 see your face. Yeah, it is apparently impossible to share my screen at the same time. So shall we start? Do you need one more minute? Let's start. Okay, okay, so very good. Let me start the recording. Recording in progress. Okay, very good. So now we will have the second lecture of Kevin Costello on twisted holography. Please. Okay, I'm so sorry for those technical difficulties. I don't know what zoom is up to today. So last time we discussed how, zoom I have a little bit more, that's not possible. Last time we discussed how an n equals two SCFT look of rise to a 2D chiral algebra. And we saw that the conformal anomaly becomes the BRST anomaly. So the first thing I want to do today is work through some examples, where we can see explicitly that the BRST anomaly vanishes. So the basic example is n equals four super young girls. Of course, as I think everybody knows, this is a super conformal field theory. So in this case, well, this is n equals two with the hyper multiply the joint representation of you. Now, when we twist, the hyper multiply becomes two fields, not one field. So each hyper multiply becomes a pair of something. So this tells us that in the chiral algebra, we're going to find a pair of adjoint value fields. X1 and X2. So X1 and X2 are both matrices. So if we write them in components, as I've done a little of the slide, find X1ij, X2kl, and the OPE takes this form. It's like the trace of the product of those two matrices. The BC ghosts are also matrices. And there will be takes the same form. So if we look at we think how things arise from the physical theory, n equals four super young girls has six adjoint valued scalar fields. In terms of string theory, these six fields describe the motion of a D3 brain in 10 dimensions. And X1 and X2 are two of these six scalars. Well, there are certain complexes in your combinations. So in this theory, the BRST charge from what we saw last time is just given by trace BCC plus trace CXX with some epsilon tensors when we contract the X's. So this is just a special case of our discussion from last time. And we can ask why does the BRST anomaly vanish? The reason is we saw last time that the BRST anomaly vanishes if the trace in the matter representation of a product of two matrices is twice the trace in the adjoint representation. Now in this case, since the matter representation is two copies of the adjoint representation, that just holds automatically. So let's look at another example. This is an n equals two theory studied by Cyberg and Wiggen in their famous work in the 90s. In this case, the gate group is SU2. And there's four fundamental hypermultiplytes. As we saw, each hypermultiplytes is two fields. So each fundamental hypermultiplytes will give four fields. And since there's four fundamental hypermultiplytes, we'll get 16 matter fields in the chiral algebra. Since these transform in eight copies of the fundamental, we can write the matter fields as i or alpha, or or ones from one to eight. And alpha is an index with a fundamental representation of SU2. So in this case, the OBE is i or alpha, is beta, is delta or s epsilon alpha beta times 1 over z, as we see. So if we think about the symmetry of this, the OBE only involves the tensor delta or s and the s indices. So therefore, it's invariant under the group SO8. So at first one might have thought that a theory, an n equals two theory with four flavors would have u4 flavor symmetry. But actually, it's enhanced to this larger group. Now the gauge fields of B and C are, of course, adjunct value. The BRST charge takes this form. Okay, so we can ask, well, maybe let me just pause one second. Are there any questions? Okay, so we can ask you this example. We know, because the literature tells us that this should be a super conformal theory, can we check there is no BRST anomaly? So the way to check this, well, let's think of the BRST anomaly as a relation that says for a matrix in SL2, which I'm calling A, we want trace of A squared in the matter representation is twice trace of A squared in the adjoint. So let's compute this. Our matrix A is trace-free. So let's diagonalize it with eigenvalues lambda and minus lambda. Now in the fundamental representation, of course, A is eigenvalues lambda and minus lambda by definition. So the trace of A squared in the fundamental is two lambda. Since we have eight fundamentals, the trace is, sorry, two lambda squared. Since we have eight fundamentals, the trace is 16 lambda squared. Let's compare this to the adjoint. The adjoint representation, we can take the spaces A, G, and F given by these matrices where H is diagonal and E and F are upper and lower triangular. The commutation relations tell us that the eigenvalues of A in the spaces are two minus two and zero. Therefore, trace of A squared is eight lambda squared. And therefore, trace, the matter is twice trace in the adjoint. Okay, so that was a little elementary computation, absence of the BRST anomaly in that case. So more generally, so we're going to be interested in holographic statements. So to have a holographic statement, we need to have an infinite sequence of n equals two SCFTs with a gauge group whose rank increases. So we've already seen this in the n equals four case, but this also happens for these examples with SO8 flavor symmetry. Sorry, once again, I'm going to try to log in on the other device too. I'm sorry for the mess with Zulu. Bobby, could you allow me to share a screen of the other device and then I can buy something so this will be better. I'll stop showing this. I'm really sorry about this. No problem. We'll try to give you host privileges. Stop sharing. Okay, it seems it's working. Okay, I'm so sorry. So now we can see also your face. Yeah, let's spend this on your ass. