Okay, so let's see So you all know MPC you have n parties and your goal is somewhat to construct such a trusted guy in the middle there The challenge of course is that certain guys are dishonest and now it's well known that many protocols You can use to to achieve this talk task. However now Let's look at this situation here. We have an incomplete network. Okay. We want to do the same We want to build a trusted party Okay, what can you do? Well It's well known what you do first you complete the network using secure message transmission And then in the second step you can do MPC on top Now if you look at classical secure message transmission protocols What they have in common is that the network graph of the incomplete network is assumed to be public knowledge Okay, and the protocol may depend on that Knowledge now if you consider certain networks like real-life applications the network might contain Confidential information for instance if your network is based on social a social network. Okay, so you know certain guys and you talk with those Well, maybe you want to hide that information. Okay, so Therefore it would be bad if you would use a classic secure message transmission protocol if you want to hide this information This led to the notion of topology hiding communication, which was introduced at last year's TCC And the goal there is okay. You still want to complete the network, but this time the graph is the secret Okay, so no one should learn it Now it's so if it's clear that Somehow parties need to know at least some information about the graph. Okay, for instance, you need to communicate with your neighbors You know your friends. Okay, so For instance, if you consider that guy here He knows Those his neighbors here, but he should not learn more about the graph, right? He should not learn that for instance Those two guys here know each other Okay, so more formally. Here's our communication model We have this those end parties up to T of them are dishonest we look at synchronous communications of protocols proceeding rounds This incomplete network. Okay, it consists of private secure channels Then we have the network graph, which is from some family of graphs. Okay They should be at least connected. Otherwise, it's hard to complete the network and It could be an arbitrary graph out of this family of graphs. Okay, and kind of this family That's all the information, you know about the graph As I said before you can communicate with your neighbors and this means that the adversary knows the neighborhood of all dishonest parties Okay, so The security we consider a simulation base. So the idea is in the real world. You have the incomplete network You you run your secure message transition protocol and here on the right in the ideal what you have is you have the complete network and Additionally have still given the incomplete network and you have to simulate whatever you see in the real Using this ideal world here The idea is that the incomplete network is completely independent of the graph of the graph of the network You have given and so the only information the adversary learns in the ideal world is actually kind of the neighborhood of dishonest parties Okay, so that's kind of the idea of this topology hiding communication So what is known well as I said it was introduced at last year's TCC by Moran et al And so they looked at the feasibility of this topology hiding communication in various models So here's bad news if you consider unbounded adversaries. It seems not to be possible. There will be some leakage In general The same is true if you look at fail stop adversaries Okay, so kind of an adversary who can just make its party stop at any time The issue here is that if you have a cut vertex in your graph for instance this guy here So the adversary can actually find out whether guys cut vertex or not Okay, just stop it and then see whether you can still communicate over the whole graph or not Okay, so these limits the possibilities we have So what they show is actually that if you have computationally bounded semi honest adversaries, so curious adversaries Then it's possible. Okay, you can achieve static security against an arbitrary amount of corruption The idea of their protocol is to use MPC in MPC in MPC in MPC. Okay to hide the network topology They use this recursive MPC approach like virtual parties However, this makes it a bit Involved and also inefficient.  For instance, what they they need is Public key encryption in a white box manner in an MPC. So you need to do Encryption and decryption of in the MPC. Okay, and then you do this recursively Okay, so the question is can you do this with less heavy tools? Okay, and for instance the natural question is can you do it just with encryption? Okay, we want to hide information. So let's encrypt and Our results are well, we have the same setting here. So computationally bounded semi honest adversaries and In this work here, we present a protocol which is just based on the DDH assumption We only use black box encryption Okay, so we do not have any MPC in MPC. It's just plain encryption and sending around messages And this leads to a simple protocol structure. Okay, and on top of such a protocol We then also describe applications. I mean classical application as before is MPC as was introduced in the TCC result Well, we especially focus on in our paper is topology hiding anonymous broadcast. So this is a broadcast way not only Hides the sender of the message, but you also hide any information about the network. So that's kind of complete anonymous communication Now in the rest of this talk, I want to focus on this protocol I want to give you some idea of how you can achieve topology hiding communication just using a threshold encryption Okay, but first let's start simple. Let's try to do this Communication so first of all as we are in the semi honest setting it's enough to actually construct binary broadcast Okay, once you have binary broadcast you can broadcast public keys and then you can complete the network Okay, so from now on we just want to focus on binary broadcast Okay, so here's the first attempt. How can we do this? Well, it's simple. Okay We have a sender here with its bit and we want to broadcast that value. So simply flood it Okay, it sends it to all its neighbors and so on and so on until you do this like a diameter many rounds You're done. Okay, then everyone in the network has actually learned this bit However, it's quite clear that this approach is not topology hiding at all Okay, for instance, if you consider that guy here, he can actually learn its distance to the sender by just Timing we have rounds. Okay timing the number of rounds until it gets the bit Okay, it's even worse. He can actually find the direction to the sender Okay, he now knows that those two guys sent him the bit So he knows the shortest path to the sender is actually over this or that note Okay, so this means the message pattern in this case leaks information about the network topology Okay, so can we do better? Yes, we can It's simply by using an ore flooding Okay, so in this case when you not only have this bit of the sender, but everyone has a bit Okay, sender has its bit B and the others have the input bit zero And so what we do now in this ore flooding is the following in each round You just send your current bit to all your neighbors Okay, and you compute the ore of the received bits. Okay, for instance, if you consider that guy here in the middle He receives all the bits of its neighbors and from the sender. He actually receives the real bit Okay, it computes the ore and this gives him the actual bit of the sender and you now repeat this process Enough rounds and then everyone has actually the bit of the sender So in the end what we do in this or flooding we compute the ore of all input bits However, it's quite clear that this is also not topology hiding because if say the sender has an input one Then you can do the same timing attack as before you just count the number of rounds until you see a first one And then you know the distance to the sender Okay, so not only message pattern, but also the message content leaks information about the topology And that's where we can now go and use encryption. Okay, so here's a third attempt Now assume that out of the nowhere suddenly parties have a public key Okay, everyone has the same public key and there is some trusted guy here at the bottom who has the secret key Okay, no one else knows the secret key. So what you can do now is you encrypt your message Okay, and now we do the same as before we just do this or flooding using encrypted values Okay, so you send around cipher text you compute the ore on the cipher text And if you do this in off rounds in the end what you have is a cipher text here Which contains the ore of all bits? Okay, and then you can send this cipher text to the trusted party the trusted party decrypt it for you and voila You have binary broadcasts Okay, there's there are some issues here. Okay first we have now assumed setup. This is bad We would like to do this without setup Second to compute the ore on us on cipher text. That's not so easy. Normally cipher texts are homomorphic with respect to addition So it's not so clear how we can do this or computation there and third obviously We do not have a trusted party which decrypts the values. Okay So how can we solve those problems? Well, the answer is our main tool is threshold encryption Okay, so here's the classical way of one looks at threshold encryption So you have a trusted party which computes you a public key and key secret key shares Okay secret key shares are given to the shareholders and the public key is given to some guy who wants to encrypt values Okay, but the guy has a message he can encrypt it and give it to someone else who has now the cipher text And now to decrypt what you need to do is you need to send the cipher text to the shareholders Okay, which then compute you what we call a decryption share and Using all those decryption shares you then can get back the message. Okay. Now in our case the threshold It's it's a full threshold. So n out of n threshold So you need all decryption shares in order to decrypt the message However, we cannot use directly this traditional threshold encryption because we do not have such a trusted guy who Distributes keys. So what we need is a decentralized way to generate the shared key So instead of having a trusted party what we need is we need a local key generation So part is the key the shareholders here. They can actually compute their own key keys Okay, public key and a secret key and what you can do then you can send around those public keys You can homomorphically add them up. Okay, and what you get is then this kind of composed public key Okay, and with this public key you can then do the same thing as before and We also assume same similar way that decryption shares are also homomorphically. Okay So now we no longer have a trusted party, but we can now decentralize generate the public keys And we can also decentralize do this decryption Okay, now, how can you achieve that? That's I mean, that's a nice property But how can you do that? Well, you can use do that using a DDH based encryption So Ilga Mol style threshold encryption will do the job Okay, this is a special property if you look at by encryption. It's not so clear if you have this property So that's quite cool that we have this type of encryption. So how can we now do the setup? Okay, so ideally as