 The honor and pleasure to have Professor Tom Rocha from MIT visiting here, visiting ICTB for actually for the first visit for another few days. But then he will be back in August for another couple of weeks. So hopefully you will also have occasion to ask more questions later. Because actually the school which we have in plan for August is actually going to be around the topics that Professor Rocha has mastered over the past 30 years. So apology and joy, the kind of school and conference resizing on this kind of topics. So that's one of those mathematicians who really need a very big introduction, everybody. So he has been professor at MIT since 1996. He has been the recipient of many important prizes by the Febren Prize in Geometry, the Nobel Prize for the monograph that he wrote with Peter Kroleimer. He is a member of various academies of science around the world. And I'm still affectionate to the fact that all these recognition prizes come after theorems. And a little autobiographical, I remember the shock as a graduate student when Tom, together with Peter Kroleimer, proved the Tom conjecture, which is one of the most beautiful. Not me, Tom. Sorry, it wasn't me, Tom. The Tom of the kitchen. No, yes. There's an age of different. I don't know, maybe there's not people. No, there's an age. But it was a Tom conjecture by René Pomp and Tom Rocha and Peter Kroleimer proved it in 1997, 1998. And it's one of those theorems that really for, especially when you're a graduate student, give you the energy to study even more. It's a beautiful statement. The more it curves, minimize the genus in CP2 among all embedded surfaces. It's one of those things that say, what? And you go back to the library and study more. So, okay, it's a great pleasure. It's a great pleasure to have you here. It's time to start. So, I leave us down the ground and the fact that it's more deformations on the homology of being of the model space of representation of the fundamental movement of the human surface come from each other. Very well accent. So, it's a great pleasure to be here in my honor. Charles, do you want to close your microphone or maybe I close your mic? No, I don't want to close it. I just want to make sure it's conformed but I just want to make sure it's conformed. No, I don't want to take a... Yeah, so it's a great pleasure to be here, to be at this beautiful place. It's amazing that we've been a friend of mine forever. And also particularly nice because some of the work will touch, I've mentioned some work in the nursing home and some work in Zagie along the way. So, wonderful. Okay, so, what I want to tell you about it is joint work with God Surprise, I think you're a contactor. I have to forget to say that because I think it's obvious but, you know, it's not being connected. I mean, it's reconnecting. I mean, it's reconnecting. I have to share again. So, I'm going to close this. We will now move to AD Newt, he said, right? So, what I want to tell you about, I'm going to start a little bit easy, but it's not so easy. Then I will step back and resolve this. The goal is that is S2N. We'll take most of the lecture to pass this statement and then hopefully have a complication. Okay, first of all, S2N is, as an overfold, it's fine, this will be great. It will be slower than everybody that we have. So, as an overfold, it has a fundamental group and so the cone angle pi over two means we're going to look at a baseline here because we're with kids and we always put base points and then we look at sphere. If it were just the two-sphere, then of course the fundamental group will be trivial, but there's a penalty for trying to slide across the overfold points. And the penalty is that, so that means that if you go around once, that's not contractable twice. It's now three times, now about four times. It's contractable twice, that would be contractable. So the fundamental group is generated by N-loops and they satisfy the relation that gamma i to the fourth is one and also with the product. Two is study representations of this group into SU2. SU2, so concretely just a bunch of the matrices in the SU2 satisfy with the fourth powers. Are all one and the product is one, but we're going to require that gamma i is, but it's square is not one, gamma i is the fourth one. So in other words, it's easy to describe what set of the matrices that satisfy this equation. I could think of SU2 sitting inside the unit, sitting inside the paternions as the unit sphere. And this description of SU2, the set of elements of order exactly four is, this is one, this is minus one, and the sphere is this two sphere, which is the unit sphere imaginary paternates. You think a paternion that's purely imaginary is squares minus its length. So these guys will all have square, have square minus one, so that's the fourth powers one. So in these terms it's equal to matrices A1, An, and S2, that's particular S2, S2i, so that the product is a innocuous thing. Now, we're going to want to look actually at conjugacy classes of representations, conjugated by elements of S2. What that means concretely over here is that we're looking at this set of points on the two sphere up to rotations of the two sphere. So now we have a kind of a nice picture of