 So the story goes back, so here's a little outline of what we'll do. This first of all is a joint work with Jesse Kass, Jake Solomon and Kirsten Vicklegren. Vicklegren, Vellish and Jay, it's really difficult. And so first I'll talk about some of the background over the complex numbers, that's the classical case, and then Vellish and Jay's innovation looking at the real case. And then I'll start, so the motivic part is to trying to get a quadratic version of these invariance of counting curves, accounting rational curves in a linear system. So that involves taking the Kanshevich modulized space of stable maps and symmetrizing it to take care of arithmetic information. And then we'll discuss the important thing is the evaluation map. We'll discuss the geometry of the evaluation map and how that gives us a description of the ramification of the evaluation map. And then we'll use the universal curve and the universal map to construct a relative orientation for this evaluation map. And at the end we'll prove this invariance theorem, which is a generalization of, to refinement the quadratic case of Vellish and Jay's invariant theorem. Okay, so in the classical case it started with the problem of counting rational curves of degree D and P2 in the following sense. So we fix some degree D, let's say bigger than or equal to three to make it a little more interesting. And then we have the linear system of curves of degree D. So L is just the line and the linear system are all the curves in P2 that are linearly equivalent to D times a line. In other words, it's just curves in P2 such that C is a degree D curve. Okay, and by looking at the defining equations of these curves, well there's a one-to-one correspondence between the curves and the defining equations modulo scalars. So that makes, and that's true for any linear system, says that this linear system is a projective space and just how the dimension of curves of degree of equations of degree D, you see that it's exactly this projective space. And also if you take a general, so for example a smooth such curve of degree D has genus G, let's call it G, and it turns out to be, well the junction formula tells you it's D minus 1, D minus 2 over 2. So you look at the difference between these two numbers and that gives you the magic number N, which is 3D minus 1. It's the difference in the dimension of the full linear system, the full projective space of curves of degree D, and the genus of a smooth member. Now if you look at the whole curves moving around, you impose the condition that instead of it being smooth that it has a single ordinary double point. That gives you a locally closed subset, you take the closure, that's a co-dimension one subset, and if you do that for exactly G of them, you get a co-dimension G subset of rational curves. So now on the other hand, if you fix, every time you fix a point and require that the curve pass through a point, that's also a co- dimension one subset. So if you require, put those two counts together, require C in DL to contain the following subset, you take N points, this magic number here, let's say general, then there are, finitely many, let's say, ND curve C in LD, containing those points, so let me just put a little subscript X star, those are the ones that contain all the points, which have normalization, so it'll have singular points to remove the singularities, which is a rational curve, in other words, which is a P1. And moreover the curve, each of these, since the points are general, each such C has exactly G ordinary double points as singularities. So that's the situation, you look at the number of, the difference between the dimension of the whole space and the genus, imposing one ordinary double point is one condition, passing through a point is one condition, you put them together and you see that if you impose the right number of conditions of the curves passing through so and so many points, then finitely many of them will have exactly G ordinary double points, I mean, if the points are general. And if you take special points, then everything goes crazy, but that's the general situation. So the basic problem is the question or problem question, problem, what is this number? And there is an answer, which was, I think originally due to consavage, there were a lot of people before that consavage money came up with an answer. And also, also, Yuan Tuan using some of the different methods, Tian Li, Li Tian, they found an answer. And so it's some formula, a formula, which if pressed, I will give you, but let me just, it's a complicated recursive formula, but that's not what we're going to talk about here, we're not going to find any formulas. So let me just mention what did consavage money do. And so there are a number of ways you can approach this problem. One way is just by looking at a degree of a discriminant locus. And but that's rather difficult. Although it does generalize nicely the higher genus. But what consavage money did due to ideas of consavage, the idea was instead of thinking of it as cutting out the rational curves in a linear system, you instead want to look at all maps of rational curves to the plane, and then cut out those which pass through your end points. It's coming, it's still the same count, but it's counting it in a different way. So what about maps? So what you do consavage money does is you look at the modular space to do some notation M zero N P two D of stable maps D P ones are rational curves of degree D N pointed genus zero curves maps to P two. Okay, so let me say a little bit about that. This is of course representing a functor. So what's the in fact we can do this more generally the nicest that you can do this for a general surface but for us, the general situation is for tell del petso surfaces. So let me say about that S E a del petso surface. So what is that if you take the canonical class? So that's the line bundle, the divisor associated to the canonical line bundle. In other words, the determinant of the differentials take its minus. This is ample. So in fact, that's a that's an abstract definition. But what it means is that S is either equal to P one cross P one, or