I said before ideally everyone has now this composed public key Okay, where shareholders all the parties have a key share. Okay, and then you could do this encryption stuff However, there's a problem. You need to be able to distribute that key. Okay, so that's kind of a circle So we have to weaken our goal. And so the idea is each party gets its own key. Okay Again, the key is still shared among everyone So in order to decrypt under that public key, you need actually all the key holders So all parties but Now every party has their own key and moreover, we do not distribute the key yet. So the key is just locally. Okay And how can you do that? Well, here's the idea If you have a graph of a diameter delta all parties are within my delta neighborhood. Okay So now you can recursively compute a public key which is shared among all parties Okay, so the idea is we do not compute one key We compute the whole series of keys. Okay, like pkk where this pkk is actually shared among my k neighborhood So k neighborhood means all parties within distance k of myself Okay, so this looks then like the following. So we have here pi in the middle who wants to have say a level k Public key. So what what do we do? Well, it's simple the part all its neighbors actually compute the level k minus one public key Okay, a public key, which is shared among their k minus one neighborhood And send it to the PI and PI and I can just Add them up and then you have a level k public key Okay, why is this topology hiding? Well, it turns out you cannot do it for one party Directly you have to do it in parallel for all parties. Okay, otherwise the message pattern will get you So here's the protocol idea So in the first round, okay, you do a level zero keys level zero keys That's just yourself you use this local key generation compute the public key and the secret key, which is now your key share How again, how do you get the level one key? Well simple send your level zero key to all your neighbors Okay, and now you receive the bunch of level zero keys You just add them up and now we have a level one key Okay So a key where the the secret keys are is shared among your immediate neighborhood and now you can continue this scheme As I said before here for level k You just need k minus one keys and if you do this long enough like diameter many rounds Eventually every party holds a public key which is shared among all the other parties Okay, so all we need to know here is let's say an upper bound on the diameter of the graph Okay, and that's how you actually compute the setup now here The next step was this or computation. Okay, so in the flooding protocol We had the ore computation during this flooding in each round you computed an ore. However, it's actually enough If you compute this or a right once, okay So the idea is each party has its own composed public key Okay, and it has an input bit the idea is the sender inputs his real message bit all the other parties input zero And so the ideal output would kind of be the ore which is encrypted of course under your own public key And the idea would be yeah, if it's zero if all bits of zero otherwise, it's one Okay, but you cannot have that. Okay, the issue is here Ciphertexts are homomorphic with respect to addition so you cannot directly get this So here we use a trick We encode the ore. Okay, so instead of saying zero one what we do is we just say, okay If all bits are zero we have zero and otherwise, it's just the random value inside the encryption And that you can get actually again in a recursive manner using this Neighborhood trick what you do is you actually compute Summs weight and sums of the bits where the weights are random Okay, so if your bits are all zero independent of the weight the whole sum is zero Otherwise, if you have at least one bit, which is non-zero you have a uniform random value in here So okay, and hopefully with The bit is non-zero. Okay, so you have a slight error probability in here, but if you have a large enough field, that's fine Okay, so this way you can now actually compute a ciphertext for each party which contains the ore Okay, you do this again in parallel for all parties to hide the message pattern Now all the remains is decryption and the idea of decryption is as follows the input is now this ciphertext containing your encoded ore and All the parties now have key shares. Okay for my public key And so we what we do is we want to decrypt this and we can do this similar to this or computation So the idea is I send out my ciphertext. Okay, which I want to do have to grip it Use when I send out this ciphertext actually you have to be careful. You cannot just send it out You have to blind it. Otherwise you leak information And what I get back are those Decryption shares which we can homomorphically add up so essentially ciphertext goes out and decryption shares come in Which allows me then to decrypt the ciphertext So in summary the protocol works as follows first you generate this setup so At the end of the setup every party has its composed public key Then we compute the ore Okay, the idea is that this ore Contains the actual bit of the sender if everyone else input zero. We are in the semi honest setting. That's fine And then we decrypt. Okay, so this way you can actually achieve binary broadcast just using threshold encryption Yeah That's it questions


Oh, so it's like My problem is easy to be just travel to the chest abdominal puncture.  The avenues of approach that ridgeline connects the flat part of the road It's a protecting there was the the reason why you're here is the biggest. Well, yeah I mean so get with and but then also you need to be talking to now You know it's a fallback firing and you got a place. There's nowhere else to it for us Just here to catch the one TC Recovering this hill sees the high ground PFC tour is here in KC TC We are here. We are protecting the spot winning first up to go down We are on standby and we're helping to show you know to show other people what we can actually do and We are here to protect the United States of America. So captain Eric P mom I'm the gladator company FSC commander. We're here today at our CTCP at KCTC. This is my distro


So yesterday we started talking about Higgs branch localization. And what we did is we chose another localizing functional. So instead of the standard thing, which is the sum over fermions that you have seen in essentially all lectures, I added this extra term where hij was some arbitrary functional of the bisonic fields in the theory. And I have the geigini over here as contracted with the killing spinners which describe this q.  And these enter over here. So with this extra term, the deformation is not positive semi-definite anymore. But as I explained yesterday, and as you've heard this morning from Francesco as well, after you integrate out the auxiliary field, which, as I mentioned yesterday, enters quadratically in here and linearly in here simply because of variation of the geigino is linear in the auxiliary field, you find an equation of motion which, when you impose it in here, is such that in fact this deformation term, at least the bisonic part, becomes a sum of positive terms again. And these positive terms, I've written on these blackboards again, we sort of looked at them yesterday. I'll just keep them there for reference throughout the talk. So I haven't quite specified hij yet. That was not necessary to ensure that this is a positive sum of squares. But there is one particularly nice choice I would like to make for hij. And this is where we finished yesterday for a u1 gauge theory with, say, one fundamental hyper-multiple, I would choose this hij to be some parameter, which is, as Francesco called it, a fake FI parameter, which is contracted with some tensor Wij, which I did not specify, but Wij is essentially on the round three-sphere. It's a symmetric matrix with off-diagonal entries, one over the radius of the three-sphere. So we have this fake FI parameter, and then we have things that typically enter into the vacuum equations in over flat space n equals 2 supersymmetric theory. Or if you like, this is the moment map operator. So this is where we were yesterday. And in fact, for hij equals to 0, so when this thing is not quite turned on yet, of course, we just have the old Coulomb branch localization, locus, and the entire computation we performed in the first two days goes through for that particular case. So the next thing we would like to consider now is, of course, what happens if you actually do have this hij turned on. So the first set of configurations I would like to consider are what I will call deformed Coulomb branch configurations, which I will characterize by the statement that the hyper-multiple scalars still vanish. So I'm just going to declare for this particular solution to the localization equations that the hyper-multiple scalars vanish. Of course, there will be more general solutions. We'll get to them later. But at the moment, let me just declare hyper-multiple scalars vanish. Then I don't need to worry about any of these equations. Since, of course, they're trivially satisfied. But I do need to worry about these equations. So when the hyper-multiple scalars vanish, you see that this extra term here is gone. But hij is still something non-trivial. It's proportional to this fake evi times this particular matrix wij. So that hij enters in here, and we need to solve these four equations. So we can do that without all too many details. I will just tell you the answer. Maybe you can check as an exercise that it is indeed a solution. So the gauge field A is proportional to kappa, where I defined kappa over here. Kappa is a one form, which is essentially the one form you get by lowering the indices on the keeling vector, where the keeling vector is obtained as the sandwich of keeling spinners. So kappa is some well-defined one form we construct from the keeling spinners. And the gauge field I claim is proportional to that one form with less proportionality factor, precisely this fake evi parameter. Note is also that this configuration satisfies the Bianchi identity. It's of course important that we do not violate the Bianchi identity, so indeed you can check that this is true. But this is not the only part. If you just turn on this A, you will pick up all kinds of terms in these equations, which are non-zero on odd first site. And indeed, we need to turn on additionally phi 2. So far, we had phi 2 to be 0 as over there, but now it needs to be turned on in the following way, where zeta is, again, this fake evi parameter. And now rho is a coordinate on S4, where I'm thinking of the S4 as an S3 fiber to over an interval. It's just the usual spherical coordinate. So we have S3s that grow and shrink again, and rho is that coordinate. Like on S2, it would be theta. So I have that particular coordinate explicitly in phi 2. And finally, phi 1 can still be its usual constant, where I should say that this constant, of course, commutes with zeta, because zeta was a u1 value thing. Maybe I haven't quite specified that. But this always sits in the u1 factors of the gauge groups. So phi 1 can still be a constant. And finally, dij just follows from the equation of motion over here. So it is minus 1 half times hij. I turned these off, so it just picks up this piece. zeta over l minus ia wij. So OK, we found this new solution. And you notice that when you turn off zeta, when you send zeta to 0, this new solution just smoothly goes back to the old