 Next speaker is Matilde Mekang, who will tell us about the homogeneous spaces in characteristic two and three. Thanks very much. Yeah, sorry for the start. So I'm a second year PhD student and thanks for the invitation. I work with Mathieu Romani in Rennes and Michel Brion in Grenoble. So we're going to talk about in this 20 minutes, we're going to talk about homogeneous spaces in characteristic two and three. So we're going to focus on some notation, which everyone is probably used to, and then a little bit what happens in prime characteristic and why we should be careful, then what is known. So stating some results that are already known, then the questions that we asked ourselves, and then I'll give one results and if time permits a couple of ideas on how to prove it. So first of all, we fix K, an algebraic leaf closed field. For now, we don't fix any characteristic. And we fix G, B and T, a semi-simple algebraic group, a Burrell subgroup contained in it and a maximum torus. And we want to study parabolic subgroups. So subgroups that are in between a Burrell subgroup and the group G, the notation and then when we consider the Lie algebra and the T action on it by conjugation, we have the following, the composition. So we're going to denote as G gamma, the root spaces, and phi is going to be our root system. I'm going to denote as gamma any root and as alpha, a simple positive root. So delta is going to be a basis and alpha is going to denote in everything that follows a simple root. The aim is to study projective varieties, which are homogeneous spaces under G action. This translates on the other side into the study of parabolic subgroups. And since each parabolic is conjugate to a parabolic containing B, we can fix the Burrell subgroup and just study the parabolic that contain B. So this is what we're going to do. So something about prime characteristic, the behavior is not going to be the same as the one that we're used to oversee. This is mostly because of the Frobenius homomorphism. So this is something that we can see as a map from G to itself, which is a homomorphism of group schemes. So it is the identity as a topological map. And on functions, we can see it as something that sends a function to its B power. So here we fix P as prime number, which is the characteristic of our field. And so we introduce the notation of G1 and more generally Gm as being the kernel of this morphism. This is something that cannot happen in characteristic zero, meaning that we have a normal subgroup whose support topologically is trivial. It is an infinitalimal subgroup, but it is not trivial as a group. It has no trivial functions. And also something which is also very weird is that its Lie algebra is the same as the Lie algebra of G. And you can see a sort of proof, the easiest way I have found to see it on the right. So the Lie algebra is the tangent vector at the identity that belonged to the Frobenius kernel. But if you develop the condition of being in a Frobenius kernel, we get a trivial condition. So this is satisfied for any tangent vector because we see epsilon squared as equal to zero. So epsilon to the pth power is going to be equal to zero. And the Lie algebra is going to be the same. So one example which explains pretty much everything that happens next, the knot, I think it's a bit sloppy. It means notation. So we fix alpha as simple root and we associate to alpha, p alpha, which is a maximum parabolic subgroup, maximum reduced because we have seen that we can have the Frobenius kernel is an example of a non-reduced group scheme. So in characteristic p, we might have no reduced group schemes. And this is seen as the reduced parabolic subgroup, which has this levy subgroup, the one having root system delta without alpha. So this is an example on the right where we fix the group is SLN. And we fix alpha M as a simple root and then we take the drawing on the right represents the parabolic subgroup which is associated to that root. And we can see an example if we are in prime characteristic, this is an example of a parabolic subgroup whose reduced parabolic subgroup is the one associated to alpha 2, so the second root, which is not reduced. And how do we get the non-reduced mass we multiply by a Frobenius kernel? So that's by adding the condition on the two coefficients g and h to the p power r equal to zero. And this is now reduced and its reduced subgroup is p alpha 2. So what do we know about general parabolic subgroups in characteristic p? First of all, what do we know about reduced ones? So all of them if we are working over c. All parabolic subgroups over c are of the form so we fix a subset of delta and we intersect all of the parabolic subgroups which are associated to those roots and we get all possible parabolic subgroups this way. Then we pass on to positive characteristic. So from now until the rest of the talk, we're going to fix p, a prime integer and it's going to be the characteristic of our field, we're not going to work in over c anymore. The result is the one from Wenzel. He proved that if we have the following condition, so either p is at least equal to 5 or the group is simply laced which means that if we look at the Dinkin diagram, it does not have any multiple edges so it's going to be type a, d or e. So under this assumptions then any parabolic subgroup, it's going to be of the form so we fix again i as a subset of delta and we just add a little bit of non-reducedness so as we have seen in the previous example in this slide, we just modify this by a little bit our parabolic, we multiply by a Frobenius kernel of some height so we fix integers m, alpha and then we multiply this by p, alpha and we intersect this over delta minus i and he proved that any parabolic subgroup is of this form under this assumptions. So this is what was known and we asked ourselves if we do not assume if we take the other cases so either p is equal to 2 or p is equal to 3 and also our group is not simply laced so it's one of the diagrams on the right so it's either of type b, c, f, 4 or g2. The first natural question would be are we able to classify all parabolic subgroups and this is not been done yet, I still don't know how to do it and then an easier question would be can we classify at least all parabolic subgroups whose reduced subgroup is a maximal one so this we have done it, it's not being published yet, it's still work in