 Repeat, I'm happy to introduce today's speaker at the Schubert seminar, Aakush Malsangus from Budapest, and you see his title. Aakush, go ahead. So I would like to thank the organizers for the opportunity to speak at this seminar. So yes, I'm going to talk about Schubert calculus over the reels, in particular about lower bounds. This talk is based on joint work with Lancelot Faye, Thomas Hudson, and Matthias Vanth. So I'm going to talk about, sort of in four topics. So first I'm going to say a few words about the numerative geometry and the particularities of the real case. Then I'm going to say a few words about the existence of fundamental classes. I think that will be a good point to take a five minute break. And I'm going to conclude the talk by talking a little bit about the co-mology rings of red flag varieties and their applications to lower bounds in real Schubert calculus. So the simplest enumerative geometry problem we encounter is already in high school, in some sense. So it's counting the roots of quadratic polynomials. So we can think of counting the roots of a quadratic polynomial as an enumerative geometry problem. And over the complex field, by the fundamental theorem of algebra, the roots of a generic polynomial are always that there's always two roots. And over the reels, the answer depends on the sign of the discriminant. So if the discriminant is negative, then we have zero roots. And if the discriminant is positive, we have two roots. So we can think of the space of polynomials as given by their coefficients. So it's a three-dimensional space. And the discriminant there forms a surface. So this is a double cone. And inside the upper and the lower semicone, so in the interior of the upper and lower semicone, the number of solutions is zero. And outside of this double cone, the number of solutions is two. So even though this example is completely elementary, it's nice from the point of view that it already, some of the general phenomena that occur in real enumerative geometry problems, appear on this problem. So here's a sample of some enumerative geometry examples. So for example, counting the lines on a cubic surface, the number of conics tangent to five given conics, the number of lines on a quintet three-fold, and the general problem of Schubert calculus. So for example, the simplest problem we usually encounter is how many lines intersect for given lines in p3. So this can be formulated as intersecting for Schubert varieties in harassment 2.4, which are indexed by the one box partition. More generally, we can ask how many k planes in an intersect some given or l planes in some given dimensions. So we would always like such an enumerative problem to have a finite number of solutions. So we expect, so we would like to have the intersection to be zero dimensional. So this gives a condition, so namely that the total co-dimensional of the Schubert varieties should be equal to the dimension of the rest man. So this gives a relation between these parameters. So there is also a very particular instance of this problem, the problem of that subspaces. So this is a choice of these parameters, so these choices of parameters. And in this instance, the solution has a very, this problem has a very explicit solution. So I'm going to say a few more words about this later. So in all of these problems, what's common that there is a configuration space of the problem. So for example, given polynomial in the case of quadratic polynomials, and there is a subset of degenerate configurations. So these degenerate configurations are usually given by some algebraic equations. So as such, they have complex co-dimension one, which means that they have real co-dimension two. And in case the configuration space is connected, then it's not hard to show that if you throw out two co-dimensional subsets in a connected space, the space is still going to remain connected. So just think of a curve in three space. And since the solution function is a locally constant function, this implies that the number of solutions throughout the configuration space is constant. Now in the real case, the situation is different because now the algebraic equations are real algebraic equations, and they have only real co-dimension one. So that, and that can easily happen that you separate the space into disconnected component by a one co-dimensional space. So just think of the example of the quadratic polynomials or just a rough sketch here. So over these different components, the number of solutions are constant, but as you pass from one component to the other, the number of solutions can jump. So this is what we saw in the case of quadratic polynomials, for example. So these connected components are often also called chambers. So in the case of a real enumerative problem, the answer is a list of numbers as opposed to a single number. So it's a natural question to ask, okay, can we give an upper and lower bound to this list of numbers? So there's an obvious upper bound, which is given by the number of complex solutions. And it's a question whether this upper bound can be achieved. And in this talk, I'm