 Okay, then I am very, very happy to introduce Felix Lezebnik from the University of Delaware. He was, I would say, one of my advisors when I was there. So it's really great to hear his voice again. And he's going to tell us today about crafts without short cycles and asymmetric lifts. So take us away, Felix. Thank you very much, Stephen. Thank you, everyone, for the invitation. I'm sorry that you cannot see me, but as long as you can see the slides, that's the most important. Okay, the title of my talk is Graph Without Short Cycle and Asymmetric Lifts, but most of the talk, I will just do some survey and you will see many pictures, actually. So I want just to say that everything I mentioned and more is presented in these two surveys. One is this one, two with Boulder, and another is with Shwin Sang and Yevong and Seventeen. The last paper is on, this paper is on Archive and it's submitted. It has kind of a survey from Seventeen to 21, whatever happened. All right, so let me begin. So motivation for this talk is this. Motivation will be constructing a new generalized foregone. I will tell you immediately that we did not succeed yet in doing that. But that's the main motivation and many problems I have been working lately on and off. Really, these problems came from these attempts. Another smaller things are lifts that result in graph with small automorphism groups and the need of three variables for defining some algebraic graphs. So that will become clear after I start. Okay, so first of all, what are generalized M-guns? So suppose N and S are integers, N is at least two, S is at least one. A generalized M-gun of all the rest is a bipartite S plus one regular graph of the emitter N and goes to N. So this definition, so it's bipartite graph, S plus one regular with the emitter N and goes to N. This definition is a little restrictive. In many books, generalized polygons are defined as regular bipartite graphs. So vertices on one partition have degrees A, T plus one and on another S plus one. But for our purposes, that will be the definition. So they will be regular bipartite graphs with this diameter and this nurse. So for S equals one and N is at least two, the usual even cycle gives you an example of that. And if N is two, so the emitter is two and the nurse is four, then complete bipartite graph gives us example for arbitrary S. But when S is at least two and N is at least three, then it's not easy to construct such graphs. And so there are only three values of N for which they constructed. And they were constructed by Jacques Teetz in 1959. And he proved that they exist for each prime power S for N equals three, four and nine. So that's the result of Teetz. Three guns, so three, four and six, yeah. So for three guns, these are bipartite graphs which have diameters three and your six. They are exactly point incident graphs of projective planes of order S. So they, of course, were known before Teetz. But these are, in his language, generalized three guns. And if degree, if order is S plus one, yeah. If order is S, so degree is S plus one, they have so many vertices in each partition. And they are asymptotic for the number of edges in terms of number of vertices becomes this when V goes to infinity. Generalized four guns are a little much less known. And again, their analysis, if the ammeter four and girls say it, you can count that there will be so many points in lines. And then, you know, they are asymptotic for the number of edges in terms of vertices as this. And generalized hexagon of order S will have in each partition so many vertices diameter six goes 12 and that will be the asymptotic for the number of edges. Now, here is Jack Teetz. He introduced this notion in this language of incident structure when he studied simple groups. Later, he developed it into the theory of building which uses this type of terminology. Incidents. He did it in the following way. So suppose we have a finite group and we have two subgroups, P1 and P2. We can define the following bipartite graph or incident geometry. So take as a set of points, say all left cosets of G with respect to subgroup P1. And another set will be all the cosets with respect to P2. So these are our set of vertices. And you join vertex in one partition with the vertex of another partition, one coset with another, if their set theoretic intersection is not empty. So that's the definition. Now, this can be done for every group and you will get nice graph with a lot of symmetry. So this is a huge source of examples. But if you apply this to some special groups, which are called finding rank two simple Chevrolet groups of li-type, and two very special subgroups of them, which are called maximum parabolic, then the graphs you obtain by paragraphs will be very nice. And actually, they will be exactly generalized three guns, four guns, and six guns. So that's how they appeared in the work of Teetz. Now, about five years later, Fyton-Hiedmann published a great theory. They proved that there are no other generalized M guns. So M guns, if N is not three, four, and six, we cannot have a generalized M gun. So they just don't exist. Again, if you used to generalize the polygons in some other setting, I repeat that we have a restricted definition. For us, they are just regular graphs. Every point has the same degree as plus one. So in this