 All right, so welcome back, everyone. We're very happy to have Kristin Wickelgren from Euclid University telling us about an arithmetic count of the lines meeting four lines in space, an introduction to A1 and relative geometry. Thank you very much. It's a pleasure to be here. The arithmetic count of lines meeting four lines in space was actually suggested by Leonardo Malaccia. So it's especially fun for me to get to talk to you about this here. And I am very grateful for the suggestion. It's an enrichment of this classical question, which is how many lines meet four general, meaning no two of them intersect, lines in P3? Although this question is classical, the part of this that's new in this portion of the talk, this is joined with Padma Vati, Srinivasan. The classical story goes like this, for example, in 3264. So answer. Let's take three of those lines. Here's P3. Let's give ourselves line one. And line two, not intersecting. And line three, not intersecting that. And let's first consider the lines which intersect lines one. Well, so let's first consider the lines which meet or intersect these three like so. So for any point P, for any P on line one, there's a unique line through P that meets both line two and line three. And we can see this by projecting away from P and looking at line two and line three in the P2 of the sky. And then they'll intersect a single point line through P intersecting line two and line three. So we look at the P2 in the sky of P and the projection from P of line two and the projection from P of line three. They have to meet at a certain point. It's something like from viewing from P, it's something like there and there on the line three. And then we have this unique line here, this line like that. So if you take the union of all the lines we get from this procedure, then they'll form a surface of the union of all the LP. And we can figure out that this is a degree two surface. A degree two surface. So the proof is that the surface contains L1, L2, and L3. And the dimension of degree two polynomials, so the dimension of H0, P3, O of 2 is we need to and split into four groups is 10. And so those are all the degree two polynomials. We need it to vanish on these folks. So that's the conditions ally. This is the P1, O of 2. And there are three polynomials of degree two on P1. So we get three conditions per line. What is three conditions per line? So there's a dimension one, dimension one space of degree two homogenous polynomials vanishing on L1, L2, and L3. So this, let's say, F is a generator of this one dimensional space, say F is a basis. So it's a degree two polynomial generating this one dimensional space. So the zero locus F equals 0, which we're going to claim is this union of all these lines. This contains three points of any LP. That means since it's degree two, that means, in fact, it has to contain all the LP. And since these two of these can't intersect, this containment has to be inequality because we can't have this break up into some sort of degree one piece. So LP does not intersect LP prime or P not equal to P prime or else L1, L2, L3 would lie on a plane. So then we get this equality. And we have this description of all of the lines meeting three of our lines. So we need to figure out which meet the fourth. And so which LP, those lines meeting three, also meet L4. And the answer is the intersection of Q with L4 or those meeting. And since there are two points of this intersection, the two lines in Q containing those two points are our answer. So by Bezuse theorem, Q intersect L4 is two points, giving us two lines. And this holds over C or over an algebraically closed. Or I guess L4 could be tangent to Q. OK. So let's relax the assumption that we're in an algebraically closed situation question over a field. And we'll take a general field K. And let's assume the characteristic of K isn't two. So I mean, what happens here? You can have the possibilities are two lines over K, a conjugate pair of lines over a quadratic extension of lines. K, a giant square root of A. Or L4, again, could be tangent to Q. We're going to package this information. In a way that'll give us one answer for the count, a way to get an invariant count that also generalizes to give a whole bunch of invariance of number of results over fields that aren't algebraically closed. So let's give another point of view on our two. So let's consider all the possible lines. So across monion 2, 4 is all the lines in P3 lines. And P3, for equivalently, dimension 2 sub-spaces, a four-dimensional space. Let's give our subspace a name. So W, it'll correspond to a point. And then we can view it as a subspace of its field of definition. So this would be a point of the gross monion valued in K and W. So let's choose a basis. Choose some basis E1 through E4 of, I guess, I'm going to put my line, my original line, over K. So let's do that. Such that L1, one of the lines in our configuration of four, is the span of E3, E4. And we'll let by 1, by 2, by 3, by 4 be the dual basis. Then we're going to consider some other line. It'll be span of some other basis. Hey, it may be not defined over K, but it'll be linearly independent for