 OK, so thanks a lot for letting me speak at this conference. And also thanks a lot to you for coming. I realize it's the last hour, and you must all be pretty tired. So I try to not give it too hard talk. And also, I have to announce that at the end, my talk will be a little bit experimental in the sense that I cannot answer the main question I raise. Anyway, let me tell you what the talk is about. The talk is about graph complexes, which have been maybe not introduced, but at least popularized by Maxim Konsevich. So graph complexes, in general, are complexes of vector spaces, graded vector spaces with a differential, where? Excuse me? You think that he produced it. So OK, yeah, so there's ever some of some you introduced. And I have to say in my defense that I titled my paper, Maxim Konsevich's Graph Complex. And so on, so I. But yes, so the complexes I'm going to speak about today were introduced by Maxim Konsevich. So these complexes in general, so the underlying vector spaces are spaces of linear combinations, or of series of certain isomorphism classes of graphs. And you can put a differential on these spaces, and the differential is usually defined by either contracting an edge, or duly by splitting a vertex. Edge contraction or vertex splitting. And in the literature, there's a whole zoo of that graph complexes, because there are a lot of different types of graphs you can take, ribbon graphs, or ordinary graphs, or directed acyclic graphs, or otherwise decorated graphs. And these classes will give you different complexes a priori. I will talk today about the simplest version, where you just have ordinary graphs without any decoration. And these types of complexes were introduced by Maxim Konsevich. So let me give you a little bit more precise definition. Let me define a z, let's say, nk, the set of graphs. So they're not isomorphism classes of graphs, set of graphs that they are connected, and the vertices are at least wellened. And I say that the vertex set are the numbers from maybe, if I may change it, make a capital N, because I will use the lowercase n later, numbers from 1 to n. So you have n-numbered vertices, and k-directed edges. So each edge is just a pair of numbers. So you have a list of k pairs of numbers, and this is what I call a graph. Now let's also assign to each such graph a degree. And the degree, of course, is just a set, but the degree will be used later when I construct a vector space formally spent by these things. So the degree of a graph, this is some number n, which I fix, n times number of vertices minus 1, minus n minus 1 times the number of edges. So some number little n, which is a degree of every vertex, times number of vertices minus 1, minus number of edges, times degree minus n minus 1. So this is the degree of one edge. Then you have an action. So this is n for a sum n. Then small is equal to n. Possibly not equal to n capital N. OK. It's like for E n operator. Yes, so this n is the dimension. If you wish, these are vacuum graphs and some field theory, topological field theory, this n is the dimension. So maybe I should call it d, but there will be an E n operator. And usually you say E n operator and not E d operator. So this little graph and curve, the set of graphs is acted by some group. So the metric group in n letters acts by just permuting the labels on the vertices. Then you have action of sk by permuting the labels of the edges and an action by s2 to the k by permuting the direction of edges. Then we define the graph complex. The graph complex subscript n will be defined as a product over all capital N and k of a vector space, spanned by the set gra and k. And here without marking it in the notation, I assume that a graph with so many edges and vertices is put in this degree. And then I take invariance or coin variance. It doesn't matter much, just two different rotations. So coin variance, let's say, under this group action. And here I fix certain sign conventions. I'm sorry for being a little bit pedantic about the signs. Usually just skip over. But it happened to me many times that later people would ask the question anyway, so I just do it. So here I say that I let Sn act by permutations of the decorations of the vertices with sign if this little n here is odd. So you give the vertices an odd degree, so it's natural if you permute the labels on the vertices that you get a minus 1 if the permutation is odd. And these two things act by permutations of this k acts by permuting edges with plus minus if n is even. That also makes sense because we assign, in this case, an odd degree to the edges. So if I permute to edges, I should have a minus 1. And one piece of the sign convention is that if n is odd, the permutation of the directions of the individual edges also comes with a certain sign. For the moment being, if you don't care about this so much, you can just ignore the details about the signs. But remember that here we define two different complexes, essentially one for n even and one for n odd. So the signs matter modulo, sorry, the n matters only modulo 2 with respect to the signs. Let me give you my n deep, and then small, so a nice n. Small n, yeah. So lower case n matters. And your k is changing form. Your k is changing form. It's changing form. Our kids write in the tail of change, age, and so on. Yeah, like this. S e to the power k. Yes, it should be k. So maybe, yeah, the students