regard, that's what their theorem says. And the proof of the theorem was the first really non-trivial and beautiful application of spectral techniques. This proof is not well known. It's published, of course, but it takes time to get through it. The original proof, I think, is harder. The proof I like more, for example, can be read in the book by Brauer, Cohen, and Neumeyer. So that's this time I stop about generalized M guns, and I will try to formulate another motivation for this type of graphs. And they come from Turan-type problems. Of course, probably everyone knows what Turan numbers are. So if we have HGB graphs, and H is a subgraph, if set of vertices is subset of set of vertices, and set of edges of H is subset of set of edges, and we will say that if G contains no subgraph isomorphic to H, we will call it H3, and write that. And then we can introduce this function of number. So for V and H, we consider the maximum number of edges in a graph on the vertices which is H3. So among goal graphs on the vertices which are H3, we take graphs on the largest number of edges. And similarly, when we have familiar graphs script H, then we introduce a similar function, which is the greatest number of edges in a graph on V vertices which contains none, which contains none of the graphs from this family script H. So when V and H are fixed, these are numbers, and they are called Turan numbers. So when V changes Turan function. So maybe I will skip this slide for a moment. These are examples of Turan numbers, but I, interrupt me please anytime. If someone not very familiar with Turan numbers that I can comment a little bit more, but since it's recorded, you can just see. So they're not known well for many, many H. For example, C8 is one of the first examples where there is no even magnitude known. In many other cases, magnitude may be known or inequalities, for these numbers are present, but exact results are very, very rare here. So the main theorem of Turan 41 was that if you forbid complete graph on R vertices, then among goal graphs on the edges, the multi-partite Turan graph, R-partite Turan graph is the best instruction and it gives so many edges. Then Edgerstone generalized this for arbitrary graph H, and they express the main term in asymptotic by using chromatic number of H. Then in 20 years it was a little generalized but by Ertigian Shimanovic, when we had forbidden family of graphs, and then you just pick the smallest chromatic number and this number governs the coefficient in this asymptotic. So it's V-square times the number. But when chromatic number of forbidden graph is two, like even cycle or complete bipartite graph or a tree, then this formula just gives little of V-square because when you divide by V-square, you will get here zero. So the limit will be zero. So little of V-square, it's not a very precise information and people try to improve this. And this type of problems where forbidden graph is bipartite or at least one forbidden graph is bipartite, they are called degenerate Turan type problems. And there is a nice big survey by Furidi and Shimanovic in 2013. So that's Turan, Polarjash, Artur Stone, Zoltan Furidi, Miqvash Shimanovic. Now suppose case fixed positive integer and what can be said about these two numbers? How are they coming together? Of course, you know, this number is smaller than this number because you forbid much more so you could put less edges on V vertices. So this number is less. Actually it will be always strictly less. Equalities here won't happen, but in all examples known when the magnitude as a function of V is known of these numbers that the magnitude is the same, coefficients only change and coefficient may depend on K. So this is a very nice open question that if we consider the ratio of this number to this, is it possible to bounce it from below by a positive constant depending on K? So that's a very nice question. Do these numbers have the same magnitude or not? And as far as I know, it's still open. Now about the upper bounds. The upper bound here come from the unpublished result of Ergjash and the result of Bondi and Shimanovic who really proved a little bit more than that. But that's the upper bound. It's constant dependent on K, V to the power of 1 plus 1 over K. So, of course, having this upper bound, people try to construct lower bound for, say, this number of the same magnitude. And that was hard. So simple probabilistic argument gives exponent instead of 1 plus 1 in lower bound it will be this. Then a little better probabilistic argument gives you a little bit greater exponent in the lower bound. Then Margulis-Lubotsky-Philipsavnik independently, Margulis and Lubotsky-Philipsavnik, they constructed graphs such that the exponent in the power of V is 1 plus 2 thirds times 1 over K roughly. Here it's 1 over K, here. And this is roughly 1 plus 2 thirds 1 over K. And that was explicit construction, you know, it's famous Ramanujan graphs. We could do explicitly a little bit better than this probabilistic argument of Ergjash, but it was constant here is 1 half times 1 over K, so it's worse than that. But lately, with Stimenko and Boulder, we could get 2 thirds times 1 over K, but we have a little bit better here. Epsilon here is 0 or 1, so our bound is better than bounds of Margulis and Lubotsky-Philipsavnik. And it was obtained by the