it's P two blown up at at most eight points. And the eight points have to be sufficiently general. They can't lie on too many curves of small degree. Okay, so one classic example is the cubic surface in P three and cubic. Well, that's interesting. What's the where is the three coming from the cubic surface? You have the degree of the del petso, which is the self intersection number of this ample class. So the cubic surface is embedded in P three by minus its canonical class. And that means when you take minus the canonical class squared, which is of course the same thing as canonical class squared, it gives you the degree of the embedded surface, which is three. So for example, for P two, this number is nine. And it goes since you blown up at most eight points, it goes all the way down to one. The idea is these surfaces become more and more complicated as the degree gets smaller. Okay, so that's one number associated this. And then you also fix a so it's called a curve class. Curve D, an effective curve. So just some curve on the could be a positive sum of sum with positive integer coefficients of irreducible curves on S. And let's let D equal the degree with respect to the canonical embedding. So it's just this intersection number. And then we have this magic number and will be minus this, this degree. So D minus one. And for example, if S is equal to P two, then minus K s is three times a line. So that DS is nine, as I said. And if we take D to be D times L, then well, then what's this end this end is exactly this number, 3d minus one, which came up before. So it turns out this is the magic number. So what that says is in general, if you take n general points, x1 up to xn on S, then the set of C in the linear system of D, in other words, curves which are in the same class, if you in this case, it's the same homology class or the same linear equivalence class, it doesn't matter as D such that C contains all these end points. You look inside there and C is rational, the normalization is rational, is a finite set. And what you want to do is with, you know, so let's say the order of this set, let's call equal to NDS. So this little D last time was the case where S was the P two. So this was the example, but this was the year. This is the general case. This is general. All right. So, and that's the problem. So how does this have to do with maps? So you take, now you let m bar zero n SD, this is equal to the module I, space or stack will be, it'll actually be a scheme when we throw away enough things which we don't care about, stack, whatever, of stable maps. So what is this? So let's see. It's the set maps F from some P. I'll say what the P is in the second to S together with N points, where here P is a sort of semi, it's a semi stable genus zero curve. So it's a tree of P ones, P ones where they're joined together. There's no three of them through a point. So something like this could have several connecting one component. So that's a tree of P ones. The Xi are in P, but in the smooth locus, they don't wear two curves across the Xi's are not allowed to be there. And F is a stable map. So well, F together with the Xi's are stable. So what does that mean? It means if you consider the F as a map and you look at automorphisms that F that fix the Xi's are only finitely many. That's the stability condition. And let's see. What's the D? If you take the push forward of the fundamental class of this curve, this is in the linear system D. So it's homologous to D. Okay, so that's a, that's what a point of this thing is. And in general, you have to say what families are, but let's not do that. Good. So that's that. And what does that have to do with our question? Well, we have this evaluation map from this. So the main point of this, this is, this is a proper gadget. It's compact. That's one important fact about this. So we have this map, evaluation map going to S to the N. Where what does it do if you have your F together with your points on your curve? F is a map from P to S. You send it to the tuple of points, F of X1 up to F of X. Oh, I should have said, of course, the Xi's are in the smooth locus and they're distinct. Distinct. Sorry. Okay. So I hope that's clear. There's a question. Can you see it or I can read it? When you write a tree of P ones, it looks like you're working over an algebraically close field. Yes. So I'm talking these are the geometric points. Exactly right. So yeah, so this is what a point of this thing looks like over an algebraically close field. But I'm not assuming the base field is algebraically close. That's exactly right. Thanks for the question. Hope that clarified. Okay. So but I am going to assume for the talk today, assume that the characteristic of K is zero. I'll say a word at the end about what to do in positive characteristics. Okay, so we have this evaluation map. That's just a little side thing. And the point is that if you fix, if you say that this is equal to P1 up to PN, in other words, you fix N general points in S. What's the fiber? Well, the fiber are all the rational curves, which are in the linear system D that pass through those points. So what that says is that the for a general point, the number of points in the fiber is this NSD. So it says that this NSD is just equal to the degree of this evaluation now. Another way to say it is if you take the evaluation map, and you look at it a map at a map on cycles or on varieties or whatever, if you take this and apply it to the fundamental class, this turns out to also be irreducible in most cases. Let's not worry about that. Take the sum of the irreducible components, they're all going to be of the same dimension. You take the push forward of this thing. It's actually equal to this multiple NSD, so the degree times the fundamental class of S to the N. Okay, so we can count this guy as a degree of a map or as measuring this integer invariant, which tells us what the push forward of the fundamental class is. Alright, so let's look at Valchin J's real story. All right, so when I say general point, what this means is there's some Zariski open subset inside of S to the N where this number is actually equal to the