 Half of my talk will be devoted to the notion of hyperkelechones in general, and then I will move to discuss brains and to describe the work of two mathematicians from the 70s about the descriptions of a particular class of hyperkelechones. Now, what are hyperkelechones? They show up in theories with eight supercharges, in particular on Higgs punches, and this Higgs punch can be classical, or it can be also of some interacting CFT. In the dimensions 3D up to 60, and there is a very standard way of constructing them using hyperkeler equations. In other words, these are F and D term equations, and a lot has been discussed on that. Today I won't discuss this topic so much. They also show up in theories with eight supercharges on Coulomb branches, and in this case it's going to be just in three dimensions. So three dimensional n equals four, and you may have seen that there has been some excitement about it, in particular in algebraic geometry in the work of Nakajima. Now, Coulomb branches are quantum moduli spaces. As usual, Coulomb branches receive quantum corrections, and you have to take into account all of these monopole operators which show up. And I consider this as a different construction. There is the hyperkeler quotient, and there is the other construction of Coulomb branch. And if you have a CFT, then perhaps you could construct it using some gluing techniques, but that's also under hyperkeler quotient. Today, what I'm going to do is to look at this class of all hyperkeler cones. So I'm going to concentrate on this hyperkeler cone and see what we can say about them in the most general way, and we will start with the geometry, or perhaps a more algebraic description. And so since it's hyperkeler, there's always an SU2, and we will denote it by SU2R. It acts on a triplet of complex structures which transform in the adjoint representation of this SU2. And what we are going to do, the approach we will take is that we will pick a U1 subgroup of this SU2, and with respect to this subgroup, pick a single complex structure and define the notion of holomorphic functions. So physically that means that I pick subalgebra or four supercharges, and it has a U1R symmetry, the subgroup of SU2, and I'm going to think about holomorphic functions with respect to this particular U1 symmetry. And so all of these holomorphic functions are now going to transform in irreducible representations of SU2R, and not just in irreducible representations. There are actually going to be the highest weights, the corresponding representations, so in general if you'll have a collection of operators which are transforming in a given SU2R representation, the top component will be holomorphic or will be chiral physically, but the lower components will not be holomorphic anymore. So if we restrict to holomorphic functions, we'll just get the highest weights. So the degree of the holomorphic function is going to correspond to the spin under SU2R. OK, so let's use this data, and now think about all possible holomorphic functions. We're going to have a chiral ring or the ring of holomorphic functions, and it's going to be finitely generated. So we will take the generators, GI, to be in certain representations of SU2R, and we are going to look at the possible representations we can get. So what could those representations be? Let's be incremental and look at the simplest possible representations. So if the representation is the spin zero, the trivial representation, then there's only one operator, and actually it's not even a generator. It's just the identity operator, and every quantum filter will have precisely one such thing, and you could say that if the spin is zero, its identity is spin zero. So that's fine. Now take the next representation. If the representation is one half, then physically it corresponds to a free field, chiral multiplet, and in fact when we have some generators of the chiral ring which have spin one half, there will always be an even number of them. Let's call the number of such generators to n, and then I will know that the quantum filter will have a decoupled sector of a collection of two n-free chiral multiplets or n-free hypers. The corresponding hypercalacone will contain a factor of C2 to the power n, or sometimes it is noted as the quaternionic flat space h to the n, and this will be a factor and there will be another factor which is non-trivial generated by operators of spin higher than one half. So let's concentrate on those which have r equal to one and above. Let's move to r equals one. What happens there? By supersymmetry, the generators transform in their joint representation some global symmetry. Geometrically, this symmetry is going to be the isometry of the hypercalacone. Physically, it's going to be the global symmetry which acts on all other operators and they transform in representations of this global symmetry. Now, this is a very important crucial point