techniques I will be described a little bit later. So these are about lower bounds. Now that's Andrian Bondi whose picture you haven't seen yet, and that's Margulis, Lubotsky-Philips and Savnik. So these are people who worked on that. That's Vasily Stimenko and that's Andrew Boulder. Now again, so let's again look at this upper bound. So one question is, for which K there is a lower bound which matches in magnitude this upper bound? You see, because what we did here, we talked here about general bound, the ones which exist for every K, but for some particular K, the lower bound can be better. So if we try to match this, then exactly these graphs which we touched, generalized, three gun, four gun and six gun, they really provide the same magnitude as the upper bound. So for K equals 2, for K equals 3 and for K equals 6. So they are kind of give you magnitude twice the best lower bound. Constants are not matched here. The only thing when constant is known that's when K is 2, but it still doesn't match, sorry. If you just forbid C4, then it's known, but if you forbid C3 and C4, it's not known. The upper bound, constant and upper bound, when you forbid C3, C4 and the lower bound, they are different constants, but by the multiple 1 over square root of 2. So I will not give you examples of projective plane, but of course, someone, you know, we gave definition of the foregun, but what is foregun at least one example. And it's not that easy to write and analyze. For QR, for example, only foregun is known. And that is, you know, something very, very different from three guns. For three guns, projective planes. If order of projective plane is prime power, which is not prime, and at least nine, then usually there are many, many non-isomorphic projective planes. So the extremal constructions for graphs, you know, without C3, C4 and C5 are non-isomorphic graphs. They are all, you know, Q plus 1 regular, all have diameters 3 and GUR 6, but they are not isomorphic. On the other hand, only one foregun is known. And it's description due to Clark Benson, and that's the description. So you take a projective four-dimensional space of the same as, you know, usual vector space of dimension 5 over fq, and you consider five tuples and you identify the proportional one, collinear one. So you obtain this. So you can think that, you know, that's the definition. That's the definition. So you take a vector which satisfies this condition, you spin a line. So points in this geometry will be one-dimensional subspaces lines in fq5. And lines will be two-dimensional subspaces, such that all vectors satisfy this. So they are called totally isomorphic. So it means every vector satisfies this, but it's two-dimensional. So, and adjacency is by container. They seem to be not bad description, but, you know, it's high-dimensional. And it's known that if you go to another non-singular quadric, then all of them will be projected the equivalent. So you will not get a new generalized foregun. Now, now some of you may remember that actually there are two. There is this q4 and there is another one, which very often is written as Wq, symplectic one. As geometries, they are not isomorphic. One is a dual of another. But as bipartite graphs and our definition for generalized and GANZA, in terms of graphs, they are isomorph. So really, there is only one infinite series of generalized foreguns. And that's, you know, was a motivation. Just after, you know, we learned about this. And actually, Ustimenko tried to do this even before we started working with him. He tried to construct new generalized foreguns. But that the striking difference, you know, here you have infinitely many extreme objects and here you have, you have essentially only one infinite series. Now, Clark Benson was a graduate student of Walter Tate, Cardnell at that time. And he was interested in this problem because Andrew Gleason pointed out to him that probably these generalized foreguns are related to minimal graphs, which were introduced by Tate, you know, with those more bounds and bound. And so Benson really was interested to describe those minimal graphs. And so they are, you know, called cages now. They are yours eight q plus one regular graphs on the smallest possible number of vertices. So that's minimal. So that was motivation for Benson. But of course, fight, you know, work with group theoretic description of this generalized foregun. So that was probably another way just to present it without any group theory. And that's what it is. All right. So you see, there are not many generalized end guns. They exist only for n equals three, four and six. Now, so what we will consider will exist in much greater number. They will, we will call them at the beginning, biophine parts of generalized end gun. What are biophine parts? So suppose you have a generalized end gun, that's biophine type graph. And you pick an edge x, y, and you remove this edge from the graph and you remove all vertices that are at distance at most n minus one from both x and y. So you are an end gun. It's a biophine type graph. You pick an edge and then you consider all vertices which are at the distance at most n minus one from x and y. It's very easy to understand that what you really remove edge and all those vertices, you are removing a spanning tree of generalized end gun. There will be no cycles