solution we had over there. This is a billion solutions. Since, again, zeta lives in these u1 factors of the gauge group, so nothing changed the conclusion that on top of this solution, we still have instantons at the north pole, and similarly, anti-instantons at the south pole, just like we had over there. So good, we have found a first non-trivial solution to the new localization equations over here. But there are, of course, not all solutions that these equations allow for. There will be further solutions where I do not impose any more than the hyper-multiple scalar fields vanish. In particular, I can impose that what I call hij sw vanishes. We recall that to define this term, I split zeta as a sum of zeta-vac plus zeta sw, where zeta, zeta-vac, and zeta sw all have the same sign. And we'll fix zeta-vac in a second. But anyway, this condition now means that written out is the same as over there. But instead of having zeta, we have zeta-cyberg-withen. OK, so is it clear? I didn't do anything. I just defined zeta to be a sum of two zetas, where I declare that they both have the same sign. This type of a split will be clear in a second, why I do it. And then I define hij sw as that part of hij, which talks to one of these two parameters. So I just added zeta-vac over lwij. OK, fine. So I do this. So if I set this piece of this deformation h to 0, then of course, I still have, in the full hij, I will still have minus dvac over lwij. So I'll still have this piece. If I set this to 0, then of course you see that as far as a vector multiple equations go, I will get a solution of this type, but with this particular value plugged in. So as far as a vector multiple equations go, I will have, say, for a over there, I will get now zeta-vac and so forth. So this is not the interesting part. What is interesting now is that this equation, setting hij to 0 actually has non-trivial solutions as far as a hypermultiple. Let's go. Indeed, if I fill in the entries here and the same seed over here, if I fill in these entries, you may recognize these equations as the vacuum equations. I've been announcing all along that that was the aim of choosing this particular hij. So setting this equation to 0 is like imposing vacuum equations. And in fact, it's not just that equation that we need to take care of. It's also this equation, which you should recognize from the lecture this morning. We have an equation like phi q is equal to 0. Then we have equations like that, which in particular contain an equation that roughly speaking looks like this. And this should really, it's the same equation that Francesco had this morning. But now, since I'm in Ennico's 2 theory, I have two additional equations coming from these off-diagonal components. So I was sloppy here. There's some factors which I ignored. So there are two off-diagonal components, which since these W was, well, there are two diagonal components. W was off-diagonal. So additionally, I have equations like q tilde is equal to 0. So at the end of the day, these are just vacuum equations. For the 4-D Ennico's 2 theory, where I have turned on an FI parameter. So the picture is exactly the same as Francesco had this morning. You have your Coulomb branch, which is parametrized by phi 1. And at certain points in the Coulomb branch, there is a possibility of having a Higgs branch. So because at certain points in the Coulomb branch, this equation is satisfied. So remember that this equation is really to be read as having also the mass terms, because the masses were background values for vector multiple that's coupled to the flavor symmetry. So we really have phi 1 plus masses. So it's exactly the same picture as Francesco had, where roughly speaking, at these points, one of the eigenvalues of phi 1 equals minus one of the eigenvalues of this mass matrix. So we have these specific points on the Coulomb branch, where these Higgs branches open up. It's clear from here that it's only happening at point. And it's clear from here that there are indeed some solutions. If zeta is positive, then, for example, we can set q to be square root of zeta, or at least. Yeah, I'm thinking of it as in the fundamental. One can generalize, but I'm just sticking to the simplest example of u1 with one fundamental hyper-multiple, but it's general. So when zeta is positive, you see that there is a solution when q is equal to square root zeta. When zeta is negative, there is a solution where q tilde is equal to square root zeta. And you see that not the both q and q tilde can be non-zero at the same time. So if I parameterize, if you add it to the operator, so that's in the operator, if you talk to them, does this zeta fit the total zeta? Yeah, it's the total zeta. The sum? It's this sum, yes. I see, OK. Because, yeah, I haven't quite said so, if you turn on this kind of vacuum configuration, you're not quite done solving these equations. You can check that just plugging that in. You don't find a solution. In particular, you see that q sits here without the derivative. So if I set it to be constant, I should not expect this equation to be satisfied. And indeed, it is not satisfied. And that is why I defined this split. I'm going to tune this zeta vacuum precisely such that these equations are satisfied when I choose such a vacuum configuration. So again, imposing the vacuum equations, that was a good thing to do because of the equation over there. We had vacuum equations, so we imposed them. But the vacuum configuration is not a solution of these two equations, because you see here these q's just sit here, and they're not acted on by derivative. So if you set them to be constant from that equation, you should not expect this to be solved. But as I said, the vector multiple equations were solved by having a proportional to this zeta vacuum, where all the other ingredients are still written here, where you should replace zeta with zeta vacuum. If you take that configuration, you plug it in here, and you also plug in here the new vacuum solution for the hypermultiplets. Then you can figure out that if you set this parameter zeta vacuum to be equal to something, to 6, sorry. So if you set it equal to 6, then those equations are solved. If you are in the case where the total zeta is positive, otherwise you should set it to minus 6. So this is the case that corresponds to giving a non-trivial Vef to the hypermultiplet q scalar. This corresponds to the case where you're given a non-zero Vef to the hypermultiplet q tilde scalar. So just to be a bit more concrete, I mentioned it quickly in words what precisely the solutions are to these vacuum equations. But let me be a bit more precise. So first of all, let's look at this equation. And let's now maybe upgrade to a UN theory with fundamental flavors. So I'm going to choose the masses sufficiently generic. The masses are external parameters. So let me just choose them to be sufficiently generic such that all the components of phi are uniquely solved in terms of the components of m. For u1, it's obvious that this is the case, since then this is just literally a sum of two numbers. So you just choose phi 1 to be minus m. In general, you should think of this as a matrix acting on the gauge indices of this. This is a matrix acting on the flavor indices. You can try to solve it. And I will make sure what I will demand of phi 1 is uniquely solved in terms of the masses, which is essentially what I did already over there in that picture. Now if you do so, if the phi's take are uniquely solved in terms of the masses, and if they are distinct, then it's clear that you have broken your un gauge group, which I'm using now for my example, to a u1 to the n. Just because all these guys take separate values, and the gauge group is broken to u1, which act on the separate entries. But second, this u1 to the n is broken by the standard Higgs mechanism as soon as I also impose these other equations. So as soon as I impose the equations that state that this constant number, which lives along the u1 of un, is equal to this difference of positive definite things, then having the non-zero zeta here means that one of these two terms needs to be 0. That means that these guys get a Vef. And that Vef breaks this u1 to the n through the Higgs mechanism. So u1 to the n is broken through the Higgs. So all in all, we find discrete vacua just like in the picture. We have discrete points in the Coulomb branch, where Higgs branches pop out. And this is sort of the general logic. First, you break the gauge group to u1 to the n, just because you need to turn on non-zero entries in this phi1.  OK, so that was the second configuration. Now we already have two types of configurations. We have this different Coulomb branch. And we have these types of discrete vacua. And the third type of configuration that we can find is obtained by relaxing this condition that HijSw is equal to 0. Let's relax this condition. And let's, for concreteness, choose zeta to be positive. So when zeta is positive, I'm giving Vefs to the hyper-multiple scalars, q, and not to the q tilde z. Now, when you do this, just like this morning Francesco obtained vortex equations, we obtain some, well, these equations in these particular cases when you supplement all the information. And these are not analytically solvable, just like the vortex equations cannot be solved analytically. But we can gain some insight by studying various limits of the equations. You can look close to the north pole, look close to the south pole, far away from these two points. And if you do so, the following picture emerges.  Let me tell you the picture as it emerges on the squashed force sphere, just to be general. So this is the squashed force sphere. And inside the squashed force sphere, you can define two spheres. These three corners, they define for you one two sphere, which whose angular coordinate I will call phi. So what I mean by these three corners define something, I just mean that the locus where x3 and x4 is equal to 0 defines a two sphere, well, actually a squashed two sphere. So this equation defines a two sphere. Similarly, these three coordinates define a two sphere. And notice that these two spheres are actually set wise. They are kept fixed set wise by these two rotations to which the q will use square. So q squared to some j12, some j34, with some coefficients, and some more stuff. So this j12 rotates these two corners, j34 rotates these two. But the two spheres I have defined here are set wise invariant under these rotations.  