one of the main players in the story. And so we're going to call this Rn, it's just a bunch of points and points. Feel free to ask questions, restrict ourselves to end the story simple when N is odd. When N is odd, what you can prove is that Rn is a smooth manifold dimension one minus six. Where does that come from? Well, the two end is pretty clear, two end is two end points from the two sphere, right? Where does the minus six come from? Well, actually three comes from this equation. We're asking this product of points is the point one in the three sphere. So that's three equations fixed in the point of the three sphere and another three comes from the fact that S03 is acting on this. So that's the dimension and of course there's a lot of standard details to check to see that that heuristic future that Chris mentioned. But let's look at this a little further. So N is one, just asking for something on the two sphere that's equal to R3. What you can check pretty easily is that R3 has to be conjugate to i, j and minus k. So the product of the three standard imaginary fraternities is minus one. So if I flip the side of one of them, one, and you can check easily of the conjugacy, this is the solution. So this is a point R5. So dimension is formula. And if in fact dimension, what is it? The dimension, if it's one, it's minus four. It's empty, and it's three, it's zero. It's a point where it's five. It's dimension is four, and it's CP2 blown up at five points. As you go on, all the direction goes by this formula so, et cetera. And what you get here are about the big geometry that we recognize. This is a final manifold, and in fact, all of these guys are the underlying smooth manifold of interesting family of final manifolds. They have B2, N plus one. B, N plus two, N plus one. So this is, I think it was five here, B2 is six. That pattern repeats. But just a question to ask, if you're a pathologist, then it's the homology. This say, it's rational coefficients. So there's a long story about computing. I'll do it this way, you can do it. Well, what about the rest? So how about the story that one can do for all the four points? Look at the answer for some of higher genus. There's so many people that have contributed to this area. If they want to take the rest of the board, their names, but being involved in this are, well, first of all, there's not as much, so Shadby, at some point, you may have thought about this case, I mean, as in the context of algebraic geometry, then John Saragate, Seepert and Kern, Baranowski, and King Moustad, that is Kieran, Keen. I'm forgetting other people, but a lot of people that contributed to understanding this. That's mostly, most of those people worked in the case of no four-fold points in higher genus. Hans Bohn, Eric Klassen, Ethan Street, who's mentioned that I come up a lot with this case. Okay, so let me explain that, at least computing the rank of this particular group is actually surprising the straight forward, completely uninterrupted, in fact. Well, you know, all of the intrigues always in the eye of the holder, that you can argue via Morse theory to think about this, and the idea is the following. So we've got our points, let's look at the last two, so we put the base point and take this loop that goes around the last two, and then we take the trace of the image of the representation. So concretely, the bunch of matrices, just the trace of the product of the last two matrices. Okay, now this product, these were in the two-sphere, but the product's in SU2, so the trace is actually in the interval minus two. Two, here's a, and up the time we can see, I can assume that this is equivalent to i, and the same time you can see this guy is going to be cosine i plus j plus sine theta j. And so you see that the trace of a n is, that's just going to pick up this cosine factor, which is a factor of minus two, that is two cosine theta, and so you see that what happens when you have at the maximum and minimum are achieved, when you have a orthogonal, and what happens when you have that, that means that the product of all the matrices is one, so that's min plus n minus two of these guys is equal to plus or minus one, because that's one. So in other words, the maximum and minimum are copies, so when the product is one, for sure that's a copy of rn minus two. Actually when the product is minus one, well I can just, because these are all, this is an odd number of these guys, I can switch the sign of all of them. So that's also a copy of rn minus two. So that's an inverse of minus two is equivalent to f inverse of two is minus two. Another thing to notice is that there exists a connection. This is, this argument appears in some papers of Thaddeus paper of perfect classes in different contexts, but anyway, there's a circle action here. Where does that come from? So if you're in this interval, then a n minus one, let's fix it to the, up to kind of, you can fix it to the i, I can fix a n, a n that's going to be non-trivial. This is going to be non-zero. Now what you see the next level points, right? Now without changing the trace, I can rotate around the k-axis. That induces an action on, this was an action on the, the libelized space. It's fixed points are easy to see. The