Arakelov geometry on degenerating curves
Faltings, Gerd
We investigate the asymptotic of Arakelov Green functions and metrics, and of the delta-function, if a smooth Riemann surface degenerates to a stable curve.
 Thank you. So I'm supposed to say something about Gabba. So he likes to do technical things and he even liked 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, 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 they should have big intersection. And of course, so GPQ is sort of norm 1, PQ or metric. And delta, delta C is a diagonal. Okay, 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 Kieloff said if we choose a measure mu, mu is 1-1 form, and the integral, well, now C has become a Riemann surface, is 1. Then we look at Hermitian line bundles. So Hermitian. And the moment it doesn't have to be, with curvature is degree L times mu. So we know that the curvature integral gives the degree. So if it's a multiple of mu, it has to be this multiple. And we look at the line bundles which we call admissible metrics. And then we know that such a metric is unique up to scaling. Okay, 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 they are the unique metric up to scaling. And so we have a metric on O of delta with curvature. Maybe I'll mu tensor 1, just 1 tensor minus some alpha j1. And then if I change these on the diagonal alpha... I hope I got this right. There's a alpha j on an orthonormal basis. No, sorry, this has to be basis of... This is true because sort of this is the harmonic representatives for the diagonal. And if we have such a metric, we get a metric on the differential. Maybe I'll 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 for the curvature is twice minus 2 mu. And if we want this to be admissible, I think I should put the 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 times some alpha j. And this is called the Aracheloff measure. And if I do this, one can do some things. For example, one can define volume forms on a gamma cl. If l has an admissible matrix, which sort of satisfies the... Obvious scaling condition 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 that are omega and the differential. It's given by the square integral. So normalize. Square integral. And the quenches. And then one knows that on l of degree g minus 1, so l has degree g minus 1. Then this is independent of scaling because we get the power of the Euler-Ponkery 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. And the Jacobian. What did you say about independent scaling? 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, what do you use the correction by the analytic torsion or what not? No, 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, okay l of. If you have an admissible line bundle at 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. And this. Will you need a metric on l minus phi? Well, I require on the determinant. No. Well first of all, well this also has a metric because l of minus phi is a metric given by the. By Akeler. By Akeler. And then you want this to be compatible and this gives you essentially a metric on the all of these. The 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 one. And for this one, well I mean there you have no scaling problem. 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? I mean if you have all of the divisors and 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 under the term and once you, but this for all line boundaries is an admissible metric. And so there is a unique choice on the category of those things up to a scalar. Yes. A common scalar. And so you don't get rid of the common scalar. No, you have the omega. The h one of omega is canonically trivial. Okay. So you have this one and then from the curvature then you see that the volume is sort of, I mean it's a log volume is proportional to minus log of norm of the theta function. So on the Jacobian of degree g minus one you have the canonical theta function and you can normalize it for example that the square integral is one or whatever. And then so it's up to a constant plus a constant. And this constant I call so define minus delta over eight. 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. 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, so the norm of theta as determined up to a factor and you can normalize it such that the square integral is one and then you get a value for this one. Square integral of the section, of which section? Of theta, I mean of the norm of theta. If you multiply the norm by the scalar and the square integral is also multiplied and then if you normalize it. The norm of O of theta? Yeah, O of theta. Okay, so you have this one and then you can prove for allosomatic surfaces. And the measure on Jg minus one is normalized with the volume is one for example, yeah. For an osmotic surface you get Riemann Rochseorin and you get a sort of delta formula. That's the arithmetic and the denouce is sort of omega squared plus delta over 12. That's say for semi-stable. Where omega is sort of the relative differential, so the dualizing sheaf is the admissible metric and you have this omega squared by the R-Kilov 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 I mean plus infinite and the delta v is the number of singularities over v. So if you have a semi-stable curve, so engineeringally 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 could count the number of geometric points or the number of points. Well okay, you would have to look at the extension field. And when you have real and complex places that influence the formula that is you take the sum of embeddings and say that it's twice. And then you get the sum of these over these infinite places. Sorry is the delta the same in both places? Which delta? You have delta on the curve. No no, delta is in the variant of Riemann surface and at all infinite places you have different Riemann surfaces in the Riemann variant delta and the sum occurs here. But the complex conjugate is the same delta I think. And then the question was, well 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 on, I take the modular stack and well. 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 log of the norm of one or something. So this is the problem I wanted to solve. And for this one I study degenerations. So this means if I have Riemann surface and if it approaches the boundary I want to look what happens. So study. So if I have a family of Riemann surfaces, 0, my S. I think G can be 1 but it shouldn't be 0. And if the K is 0 do you get anything, does what you wrote make sense for G for 0 or not? No it doesn't make because you have to divide by G for the Achille of measure. I think it seems related to the fact that you have minimal models in Venus, bigger equal 1. Okay, and so you have a pi by C0 and you want to study the Arachyl of 1. So you want to study asymptotic behavior of Arachyl of data. And I should say if I have a curve C0 I get a graph gamma which corresponds vertices of E, so these are the reducible components of C0 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 so what I want to say this here. And then sort of each surface small s, so each fiber is a union of maybe C0V, union C0S, C0E where these are sort of curves with this removed and these are sort of annuli. So I do a picture so you have several curves and they are sort of they are annuli. And so the degeneration happens if these annuli are squeezed. And so to do the asymptotic and variant I consider them either sort of on these components or on the annuli which means I sort of write something distinguish between them and the annuli have equation u times v equal te. Let's say and u v less equal 1 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. And then I look at the GPQs, sorry this is the picture I will need later. And I look at the GPQs that 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 a bounded uniformly NS bounded function. And then this means sort of up to a function, it means I sort of need the reference model for these GPQs which is sort of standard, I mean well if p lies here and q lies there then I want that sort of a local equation times something and so on. So you allow also a component which intersects itself, I suppose in the special fiber. Yeah but then sort of it has an edge here. Okay. Up to bounded function and the result will be, but they are sort of constant, I mean depending on s, so the result would be that they are sort of constant, so constant depending on s, s and v on the curve, 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 maybe I write linear interpolation by which I mean sort of it's a linear function in log u, log v, so in this coordinates, so they have one 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 means when the two points line the same c0v or one point in one and the other in another. No, no, I mean while it's, while it's both is true, I mean. But constant up to bounded function is the same as zero, so this is what you mean that is it's uniformly bounded in your sense? No, no, no, but the constant depends on the parameter s and the bounded function is bounded uniformly in s. Okay, constant namely function of s and v. Yeah, but this why I wrote depending on s and v. Okay, okay. And then when you have points in one in c0v and other in c0w then it depends on s, v and w. Yes. Okay. So one exception if you have the same annulus, so if you have the annulus cos, well if you have the same one then it's not violin, yeah, but there's a correction term and so correction term. So I write this thing also maybe usually it's about the minimum of p times one minus q and q times one minus p. 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 on the product of two annulis this is not a Cartier device and you have to do a blow up and this way you get this one. What is the p and q? P and q are the coordinates on the, what sort of these log u which I did write, I did write this linear interpolation. So the linear interpolation means that let us say one point is in the annulus and one point is in the interior and then you get a u. No, no, I mean 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, yeah but when you say the linear interpolation before the exception stuff, so what does it mean? It means we are referring to the case where one point is in 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 back above this. Yes. 