So essentially this circle and this circle are parameterized by phi and chi. And you can also notice that the intersection of these two two spheres is precisely two points. It's the north pole and the south pole of my S4, OK? So why did I introduce these coordinates phi and chi? I introduced them because now I would like to define some winding numbers for the hyper-multiple scale. It's just like you define winding numbers for vortices. I will define windings for my more general equations over there as follows. I will say that q has winding phi with winding number n. And in the chi circle, it has winding number n. So this is just winding behavior I declare. And you notice that far away from the core, if I put, again, this square root zeta in here, then it will at least solve these vacuum equations. Far away, just like in the vertex case, you should expect that q goes to a vacuum, but now this vacuum has additional windings. So anyway, I define q to have these windings. And then the picture that you get by analyzing in particular these two equations is the following. So we have these two two spheres. Let me draw them. So this particular two sphere is embedded in S4. But locally, at each point on this S4, you can think there's a little plane leaving there. So S4 locally looks like R4. So any point on S2 looks locally like R4. I have two directions inside. S2, I have two other directions, which define my little plane over here. We call it R2 orthogonal. In fact, R2 orthogonal, where the angular coordinate is precisely going to be this kai. And the behavior you find by analyzing these equations is that in this plane, there lives a vortex-like solution. So this plane, of course, it can move on this two sphere. But wherever it sits on this two sphere, you have a vortex-like solution in that plane. So differently, the core of this vortex is this full two sphere. Is this clear? So the vortex solution Francesco drew this morning how the profiles look. You just, for q, you had something like this. So this is radius. This is like absolute value q. So in the core, it's zero far away. It will be square root of the phi et aliopolis parameter. And I'm just saying that this core wraps this entire two sphere. Similarly, so let's say that this was, for example, when I turn off m. For m equals to zero, this is the exact behavior of the solutions. And similarly, for n equals to zero, you find vortices defined in orthogonal planes, which carry winding number m. And these vortices have winding number n. So for Francesco this morning, his vortices, they were just like living at the North Pole, essentially. Their core was just a point. For me, their core is a full two sphere. And I have two sets of them. And in fact, when I relax these conditions, when m and n are both non-zero, I will have a vortex-like behavior on this plane. I will have a vortex-like behavior on this plane. And where these two spheres intersect, that is precisely at these two points, you glue them together in some smooth way. If you would like to see how this gluing works precisely, I can only refer you to the paper. It's somewhat complicated, but you can really see how they are smoothly glued together. So in flap R4, can you have the political immediate propagation? Sure, yeah. Non-zero m and n. Yes. This vortex-like behavior on this plane? No, for non-zero m and n, you have a vortex that wraps these two spheres. You have a vortex that wraps these two spheres. But these two spheres, they intersect at two points. And at these two points, these two vortex-like solutions are glued together in some smooth way. So maybe I should emphasize this. These are smooth solutions. That's the reference. Right. If extra charge is located at the intersection of the nodes, you see it at the most water sources. Sorry, say you? So you have to avoid defining numbers from each head and head. But is there any extra charge at the intersection? At the intersection, something interesting happens. Yes. Right. But there is no positive charge or something, so. There's something like a binding energy. There's an extra configuration at CISTER, which probably absorbs whatever winding energy you have in mind. So the picture I just gave you was really thought of as, when zeta is sufficiently big. So it's positive, but this picture is really valid when zeta is sufficiently large. Like, say, you take the limit to infinity. Because in that limit, these configurations, they essentially squeeze to have zero size. So roughly speaking, this size here goes like 1 over square root zeta. So if you take zeta to be sufficiently big, then the vortex solution is supported like in a very small neighborhood in this R2. And then you can really find these two solutions. You can find that they are smooth around the north pole. But in fact, you can do a bit better. Even though we cannot solve things analytically, when you go away from this type of a limit, we can find analytical bounds that must be satisfied. So when you relax zeta, when zeta is just finite, for zeta finite, you see that these vortex solutions, they acquire a finite size. They start taking more and more space. But of course, S4 is a finite manifold. So you do not have all the space you can ever wish for. So you should expect that if I had been more precise, I would have told you that this is really a square root of some winding number divided by zeta. So if the winding number becomes sufficiently big, you should really expect that things go bad. Or if zeta is sufficiently small, you should expect that you cannot support any arbitrary vortex on this compact space. And indeed, that is true. You can use these BPA's equations to arrive at an exact bound for the vortex numbers. You can find that this winding number M plus the winding number N should be smaller than or equal to zeta, where zeta is this fake FI parameter divided by 12. So this is an exact bound. It tells you that for a given value of zeta, there is only a finite number of windings that can never be supported on the manifold. And the reason is, again, because S4 is a compact manifold and these fortresses, they require more and more size to be defined. So at one point, you reach some tension between these two statements. And the tension is precisely characterized by this bound. Is this clear? On flight R4, there's no such bound. Because, OK, you see, this zeta is really multiplied with the volume of S4. And the volume of R4 would be infinite, so you don't see these kind of bounds. How far we can inflate? I mean, I don't consider the need of half ground. Right. OK, if you have. OK, I don't know what happens in the omega background. But I would still expect, well, there would be an equivalent volume sitting here probably. Anyway, I don't know. So OK, let me summarize a bit what we have done so far, just to keep track of the various solutions. So we have these parameters zeta which we can tune and we are currently tuning it towards a positive direction. When zeta was equal to 0, we just found the original Coulomb branch localization locus. Then we started turning on zeta. And we found that for any value of zeta, we have a deformed Coulomb branch solution. Moreover, we find that as you increase zeta, for certain values of zeta, suddenly this bound is satisfied. This bound is saturated. At that point, one of my Higgs solutions becomes available. And if zeta then grows even more, this Higgs solution smoothly becomes a vortex-like object. So there are certain points along this line. Precisely every time when this bound is satisfied, when extra smooth solutions start branching out, they start off as just some Higgs vacuum and then they smoothly go into a vortex thingy with winding m and n, depending on which thing you just satisfied. So this is at the moment our picture. We have Coulomb branch solutions. We have a deformed Coulomb branch solution anywhere defined for any non-zero zeta. And for certain values of zeta, we additionally pick up these vortex-type solutions. So at that point, we are getting new solutions, but still the answer is the states. Right, we'll see how that happens. Yeah, we'll see. It's a nice picture. But before we see that, I want to describe one more set of solutions. So as you pointed out, at the North Pole and South Pole, you may expect something more interesting to happen than what I have described so far. So on top of these smooth vortices with winding m and n, you should expect perhaps that at the North Pole and the South Pole, some additional point-like objects become available. Just like in the original Coulomb branch localization, at the North Pole and South Pole, point-like instantons were available. So at the moment is a question. Are there additional things to be considered? And the answer is yes. So let me describe that answer. So locally around North Pole and South Pole, we're really looking at C2. C2 is, of course, a complex manifold. And you can see that reflected in, as Guido was describing this morning, in having some almost complex structure, which in fact is integrable, so which satisfies the Nain-Hust tensor is equal to 0. So I will not call it a horrible tensor, but just the Nain-Hust tensor, which is how you pronounce it. So we had this almost complex structure, which you can define using the Killing speeders in much the same way as Guido was doing this morning. But since, of course, globally on S4, as Guido was mentioning as well, it's not even an almost complex manifold. So globally, you should not expect to be able to define such a thing. But in the patch containing the North Pole and separately in the patch containing the South Pole, you can define such a thing. It will be an integrable thing, meaning that the Nain-Hust tensor is equal to 0. So OK, we have this object. And in fact, around the North Pole, it is self-dual. Again, much like this morning in Guido's lecture, he found some self-dual and some anti-self-dual objects. Around the North Pole, we find the self-dual object. And with this object in place, we can try to analyze these equations locally around the North Pole. Before doing so, let me just do land on conventional notations, define alpha to be just this hyper-multiplet scalar, which compared to this almost complex structure is, of course, just a 0, 0 form. It's just a function. But I can also define a beta, which is essentially the other scalar. In fact, it's complex conjugate. But there is some dressing here, which is not terribly important. What is important is that this is now a 0, 2 object. So it's an anti-hologmorphic 2 form. The 2 form stuff comes essentially from this theta tilde. But anyway, roughly, alpha is q, beta is q tilde dagger. Now, with these 2 objects defined, if I look at these equations, I look at those equations, I focus at their behavior around the North Pole. Using this complex structure, I find the following equations. I will explain in a second. And finally, we have these equations. So let me define some symbols. First of all, I have split off the vacuum solution. So we had this vacuum solution, which was defined as a deformed Coulomb branch solution with a particular value for the fake evide parameter, something like 6.  