going to talk about the lower bounds to this possible list of solutions. Okay, so one way to solve such an enumerative problem is using a co-homological computation. So first, I would like to review the complex case. So in the complex case, what happens is that since we are in the complex case, every subspace that occurs in the problem has some canonical orientation, which comes from the complex structure. So for example, there is a fundamental class of a point, which is one of the generators of the top degree co-homology. And also the Schubert varieties in the problem come with some canonical orientations. So we can do the following simple computation. We can take the product of these Schubert varieties. And since if the Schubert varieties are transversal, then this is equal to the fundamental class of the intersection. And if we chose the dimensions correctly, then the intersections should be zero dimensional. So this is just a bunch of points. And since everything is canonically oriented, all the points come with the same sign. And in the end, we just get a count of the number of points in the intersection. Now, in the real case, the situation is less simple. We have to first of all assume that there's some sort of orientability, choose some orientations. And such a choice of an orientation gives rise to a fundamental class in the top degree co-homology of the Cresc Manian. And also, it gives rise to some orientations of the Schubert varieties if they have fundamental class. And then we can do the same computation. But in the end, since there's no canonical choice of orientations, every point in the intersection comes with a sign. So in the end, what we get from this computation is that we get a lower bound to the number of solutions. So geometrically, the situation is following. So we have some ambient manifold of variety M. And we have two subvarieties, S1 and S2. And we take some orientations of the ambient variety. So you orient the ambient variety, you orient the two subvarieties. And then at each point of intersection, what you do is you compare the orientation given by the two varieties to the orientation of the ambient variety. If these two are the same, then you associate a plus sign. And if they're different, you associate a minus sign to that point of intersection. And you do this for each point in the intersection. And then you sum over these points. And this gives you the sign count of the intersections. So there's a nice viewpoint that emerged in recent years. So this is the arithmetic point of view. So this was introduced by a paper of Kasan Bickalgram. So the idea is that these problems are all algebraic. So it makes sense to ask them in the algebraic category. So fix the ground field K. And so there is an analog of the homology with integer coefficients, which is called the Chauvet ring. And under some orientability assumptions, for a smooth variety, the top degree Chauvet ring is isomorphic to the grid and the grid ring of the ground field. So this is sort of an analog of the statement that the top degree homology of an orientable space is just C. This grid and the grid ring can be thought of as the K theory of non-degenerate quadratic forms over the ground field. So by this, I mean that the non-degenerate quadratic forms form semi-ring. So you can take direct sums and tensor products of quadratic spaces. And as such, you can apply the grid and the counter to this semi-ring and you get the grid and the grid ring of the ground field. So over the complex field, the only invariant of the quadratic form is its rank. So the rank gives an isomorphism of the grid and the grid ring of C to the integers. Over the reels, by Sylvester's law of inertia, there's two invariants of a quadratic form, its rank and its signature. So the grid and the grid ring of the reels is isomorphic to Z cross Z. And this ties up to the previous computations. So the rank in this situation is going to give the upper bound and the signature gives the lower bound to the number of real solutions. What's nice about this viewpoint that is also gives us some information over finite fields. So there, the grid and the grid rings turns out to be isomorphic to Z cross F2. So this also gives some additional information in values of F2 of the numerative problem. So this point of view is often also called enriched count of such an numerative problem. And many people have worked on these this point of view in the past few years. So without any claim of completeness, here's a few nice papers. So returning to the list of problems that I mentioned earlier. So one of the problems that's very well understood is the lines on a cubic surface. So this was already determined by Schlaefli in the 19th century. So he determined that these are the possible numbers of lines on a smooth cubic surface. And so you can see different topological types of the cubic surfaces on top with different number of lines on them. So Sagran noticed in the 50s that one can define two types of lines on a cubic surface. So there's hyperbolic and elliptic type. If you take the difference of these two types of lines, you always get