progress but we know how to do it, I am not going to talk about it, we can talk about it later I think we'll have some time because it involves other tools and other definitions which we don't have time to explore and another even easier question would be to classify all quotients so homogeneous projective spaces of this form g over p such that their picar group is a rank one and this is what we're going to talk about in a few next slides, why is this question related to the previous one because if the reduced subgroup of p is given by the intersection of the ones associated to alpha 1 and alpha m for some simple roots alpha 1 to alpha m then using Brouhat cells and Brouhat the composition we can see that this picar group is a rank m so studying the parabolic subgroups which is reduced is maximal are exactly the one which give as a quotient a variety which has picar isomorphic to z so and the third question is a little bit easier than the first one because we might be able to express x, x in two different ways as a quotient so classifying varieties is easier a priori than classifying parabolic so this is what we're going to talk about now the the main result the main theorem is that so we fix a projective variety which is how much you use under a faithful g action the faithfulness assumption does not impact the classification of varieties if the action is not faithful which is replaced g by its image in the automorphism group and then we assume that the picar group is isomorphic to z then we can show that the variety is isomorphic to g over p with g a maximal reduced parabolic here I have not put any assumption on the characteristic so in characteristic zero it is trivial because any the reducedness is trivial because any parabolic is smooth by Wenzel's theorem we just add a Frobenius kernel so there is an isomorphism of variety with with a parabolic subgroup which is reduced as a stabilizer and so the non-travel part is to prove this when we are in very small characteristic I just give an example to understand how this works in type A if g is SLN and we take just a quotient of g by a parabolic subgroup without assuming that the action is faithful and we assume that the picar group is isomorphic to z this is equivalent to asking that the reduced subgroup is p alpha for some root which is by Wenzel's theorem equivalent to saying that p is a Frobenius kernel of order some m multiplied by p alpha so we get this action so it's just g the variety is isomorphic isomorphic isomorphic as a variety to g over p alpha it's just that the action is twisted by the mth power of the Frobenius morphism but this shows why the result is true at least for SLN are there any questions so far if not I can take a few minutes to give an idea of the proof so there are mainly two steps the first one consists in showing that so we take x negation of g by p and we have the assumption on the parabolic subgroup if p is non-reduced then we can show that the Lie algebra of p is either or the entire Lie algebra of g which I recall is equal to the Lie algebra of the Frobenius kernel or some other condition which we cannot explore today because I have no time to define it it's just there is a similar reasoning for this condition so n is going to be a subgroup which is smaller than the Frobenius kernel it's normal and we can show that the other case is that the Lie algebra n is contained in the Lie algebra of p and the subgroup does not exist in all cases so there are some more hypotheses in half of the case it exists and the other half it does not and then how do we conclude from this we use the equivalence of categories between p Lie subalgebra of the Lie algebra of g on one side and then on the other side the subgroups of the Frobenius kernel so which are called the subgroups of height one of g and if we restrict that on one side ideals and on the other side to normal subgroups this stays a bijection and we have that under this equivalence of categories on one side we have the Lie algebra of g and on the other side we have the Frobenius kernel so using this result we know that the Lie algebra of p is equal to the Lie algebra of g if and only if the Frobenius kernel is contained in p however we assume that the beginning and the statement of the theorem that the action was faithful so the stabilizer cannot contain a normal non-trivial subgroup inside of it so this gives us a contradiction which means going back to what we assumed that p must be a reduced subgroup actually so a few words to say how do we prove step one so this is what we aim to prove first we consider the levy subgroup of the alpha and then this acts on the vector space given by the quotient of the Lie algebra of g by the Lie algebra of p alpha and inside of it we have some module which is given by the quotient of Lie p by Lie of its reduced subgroup and we know that this last sub vector space is not trivial because we assume that p is non-reduced and so we studied this representation we can ask I mean if this is already under reducible representation then we are done but it's not always the case but it's very concrete so it's a matrix action on some vector space and we can calculate it. Five minutes. Okay thanks and then so the second step consists in using this lemma on roots which is due to Chevalet if we consider two roots gamma and delta whose sum is also roots a root so in particular we take delta which is different from plus and minus gamma and then we consider the delta string which goes through gamma and we denote as gamma minus r delta the last root so if we go to r plus 1 then it's not going to be a root anymore and on the other side the same thing so if we go to q plus 1 plus 1 then it's not going to be root anymore then the lemma says that when we do the bracket of the corresponding root spaces we're going to get the structure constant it's going to be plus or minus r plus 1 and if these numbers are integers and they go from 1 to 4 so in this step we see where the characteristic comes into play because if the characteristic is 2 or 3 then some of those coefficients might vanish so if we try to use this lemma then we need to be really careful because in some cases this might not give what we want because it might vanish so we need some additional arguments and so on so thank you for your attention and do you have any questions or things I should go over again? Yeah very nice talk, any questions?