 and he's gonna tell us about twisted holography. So I will just start the recording. Recording in progress. And please Kevin, take it away. Okay, thank you very much. And I'd like to thank the organizers for the opportunity to give these lectures. I'd also like to apologize that I wasn't able to make it in person. I'd appreciate it if I'm not entirely sure the level and background of the participants. So I really appreciate any feedback about how fast I'm going and the level of the material. Interruptions are very welcome. So what I wanna talk about is of the title is Twisted Holography. So what I'll talk about is based on work by many people. And it's a weight similar to in Matthias' lectures. Twisted Holography is a weight of getting really sharp exact results from a supersymmetric subsector of ordinary Holography. Matthias' lectures, he's not considering a supersymmetric subsector, but they're also exact results. So let me remind you a little bit about the background. So the original ADS-CFT correspondence, you know, we've all heard a great deal about it, but it's important to recognize that the original correspondence is still very much conjectural. So it's conjectural duality between N equals four super young mills and type two B super gravity on ADS five times S five. So the reason, once you see this as conjectural, is that both sides of the duality are incredibly difficult to understand. So on the gauge theory side, N equals four super young mills, we're supposed to be studying strongly coupled gauge theory. You know, strongly coupled constructing rigorously or in any sense, a strongly coupled gauge theory is probably the most difficult problem in mathematical physics. It's one of the clay millennium prized problems. So this is really out of reach of any mathematical techniques. And on the other side of the duality, when studying type two B strain theory on ADS five times S five, this is difficult for two reasons. If one wants to take the super gravity approximation, and we're studying quantum gravity, quantum gravity is of course hard because it's not normalized on so ill-defined of the quantum level. On the other hand, we might try to think of it as a string theory. String theories are a way of defining quantum gravity, but it turns out that string theory in ADS five times S five is also very challenging. The reason that I want to get into you, but it involves the fact that this is a Riemann-Riemann background. So of course, people have done lots of checks and there's like very heroic, extremely difficult computations of both sides. So it's very plausible that this works. But what I'm going to talk about today is a sub-sector of all of this where both sides are really easy to understand and you can really get your hands on things and check things cognitively. So precisely on both sides, we're going to select certain states which are preserved by supersymmetry. Now on the gauge theory side, why is that easier? You'll find it has no coupling constant. The gauge theory is basically free. It's very, very simple. And on the gravity side, we find that the gravity side is a topological string which is also much easier to understand than the physical spring. So in this context, in these sub-sectors, we can perform exact computations to try to match both sides. And as I mentioned, the ADS-3 etiography in Gabriel's talk is a different, not unrelated context where exact results can be obtained. So there might be some parallels with the Matthew Isis talks. Easy. Let me be able to see, you guys can see a little bit more of the text. Is this legible or is it too small? No, it's good. It's not too small. Can I zoom in a little more? I think maybe it's the good size. Maybe if you make it smaller, it would be a little bit too small. All right. I didn't want to just kind of be scrolling too fast. So the starting point of all of this analysis of twisted supergravity is the concept of a twist of a supersymmetric gauge theory. This was introduced by Witton in the late 1980s from pretty famous work in which he related n equals two gauge theory to the Donaldson-Darran's report manifold. So I'm going to start by reminding you of how one builds the twist of a supersymmetric gauge theory. So the example we'll start with will be a four-dimensional quantum field theory of n equals two supersymmetric. So let me remind you what it means to have n equals two supersymmetry. The group spin four is, well, in Euclidean's signature, it's SU2 times SU2. So each copy of SU2 has its fundamental representation. And these are the two spin representations of spin four. They're of complex dimension two. Quip. I'm going to use Penrose's notation for spinners. Spinners in S plus will have an index, a Greek index alpha with this. And in S minus, they'll have a dotted index like that. When we have an n equals two theory with n equals two supersymmetry, we're going to have two pairs of supercharges. Q alpha i, where i goes from one to two. And Q alpha dot i, where also i goes from one to two. So we should recall that the vector representation, complexly vector representation is a tensor product of these two spin representations. So in Penrose's notation for spinners and gamma matrices, instead of having an index for vectors, we write a vector as having a pair of indices, one dotted, one un dotted. So X alpha alpha dot indicates a vector. Now the commutation relations with the super symmetry algebra is simply Q bar is delta ij times translation in the direction of this vector. So what does it mean to twist? If I have a theory with n equals two supersymmetry, well, I have two pairs of supercharges with this extra index i. And we want, so the super algebra has a symmetry rotating that index. It's an SU2 symmetry. It's called the