains the fact that you get a zero and the contribution of Andres to understand that all possible exceptions come from this kind of poles. Okay, so that's the theorem. And yeah, I, yes, maybe say it again. So it's a really a very powerful result in the transcendence theory which is completely false if you do the same thing with the G functions. So in general for G functions, you will have many algebraic relations that are not at all explained by the relations between the functions but for E functions is true and it's powerful because it's much easier to prove that a function is transcendental than to prove that the given value is transcendental. So in the case of the exponential, you can, for example, simply observe that the exponential is periodic. And from this, you get, for example, if E of X, Y is a polynomial of minimal degree in Y that annihilates the exponential, then you evaluate that all integer multiple of two pi i. So you, as a polynomial in Y, you will get a polynomial with infinite roots. So it needs to factor and then you find something of smaller degree than annihilates the exponential. And the middle in demand theorem we started with is the particular case where the functions f i are exponential of alpha i times z. So in this case, the matrix A is a diagonal matrix with alpha one, alpha two in the diagonal. And it's easy to see that all relations between the functions come from Q linear relations between the exponents. So it's really a generalization. Okay, so let's go back to Burgess hypothesis. So in the case of the Bessel E function and the Bessel differential equation, the only pole is at zero. Yeah, you can maybe see this from the equation. Where is it? Here, so you just get a pole at, well, the differential equation has a singularity at zero and infinity. So the only thing to exclude is alpha equals to zero. And the theorem says that otherwise and the value of the Bessel function is transcendental. And in particular, we have now approved the theorem that there are no common zeros other than zero to Bessel functions of different order because we saw that if such a zero exists then it must be algebraic. But the Bessel function takes transcendental values of algebraic numbers and zero is not transcendental. So there are no common zeros and it was a big success of Siegel but he doesn't even say it. So he says, well, we have proved this theorem and then there are no common zeros and that's it and he moves to the next topic. Okay, and yeah, this story seems not to be very well known because if you type a Burgers hypothesis on Wolfram Alpha, for example, it tells you that it has been proof for differential for Bessel functions of order one, two, three and four. So it's maybe time to get updated. Okay, so let's move to the new results now. So I gave you a few examples of E functions and just after introducing the notion, Siegel presents the case of hyper geometric functions. So this power series, I'm going to write them like this without the factorial now. So I told you that being solution of a differential equations means that the coefficients satisfy a recurrence relation with polynomial coefficients and hyper geometric functions are the case where this is a recurrence of length one. So it means that a n times a polynomial is equal to another polynomial times a n plus one. And in other words, as a function of n, the question of two successive terms is a rational function. And I'm going to take this polynomials with algebraic coefficients. This is for the general definition of hyper geometric function. And then the situation where you get a nif function is when p and q have rational roots. So you might have an algebraic leading term but the roots need to be rational. And then the degree of the polynomial p is strictly smaller than the polynomial q. And instead of considering just this power series, you evaluate that set powers to the power q minus p. This is similar to what we did here for the Bessel function. For example, here. So you see that you have, well, actually the variable appears to the power two n. So this is the same two n in set to the q minus p and it's important to get precisely a nif function. And yeah, so this is maybe the important property that the coefficients are given by a rational function. But then you can explicitly write down what you get. Shana, yeah, I'm sorry. Yeah, so can you scroll back one page? Yeah, this one. Thank you very much. So the relation in n plus one is that related to the Frobenius isocrystal properties that it's really Frobenius isocrystal and not the power of Frobenius. I mean the step is one and not two or three or four. Oh. You have the whole Gauss-Mann in here, don't you? Not a factor of the Gauss-Mann. No, yeah, the differential equation will be of the degree of order, the degree of q. But it usually appears as a factor and not the whole thing. What is the answer? Is that a factor or is that the whole Gauss-Mann? Usually it's a factor. Oh, it's a factor. Yeah, for example, the most classical case is q will be, boys, you get a differential equation for the two, but then it appears on a family of curves of higher genus. And this is still one. So step n, n plus one is still one. Yeah, yeah. So suddenly it's likely not going to be the explanation. Yeah, maybe not. What you do here when you put set to the power q minus p is to get the slope one and then your differential equation is a Fourier transform. Oh. So maybe this, yeah. Well, here I'm doing at the same time, this time I'm talking mostly about the regular case. But that one will be a Fourier transform of the regular one. And it's, yeah, probably it's not the explanation, but I'll have to think about it. Oh, I