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 if two are in the same annulus except that I have this correction. Are you going to interpolate between the two edges of the annulus? And so if you have two points in different annulus I also do this. And so there is this one correction term. This implies that we give an annulus between let's say kappa, kappa, lp and lp. There are four numbers originally in this, this, this, this, this, this, this. So when you want to incorporate linearly. Yes, then you want a bilinear function and this you have to know on the four ends, on the four corners. I mean also with x, y, okay. Okay, 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. By the graph. So they can be computed from the graph G. And okay, so maybe how do you do this computation? So how do you compute the arcale of these? This can be done as follows. If you have a Riemann surface and you have two points, you have a sort of harmonic function. Hpq of z, which is sort of poles. So we have sort of poles with residues plus minus one. This means it's sort of, these points, if I do a local coordinate, it's asymptotically like plus or minus like z. And this is unique up to a constant. And how do you get this? With residues plus minus one, this you know, and then it's unique up to adding holomorphic differentials and you can acquire 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 Hpq, you take sort of the infinite, indefinite integral of alpha pq. So this is unique up to a constant and because all the, sort of the real part of this, all the real part is independent of the parts because your residues are there. And then you have this gpq, maybe our gp of z. I'm sorry. So you take the integral, maybe, so a z of q, what? And you get this one, you take the integral of Hpq of z times the mu of p plus a constant. And the constant is sort of such that the mu integral is zero, so integral over mu. So, yeah, that's the same. Okay. I guess the letters are not the different. So the g, z, so q is fixed. Yeah. Yeah, I integrate over one variable and. So what does it, z is a variable? How do you enter? Ah, z is a variable. So q depends on three variables on p and q and on the variable z. Okay. And you take an Hpq. So the Hpq 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 a right curvature for the gpq. And then you have to add a constant to make the mu integral equal to zero. And mu integral 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. For a fixed z. Yeah. 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 R, well, how I do this on a degenerate curve. So first of all, I have to know something about the R-Kilov measure. So on C0, 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 Cv omega Cv. Then omega, well. And then I have the residues. I have sort of the, see, to the edges. I have the residues at these. And now, anyway, more rigorously, the notion of a graph is like in Serres's book, that is, you have the set of edges is really some, you have the set with an involution, so it is equivalent because you need the retains. Yes. And you have the, 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. Yeah. In some way. I don't know how it is in Serres's book, but. Okay, yes, somewhere I said, in the evolution of the direction of the. So talking about the graph, I forgot once. The notion of what the graph is because this, if you look at different books, what should they say is a graph slightly different depending on what you read. So. Yeah. Yeah. Unoriented multigraph. What? Unoriented multigraph is, I think, the nicest thing to do. But I say it's oriented. And it's no multigraph. It's a single graph. Well, 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-carnal. And usually I call this differential d. And well, and the residues of a differential are lie in x to the r. So this maps over xc. 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 that it's a boundary. So the boundary map sends h to the two endpoints, the difference of the two endpoints. And because I had the edge oriented, well, you can figure out whether you want the beginning minus the end or the end minus the beginning. OK, so I have this one. And so the differential is from a nice family. Well, and they are similar on the singular fibers. You have this residue map. So it generates on each. So there are the residue maps. And then you get the base of the differential. You get the differential on the curves. And you get the differential which have non-trivial residues. So then you have a co-carnals 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 a dozen of poles, then it's about the square integral on the component. But if it has poles, then the square integral is sort of concentrated on the annular. And so it's about the sum of Se times residue E squared. So if you integrate over an annulus, you get some more of this one. If you look at the residue. So by the way, the residue is defined also on the other fibers by sort of integrating once around the loop here. Okay? An integral of what? Of a one form. It 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 Te to one. You get what is it, dx over x and the sum of log Te, dx over x. And this gives you 2 pi minus 2 pi times log Te. And the other terms, so in the differential, the u over u term is the one which dominates the integral. And so here you are doing the integral kind of in the family of the mobilized family. Yes, yes. You choose a lifting of something in the special section of the conund of the dualizing shape on the special fiber you extended 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, 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 the basis of the differential on each curve. And it has a residue map 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 SE are big and 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. Yes. You want to see to the… you want to look… what do you say about the x? You say that the… The x is the space, is the space of loops in the graph. Yeah, but what is the relation to the… X lies in C to the E because the kernel lies in this one. Are you claiming that this is the image of the residue map? What? Yeah, it's the image of the residue map of this x. Okay, 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 I get an orthonormal basis. I take an orthonormal basis for x with norm xE, so xE squared, y. This one, and I take an orthonormal basis. And then I sort of get an orthonormal basis of the whole space by combining them. And while I can sort of lift them, I mean if the xE becomes big, then the xE 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 an orthonormal basis and then they are kilo. Okay, so this is one thing. Then I have to define this hpq. So I have to find… What is the last comment plus our kilo? I get the our kilo of measure if I know an orthonormal basis. So here it's an orthonormal basis for the space of the differentials. What do you mean you get the… The alkyl of mu is on the fiber CS. So the alkyl of mu has 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 alkyl of measure of these curves, except I have to multiply with the genus divided by the total genus because for the alkyl of measure I have this one over 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 the genus zero… well, on the genus zero components the gv is zero so the alkyl of 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. But anyway sometimes when they do the generation of curves they assume like the stability condition, the rational curve is at least three… Yes. Do you assume something like this or does it… does things break down? 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. Okay, 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 to condition that periods are purely imaginary. So then how to get purely imaginary… 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 anuli, 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 correspond to loops in gamma. I think I called this gamma now, no G. And sort of I mean and the integrals over this one are the residues so I want the residues should be real. So then you have these interior loops and this by you can see that the anuli is the 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 at alpha cv and then they are sort of the most interesting ones are the loops in x, the loops in x while in the graph gamma which correspond to x. So I said I have differentiate this given residue and I can lift them uniquely that the interior periods are purely imaginary. So I lift x to differentiate with purely imaginary maybe a right interior loop. And then well and sort of and I also should… so this is sort of the residue should be in the real base 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 the loop is sort of a linear form on the xe, so integral around the xe is a linear form on x and it differs from the sum r. So differs from sum se times xe times ye. So I have a linear form on x by bounded function by bounded linear form. You have a linear 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 linear form. And the integral is sort of dominated by the integral around the interior of the loops which sort of give you these terms and then there is sort of the integral which sort of this part which goes from one end from the loop to the other but this is bounded. So this is a small function. And then you sort of want to want this form to be zero. So we have a linear form on x and well and you have this non-degenerate inner 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 over which is it xc or xr or? No I mean I have only real parts because I want 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 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 on x and this linear form comes from x. x itself gives linear forms which are about this one plus a bounded function and so this means I can correct it by sort of taking the orthogonal projection from a linear form. Well I mean this linear form is determined by inner product with an element of x and then the correction is this one plus some bounded term uniformly in x. So this gives me the… So this gives me the… Five more minutes. What? Five more minutes. Yeah sorry. I don't know whether answering question is added like in soccer. So this way you can construct your differential HPQ and sort of and these orthogonal projections are given by the graph and then you can compute the GPQ 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 of rational functions are sort of homogeneous rational function in the homogeneous of degree 1, 1 in the SE. So this means to the values of these g's on the component also to the delta function but unfortunately they are not linear forms so unfortunately… So delta is not a linear combination of the SE and so this means there is no metric on the or of the boundary of the modular space which gives this delta function. So it doesn't… At least you can compute things here. I should say I have a student Robert Wirmes who was supposed to do this but he did something else which happens with 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 he has questions. In this case we have what you call the linear interpolation. 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… is there some intuition behind that? They seem like even on the limit and as far apart. 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 is sort of… I had this d to the v and a d and I have an adjoint and I can also scale the inner product well it is natural inner product so scale product ys and then you sort of get a Laplace operator and there is a similar thing sort of if you have a… now you have a sort of a scale of measure here so you have sort of you put v over g 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. You want to solve the equation delta of g is sort of a point minus p… my mu minus p I should say. So the mu is total measure 1 and if you have a point also then you can… then this is total measure 0 so it's the image of the Laplace operator and this gives you a Green's function 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 at 5 to 12 but let's thank Paul Tins again.