or symmetry. So we're going to assume that our field theory also requires this entry. So the twisting procedure says that we're going to change the spin of our fields by using the SU2 or. Now recall that spin four is a product of two copies of SU2. So I'm going to change the spin of the fields by instead of making SU2 acting SU2 plus active in one way, I'm going to make it act by the diagonal map between the original SU2 plus and the or symmetry. So what does that mean? That just means that, or is before I had an index i, my super symmetry algebra, I'm now going to make that index transform under the rents group. So it transforms as the index alpha. So what do we're supercharges that become? Qi alpha becomes Q alpha beta or again alpha beta on our spinner indices and Q bar alpha dot i becomes Q bar alpha dot alpha. We'll leave that in a minute, but you notice that Q bar now transforms as a vector because it has two spinner indices of opposite elicits. So what this means is that because the index now transforms as a spinner index, we can build a super charge, which I'll just call a Q, which is a scalar. And it is automatic that in the supersymmetry algebra this is a nil put in super charge. The reason is, well, I take Q squared, the square of any supersymmetry is a vector, but this is also a scalar. So it must be zero. So the key point of the twisting procedure is that given a super charge Q of square zero, we change the theory by replacing the original Hilbert space by the Q called monogym. So the twisted Hilbert space is defined to be those states which are in the kernel of Q modulo those which are in the imitative Q. And we do this for everything else. We do this for local operators. Of course, if it's a CFT, that's the Hilbert space on the three square. It's a bit more challenging, but it is also possible to formulate this definition for extended operators. If one reads the older literature on twisting, the part which is really emphasized is the first step. This part here, changing the spin of the fields. But for me, I don't think that goes very much. In fact, it does absolutely nothing. All that's happened is step one, when I've changed the spin of the fields, especially if I'm on class space, is I'm just calling things by a different name. Instead of having an index I, I decide to call it alpha. I still have the same number of fields. My Hilbert space is still the same. My Hamiltonian is the same. I have all of the complexities are there. However, step two, where I've replaced my Hilbert space by the Q-colonology is a really radical simplification. And this is what allows you to do computations the twisted theory. What I've described is the Donaldson-Witton twist of N equals two supersymmetrications. It turns out to be a topological twist. So let me try and explain what that means. Although maybe, unless anybody has any questions about the material so far. Sorry, sorry if that was awkward. Hello, can you hear me? Yes. So you have the index I for the SU2R and an index alpha for the SU2 plus. Why are you allowed to call them the same letter and then treat them as if they're in the same space? Because you take a trace at some point in defining Q, right? The scalar. Right, that's right. So what I've done is I've changed the way SU2 plus accident. So the charge of SU2 plus is the original charge it had before twisting plus how it transforms under SU2R. So that's, let's scroll up a little bit. So I'm taking this step. So my new action of SU2 plus is the original action of SU2 plus but then I embedded diagonally in the product of these two groups. Yeah, I see, I see. Okay, so let me move to, unless there are further questions let me move to explaining what it means to be topological. So the hallmark of the topological theory is that the energy momentum tensor vanishes. So energy is the action of translation in time on the Hilbert space. And momentum is the action of translation in space. So let's explain why these operators vanish. When I twist my Q bar supercharges, the index I becomes an index alpha. So it looks like this, Q bar alpha dot alpha. So it transforms as a vector. And if one looks at the original commutation versions, we take our supercharge Q called Q is like, when I can commute it with Q alpha dot alpha, well, the only thing it can really become if it thinks that one thinks about the symmetry is the operator of translation by the space time vector X alpha dot alpha. So what this means is that on the Q-core homology, this translation vanishes. So I didn't write up the equation. Does that? So if I take a state psi, which is Q closed, then it's derivative. If I hit it with translation in some direction, it's like this. Because here, if I'm the term where I move the Q past the Q bar is zero because Q psi is zero. And this is equal to zero in commology. So that's why we're actually with this for local operators, the same thing holds. If I have a Q closed local operator, then if I try to move it around, it's Q exact. So what this means, if I'm studying the correlation functions of Q closed operators, they're independent of where I put them. So let's see this explicitly. I'm assuming all my operators here, I'm assuming these guys are Q closed. And I differentiate with respect to the position of the first one. Now, I move the derivative inside the bracket and I replace it by Q, Q bar, the Q, Q bar of the first one there, and then the other ones. And because Q is a symmetry, I can distribute Q over the other operators instead. But they're all Q closed, so that's zero. So this is a summary of Britain's construction in the late 80s. By performing this topological twist procedure, we find special operators of correlation functions which are independent of position. This is a very unusual