fixed points are when all of the points are either plus or minus k. I mean, you get the plus and minus signs, but we're going to rotate around the k-axis. And the fixed points of that action are where all the points apply to the sort of north and south pole in the k-section, right? So there are two n minus one fixed points and this fits into a kind of general story. Actually, this action is up to scale. In fact, if I take cosine inverse of f, that explains why it's open. Cosine inverse of f is actually for a natural symplectic form on this manifold cosine inverse of f is a Hamiltonian that generates the circle action. And it's a sort of general fact in the story that then in that case, the critical points of the function are the fixed points of the circle action. The fixed points, the critical points all have even index. This goes back to my idea of the triangle. Anyway, this is all to say that there's an inductive formula there for these dimensions. And in fact, for the function of the polynomial, I'm not using that because I make a double mistake. This isn't quite true. It's actually, it's a two-seater bundle over the sky because I actually get to pick, I had to do some conjugation. That's the range of the conjugation. So in any case, there's an inductive formula. The three-fold formula is this in the second. That's the inductive formula. This, so the maximum and minimum two-seater bundles over Rn minus two. So this is what we call the polynomial for Rn minus two. That's the polynomial for the two-sphere from the minimum. This comes from the maximum. So it gets a boost from the co-index. That's the contribution. And all together, this you can use the, I mean, you call the many numbers but in particular the convention of the sky. So anyway, so that's an entry as it were. But now we really want to understand the comology ring to do that first thing we want to do is understand that there are natural comology classes on the space that I've generated through the ring. And for that purpose, think of this, the geometric origin of these guys and essentially what happens, so we'll know where I'm going online, but there's, well, let's say R3 bundle, R3 bundle over the sky, which comes from by the representation of the fundamental group and that gives me a flat bundle over the two-sphere. So the fiber over a point here is the flat bundle. Think of these as in R3 bundle. And so I can take the quadratic classes, universal bundle, convention, minus a quarter of this guy and look at its couldn't components. So there's going to be a class that's more of a body form of the sphere. But there's some other interesting comology classes coming from each of the orbital points. And each of the orbital points, this R3 bundle is away from the orbital points as an honest R3 bundle at the orbital point. This action decomposes this R3 into a line and a complementary two-plane. That two-plane still has a residual action on it, but there's a, once I take the quotient, there's a line bundle, this is point I, R2 bundle, sitting above each point. And so some of the classes, which are minus the quarter of the class, is we just line bundles over the part of our space, not over the there. So now you can check that these guys sphere all square the same thing because, so there's a natural algebra, a natural ring that acts, just the ring generated by these guys and restrict those guys to the, to the on-surface, the marginalized space. And this is that algebra generated by the guy's nodular relation to the space spirit. You can prove it is that the ordering of the sky, cyclic generator, the sky, everything's a multiple, multiple algebra elements, that generator. And there's some magical kernel and that's the thing you want to get your hands in, the kernel of this map. It was muffled with an observer, a natural guess as to what the kernel is. That goes by observing, appeals to a similar data, this is a marginalized space, convention with parabolic weights. So that part of the same quarter as this case, what that means is to have the same end points. And now I have a whole morphic two vector bundle over this guy. And at each point, at each point I fix up a line in the fiber. So I've got this whole work bundle and I want to kind of understand stability, stable parabolic bundles. So I want to understand in some sense how rigid this data is. Now what I'm going to measure it is I'm going to probe this pair of data, and I'm going to probe it by mapping in a line bundle. In a line bundle, let's assume the map is injected everywhere, it's going to be incredibly important, but if it's injected everywhere, I'm sorry, as soon as it's injected at these two points, then I can ask, is the image equal to the given line at the point F is a sub, two spheres, so this is just a sub point and it's a degree. And then you know what I'm going to look at, minus one to the degree, sorry, minus one to the dimension. So for the parabolic degree, I get a degree and it gets a bonus, to notice the way the signs work out, it gets a bonus if it hits the line and it gets a penalty if it misses the line. That's the way it works. And then this guy's stable is less than a half. The ability is now, and the Stem of Narcimland Method identifies this modular