A combinatorial description of knot Floer homology (contd.)

0:00 So now let's write down the complex. So the pillar complex is generated by intersections of this tori, of the product of the alpha and the product of the beta, meaning that it's generated by things like x1, x2, xn, and zn. Sigma is the permutation of Xn, and Xij intersects beta. So each alpha intersects a curve of each beta at one point, and a generator is a collection of n points. I'll show it in blue. Basically, this is called the hypercash math law. Right, but this is the same as the symmetry group on any generator, so it has n calculators in it. Okay, so I also know what is the differential, so the differential, so in order to define the differential, let me...

2:30 The differential basically counts empty squares, so you want to count things like this, and disappear only when two generators differ by a transformation. So let me call something like this n. So if X and Y in X differ by a transposition, then they are connected by two rectangles, R1 and R2, meaning, right, so there is one, so X and Y differ by a transposition, some point in X and this is some point in Y, and you have some other points at different times. We have one rectangle here, R1, and another one that goes around the course. But this is still a rectangle if you look at it as a course. So these are potential differentials. So let's denote RxY to be the set R1, R2. These sets are either the same or they work if they differ by more than one aspect. This is the set of rectangles connecting X and Y. There are two more of them. Yes, there are two more, but this connects Y to X. So, yeah, it's just some more information. How do you tell which way is that?

5:00 How do you know which way is that? You want to have distribution. Like, if you go from X to Y on the alpha curve, you want the domain to be on the left. So now we can define some quantities, so if r is some rectangle like this, or in the grid diagram, then we can define d of r like that, that is sometimes in black dots, and w of r if it's a number of white dots, and we can also define the number of generators if we have a particular x, like in this case. The rectangle and our generator X, our X is the blue dots. We can count how many of them are inside, just in the interior. That is one. The number of generator points in X, again inside R. So we have a differential, we have one pseudo-homomorphic derivative of it, if and only if we have an m-derivative. What were the blue dots again? Oh, right, so this depends on some vector x, some thing in here, like a collection of blue dots.

7:30 But are these x? Yes, they are the x. Oh, and that's where that depends on x, so x is the name of the generator. Yeah, some element, yes. But basically, for this to be a differential, you don't want anything inside. The differential of x is the sum over y's of y. You can have a little y here. And you have a differential only if they differ by a certain position. So if this is not empty, then you can have a multiple. You can round this, so it counts, by one. All of these things are zeroes. So PR is zero, W is zero, and all of this is two times one, or it becomes zero, and this, so now you can take the homology of this, homology, Spencer, Webb, and minus one copies of the two-dimensional spectrum. I'm working on room mark two. So there is a sign. Well, there is a sign and I don't know what it is. Oh, here it's Mach 2. Is Mach 2 enough to get the genus? Yes. No, it should be defined over C. I mean, the usual math homology is defined over C, but the version with more than two base points, well, we only define it over Mach 2.

10:00 So, that's given off the code for now. So it's the same as Geo-X, right? If you have something that, if X and Y do not differ by a transposition, then this is M-O. If they do, then it's the same as Y. Okay, so right, so maybe I could say the two gradings. So we have Alexander and Maslow gradings. Amplified, differ by a transposition, two permutations multiplied by one by an inverse. Yes, yes, yes, yes, exactly. Basically, yeah, they differ in exactly two positions. So the Alexander and the Maxwell gradients, so let me just define relatively if x and y differ by a transposition, Let's pick some R, or a tango between the two, then the relative, so, right, so to define the relative Alexander gradient, you just have to say what happens when the supervisions, it is, it is V of R minus W of R, so you can count black and white dots, and you get the Alexander polynomial, and for the homological gradient, black and white dots, and you get the Alexander polynomial, and for the homological gradient, black and white dots, and you get the Alexander polynomial, and for the homological gradient, black and white dots. The number of generators, the number of... So when you have a differential, then all of these quantities are zero. So this is one, and this is delta, preserves A, and decreases N by one.

12:30 So it's a differential, and it preserves the equation, the Alexander equation. I do? Oh, well, all right. Well, maybe I should make some remarks. Why is that? Well, basically, if you have something in D and you compose it by something else in D, you get either this, or you basically get two squares. This is the first time you apply D, And these come in pairs. You can also do this 1-2 or 1-3. What I'm saying is that all contributions to this work come in pairs, so then math 2 is 0. So you can check this from here, and you can also check that it's not invariant. And this is true because it was known for Floer homology, but you can check it by the sequence of...