number for any of those points. And if you're working over C, so this is this, let's say, this is working over, when you're counting, we should be counting on geometric points. So let's say we're working over C and just taking closed points, then the subset of S to the N, any Zariski open subset, which is not empty, is connected. So it means that this number really doesn't depend on the choice of the points as long as they're general. Okay, so what happens in the reals, this is no longer the case. So if you have a counting problem, similar counting problem over R. So let's assume that our surface S is defined over R. So it's defined by equation, real equation. So we can think of it as a complex manifold, but also a real manifold, the way I'm looking at the real points. And let's assume that the chosen general collection of points is real. So in other words, that we can break it up. So let's say P1 up to PR are actually real points of S. And then the remainder are in complex conjugate pairs. So we have PR plus one PR plus one conjugate of two, say PS, PS conjugate and total number N, R plus 2S, N. So that's a real configuration, some real points and then some complex conjugate pairs of points. And then you can ask for real curves containing those points. So the question is how many, though we can take this to be real and general. So it's general in a complex sense. How many of the NDS curves that we talked about, these rational curves contain, containing PIs are real. And the answer is it depends. There's not one answer. Because if you think about this, well, okay, these complex conjugate points, these move around in some open subset of S to the, you know, you can think of just taking one of each pair. They move around in some open subset of S to the S, sorry, with the two S's. In other words, just the complex locus minus the real locus. And the real points move around in the real locus of S, but they have to be general. So it means they avoid some real zarisky, some real algebraic subset, and that real open subset may not be connected. So there's, so the wall crossing, the parameter space is not connected. So Velschenjie found a way to make the number independent, essentially, only depend on things being in the same topological component and having the same number of real points. So what Velschenjie said is the following. You consider one of these real curves. So let's take one of these real rational curves and I'll draw a typical picture of it. It will have ordinary double points and there are two kinds of ordinary double points over the reals. You have the usual kind. So here's your typical C that we're going to be counting. And here the local equation would be something like x, y equals zero plus higher over term, sorry, here at this point. You can see the two tangents, but then there are also real singular points, which look like x squared plus y squared equals zero, here at this point. The isolated point. So Velschenjie said, well, we count this one, let's call this one, this point p, this point q. We say that mp is equal to plus one in this case, and mq is equal to minus one. And he defines the mass of the curve to be the product over all the real points p in the singular locus. This is assuming the singularities are only ordinary double points of these mps. So this is plus and minus one. And then you have this Velschenjie invariant of the configuration of points. So what's the Velschenjie invariant? It's the sum of these masses over all real curves C that we're considering, the ones which are real rational curves in the curve class D, really maps, but since the map is giving the curve such that the C contains all these points. Okay, and what he showed the Velschenjie's so invariance theorem states the following that this number number, let me just abbreviate this p dot depends the number of real points and the collection of classes, the real points in telling you which component it is in pi not of S of R. Well, of course, since a real rational curve, the real points are connected, two points have to be in the same component. So it's just really telling you which connected component of S they're all in. If two of them are in different components, then this number has to, of course, be zero. There aren't any curves. This is i equals one to R. Okay, so that's Velschenjie's theorem. Alright, so what we want to do today is get a quadratic version of that. So our goal is a quadratic and purely algebraic version. Alright, so maybe I should say that Velschenjie originally proved this using symplectic methods. But then Eatonburg, Karlamot, Karlamot and Schusten gave a essentially algebraic, mostly algebraic discussion, which we rely on a lot there, their work we rely on a lot to give this quadratic version. Okay, so what's the what's what's the goal? We want to find a suitable on a suitable parameter space. We want to find a section of the Groben-Degbitt-Schief on a suitable parameter space, parameterizing the configurations of points that we want to allow our curve to contain, such that, let me call that QDS, that's quadratic form, such that if we take QDS and evaluate it on some configuration of points, and we take its rank, this should be this number NDS. And if the configurations of points are real, just write it that way, then this QDS should be a real quadratic form. And its signature should be this Velschen gene. So that's and then ask what we can then try and ask what the arithmetic additional information is, if any. Okay, so that's our goal. And so how do we start that? Well, first we want to make what's the suitable parameter space. So you see in the case of the complex numbers, if you have n points, they're just n points, you could, of course, reorder them, it doesn't change the problem anymore. But all the points will be complex points. But you saw already in Velschen J, you had some of the points were real, some of their complex. And if you think about it a minute, this condition of being a real configuration is just saying that your point P is a real point in the nth symmetric power of S, which actually, you also have to assume that all the points, it's general, so they have to be