always complain when I write this k in my lecture. So maybe, so let's just give some example. So elements of my complex are just some linear combinations of these morphism classes of graphs. For example, this one plus, I don't know, 3 half times this one, let's say. But one thing to notice is that if the graph has an odd symmetry, and odd symmetry using these sign conventions there, then the graph is 0. So for example, for n even, we know that the edges are odd. And this graph has a symmetry by interchanging two edges. So this has an odd symmetry, and this graph will be actually 0 in GCN for n even. So it's important to keep in mind that not all isomorphism graphs can appear, but only those with. OK, let's say k is, for all that I will talk about, let's say the characteristic of k is 0. But you're right for characteristic, not 2. Anyway, so the important thing to notice is that whether n is even or odd gives you a different complex. So different isomorphism classes will be present in this complex and others not. And then there's a differential, let me just give you the formula. So if you want to apply the differential to some graph gamma, what you do is you sum over all vertices of your graph. Then here's your graph. So here's the vertex you have. Then you delete this vertex. In place of this vertex, you put two vertices connected by an edge. And then you sum over all this of reconnecting these edges to these two vertices. And these designs above have been chosen such that delta squared is equal to 0. So you can check that with these nine conventions, this equation holds. And then the graph homology is just the homology of this complex GCN with differential delta. Is there any reason why in a definition take direct product and not direct sum? Yeah, there's some reason, because I want certain statements that I'm going to present later to be true. But there's no reason in the sense that I could have equally well taken a direct sum. So we will see, maybe it's a good remark. So we want remark. So this differential does not produce bivalent vertices. So vertices are valence, too, if they were none before in the graph. So delta does not produce bivalent vertices if they were none before. So basically, let's say here, I draw some portion of the graph. And now you split the vertices on principle, you could produce something like this. But an equivalent picture is also obtained, or the same picture is also obtained by splitting this vertex. And if you compute the signs here, you see that there are always such that these two terms cancel out. So it does not produce any bivalent vertices. And there is some result or proposition or theorem, which was already in Max and Conceivages paper, that if you compute, if you want to compute this homology of this graph complex, it can be written as the homology of the at least trivalent part. So all vertices are required to be at least trivalent, plus the parts that are spent by graphs with all vertices bivalent. So these we call, let's say, wheel or loop graphs. Let's call this loop graphs LLM, and over here we have M vertices. And here you have some more product that doesn't really matter. I mean, I don't want to get into this. Let's say products, we don't have a discussion. So M equivalent to N plus 1 mod 4. So you have those wheel graphs and the completely trivalent piece. Furthermore, the grading, it's that graded, yes. So the grading is the degree is N times the number of vertices minus 1, minus N times minus 1 times the number of edges. That's the degree of. So for me to take direct product, it's not a gradient, right? Well, you can take a, let's call it a complete grading. I mean, if you're purest, you will call it a filtration. But in the definition of a graded vector space, we place direct sum by direct product. So the statement I'm going to make now will probably settle this. So this delta preserves the Euler characteristic of the graph, right? I mean, betting number of the graph does not change if you split one vertex. So that means that your complex decomposes into a direct product, according to my definition, of pieces for each Euler characteristics. So Gc N greater or equal to 3 to product of Gc N greater or equal to 3 comma E, where E denotes the Euler characteristic. And these pieces are actually finite dimensional. So actually, you have a direct product of finite dimensional complexes. And so the difficulty is whether you take direct products or direct sum are then relatively non-severe, right? So if you take a chronology of this, you take a chronology of every finite dimensional piece. And then afterwards, you can take either direct sum or direct product. It doesn't matter much. And the main problem I want to talk about is this. What's the graph chronology? It has been studied for a while, but currently, nobody knows. We do not even have a conjecture of what it is. And I'm going to talk about some constraints. I used to have a conjecture, but unfortunately, it turned out to be wrong. So I'm going to talk about some hints only. But let me maybe state one thing. So why should we care? So here I presented it as some combinatorial games that you can play. But many people considered it and why is it relevant to more grown-up mathematics, let's say. And the answer is that these graph complexes, so the GCN governs the deformation theory of EN operas. So more precisely, I write it a little bit sloppily