apologize for the interruption. No, no, thank you for the question, Ellen. Yeah, okay. So, yeah, I was saying that you can explicitly write down a formula and you get the questions of pohamer symbols which is a generalization of the factorial. So you do x times x plus one to x plus n minus one. So if you do the pohamer of one, you get n factorial. And yeah, there are p rational numbers upstairs, q rational numbers downstairs. This should be a not a negative integer, sorry. So because otherwise you get zero in the denominator and then you put the variable to the power q minus p and then you still have the freedom to multiply by some lambda, okay? So, and Siegel proves that this is an e function. So the differential equation follows very easily from this relation, it is of degree of other q. It's not hard to bound the absolute value of the coefficients, so they are rational numbers. And where there's a bit of arithmetic content is in the denominators. So essentially you need to play with the periodic evaluation of factorials. And at the end, it relies on some weak form of the prime number theorem. For example, in the shape that the less common multiple of one to up to n is at most three to the n, let's say. So he proves that these are all e functions. And then he asks the questions, is it true that, yeah. So from the operations, some derivative products and so on you get many, many more. And then he asks the question whether it is true that all e functions can be obtained this way. So as polynomial expressions in hypermetric functions. And this happens to be true if the equation e for so further one or two. So this is the case of the two functions we saw, the exponential and the Bessel e function. It was proved by Gorel of in 2004. And then in 2020, there was a very beautiful paper by Fischler and Rivo Al, where they prove that if one assumes the Grotendig period conjecture or maybe a variant of it that I will explain a bit later, then it should be false starting from e functions which are a solution of a differential equation of order three. So there's such an e function which cannot be written as a polynomial in hypermetric functions. And let me maybe say that what is very easy is to construct an e function which is not literally equal to a hypermetric function or to a multiple of it, because the property that the successive questions are rational function of n, this is very special. So you can easily produce one that doesn't satisfy this but then you need to take polynomials, products, sums, you can vary all these parameters here, pq, a, b, and then you lose control over this rationality, okay? And yeah, so this is my collaborator, Peter. And here we are working in the Island of Tattoo on this question. And we managed to prove, I mean, actually during COVID time that the answer to Seagal's question is negative. So there are e functions that are solution of a differential equation of order three, the minimal possible and that are not in the algebra generated by hyper-parametric. e functions are not even algebraic over this algebra. And actually most of them are of this type but it's not one of these generic situations where it is hard to write down an example. So there's a machine to produce examples that start with a polynomial of degree four and just to give you one. So this power series where it's a power series set to the end and then the coefficient is given by this which is, well, still very close to Po-Hammer symbols and factorials but you mix them in a bit of a complicated way. So this is an e function and it's not of the, it's not a polynomial expression in hyper-parametrics. Okay, so the way we, now I'm going to change a bit the topic for a while. So the, this is the geometry part of the title. So the way we got to this question and we could solve it is because we were studying for other reasons, exponential periods. What's the reason for the theorem? Yeah, I'll try to explain to say a few words about why this is true at the end of the talk. So now I'm going to try to explain how we found this function, okay? So we will have a look at exponential period functions and then at the end I will say a word about why it is not hyper-parametric. Thank you. Yeah, so we were studying for other reasons, exponential periods. So let me say very quickly what periods are first. So periods are the complex numbers that you can write down as integrals. Well, I need to be more precise about the integrals of what. So you integrate an algebraic differential form with omega with algebraic coefficients and you integrate on a subset of C to the N which is defined by, which is semi-ial algebraic which means it's defined by polynomial inequalities and you also want these polynomials to be with algebraic coefficients, okay? So two examples are the number pi when you express it at the area of the circle of radius one or for example, theta of three, the special value of the Riemann's theta function, if you develop the geometric series and integrate three times, you find that this is the integral on the unit cube of one over one minus x, y, z. And from a more sophisticated point of view which is the one which actually allows you to prove anything about these numbers, what we are doing is we are looking at the coefficients of a pairing between two vector spaces associated to algebraic varieties. So you can see an algebraic variety x over q. So this will be given by the shape of your integral. For example, if you are looking at this one, well, it's an integral in three variables. So typically you will start with a three, the affine space of dimension three and then you want a differential form. So you will remove the locus where it has