there because the shortest cycle has lengths to n. And really, you will be removing so many vertices in each partition. So what will be left in each partition will be this number of vertices. And what you will obtain is this, to argue, will be a bipartite graph, which partition will have so many vertices. That's exactly as many in magnitude that in the whole generalized end gun because what you remove has smaller order. And the diameter will go up. The diameter will become n plus one. But the girls certainly, because it's a subgraph, the girls cannot become less, but it really will still be two n. So you will not affect the girls by removing this spanning tree from it. Now, when you again take this kind of simpler graph a little bit, it's as regular. Then you compute that the number asymptotic for the number of edges in terms of number of vertices is as good as for the whole generalized end gun. So from the point of view of Turan theory, studying these big subgraphs a little bit easier than the studying the whole generalized end gun. So that's, by a fine parts, are as good actually as the whole object for the Turan type problems. Now, I did not give you definition of generalized six gun. And I didn't do it because it's not easy. It's really messy. It can be done only also by using this form. But when you define lines, it's really messy. It's all in Benson's paper. So here I couldn't find a picture of Benson, sorry. But he worked at the University of Arizona after he graduated for many years. Now, what you see here will be a description of these biofine parts. So graphs which are presented here in blue are really biofine parts of generalized three gun and generalized four gun. Since there are many generalized three guns, I picked just this classical one. That's projective plane of other Q. So it's a bipartite graph. So one partition points, these are one-dimensional subspaces in the three-dimensional space over FQ. And lines will be two-dimensional subspaces in the same space. And the condition of adjacency is bicontainment. So it turns out it's very easy to check that that's the description of this biofine part of classical projective plane. So it has Q square points and Q square lines. And a point and a line are joined by an edge, only this relation holds. So it's easy to explain that that's exactly what happens when you do this procedure of going to biofine part in projective plane. The graphs will be six. If I take, yeah, and here I wrote the biofine part because, you know, in this graph, if you consider Lady graph or point line bipartite graph of this PG2Q of this geometry, it's edge transitive. So it doesn't matter what edge you take. So that's why the biofine part in this particular. Now, when you take this particular generalized quadrangle, what we saw described by Benson, also it can be explained that as bipartite graph, it's edge transitive. So here also, when you construct biofine part, there will be only one. In general, if you take projective plane, which is not flag transitive, then it depends what you are left with after you delete different edges. The result should not be isomorphic. Right, but here they are. Now, and this, no matter what edge you take, that will be the description of this biofine part. So how does it work? You know, it's a foregun, so it will be QQ vertices in each partition. Each partition can be thought as a three dimensional vector space or whatever. And we take a point and we join it to the line. If the following two condition hope, the sum of the second coordinate of a point in a line is product of the first and the sum of the third is product of the first coordinate of a line and second coordinate of a point. So this is a very simple description for the biofine part of the generalized quadrangle. I did not give you any model or example of generalized six-gun, but it exists and it exists only one. And if you write it's biofine part, it will look like this. It will look like this. It will be Gorsk well. So there is some similarity in the way here they are presented. People believe that there is only one generalized hexagon, but about generalized quadrangle or foregun opinions differ. Some people believe that there is no other than this one for RQ. Q is out here, by the way. Sorry. And other people think that there must be more, but that's where it is. Another thing which will be useful to us. So suppose we have two graphs, H1 and H2. And suppose phi will be a function from set of vertices of the second graph to the set of vertices of the first graph, such that every edge xy will be mapped to the edge. So then phi is called the homomorphism from H2 to H1. And we will write it H2 maps to H1, so phi is the homomorphism. So suppose we have a homomorphism from H2 to H1. And suppose it's surjective. So the image will be the whole set of vertices. And suppose that for every vertex V in V2, or every vertex of graph H2, the restriction of phi to the neighborhood of V in H2 is bijection onto the neighborhood of phi of V in H1. Then phi is called a covering map. And H2 in this case called the cover of H1, or it's called a lift of H1. So similarly to topology, it's just local homomorphism. So that's what happens for covering map. Example, suppose here you have two graphs. And H and C, C stands just for cover. So C really is a lift of H, C covers H. To see