like this. So you can take some cofibrant replacement for this EN operat. You can take the complex of derivations and you compute its homology. And this can be expressed through this graph homology. So it's some completed symmetric product space in one class plus the graph homology up to some degree shifts that I will just insert. So here you have a degree shift by N plus 1 units. And then you have to shift back afterwards. So here we take derivations in operas. That's kind of intentional because I don't regard them as very important. But let me know. And then you shift again by N plus 1 units afterwards. So S plus means it's completed and you don't have a 0's term. You don't have a K term. Does this require N is at least 2? Yes, certainly. The deformations of the associative operata are relatively trivial, so I don't. And there's quite a recent result which taint in collaboration with Victor Turchin and Benoist Fress. Now this looks a little bit odd that you have this big product. And in fact, if you consider derivations not in operas but in Hopf operats, be a little bit careful how to define this, then this is just the primitive part of what you see here. So this is the main motivation for studying this graph complex because these N operats are relatively abundant. So whenever you have an N operat in the theory, you're expected to see some shadow of its deformation theory in what you're doing. Yeah, so I should be a little bit careful. So these N operats exist in various versions. First of all, it's a topological operat. You can say, and then you take chains on this operat. So here, by chains, I mean rational chains, I'm not aware of that or anything. And this will be an operat in dG vector spaces. And there you take a co-fibre replacement, take the derivations in operats. But from the topology, because it's a topological operat, you get a little bit extra. You get a co-product in each arity. And if you also consider this and then your space of derivations changes, so this is from the point of view of rational homotopy theory, this is a better object. But this is more complicated to compute. And the question is, if we want to attack this, so we want to compute this open problem, how do we do this? How to get information? So one relatively simple source to get information is that you can derive certain degree bounds. And for each of these pieces of fixed Euler characteristic, you can get an upper and lower bound where the homology lives. So in general, just from the fact that the graphs are at least trivalent, you get an upper bound. That's kind of trivial because there's just no, well, the complex is just 0 above a certain degree. And so the bound you get for upper bound for H of GCN greater or equal to 3 is fixed Euler characteristic. And the upper bound, if that's a degree, has to be less or equal to E minus 1 times 3 minus N minus 3 if I have computed this correctly. So here you will say it's very trivial. But already this gives us some interesting information. If N is at least 3, this is already negative degrees minus 3. But that already tells us that these EN operas for N at least 3 have very few automorphisms. And there are almost no derivations in degree plus 1. So plus 1 is where the obstructions land if you want to construct maps, for example, the formality morphism. This bound already tells you that the only possible classes there can come from these wheel classes. And you can check that at least. E is negative, H is positive. E is positive. No, E is positive. I thought H is positive. Sorry, I'm negative. First passing number? I had it. Let's say first passing numbers. Yeah, absolutely. My calculations I usually define as E is the number of edges minus number of vertices. But you're completely correct. This is only correct up to sign. So this is interesting already. But you also get a lower bound. And to obtain the lower bound, what you actually do is you take this more or less as a definition and you use a different model for this EN operator. If you use a different model, you get a lower bound. I will not discuss it. So second source of information is you can always do computer experiments. Can you turn on the beamer? So the results I'm showing you are computer experiments mostly by Barnett, Tan, and McKay. And some are run by me. I also wrote a program. I don't know which of the numbers were computed by them and which by me. But let's just look at it. And of course, invalidating the Bible codes. This was invalidating the Bible code. I'm not. Ah, OK. Yeah, that may be. So here I have taken n equal to 2. As I already told you, we only have two cases to consider, n even or n odd. So I just took two representatives of an even and odd number. So for an even number, I allowed myself to take 2. And for an odd number, I will take 3 in a second. So here this is a picture for n equal to 2. So these are derivations of the EE2 operator, if you wish. So this accounts the genus. I mean, if you're a graph theorist, then genus is not the right thing. So what I mean by genus is the first betting number. For me, genus is defined to be the first betting number. So here you have one loop, two loops, three loops, four loops, five loops. So for physicists, it's a loop order. So you have all these loop graphs we had before. They are spaced by four units. They live in loop order one. And then you have many things in higher loop orders. What you