because, in fact, people in this room and not here have used this to say the following. If, for some reason, I know to identify the global symmetry of a quantum system, then that means that I should look at operators of spin one and I know that they should transform in the joint representation of this global symmetry, so I have to account for all of them. Conversely, if I have any means to identify all operators at spin one, then they will have to transform in a joint representation of some group and this will allow me to identify which particular global symmetry the physical system has. This is a very crucial tool which has been used, I would say, as we speak, in various quantum systems. This is very important. But now, here's a punchline that I suspect that not too many people know and it's the following. Suppose the generators of the chiral ring are in spin one representation. Then this implies that the cone, so what I mean by this is that no other generators are in higher spin. So we took care of this, we took care of one-half this decoupled. Suppose that what I'm left with has only spin one and not more. Then the modular space, the hypercalic cone, is an impotent orbit. So this puts an impotent orbit into a very special position, very special place. It just tells us that if we want to understand the physics of modular spaces of certain supersymmetric gauge theories or certain brain systems, then we first need to understand the physics of spin one operators and, or in other words, the physics of an impotent orbit. So the rest of my talk is going to be devoted to an impotent orbit. And the nice thing about an impotent orbit is that it's been classified for us by embeddings, the group SU2, whatever your global symmetry is, G. G here is the global symmetry that I talked about over here. What is an impotent orbit? It's just simply you take an impotent element in the algebra and take the orbit under the adjoint action. And nicely enough, those spaces form a hypercaler spaces. The space of all such orbits for a single element is a hypercaler. In fact, there is a subtle point here. I'm going to take a closure of an impotent orbit. And this is important since whenever I have an impotent element, I can describe it by, you know, if my group has matrix presentations, then I can describe the nilpotent element, let's say for a type by some collection of Jordan blocks. And when I conjugate by the adjoint, this matrix has a certain rank and certain degree of nilpotency. Adjoint conjugation is not going to change, not the rank condition and not the nilpotency, sorry, not the nilpotency. So for this reason, I need closures. So if I take closures, I can allow for those elements to be zero. And once I set one of the elements to zero, I can reduce the rank, reduce the degree of nilpotency, and I'll get a cone. So the closure of nilpotent orbit is a cone, is a hypercaler cone of the type that we are interested in supersymmetric gauge theories and in-brain systems. Any questions so far? Yes, yes, it's a singular space. So this goes under the name of a symplectic resolution. And there are some ways to do that. So for example, if you take for nilpotent orbit of type A, the symplectic resolution, it's, so in the math, math literature is called a springer, springer resolution. And in fact, it's just a cotangent bundle of some coset. And given the type of nilpotent orbit, there is a simple way of computing the subgroup H. And this is the desingularization of this object. We will be mostly interested in the cone in the non-singular case. And in fact, it turns out that not all orbits have a symplectic resolution. So that, in fact, that makes it quite interesting. Yes. So if we know, soon I will show certain constructions and I will show how we can identify the global symmetry on either Higgs branches or on Coulomb branches. And typically to identify the global symmetry, you need a little bit of physical, either intuition or physical, physical arguments to identify this. We will see how it goes and then you'll tell me if you're satisfied with the answer. More questions? So let's take the case where G is of type A. And here the classification of nilpotent orbits is just embeddings of SU2 into SUN. And this goes back to the work of Dinkin. It's just parameterized by partitions of N. And let me just give you an example. Take N equals to 2. So that's embedding of SU2 into itself. There are two types of partitions, 2 and 1 squared. And 1 squared is the trivial partition. The corresponding orbit is just going to be a point, the origin. But the second one is non-trivial. And just think about the space of all 2 by 2 matrices which have 0 trace and that squares to 0. And a quick computation shows that this space is C2 mod Z2. And this is the closure of the orbit 2, this non-trivial orbit over here. Okay, any questions? Yes, yes. Is that unrelated to the Schmitt nilpotent orbit? I don't know of the