15:00 I guess they were invented by Cromwell and also studied by Winnicott. Relate any two diagrams of the knot. Any two grid diagrams of the knot. So kind of like the Reitermeister moves. And you can check that just purely combinatorially if this gives an invariant of the knot. Right, there are other versions. There's another version which is called HSK+, which contains more information. And in particular, this is HF+. So these are versions of Heegaard Floer Homology for surgeries on knots. So this is all now combinatorial. And there is some related work of Shichori Shorikawa and Gerson Wang to prove that they found a somewhat different way of showing that HF had of all three manifolds. Take what they gave for combinatorial description. So what's not yet done, so not yet done would be not worthy, I mean this thing's not yet done, but you want to find a formula for them. This is a cross version for knots, for 3-manifolds. So this was done for knots, and the half version was done for 3-manifolds, but this one is going to be more complicated, and this is what you need to get in variance. So if you do that, the next step is to define the Orpah-Sovereign diet for 4-9-quad, which are fundamentally the same as hyper-Witten, but for this you put in 8-plus. And other questions?

17:30 Yes, yes, yes, yes. Right, yeah, the problem is just to define, to define the sign for the differential. Like if you have a rectangle, what's the sign for it? But if you just count all your characteristics, it is the Alexander polynomial. The permutation signs don't work. Oh, so you have to use some other sign. Yeah, I don't know what they are. It's an open question. But it's more complicated than the permutation. Personally, I think the formula you get this way for the Alexander Polynomial is the prettiest one there is. And I'll try to justify it in my talk in one slide. Can you compute this concordance environment from this complex? Yeah, good question, yes. So this is, you have to use this illustration. So you have to use more than just what I said here. I'm sorry, what of this is in your preprint, what is not, what is, and where it is? Oh, everything is in the preprint except this, except this, so this is not. I'm going to tell you what's here. The crumb of the nickel. Are you writing? Here. Yeah, and I should say that I'm also trying to work with Dylan Cushman. How about a two-minute break, and we'll start again.

20:00 Okay, so then we'll get the Alexander phone over here. I think there's some of them. Yes, it's beautiful. So, that's the only statement I have. Thank you for your attention. Thank you for your attention. It's not sort of ready-for-the-public, you know. The pubs that I show now.

22:30 No, no, no. It's on the web. It is ready-for-the-public. It's not... It's kind of a cramped thing, you know. It's not something that you see. It's not something that you want to see. It's not kind of clear. No, that's all right. I'll play with it. Are you a removable disk head? I must be a removable desk head. No, you're not a removable desk head. You're a removable disk head. Sorry. You should. I'm sorry, what? Any chance of doing dinner this evening? Oh, are you already? That would be great, because tomorrow we could get together and have a time of our own. And also, I could listen to your recapitulation. It looks like a very interesting conversation to have during lunchtime. Okay. Let's start again. Oh, sorry. So before we start, Dr. Fannin is going to be on in a minute. Namely, we are going to revise our list of problems on virtual knots. It's just on the web. And so if you have any interesting problems related to virtual knots... We would like them, we would like you to communicate them to us and we will get them into some problem solving. I think you all have our email addresses.

25:00 Yeah, when I say virtual I mean virtual and related things, welded knots or if you come up with a new definition of virtuality, whatever. Okay? And as I said, I think it's easy for you to reach us, so we won't write down email addresses. The next talk is Young-Woo Rang, and he's going to talk about the Khovanov-type homologies for graphs. Thanks to the organizers for their invitation. I feel kind of nervous right after this exciting talk. So, I want to summarize some work that I have been doing in the past few years with various collaborators in analysis and categorification of a common homology theory for graphs. So, specifically, we will talk about homology theories for chromatic polynomial, for the Tufts polynomial, for the Ulaanbaati-Wealdom polynomial, So starting with the chromatic polynomial I want to give a quick review of what the polynomial is. The chromatic polynomial was defined, the motivation was trying to solve the four color projection. Basically you define for each integer lambda, for each integer lambda you define the number of colorings or ways of coloring graphs. So that adjacent vertices have given a color. So that's a key denominator. And it is well known that it has a state sum. In general, the general principle for Khovanov's theory is that when we have a state sum, we have two n-terms and we have plus-minus signs there, then most likely we can find some way to characterize that. We can find some cohomology theory corresponding to that. It is also well-known that this economy has an opportunity to be competitive. So, to explain the definition of the topology theory,