should recognize or see is that in degree 0, we have many classes. Then there are some on this line, and there's one on this line. And here, unfortunately, we cannot go on much further because it takes too much computer time. Actually, the thing, so what takes most of the time is not enumerating the isomorphism classes of graphs. So setting up the differential, but actually solving the linear system. So that's a bottleneck. It's not computing automorphisms of graphs. Let me also show you the other slide here. In loop order one, you have again your loop graphs here. I omitted some numbers. So they are spaced by four units. Then most of the homologies is in degree minus 3. And then here, there are some things in minus 6. And these numbers are different. So it's not just the same. And there's also no obvious map between these two. I'm confused, Thomas. This seems to contradict your upper bound for b equals 1. But here, n is equal to 3. Right. And then minus 3. The degree is less than or equal to minus 3. And it's only for the non-loop part. So here's at least trivalent part. Oh, and there's wheels. All the left colleges. Yeah, these are just the wheel pieces. Those we already have. And here I indicated the bounds actually by this red line. So everything beyond these red lines must be 0 except for this wheel piece. And can you switch it off again? Maybe. I'm sorry. Can you switch it on again? Because there's one thing I wanted to show. So one thing we can explain is that line over here. I want to show that this line is just this Grotendijk-Teichmuller Lie algebra. So compute h0 of GC2 is the same as the Grotendijk-Teichmuller Lie algebra. And instead of giving you a definition, I tell you a conjecture about what it's supposed to be. So it's supposed to be. It's called the Lien-Drinfeld conjecture, or Lien-Drinfeld-Ihara conjecture. It's supposed to be the completion of a free Lie algebra in generator sigma 3, sigma 5, sigma 7, and so on. And there was a very remarkable theorem due to Francis Brown, who recently showed one half of this conjecture. The Grotendijk-Teichmuller contains this. And if you look at this line over here, so here there's sigma 3. I mean, the number here is the sigma 3, sigma 5. It coincides with the loop order. So the sigma 3 is here, sigma 5. Here's sigma 7. Then you can make a bracket, sigma 3, sigma 5. It's of loop order 8. So it lives here, and so on. I didn't try to co-collect commons, which was the generators of the cell. So they can make complex free Lie algebra. You can see the generators come all the way to the gates. So I can give you an explicit formula for these elements in the graph complex. No, no, no. I mean, because it should be simple, complex. These come all just in sigma 5, sigma 7, yeah? Because for any Lie algebra, you can make this, you can do bar-constructions. And people plot these graphs in this bar-construction. Yeah, I tried that, but I couldn't show that it's actually just a generator. So the Chevrolet complex is relatively difficult. I don't know. But maybe I was too stupid. I don't know. But you not only have GRT, I love it. You have Pazen, you have 3, 7, 6, 7. Yes, you have things above. And you have 1, 1 famous conjecture due to Maxim Konsevich and independently Drindfeld I said, here in the first line, you have all 0s. Fortunately, you couldn't show that, so you had to create this approach to deformation quantization that I did my PhD thesis on. Now can you switch it off? So if you want to get some further information, you can get it by the following route. So on these graph complexes, you have some extra algebraic structures. The most trivial you see, or I mean not the most trivial, but one you see by this realization as derivation. So there's a Lie algebra structure. And it is obtained as follows. So if you have two graphs, gamma and gamma prime, and you want to take the bracket, you take the commutator with respect to some freely product. And to compute this, what you do is you sum over all vertices of gamma. Here's gamma, here's the vertex v of gamma. And you delete this vertex, you plug in gamma prime instead, and you reconnect the edges in all possible ways. Well, it uses the same algebraic structures, actually. So the dual thing is just you identify a subgraph, you contract it, and you take it out. It's just that here we encode the same structure a little bit differently. Here you have a DGD structure, the concrimer. They set up some Hopf algebra. So for me, this is a little difference. So the main difference is that here this is for topological field series. And the difference is that you have a differential. You don't have a differential in the concrimer story. So this gives you a way to cook up at least co-cycles from known co-cycles by just taking the brackets. But you have a priori no way of knowing that these co-cycles are non-trivial. Actually, the differential can be written as a bracket with the Morakato element, which is a two vertex graph. And one more thing you can notice that there's an extra Morakato element, let's say, up to a degree. Namely, you can consider this graph just a tadpole, if you wish, for physicists. You can check that. Yeah, so it's a tadpole who narrowly escaped the fish. And you have these equations over here. And this means that you can define a new differential, which is just the sum of these two. And to make it the right degree, let's take for my even