distinct. So there's an open subset of the nth symmetric power, which is just you take s to the n, remove all the diagonals and take the portion by the symmetric group. So that's our parameter space. We're going to take points in sim n of s, zero sits inside sim n of s. And this thing, of course, is you take s to the n modulo, the symmetric group. And this thing you take s to the n minus all diagonals, and modulo, symmetric group. So you can think of this thing as a sort of unordered configuration space. So this will be our parameter space. So a point. So what's a point in this thing, if you take P in there, it'll be a sum of close points, let's say in sim n of s over some field K, where they want to R, P1 up to PR are close points in s, base extended to K, such that the sum of the degrees over K of the residue fields, let's say the PI is equal to n. So over the algebraic closure, and they'll be distinct. So that's what this that's what this guy does. And this guy here says the P is our close points and they're distinct. So over the algebraic closure, I get n distinct points on s over the algebraic closure. All right, so that's our parameter space. And how do we, what's the modular space? Well, we have our evaluation map from the consavage modular space. This is defined over our algebra over our base field K doesn't have to be algebraic close, algebraic closed. This maps to s to the n. We have our symmetric group acting here. And of course, we can lift that action here by just leaving the map alone and permuting the points. So the evaluation map then will be sigma n invariant. And that gives us the quotient spaces or stacks, if you want to fancy it's called a quotient space, this thing here. This is the quotient space we just described as the symmetric and symmetric power of s. So we have evaluation map passes to there. And inside here, we have this unordered configuration space. And we just pull back to give us the object that we want to look at. This notation, that's the pullback. Right, so it's just, yeah, so not only are the points up here distinct on the curve, their images in s will be distinct. All right, so that's the basic gadget. And we have this map everything. This is still a proper map. All right. And just by counting dimension, we already know that these two things have the same dimension. So it's a proper generically finite. So what do we want to do remember here in this situation, we had this n d s was equal to the degree of this evaluation map. In other words, the push forward of the fundamental class upstairs was n ds times the push forward times the fundamental class downstairs. So now we'd want to do growth in the version of that. So what we'd like that growth the quadratic analog is, we'd like to define our quadratic form. I guess call this n. Should I call it n s? Yeah, n ds q. Yes, that's what I did call it. q ds should be equal to take the push forward of well, the the fundamental class is just the constant quadratic form one on the modular space. We just abbreviate and call it m bar sigma. And this should be viewed as as a section on this sim n s zero of the sheaf of growth and dig bit ring. So just locally, it's given by a quadratic form and they patch together on overlaps. That's what this guy should be. And that's really all I need in order to be able to evaluate it on point. I don't need a global quadratic form in order to be doing that. Okay, so that's what we'd like. And what's how do we do this? Well, that's the basic problem. Here's the main problem. And just saying let's just do that the main problem. In defining this thing is that this sheaf gw is in the terminology I introduced right at the first lecture, it's not orientable. It's not oriented, orientable in the sense of a one homotopy theory. But it so in other words, you don't have push forward maps for this sheaf, or even for finite maps, even among smooth schemes. You need an orientation. So what kind of orientation it is, what I call s l c oriented. So this is actually easy to see from the point of view of growth and deep duality theory. But in the interest of time, let me just skip that. Let me just say what this means in order to solve our problem. So to make sense, or even have of this EV lower star, so EV lower star, in other words, it is SL oriented. So remember, you have a canonical push forward from this guy here, assuming this is smooth. So we will get to that in a minute. But it has to be twisted. It has to be twisted by the relative dualizing sheaf. So let me call that omega of the map. So there is a push forward map here. Should put a little zero there, I guess, sim and s zero gw. So that's where we have a push forward map. But unfortunately, you know, one has values in the untwisted sheaf. So what do you do? We have to identify this one with the untwisted one. So what we need is, since it's s l c oriented, what we need, that means you can ignore squares of invertible sheafs. So we need is an isomorphism. So it's called an orientation, isomorphism row of this relative dualizing sheaf with the square of some line bundle. And then that will enable you to define this push forward map by using this guy to you just do the following you take on the sheaf level, you just take gw and then that'll be isomorphic doesn't ignore squares. I'll say a little bit about this next time next time, but a little later if I have time, you can safely ignore squares and then you apply row to get over to gw of omega, the evaluation. And you just send. Well, let me just say what it is. If you just have a one dimensional quadratic form, let's say lambda goes to lambda squared on the structure sheaf. Well, then you map l to l squared by taking a section lambda l and sending it to the square. Or even if you put you, you put u times this right. So here, lambda is set of a section of oh, it's a section of l and it squares and l squared. So that's how you go from quadratic forms to quadratic forms of values in the square. You just multiply the source by l on the target gets multiplied by l squared. And then of course, by a row, then you compose with row going to omega. And that's how you end up over here. Okay, so that's what