this, these colors have nothing to do with graph coloring. It's just to see if you take graph C, then these two yellow vertices will map to this yellow. These two red will map to this, blue to blue, and so forth. You can easily check, looking carefully, that what you get will be a covering map, when vertices of the same color are mapped here. And what is interest, or this will be lift. This graph, this graph C, will be lift of this graph H. It's interesting that this triangle becomes a set of two triangles. But this triangle becomes a six cycle. So this is a very simple example which shows that lift can destroy a cycle. So it was this triangle, now you have this. And that's a longer cycle. So lift can map cycles to longer cycles. And terminology, when you have the covering map and U is the vertex of B1 of this graph, then phi minus 1 of U is called the fiber of U in H2. And if all fibers have the same cardinality, then we say that H2 is really a lift of H1. So that's our lift of H1. And here we come to these constructions of biophine part of generalized polypons. So we said that this graph H1 here is a biophine part of the classical projective plane. And this is the biophine part of the generalized foregun. You see that the first equation is the same here. And actually, the second graph is few lift of the first graph with a very simple reason. If you consider set of vertices of this graph H2, it will be this, P3 union with L3. These are two three-dimensional vector spaces over F2. And here the union of two-dimensional vector spaces over F2. And consider this map on this set of vertices. Just every vertex is truncated by just to raise the last coordinate. So point x, y, z will go to point x, y. And line x, y, z will go to line x, y. And you can very easily see that this will be a covering map. And you can very easily see that what you get will be a Q lift. So this graph is Q lift. I didn't explain this yet, but these graphs are Q regular. Of course, that follows from the definition of biophine parts because we take out a tree there. But another way to see it is that they are Q regular Just take a vertex and how many neighbors does it have? To construct a neighbor of this vertex, we use these two equations. We give L1 an arbitrary value. That will allow us to compute L2 uniquely. Now, if we know L1 and L2, it will allow us to compute L3 uniquely because P1, P2, and P3 is fixed. As a fixed point, we will count in neighbors. So choice of L1, any choice of L1 create exactly one neighbor. And there are Q choices for L1. That's why there are Q neighbors. Absolutely the same as here. And this localize or morphism is easy to check because the choice of the first coordinate of a neighbor defines this neighbor completely. So that's why this will be a Q lift. And this Q lift is remarkable. The girls of this graph is six. And there are many, many six cycles on this graph. But the girls of this graph is eight, generalized part of generalized one. So every six cycle you can find in this graph will be destroyed under this lift. Will be destroyed under this lift. It has to be checked. But it's really just trivial verification with these equations of adjacents. There are no six cycles here you can check. And here there are millions of six cycles. All right. Now this scheme can be generalized. Here I will take just f of P1L1, any function of two variables here. And I will construct graph gamma QF. F is defining function. And here I can start with two functions, F and G. And I can construct a graph like that. Again, again, this graph will be Q regular. If you fix point P1, P2, P3 and then you give L1 any value then L2 and L3 will be uniquely computed after that. So it's Q regular graph. And again, this truncation of the third coordinate gives you the covering map from bigger to smaller. So how can one try to construct new generalized polygons? Okay, started like you had before but instead of the second equation, write something more general because we know that P1L2 works here but you can take any function. And you just don't think you want us to get your save. Suppose you will get your save then you will try to attach the tree. We believe it's easy to do but no one could construct a graph of yours aid by using function G which is not isomorphic to the graph which has P1L2 here. So no one could still do it and no one could prove that this doesn't exist. Of course, having four variables here, it's not needed. You see, because from the first equation L2 can be written in terms of P1L1 and P2. So really there are three variables here. Therefore, I will just rewrite it this way. And that's what I was doing and other people were doing trying to find this function G to get gursate with the hope if we get this then probably attaching tree will be easy after that. And we could get a new generalized foregone. So far we couldn't do it. What was great about this problem and a lot of computer simulation was done still with computers, I think much more can be done that what we did. But what is interesting here, you try to find graph G in some simple form and check it and you get very interesting problems. Just problems you want to work on immediately. You want to forget the original question which led to that problem and just work on that and I will show you some. So that's the status. Now, no such function G has been