 Thank you. So I'm supposed to say something about Gabba. So he likes to do technical things and he even likes to do almost mathematics when nobody else wanted to touch it. So I want to talk about our Kieloff geometry, which is an old topic. So if you have a curve over the integers of a number field, then you want to do an intersection theory for divisors. And you need an intersection number at infinity, infinite places. And this is done if you have pq in the complex. You want the intersection number to be minus log gpq. Where gpq is something like a sort of log of gpq. Where gpq is something like the distance between p and q. So if p and q are close, then the distance is small and minus log is big. And so the head should have big intersection. And of course, so gpq is sort of norm 1. pqc is a diagonal. And of course, there are many such choices. And Kieloff made the choice that if you have a measure mu, well, I mean, there are many such metrics. But at least for such metrics, we know something. We know that the curvature is a 1-1 form which represents the class of the diagonal. And so it's restricted. And our Kieloff said if we choose a measure mu, mu1-1 form, and the integral, well, now c has become a Riemann surface as 1. Then we look at Hermitian line bundles. So in the moment, it doesn't have to be. L with curvature is degree L times mu. So we know that the curvature integral gives a degree. So if it's a multiple of mu, it has to be this multiple. And we look at the line bundle which we call admissible matrix. So admissible matrix. And then we know that such a metric is unique up to scaling. OK. And so the GPQ should give a metric on the diagonal if I restrict one factor on O of Q. And then, well, if one requires that these metrics are admissible, then there's a unique, so there exists a unique metric, unique up to scale delta with curvature. Maybe I'll mu tensor 1. Just one tensor mu minus the changes on the diagonal. I hope I got this right. There's an alpha j on an orthonormal basis, but no sign that this has to be basis of. This is true because sort of this is the harmonic representatives for the diagonal. And, OK. And if we have such a metric, we get a metric on the differentials. So then get metric, differentials, omega c by the fact that this is delta. And the curvature of this metric is gotten. I take the negative of this and restrict this to the diagonal. And if we want this metric to be admissible, so the curvature is twice. And if we want this to be admissible, I think I should put a factor I over 2 for the inner product. And so this is admissible if mu is a multiple of this. So if mu is I over 2g. And this is called the Arkelov measure. So Arkelov. And if I do this, one can do some things. For example, one can define volume forms. So one can define volume forms. If L has an admissible matrix, which sort of satisfies the obvious thing on the determinant of this one, which satisfies obvious scaling conditions. And also if one has short exact sequences, one gets something. And this is up to scalar. And then up to scalar. And again, we can normalize the volume forms on the differential. It's sort of given by the square integral. So normalize. And then one knows that on L of degree g minus 1. So if L has degree minus 1. Then this is independent of scaling because we get the power of the Euler-Ponkeret characteristic, which is 0. And then this determinant is sort of O of minus theta. So the determinant is isomorphic to O of minus theta on the Jacobian. What did you say about independent of scaling? So then the metric here, if you have a line bundle of degree g minus 1, which is an admissible metric, then on the determinant of homology, it's independent of scaling because you get the power of the Euler characteristic. To define the metric on the determinant of homology, would you use the correction by the analytic function? Well, it can be written as the square integral correction by analytic torsion, but I use a different one. I use sort of if you have a line bundle. If you have an admissible line bundle in the point, you have this exact sequence. And this thing has a metric. And so the difference of homology of these two has a metric. And I require that this is compatible. You need a metric on L minus P? Well, I require on the determinant, no. Well, first of all, this also has a metric because O of minus P is a metric given by Akeler. And then you want this to be compatible, and this gives you essentially a metric on the O of these, where D is the divisor. And then you show that it factors over the isomorphism class, and it's enough to do this for degree g minus 1. And for this one, well, I mean, there you have no scaling problems. And for this one, you compute the curvature of this metric, and you show that it's induced from the Jacobian. You mean the curvature is a form on the curve? No, the curve, I mean, if you have O of a divisor, then the curvature is, so the divisor is parameterized by C to some big number. Then it's a one-form on C to this big number, so it can be written well. And you can compute this curvature, and you see that it's induced from the Jacobian. Okay, that is answer your question. So this shows that your rule specifies a choice of metrics on the determinant. Yes. Once you, but this, for all line balance admissible metrics, and so there is a unique choice on the category of those things, up to a scalar, a common scalar. So you don't get rid of the common scalar. No, you have the omega. The H1 of omega is canonically trivial. Okay. Yeah? So you have this one, and then you, from the curvature, then you see that the volume is sort of, I mean, that the log volume is proportional to minus log of norm of the theta function. So on the Jacobian of degree G minus 1, you have the canonical theta function, and you can normalize it, for example, that the square integral is 1 or whatever. And then, so it's up to a constant. That's a constant. And this constant I call, so define minus delta over 8. So this is, delta is just a real number, which is an invariant of the Riemann