27:30 we try to look at the sums carefully and organize the terms in the following way. So if you take the empty eggs as a three box, and then you have one egg there, you have this egg, or that one, or the third egg, or you can have two eggs, or you have all three eggs, and then you simply take a number for the number of components, and then you take the summation as the polynomial. So the kind of representation is pretty natural there. You basically assign, well, you fix a graded algebra of the ones that still have some obvious conditions, which means the dimensions at each level is graded in the final, so that one grade seriously is well defined, the Q dimension is well defined. So what you want to do is to, for every component, remember every component contributes a lambda. So you just assign a copy of the algebra to these components and then take a tensor product. This will, if you take the two dimensions, this is exactly lambda to the k power. And take the direct sum, then you have the i's, chain moves, and now you want to define the differential from c i to c i plus 1. In order to define the differential, you need to understand the math going from this picture to that picture and so on, from one to the second. So what happens is that sometimes the number of components will remain the same from here to here. They're both two components. You define the math to be the identity math, but sometimes you could have two components that are connected by an edge. In that case, you have two copies of algebra A go to one copy of A. So you need a map from A tensor A to A, and here you need a multiplication map on the algebra A.

30:00 And it turns out that this is all you need to have a homology theory. You want to prove that this does define a differential and homology groups are well-defined for graphs. All right, so this is the definition for the homology theory. It has some properties, so the most interesting property is the This can be considered as a categorization of the deletion-contaction rule. Remember that rule relates the graph G to the graph obtained by deletion, the edge of E, and also contraction E to a point. Consider those three graphs, and then there are some natural maps. You can prove the chain maps, going from here to here, and also from G over E to G. Basically, this is just an inclusion map, and this is induced by the inclusion map. This one you just expanded the E from one point to the whole X. You can show this is a short exact sequence, and so bystander of cohomological algebra, this will yield a long exact sequence on the cohomology groups. We use a map between graphs. The map is from a subgraph to a logic graph. It just means every vertex here is in here, every edge here is also here. Then we use a map on the homology groups. There are some relations with Khovanov's non-homology. Some of the relations are written down in the paper by Joseph and Laurie.

32:30 We know who is my collaborator in defining the grammatical homology group, myself, and also Joseph has found some connections with writing the homology group, and this has in particular motivated Khovanov to rewrite his homology group, to rewrite the Khovanov-Rozansky homology using this new language. Without using matrix factorization. So I would say this does have some applications. Alright, let's look at the next polynomial, the Tufts polynomial. Remember this is a two-variable polynomial defined by Tufts. So you can consider these to be the scaling relations that define the Tufts polynomial. So again, it has a division and contraction rule written like this. But there are some complications if the edge is a isthmus or the edge is a loop. So using those skin relations, eventually you are reduced to a graph without any edge. And you simply define that to be 1. Using those skin relations, you can write down a state sum. So you have a state sum here, you have the n terms. You don't see the plus minus sign yet. You can always create one, but I think the main problem is that the powers here don't have a topological meaning. So let me explain these terms. rs is the number of vertices of the graph g minus k, s where k is the number of components of the spanning graph coming from the edge set. This term, this summation is not entirely convenient for us to write down a cohomology theory. What we did was rewrite the summation into a different format, which is more convenient for us. So you can rewrite the topology in the following format.

35:00 So basically up to some factor, up to this factor here. It is essentially another polynomial which we call T-hat. T-hat has the following nicer form. So you have two to the nth terms plus minus sign here. And then you have something to the b0 power, something to the b1 power. Here b0 is simply the zero's value number. b1 is the first value number. Now we are topologists. I mean, I'm a topologist. I know some people here are algebraic. But for topologists, these terms are much easier to handle than those other terms, like RS and combinatorial terms. So this makes more sense to me, and it will help us to write corresponding homology groups. I'm not going to go through the proofs here, it's not too hard. All are characteristic equationally. Okay, so essentially you have a state-sum term involved with zeros and the first value number, and that's pretty much all you need. So we're going to define a chain complex where every chain group is a bi-graded module, V-module. And you want all the characteristics equal to the T-hat, which is essentially the task polynomial. Well, you need a bi-graded, so you consider two algebras, A and B. B is really just a module, you don't need an algebra structure, you just need a module structure. They are both bi-graded, so there are two directions. The degree of x is defined to be 1 comma 0, and the degree of y is 0 comma 1 in the other direction. Write 1 plus p here. Sometimes I'm using the letter 1 plus x. It doesn't matter which letter it is. So, for every s, subset of the x and e, you have this Spanning subgraph, which is the v, e, union, s subgraph, Spanning graph.