found so far. Now, but there is another thing here. When you look at these graphs, at these graphs, you see this is Q the same as here. The first equation has P1L1 on the right. The second has G sub one and here the same P1L1. So it's a bifine part of projective plane and that's G2. So we consider two lifts of the bifine part of projective plane. We consider two lifts. And if G1 is not G2, they can be isomorphic. That's a big thing, you know. For example, these graphs which one, two, three, four, five they are all isomorphic. You see, they all I underlined this P1L1. They are all lifts of this bifine part of projective plane and the classical project. But these equations, this has three variables. This has two variables. This has two variables. This has actually only P1L1 and only P1L1 is sufficient to write that graph. So it's not easy, it's not hard to prove it all. So when we are here, when we are here and we have this generalized bifine part of generalized foregone, instead of P1L2, I can write P1L1 square and you will get isomorphic graph. It's easy to check. It's not obvious, but it's easy to prove, to prove. So functions can be very, very crazy looking and graphs can be isomorphic. So maybe this suggestion, because there is one classical generalized foregone for which bifine parts can be written only in terms of P1L1, then the question is maybe instead of taking this function g of three variables, maybe we can just take function h of two variables in the second equation of P1L1 and try even, this is smaller class, of course, than class of functions of three variables. It's contained in that, but that was natural to try. And again, we couldn't find such function h, but we couldn't prove that it does not exist. And that led to another question. Given this graph, and here the second equation is written by using P1L1, P2. Is it possible to find function of two variables such that this graph with three will be isomorphic to this graph with having h here for sufficiently large, for sufficiently large. So in some sense, is it needed to have three variables in this graph? The expected answer is no, because if you count how many functions of three variables are from FQ to FQ, that's the number, and here it's much smaller number. So probably there are g's for which you cannot find h, but somehow we couldn't make this counting argument clear. So what parameters really distinguish this graph if you have only two here and if you have three variables here? And it turns out that one of the best parameters for us turned out to be group of automorphism. And this graph, which has only two variables in the second equation, will have this group as a subgroup. So that's additive group of vector space of dimension two. But this graph, when you have three, it may not have this type of subgroups. So that was the, and we found this really by using computers. No, this was clear, but how to find the graph which has one property, but not another, that was not clear, explicit example. Why this is true? Because if you take graph, this graph, when the second equation has only two variables, it should be comma here, P1 comma L1. Then this will be an automorphism. Every, so fix A and B, two elements of the field. Add A to P2, add B to P3, and then subtract A from L2 and subtract B from. Now, when you add second coordinate, it's the same as it was, and it's the same as it was, but the right-hand sides are using only first coordinate. So this should become a different P1 and L1. So that's obvious, that's obvious that if you have this graph that it must contain this group. So in particular Q square has to divide the order of this graph but when you have three and here two and three correspond to number of variables, then you can easily say that this is an automorphism, but you cannot add constant to P2 and subtract it from L2 because P2 will be substituted inside of function G and it may not work. You, so an often it doesn't. Therefore, this graph for sure has only one, has smaller subgroup. It will be this F sub Q here and Q divides this. Okay, all right. So sometimes a Q lift of a graph has larger order of automorphism group than the graph. If you lift to the graph, we don't need this one, has larger order. For example, if you start with prime field of P elements, then if you just analyze automorphism group or by a fine part of projective plane, it's not too hard to see that that's the answer, but when you lift it, actually the number is bigger, P is odd. Actually, every automorphism of this graph can be lifted to the automorphism of this graph, plus this graph will have a little bit more automorphism. So that's, so the group can grow when you lift. On the other hand, the group can become much smaller. And we take the same by a fine part of classical projective plane, but function G is, you know, this. And now we can prove that the automorphism group of this graph is small. This graph appeared in the dissertation of Ben Nassau. But he couldn't prove that this graph has only P-automorphism. So this was done by Tarantuk and myself, and the theorem is this. If P is one, what's three, then the group of automorphism of this graph are, are stands for rigid, it's small. You see, we have to have those P-automorphism, that's addition on the third quarter, but that's inevitable from this type of construction. We will have