And similarly, for phi 2, phi 2 took some value on the deformed Coulomb branch. I split it off, and I defined the difference to just be delta phi 2. So delta phi 2, you see, appear here. And this little a appears as a subscript, by which I mean that these objects are gauge covariants. And this is the field strength with respect to a, field strength with respect to a. So let's go through these equations one by one. So this is the Tobol derivative. It's gauge covariant. It acts on alpha, where alpha was this 0 comma 0 form. This is the dual. So you see this subscript is just a star. It eats the 0 comma 2 form. It makes it into a one form. I have the 0 comma 2 part of the field strength. It is equal to this bilinear in alpha and beta. I have the component of the field strength proportional to J tilde, where J tilde was this self-dual thing. So you see it's obviously proportional, because I have the J tilde sitting here, where I have, what roughly looks like the vacuum equation yet again, inside. And finally, I have equations like so. So the first thing to observe when you find these equations locally around the north pole is that there are a little bit too much. There's too many equations on the blackboard, simply because equation one and equation two, when you take them together, when you take these two equations together, you find that either alpha is equal to 0 or beta is equal to 0. Maybe I should do this little exercise. It will slow me down a bit. So we find these equations around the north pole, but we observe that the first two have some consequences. Indeed, if I take the first equation and I act well. So the first equation says that this is true. Now, let me act with another double derivative. Then I find something like this. But here, this is all in form language. So this is essentially an anticomitator of two covariant derivatives that we know is proportional to the field strength and more precisely to 0,2 component of the field strength acting on alpha. But now you see that from there the 0,2 component is precisely equal to i over 2 alpha bar beta. So this is really equal to minus a half alpha squared beta. So this was easy enough. And now the usual trick is if you see an equality like this, you integrate it over your manifold, which in my case maybe is just R4. So you integrate it. The left-hand side becomes what? You integrate it against the complex conjugate of beta. This is really just after partial integration. This is really just the norm of this object. This is, of course, a positive definite thing. But from the right-hand side, we find if I multiply with a beta over here, you find something, well, up to a factor of a half. You find the norm of alpha squared, the norm of beta squared. As, they're all little a's. So little a was here. So I split off the vacuum solution. I have everything else. That's a little a. So all these derivatives are catecholarian derivatives. So you see here is something positive definite is equal to something negative definite because of the minus sign is equal to something positive definite. So that better means that both sides are separately 0. So in particular, this implies that alpha is equal to 0 or beta is equal to 0. And that this piece is equal to 0. And since this piece is equal to 0, you can go to the first equation and find that also this thing should be 0. This is what's the sign of 0. So in the first time, you have minus sign. So you can go to the big bar. Hey, look here. No, it is important. I agree. So good. We reached this conclusion. So in these equations, they're really a bit too much. So we have either alpha is equal to 0 or beta is equal to 0. But we were all along considering the case where zeta was positive. And for zeta positive, we really like to give q of f. This is reflected in here the far away. You expect the field strength to be 0. If zeta is positive, then from here, we already know that delta phi 2 is equal to 0. So if zeta is positive, then you better be alpha that gets f of f instead of beta. And since one of the two has to be 0, yeah. So it has to be alpha. So we can set in these equations, OK, this equation. In particular, we can kill this equation. Either alpha or beta is equal to 0. We chose alpha to be non-zero, which implies that delta phi 2 is equal to 0. Or at least delta phi 2 is equal to 0. This thing is equal to 0. And let's keep these equations as they are. And now you can believe me that these are precisely the Cyberg-Witton monopoly equations. Sorry? 0 is equal to 0. Are you going to explain them? I didn't catch it. Are you going to explain 0 is equal to 0? No, these are the equations. And then the equation for aberrant theory? I wrote them down for u1. Typically, you write them down for u1 theories, yes. Just because the idea of Cyberg-Witton is that you sit in some Coulomb branch vacuum, you flow to the IR, you have some u1 theory. But you can generalize these if you like. Now you get what it goes on the name of generalized monopoly equations. I mean, there's a generalization to non-abiding gauge groups. But I'm happy with the u1 case, since you remember that one of my vacuum equations broke the gauge group to its u1 to the rank of the gauge group. And I'm just focusing on one of these u1 factors. So it's enough to look at u1 for my case. OK, so these are the equations. And at the south pole, you can play an entirely similar game. You find also Cyberg-Witton monopoly equations. So we should, in principle, include these guys. And in particular, we should include, as for the instantans, we should include an integral over their modular space. There's one comment I would like to make about solutions to these equations. Locally, in this case, around the origin, you can show that there exists a solution that goes like this. So again, we have windings m and n, which are accompanied by the relevant radii in the two orthogonal planes. So you have a solution like this locally around the origin. You can complete it to a solution all over the place. It's essentially like the vortex, which is, in fact, the comment I wanted to make. If you set this as a ansatz for the solution to these equations locally around the origin, you're just going to find vortices. So Cyberg-Witton monopoly equations are essentially a generalization of the vortices. If you restrict attention to one of the planes inside R4, these equations just become vortex equations. I don't want to say much more about these equations. They're just there. There's extra configurations at North Pole and South Pole. What I would like to get to now is to picture how all the solutions I found so far are glued together in some coherent fashion. So let me start computing the partition function in my Higgs branch localized setup. So again, it cannot be repeated enough. Hij was introduced through a q-exact term. So that means that whatever answer I find for the partition function, it has to be given by the same result as the original computation. So the representation may be different, but it should really be equal to this Coulomb branch integral, which we found earlier. So in particular, the answer will be the same. So let's see how it is even possible that all these complicated configurations give an answer which is the same as the original integral. So on all these configurations, you should be computing the classical action. You should be computing the one-loop determinants. To compute the one-loop determinants, I'm just going to assert that it is possible to evaluate the one-loop determinants even on these somewhat strange-looking solutions. Even though I don't know analytically how the solutions look like, I have enough information to actually do so. Because these index theorems, which you will learn about next week, tell me that it is sufficient to know what the behavior is locally around the north pole and the south pole of the solutions. And that type of behavior I have access to. So this we wrote before. This equation is exactly the same. There's this upstone function, product over all non-zero roots. But here, now the a hat is really evaluated at the north pole, but it happens to be so that it's equal to the evaluation at the south pole. This doesn't have to be. And if it's not, then this thing splits up in some interesting way. It's equal to what I call a hat over here. And this expression is something like this. So it's a very explicit expression. You see the gauge field sitting here. You see this killing vector. You see the two scalars, phi and phi tilde. You see these pyliniers, ezenes, tilde, which are defined somewhere here. It's a very explicit expression. In this expression, I know how to evaluate, even on the solutions I don't know analytically, I still know enough to evaluate them at the north pole and the south pole. So this type of expression, you evaluate it. You shove it in here, and you have computed the one loop determinant of the vector multiplet. And similarly, for the hypermultiple, you have this product over weights of the representation in which they transform of the epsilon function. If you like, you can dress it with some masses. OK, so whenever I say I claim that I know how to compute one loop determinants on these configurations, this is what I really have in mind. These results come from an index theorem. They only care about what happens at the north pole and the south pole. What happens at north pole and south pole is, in fact, equal. And it's equal to this thing, which I know how to evaluate. OK, so those were a lot of words. But let's put them to practice now. Let's start with these deformed Coulomb branch solutions. What do we need to do? Evaluate the classical action on the deformed Coulomb branch configuration, and evaluate that equivalent gauge parameter, and then plug it into this one loop determinants. So we do the computation. We take the vector multiple action. We evaluate it on this deformed Coulomb branch, and we find that this thing is actually remarkably simple. It is just given by the following expression. So notice when zeta is equal to 0, we just find the old expression, which of course has to be so, because when zeta was equal to 0, we just had the old Coulomb branch configuration. So we just find this shift in A in the evaluation of the classical action for the classical Yang-Mills action. The hyper-multiplet action doesn't do anything on a deformed Coulomb branch because, of course, we set it to 0. All the hyper-multiplet fields were 0 in this configuration. And now the true miracle, almost, is that if you evaluate A hat, this parameter that enters into the one loop determinants, you evaluate this parameter on this deformed Coulomb branch configuration I gave earlier, you find the following. You find that it is A. This is just the constant value of phi 1 plus this imaginary shift, which is exactly the same as this shift over here. So now slowly, the picture starts to become clear. We are supposed to integrate over phi 1 or just this A, this constant value of phi 1. And we're supposed to integrate it along the real line. This was the old Coulomb branch localization result. Now you see there from this one loop determinants, in particular from the hyper-multiplet one loop determinants, the integrant has various poles. These poles, they sit somewhere in the complex plane. There's like a double infinite tower of poles located at specific points where A is like minus m, minus some other m. So at these points, you have this double infinite tower of poles from the one loop determinants of the hyper-multiplet. OK, so this is the analytic structure roughly of the old Coulomb branch result. Of course, we also had instant on partition functions, which naively may give you also some poles. But the poles of the instant on partition function precisely cancel against the zeros of the vector multiple of one loop. So there are only these perturbative poles in this integrant. And over here, there's also towers of poles. So this is D from Coulomb branch. What this does for you is it simply shifts the contour in the imaginary direction. So it's crucial that there is this I here, and it shifts this contour, it shifts it up if zeta is positive by an amount proportional to zeta. So this is a D from Coulomb branch configuration. It's the integral over this line of exactly the same integrant. Is this clear to everyone? But now we can start having fun. When zeta is 0, of course, we get the same answer as we used to. If we slowly increase zeta, we will keep finding the same answer as long as we don't cross poles. But as soon as we start crossing poles, as soon as we reach this point, you can guess which zeta value this is. This is precisely zeta vac. So as soon as we reach this special value for zeta, all of a sudden these Higgs branches start to pop up. Or I said differently in terms of my bound. This is the first value of zeta for which I find the bound to be saturated. So my bound