Mark, there's a question of Shankelson about materialization and the link between the SLC orientation orientation of KO. You can does any realization under better it does this recover the orientation of it's not really KO. It's like a sheet version of KO. But if you do this for the motivic KO, then yes. Okay, was that? Yeah, thanks. I should just put this to the side. I should. You're a welcome shot. Okay, so let's see. So that's where we are. That's what we have to do. And now, oh, yeah, that's the problem main problem, but there's a positive note. And this is why we use GW instead of KO. GW is an unrammified sheaf. This makes life incredibly easy, or easier. What does this mean? It means if I take Y smooth over my field K, and I take F inside of Y co dimension, at least two, then the restriction map from H zero on Y GW to H zero on Y minus F GW is an isomorphism. So what that means is I can safely throw away. Let's go back to the picture here. I want to study this evaluation map here. And this is generically finite. And I want to get something downstairs. So downstairs with this unrammified property, I'm free to throw away anything I like of co dimension, at least two. And then I pull it back, take the inverse image of the evaluation map and throw away the corresponding closed subset of the modular space. And after doing that, what that says is so after. So here's a little geometry after removing such an F from this sim n s zero and then removing the inverse image from the modular space, then this modular space is smooth, smooth scheme. And this evaluation map is fine. So there are no stacky or singularity related problems after doing this. And so throughout what I'm going to do instead of keeping track of these Fs, I'm just going to continually allow myself to do this and just not change the notation. So the modular space will get smaller and smaller. And I'll actually construct my form on some open subset of this sim n of s zero. But in the end, it'll extend uniquely to the sim n of s zero whole thing. Okay, good. Let's see. So that's my basic fact. So now let's see what we have to do in order to do this. So this brings us to the geometry of the evaluation map, and of course, the modular space. Okay, so what's the geometry? Well, there's again, since I'm allowed to throw away co dimension two things, the geometry is on telling me about the generic, the open subset, the general open subset and some divisors. And it turns out there are only a few divisors that are relevant. So here they are, you have D cusp. So what is this, you take the set of all maps f from some P and tree irrational curves to s together with a configuration of points x, let me just call that x dot, such that well P. So again, this is over an algebraically, this is over a geometric point. So P is irreducible and smooth. So it's just a P one. And f from P to its image f of P is by rational. Okay. And so it's not too far off from the isomorphism. And in general, these will always have singularities a number of ordinary double points, but this has one non ordinary double point. So f of P has a single non ordinary double point singularity, namely an ordinary cusp ordinary cusp. So that's the set. And then I take its closure. And that turns out to be a divisor. So what's an ordinary cusp, it looks like this. And it has local equations and analytic coordinates y squared equals x q, which is the same thing, except I take an ordinary tack node. So what's an ordinary tack node that looks like this smiley thing. So in other words, y squared equals x to the fourth, local coordinates. And then of course, the closure of that set just as before. And then I have the trip. Same thing. Except I allow a single ordinary triple point. What triple point as you might imagine is something that looks like this. So that's something like x times y times x plus y or x minus y I guess is what I drew here equals zero. And then one more d two, that's not the same thing. It's where my P is two components P one union P two, P is are both over the outbreak closure P one. So it means there are two curves joined together crossing normally like that. That's what the P looks like. Like an x. And I assume as before, F from P to F of P is irrational. So generically isomorphism. And F of P doesn't have any funny singularities has only ordinary double points. So in particular around this point here, the map is really an isomorphism. Okay, so those are the Oh, and then I have those are the divisors and then I have the interesting open subset, Delta. This is the set of all f's together with the points. Let's ignore the points such that if I take any G from P to S with the points, living in the inverse image of the image of F, then it's as good as you could possibly imagine, then P is a P one G from P to G of P is irrational. And G of P has only ordinary double points. So this is this is the good one. This is where everything is as good as it can possibly be. All right. So the first there are the main theorem about this guy, let's see. So let me see if I can keep too much of it there. But okay, so here's the theorem. Let's see. So you need an assumption, you need to assume that this DS is at least four, or that's the self intersection of the canonical class. It's the embedding degree or DS is equal to three. And the degree of the D is not seven. Seven is not six. Sorry. Not six. Yeah, six is unlucky. Or DS equals two and D is at least seven. And there's some other condition of DS is one, but you don't want to hear it. Then, first of all, the main point is that if you take the whole space minus these divisors, cusp, d attack, union, d trip, union, d two, this is equal to this good space. So that's all there is. And again, after removing co dimension two things. And secondly, the evaluation map is un-ramified, the whole space minus D cusp. D cusp is the ramification law. And D cusp, so let's see, this way, this thing has simple ramification D cusp. So in other words, if T is a local defining equation for D cusp, it looks like T goes to T squared, just like this smallest amount of ramification you can have. So what does that imply? Well, we're in characteristic zero