them, but we don't have any other automorphisms. And because of that, the, because of this, the order of this group is divisible by P, but not P-square. So it cannot be of this type. It cannot be of this type. So this graph, you know, has function which has three variables, but it cannot be rewritten as a function which has two variables, which is not surprising, as I said, because, but this is explicit. And what was another not surprising that it was hard to prove. We checked it for all prime powers, even up to 41, and program shows that the group of automorphism has only P elements, so there are only trivial automorphisms, those translations on the second, third quarter, but this graph, but the proof was not easy. And I mentioned the paper already, it's an archive, and it's a little technical, but the idea of the proof is to show that this graph, there are point lines, there are two very special lines, and you take this three neighborhood of a line in this bipartite graph, and there are two lines, specific lines, such that one line has three neighborhood larger than any other line, and second line has second largest than any other one. So these are very two special lines, so every automorphism must map them to themself, because they are winners, they are absolute winners, those two lines. But having only two lines fixed by automorphism is not enough to say that you can fix everything else. And that was a lot easier for now. So that's what I want to say, that's Benjamin Nasser, and that's Vladislav Tarantyevich. Now, I'm saying now open problems and I will finish. I was saying that the whole subject is great fun because you get problems, you never kind of think about fun relating. Of course, one main problem is this, just can we lift this classical plane to something, to graph of pure state? Can we make it by Q? That's a big question, it's not answered, and we cannot answer them all. But trying different, you know, H is usually to prove that we cannot, we cannot, we came to some questions about permutation polynomial. So suppose F is a function from FQ to FQ, and K be positive integer, you can consider this function. And the question is, when it's bijection, it's very easy to check that if K divides Q, meaning that K is power of P, then it's a bijection. But if K doesn't divide Q, it's not a bijection, and we couldn't prove it. We tried very hard, and I ask, you know, all people who do, who work in finer fields and with permutation polynomials, and it's still open. So, and I really spent a lot of time in that. Another function is this. This function is a little preter, the first one, because if you look at its structure, it's what is written here is a discrete derivative of monomial X to the K. So you take monomial and you multiply it by its derivative. So it's like Y times Y prime. It's not by design, it just turned out this when we studied some six cycles, some graphs. So that's, you know, kind of, and there are many questions of this time. Another big question is this. Suppose you go to four dimensional vector spaces your points and lines, and you want to find functions of two F3F4, two variables, four variables, six variables, such that when you consider this graph, it will have no eighth cycle. If you manage to do this, then you will settle, you know, this 60 plus year old question. What is the total number of the eighth cycle? Because in this graph, the asymptotic of the number of edges as function of number of vertices precisely this. It will be Q regular graph. That will be this. So, so far we, I cannot do it. Another people who try could not do it. That, and what is interesting here is that you try different thing, it doesn't work, doesn't work, doesn't work. But even to show that it doesn't work for some very, very particular functions, these are very enjoyable problems. But of course, it's interesting whether they exist or not. But that's the closest idea I have, for example, attacking this question. I have no other ideas how such graph can be constructed, but this, you know, didn't work. Codexon, Cron and Talon won't. Very recently just came to this question, which I wanted to present to you. They submitted the paper. Now, instead of finite field, take real numbers. So these are exactly the same graphs as we discussed before. They have two equations. The first equation P2 plus L2 equals that and P3 plus L3 equals that. But now we do it over real numbers. So graphs are infinite. And the question is, are these graphs isomorphic? When you have L1 square here and L1 to the fourths. Over complex numbers, they are not. Over finite fields, they are not. But over reals, we cannot prove it. And we cannot understand really what we, I say, that's their problem. They came up with this question, but I tried a little. So these are infinite bipartite graphs. Cardinal, just Valency discontinue, of course. And what, yeah, they have, both of them have four cycles. So they are not graph, you know, so these graphs, no, not four cycles, six cycles, I'm sorry. They cannot have four cycles because they begin with projective plane and that is leaf of them. So projective plane has no four cycles. So that is the question. So these are these three young men who came up with this very interesting question. They could answer some other questions over R. And then the last problem, that's the whole