 Thank you. So I'm supposed to say something about Gabba. So he likes to do technical things  and he even liked 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, 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 they should have big intersection. And of course, so GPQ is sort of norm 1, PQ  or metric. And delta, delta C is a diagonal. Okay, 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  Kieloff said if we choose a measure mu, mu is 1-1 form, and the integral, well, now C  has become a Riemann surface, is 1. Then we look at Hermitian line bundles. So Hermitian.  And the moment it doesn't have to be, with curvature is degree L times mu. So we know  that the curvature integral gives the degree. So if it's a multiple of mu, it has to be  this multiple. And we look at the line bundles which we call admissible metrics. And then  we know that such a metric is unique up to scaling. Okay, 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 they are the unique metric up to scaling.  And so we have a metric on O of delta with curvature. Maybe I'll mu tensor 1, just 1 tensor  minus some alpha j1. And then if I change these on the diagonal alpha... I hope I got  this right. There's a alpha j on an orthonormal basis. No, sorry, this has to be basis of...  This is true because sort of this is the harmonic representatives for the diagonal. And if we  have such a metric, we get a metric on the differential. Maybe I'll 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  for the curvature is twice minus 2 mu. And if we want this to be admissible, I think  I should put the 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 times some alpha j. And this is called the  Aracheloff measure. And if I do this, one can do some things. For example, one can define  volume forms on a gamma cl. If l has an admissible matrix, which sort of satisfies the... Obvious  scaling condition 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 that  are omega and the differential. It's given by the square integral. So normalize. Square  integral. And the quenches. And then one knows that on l of degree g minus 1, so l has degree  g minus 1. Then this is independent of scaling because we get the power of the Euler-Ponkery  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. And the Jacobian.  What did you say about independent scaling?  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, what do you use the correction by  the analytic torsion or what not?  No, 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, okay l of. If you have an admissible  line bundle at 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.  And this.  Will you need a metric on l minus phi?  Well, I require on the determinant. No. Well first of all, well this also has a metric  because l of minus phi is a metric given by the.  By Akeler.  By Akeler. And then you want this to be compatible and this gives you essentially a metric on  the all of these. The 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 one. And for this one, well I mean  there you have no scaling problem. 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?  I mean if you have all of the divisors and 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 under the term and once you, but this for all line boundaries is an admissible metric.  And so there is a unique choice on the category of those things up to a scalar. Yes. A common  scalar. And so you don't get rid of the common scalar.  No, you have the omega. The h one of omega is canonically trivial. Okay. So you have  this one and then from the curvature then you see that the volume is sort of, I mean  it's a log volume is proportional to minus log of norm of the theta function. So on  the Jacobian of degree g minus one you have the canonical theta function and you can normalize  it for example that the square integral is one or whatever. And then so it's up to a  constant plus a constant. And this constant I call so define minus delta over eight. 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. 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, so the norm of theta  as determined up to a factor and you can normalize it such that the square integral is one and  then you get a value for this one. Square integral of the section, of which section?  Of theta, I mean of the norm of theta. If you multiply the norm by the scalar and the  square integral is also multiplied and then if you normalize it.  The norm of O of theta?  Yeah, O of theta.  Okay, so you have this one and then you can prove for allosomatic surfaces. And the measure  on Jg minus one is normalized with the volume is one for example, yeah. For an osmotic surface  you get Riemann Rochseorin and you get a sort of delta formula. That's the arithmetic  and the denouce is sort of omega squared plus delta over 12. That's say for semi-stable.  Where omega is sort of the relative differential, so the dualizing sheaf is the admissible  metric and you have this omega squared by the R-Kilov 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 I mean plus infinite and the delta v is the number of singularities  over v. So if you have a semi-stable curve, so engineeringally 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 could count the number of geometric points or the number of points.  Well okay, you would have to look at the extension field.  And when you have real and complex places that influence the formula that is you take  the sum of embeddings and say that it's twice. And then you get the sum of these over  these infinite places.  Sorry is the delta the same in both places?  Which delta?  You have delta on the curve.  No no, delta is in the variant of Riemann surface and at all infinite places you have  different Riemann surfaces in the Riemann variant delta and the sum occurs here.  But the complex conjugate is the same delta I think.  And then the question was, well 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 on, I take  the modular stack and well.  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 log of the norm of one or  something.  So this is the problem I wanted to solve.  And for this one I study degenerations.  So this means if I have Riemann surface and if it approaches the boundary I want to look  what happens.  So study.  So if I have a family of Riemann surfaces, 0, my S.  I think G can be 1 but it shouldn't be 0.  And if the K is 0 do you get anything, does what you wrote make sense for G for 0 or not?  No it doesn't make because you have to divide by G for the Achille of measure.  I think it seems related to the fact that you have minimal models in Venus, bigger equal  1.  Okay, and so you have a pi by C0 and you want to study the Arachyl of 1.  So you want to study asymptotic behavior of Arachyl of data.  And I should say if I have a curve C0 I get a graph gamma which corresponds vertices of  E, so these are the reducible components of C0 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 so what I want to say this here.  And then sort of each surface small s, so each fiber is a union of maybe C0V, union C0S,  C0E where these are sort of curves with this removed and these are sort of annuli.  So I do a picture so you have several curves and they are sort of they are annuli.  And so the degeneration happens if these annuli are squeezed.  And so to do the asymptotic and variant I consider them either sort of on these components  or on the annuli which means I sort of write something distinguish between them and the  annuli have equation u times v equal te.  Let's say and u v less equal 1 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.  And then I look at the GPQs, sorry this is the picture I will need later.  And I look at the GPQs that 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 a bounded uniformly NS bounded function.  And then this means sort of up to a function, it means I sort of need the reference model  for these GPQs which is sort of standard, I mean well if p lies here and q lies there  then I want that sort of a local equation times something and so on.  So you allow also a component which intersects itself, I suppose in the special fiber.  Yeah but then sort of it has an edge here.  Okay.  Up to bounded function and the result will be, but they are sort of constant, I mean  depending on s, so the result would be that they are sort of constant, so constant depending  on s, s and v on the curve, 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 maybe I write linear  interpolation by which I mean sort of it's a linear function in log u, log v, so in this  coordinates, so they have one 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 means when the two points line the same c0v or one point  in one and the other in another.  