surface. And I call it this way, this strange way because of the following. So delta is the, the log of the value delta is what? You take the theta function and it has a canonical, well, it has a norm with invariant, means the O of theta has a norm with invariant curvature. And you can, and so the log theta, the norm of theta is determined up to a factor and you can normalize it such that the square integral is 1 and then you get a value for this one. The square integral of the section, of which section? Of theta, I mean, of the norm of theta. If you multiply the norm by a scalar then the square integral is also multiplied and then if you normalize it. The norm of all of theta? Yeah, in all of theta. Yeah, okay. So you have this one and then you can prove for allosmetic surfaces, surfaces. And the measure on J, G minus 1 is normalized with the volume is? One, for example, yeah. For allosmetic surfaces you get the Riemann Roch theorem and you get a sort of delta formula that the arithmetic genus is sort of omega squared plus delta over 12. It's a for semi-stable. Where omega is sort of the relative differential or the dualizing sheaf with the admissible metric and you have this omega squared by the Arkelov theory and delta is the sum. So delta is the sum of all places. Delta v. Well, at finite places you said maybe I write log qv times delta v, plus infinite. And the delta v is the number of singularities. So if you have semi-stable curves and generically it's smooth but there are finitely many singular fibers and there you have double points. And well, if you have a regular semi-stable model you count the number of points or if you have a stable model you would count with multiplicity and you get this one. And you also could count the number of geometric points or the number of... Well, okay. You would have to look at the extension field. Okay. And when you have real and complex places does it influence the formula that is your technique? Some of our embeddings in C is 12. Yes, yes. And then you get the sum... sort of sum of this over this infinite places. Sorry, is the delta the same in both places? Which delta? Delta on the first line. No, no. Delta is in the variant of Riemann's surface and at all infinite places you have different Riemann's surfaces and you have invariant delta and the sum occurs, yeah. But the complex conjugate is the same delta, I think? Yes. And then the question was this suggests that the infinite places the delta should somehow measure the singularities at infinity and then the question was... Well, the question was is delta related to a metric? I take the modular stack and... well, of all semi-stable curves and I look at the divisor which gives the singular curves and whether there is a metric so there should be something like look of the norm of one or something. So this is... this is the problem I wanted to solve and for this one I study degenerations so this means if I have Riemann's surface and if it approaches the boundary I want to look what happens. So study. We use the delin map for the complication. Yes. So if I have a family of Riemann's surfaces C or my S. Whenever G here is at least two or it can be one or what is... I think G can be one but it shouldn't be zero. And if the k is zero do you get anything? Does what you wrote make sense for G for zero or not? No, it doesn't make because you have to divide by G for the Arkelov measure. I think it seems related to the fact that you have minimal models and genus bigger equal one. Okay, and so you have a pi by C zero and you want to study the Arkelov... well, want to study. So want to study. Asymptotic Arkelov. And I should say if I have a curve C zero I get a graph gamma which corresponds vertices of E. So these are the reducible components C zero and these are the double points and an edge joins two reducible components if the double point lies in between them and I usually orient the edges but I won't say this here. And then sort of each, so for small s so each fiber is a union of maybe C zero V union C zero S C zero E where these are sort of these are curves with this removed and these are sort of annuli so I do a picture so you have several curve and they are sort of their annuli maybe there is a different curve here and it may go on and whatever and so the and the degeneration happens if these annuli are squeezed so yes, squeezed okay and so and so to do that as in topics and variant I consider them either sort of on these components or on the annuli which means I sort of have something distinguished between them and the annuli have equation have equation U times T equal TE let's say and U less equal to one and sort of in the TE are parameters which measure the singularity and I assume that they are small and usually I also take the invariant log minus log of TE I call SE so that I don't have to write negative numbers too often okay and then well and then I look at the GPQs oh sorry this is so the picture I will need later and I look at the GPQs let's say the P and Q are points either in the same or in different components and which behave nicely and then I look at the asymptotics and asymptotics means I want to study them on the curve up to a uniformly bounded function so asymptotics means up to bounded uniformly and then this means sort of up to a function it means I sort of need a reference model for these GPQs which is sort of standard I mean well if P lies here and Q lies there then I want it as sort of a local equation time something and so on so you allow also a component which intersect itself I suppose in a special time but then sort of it has an edge okay up to bounded function and the result will be but they are sort of constant I mean bending on S so the result would be that they are sort of constant so constant depending on S S and V on the on the curves on the interior of the curve meaning I guess that it has bounded variation on them and on the annuli so on annuli so mostly I write linear interpolation by which I mean sort of it's a linear function so in this coordinates so I want linear