problem of Dominic Vecain and Lassie Seke and that problem to study this number over the family of all bipartite graphs with partitions of size M and M. Lassie, do you remember this problem? Yeah. So here we want to find bipartite graphs with partitions of size M and M that maximize the number of edges if we don't have C4 and C6. So these are, this is family of all bipartite graphs. In particular, when M is this, Erders thought that these graphs, this number should be at most linear. And the reason Erders thought about this, because if you use just very, very simple probabilistic method, it will show you that you could expect the number of edges to be linear. So probabilistically, you will get linear. But Vecain and Seke, they just proved this conjecture and they really constructed graph on so many with so many edges. And without C4 and C6, when M was of this type. And then with the steaming can welder, we could improve the lower bound a little bit. And we used some Q square lifts of induced subgraphs of this biofine part of projective. It relates to another quadrangle, but it's not without, it's a bi-regular quadrangle. There are such, but this, in view of these equations, which I wrote, that will be exactly this and I can write equations precisely. And that produce better lower bound. Now, the question is here, narrow the gap. I think for this problem, this upper bound and lower bound are too different. That's kind of, you know, C4, C6 forbidden but the difference is so big in what is known. I would say that this upper bound should be reduced, but I certainly don't know how to do it. You reduce to what I don't know why. And that's Dominique and that's Lassie. And that's all. Thank you very much for your attention. Thank you very much, Felix. If we could all thank Felix in some way. Thank you. Thank you for citing me. Oh, my pleasure. My pleasure. I like this picture. I didn't write, you know, the year you were born, I couldn't find it on the web. So... 55. Yeah. So you are a young fellow. All right. Does anybody have any questions for Felix? Ace, I have a little question. This permutation polynomial question. What's going on there? I mean, is this, if K does divide Q, is it easier to describe what permutation is? That's very, if K divides Q, so K is the power of P because Q is power of P. Yeah. So then when you use Frobenius automorphism, so X will be to the power, say, P, P squared or P cubed. So that will become trivial. That will become trivial. You see, because this will be X plus one to the, say, suppose K is P, then it will be X plus one to the P. It will be X to the P plus one to the P minus X to the P. And it will be just X to the P times one. And X to the P, that's permutation polynomial. The same piece squared. So it's in one direction, it's trivial if K divides that. But if K doesn't divide that, it's so far, we couldn't do it and spent a lot of time on other people. I try to popularize this problem, especially because it's like function, monomial times its derivative. But as I say, it's not by design. It's just, we try to try to try some functions H and try to make your state and we came to this question. The same with this polynomial. It's not as good with this. But there are a million of simple questions, which sometimes, you know, as I say, you forget what you're doing really, and you just want to like this question. I want to see what am I missing here. Are they are some more, you see, computer is useless here. Unfortunately, over finer fields, you just put in the machine and it tells you, but here, how you do it? Over algebraic closed field of characteristic, algebraic closed field of characteristic zero. They are not, I assume. Complex numbers in particular. Over rational, I'm 100% sure they are not. I didn't try actually, I'm sure. But over reals, I don't know. See, what's very, yeah, okay, I don't want to take more of your time. But with reals, you know, what is interesting that if you take additive group of real numbers, additive group, and you take additive group of two-dimensional vector space over real numbers are two. These groups are isomorphic. These two infinite groups are isomorphic. Just additive group of reals and additive group of two-dimensional vector space over reals. They are isomorphic additive groups. And so, I'm not sure what's going on. I'm still here. I can see you, but you cannot see me. Any other questions for fields? If not, then thanks again for a wonderful talk. And I will, I think we'll end there. And I'll see everyone else next week. Felix, maybe. Thank you, Stephen, very much. Yeah. What did you want to say? I was going to say maybe we'll get a chance to Zoom chat at some point. Of course. We'll be very happy. I just want to end one more sentence. I forgot to say that similar equations define directed graphs too. Only they're not bipartite. You just take vector space and you write this type of equations. And in the references I gave, I think digraphs are much more interesting because digraphs even with one equation produce such a variety of non-isomorphic example, much, much greater than say bipartite graph with one equation. So I just wanted those who need some interesting digraphs which appear naturally in this, in this setting. Let me try. Thank you. Sorry. Thank you very much. Thank you. Thank you. Nice to see everyone. Bye-bye. Bye.