two is not divided the characteristic. So it says if you take the relative canonical sheaf, this is everything is smooth. So this is an invertible sheaf. You have a canonical map, you have a canonical section of this thing, which is just, you take the differential of the evaluation map, it's a canonical section. And this says its divisor of this map is exactly one times D cusp, right, because the derivative of T squared is two T dT. So the divisor of this thing is exactly D cusp. And that says, if you compare this with the canonical map of the structure sheaf to the structure sheaf twisted by the divisor D cusp, you get an isomorphism here. So that identifies exactly what the canonical sheaf is, it's just the structure sheaf twisted by D cusp, isomorphic to that. All right, so that's half of the battle, we've identified what the canonical sheaf is. Now we want to find something who's square is isomorphic to this, it's not apparent because, well, I don't have two D cusp, I have one times D cusp. So I can't use D cusp, I have to use something else. So what is the other thing? So that's the next part, the orientation, or really from D cusp, but whatever to elsewhere. Okay, so let me say a little bit about that, I'm kind of running out of time. So I won't say too much about it. But the idea is follows, the this uses the universal map, curve and map, right? That's what this modular space is, it's a modular space of maps and curves. So what is that? We have our modular space. And then over it, we have the universal curve, the fiber of this thing is the P that the F is mapping to. And then we have the map F. And then we have the image curve, let's call that C, the image curve lives inside of the parameter space, cross S. Okay, of course, this sits over here by just P one, projection P, that's the story, if I take a fiber of this over a point, F, because I have all the sections of the X's, I'm not going to worry about those, take a fiber of this over some map, then I just get the map of the P corresponding to F to its image curve, F of P, that's what the C is. C is just F, P, if you like. All right. So let's look at this. Let's take the fiber over some good F, again, geometric point. So these are the ones where it's as good as possible. And then of course, we have these marked points, but we're not going to worry about them. So here we have our point F, the fiber of P F is a P one. And it's mapping to CF, such as just F, P. This is sitting inside of S. All right, so this thing contains its singular locus, C F sing. And remember the condition that you're in this delta is that in particular, the this image curve, this map is birational. And the image curve has only ordinary double points. So if you make a picture of this, here's the image curve, here's my CF. And then the curve P is just a smooth P one, here's P F, which is P one. Okay, so I have all these singular points, let's call this one one, is there a bunch more? This one is let's say y q, they're q of them, for instance. And of course, over y one, I have two points, I have one prime and one one double prime mapping to that. And similarly, all the way down here, I have two points here, y q prime, y q double prime. Alright, so now if I move around and close things up inside P, and inside C, what does that do? I get the following picture. Okay, let's see. So now I have P. Here I have P contains the part over the good locus P delta, this maps to C delta. This contains, let's say it's relative singular locus, those are these double points here on each fiber. And then over that, inside of here, I have D delta, which is the inverse image. So it's a two to one map for each of these singular points. I have the two inverse images on the smooth curve. And then I just close this guy up in here, and, and take its normalization. So this thing will be the closure of D delta, then normalize. Okay, so let's call this map. And this thing, of course, maps to the modular space. So we have this composition pie. And so how does this look? Here's a little bit of geometry. Well, we're interested in this over two points. This is interesting. It turns out that nothing much happens over most of the devices, only two devices over which something happens. So let's take the generic, a generic point inside of D cusp. So, and then see cusp, and then we'll have the whole generic point of the modular space specializing to this. So that gives me my curve here, here's C eta. That's the image curve, and I'll have some other ordinary double points. But then it will generate to this curve, C eta cusp, which has this particular cusp here. Okay, now what happens on P? Here's P eta, here's P eta cusp. On P eta, these two points give me two points, this point gives me two points here. But this point only gives me one point here. And then local calculation says it looks like this. So that's the geometry near a cusp. So you see that this thing is ramified to order one over D cusp. Now what happens over, see, you know, so a few minutes. So what happens on D tack? So again, we have the same thing, eta tack. Here's eta, the generic point. Here's, so C eta tack, what does a tack node look like? It looks like this. Okay, this is C eta tack. And you can see what happens if I deform this, it looks like this. I mean, this is just a picture, but you can write down the equations, right? So here's a point y zero, I have two points here, y one, y two, they're converging to y zero. And here, that means I have four points. So I have a y one prime and a y two prime and then I have a y one double prime and a y two double prime. And what do they do? They just do this. You can again take local equations. So again, it's ramified to order two, but it occurs twice. So what does that tell me? Well, it tells me that the discriminant of this map pi, remember, pi is this map here, has divisor d cusp plus two d tack. Now, the discriminant is always a map. So since I'm almost out of time, I won't be able to tell you this discriminant is always a map to a square. It's always a section, a square of some bundle. This is just a general fact. And so since the divisor of the discriminant is, okay, so that says that the