No, no, I mean while it's, while it's both is true, I mean.  But constant up to bounded function is the same as zero, so this is what you mean that  is it's uniformly bounded in your sense?  No, no, no, but the constant depends on the parameter s and the bounded function is bounded  uniformly in s.  Okay, constant namely function of s and v.  Yeah, but this why I wrote depending on s and v.  Okay, okay.  And then when you have points in one in c0v and other in c0w then it depends on s, v and  w.  Yes.  Okay.  So one exception if you have the same annulus, so if you have the annulus cos, well if you  have the same one then it's not violin, yeah, but there's a correction term and so correction  term.  So I write this thing also maybe usually it's about the minimum of p times one minus q and  q times one minus p.  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 on the product  of two annulis this is not a Cartier device and you have to do a blow up and this way  you get this one.  What is the p and q?  P and q are the coordinates on the, what sort of these log u which I did write, I did write  this linear interpolation.  So the linear interpolation means that let us say one point is in the annulus and one  point is in the interior and then you get a u.  No, no, I mean 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, yeah but when you say the linear interpolation before the exception stuff,  so what does it mean?  It means we are referring to the case where one point is in 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 back above this.  Yes.  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 if two are in the same annulus except that I have this correction.  Are you going to interpolate between the two edges of the annulus?  And so if you have two points in different annulus I also do this.  And so there is this one correction term.  This implies that we give an annulus between let's say kappa, kappa, lp and lp.  There are four numbers originally in this, this, this, this, this, this, this.  So when you want to incorporate linearly.  Yes, then you want a bilinear function and this you have to know on the four ends, on  the four corners.  I mean also with x, y, okay.  Okay, 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.  By the graph.  So they can be computed from the graph G. And okay, so maybe how do you do this computation?  So how do you compute the arcale of these?  This can be done as follows.  If you have a Riemann surface and you have two points, you have a sort of harmonic function.  Hpq of z, which is sort of poles.  So we have sort of poles with residues plus minus one.  This means it's sort of, these points, if I do a local coordinate, it's asymptotically  like plus or minus like z. And this is unique up to a constant.  And how do you get this?  With residues plus minus one, this you know, and then it's unique up to adding holomorphic  differentials and you can acquire 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 Hpq, you take sort of the infinite, indefinite  integral of alpha pq.  So this is unique up to a constant and because all the, sort of the real part of this, all  the real part is independent of the parts because your residues are there.  And then you have this gpq, maybe our gp of z.  I'm sorry.  So you take the integral, maybe, so a z of q, what?  And you get this one, you take the integral of Hpq of z times the mu of p plus a constant.  And the constant is sort of such that the mu integral is zero, so integral over mu.  So, yeah, that's the same.  Okay.  I guess the letters are not the different.  So the g, z, so q is fixed.  Yeah.  Yeah, I integrate over one variable and.  So what does it, z is a variable?  How do you enter?  Ah, z is a variable.  So q depends on three variables on p and q and on the variable z.  Okay.  And you take an Hpq.  So the Hpq 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 a right curvature for the gpq.  And then you have to add a constant to make the mu integral equal to zero.  And mu integral 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.  For a fixed z.  Yeah.  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 R, well, how I do this on a degenerate curve.  So first of all, I have to know something about the R-Kilov measure.  So on C0, 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 Cv omega Cv.  Then omega, well.  And then I have the residues.  I have sort of the, see, to the edges.  I have the residues at these.  And now, anyway, more rigorously, the notion of a graph is like in Serres's book, that  is, you have the set of edges is really some, you have the set with an involution, so it  is equivalent because you need the retains.  Yes.  And you have the, 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.  Yeah.  In some way.  I don't know how it is in Serres's book, but.  Okay, yes, somewhere I said, in the evolution of the direction of the.  So talking about the graph, I forgot once.  The notion of what the graph is because this, if you look at different books, what should  they say is a graph slightly different depending on what you read.  So.  Yeah.  Yeah.  Unoriented multigraph.  What?  Unoriented multigraph is, I think, the nicest thing to do.  But I say it's oriented.  And it's no multigraph.  It's a single graph.  Well, 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-carnal.  And usually I call this differential d.  And well, and the residues of a differential are lie in x to the r.  So this maps over xc.  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 that it's a boundary.  So the boundary map sends h to the two endpoints, the difference of the two endpoints.  And because I had the edge oriented, well, you can figure out whether you want the beginning  minus the end or the end minus the beginning.  OK, so I have this one.  And so the differential is from a nice family.  Well, and they are similar on the singular fibers.  You have this residue map.  So it generates on each.  So there are the residue maps.  And then you get the base of the differential.  You get the differential on the curves.  And you get the differential which have non-trivial residues.  So then you have a co-carnals 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 a dozen of poles, then it's about the square  integral on the component.  But if it has poles, then the square integral is sort of concentrated on the annular.  And so it's about the sum of Se times residue E squared.  So if you integrate over an annulus, you get some more of this one.  If you look at the residue.  So by the way, the residue is defined also on the other fibers by sort of integrating  once around the loop here.  Okay?  An integral of what?  Of a one form.  It 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 Te to one.  You get what is it, dx over x and the sum of log Te, dx over x.  And this gives you 2 pi minus 2 pi times log Te.  And the other terms, so in the differential, the u over u term is the one which dominates  the integral.  And so here you are doing the integral kind of in the family of the mobilized family.  Yes, yes.  You choose a lifting of something in the special section of the conund of the dualizing shape  on the special fiber you extended 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, 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 the  basis of the differential on each curve.  And it has a residue map 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 SE are big and 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.  Yes.  You want to see to the… you want to look… what do you say about the x?  You say that the…  The x is the space, is the space of loops in the graph.  Yeah, but what is the relation to the…  X lies in C to the E because the kernel lies in this one.  Are you claiming that this is the image of the residue map?  What?  Yeah, it's the image of the residue map of this x.  Okay, 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 I get an orthonormal basis.  I take an orthonormal basis for x with norm xE, so xE squared, y.  This one, and I take an orthonormal basis.  And then I sort of get an orthonormal basis of the whole space by combining them.  And while I can sort of lift them, I mean if the xE becomes big, then the xE 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 an orthonormal basis and then they are kilo.  Okay, so this is one thing.  Then I have to define this hpq.  So I have to find…  What is the last comment plus our kilo?  I get the our kilo of measure if I know an orthonormal basis.  So here it's an orthonormal basis for the space of the differentials.  