space representation with the modular space of stable parabolic bundles. So we have all these, and you take these as these interesting monomorphic bundles, and the thing to take away from the stability condition is that some kinds of living inside our bundle. And what that means is we can construct a constraint that a co-mology class that is always vanishing on the modular space. So when I think about an interest of the parabolic, if I weren't thinking about parabolic bundles, now we think about this, that's still the notion of stability for active bundles. What we're saying is that if this condition is violated, if the degree's too big, then this is zero. And that would imply that H1 is a vector bundle. There's an extra condition. That is a little bit slower. Anyway, let me just say using stability as well as N, we can compute the turn class of this guy using Riemann-Roch, it's turn classes. Turn class above the dimension is zero. Those are the relations. So it's an interesting story in all of these cases to work out the combinatorics of exactly which guys appear at destabilizing line bundles that violate this condition. So there's a lot of those. Each of those gives you a relation. It's interesting to see what the minimal set of relations, et cetera, is. That's a well-worked out story. But let me say one interesting thing about that in this case. The pen is passed inside P subset once you add, which is, it is a set of hits, P i is equal to. And then it turns out that the first relation started to mention M. There's actually a nice formula for these guys, which depends on this eta minimally destabilizing C M of this bundle, which depends on the degree of that. And eta, set of hits, it is one over and factorial. It's the product, B eta minus one. This one is actually new. I mean, it's like, you know, Zagate has a beautiful generating function for these guys, but I don't think you'll notice that. Here's another formula for these guys. This is also equal to, so what's happening in this formula? Just some combinatorial factor, there's some polynomial of homogeneous polynomial degree J, these guys, I mean, co-homological degree. And then this is the symmetric function in M minus J, sorry, M minus J symmetric function in N variables where the variables are put in the deltas with a sign according to whether it's a hit or miss. So, nice concrete formula. And these are unsatisfying recursion relation, which is the recursion relation that I learned about from C that it sounds like many other people discovered it. Same time, the kind of satisfy the seclusion relation if you formally put in equals one and look at this, it turns out that these are orthogonal polynomials. Cents, classical sense of orthogonal polynomials with respect to the measure of the gene delivery sequence. So the coefficients of sense, essentially. Now, in the next two minutes, I'll tell you the rest of my talking words. So, it turns out, so in this particular case, it's due to Ethan's street, but it follows work of C, C, C, C, C, C, C, et cetera, okay, so that point of the genus that actually these generate the ideal of relations to N by two plus one, that's the choice. It's a very hater that gives me a bunch of homology classes of degree F and those homology classes generate the ideal of it. Interesting thing, so that's great, that's great. But Ethan's street, it was asking himself in August, so another story for instant time homology, whatever that is, but instant time homology provides an interesting way of deforming its relations. So, it turns out that this instant time homology sort of, in this case, I think maybe it's still conjectually, but in the case of my genus, it's also said that the quantum homology of the representation variety, so there's a deformation of the homology, right at that way. And the way people, way you can speak, computed these things, computed the deformation, was quite difficult. And Peter and I wanted to compute something even harder, which is to, it turns out that the space of connections when you do this in time for homology has not revealed from the middle group, and so you can look at local coefficients, so homology of local coefficients. And we wanted to compute for our own nefarious purposes that those homology groups with local coefficients, now we can do it. And the interesting thing is that, one of the things that, because we're interesting when you start doing local coefficients is that for the homology, what's just a Z module, but now when you do local coefficients, it becomes a module over a one polynomial rate. So now it has suddenly a lot more structure, and it gives you, in this case, a family of interesting algebraic curves, which seem super cool. It makes the for homology in this form of the complex that knots it makes. So this is product case, it gives you an interesting curve. If you have a two-sphere embedded in the earth, three-manifold hitting the knot in a certain way, the for homology of that knot becomes a module over that curve, so it's an interesting vector bundle over the curve. And anyway, a lot of interesting things to do, but that's for some of it. So yeah, good? Yes, good. Thank you. Do you have any comments or questions, or so all of those? Or some mind, of