E3, E4, linearly independent. So the L1 and L will meet exactly when the, if we take these folks, these dual vectors, which cut out L1, wedge them together and evaluate them on E3 wedge E4, we get 0. So we can identify the lines L, meaning L1, as the zeros of, say, this equation, which is a section of a line bundle. So let's write down this equation. We have a tautological bundle over the Grasmanian, and what will the tautological bundle on the Grasmanian over the point corresponding to W, we can stick the fiber SW, is W. So there's a tautological bundle, and phi 1 wedge phi 2 is a section of S dual wedge S dual, which is the line bundle over the Grasmanian, whose fiber at W is a W dual wedge W dual. So we have a global section, phi 1, phi 2, and H0 of the Grasmanian, given by its value at W is phi 1 viewed as a polynomial, it's a polynomial on the entirety of the vector space, and we restrict it to W. So phi 1 by 2, its section W is the element of W dual given by phi 1 restricted to W wedge phi 2 restricted to W. So the zeros of this section tell us whether we meet the first line, and we can form an analogous section telling us whether we meet all four lines, form the analogous section sigma of four copies of the dual tautological wedge, the dual tautological. And then the lines meeting all of our lines are precisely the zeros of this section. Our points or L is the projectivization of our dimension two space, which is a zero of our section. And in particular, the number of lines meeting all four is the number of zeros of our section. And the Grasmanian is four dimensional. To make a line, we need to know which points it hits two different planes at. So that's two plus two is four. So we can see why we might expect this number to be finite. We have four dimensions of Grasmanian and four equations from our section of this rank four bundle. So we've reduced to counting the zeros of a section of a line bundle. So we've reduced to counting the zeros of a section for a four line bundle's direct sum of a vector bundle of a rank equal to the dimension. Over C, there's a tool from topology for this. It's the top turn class or the Euler number. So over C, this number of zeros of a section provided that section has simple zeros or you count with some multiplicity is the top turn class, which is a number because we're in the top dimension or equivalently the Euler number of the vector bundle. In particular, it's independent of this section here because we've lost it in the fact that topology gives us a top turn class. And then if we know the homology of two, four, we can compute the Euler class of this particular bundle is two. And what we'll do is generalize the Euler class to work over an arbitrary field. It's useful to think about what happens over R. So over R, we also have an Euler class for the real points and it becomes important to orient your vector bundle. So over R, there is an Euler number of a relatively oriented rank R vector bundle. And it counts the zeros of a section with a sign given by the orientation. So we have a degree map. If you have two spheres of the same dimension and a map between them, the homotopic class has an invariant called the degree. And if you had a section, let's call it sigma. So with an isolated zero, w, isolated zero of sigma, we can choose local coordinates and a local trivialization in a way respecting the orientation and local trivialization of v and then some sort of restriction of sigma to an open set looks like R to the R and if we take a small ball around our point w viewed in there, we can get a map from a sphere to a sphere. So this leads to a degree of a local degree of sigma at w as the degree applied to a map on the restriction of a small ball. And then the Euler number is the sum of these local degrees. And this is independent of the section because there's also a homological interpretation or construction of this Euler number. If we want to compute it with multivariable calculus, then for a simple zero, w, simple zero, we write out what our equations meant in local coordinates and then we can form a Jacobian as we have these sigma i. And if you've got coordinates xj, we can take this determinant and evaluate it at w. And if it's a simple zero, this Jacobian is non-zero. And if it's positive, it preserves orientation and the local degree is a 1. And if it's negative, it reverses orientation and the local degree is a minus 1. And we add all these up and we get the Euler class. A remark is that when you have a sum and of odd degree, so we did it in this case, we had a bunch of sum ends of degree 1, then changing one of the, changing the section to be minus on one of its coordinates. So we have this real vector bundle. This is relatively oriented. And if we change our section, which was sigma 1 through sigma 4 to some new section, which was minus sigma 1 to sigma 4, then one row of our Jacobian changes sign. So what we'll get is that E v sigma prime is minus E v sigma because all the local, because all the Jacobians changed sign. But since we know both of these equal E v, this implies that the real number here is 0 or is 2 torsion. But we have no 2 torsion because we're in top degree. So something