interpolation and so if this is always the case then sort of the interesting function are given by the values on the components but there is one exception so one exception but when you say constant on C0V when the two points lie in the same C0V or one point in one and the other in another no no I mean well it's both is true I mean but constant up to bounded function is the same as zero so so this is what you mean that it's uniformly bounded no no but the constant depends on the parameter S and the bounded function is bounded uniformly constant namely function of S but this why are all depending on S and V and then when you have points in one in C0V and other in C0W yes so one exception if you have the same annulus so if you have the annulus well if you have the same one then it's not my linear but there is a correction term and so correction term so I write this thing also maybe usually it's about the minimum of P times Q and Q times one minus P one so I mean these coordinates log U they sort of go from zero to one on both of them and I get the correction term which is a multiple of this one which is non-linear in P and Q and this I think is related to the fact that if you consider the diagonal then then on the product of two and you lie this is not a Cartier device and you have to do a blow up and this way you get this one P and Q are the coordinates on the what sort of these log U which I did write I did not write this linear interpolation so the linear interpolation means that let us say one point is in the one point is in the interior and then you get so the log U I normalize so that it runs from zero to one and the log V also and then I call this P and Q and then I get this multiple of this function you get when you say the linear interpolation before the exception stuff what does it mean it means we are referring to the case where one point is in an annulus and one point is not no no I refer to the fact that both points are in the same annulus before I was asking three lines about both this when you write the linear interpolation yeah I said I mean that the function depends linearly on the interpolation I mean this is also allowed the two are in the same annulus except that I have this this correction you interpolate between the two edges of the annulus and so if you have two points in different annulus I also do this yes yes and so there is this one correction term this implies that if you have an annulus between that's a component there are four numbers originally this this this this this this so when you want to interpret linearly yes then you want a bilinear function and this you have to know on the four ends on the four corners but also with yeah okay x y okay okay so I so this is the result and this is related to this and well and the things and these values are sort of given by the graph so then it's given by graph so they can be computed from the graph t and okay so maybe how do you do this computation so how do you compute the alkylov t's this can be done as follow if you have a Riemann surface and you have two points you have a sort of harmonic function h p q of z which is sort of poles so we have sort of poles with videos plus minus one this means it's sort of near these points if I do a local coordinate it's asymptotically plus or minus like z and this is unique up to a constant and how do you get this you take alpha p q a differential q with residues minus one this you know and then it's unique up to adding a holomorphic differentials and you can acquire it to have purely imaginary periods because if you have holomorphic differential you can prescribe the imaginary parts of the sort of the real parts of the periods and you can subtract this so you get this one and then you take the h p q you take sort of the infinite indefinite integral of alpha p q so this is unique up to a constant because all the sort of the real part of this so the real part is independent of the path because your residues are there and then you have this g p q maybe a g p of z I'm sorry so you take the integral sorry z of q this one you take the integral of h p q of z times the mu of p plus a constant and the constant is sort of such that the mu integral is zero so the graph is gamma or g u or the gamma there when g e it is the same okay I guess the letters are not the different so the g z so q is fixed yeah I integrate over one variable and what does it z is a variable how do you enter z is a variable p q depends on three variables on p and q and on the variable z you you take h p q so the h p q is sort of curvature delta p minus delta q and if you integrate over p then you get curvature mu minus delta q so you have sort of something like curvature which is right curvature for the for the g p q and then you have to add a constant to make the mu integral to zero mu integral is relative to z or what yeah well you could do any well you could for example do the integral over z or you could also do the integral over q fix that and the both conditions are equivalent yes so this one and so this is how I got this and now I have to talk about the well how I do this on a degenerate curve so first of all I have to know something about the Arkelov measure so so on c zero on the fiber I have the differentials and I have either the differential which have no poles at the double points so this is the gamma c v omega c v omega and then I have the residues I have the sort of the c to the edges I have the residues at the ease and now more rigorously so the notion of a graph is like in Saer's book that is you have the set of edges is really some you have a set of an involution so it depends because you need a return so you have so when you write this you may have to take this into account because you have the well I said you orient all edges okay I don't know how it is in Saer's book but yes somewhere I said there is an involution in the direction of the so talking about the graph I forgot one I don't know what the graph is because this if you look at different books what do they say is the graph slightly different depending