Schmitt nilpotent. Well, I mean, this is just how a period is degenerate. Yes. I will be very interested to learn about this. I'm not aware of this. So I'm not sure I can comment now. But can you replace G with the variable? I need a continuous space. I'm not sure what you mean by replacing. You see here, this space is a two-dimensional space. You'll have to pick your elements very carefully. You'll need to pick three elements and then act on them discretely. Sorry, two. Not sure. But let's try to clarify this. So this leads us to the work of Kraft and Pochessi, which this goes back to the 70s, late 70s, where what they said is that wait, we can construct a quiver. Actually, they didn't call it a quiver because back then the notion of quiver was not introduced yet. However, they did quivers before quivers were called like that. And what they said is that suppose that I take a partition. So let's characterize it by a young diagram. I can construct a quiver by taking denoting each column. Let's call it CI. So it will be ranked from one until the number of columns. And CI is going to be the number of boxes in the column. And what they said, they construct the quiver. The quiver will have as many nodes as the number of columns. And each node will have number N and I, N1, N2, and so on. And those NIs are just the sum I from, let's call it J from 1 to I, C, J. So conveniently, they told us how to construct a quiver. So let's do a simple example for the case of N equals to 4. I will have the maximal partition 4. And that will correspond to a quiver, which looks like 1, 2, 3, and 4. The corresponding young diagram is this. And so you see that the first column has one box, the next two will have 2, the next three will have 3, and the last will have a 4. So they told us how to construct a quiver. 3, 1 will be this partition. And the quiver is going to look like 2, 2, 1 node, 1 flavor. And 2 and 1, the quiver looks like 1 and 4. And the partition is this. And then there is the trivial, right? And what they also told us is that if I take the hex branch of this quiver, is the closure, the corresponding cone. So this is, they gave us a prescription how to construct quivers. And I will deal in this talk just with A type for simplicity, even though other types of nilpotent orbits, including exceptional, are extremely interesting. And there's plenty of physics one can extract from those guys. Now that we have a hex branch quiver, by this I mean a quiver such that the hex branch is a nilpotent orbit, we can apply a three-dimensional mirror symmetry, which is easy. What you need to do is to embed this system into brains. So here's an example. One, two, three, and four. Like that, those are, the lines are NS brains. These lines are the three brains. And these are the five brains. So now it's the first time I mentioned brains. You are in type 2b, you get this collection of brains. And this is the brain system that characterizes this quiver. U1 times U2 times U3, and U3 has four flavors. You do this, and what you do is you move the brains such that they realign and they are in between the five brains. So when you do that, you need to make sure that you don't have configurations which break supersymmetry, and you need to account for the right number of brains. You have brain creations. When the dust settles, you end up with a picture which looks like this. And except that now these are the five, these are the three, and these are NS. So it's kind of a dual picture. And note that the only difference for this particular quiver is that the picture is mirrored, perhaps to reflect this point. But that's not common, actually. It only happens for the maximum input and orbit. So you end up with a picture like this, and when you do S duality, you can now replace d5 by NS, NS by d5, the three remains, and you get the mirror quiver, and you will say that the Coulomb branch of this mirror quiver is the corresponding nipotent orbit. So I'll give you a prescription how to compute it, given a partition. So draw an A-type quiver which is framed. That's the math, a word, or a flavor. The flavor is determined by the partition, and also this is not just framed, it's also balanced. So let's do examples. How do I determine the flavor according to partitions? I just draw the diagram, and I give the nodes labels. I give them labels one, two, and three, and in order to distinguish from those labels, I'll just circle them like that. We could use different colors, but we don't have the time for that. So these are just labels, one, two, and three. And if I have a partition, so let's take this partition here, I just count the number of columns which have one box, and there are four of them, and so that tells me that the flavor is going to be four types of the partition one, and the rest of the quiver is determined by balancing. Balancing means that every node has a number of flavors equal to twice the number of