What do you mean you get the…  The alkyl of mu is on the fiber CS.  So the alkyl of mu has 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 alkyl of measure of these curves, except I have  to multiply with the genus divided by the total genus because for the alkyl of measure  I have this one over 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 the genus zero… well, on the genus zero components the gv is zero so the alkyl  of 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.  But anyway sometimes when they do the generation of curves they assume like the stability condition,  the rational curve is at least three…  Yes.  Do you assume something like this or does it… does things break down?  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.  Okay, 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 to condition that periods are purely imaginary.  So then how to get purely imaginary…  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 anuli, 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 correspond  to loops in gamma.  I think I called this gamma now, no G. And sort of I mean and the integrals over this  one are the residues so I want the residues should be real.  So then you have these interior loops and this by you can see that the anuli is the  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 at alpha cv and then they are sort of the most interesting ones are  the loops in x, the loops in x while in the graph gamma which correspond to x.  So I said I have differentiate this given residue and I can lift them uniquely that  the interior periods are purely imaginary.  So I lift x to differentiate with purely imaginary maybe a right interior loop.  And then well and sort of and I also should… so this is sort of the residue should be in  the real base 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 the loop is sort of a linear form on the xe, so integral around  the xe is a linear form on x and it differs from the sum r.  So differs from sum se times xe times ye.  So I have a linear form on x by bounded function by bounded linear form.  You have a linear 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 linear  form.  And the integral is sort of dominated by the integral around the interior of the loops  which sort of give you these terms and then there is sort of the integral which sort of  this part which goes from one end from the loop to the other but this is bounded.  So this is a small function.  And then you sort of want to want this form to be zero.  So we have a linear form on x and well and you have this non-degenerate inner 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 over which is it xc  or xr or?  No I mean I have only real parts because I want 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 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 on x and this linear form comes from x.  x itself gives linear forms which are about this one plus a bounded function and so this  means I can correct it by sort of taking the orthogonal projection from a linear form.  Well I mean this linear form is determined by inner product with an element of x and  then the correction is this one plus some bounded term uniformly in x.  So this gives me the…  So this gives me the…  Five more minutes.  What?  Five more minutes.  Yeah sorry.  I don't know whether answering question is added like in soccer.  So this way you can construct your differential HPQ and sort of and these orthogonal projections  are given by the graph and then you can compute the GPQ 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 of rational functions are sort of homogeneous rational function in the homogeneous  of degree 1, 1 in the SE.  So this means to the values of these g's on the component also to the delta function  but unfortunately they are not linear forms so unfortunately…  So delta is not a linear combination of the SE and so this means there is no metric on  the or of the boundary of the modular space which gives this delta function.  So it doesn't…  At least you can compute things here.  I should say I have a student Robert Wirmes who was supposed to do this but he did something  else which happens with 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 he has questions.  In this case we have what you call the linear interpolation.  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… is there some intuition behind that?  They seem like even on the limit and as far apart.  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 is sort  of… I had this d to the v and a d and I have an adjoint and I can also scale the inner  product well it is natural inner product so scale product ys and then you sort of get  a Laplace operator and there is a similar thing sort of if you have a… now you have  a sort of a scale of measure here so you have sort of you put v over g 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.  You want to solve the equation delta of g is sort of a point minus p… my mu minus  p I should say.  So the mu is total measure 1 and if you have a point also then you can… then this is  total measure 0 so it's the image of the Laplace operator and this gives you a Green's  function 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 at 5 to 12 but let's thank Paul Tins again.
G__Faltings_-_Arakelov_geometry_on_degenerating_curves_J7qFP-cwhQY_0_383.png G__Faltings_-_Arakelov_geometry_on_degenerating_curves_J7qFP-cwhQY_384_971.png G__Faltings_-_Arakelov_geometry_on_degenerating_curves_J7qFP-cwhQY_972_3575.png G__Faltings_-_Arakelov_geometry_on_degenerating_curves_J7qFP-cwhQY_3576_9635.png G__Faltings_-_Arakelov_geometry_on_degenerating_curves_J7qFP-cwhQY_9636_10655.png G__Faltings_-_Arakelov_geometry_on_degenerating_curves_J7qFP-cwhQY_10656_10715.png G__Faltings_-_Arakelov_geometry_on_degenerating_curves_J7qFP-cwhQY_10716_13415.png G__Faltings_-_Arakelov_geometry_on_degenerating_curves_J7qFP-cwhQY_13416_15155.png G__Faltings_-_Arakelov_geometry_on_degenerating_curves_J7qFP-cwhQY_15156_16847.png G__Faltings_-_Arakelov_geometry_on_degenerating_curves_J7qFP-cwhQY_16848_21695.png G__Faltings_-_Arakelov_geometry_on_degenerating_curves_J7qFP-cwhQY_21696_23027.png G__Faltings_-_Arakelov_geometry_on_degenerating_curves_J7qFP-cwhQY_23028_23207.png G__Faltings_-_Arakelov_geometry_on_degenerating_curves_J7qFP-cwhQY_23208_23711.png G__Faltings_-_Arakelov_geometry_on_degenerating_curves_J7qFP-cwhQY_23712_24443.png G__Faltings_-_Arakelov_geometry_on_degenerating_curves_J7qFP-cwhQY_24444_25427.png G__Faltings_-_Arakelov_geometry_on_degenerating_curves_J7qFP-cwhQY_25428_26327.png G__Faltings_-_Arakelov_geometry_on_degenerating_curves_J7qFP-cwhQY_26328_27947.png G__Faltings_-_Arakelov_geometry_on_degenerating_curves_J7qFP-cwhQY_27948_29879.png G__Faltings_-_Arakelov_geometry_on_degenerating_curves_J7qFP-cwhQY_29880_31499.png G__Faltings_-_Arakelov_geometry_on_degenerating_curves_J7qFP-cwhQY_31500_31655.png G__Faltings_-_Arakelov_geometry_on_degenerating_curves_J7qFP-cwhQY_31656_31751.png G__Faltings_-_Arakelov_geometry_on_degenerating_curves_J7qFP-cwhQY_31752_31931.png G__Faltings_-_Arakelov_geometry_on_degenerating_curves_J7qFP-cwhQY_31932_32231.png G__Faltings_-_Arakelov_geometry_on_degenerating_curves_J7qFP-cwhQY_32232_34415.png G__Faltings_-_Arakelov_geometry_on_degenerating_curves_J7qFP-cwhQY_34416_36359.png G__Faltings_-_Arakelov_geometry_on_degenerating_curves_J7qFP-cwhQY_36360_37607.png G__Faltings_-_Arakelov_geometry_on_degenerating_curves_J7qFP-cwhQY_37608_38327.png G__Faltings_-_Arakelov_geometry_on_degenerating_curves_J7qFP-cwhQY_38328_38639.png G__Faltings_-_Arakelov_geometry_on_degenerating_curves_J7qFP-cwhQY_38640_39227.png G__Faltings_-_Arakelov_geometry_on_degenerating_curves_J7qFP-cwhQY_39228_40127.png G__Faltings_-_Arakelov_geometry_on_degenerating_curves_J7qFP-cwhQY_40128_40727.png G__Faltings_-_Arakelov_geometry_on_degenerating_curves_J7qFP-cwhQY_40728_41639.png G__Faltings_-_Arakelov_geometry_on_degenerating_curves_J7qFP-cwhQY_41640_42275.png G__Faltings_-_Arakelov_geometry_on_degenerating_curves_J7qFP-cwhQY_42276_43091.png G__Faltings_-_Arakelov_geometry_on_degenerating_curves_J7qFP-cwhQY_43092_43487.png G__Faltings_-_Arakelov_geometry_on_degenerating_curves_J7qFP-cwhQY_43488_43595.png G__Faltings_-_Arakelov_geometry_on_degenerating_curves_J7qFP-cwhQY_43596_44879.png G__Faltings_-_Arakelov_geometry_on_degenerating_curves_J7qFP-cwhQY_44880_47507.png G__Faltings_-_Arakelov_geometry_on_degenerating_curves_J7qFP-cwhQY_47508_49943.png G__Faltings_-_Arakelov_geometry_on_degenerating_curves_J7qFP-cwhQY_49944_50435.png G__Faltings_-_Arakelov_geometry_on_degenerating_curves_J7qFP-cwhQY_50436_52211.png G__Faltings_-_Arakelov_geometry_on_degenerating_curves_J7qFP-cwhQY_52212_53567.png