looked like so. If you said m and n equals to 0, you find that precisely when zeta is equal to 6, which I claimed was indeed the value of zeta vacuum, you find that this is saturated. This saturation indicates that there is a new configuration that becomes available and it's precisely this Higgs branch faculty found. So we should expect that the residues of these poles that we start crossing should precisely correspond to the contribution of these extra vortex-like solutions together with these cyberquita monopoles. So we keep pushing up zeta. We start crossing more and more poles every time this type of a bound is saturated. And we need to include more and more contributions from these vortex-like objects and these cyberquita monopole type objects. So let's see if that is, well, maybe I just assert that it is true. There's no point in trying to write the same equations where you can see that these residues are indeed there. Just the fact that you get this residue has to do with the location of the pole. The poles sit at a specific value for phi, namely the value such that this is equal to 0. This means that that particular hyper-multiplace is effectively massless. So it used to have a potential, but now this potential becomes, well, it flattens out. So you have this 0-mode. You need to remove the 0-mode. And this is precisely what the residue does for you. So when you push this line up, you pick up all these residues simply because you have a 0-mode that you need to kick out. Now, I should mention that for this picture to be completely consistent, there is one statement that needs to be true, namely that if you have the instant-on-partition function evaluated at such a special value for a, parameterized by these two types of windings, that this thing needs to be equal to whatever my Cyberg-Whitham monopoles give. So we have a Cyberg-Whitham monopole of two winding numbers defined by m and n. And it must be so that if you evaluate the instant-on-partition function on a very special value of a, to be precise, this value is precisely the value in the Higgs vacuum plus i times mb plus nb inverse plus q over 2. When you plug in this value for the gauge equivalent parameter in the instant-on-partition function, the claim is that that should produce for you what maybe you should call the Cyberg-Whitham partition function. It's the non-perturbative piece you would find when you try to compute the partition function in some omega-deformed background where you have turned on some evi parameters such that configurations you have are Cyberg-Whitham monopoles. So this is an assertion, or it's sort of a consistency requirement for this picture to be true, but there is no independent verification of a statement like this. So essentially, it's an open problem. It's an open problem in general. But there is one, or there are two limits where it is easy to check or relatively easy to check. When m is equal to 0 or n is equal to 0, you can evaluate the instant-on-partition function on this particular value for the gauge equivalent parameter. And you find precisely the vortex partition function of the vortex theory described by this winding m. So we have a vortex configuration in 4D. It lives in some plane. Its core lives along some R2. And we have the same picture as we had for the instant-ons while the vortex moves in this R2. It can explore its modelized space. So really, you have on this R2 the modelized, the world volume theory of these vortices. And that world volume theory, in fact, also possesses vortices. So the vortices of the vortex theory are precisely captured by this object. Similarly, when you set m to 0, you find a vortex partition function popping out. And more generally, the claim is that you find this non-perturbative cyberquit and partition function. So on the right-hand side, does it always correspond to a single-prosciption where there could be multiple-prosciptions used originally for it? So you have to use it. When you cross the abstraction of my own data, there is a new classic solution. So is it always a single solution? Like, for example, sometimes it looks like that in the case of the contours, you might cross two poles, for example, three poles. So that's kind of a thing. There is a one-to-one correspondence between the single poles or the state-of-the-art sum. If it's U1, it is just, I think it's 1, 1. But if it's like UN, you need to split this up. You need to partition this for the different directions in the maximal torus. So then you get this, yeah. So really, OK, this is schematic maybe for U1, and maybe I shouldn't have. Just to make sure, to some extent, your logic shows that that should be the case. But you both go to questions of verification directly. Exactly, yes. Just try to compute this object just like one would compute the instant-on-partition function. Maybe as some integral over the modelized space, but who knows what the modelized space is, when you consider these things on C2 with some omega deformation.  Now we're really starting it on C2. So yeah, I suppose it has an open problem, but probably it's true. I mean, it should be true. At least in this limit that I described, it's manifestly true. This thing becomes a vortex partition function of the vortex theory when you set one of these numbers to zero. What do you write as the C SW after you contain the configuration from vortex-like configuration after you go for this stream? Well, these smooth vortices, they're smooth, right? They're smooth configurations. I don't need to do anything weird. I just use these one loop determinants, and I evaluate those guys. I evaluate one loop determinant of quadratic fluctuations around these smooth configurations. I can do that not by hand, but using an index theorem, and I will just get the result that I read up from these expressions. But I could use it in some limit. In some limit, like n equal to zero, the C SW should be just a vortex function. It becomes a vortex partition function, but not of these smooth vortices. They're vortices of the vortex theory. OK, yeah. So there's two sets of vortices in time. Yeah. But yeah, this statement has been verified in various examples. But in principle, there's a machine able to define the modularized space intersecting all the streams. On top of that, I'll read them all. Not to define the modularized space, you should know that you can do some reading information without a commotion, because you have faith in it. Sorry, I missed the last part. In order to define the modularized space, you should know that you can do some commotion. It's probably true, yeah. But to compute this beast, you do need to compactify things properly. OK, so are there any questions about this picture? I mean, I can give more details, but. So what happens if zeta is negative? Well, when zeta is negative, then essentially all the same words go through. But instead of giving q of f, I start giving q tilde i of f, there's a similar threshold over here, where zeta vacuum is minus 6. And then it's the same story. OK, so we have this picture. But at the moment, for any finite value of zeta, we're going to get some mixed representation. We have to perform this integral over this shifted line. And we have to add in some residues of some of these smooth vortices together with their cyberquit and monopole dressing. So what would be more useful, perhaps, is to send this thing all the way to infinity, such that this deformed Coulomb branch integral, well, it's add infinity. And if we're lucky, the contribution from infinity is equal to 0. And then we really just get a sum over all these types of configurations. So let me write that in some equations. So if we send zeta to infinity, we should, of course, verify that the contribution of, well, it's not even an arc. It's really a line. Of this line at infinity goes to 0. Let's assume that it is the case. Then the result really looks as follows. And this is where, really, the Higgs branch localization shows that it is a Higgs branch localization in its representation as a sum over Higgs vacua. So we're going to sum over the Higgs vacua, these classical solutions to the vacuum equations. We find the classical action evaluated on the Higgs vacuum. We find the one-loop determinant evaluated on the Higgs vacuum, where the prime means remove the guys which actually are diversions precisely because of the zero-mote I explained and replace those with a residue. And then finally, we have a contribution from all these re-summed, or yeah, we have all these smooth vortices or vortex configurations with winding m and n. When you send zeta to infinity, their size shrinks to 0. You have them all because this bound is not valid. Well, it's still valid, but it doesn't have any meaning anymore since zeta is infinity. So we can sum over all these configurations so we can sum over its one-loop determinant and this cyber-quitten beast. So internally, this big re-summed object is really a sum over unrestricted integers m and n of some classical action, which I will not try to specify, some perturbative piece and some cyber-quitten piece. And now just to wrap up the discussion, if you pick integers m and n, or let's in particular pick n equals to 0 and let's pick any m, then this piece, if everything I've been doing is consistent, this should be the s2 partition function of the m-vortex theory. So why does it have to be the case? Remember that we had these vortices living in the planes and wrapping in s2. So the core of the vortex is the s2. So as we move along the s2, the vortex will wander around in this modulite space. Set differently, on s2, we should have the world volume theory of this m-vortex. That's what I wrote here. And we should include the partition function of that object. And indeed, you can verify explicitly that if m is what it ever is, but n is equal to 0, this is precisely the s2 partition function. Now, it doesn't look like a partition function at all since you just have a product of things whereas the integral. But this is what Francesco explained tomorrow. When you do Higgs branch localization on the two sphere, you will just get an expression like this. You will get some classical actions, some perturbative stuff, and then two vortex partition functions, one at the north pole and one at the south pole of the s2. So again, these are the vortices of the vortex theory. And I already mentioned before that this thing precisely becomes the vortex partition function. So this really is a s2 partition function, but it's expressed in its Higgs branch localized form. Are there any more questions? I thought you have a different way are you going to talk to the breakers? Yeah, so if you like, yeah. So what you can do is if n and m are both non-zero, you can still compute this because I gave you, I told you how to compute this thing. It's just an evaluation of the instant on partition function at the specific value of the gauge equivalent parameter. And then you find that this thing is really a partition function of two intersecting s2s. So you have this vortex theory on s2 number one. You have a vortex theory on s2 number two. They intersect at north pole and south pole. And that is what this thing computes for you, the intersecting partition function of these two vortex theories. But the system you start with does not have a surface operator. No, but you can think if you just take this configuration and you float to the infrared, it becomes essentially a rigid object, which is an intersecting surface operator. So here we were