columns. So then the answer will be three here, two, and one. So we see that we get this quiver. So quickly I will draw the mirrors. For this case, maybe I'll do it over here. Let me raise it over here, and I will draw the corresponding mirrors. So this is self-mirror. This one, again, I draw an A-type dinking diagram. I have one flavor of two and two of one, and the balancing is that one, two, two. Over here I'll have a dinking diagram of type A for SU4. So that's A3. I have two flavors of two here, and the balancing gives me one to one. And finally, the last node, 111, and one flavor here, one flavor here, as the diagram tells me. And we can read from this the dimension of the Coulomb branch. It's just rank, so the quaternion dimension of the Coulomb branch will be given by just summing those numbers, 6, 5, 4, 3, and there will be the trivial quiver where we have zeros, and here you have a trivial quiver, just a point which has dimension zero. So this is the list of nilpotentorbits of SU4. Any questions? Craft and Pochessi? For sure. Have a look at the paper. They knew it before many other people thought about hyperkelequations. I asked Martin about it. He wasn't aware of the work. But we still need to know whether Hitchin was aware or not. Any other questions? Okay, now, of course, Craft and Pochessi did not think about the Coulomb branch. So for them, these type of quivers did not exist. These quivers certainly show up in their work. Here we have a hyperkelequation. Here to determine the Coulomb branch, I need to understand the physics of monopole operators and the way they form relations. I figure this out, and then I should find that the modular space is the closure of the corresponding nilpotentorbit. Sorry. So that's nice. And the next thing that we want to do is to try and understand all of these spaces as a general thing. And that leads us to the title of the talk, the Craft and Pochessi Transition. And here, Craft and Pochessi use an idea by Briscone. I think it's with the C, maybe K, right? So a leading German algebraic geometry. Here the work is done in early 70s, maybe 71 or something. And he says that, okay, if I want to understand these spaces, then perhaps I should understand differences between them. And he looks at the maximal cone, the one which is six-dimensional, and he compares it with the one below it, which is five-dimensional. In fact, all cones are contained in the maximal one. That's a theorem. And then he says, okay, what's the difference between these two? Can we understand the difference? And by this, when we understand the difference, we understand the structure of these spaces. And he comes up with the feature. And here you see that the difference is one-dimensional, one complex dimension. And in one complex dimension hyperkeler, we have a classification. So it could be one of C2 mode gamma, where gamma is classified by ADE. And so it's very suggestive that for the case of SU4, the difference will be A3. So it's C2 mode Z4. And indeed, that's what it shows. So the difference here between the top nilpotent orbit and the one next to it is C2 mode Z4. And I will show you how we can understand this using brains, because after all, we want to extract the physics from this geometry. And so, Kraft and Pochese say, okay, why should we stop and compute the difference between the first two, the highest two? Let's move on and continue to compute differences for all orbits. That's what they do in their work in the late 70s. And you see that the differences here are one, one, one, and three. So maybe we could figure out what are the differences in this case, but in the last case, that's a question. So you look at the cone and you look, so this orbit is contained in the closure of this one. And so what you do is you look at the transfer slice to this. So in a sense, let's say this is the closure and here's the orbit. So you just look at the transfer's directions and it turns out that this is simple enough. And in fact, using brains, we will see how simple it looks. And I will use this picture here. And what I'll do is to consider the Higgs mechanism. And by the Higgs mechanism, I'm going to look at this picture here and just look at one brain and consider the rest as spectators. So transfer slicing here means that I just ignore those other five and I just look at one of them. And I see that if this one touches the other four, let's draw it in this way. Let's just align them. What I get, this looks like the quiver of U1 with four flavors for which we know the Coulomb branch. What's the Coulomb branch? See two more, Z4. And in fact, I'm going to need one more Coulomb branch which will look like this. Suppose that I have a collection of brains, all of them stretch like that. One at the one flavor at one side, one flavor at the other side. So the quiver will look like an analog of this one, the minimal orbit. And the Coulomb branch has as many dimensions