G__Faltings_-_Arakelov_geometry_on_degenerating_curves_J7qFP-cwhQY_53568_53735.png G__Faltings_-_Arakelov_geometry_on_degenerating_curves_J7qFP-cwhQY_53736_56027.png G__Faltings_-_Arakelov_geometry_on_degenerating_curves_J7qFP-cwhQY_56028_57251.png G__Faltings_-_Arakelov_geometry_on_degenerating_curves_J7qFP-cwhQY_57252_57947.png G__Faltings_-_Arakelov_geometry_on_degenerating_curves_J7qFP-cwhQY_57948_58199.png G__Faltings_-_Arakelov_geometry_on_degenerating_curves_J7qFP-cwhQY_58200_58391.png G__Faltings_-_Arakelov_geometry_on_degenerating_curves_J7qFP-cwhQY_58392_62495.png G__Faltings_-_Arakelov_geometry_on_degenerating_curves_J7qFP-cwhQY_62496_63239.png G__Faltings_-_Arakelov_geometry_on_degenerating_curves_J7qFP-cwhQY_63240_67919.png G__Faltings_-_Arakelov_geometry_on_degenerating_curves_J7qFP-cwhQY_67920_71291.png G__Faltings_-_Arakelov_geometry_on_degenerating_curves_J7qFP-cwhQY_71292_71723.png G__Faltings_-_Arakelov_geometry_on_degenerating_curves_J7qFP-cwhQY_71724_73499.png G__Faltings_-_Arakelov_geometry_on_degenerating_curves_J7qFP-cwhQY_73500_73787.png G__Faltings_-_Arakelov_geometry_on_degenerating_curves_J7qFP-cwhQY_73788_75287.png G__Faltings_-_Arakelov_geometry_on_degenerating_curves_J7qFP-cwhQY_75288_75779.png G__Faltings_-_Arakelov_geometry_on_degenerating_curves_J7qFP-cwhQY_75780_75839.png G__Faltings_-_Arakelov_geometry_on_degenerating_curves_J7qFP-cwhQY_75840_79235.png G__Faltings_-_Arakelov_geometry_on_degenerating_curves_J7qFP-cwhQY_79236_80423.png G__Faltings_-_Arakelov_geometry_on_degenerating_curves_J7qFP-cwhQY_80424_82043.png G__Faltings_-_Arakelov_geometry_on_degenerating_curves_J7qFP-cwhQY_82044_83279.png G__Faltings_-_Arakelov_geometry_on_degenerating_curves_J7qFP-cwhQY_83280_83615.png G__Faltings_-_Arakelov_geometry_on_degenerating_curves_J7qFP-cwhQY_83616_85727.png G__Faltings_-_Arakelov_geometry_on_degenerating_curves_J7qFP-cwhQY_85728_86567.png G__Faltings_-_Arakelov_geometry_on_degenerating_curves_J7qFP-cwhQY_86568_87563.png G__Faltings_-_Arakelov_geometry_on_degenerating_curves_J7qFP-cwhQY_87564_87743.png G__Faltings_-_Arakelov_geometry_on_degenerating_curves_J7qFP-cwhQY_87744_89411.png G__Faltings_-_Arakelov_geometry_on_degenerating_curves_J7qFP-cwhQY_89412_89771.png G__Faltings_-_Arakelov_geometry_on_degenerating_curves_J7qFP-cwhQY_89772_90695.png G__Faltings_-_Arakelov_geometry_on_degenerating_curves_J7qFP-cwhQY_90696_91775.png G__Faltings_-_Arakelov_geometry_on_degenerating_curves_J7qFP-cwhQY_91776_92219.png G__Faltings_-_Arakelov_geometry_on_degenerating_curves_J7qFP-cwhQY_92220_92471.png G__Faltings_-_Arakelov_geometry_on_degenerating_curves_J7qFP-cwhQY_92472_94163.png G__Faltings_-_Arakelov_geometry_on_degenerating_curves_J7qFP-cwhQY_94164_94319.png G__Faltings_-_Arakelov_geometry_on_degenerating_curves_J7qFP-cwhQY_94320_95519.png G__Faltings_-_Arakelov_geometry_on_degenerating_curves_J7qFP-cwhQY_95520_97487.png G__Faltings_-_Arakelov_geometry_on_degenerating_curves_J7qFP-cwhQY_97488_97907.png G__Faltings_-_Arakelov_geometry_on_degenerating_curves_J7qFP-cwhQY_97908_98471.png G__Faltings_-_Arakelov_geometry_on_degenerating_curves_J7qFP-cwhQY_98472_98615.png G__Faltings_-_Arakelov_geometry_on_degenerating_curves_J7qFP-cwhQY_98616_99023.png G__Faltings_-_Arakelov_geometry_on_degenerating_curves_J7qFP-cwhQY_99024_101723.png G__Faltings_-_Arakelov_geometry_on_degenerating_curves_J7qFP-cwhQY_101724_103763.png G__Faltings_-_Arakelov_geometry_on_degenerating_curves_J7qFP-cwhQY_103764_104507.png G__Faltings_-_Arakelov_geometry_on_degenerating_curves_J7qFP-cwhQY_104508_107481.png
Thank you. So I'm supposed to say something about Gabba. So he likes to do technical things and he even liked 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, 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 they should have big intersection. And of course, so GPQ is sort of norm 1, PQ or metric. And delta, delta C is a diagonal. Okay, 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 Kieloff said if we choose a measure mu, mu is 1-1 form, and the integral, well, now C has become a Riemann surface, is 1. Then we look at Hermitian line bundles. So Hermitian. And the moment it doesn't have to be, with curvature is degree L times mu. So we know that the curvature integral gives the degree. So if it's a multiple of mu, it has to be this multiple. And we look at the line bundles which we call admissible metrics. And then we know that such a metric is unique up to scaling. Okay, 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 they are the unique metric up to scaling. And so we have a metric on O of delta with curvature. Maybe I'll mu tensor 1, just 1 tensor minus some alpha j1. And then if I change these on the diagonal alpha... I hope I got this right. There's a alpha j on an orthonormal basis. No, sorry, this has to be basis of... This is true because sort of this is the harmonic representatives for the diagonal. And if we have such a metric, we get a metric on the differential. Maybe I'll 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 for the curvature is twice minus 2 mu. And if we want this to be admissible, I think I should put the 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 times some alpha j. And this is called the Aracheloff measure. And if I do this, one can do some things. For example, one can define volume forms on a gamma cl. If l has an admissible matrix, which sort of satisfies the... Obvious scaling condition 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 that are omega and the differential. It's given by the square integral. So normalize. Square integral. And the quenches. And then one knows that on l of degree g minus 1, so l has degree g minus 1. Then this is independent of scaling because we get the power of the Euler-Ponkery 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. And the Jacobian. What did you say about independent scaling? 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, what do you use the correction by the analytic torsion or what not? No, 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, okay l of. If you have an admissible line bundle at 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. And this. Will you need a metric on l minus phi? Well, I require on the determinant. No. Well first of all, well this also has a metric because l of minus phi is a metric given by the. By Akeler. By Akeler. And then you want this to be compatible and this gives you essentially a metric on the all of these. The 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 one. And for this one, well I mean there you have no scaling problem. 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? I mean if you have all of the divisors and 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 under the term and once you, but this for all line boundaries is an admissible metric. And so there is a unique choice on the category of those things up to a scalar. Yes. A common scalar. And so you don't get rid of the common scalar. No, you have the omega. The h one of omega is canonically trivial. Okay. So you have this one and then from the curvature then you see that the volume is sort of, I mean it's a log volume is proportional to minus log of norm of the theta function. So on the Jacobian of degree g minus one you have the canonical theta function and you can normalize it for example that the square integral is one or whatever. And then so it's up to a constant plus a constant. And this constant I call so define minus delta over eight. 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. 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, so the norm of theta as determined up to a factor and you can normalize it such that the square integral is one and then you get a value for this one. Square integral of the section, of which section? Of theta, I mean of the norm of theta. If you multiply the norm by the scalar and the square integral is also multiplied and then if you normalize it. The norm of O of theta? Yeah, O of theta. Okay, so you have this one and then you can prove for allosomatic surfaces. And the measure on Jg minus one is normalized with the volume is one for example, yeah. For an osmotic surface you get Riemann Rochseorin and you get a sort of delta formula. That's the arithmetic and the denouce is sort of omega squared plus delta over 12. That's say for semi-stable. Where omega is sort of the relative differential, so the dualizing sheaf is the admissible metric and you have this omega squared by the R-Kilov 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 I mean plus infinite and the delta v is the number of singularities over v. So if you have a semi-stable curve, so engineeringally 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 could count the number of geometric points or the number of points. Well okay, you would have to look at the extension field. And when you have real and complex places that influence the formula that is you take the sum of embeddings and say that it's twice. And then you get the sum of these over these infinite places. Sorry is the delta the same in both places? Which delta? You have delta on the curve. No no, delta is in the variant of Riemann surface and at all infinite places you have different Riemann surfaces in the Riemann variant delta and the sum occurs here. But the complex conjugate is the same delta I think. And then the question was, well 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 on, I take the modular stack and well. 