on what you read so what? unoriented multi-graphers I think the most is thinking but I say it's oriented and it's no multi-graph it's a single graph for the graph I have the homology I have a map from z to the e to z to the vertices which is the e and the kernel I call x which is h1 of d what did I say I had this gamma and the co-kernel I call this differential d and well and the residues of a differential lie in x to the r this maps over x z and then it's exact this is because if you look at the sum of the residues and the components and it's 0 and this means just it's the boundary so the boundary map sends h to plus to the two end points the difference of the two end points and because I had the h oriented well you can figure out whether you want the beginning minus the end or the end minus the beginning okay so I have this one and so the differential form a nice family well and there is a similar on the singular fibers you have this residue map so it degenerates so on each well so there is a residue map and then you get a basis of the differential you get the differential on the curves and you get differential which have non-trivial residues so then you have so so the cokernel is isomorphic to this one and then if you do the square integral of a differential so if you do the square integral then if the differential is holomorphic on doesn't have poles then it's about the square integral on the component but if it has poles then the integral is is sort of concentrated on the annular and so it's about the sum of s e times residue e squared so if you integrate well over an annulus you get some of this one if you look at the residue so by the way the residue is defined so on the on the other fibers by sort of integrating once around a loop here an integral of what? of a one form this is the sum of so if you want to integrate the u over u squared over this annulus then you sort of take polar coordinates then you get 2 pi times integral maybe of log t e to 1 you get what is it d x over x and the sum d x over x and this gives you 2 pi and the other terms so in the differential the u over u term is the one which gives which dominates the integral and you get this one so here you are doing the integral kind of in the family of all the mobilize yes yes you choose a lifting of something in the special section of the canon of the dualizing shape on the special fiber you extend it to I mean on the other fiber I still have a residue map by doing this integral and I can do the square integral well and I can have a nice algebraic family over my base s because the omega is a vector bundle and I can take a basis near s and then I get a basis of the differential on each curve and it has a residue map to this to this x and the kernel I can also consider and for something in the kernel the square integral is about the square integral over the components and for something and for the residue then the square integral is dominated by this one because I assume that the s e are between I want to look at the asymptotics if I get close to the singular curve so by the way I don't understand you have this exact sequence you want to see to the if you want to look what do you say about the x you say that the x is the space is the space of loops in the graph but what is the relation to the the x lies in c to the e because the kernel lies in this one you claim that this is the image of the residue map the image of the residue map is the x and so and so well I want to find an orthonormal basis and then I get an orthonormal basis I sort of take a basis for the inner product on x so so I take so I get an orthonormal basis I take an orthonormal basis for x with norm x e squared this one and I take an orthonormal basis and then I sort of get an orthonormal basis of the whole space by combining them while I can sort of lift them I mean if the s e becomes big then the x e becomes small and I can sort of lift them that they are small on these components so the integral is like this and so this gives me an orthonormal basis so this gives so it gives an orthonormal basis and then there are kilo kilo so this is one thing then I have to define these h pq's so I have to find so I have to find pq what is the last comment plus arachelov I get the arachelov measure if I know an orthonormal basis so here it's an orthonormal basis for over in the family of the space of differential but what do you mean you get the arachelov mu is on the on the five on the fiber C s so what is the so the arachelov mu has sort of some part which lies in these components and some part which is concentrated on the anuli and the rest is sort of small and on the component I sort of get the arachelov measure of these curves except I have to multiply with the genus divided by the total genus because for the arachelov measure I have this one over one G and what are the stability or semi-stability you have on the family I mean about genus zero components you assume something or not I mean genus zero well on the genus zero components the GV is zero so the arachelov measure is sort of zero the component and then I get an additional contribution which sort of lives on these anuli and this is sort of controlled by this X with this inner product sometimes when they do the generation of curves they assume like the stability condition, the rational curve with this list do you assume something like this or does it break down when you I think certainly semi-stable is enough but I standing up I don't want to well I guess semi-stable is enough but otherwise I don't want to commit myself out of it so we have to find these alpha PQs and you sort of can find nice algebraic families let's say if you have two points PQ in there you can find nice algebraic families but you need the condition that periods are purely imaginary so so then how to get and sort of the periods are sort of I mean the infiltration there are sort of three types one is sort of the loops around the annular the second one is sort of loops in here in the curve CV and the third one are sort of periods which come