number, let's take 0 instead of 2. So we consider it dc0. We define the differential bracket with this Morakato element over here. And then relatively, not so deep statement, let's just call it proposition. It's that h of dc0 with this extra differential, let's call it delta tilde. It's one-dimensional with just this one class represented by the tadpole. So it means that that adding or deforming the differential like this kills almost all comology. But this gives you information, because in order to compute this, you may take a spectral sequence the other direction, that in the first page of the spectral sequence you take, for example, take the spectral sequence on the genus. And then this changes the genus by 1. That leaves the genus invariant. On the first page, you just have the ordinary comology. And then on the extra pages, you will see things that turn by turn kill all comology except for this. So what you learn from this is that all comology classes that are there have to come in pairs. They have to come in pairs, and they have to kill each other out. So by taking spectral sequence on genus, quantity classes have to cancel in pairs for even n. First of all, there's an analog story for odd n. Can you turn it on yet? So let's see how it goes. And I can just tell you by just looking at the degrees and where these classes live. You see, here you have your Grotnik-Taichmanou-Lea algebra, which is conjecturally free. Here you have the real classes. Now what you can show is, let me maybe write this as a proposition or remark, the loop in the spectral sequence, you can show that these loop classes, so the things we have on this line, exactly have to kill the generator of GrT. These are found quite a nice picture, actually, quite pretty. So we know that, for example, maybe I can give you one example to make you believe it if I have enough time. So I have time until 15 minutes past six. OK, that's very good. To motivate this remark, just make one quick calculation. So the sigma 3, just believe me, is represented by this graph over here. We can start with the 5 field. This is the first 1, 2, 3, 4, 5 non-trivial one. And let's just check where it goes under the spectral sequence. So this is not closed under this extra differential coming from here. So if I break it with this, I add an extra edge. So I get up to a pre-factor, which I omit. I get this. Now, this is exact unless the other differential. And you can check it's up to pre-factor. It's killed by this. And now, if I apply this differential again, I just get this. So I've just shown that this 5 loop kills this sigma 3. Anyway, so here you have these loops. They kill the generators. And you can see that in GRT, not everything is a generator. You have a lot of brackets also. And these brackets are killed by whatever is above. So for example, here's the first bracket. This is the bracket of sigma 3 and sigma 5. And you can check it's killed by something in degree 3. Actually, this is something you can give a formula, right? So sigma 3 is killed by this 5 loop L5. So the bracket will be killed by L5 bracket with sigma 5. So this is what you see here. And similarly, you can just check and these classes all match. So these classes here correspond exactly to brackets in here. Also here you have a higher bracket. I think this corresponds to this. I mean, I don't really know because this should correspond to the commutator. This should kill the commutator of sigma 5 and sigma 9. So from this picture and this extra differential and this result, you get a very nice picture. You can explain all the graph homologies that has ever been computed. So very nice. We had a very good conjecture. I already started writing a paper about it. But yeah, I will talk about it in one second. Unfortunately, not in a very positive way. But let me also spend two minutes about the odd case. So this was all for the n even case. What about the odd case? So on the line 0, you have these classes which are raised to gt. And so they relate to some numbers, like multiple value numbers. And so is there any relation to any kind of periods? It's not a number. So you should be given a form of high degree of rule dimensional cycles. It should be a sum no need to co-object in the higher case. Maybe it's sort of a geometric mechanism. So it's just combinatorial game, what's the motive for that? Higher cycles, I can't say. I mean, there are some. So for n odd, you actually have a very similar story. Namely, you can deform your original Maura-Katoa element in this manner to another Maura-Katoa element. So here you have sums of all graphs with two vertices and an odd number of edges connecting them. So for example, for j equal to 0, you just recover this graph, which was given the Maura-Katoa element, which yields the original differential. And the rest will be some perturbation. And again, the nice thing is you can show a proposition that if you take a homology of GC1 with this other differential, this is also one-dimensional and spent by, well, I mean, this is just a leading order term, but spent by one class. And you can check that also for n equal to 3, you observe these cancellations. So I mean, if you know that this is 0, then by the same trick, taking some spectral sequence, you will see that all classes come in pairs. And for example, you have these loop classes here. And I can check up to a certain degree that you can just rule