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 log of the norm of one or something. So this is the problem I wanted to solve. And for this one I study degenerations. So this means if I have Riemann surface and if it approaches the boundary I want to look what happens. So study. So if I have a family of Riemann surfaces, 0, my S. I think G can be 1 but it shouldn't be 0. And if the K is 0 do you get anything, does what you wrote make sense for G for 0 or not? No it doesn't make because you have to divide by G for the Achille of measure. I think it seems related to the fact that you have minimal models in Venus, bigger equal 1. Okay, and so you have a pi by C0 and you want to study the Arachyl of 1. So you want to study asymptotic behavior of Arachyl of data. And I should say if I have a curve C0 I get a graph gamma which corresponds vertices of E, so these are the reducible components of C0 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 so what I want to say this here. And then sort of each surface small s, so each fiber is a union of maybe C0V, union C0S, C0E where these are sort of curves with this removed and these are sort of annuli. So I do a picture so you have several curves and they are sort of they are annuli. And so the degeneration happens if these annuli are squeezed. And so to do the asymptotic and variant I consider them either sort of on these components or on the annuli which means I sort of write something distinguish between them and the annuli have equation u times v equal te. Let's say and u v less equal 1 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. And then I look at the GPQs, sorry this is the picture I will need later. And I look at the GPQs that 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 a bounded uniformly NS bounded function. And then this means sort of up to a function, it means I sort of need the reference model for these GPQs which is sort of standard, I mean well if p lies here and q lies there then I want that sort of a local equation times something and so on. So you allow also a component which intersects itself, I suppose in the special fiber. Yeah but then sort of it has an edge here. Okay. Up to bounded function and the result will be, but they are sort of constant, I mean depending on s, so the result would be that they are sort of constant, so constant depending on s, s and v on the curve, 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 maybe I write linear interpolation by which I mean sort of it's a linear function in log u, log v, so in this coordinates, so they have one 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 means when the two points line the same c0v or one point in one and the other in another. No, no, I mean while it's, while it's both is true, I mean. But constant up to bounded function is the same as zero, so this is what you mean that is it's uniformly bounded in your sense? No, no, no, but the constant depends on the parameter s and the bounded function is bounded uniformly in s. Okay, constant namely function of s and v. Yeah, but this why I wrote depending on s and v. Okay, okay. And then when you have points in one in c0v and other in c0w then it depends on s, v and w. Yes. Okay. So one exception if you have the same annulus, so if you have the annulus cos, well if you have the same one then it's not violin, yeah, but there's a correction term and so correction term. So I write this thing also maybe usually it's about the minimum of p times one minus q and q times one minus p. 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 on the product of two annulis this is not a Cartier device and you have to do a blow up and this way you get this one. What is the p and q? P and q are the coordinates on the, what sort of these log u which I did write, I did write this linear interpolation. So the linear interpolation means that let us say one point is in the annulus and one point is in the interior and then you get a u. No, no, I mean 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, yeah but when you say the linear interpolation before the exception stuff, so what does it mean? It means we are referring to the case where one point is in 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 back above this. Yes. 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 if two are in the same annulus except that I have this correction. Are you going to interpolate between the two edges of the annulus? And so if you have two points in different annulus I also do this. And so there is this one correction term. This implies that we give an annulus between let's say kappa, kappa, lp and lp. There are four numbers originally in this, this, this, this, this, this, this. So when you want to incorporate linearly. Yes, then you want a bilinear function and this you have to know on the four ends, on the four corners. I mean also with x, y, okay. Okay, 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. By the graph. So they can be computed from the graph G. And okay, so maybe how do you do this computation? So how do you compute the arcale of these? This can be done as follows. If you have a Riemann surface and you have two points, you have a sort of harmonic function. Hpq of z, which is sort of poles. So we have sort of poles with residues plus minus one. This means it's sort of, these points, if I do a local coordinate, it's asymptotically like plus or minus like z. And this is unique up to a constant. And how do you get this? With residues plus minus one, this you know, and then it's unique up to adding holomorphic differentials and you can acquire 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 Hpq, you take sort of the infinite, indefinite integral of alpha pq. So this is unique up to a constant and because all the, sort of the real part of this, all the real part is independent of the parts because your residues are there. And then you have this gpq, maybe our gp of z. I'm sorry. So you take the integral, maybe, so a z of q, what? And you get this one, you take the integral of Hpq of z times the mu of p plus a constant. And the constant is sort of such that the mu integral is zero, so integral over mu. So, yeah, that's the same. Okay. I guess the letters are not the different. So the g, z, so q is fixed. Yeah. Yeah, I integrate over one variable and. So what does it, z is a variable? How do you enter? Ah, z is a variable. So q depends on three variables on p and q and on the variable z. Okay. And you take an Hpq. So the Hpq 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 a right curvature for the gpq. And then you have to add a constant to make the mu integral equal to zero. And mu integral 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. For a fixed z. Yeah. 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 R, well, how I do this on a degenerate curve. So first of all, I have to know something about the R-Kilov measure. So on C0, 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 Cv omega Cv. Then omega, well. And then I have the residues. I have sort of the, see, to the edges. I have the residues at these. And now, anyway, more rigorously, the notion of a graph is like in Serres's book, that is, you have the set of edges is really some, you have the set with an involution, so it is equivalent because you need the retains. Yes. And you have the, 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. Yeah. In some way. I don't know how it is in Serres's book, but. Okay, yes, somewhere I said, in the evolution of the direction of the. So talking about the graph, I forgot once. The notion of what the graph is because this, if you look at different books, what should they say is a graph slightly different depending on what you read. So. Yeah. Yeah. Unoriented multigraph. What? Unoriented multigraph is, I think, the nicest thing to do. But I say it's oriented. And it's no multigraph. It's a single graph. Well, 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-carnal. And usually I call this differential d. And well, and the residues of a differential are lie in x to the r. So this maps over xc. 