discriminant compared with the canonical map gives you an isomorphism here. And so what does that do that tells you that gives you this isomorphism. This is isomorphic to O twisted by d cusp. And this isomorphism tells you that this is isomorphic to M twisted by minus d tack tensor two. Well, there we are. This is my L tensor two. And this is my row. Okay, so in the remaining, so then we can make the following definition. We define this q d s to be the valuation or star of this one. And we'll make the computation that indeed, if we take some x in sim and s real, and the signature is q zero d s x is equal to the version j invariant x. It's an easy computation. And of course, the rank is just the degree of the map. So the rank of this q d s is n d s. So let me in the remaining minute, let me state the invariance theorem says the following. So suppose we have two points, both in sim and s zero of some field k. And let's assume let's let k suppose suppose that the fields are the same. So they have the same residue field type after reordering. So if this is true, and if the class of Xi is equal to the class of Y i in the a one pi not of s evaluated k i, then the q s x is equal to the q d s and y. So this recovers version j's theorem. And it's exactly time to stop. So I'll stop here. Thanks very much. Thanks a lot, Mark for the series of talks. And we will start with questions now. So I have questions to start. So first, it seems that the orientation you choose is really important here. Yes, I mean, and how is it shows up? How is it show up in the, in the computations, the choice of orientation that you make? Well, if you had a different orientation, so we had these two things that we wanted, we wanted to recover the Vell Jean J invariant. I mean, the rank is the degree is automatic, but we wanted to recover the Vell Jean J invariant. And in fact, the idea, the original idea I had before this joint paper was to look at the local defining equation which gave us the Vell Jean J invariant. So remember that the Vell Jean J invariant was find if I had a curve that look a real curve that looked like this, it looked like x squared minus y squared equals zero, then that gave us a plus one. And if it looked like this, looks like x squared plus y squared equals zero. And it gave us a minus one. So now, on the other hand, if you take the local defining this equation, you view this, you take the differential of this thing, then you have the local Euler class. And it's exactly the opposite of this over the reels. So my first idea was to use the local defining equation, take the local, sorry, this is this is the point zero. You look, take the local Euler class of the differential of the equation, then you could use this for your M zero. That was the idea. And that gave you exactly the opposite of this thing. So that that worked. At least I worked it out for P two. But then we had this idea and that instead of that, you can take the remember we have this double cover, right? If you have this guy here, then you have the singular locus, you have this double cover. So this is my point y and this is y prime y double prime. This gives me a double cover k y inside of k y prime say choosing one of the roots. And if you take the discriminant of this extension, this is exactly equal to minus local Euler class of the point. So this discriminant, if you factor the, we got the orientation by taking the discriminant of the double points upstairs, all the way down to the modular space, you can factor that is first mapping to the singular locus and then mapping to the modular space. And since this original degree is two, that will give you the discriminant of this map times the some discriminant coming from the rest of it, but square. So that says that this discriminant that we wrote down is exactly giving you minus this thing, which is exactly the same as the ocean. So that's how we were motivated by it. But there's nothing that says you couldn't find another one. Okay. So very, very, can you see if you have a question? Oh, yeah, I can. What goes wrong in positive characteristic? Well, not. Yeah. So what goes wrong in positive characteristic is the geometry has been much less investigated. And there are lots of problems. I mean, this was an extended exercise in deformation theory. And a lot of things go wrong in positive characteristic deformation theory. For example, if you just look at an equation like y squared equals, let's say a high order tack node. Well, once the characteristic starts dividing this exponent, then you run into trouble. And you have to remember at this basic geometric property that the divisors that I wrote down were the only ones of interest after you throw away things of co dimension two. And you have to get rid of all these other possibilities first in order to achieve that. And you have to understand the local deformation theory of these singularities. And the same thing with the high order cusps. So once the characteristic starts dividing that, then you run into trouble. But there's this nice trick to go from characteristic zero to characteristic P. If you just assume that your modular space in characteristic P is generically nice. So if you assume like, which is the case for P two, for example, assume that there's some dense open subset, which lifts smoothly to characteristic zero, then you can just take the degeneration. Yeah, and you assume maybe there are no other components of high dimension. So let's say you assume that the modular space in characteristic P that you're interested in is irreducible and has a dense open subset that lifts nicely to characteristic zero. Then you have this section, you can define this same gadget on the open subset, you can make the same definition, you have all the tools you need to define the orientation to define the quadratic form, but just on some open subset. And then of the symmetric power. But then if you look at the total family, what's missing? What's missing is something co dimension one in characteristic P, but on the total space, it's co dimension two. So in fact, you have this beautiful theorem of Pauline, a young one that says that the golden dig vit sheet