out that they cancel anything below the line minus 3. So they cancel something here on this minus 3 line. Ah, sorry, this is for n equal to 3. So here I choose a different representative of an odd numbers than this one over here. But it should not confuse you. And one question or one conjecture is that these wheels always cancel something on this line minus 3, but I can't show it. So I don't have an analog of this remark in the odd case. Although the question is, I mean, in the even case, the guys that are canceled by these loops, they have the special meaning. These were conjectually generators of this group. What's the special meaning of the classes that you find here? I don't know. Given that you have an upper or infinite sum there, is it even clear that it cancels the lonely maps to a finite character? Yeah, you have to be very careful that you complete right. And yes, I mean, it's a valid concern, absolutely. But if you just compare these things and you see these kills something on the minus 3 line, minus 3 line kills something below. And this minus 3 line, is it related to a significant variance? Not, yeah, of three manifold, so of rational homology field. This is what this is, yes. You can understand the cycle. It's finite of invariance of 2. So this is dimensions of dimensions? Yes, these are the dimensions, yes. And so these things are well studied. But I think there's no formula for what the dimension is, right, there's no. Anyway, so here we have a nice conjecture of what this graph of homology is. So in every case, you have one kind of, yes, your first conjecture that I raised. In every case that you have one dominant degree where most of the homology is, plus you have certain things that, I mean, they are partners in the other dimension. So here you have, it never happens that, for example, something below cancels something below. It always happens that something cancels something on this line, or something on this line cancels something else. So you kind of have two copies of whatever is here. So that was my conjecture. But in five minutes, and maybe I can make, no, I don't make one more conjecture, one more remark. But then there's a different source of information. Namely, you can count graphs, right? Graphs are just some combinatorial objects. And the combinatorialists, they have been, I mean, they have invented methods for counting graphs long ago. So other source of information, one can count the graphs. So count the dimensions of these graph complexes and compute the Euler characteristic. Maybe one remark is, of course, not my idea. So this Euler characteristic has been previous work, for example, for other types of graph complexes. So for the ribbon graph complexes, a nice formula for the Euler characteristic has been obtained by Ketzling-Tapranov. And for the graph complex, we care about formulas for, how to call it, a virtual Euler characteristic. So virtual Euler characteristic means that you count a graph by weight 1 over the symmetry group of, 1 over the size of the symmetry group of this graph has been computed by Maxim Konsevich also. But now we don't want this virtual Euler characteristic. And the reason is that there are far too many graphs with symmetry. So it's true that in high dimensions, of course, most graphs don't have symmetry. So the number of graphs with symmetry is roughly, I don't know, square root of the number of all graphs. But the homology is much, much smaller than this. We will see some tables soon. So we don't want this counting by 1 over the symmetry group. But really, we want to count these isomorphism glasses of graphs. And you can do it, given that I have three minutes of time. I think I will just give you the formula. So here we make a generating function for the number of graphs, essentially. I mean, the notation here I took from our paper doesn't quite match with what I said before. So read number of isomorphism classes of graphs without these odd symmetries and without vertices of well and less or equal to 2. And by the standard combinatorial tricks, essentially, you can compute this generating functions. They are relatively complicated. But if you compare with this Polyas function for counting the number of simple graphs, it's very similar. So you have the sum over partitions. And then in this Polya function or Polya formula, there is one piece which is kind of linear power in these j's. And one piece which is quadratic in these j's that forms a petition. And the only difference is that here in the simpler formula, there's just some very simple polynomial, like 1 minus st to the something. And here it's replaced by some, let's say, in a physics language dressed version of edges using this cube Pochama symbol. And then there's some vacuum piece, if you want to call it like this. Anyway, I can't see much in this formula, so I just skip over it. But you can feed it to the computer and say compute me the number of the dimensions, compute me the Euler characteristics. And this is what comes out. But then you need many more variables, can you take some more numbers? Yes, here you have sum over many variables, so it's computationally expensive. So I hope it's not the best formula you can find. Actually, I hope you can find a simpler thing. Is odd the n-odd case? Or is it? Yes, odd is n-odd. Yes. And this is what the computer finds. So we computed it as though this B is a loop order. We computed