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 that it's a boundary. So the boundary map sends h to the two endpoints, the difference of the two endpoints. And because I had the edge oriented, well, you can figure out whether you want the beginning minus the end or the end minus the beginning. OK, so I have this one. And so the differential is from a nice family. Well, and they are similar on the singular fibers. You have this residue map. So it generates on each. So there are the residue maps. And then you get the base of the differential. You get the differential on the curves. And you get the differential which have non-trivial residues. So then you have a co-carnals 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 a dozen of poles, then it's about the square integral on the component. But if it has poles, then the square integral is sort of concentrated on the annular. And so it's about the sum of Se times residue E squared. So if you integrate over an annulus, you get some more of this one. If you look at the residue. So by the way, the residue is defined also on the other fibers by sort of integrating once around the loop here. Okay? An integral of what? Of a one form. It 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 Te to one. You get what is it, dx over x and the sum of log Te, dx over x. And this gives you 2 pi minus 2 pi times log Te. And the other terms, so in the differential, the u over u term is the one which dominates the integral. And so here you are doing the integral kind of in the family of the mobilized family. Yes, yes. You choose a lifting of something in the special section of the conund of the dualizing shape on the special fiber you extended 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, 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 the basis of the differential on each curve. And it has a residue map 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 SE are big and 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. Yes. You want to see to the… you want to look… what do you say about the x? You say that the… The x is the space, is the space of loops in the graph. Yeah, but what is the relation to the… X lies in C to the E because the kernel lies in this one. Are you claiming that this is the image of the residue map? What? Yeah, it's the image of the residue map of this x. Okay, 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 I get an orthonormal basis. I take an orthonormal basis for x with norm xE, so xE squared, y. This one, and I take an orthonormal basis. And then I sort of get an orthonormal basis of the whole space by combining them. And while I can sort of lift them, I mean if the xE becomes big, then the xE 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 an orthonormal basis and then they are kilo. Okay, so this is one thing. Then I have to define this hpq. So I have to find… What is the last comment plus our kilo? I get the our kilo of measure if I know an orthonormal basis. So here it's an orthonormal basis for the space of the differentials. What do you mean you get the… The alkyl of mu is on the fiber CS. So the alkyl of mu has 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 alkyl of measure of these curves, except I have to multiply with the genus divided by the total genus because for the alkyl of measure I have this one over 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 the genus zero… well, on the genus zero components the gv is zero so the alkyl of 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. But anyway sometimes when they do the generation of curves they assume like the stability condition, the rational curve is at least three… Yes. Do you assume something like this or does it… does things break down? 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. Okay, 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 to condition that periods are purely imaginary. So then how to get purely imaginary… 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 anuli, 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 correspond to loops in gamma. I think I called this gamma now, no G. And sort of I mean and the integrals over this one are the residues so I want the residues should be real. So then you have these interior loops and this by you can see that the anuli is the 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 at alpha cv and then they are sort of the most interesting ones are the loops in x, the loops in x while in the graph gamma which correspond to x. So I said I have differentiate this given residue and I can lift them uniquely that the interior periods are purely imaginary. So I lift x to differentiate with purely imaginary maybe a right interior loop. And then well and sort of and I also should… so this is sort of the residue should be in the real base 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 the loop is sort of a linear form on the xe, so integral around the xe is a linear form on x and it differs from the sum r. So differs from sum se times xe times ye. So I have a linear form on x by bounded function by bounded linear form. You have a linear 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 linear form. And the integral is sort of dominated by the integral around the interior of the loops which sort of give you these terms and then there is sort of the integral which sort of this part which goes from one end from the loop to the other but this is bounded. So this is a small function. And then you sort of want to want this form to be zero. So we have a linear form on x and well and you have this non-degenerate inner 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 over which is it xc or xr or? No I mean I have only real parts because I want 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 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 on x and this linear form comes from x. x itself gives linear forms which are about this one plus a bounded function and so this means I can correct it by sort of taking the orthogonal projection from a linear form. Well I mean this linear form is determined by inner product with an element of x and then the correction is this one plus some bounded term uniformly in x. So this gives me the… So this gives me the… Five more minutes. What? Five more minutes. Yeah sorry. I don't know whether answering question is added like in soccer. So this way you can construct your differential HPQ and sort of and these orthogonal projections are given by the graph and then you can compute the GPQ 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 of rational functions are sort of homogeneous rational function in the homogeneous of degree 1, 1 in the SE. So this means to the values of these g's on the component also to the delta function but unfortunately they are not linear forms so unfortunately… So delta is not a linear combination of the SE and so this means there is no metric on the or of the boundary of the modular space which gives this delta function. So it doesn't… At least you can compute things here. I should say I have a student Robert Wirmes who was supposed to do this but he did something else which happens with 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 he has questions. In this case we have what you call the linear interpolation. 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… is there some intuition behind that? They seem like even on the limit and as far apart. 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 is sort of… I had this d to the v and a d and I have an adjoint and I can also scale the inner product well it is natural inner product so scale product ys and then you sort of get a Laplace operator and there is a similar thing sort of if you have a… now you have a sort of a scale of measure here so you have sort of you put v over g 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. You want to solve the equation delta of g is sort of a point minus p… my mu minus p I should say. So the mu is total measure 1 and if you have a point also then you can… then this is total measure 0 so it's the image of the Laplace operator and this gives you a Green's function 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 at 5 to 12 but let's thank Paul Tins again.
semi-stable curve, so engineeringally it's smooth Riemann surface and if it approaches the boundary up to a bounded uniformly NS bounded function. function on the graph and the asymptotic of the