this number three. So it was much later by Finocci and Carlamo, Volcanic and Telemann, who noticed that this has a topological interpretation. So that similar to what we saw earlier, so you have to fix some orientations, and you compare these orientations at the points of the intersection. So those can be identified with the hyperbolic and elliptic types defined by Sagran. And there's this recent paper by Kastan Vikagran, which gives this arithmetic point of view for this problem. Now, so this is an example which is really well understood, but as we pass on to the following examples, the situation is less nice. So for example, the conics tangent to five given conics, so there's an upper bound given by 3,264, a lower bound given by zero, and in between the number of solutions is not entirely clear what the possible solutions are. So the upper and the lower bound can be achieved, but in between it's not clear what happens. So there's a problem of lines on a quintic threefold. So there's an upper bound given by this number, lower bound given by 15. And again, in between it's not clear what the possible number of solutions are. So the problem of lines intersecting four given lines, there's two possible solutions, two or zero. So this is not very hard to show. And so this is a simpler Schubert calculus problem we usually encounter. And then there is the problem of general Schubert calculus, where, well, in this general point of view, not much is known. So it is known that there is an upper bound given by the number of complex solutions. In the case of Grassmannians, it's shown that this upper bound can be achieved. But as far as I know, for example, for flag varieties, this is not known. And in this talk, I'm going to talk about the lower bound on the number of solutions. So this problem that I mentioned earlier about subspaces, this is nice because here one can give a complete list of all the possible number of solutions. So I would like to say a few words about this. So the idea here is that to a given configuration of the problem, one can associate a general linear map. And the solutions to the problem then correspond to the p-dimensional invariant subspaces of this map. Akush, can you recall what balance subspaces mean with the number? Yes. So this is this problem. So we are interested in, I'll just scroll back a little bit. So we are interested in this type of problem. So K planes in A and intersecting some four given out planes in p-dimensions. And these are the particular choices of the parameters. So there's four. So maybe I should write it. So there's two p-planes in 2L, A2L intersecting four given out planes in p-dimensions. So it's a two-parameter family of Schubert problems. So a very simple one, but so it's a two-parameter family of Schubert problems. So here I just really want to give the main idea of this elementary way how one can view this Schubert problem. So the idea here is that you can translate this problem to the following one that you have to count the p-dimensional invariant subspaces of a general linear map. And invariant subspaces decomposes a direct sum of minimal invariant subspaces. Over the complexes, these minimal invariant subspaces are just eigenspaces. So there that means that you just have to build p-dimensional subspaces from one-dimensional subspaces. So that's L choose p. So there's, you have L eigenvalues and you have to choose p eigenvalues in order to determine p-dimensional invariant subspace. Now in the case when we're over the reels, there's two possibilities. These minimal invariant subspaces are either eigenspaces or they are two-dimensional invariant subspaces, which correspond to complex conjugate rules. So you have to build p-dimensional subspace from one and two-dimensional subspaces. So here's a specific example. So for example, and yes, so you have either all of the eigenvalues are real or some of them are coming complex conjugate pairs. So this depends on the roots of the characteristic polynomial. So if there's no complex conjugate roots, then you can, for example, if L is equal to four and p is equal to two, then you can pick any two of the four eigenvalues that gives you six possibilities. So if you have one complex conjugate pair of eigenvalues, then you can pick either these two or you can pick the other two as these two-dimensional invariant subspaces. And the case is similar when you have two pairs of two complex conjugate pairs of eigenvalues. Now if you have, for example, you have L equals four and p equals three, then when there's no complex conjugate pairs, then you have four choose three possibilities. And if you have one complex conjugate pair, then you can choose either this eigenvalue as the third one on this one. And if both, if there's two complex conjugate pairs, then you cannot choose three eigenvalues from two pairs of complex conjugate eigenvalues. So what you get here is, in general, you get a complete list of the possible solutions. And so this lower bound is either alpha. So this number or zero, depending on the parity. So this is an example to illustrate this. So in the next part of the talk, I would like