looking at dynamical objects, but if you freeze them out, they become surface operator. Moon dynamical for surface operator. That is theory, that is story, general. If I additionally introduce surface operators. OK, I don't know. If there's no more questions, then I'll change topic to something hopefully easier and more fun, namely the AGT correspondence. So now we have seen two ways to compute the s4 partition function. We really know how to compute this object. I already mentioned that next week, so I will teach you how to use this object to compute correlators of n equals 2 carol operators. But what I would like to do now is to explore the AGT correspondence, which relates this object to something that at first sight doesn't seem to be related at all to this partition function, namely a two-dimensional CFT, which can be either Liouville or Toda if you consider class s theories. So OK, I introduced two things. I will explain them one by one. I will first explain to you what are class s theories. And then I will explain a little bit what is Liouville theory. I will not talk about Toda. And then we'll see why you should expect that there is a relationship between these two and give some meat to that relationship. So let's start by introducing theories of class s. And I will restrict myself to theories of class s of type A1. It's the simplest class of class s theories. It's sufficient to explore what AGT proposed without having to deal with too much complications. So the canonical reference for anything having to do with class s theories is Davide Gaiotto's original paper, which I will sort of follow for these lectures. So let's just start by considering four free hypermultiplets. As you remember from the first lecture I gave, four free hypermultiplets, they carry a flavor of symmetry, USP 8. So USP 2 times the number of free hypermultiplets. So let's remind ourselves why this is the case, or why this at least could be the case. Four free hypermultiplets contain 16 real scalars. Each hypermultiplet has four real scalars. So in Toda we have 16 real scalars. 16 real scalars you can try to rotate with an SO16 symmetry group. And indeed, you can rotate them by an SO16 symmetry group. But this symmetry group really contains part of the R symmetry. So we should split off the R symmetry explicitly, and then the commutant is precisely USP 8. So this is just some standard group theory fact that SO16 splits into SU2, which in this case is the R symmetry times USP 8. So USP 8 is a flavor symmetry of four free hypermultiplets. But really I would like to think of this USP 8 as in a similar decomposition as here, as SU2 times SO4. And then I'm also going to use that SU2 is, SO4 is really SU2 times SU2. So I can study a subgroup of this full flavor symmetry of the free hypermultiplets given by SU2 times SU2 times SU2. So in some picture, I could try to describe four free hypermultiplets by emphasizing these three factors in the flavor symmetry group by some trivalent vertex. So each leg tells you there's an SU2 flavor symmetry group. So this is just a picture I'm inventing to describe four free hypermultiplets making manifest part of their flavor symmetry. What was important about this subgroup of USP 8? At the moment, nothing is important. It's just cute that there's three SU2s, and I'm going to invent a picture for them. You'll see why it's useful in a second. In fact, I can be a bit more artistic. I can fatten these lines a bit and draw some kind of a picture like this, which, if I round the needle a bit, is really a sphere with three punches. So these are just pictures where either the legs or here the holes describe an SU2 factor in this flavor symmetry of the free hypermultiplets. So this was really a simple theory for free hypermultiplets. Let's consider a slightly more interesting theory. Let's consider an SU2 theory with four fundamental flavors, with four fundamental hypermultiplets. Notice that the matter content of this theory is precisely such that the beta function is equal to 0. So the perturbative beta function is equal to 0, but we're looking at n equals 2 supersymmetric theories. So if the one new beta function is 0, it's 0 all the way, also non-perturbatively. So this thing is really an SCFT, just like free hypermultiplets are an SCFT. And recall from, again, the first lecture I gave that when you have four hypermultiplets transforming in a pseudo-real representation, so the two of SU2, the fundamental representation of SU2 is pseudo-real. We have four objects transforming in a pseudo-real representation that implies that you have SO8 flavor symmetry. So the naive U4 flavor symmetry, which you would assign to four hypermultiplets, is enhanced to an SO8 flavor symmetry. So the standard quiver description for this theory is as follows. We have an SU2 gauge group, and we couple four hypermultiplets, where in the back of our minds we know that the naive U4 is really enhanced to an SO8. What I will do now is, again, maybe a little bit arbitrary at first sight, I will split these four hypermultiplets in two groups of two. So nothing happened. Really, each of these two groups carries a naive U2 flavor symmetry, which is embedded in this U4. But really, because we have an SU2 gauge group, there are SO4 flavor symmetries. So each of these U2s is enhanced to SO4 precisely, because the two of SU2 is pseudo-real. So I can invent a new picture that looks like so, where the SO4 flavor symmetries that are acting on each of these two hypermultiplets is represented as SU2 and times SU2, where I have, well, I invented this type of a picture to indicate that flavor symmetry a bit more explicitly. From this U2 over here, I made explicit the SO4 as SU2 times SU2. Now, let me simplify this picture slightly. I can draw it maybe like so, where the gauge group I just replaced by this disconnected line, and I have this trivalent vertices sitting here again. But really, what you can do now is let's split this for a second, and I realize that this is precisely the description I have over there. And when I glue them together, all I'm doing is gauging the diagonal SU2. So each of these legs that stick out here carry an SU2 flavor symmetry as over here. And when I glue them, what I really have done is just gauge the diagonal subgroup of that SU2. I've literally done this picture when I connect these two objects like this. Is this clear? It's sort of an important point. And then, of course, again, I can be my artistic self and draw it as some fattened version of itself, where the long tube indicates gauging, gauging of the diagonal subgroup. Maybe I should stop here. Tomorrow, we'll continue with these pictures, and we'll see what it all means in terms of 2D CFTs.


everyone I hope you're welcome back to my channel or if you're new here my name is seppi and in today's video I'm bringing you along for my weekend off vlog so I've got some time off it is the bank holiday and I basically thought I'd start by showing you guys what I'm having for breakfast today doing some shooting which is why my makeup is so OTT but I followed a makeup tutorial by someone on youtube called hindash and I really really really like how it came out was like the summery look and for breakfast I'm having toast with avocado I don't know if you can see that I'm having toast with avocado and egg and latte I'm joined by my lovely mom hello everybody so we're gonna have some breakfast and then we're gonna go shoot some content for the gram my bun is like slanted but yeah I'm gonna have this breakfast and then I'll catch up with you your makeup is really thank you I was saying I followed a makeup tutorial from a guy on youtube called hindash wow he's Arab and he lives in Dubai and he is incredible thank you it looks a bit harsh when I hold the camera close but in this lighting it looks a bit better and in photos it's really good for photos guys and like parties and like summery time but the weather in England is so rubbish guys it's far from summer so guys I've just arrived for my laser hair removal appointment I'm actually really excited to vlog this for you guys because I've shown it on my instagram before and you guys had a lot of questions about whether it's painful how much it costs etc so I'm gonna go inside and introduce you guys to the lovely ladies who do my laser and we will answer like the FAQs that I got last time that I got laser that I shared on instagram so yeah I'm really excited it's completely pain free so it's like chillaxing vibes which I'm looking forward to so guys I'm now literally laid down relaxing like I told you getting laser is actually a relaxing experience because a lot of you guys asked me last time if it's painful and it is absolutely not painful the only part of my body that we put numbing cream on my underarms and bikini area and that is in order to ensure that it is definitely pain free because that's like more you're more sensitive areas but last time I honestly did not feel like a thing like I was just having tea chilling like chatting so today I want to introduce you to someone in the tea this is Monica so Monica and how to do my laser me here at bliss and we're gonna answer well I'm gonna ask Monica some of the questions that people asked me last time and we are gonna like talk through the answers to them basically is that okay Monica okay so the first thing everyone asked last time is what machine do you use and like how is it painful is it not painful okay so we use the infinity diode laser so it's actually one of the main things that are out there at the moment actually for that reason this is a diode laser it's very refined and it just focuses on the actual hair that's what draws the heat in the head the color of their follicles and that laser will just zoom into those so it's very safe pain free it doesn't damage any of the surrounding tissues yeah I didn't see like any sort of damage to my skin after the first time it was completely pain free and this is the machine as well just in case you guys want to see it the next question that I got is how long does it take to do like the whole body laser okay so whole body about an hour and a half there's obviously some set of times and obviously we're getting prepared and a bit of off to care but about an hour and a half to hours maximum and how many sessions in total to be like hair free all right so the recommend at least six and then we sort of assess to see how it goes and see how the hair is responding to the treatment yeah but how about the room between six and eight okay one last question how much does it cost to get laser okay so small areas start from about point of camera first session and if you've got a whole body and you're feeling pretty much everything that's 400 pounds okay so you can find their prices on www.theblitzgroup.co.uk that'll be across the screen or you can just DM them I'll write there at the screen as well this laser clinic and they will answer all of your questions on there but we are just gonna go ahead and start the laser 45 minutes later and I am done my full body is lasered if I've even been less than 45 minutes because what you guys didn't see at the start of this clip was I actually locked myself out of my car and Hannah helped me climb through my boot to get my keys again so that took up some time but yeah literally 45 minutes later and my whole body is done I'm lasered and my next session will be in four weeks and as Monica said it takes about six to eight sessions to be completely hair-free I absolutely don't my experience