poles. So your x in this case will be a three minus the zero locus of, well, the locus where x, y, z is equal to one. Yes, so this is your algebraic variety. And then you have two vector spaces, the cube vector spaces in this case associated to it. So one is algebraic, the ramp homology, where you do like in differential geometry, except that your differential forms have polynomial coefficients. So it's the homology of the complex of scalar differentials. And you can also, so this is purely algebraic. And you can also consider the complex points of your algebraic variety. That will form a nice topological space and then consider the singular homology of this space. So in this setting is usually called Betty homology. And you will have a pairing where you take a class of differential for omega and the class of a form sigma and you integrate omega alone sigma, okay? So if x is smooth and affine, all I'm saying is totally correct. And then by a theme of growth and dig, this is a perfect painting. And if, well, in the either non affine or non smooth case, you need to be a bit more careful what it means to take the class of a form and integrate and so on. So in such a way, you get a class of complex numbers and it will essentially be the same that we defined before except some, well, minor subtleties. Like for example, here you are integrating on the unity hypercube, which has a boundary. So it's not exactly a cycle in singular homology, but all this can be solved and then you get exactly the same class of numbers. And one advantage of this point of view is that you expect to have some kind of Galois theory of periods that will generalize that of algebraic numbers. So in general, these numbers will be transcendental, but you can imagine to act on such an integral. So G will be a linear automorphism of beta homology and you can imagine to act on the integral by simply acting on the cycle. And the problem of this is that it is a well-defined action on the writing as an integral, but of course, usually you can write a complex number in many different ways as an integral and this will depend on the way of writing. So the grotendic period conjecture that I mentioned in one of its incarnations tells you what are all the possible ways of writing the same period as different integral representations or how to pass from one to another. And this is very totally out of reach, but it's amusing that we can use this conjecture to actually define this Galois group and to get an action which is well-defined at least on symbols. Okay, so that's the very short summary about periods. Many numbers we like are periods, but some other numbers that we also like very much are not expected to be periods. So one of them is E that we already had and another one is Euler's gamma constant, which is usually defined like this as the limit of differences between the partial sums of the harmonic series and log. And we, well, there are good reasons to think that they are not periods in the previous sense, but they are both exponential periods. So exponential periods is a generalization of that notion where you integrate omega along sigma as before, but you are also allowed to put the exponential of an algebraic function. So the new ingredient is that there's an algebraic function F, omega is still an algebraic differential form, and sigma is what is called a rapid decay cycle, which is something that the main property is that it makes the integral convergent. So typically you, if sigma has a boundary, you want the real part of F to be very positive along this boundary, because then exponential of minus F decays rapidly, which means it decays faster than omega, which is a polynomial differential form and then the integral converges. A typical example of this is the square root of pi, which is not expected to be a period either, but that you can write as exponential integral of the Gaussian integral, so integral of exponential minus X square on the real line, which is a rapid decay cycle for this function. And in the same way as before, it's possible to think of exponential periods as the coefficients of a pairing where now you have the RAMc homology, which is associated to the variety, but also to a function and a homology group, which is this rapid decay homology. And the pairing is given, so classes here are non-compact cycles, so you write them as limits of sigma t's. There should be a t here, but they only go to infinity in the directions where the real part of F is very positive. And there's a question. Yes, the notion of rapid decay cycle depends on the function. So actually what we define is this space. So given the variety and the function, this will be unbounded cycles on the complex points of the variety and the directions where they go to infinity depend on the function. And again, this is a perfect pairing, so this is the first proof by Deligne for a polynomial on A1 and then Bloch and Eno for curves in a more general setting. And in general, it's at the end of in Rukeol. So this is exponential periods. And now I try to go back to E functions. And one nice thing about exponential periods that you don't have for useful periods is that it's very easy to deform them. What I mean by deform is that starting from exponential of minus F times omega alone sigma, you can introduce a variable Z and simply consider exponential of minus Z F. This is of course the case where F is equal to zero and then you are back to the wall of classical periods and they are just writing down the constant function given by a period. But in general, F is non zero and then you will get a function. And since the important property that defines the rapid decay cycle is the real