feature and it's a hallmark of having a TFT. Soon after Whitman's construction, people thought about twists which were not topological. That's gonna be the focus of my lectures. So in the mid 90s, there was work by Johansson, Necrosal and others. Where one studies the holomorphic twist of a theory with N equals one supersymmetry in dimension four. So this particular example is not gonna really be the focus of these lectures, but I'm just gonna spell it out to demonstrate how twists can be holomorphic tricks. Now, if one only has N equals one supersymmetry, then we no longer have the index I in the supercharges. We just have, the supercharges are just consist of two spinners one of each other's. So here, whatever we do when we twist, you can see it's not going to be possible for the twist to have the full spin four symmetry we started. Whatever supercharges like we'll break that. So we're gonna break the supersymmetry to SU two minus or SU two minus rotates the Q bar off the dot and fixes the Q alpha. So I'll take my supercharge Q to be Q one. So just one of the two components of Q. And then I'm gonna write holomorphic coordinates on space time by Z alpha dot is X alpha dot two and Z bar alpha dot is X alpha dot one. So what's going on is that once I've broken the Lorentz symmetry group to SU two, that SU two preserves a complex structure in space time. And here you can think about SU two bar. The key thing we needed earlier was that Q bar would give me all of the translations. Here what we find is that Q, Q bar only gives you two of the four translations. Those in the directions I call the Z bar. So now what does this tell us? The same argument tells us that correlation functions are independent of Z bar. So D by D Z bar is A alpha one, I differentiate the Z bar direction first coordinate. By the same argument, I bring this inside. I replace it by Q, Q bar and I move Q around. That's gonna be zero. So the correlation functions are holomorphic functions of the position. So this equation here is, this is the Cauchy-Meyne equation. So once I had some time, it's interesting to study the behavior of these holomorphic twists. There's lots of fun things we can say about them. And general, the most general behavior would be something that's partially holomorphic and partially topological. But let's not get into that. This is mostly an example. Now I wanna move to the kind of twists where we're gonna be focusing on to do twisted along the selectable. So I wanna focus on a construction which seems a little odd at first, whereby one starts with a four-dimensional theory and N equals two theory, which you assume is super conformal. And then you turn it into a two-dimensional chiral theory. So this is a little bit funny. It's a little bit of a different thing than we did so far. There are two approaches to this. One is, which is perhaps the most widely studied. Instead of taking Q to be a supersymmetry, you take it to be in the larger algebra of the super conformal symmetries. And the second approach, which gives you the same answer, is to use a version of Necrossov's omega background construction. Actually, I'm gonna spend a bit more time in this one. So this, well, Necrossov did really kind of SGL version of this. This particular omega background construction goes to G1O and Gianniaghi. This construction is a bit more technical than what I've described so far. So I'm not gonna get all the details. So mostly what I want to do is explain the answer. You give me an N equals two theory, we will be able to figure out what it's, what the two-dimensional chiral theory is, and then we can do computations. Let's take an N equals two theory in dimension four. Actually, I should, Paul, are there any questions from here? Hi, I have some questions from the beginning. I got a little bit lost. So by the twisting, do you mean working on the cumulogy of the Hilbert space? Yes, we're in the cumulogy of the Hilbert space and the space of local law. Okay, and I think that after that, you mentioned that after performing this twisting, we preserve a complex structure, right? What does it mean physically? I think it depends, so this was the case with N equals one supersymmetry and the complex structure arises by noting that if I take correlation functions of Q closed operators, they only depend on half of the coordinates. So in Euclidean signature, they depend on certain complex combinations of the coordinates which are called Z and they don't depend on Z bar. So the complex structure arises just from that equation. Let's see. Is that, is that up over here? Yes, more or less. So the fact that our correlation function is holomorphic, is that a consequence of the twisting and a consequence of this preservation of the complex structure or are they related at all? They're related, the preservation of the complex structure, the fact that spin four is broken to SU two, maybe we should put that aside for now. And then the most important thing is that the correlation functions are holomorphic functions and that statement is just a formal consequence of the supersymmetry out. It's just that D by D Z bar is the commutator of Q is something, so that's all. Okay, thank you. So in the N equal one example, can you say in a few words what is the set of operators that you are left with because I guess it's larger than the color ring of the theory? Yeah, let's say we start, we find some black paper, N equals one parallel to that, one bosonic field, all its derivatives in Z one and Z two and one fermionic field and all of its derivatives. So if you look at the index of this, these are exactly the states that contribute to the super conformal index. Like this one gives you kind of a part of one minus Q L, one minus Q L, so it's kind of a thing. But