here I want to see 10 out time recommend it and definitely so guys I'm in Zara in Knightsbridge I just had lunch with my friend Min who wants to say hi without a mask but we'll just show you her she's carrying my shopping too as you could tell she's already fed up with me Min and I went to uni together so so Min will have cut off in the last scene because she wanted to show you her stuff now I'm joking we're gonna take our masks off so you can have a proper introduction today in shopping Min's been like my mom everything I pick up she's like but do you need this I've seen you with this before or like if I want to escalate you're gonna fall thank you thank you so today is probably the first day that it's not been raining in ages and we are gonna shoot some pics for my instagram and we just done some shopping um and yeah I'm just gonna explore Knightsbridge say hi properly Min so as you guys can see my room is a complete mess um I don't even know how it go into the state obviously I've thrown a lot of things onto the floor to organize back into my wardrobe which is why a lot of it's on the floor and a lot of it's on my bed as well um so I'm just gonna need to organize that and get my life together so so guys I've managed to clean my room the mess on the floor my table my closet my bags are all organized in there um here is me it didn't take me until sunset to clean my room it's just I had to put makeup on and put on this top because I've opened my own d-pop store for my own clothing line so if you guys want to check it out it's remi and nato and this top is on there it is 12 pounds I believe um it's listed on there and it's just this white ribbed top which I really like and yeah if you guys want to check out the other items I'll have the d-pop across the screen but yeah I just had to take pictures of that so that took some time also I love these earrings they are from a recent boohoo haul of mine but guys forget all of this fun stuff I actually have something really big to show you guys my degree came in through the mail so I really want to show it to you guys and uh I've already opened it as you can see we see it's so weird because we meant to get your degree at graduation ceremony but we just got ours posted um so I thought I would just show it to you so you guys can see what it looks like class of 2020 booklet thing here is my degree certificate so this is the certificate and it's so exciting you guys will ask is it md what is it so in the uk we get a mbbs degree so it's bachelor of medicine and bachelor of surgery um so dual bachelor's degree um yeah that's how every single medical school in the uk gives out degrees it's just the letters that we have whereas I know that like my sister who studies in Latvia is going to get a medical doctorate so md and I believe in America it's an md as well but I think it's because Sina has to write like a 10 000 word thesis so and we didn't have stuff at that so yeah I think that's probably why but still at the end of the day you're still a doctor in terms of me getting a job uh it hasn't had an effect because we're all all of us graduates at mbbs but I thought it was so fun I thought I'd share it with you guys because I was so excited to get it and also guys can you put suggestions down below of um a really nice like frame so I don't know if you guys watched the manchester vlog so if you guys watched the manchester vlog then that you would have seen that we went out for christina's birthday and she got me this really cute frame to say thank you in this picture of the two of us which is by the way a really cute picture because on her instagram um anyway I want a frame and I really like this one I don't know if I should ask christina um where she got this so I can get this another size or if you guys have any good suggestions for frames for my degree let me know so almost doesn't have a frame on her degree either I think it's literally sitting inside an envelope but she has really nice frames for her graduation photos which are downstairs I'll show you guys actually but yeah if you guys like have any frame suggestions I want to buy one for soul and one for me and maybe while I'm at it buy sign this one as well so the three of us can have like nice pretty ones hopefully ones that like can sit up maybe we'll put it on top of the fireplace or even if it's like something that you have to put on the wall I'll show you where souls graduation pictures are we could mount it up on against the wall but obviously it's a huge achievement for all of us and we're really proud of ourselves and our parents are proud of us so we I want my degree to be up on a wall like I've worked hard for it so why not I don't think I need an excuse I just want it on a wall I want to see every day and be proud of myself so this is soul's uh solo graduation pick which is so cute and she's like holding her degree really pretty and then this is our family graduation pick for soul which is one of my favorites by the way when I when I bought this dress sign it came with me to Selfridges and she was like oh my god this dress is so hideous you're gonna look like a flower wall like what are you wearing we're gonna wear the blood I was like no I promise you it's gonna look nice and uh can I just say on the day when I actually wore it with heels makeup everything done up I looked way nicer than signer first of all because she was wearing plain black but second of all I think me and my mom were like the violences of the family because my mom's also wearing this really cute red dress I'll show you guys look at my mom like how cute does she look like she's wearing this red house of cv dress and um my dad again looks really cute um yeah so me and my mom were the show stoppers here I think um and this dress was from top shop and soul and signer wore black dresses um soul wore a black house of cv dress um for her graduation and yellow christian lebutons which were so iconic but I remember soul would always tell me oh my god you need to pick your graduation outfit so I've still got another year because apparently my uni is going to host a graduation for us hopefully um so if they do then guys I also need suggestions for graduation outfits because I want something that's like classic and timeless um but also doesn't look like I'm going to a wedding um and doesn't look like I'm going clubbing because it's so hard to get one that's like in between I'm having watermelon and mango my mom is having persian cucumber um I don't know if you guys have watched her recipes on her channel mezagedema buddy it's pretty good she's still eating but she wants to promote it to you so I just promoted it for her yes thank you very much I think I had steak for lunch guys I'm back on intermittent fasting so I think I'm gonna have chicken schnitzel for dinner yes in our house everybody's look like they are in the restaurant they order different food it's not like that they're having one food I make it for example I have to ask what do you want she said I want a steak and for dinner I want chicken steak and other one she said I want fish and I'll have fish but you're not having fish anymore because Simon said she's going to have nandas yes that's why no just I'm talking generally generally otherwise she makes zershpullo so we're going to watch our movie now and then I'm going to eat something yes so we'll see you guys in a sec so guys for dinner I am making some rigatoni and then I am adding to it some arabiata sauce and the second half of this creamy mushroom sauce that I had with my steak this afternoon so half a pack has 146 cows so it's not that bad some olives some garlic paste and some truffle pesto again from anas so I'm gonna cook that up and put some chicken Milanese in the oven for me and my mom so we're gonna have chicken Milanese with pasta with creamy tomato pasta for dinner so I'll show you guys how it turns out fun to work I actually left my camera here in the kitchen last night still want to get up to get my coffee I can film for you guys but basically I just didn't have time in the morning to even make myself a coffee I had to rush off to work so I forgot my camera for her so I forgot my camera here in the kitchen but I'm back from work now and I washed my face and I just put on a little bit of light foundation eyebrow as a lip balm because I am going to the gym it is 6 15 god has given me strength and these tight suction brew avenue sorry these tight suction bow and tea um pants to kick me back into shape because I want to seem to start hitting the gym regularly you guys I've not gone gym regularly or exercise regularly in a long time which is so bad mainly because mental health wise exercise is so good for me it just makes me feel better about myself and also when I'm there I feel good as well so yeah I really want to get back into going to the gym so let's go no more time wasting so guys I'm back home now I'm gonna go take a shower and then have dinner I'm having salmon and rice for dinner but I'm really annoyed because I've lost my airports I honestly don't have luck with airports and my new gym because one time I dropped one of my airports down the stairmaster and kind of broke the stairmaster but then the stairmaster got repaired and it just has an airport stuck in it it's really weird so I don't really have much luck with them which is so annoying but oh well I guess it goes to show like if I'm going to get wireless headphones any of the ones that go over your head because clearly I'm not competent at taking care of a slight pair of tinier like headphones does anyone else have this problem with airports let me know I've realized in this vlog there's a lot of let me know but I actually read all of your comments so um do let me know also before I end this vlog I have one cool story to tell you guys so I was at work today and my consultant said to me um oh one of my colleagues another consultant is going to come down and just quickly say hi to us because it's the first time that I'm going to see her since like the pandemic um because they work in separate buildings I was like okay cool um so she comes down um the other consultant and she says hi to my consultant and then she says hi to me she goes oh my god I recognize you I watch your vlogs your Persian bunny and honestly it made my whole day and she was like I love your vlogs and I was like so shy I can't even tell you guys I never get shy when people come up to me and say hi or like recognize me or whatever but like I was so shy I was like oh my god a consultant like recognizes me like I was so like proud of myself that she was like I watch your vlogs and like I enjoy them so if you're watching this thank you so much and um yeah it made me really really happy and it also uh finally gave me the opportunity to tell my consultant that I'm a vlogger because I hadn't told her yet I just didn't bring it out to break it um but yeah so YouTube family you guys are so special to me all of you and I just thought I'd share that like really cool intimate story with you guys because I think it's just amazing like how many people you meet through the cyber world and I just love meeting all of you and I love you so much I'm gonna go ahead and end the vlog here if you enjoyed it then please give this video a thumbs up and don't forget to subscribe it actually means a lot to me um if you want to check out my socials they are always linked down below but my instagram is persian bunny and my snap is sepi samai I love you guys so much and I'll see you next time bye