similar is going to work over any field. We're k. And this is a field of characteristic, not 2. We have an a1 Euler number. This has, in fact, we have a1 Euler numbers with respect to many homology theories. This is part of a1 homotopy theory. We'll give credit to Roland Bavatsky explicitly. Morrell has a degree map, which for a whole bunch of theories, will give you this formula here. Has a degree map of maps from the sphere to the sphere to a group that contains more information often, or it contains at least the same information as the that we'll talk about. And the spheres, in question, are we conform a quotient of schemes? And this looks like the top cell of p to the r, which would be a 2r sphere over c, or an r sphere over r, and serves as a sphere, not any one homotopy theory. And the target is this growth in the group. And this is formal differences of quadratic forms. So this is a group of formal differences. Well, isomorphism classes, so the group completion or the growth in the group of isomorphism classes of non-degenerate or unimodular is the same over a field, symmetric bilinear forms. And over a field, all of these folks can be diagonalized. So we can present this group. It has one-dimensional generators. Let's call a one-dimensional form corresponding to the matrix A, this angle brackets A. So this corresponds to the form on the one-dimensional k vector space k, k plus k to k, which takes x, y to a, x, y. And if we change basis, multiplying 1 by some non-zero b, for non-degenerate, A needs to be in k star. We can change basis, multiplying by 1 over b or something, and have that. We've also got a multiplication for a tensor product, and we can do that. And this diagonal form is the same as this one, when A, B is not zero, and A plus B is not zero. This implies that there's a special hyperbolic form, 1 plus minus 1, which is also equal to A times 1 plus minus 1, for any of these angle brackets A, that. So some examples are the over-algebraically closed fields. All of these generators are the same, and so we get just the number of generators. So we can take the rank of a form or the dimension of the underlying vector space. We've been working over across monion, which is defined over z, with vector bundles defined over z. After group completing, the very interesting unimodular forms over z are no more data than the unimodular forms than the growth and dig thick group over r. And Sylvester's theorem gives you this rank and the signature, so our generators become 1 and minus 1. And a fun one is that a tall dimension 1 fields, like finite fields, are classified by their rank and their discriminant. We're about in the middle. So why don't we take a break here, and then we'll start adding up the local degrees to make a. Well, actually, sorry. I was going to, I was going to, sorry, let me tell you the Euler number over k, and then we'll break. OK, so we're going to need a transfer map or a trace map. So if we have a tall extension, for example, of fields, we have a transfer map from forms over l-valued forms to k-valued forms. And it post-composes with the trace from Galois or the sum of the Galois conjugate. So if we had a form on some vector space, v not having to do with our vector bundle, but it's just a vector space with l, we can send the class of that to viewing v as a k vector space and applying our form and then taking the sum of the Galois conjugates or the trace from Galois theory and the stays non-degenerate from the separability assumption. So we'll need that in statements. So we have this degree that's valued in GW of k, and we'll add up the local degrees to get an Euler class. So given v to x, our rank are relatively oriented vector bundle, they over k on a smooth, proper dimension. Our k-scheme will get this Euler number, which is the sum can be computed as the sum over the points w and x, such that a section with, let's say, the zeros are isolated. There's a conological definition of this as well. And the degree of a local degree at w of our section. And using the degree map valued in GW of k, we'll get an Euler number valued in GW of k. And the multivariable calculus way to compute this, so say, suppose sigma is a section with isolated zeros. And if the zeros are simple, we can compute that same Jacobian and just stick brackets around it. So where the degree for a simple zero is a transfer from k and w down to k of this Jacobian sigma of w. Before we forgot everything but the sign and just recorded, like taking the signature of this form. And here what we'll do is remember the value of the derivative itself. And here the Jacobian is the matrix of partials. So this is ixj, where sigma 1 through sigma r is after taking some at all coordinates and restricting to a small enough open set, you can make your section look like a map from part of AR to AR. So and this is, again, it's independent of the section. And this is a viewpoint with CAS on an Euler class that's due to Barge, Morrell, and Zell, and Morrell, Levine. This is a viewpoint with Jesse CAS. Deglesion and Kahn give a lot of great punctuality for this Euler class. OK, so now I would like to take a break. Thank you. OK, well, thank you very much.