37:30 You use the same trick as before. You assign a to each component and then take tens of product. This essentially gives you the first term, which is 1 plus x to the b0 power. And then you also need some algebra, which gives you the other term, 1 plus y to the b1 power. And so you simply define, assign this algebraic object, maybe just a Z, a Z-module, tensor product, the tensor product b1 times. Then test the product all together and they just direct the sum. This is your ice chain group. Well, you need to define the differentials here. You need to study, again, just like before, you need to study what happens when you add an edge e to the edge subset. There are two possibilities, two different pictures. The first is that B joins two different components. In this case, the B zero changes. The zero's value number decreases, but the first value number does not change. So the map on the B part is identical, but on the A part you need a multiplication. The second possibility is B goes back to itself, in which case zero's value number doesn't change. So the A portion is identical map. The map on the B part, you need a map going from this to B tensile, the number of tensile vectors in 3.1. You simply use the map B goes to B tensile 1. You can use some other map. This map seems to be the most obvious map, simplest map you can use. Anyway, if you do that, then you can show that this does define a, give you a well-defined homology theory as well. All right, so that's for the task polynomial. Next, let me explain the muller-busch real polynomial, which we just heard that in the volume from Sergei's talk.

40:00 I think this is pretty much the same. For a so-called fat graph, it's a graph with the additional structure of cyclic ordering structure on every vertex. And again, we are topologists. It's easier if we can see the picture. So you can, out of this, you can construct a ribbon surface. And this basically shows the fat graph structure there. The deformation rate pattern of this is the graph, but then the ordering can be seen from the surface. So, Sergei Chemotov and Igor Peck, they wrote a paper which was discussed in the morning in Sergei's talk. In particular, they give the following summation, again you saw this in the morning talk. So, this is the three variable polynomial, so you have a summation, again you have a state sum here. X to the sum power, Y to the sum power, Z to the sum power. So for us, in order to establish a cohomology theory, we want to rewrite this in a form that is convenient for us. After some tedious algebraic work, you are able to write this in the following way. So RGXY is equal to the following. You have two set terms, and then something to the B0, remember B0 is zero's value number, and then something else to the first value number, and then something else to the following term, this is the BCS, which means the number of boundary components on the surface, and the surface that correspond to the ribbon surface that I showed you. So those are the three critical topological terms related to the polynomial B0, B1, and Bc.

42:30 Alright, so we're going to work on that. Now here we have a, we need a triply graded chain, we need triply graded chain groups. In different directions, the degree of x is 1, 0, 0, 0, y is 0, 1, 0, degree of z is 0, 0, 1. For b, you just need a module structure. For a, you need a standard multiplication structure. For c, you need a co-multiplication structure as well. To construct the graded chain complex, first we define the chain groups. These are natural. For every spanning subgraph, or equivalent, for every s, a subset of the, sorry, this shouldn't be easy, well, I guess it depends on how you call it, even if it's a spanning subgraph, you can call this s plus a, for every s, any set of this g, you do the following, for a, you just assign it to each component and then take tensor product, for b, you do the same as before, you just tensor product b one time. For C, you assign a copy of the algebra of C to every boundary component of the surface. Then you take a tensor product. And then the tensor product all together and then you direct the sum you have there. I is qian gu. You need to define the differential. Basically, you have three directions to worry about. One is the... Yeah. You want to consider when you add an x, what happens? Well, you want to know what happens to the a factor, and then b factor, and then c factor. For a factor, it's the same as before for the case for the chromatic polynomial.