course, if you want to ask them. I'm sorry it was all history, but I hope it's interesting history, because even more history. But I have a very naive question. For example, if you retranslate some of this formula, some interpretation of the, even in the examples that you told us for N equal to three, this was Cp2, none of that, five minutes. Have you learned something about Cp2 blown up at five points? Well, I think, yeah, I don't know if the, I don't know the literature, I don't know the chronocomology literature. Well, I don't know if people have computed the chronocomology of almost all of the books that arrived, probably, I don't know. Mostly, think about this extra deformation that you get from passing to the local coefficients, maybe an expert could tell me that. But I mean, we learned a lot of interesting things in the process of doing this, in fact, I like to, we started thinking about this a couple of years ago. This is one of our, this is the COVID project. I've come to describe it as the white wear. The knowledge is conveniently, this instance of the white wear is an overspold two-sphere, the teeth, you know, with 23 orbital points. And when we first tried to think about it, in the first week, we were really happy when we did it for five points where it was Cp2 blown up at five points. And that was essentially by figuring out the chronocomology, which was somehow not so hard. And then we decided that, yeah, but we should just gauge theory. And then, you know, we had, we played the written notes in particular, something, everything was just, you know, sort of barely doable without local coefficients. It all seemed very hard. And then we noticed a lot of, at some point, we noticed that it was actually less complicated than the people think. And we could just do it and we did it. And now that we have a program that can, you know, I mean, these things, you know, the rank goes quite fast. So it's hard computation eventually anyway, but we can do it up to 17 points without any trouble. I don't know if we can actually do the big white wear, although probably if we don't do it on our laptop or something else, we can do it. Anyway, so on the questions, so you, you always have the case where you have these old two points with order four. Yeah. So is there a particular reason that's what you can do? Well, in principle, you know, you know. I mean, you know, the thing is that, I mean, I, you know, if you had asked me like three months ago, I would have said it's just too hard. I mean, too hard for us, you know. But now I don't know. I mean, one of the things that, you know, so from the point of view of the instanton homology, the reason that's a nice restriction is because that's the analog, that that's the situation in which the floor homology construction is the analog monotone, the symplectic floor homology. It's monotone, the symplectic form is a multiple of the first-journ class of the complex structure. In instanton floor homology, there's a very direct analog to that, which is that there's the trans-signal function that you're using to define the floor homology, the morselology of trans-signal function. But the trans-signal function is a circle-valued function. So you get a circle-valued function and actually get a homomorphism from the fundamental group of your space into the integers. On the other hand, there's another movement. That's kind of floppy. That's the thing that depends on the combinatorial. So trans-signals as a function on the space, the space doesn't change too much, changes a bit, but the trans-signal function kind of moves as a homology, as a map, Z. The other thing, which is more rigid is the spectral flow of the Hessian of the sky, which is exactly analogous to in some type of formology. So when those guys line up, it's much easier. They don't, well, then, you know, I mean, now, of course, somebody's bothered with some normal cost frames. So you can do it. And it'd be interesting to see if you can do it. I mean, normally in this context, in the parabolic case, you'd also kind of change the parabolic structure and then look at what costs, and you think like that, but maybe I don't know whether that makes sense in your context. Oh, yeah, yeah, yeah, sure. I mean, it's a delicate thing to do, but it makes sense to think about it. But, I mean, somehow, and, you know, at this point, what actually happens, how easy it is, I mean, I didn't tell you anything about how easy, what made it easy, but anyway. So before thanking Tom again, it's our tradition now to all the old guys with the room, and we leave Tom with our diploma, the students, to have a little private chat, okay? And we all wait on the terrace. Once we're done, we are happy to offer everybody a little professional, okay? So thank you very much, Tom. Sure. Thank you very much. Oh, and then of course, that's about topology, you know. This part of the conversation will not be a call. Don't worry. Don't worry. Don't worry. Don't worry. Don't worry. I hope, but I don't know. No, I'm sorry. It's a different thing. I don't know. Maybe I haven't seen it, but we'll never do it again. So, yeah, yeah, yeah. Yeah, yeah, yeah, yeah. Thank you. Thank you. Thank you. Thank you. Thank you. Thank you. Thank you. Thank you.