if you which which correspond to loops in gamma so I think I called this gamma now see and sort of I mean and the integrals over this one are the residues so I want that the residues should be real so the residues sort of 2 pi times residues at E should be real then you then you have these interior loops and this by you can sort of correct by adding differentials on the curves which have no poles and they are sort of unique so interior loops and then there are sort of the most interesting one are the loops in X the loops in X while in the graph gamma which correspond to X so I said I have differential with given residues and I can lift them uniquely that the interior periods are purely imaginary so lift X to differentials purely imaginary interior loop and then well and sort of I also should this is sort of the residues should be in the real basic extension of X and then the loop well and then I want to add such differentials and because the interior periods are purely imaginary I don't destroy what I have already achieved so this is good and the integral about a loop sort of is a linear form on the X e so integral integral is a linear form form on X and it differs from the sum R so differs from sum S e times X e times Y e so I have a linear form on X by bounded by a bounded by a bounded well you have you have a billionaire form on X cross X where one X describes the loop and the other X describes the differential and then you get the integral around the loop and this gives you a billionaire form and the integral is sort of dominated by the integral around the interiors of the loops which sort of give you these terms and then there is sort of the integral which sort of fits this part which goes from one end from the loop to the other but this is bounded so this is a small function okay and then you sort of want to want this form to be zero so you have a linear form on X and well and you have this non-degenerate in a form and then you sort of can no no sorry no sorry this I said wrong I was too far so I have had my candidate alpha pq which I fixed in the first step that in the interior loops it has purely imaginary periods and then I take these more complicated loops and I take the real parts of these integrals and these form linear form sorry and then for X so for X sorry so this is linear form and I want to subtract something from X to make this zero because I want the real parts to be zero so are you taking now you are working on the complex or the real X is over which is it Xc or Xr no I mean I have only real parts this condition zero that's around the loops at the edges the integral is purely imaginary so I want an X with real residues then I take these takes my alpha my candidate for a form which I have to correct by adding some form parameterized by X and the real part of the integral around loops in X gives me a linear form X and this linear form comes from X well X itself gives linear forms which are about this one plus a bounded function and so this means I can correct it by this by sort of taking the orthogonal projection from a linear form sort of I mean this linear form is determined by inner product with an element of X and then correction is this one plus a bounded some bounded term uniformly in X so this gives me so this gives me the this gives me what 5 more minutes yeah sorry I don't know whether answering question is added like in soccer so so this way so this way you can construct your differential HPQ and sort of these orthogonal projections are given by the graph and then you can compute the HPQ and then you can compute all invariance so since I'm supposed to finish I should say and the final result is so the final result is that sort of the interesting invariance are rational functions so the interesting like invariance are rational functions are a sort of homogeneous rational function in the homogeneous of degree one one in the SE this means to the values of these G's on the component also to the for the delta function but unfortunately they are not linear forms so unfortunately the delta is not a linear combination of the SE and so this means there's no metric on the or on the boundary of the boundary of the modular space which gives this delta function so doesn't but at least you can compute things here or I should say I have a student Robert Williams who was supposed to do this but he did something else which happens for some of my students which was also very good so I had to do it myself are there questions perhaps from someone other than Ofer I'm sure has questions can you say in this case where you have what you call the linear interpolation so one of the points is on the annual line where can the other point really be to get a contribution well it could be anywhere but how do you I mean is there some intuition behind that because they seem like even on the limit far apart right well I mean if you have linear interpolation it's enough to have points on the different components and for the components for the graphs there is something there's sort of I had this d to the v I had a d and I have an adjoint and I can also scale the inner product well it is natural inner product so scale and then you sort of get a Laplace operator and there is a similar thing sort of if you have now you have a sort of a scale of measure here so you have sort of you put v over g on on v and then you sort of in the edges and on the edges you sort of you have a certain measure and you sort of divide it equally between the two ends so you get a certain measure and then you sort of well you want to solve a Green's function function you want to solve the equation delta delta of g is sort of a point minus p by mu minus p I should say so the mu has total measure one and if you have a point also then you can then this is total measure zero the image of the Laplace operator and this gives you a Green's function on the on the graph and the asymptotic of the Green's function is essentially given by this one as I said it sort of computed by the graph so we are going to resume and thank for things again