part of F, if you take a set to have positive real part, then the integral still converges and then you will get a holomorphic function on this half space. And these functions are not E functions for two reasons. The first one is that even if you assume they are holomorphic, it extends to an entire function. Then in general, they do not have algebraic coefficients. And we already see this in the example where you take as a variety the complex plane minus zero. Complex plane minus zero. The function is X minus one over X. You can integrate on the unit circle. So this is an exponential period function. But when you compute using Cauchy's residue theorem, there's the two pi I missing. So this is almost the Bessel function but not quite is two pi I times the Bessel function. Which makes sense because this is a situation where you can evaluate the function at set equals to zero and then you get the integral of the X over X on the circle and this is exactly two pi I. So this is one problem that arises when the function is entire. But in general, it's not even an entire function and then there will be monotronomy around zero. And what we prove with Petter, so well, the theorem is proof the in progress refers to the writing is that these kind of functions on the half plane real part of it said positive. There are actually linear combinations of E functions with monotronomy. And the monotronomy is as easy as possible. So it's quasi unipotent monotronomy. It means you have rational powers of said and integer powers of the logarithm and then E functions. So in other words, well, this comes from the fact that the differential equation that is going to solve the, well, this function solves a differential equation and that one will have a regular singularity around zero and the monotronomy is quasi unipotent. So you get linear combination of these kinds of things and this settles the question of monotronomy and linear combinations where? So they are linear combinations with coefficients in, so the coefficients can be algebraic numbers but we also saw that you need periods because this is the case where f is equal to zero and then you get a function constant equal to a period. And then the other thing you get is the gamma constant and this one appears when you have logarithms. So it corresponds to the Jordan blocks of the monotronomy and you also get values of the gamma function at the rational numbers and these rational numbers have the denominators the denominators of the rational exponents. They come from the finite part of the monotronomy. Okay, so it is a general property and it really gives a source of new functions that would be hard to come on with to come up with. So for example, you can consider the same integral as before but instead of integrating on the unit circle. So this is a case where the homology has dimension two. So one cycle is the unit circle but there's another cycle which is actually the positive real line except that you need to go a bit around zero to avoid the singularity. And if you do this you apply the theorem to this situation so you find an E function which is the Bessel function we had. It comes with some logarithm and some gamma constant and then you find another power series which is in this case is this one. So HN is the harmonic number one plus one over two up to one over N. And this is a new E function that is, well it's not obvious to figure out and to prove that this is an E function. You will arrive in this particular case but it's solution of a differential equation of order four. You need to bound the denominators and so on. So this is the geometric source of E functions. And the easiest geometric situation that you can consider is your variety is the affine line. An algebraic function of the affine line is a polynomial. So we take a polynomial. I'm going to take it monic so that the real part is just governed by the leading term. And we consider this kind of integrals. And this integral is going to satisfy a differential equation of order the degree of the polynomial minus one. Because for example, you can compute the dimension of comology and if you differentiate under the integral sign and then you get different classes in comology and after doing the degree of the polynomial you have a linear relation. So if we want something of order three, I'm going to take a polynomial of degree four. Then this will give a differential equation of order three and in this case there's only a finite monodromy. So you get a linear combination of powers of set with the denominator of four E functions. And in this case, the coefficients are just values of the gamma function at one over four and two over four and so on. And so this produces E functions. And what we had observed with better before is that the only situations where you get polynomial expressions in hyper-dromatic functions are very symmetric. So the polynomial F has a symmetry property which is in general not true and is the following. So you consider the critical values of F. So this is F computed at the zeros of the derivative. So there are three zeros. You compute F, you get three complex numbers. And if they are not on the same line or if the triangle they form is not equilateral then these functions that appear here are not going to be a polynomial expressions in hyper-dromatic functions. So then if you want to write down an example you just need to take the easiest polynomial that satisfies this. So we took X to the four minus X two plus X and you compute the integral and then this gives exactly the function I was showing to you when you expand the exponential inside the integral. Okay, and that's probably a good moment to stop. Thank you.