45:00 Either c component goes to one, in which case you need a multiplication, or one component goes back to itself, so you just use identity math. For b, you use the same trick that I just showed you for the past polynomial. So either the first value number doesn't change, in which case you use an identification map, or the first value number increases by one, in which case you use the same map as before. And finally, you want to explain what happened on the factor coming from C. Well, again, you have two possibilities. So here the picture, the circle here represents a boundary component of the surface. So you could connect a boundary component, two different boundary components. In that case you have two copies of the C goes to one copy of the C. Therefore you need a multiplication structure on C. The other case is that E goes from the boundary component back to itself. In which case you create a new boundary component here. Therefore, you need a map from C to C tensor C, and you use the co-multiplication map. And once you have all the three maps and tensor products, you can show this defines a, well, a graded chain complex. And the OLA prep curve, graded OLA curve, this one does give you the Bula-Basch-Ryudon polynomial. Two more minutes or so? Oh, you've got... I've got more than two. Four more, yes. I don't have a lot. So more recently, it wasn't something that I need to spend some more time. I had some chat with Khovanov on another polynomial called Penrose polynomial, which introduced... What's point three of your theorem? I'm sorry, this is just to say the gradient, yeah, the ordinate sum of the gradient dimension is the polynomial, the rg hat is essentially the same as the rg, which is a vr, so Khovanov explained this polynomial, called the Penrose polynomial, and suggested that there should be a way to calculate it.

47:30 I don't think this would really... Not many people. I've never heard of this before. Not well informed. So, apparently, Penrose, back in 1971, wrote an article with a strange name, a notation of negative dimensional tensors. And the idea of Penrose polynomial is implicit in this paper. His goal was to prove the four-color conjecture. At that time, it wasn't approved yet. But later, people developed a further theory on that, and they call it Penrose-Marzogna. I think this state of science is probably due to Jaeger. Is somebody named Martin? Penrose gave estates. Oh, Penrose gave estates? Yeah, somehow Jaeger wrote something on that. Anyway, so this is the state of sound here. You start with a plane graph of G, and then you do the, this is standard, do the media graph of G, so what happens is that, this is the, for example, this is G here. For every edge, you put a vertex here, and then just cross, you just draw a cross here, and every, and that vertex, then just go around every plane region. So this picture becomes this picture, okay? And then you color shade this, the complementary region, so that the infinity region is unshaded, so you have a well-defined shadiness or unshadeness. Now you have the following scan relation. For every vertex here, you either resolve this way or you just resolve the vertex that way, so the vertex is no longer here. And after you have done that for every vertex, then there's no vertex, so you have a bunch of circles. If you find a circle, each circle is an x. And then, if you just repeat this relation, this could give you the following state of the sum. Yeah, I think this is two digits, but what Penrose did was on trivalent graphs themselves.

50:00 Oh, okay. So CR is crossing a circle. Minus one. Number of crossings. Crossing. So this is called a crossing. This one is no crossing. And basically this is where the minus sign comes from. This contributes the minus sign. Okay, so you try to categorize by that, but we don't have a complete categorization of that. This part I think is obvious. You want to define the chain groups. You just define some algebra. Let's take this one for example. So you simply assign each A to each circle. And then take the Penrose product and then take the summation, so you have a C, you take a direct form, so you have a CIG. And then you need to define the differentials. So boil it down to the following picture. And if you study this carefully, there are three cases. Either you have two circles go to one, or one circle goes to one circle, or one circle goes to two circles. We've tried different things, but somehow they don't always work. And this is what we have. It's not very strong. So here you take the multiplication map on A, here you take the map that multiplies everybody by x. This kills x, but one does x. And this is the zero map. And yeah, this will give you some homology theory... The Euler characteristic should give you, this is different from, this is the Penrose polynomial, not the, the Penrose polynomial should be evaluated at a gradient dimension of a. However, because of this multiplication we use, notice this is not degree preserving, so you have to forget about the gradient of a. Once you forget that, then the dimension of a is simply 2. It's 1 and x. So you get, you get the following, you get all the characteristics in the Paris polynomial, you better add it too. This is all we did. I think that I should think about this more.

52:30 All right, I want to briefly mention some other related work. Some of them might be... First of all, Eastwood Hodges. Again, this was, I learned this paper from Kovanov. So, apparently they had a different construction. Completely a different idea from Khovanov's, a topological configuration space constructed for every graph, they construct a topological space so that the Euler characteristic is equal to quadratic polynomial divided by a given integer number. You may say this is nonsense, because this is nothing but an integer. Of course you can find a topological space whose Euler number is a given integer. But they have a canonical construction. It's a standard construction, very canonical, and they satisfy some naturality, a functorial property, and also some non-exactive sequence, similar to the one that I gave you for our program. So we don't know the actual connection with Khovanov's. There's also a paper by Jeremy Martin, who is trained as a combinatorics and algebraic geologist. If you find something called a picture space, and it has to do with the algebraic geometry and some high dimensional topology, he showed that this whole number has to do with the task polynomial. Then there are two other groups of people, Stasic and Lobo-Mofart, they did something just along the same way, but various polynomials. So, one can ask me the question, you know, why do I want to bother with, you know, that Khovanov theory for graphs? Graphs are usually to be distinguished. And so I think my answer is that in graph theory, people have a different emphasis from people in topology. People in topology like to find a powerful invariance that distinguishes, you know, different manifolds. But in graph theory, they want an invariance that... So, for those homology groups, we would like to understand what are the combinatorial means and do they have any applications at all.