by looking at words in this field and its derivatives, if you take its character with respect to the group U2 which is the symmetry because it's conformal, you'll find it's the standard index of an N equals one super conformal theory. I see. Thanks. So maybe this is a comment for the experts. I think this is really interesting because suppose you wanted to give a character a proof of cyber duality, what you really should be doing is looking at the states of the twisted theory and identifying all of their correlation functions. That's kind of accessible, but also really not trivial at all. It's the kind of thing I think of Davide as looking at this kind of question. Are there any more questions? So let me try to explain that this construction of Yankee and L. So here I wrote down N equals two super symmetry algebra again. I'm gonna do something a little funny. We're gonna take Q which does not square to zero squares through a particular translation. So I'm taking epsilon to be a small parameter. My Q adds a linear combination of like this. And if you see just by contracting the tensors, maybe the indices, my notation is not super clear possibly, but if you can just contract the tensors when you can be Q squared, you find a Q squared is epsilon times E translation. But if you look at the norm of this factor, you'll see it's epsilon times the space like transmission. So what we'll do is make this direction periodic. So really that we're working on this cylinder just coordinates x11, x2, x11 dot, x22 dot plus another plane which has coordinates x12 dot, x21 dot. And it'll be important in a moment that d by the x12 dot is Q exact. And this other plane will be where the chiral algebra lives. I'm gonna call x12 dot z and it's a homomorphic coordinate on this plane. Okay, so the result is I'm gonna take my cylinder here and I'm going to fill it in. And I'm going to fill it in so that it becomes a cigar. So at the tip of my cylinder goes along in the radius shrinks. I got this cigar geometry. And then the result is that that your n equals two theory extends across the cigar in a way which still preserves the super chart for a reason. So if we do this, we can do the following trick. So above we saw that Q squared was d by d theta. That's rotation on the cigar. If I take an operator which is at the tip of the cigar it makes sense to ask that that operator is rotation invariant. It's on charge under rotation. And on such operators Q squared is zero because Q squared is rotation. So then we can consider the Q-colonology S1 invariant operators at the tip. And this is going to be the chiral opera. So what's important for this construction is that it doesn't make sense to move my operators away from the tip of the cigar because away from the tip of the cigar Q squared is rotation and there are no rotation invariant operators. So this is why we're effectively localized to a two-dimensional theory. On the plane which lives at the tip of the cigar d by d z bar is Q exact. So for the same reason we've had before if I take this, I'm taking things which are at the tip of the cigar and I have other points EI in the plane then the z bar derivative of the correlation functions vanishes as long as my operators I'm inserting are Q-closed and rotation invariant. So in this way, we've identified in our 40 n equals two theory it must be super control, something very familiar. A two-dimensional chiral theory just like one might be familiar with from elementary studies of the bosonic strain or something like that. Sorry. Can you consider also extended operators on this cigar like a line that wraps the theta cycle and so on? Did they give you something interesting in the 2D? Or the... Yeah, that's a really good question. So if you, the answer is yes. So what happens is if instead of thinking about it as a 40 theory on a cigar we'll compactify to a 3D theory along that circle the effective 3D theory becomes topological in the bulk with the chiral non-topological boundary condition. But the, there are extended operators and these become the bulk operators in the 3D n equals four theory. From the 3D n equals four perspective they're the ones which parameterized the cool off edge. So they're kind of subtle and you realize that their 40 uplift must be the two clans rocking those that cigar. Okay, thanks. So I said, let me very briefly mention a different perspective. Proxima might have been better to explain this one, but for my purposes I kind of didn't want to get into that much detail about either way of doing it. I mostly wanted to say the answer. So the different perspective developed by the Mlemels-Sviano-Pelleris-Bristellian families they say that instead of taking an ordinary supercharge like we've been doing they take a super conformal charge which lives in the super conformal algebra which in this case is PSE 2,2 slash 2. Now what they find, things in the super conformal algebra include rotations. So in their story, rotations in one of the planes will be exact. So that implies, if you think about it, that the operators that live in the Q-chromology only exist on a plane inside of a fork. Just like we saw above the operators only lived at the tip of the cigar. Then they take the Q-chromology and it's the same chiral algebra. This perspective was historically the first one on building the chiral algebra for the Mlemels-2 theory. And there's a great deal of literature about it. So I'm sure people who are interested in it can find good things to read. But as I said, I mostly want to say given Mlemels-2 theory, what is the algebra? So let me start on this. If we have a look wrong in theory then we can just write down what the algebra complicates with the table. So we'll start with the simplest thing, which is an N