is actually even in mixed characteristic ungrammified. So then you can extend over that and you do get a quadratic section of the quadratic forms, even in characteristic P, P not equal to two under these geometric assumptions. But it's, it's again, for an arbitrary surface and divisor, it's not at all clear if these geometric conditions are satisfied, as they are. Remember, we even had to have some funny conditions on that the degree shouldn't be too low, and that the degree of the divisor shouldn't the degree of the surface shouldn't be too low, shouldn't be one, for example, or you need special arguments in case the degree is one and the degree of the divisor should be sufficiently large. So, yeah, it's, there are little technical problems with the geometry in general. So that's what goes wrong, positive characteristics. So how does this work over a number fields, for example, do you? Yeah, it works fine over number fields. But then what can you compute then because you, you know, I don't know. So, okay, so it's difficult to compute. So, yeah, the techniques you have for computing these things, they don't obviously go through to the quadratic case. So we have some ideas for how to do this, but it's still in the thinking stage. Okay, there are lots of things for the real case. I mean, people have used tropical methods to compute the real case. So it tells you, so if you assume, for example, that you're a modular space, let's say you do it for P2, so that's something un-ramified over Z. And you could hope then that the modular space, that this whole situation is un-rammified over Z. Well, if that's the case, and you would expect then that your quadratic form would be un-rammified over Z, in some sense, then it would be easy to compute. It should be just a hyperbolic form plus some number of multiples of one, and you can compute it by knowing the NDS and knowing what happens over the reals. We know what Nelson-G. So that would be a conjecture as to what the answer would be. But in general, it's hard to tell because most of these del petso surfaces won't be un-rammified over Z. Okay, other questions? Yes, maybe I will ask, this is Vanya. Maybe I will ask already a question. So in your invariance theorem, the condition on XI and YI, the condition is use the shift by not A1. Yes. But can you replace this condition by a weaker one, namely using stable by not not? Oh, yeah. Yeah, yeah, that's right. So you'd made this remark before. So I should thank you for that. And thanks for pointing this out. Yes, you can replace that with a stable. It's a sort of cruder invariant. It's more computable. Yes, absolutely. Yes, thank you. So there is an addition to a question of Stephen McKeen. Can you see that? Yes. So if your modular space minus the bad locus is not a scheme, would you want to push forward a virtual, multivit fundamental class? Well, you don't have to. So that's the that's the thing. If you were to try and do this in, in, say, KO theory, Hermitian K theory, that's a more global invariant. So this is really just the analog is right at the very beginning. If you want to compute the degree of a map, you can restrict to any open subset you like on the base. And it doesn't change, you know, as long as the source and target are irreducible, it doesn't change the degree of the map. So that's exactly the situation here, because the Grotendieg Wittschief is unrammified, you can make the you can make the definition by restricting to any open subset you like, and then, then you don't have to worry about stacky stuff on the modular stack, because you have this big open subset where it's actually a smooth scheme. So virtual fundamental class, in this case is not necessary. So this is the unobstructed in some sense, the unobstructed case. But it's completely correct. If you work in other more complicated modular spaces, where the dimension is not correct, then you need a virtual fundamental class. And we do have a very nice theory of virtual fundamental classes available in the in Motivic homotopy theory. And there's this this nice work of ideal Khan, where he does things in terms of stacks and quasi smooth morphisms. So that's really beautiful. But that since using stacks, since these are more or less required to have some kind of atrial descent properties, you really need to work with q coefficients and theories that satisfy atrial descent, which the Grotendieg Wittschief definitely does not. So that's not not a good choice if you want to use get quadratic invariance. So you really have to stick with things which are just an is navage local. And but if you stick with schemes rather than stacks, in other words, if you for there are many examples where the modular stack is really as has an open subset, which is a scheme, or even can be considered a scheme like things that come in so symmetric obstruction theories on many modular spaces are of this form. And in that case, there's a really not there's a completely general theory of virtual fundamental classes in the Motivic sense and it makes perfect good sense to evaluate those and get quadratic invariance. So and it would be, it's possible. I mean, of course, people would like to do this in greater generality to extend this these virtual fundamental classes in the or quadratic invariance to to stacks with that sort of an ongoing thing at present. Okay. So there's a remark but of Kirsten that everyone can see about computation of computation of of mark for pencil nodal curves in a pencil of cubics. So just advertisement. Yeah, there's some of the chat. I can't find on my screen for some reason. More. Oh, there we go. Okay. Oh, yeah. Right. Yes. That's right. Thanks. So yeah, so you can make there's some specific the simplest one where you can make a specific calculation is for a pencil of cubic curves. And then you can use this Riemann-Herbert's formula from last time to compute everything and you get something which is not which has ramifications. So it's not in it's not sums of ones and minus. Thanks, Kirsten. Okay, 15 minutes. It seems that we have finished all the questions. So let's thanks again, Mark for for for his nice series of lectures. Thanks a lot.