as the number of three brain segments. In this specific example, we are dealing with A5. So the number of brains here is five. That's SU5 corresponding to A4. And the Coulomb branch of this quiver is the minimal nilpotent orbit A4 SU5. Physically, it's known as the reduced one instant on molar space of SU5. And in this space, Kraft and Pochese denote by A4 will use this notation. And it turns out that this is what we need. Let's now look at the Higgs mechanism. So suppose that I get to this case, Coulomb branch of U1 with four flavors. At this point, there are additional muscle states. There are singularities in the molar space. If I think about string states between those brains, they become muscle states. And I turn into a local origin of a Higgs branch and a Coulomb branch. So both vector fields and hypermultiplets are massless in this case. And I can do something different. I can just split the brains like that. If I'll draw it in three dimensions, it would look like this. There's a brain and I just split it. I pull it along the transverse direction. I cannot do it when they are away, certainly because they don't touch. But as they touch, I can pull it back and I'll get a new branch. And that's the Higgs mechanism. That's nothing but the Higgs mechanism quantum field theory. You could just compute the muscle states and fit it with the Higgs mechanism. So this is the Higgs mechanism using brains in this particular brain system. And so suppose that I take this vacuum expectation value and I send it to infinity. I just cut this and pull it. And the remaining picture, let's take this. I started with four. And let's just take these two as spectators. And there is a brain which is moved far away. And now I can use a brain creation effect and just move those two to the other side. This brain gets annihilated. That's the inverse of brain creation. So you could think about this as coming to this by just moving it to the two sides. And so the bottom line from what I started is that I started with a configuration with four brains over here and six brains like that. That's the starting point. And then what I do, this number here is reduced by two. And I have one brain here, one brain here, and one less brain here. So I end up with this system. All the others which are spectators remain. This reduces by two. So I'm left with two. Those are these two. And one brain moves here, one brain moves here. This brain does not contribute to the flavors anymore. And I'm still left with an A3 type quiver. And this is what I get. So this is U1, U2, U2, one flavor, two flavors. U1, U2, U2, one, two flavors. So we got into this Kraft-Protcherzi transition. In terms of young diagrams, what we did, and this is also the way Kraft and Protcherzi describe it, they start from this partition and they move to this partition. They just pull the books one down. Any questions? So now we could turn it into some simple computation. I'll start describing this by so-called Hasse diagram. The first point is going to represent the maximal orbit of dimension six. Then I do an A3 transition. Namely, I do this transition with C2 mod Z4. Next, I'll do an A1 transition to the four-dimensional orbit. And as you could see, I can proceed by taking one brain and pulling this brain here and this brain here, and I get this quiver. A1 transition will continue to the three-dimensional. The difference here will be again an A1 transition. And finally, the difference between the three-dimensional and the zero-dimensional is this particular orbit, which corresponds to the minimal orbit of A3. You can look at this diagram. This Hasse diagram turns out to be symmetric with respect to the Z2 symmetry that exchanges A with capital A, lowercase A with capital case A. This is also equal to A1 and it's a nice symmetry for nearly potent orbits. Any questions? Let me finish by describing the KP transitions for A5, SU6, and we will see that the Hasse diagram gets an interesting shape. My starting point will be the maximal orbit. I draw six brains like that and this is going to be the Coulomb branch. One, two, three, four, five, and six brains over here. A complete analog of the case for A3. And I'll do the Kraft-Pochese transition. I'll get from this quiver to the quiver, which looks like this. Again, it's an A5 quiver diagram. There will be one flavor here and four flavors here. And the transition is, maybe I'll do it over here, here's the orbit, the sum of lengths is 15 and the transition to the 14-dimensional is of type A5. I'll proceed and use the rules that I depicted over here. This is one, two, three, four, and four. And now I'll get two here, two here, three, four, three, two, one. This is 13-dimensional and the transition is of type A3 because a single brain was touching four D5 brains. However, note here that I have two options. I could do a transition on this segment or I can do a transition on this segment. So the