equals two free hyperbolic. When one has N equals two supersymmetry, the smallest number of scalar fields, what has is four. So I think like this, five I, five R, this notation is kind of selecting, is breaking the symmetry in some way. And the logon for the free theory is the symptom form, like this. So what do we get when we take the chiral algebra and the way it was discussed, run this sketch? These four scalar fields, the N equals two theory become two scalar fields in the chiral algebra. Beta one and beta two. Beta and the beta I's comes from the five I's and the five R's don't contribute. These are spin half. So the OBD between the beta I's is beta zero, beta j of z, beta j of z is epsilon ij times one over z. So it's a little funny because the beta I's are bosonic, but the OBE is like that of free fermions. So this chiral algebra is called symplectic bosons, because they're bosonic fields. What I was for gate theory, the N equals two vector multiplied contributes some fields for the chiral algebra. And these are something that should be quite familiar to people who studied some theory. They're BC ghosts. Although in contrast to the BC ghosts of string theory, these are ghosts that implement gate symmetry. They're not ghosts that implement the diffume frism invariance biology. So the BC ghosts consist of BN to the spin one field and CA to the spin zero. And these are both fermionic. They have a simple OB, the B times C is delta AB times one over Z. Same OB one finds in free fermions, except there are spins one and zero. As in any ghost system, one has to add on the BRST charge. And the physical, open space or space of local operators would be the BRST chronology. So here's the formula for the BRST charge. And again, it might remind you of something you've seen in the string theory literature, except it's a little simpler. And the final thing one wants to do is consider how to couple the matter contents to the gauge symmetry. So all we do is we add the fields from the matter to the close from the gauge symmetry and build a BRST charge, which forces gate invariance. So for the matter, while if I started with a single hypermultiplet, I would have two fields, beta one and beta two. So then the most general picture is, I should take my beta I, the matter fields to be in a representation, which is symplectic. So that means some invariant tensor omega ij, which is anti-symmetric. So I'll give you some examples later. Using the fact that G-ax on the representation, we can build a current here. This is the current which generates the action of my gauge fields on the matter. And then we say the BRST charge, it's just that for the gauge theory, plus the seagulls to times the current. So we're doing kind of time, I won't have, we don't have time for questions at the end. What we find is this construction works by the BRST charge squares to zero. If and only if there's some trace identity, which is twice the trace and a joint representation of two of the outer elements is equal to the trace in the matter representation. So again, you might be familiar with such a computation from, you know, there's something like this in Poltinski. One more, in the bosonic string, BRST charge squares to zero, if and only if or the critical dimension, this computation is much easier. I'll give you the computation in just a sec. But it's the useful fact is that this is the same trace identity one needs to guarantee that a 40 equals two theory is super conformal. Let me move down to do the computation. So from the gauge sector, we have this vertex that I want to consider the OBE of this fit itself. So there's going to be a contribution from tree level involving one width contraction. This vanishes the Jacobian, the one-width contribution looks like this. Gonna have C fields on the outside. Recall the legs correspond to a width contraction. They connect to C and a B because the OBE of C and B is one over C. So let's compute this, actually like that. Connection like that. Well, because there's two contractions again, one over Z squared, zero, C, B, Z. And then let's include the flavor in this as carefully. There's going to be twice times the trace, the joint representation, TATB. The factor of two is because B and C run around the loop. So let's take this expression. We notice C is from ionic, C A zero, C B zero, trace, T A B is equal to zero because this expression here is symmetric. So if we take the OBE and expand in series, we find the first non-zero term looks like C A zero, derivative, C B, trace, P, T A. Now for the matter sector, it's going to be the same kind of calculation. The C goes to, it's going to couple to two matter fields and we're going to do a width contraction where all of the matter fields are contracted and we find one over Z, C A, del, C B is traced in the matter representation, TATB. And to be careful, this one came with a minus sign because there were fermions and the other one comes with a plus sign. I take my BRST operator. This is as no O if this condition helps. Finally, I think to ask, you might have, yeah, I should have said this first, why does a pole in the BRST operator and the OBE of the BRST operator with itself have to do with the BRST operator and no button or if I compute the commutator of its interval, we move the contrast past each other, we'll pick up a term, one of them is integrated around and a little circle around the other, like this. That said, it all goes around like that. And if we look on the right-hand side of this expression, a first order pole in the OBE of the BRST operator with itself will contribute a term in the square of the BRST charge. This is not equal to zero unless the trace identity helps. Okay, so I will stop there. Okay, let's thank Kevin. Recording stop.