55:00 I don't think it can be powerful enough to prove the four colors here. But if you can find some application graphs here, that would be nice. You can ask whether there's any relation to this morphological connection. We know, we also know there are some connections to Khovanov's and not Khovanov here, but there's more to be studied, and if you do two and a three, if you have done substantial work in two and a three, there might be some hope to give a topological reconstruction of Khovanov's theory. Without using cohomology, I would say it's a sort of long-term ultimate goal. That's all. Thank you. Question or comment? Penrose's original formula counts the number of colorings in the graph when it's planar and trivial. And fails when it's not planar. And that's interesting because the counterexamples are off the plane. There are graphs that can't be colored, they're off the plane. So there's a subtlety there. It's conceivable that something about that subtlety might be caught by categorifying. Right. I mean, the state sum certainly uses the fact that it's a planar graph. Yeah, you have to go from the plain picture, the planar picture of the graph, then draw the media graph. Other questions? So this Khovanov homology has this property that for alternating knots, it is computed from the Jones polynomial. Is there a class of graphs with similar things? Yes, so for the chromatic polynomial for all graphs. And for the cohomology theory with A being the simplest one, one comma spanned by one X with X squared equals zero, with that particular algebra, then you have, I think it's in an article, certainly his son, Michael, and myself, that you can, you know, this theorem carries over. Yeah, that's your question, right?

57:30 For algebras or is it some similar? For all graphs. All graphs are alternating? Yes, all graphs are alternating. But it's just for that special algebra. But on level 3, is it to find context activities? But only for the first three. There are three theories with that. Yes. So Joseph did a lot of computation for other algebras. It is only for chromatics, for thought and for all of us. So there should be a notion of graphs which are, I guess, I don't know, thin. No, I think the thin comes from the algebra. The algebra, the simplest one, 1 and x, is very thin. So the line exactors will imply the homology groups are concentrated on those two diagonals. That's the starting point for the least isomorphic theory. It has to do with the algebra. But for the non-opposites, multi-indices, multi-gradients, generalizations, etc. Thank you. We should move on immediately to Pat's talk. All right. Well, all right. Two minutes. ...in two minutes. So you want to go out to dinner? Yeah, that would be great. If that suits you, that would be terrific. We could have a dinner over in that same place when we have lunch, I think. Okay, well, we could sit together. That would be great. And I know that Andre would like to... Or we could go and find a good place. Whatever you fancy. I know Andre would like to tag along, because he also missed your talk on the first day,

1:00:00 so he's not getting it, and he'd be very interested. And some things about categorification in general that I'd like to... The only problem is that Andre is completely skidded, he doesn't have a beam at the moment, so he wouldn't be able to afford a restaurant, so we're going to have to treat him, or we revert to Plan A. It's quite comfortable eating in that place. Oh, I don't mind. It's great. I get more than happy with that. Just that I understand that there has very recently been... A categorical, I mean a categorical theoretic proof of the four-colour theorem. No, it was presented in two seminars, two lengthy seminars at René Guittard's category theory seminar at Jussieu earlier this spring. Yeah, yeah, yeah, it's on the web, I can refer to it. But I'm trying to think of the name of the guy. I've got the paper in fact and I've been trying to study it although frankly it's... It's beyond my hands because there's a great deal of very elaborate, so I don't think there's much to underline it. And it doesn't, a guy, I'm just struggling to record that, I'm sorry I didn't get any sleep last night at all, I haven't got there since quite this morning. One of, a guy, I think, certainly a French mathematician, not a French mathematician, a guitar... I think he was a student of guitar, he said, well, don't know, the name will come to me, I can probably check. You aren't carrying the paper, are you? No, no, no. Yeah, but if I put in a color theorem into the association line. I can probably find it in five minutes for you. In fact, I have to find it in five minutes just by the way I put it in. I'm sorry.