it up to loop order 30. And so here everything looks fine. Now the problem is, if you go to a very high loop order, these numbers grow much too fast. If you just believe this conjecture about the Grotendic Teichmüller Liegebar, this cannot explain these high numbers over here. So my conjecture, therefore, is unfortunately false. And even more unfortunately, it appears that it really, really in high degrees, you see how false it is. But the problem is, for example, here in loop order 30, so number of graphs is number with 50 digits or something. So you cannot expect to compute anything on the computer in this regime. Although something that's very interesting is if you remember the tables I had before, this n-even-n-odd case, they look very different, actually, in low degrees. So the even-and-odd theory of this n-operate seems to be very different. But if you go to high dimensions, these numbers are almost the same up to conventional sign. So the sign here is pure convention. You can get rid of it. But otherwise, these numbers are very similar. So what this picture actually suggests is that the process I conjectured before is only one piece of it. So there's one process, and there seems to be some other process contributing the bulk of the homology classes, which is the same in both cases. And another feature that is quite astonishing is the sign. So here I really pushed my student to make sure that this program was correct because I couldn't believe it. I thought before, yeah, well, high dimensions, maybe. You see, this is contributed by some family of graphs without symmetry, so they appear in both complexes. And maybe something stupid, like all trivalent graphs or something like this. But in that case, this can't explain this strange flipping of signs. So the signs, they do not flip every second time, but somehow, I don't know. So this is really not explainable. Does it move forward? What? Move forward, something like that. No. No? No. No, it's weird, though. No, that's my wrong. No, no, no. I think it's not sensible. If you, in a full positive, take the sum of pro-eurocroticism, it's something very small now, but not more than 15. If you just take it. So just take these sums. Take the sums. Get very small. Yes, absolutely. That's a very interesting thing. So it means that conjecture is that there are many classes which appear in the even and odd case. And I don't really know how. But there's signs also that don't have any regularity. Yeah, yeah. The sign also has random signs. Yes. Yeah. So unfortunately, now I don't have time to explain why it is the case. Honestly, I don't know why. No, I don't know why this happens. And this is a question I have to leave open. So that's the end. And yeah, thanks a lot for your attention. Can I ask a stupid question? Yeah. Can you explain why the signs are opposite? This is just a convention. You see there, if you. They both switch at the same time. Even if you have a convention, but they both switch at the same time. So why do they tend to be the same? If you change the convention, why are the signs the same? Yeah, that's a question. Yes. So the conjecture is that there are many classes which appear both in the even and in the odd complex. So for example, given by graphs which don't have symmetry. That would be one conjecture. But I don't know. Maybe. But I mean the number of graphs with symmetry in these loop orders is much, much larger than these numbers. So we cannot really say, I mean, of course, most graphs don't have symmetry in there. Yeah? Can you try taking the pletistic logarithm and see if that's easier to conjecture something about? No. Well, that should be made a little better. So what kind of computation resources that you used to get inspired? So you mean how long it took? So my student wrote some program. I don't know how optimal this is, but I think for loop order 30 it ran for a few days or something. Just one PC, yeah? Yeah, yeah. Just one PC. Yeah. I have a person next to me here. I think so by computer. No, no. We don't have. OK. Yeah. No, no. We don't have. We don't use supercomputers, no. Is it something about the beginning of getting a new proof of formality and you need three operas? Yes, that should follow. I mean, I didn't want to be too precise about this because it's really work in progress. And in order to, I mean, the statement that would follow is that the n operas for n, at least three, are intrinsically formal. So that means when you have a hopf operat that has homology, little e n, then it is quasi-asomorphic to the n operat. And this what should follow, there are a few technical points. So the most important thing is that you have to set up the deformation theory for these hopf operats correctly and the obstruction theory. And that's why I don't want to make two bold statements because it's technically relatively complicated. So we use Benoit-Fress formalism for this. But really it's not easy and I wanted to avoid having to answer technical questions about this at the end. But that is what it says. Yes. So you have one small thing to notice that sometimes there can be some homology in degree one, but you see n operats, they have a flip action. You can say flip one dimension and these extra classes, they are odd. And I think that shows there cannot be it. This computation via spectra sequence or...? This computation is just taking this formula and expanding. So...