corresponding Hasse diagram will have a bifurcation. There will be two 12-dimensional orbits corresponding to two different partitions. I can draw the partitions. Here the partition will be the six partition. Here I'll just move one box down and like that. And over here I'll remove one more box down. And now I have two options. I can either remove this box down here or I can remove this box down here. So I'll get two types of orbits with the transition being A1 because the free brain hits two brains in each case. C2 mod Z2. I'll proceed and it will turn out that both orbits reduce to a single one with another A1 singularity. Let's draw the corresponding quiver. Let's say that we move these two guys and so we will get the A5 dinking diagram and I reduce this by one. These remain and the brains move. Here I'll have nothing. Here I'll have a three and here I'll have a one. So this corresponds to the dinking diagram which looks like three columns of one and one column of three. In fact you see that this is going down in one of these guys but when I go further it's not going to be A1. I made a mistake here. It's going to be an A2 because now when the brain hits it hits three guys. And because of this Z2 symmetry I can conclude but you could just proceed with the Kav Pochese transition and you'll get this nice symmetric quiver where here the transition will be A2. This will be nine. Nine dimensional. Here the transition will be A1. Eight dimensional. Here the transition will be A3. So the difference is three, five and this one will be A5. So the difference is five. So here's the structure of nilpotent orbits and also here is the brain system and how it reflects this transition that Kav Pochese computed back in the late 70s. I'm ready to summarize. So what we've seen is that first of all we started the lecture by characterizing hypercalar cones and by saying that they are classified according to the spin of the generators of the chiral ring and this is something which is new. We haven't seen that is emphasized in talks in physics so far and I think we should start thinking about it in a serious way. So hypercalar cones are classified according to the spin of the operators, the generators of the chiral ring and spin one under SU2R is very important because they always transform in the joint representation of the flavor symmetry and in fact now I remember a question that you asked. In fact we see the flavor symmetry in the Dinkin diagram. These are Dinkin diagrams of type A. This is A3. These are Dinkin diagrams of type A5. They are all balanced so as long as those nodes are balanced the flavor symmetry is going to be given by the Dinkin diagram of all balanced nodes so indeed if I compute the Coulomb branch of these even though it's not visible I will always get that the flavor symmetry is SU6 in this case, SU4 in this case on the Coulomb branch. So classically when you write down the Lagrangian there's nothing which tells you that there is an SU4 here or there is an SU6. This is a special case and it's not representing but a computation using monopole operators will show that the global symmetry is always A5, A3 and so on. So back to the summary. All generators of the chiral ring are classified by spins so it makes sense to think about spin one-half, spin one, spin three-halves and so on. And to remember that if there are no other generators other than spin one the modular space becomes a nilpotent orbit and there we have a classification. So we just need to think about all members in the classification and just study them and extract the physics out of each one. We have great work to do in this case. And we think of all other hypercalar cones as built over nilpotent cones or nilpotent orbits. Why is that? Because if I'll add one more generator which has spin higher than one it's going to be a deformation, a small deformation of a certain nilpotent orbit. So I should study some sense of perturbative extensions of nilpotent orbits. Just add one more operator, two more. In fact in Higgs punch modular spaces these are always barons. So I just need to add few barons over a collection of mesons. The mesons will form the nilpotent orbit and the barons will just be slight extensions of those nilpotent orbits. So it's a new way of thinking about hypercalar cones. And finally we saw that in the subclass of nilpotent orbits type A is what we discussed today. We can have a physical description of the Kraft-Pochese transition where a single brain touches the collection of heavy brains and moves away by the Higgs mechanism, leaving us with the rest. And this is the physical notion of what Briskorn and Kraft-Pochese suggested by transfer slicing. So we have this physical interpretation of transfer slicing. All of these guys are spectators and only one brain is participating in this transition. So we have lots of fun with this transition and I hope to report on some other things. Thank you.