to talk about fundamental classes and which Schubert varieties have fundamental class. So in the complex case again, that's a simpler one. So whenever we have a smooth sub-variety, then we get canonical orientations coming from the complex structure and using tomasomorphism that determines the fundamental class of complex sub-variety. Whenever we are in the situation that the sub-variety is singular, then the singular set is a one complex co-dimensional set, which has real co-dimension two. And so throwing out the singular subset, we get a fundamental class from the first point. So since that's a smooth sub-variety in that case, and since this is a two co-dimensional set, this restriction map turns out to be an isomorphism. So this gives rise to a unique fundamental class of a singular sub-variety in co-homology. Now in the real case, the problem is that there are no canonical orientations, so we have to again assume some sort of orientability, choose an orientation, and then we get the fundamental class. And the second problem is that the singular set now only has real co-dimension one, so this restriction is no longer an isomorphism. So even if we have a fundamental class on the smooth part, it might not lift to a fundamental class on the whole space. And in order to determine this one has to compute the homology of a chain complex, which has the property that such sub-variety is a fundamental class if and only if it is a cycle in this chain complex. So I'm going to give a few examples now about fundamental classes in the real case. So first, we can take the following stratification of the sphere into the upper hemisphere, the lower hemisphere, and the equator. And both the upper and the lower hemisphere have boundary, the equator, so none of them represent the fundamental class, but if we glue them together, so if we take their sum, then they determine the fundamental class of the sphere. So if you don't choose your stratification carefully enough, it might happen that you don't get a fundamental class. And the following example is, again, the example of quadratic forms. So we can think of gl2r acting on symmetric bilinear forms by change of coordinate. And the orbits then are parameterized by the signature. This is, again, Sylvester's law of inertia. And there's three open orbits. So the interior of the upper cone, the interior of the lower cone, and the exterior of the double cone. And in order to get a fundamental class of the whole space, we have to glue together three of these orbits. So this example serves to illustrate that it might happen that you have to glue together several of these orbits in order to get a fundamental class. So next, I would like to say a few words about these chain complexes in the case of Grassmannians. So Grassmannians have the cell decomposition into Schubert cells that we all know and love. So these Schubert cells are parameterized by partitions that fit into a k times m minus k rectangle. And the incidence coefficients in this chain complex are zero or plus minus two. So this turns out to be the case. So for example, the simplest Grassmannian we can consider is a projective space. So there we have this chain complex. If you compute the homology of this chain complex, then you get that these two classes generate a z module, and there's two torsion classes in degree two. So this is the notation that I'm going to use. I'm going to denote partition in brackets and the lower index that it generates a free module and partition in brackets with a lower index two, if it generates a two torsion module. So you can do the same computation for rp4 and then you get the homology of rp4. So here you can observe that rp3 is orientable and rp4 isn't orientable. So the point here again is that not every Schubert variety has a fundamental class. So the next example we can think about is the case of Grassmann II r4. So there again you have a Schubert cell decomposition and these turn out to be the incidence coefficients. If you compute the homology of this chain complex, then you get that you get a generator in degree zero and four, and you get two two torsion classes in degrees two and three. And what's interesting about this example that in degree two, it happens that you have to glue together two of the Schubert cells in order to get a fundamental class. So these are some of the types of cycles that occur. So you can get a free z-module generators, you can get two torsion generators, and it also happens that sometimes you have to glue together Schubert cells in order to get a fundamental class. So there's something that I am spoken about. So you can consider also twisted coefficient homology. So you can consider the homology twisted by the line bundles on Grassmannians, and that allows one to define some fundamental classes of Schubert varieties in these twisted coefficient homology. So this is an extra grading, but it still doesn't solve all the problems that arise. So it happens that you cannot define fundamental classes of Schubert varieties even twisted homology. So I think this could be a good point to take a five minute break. Indeed. Thank you.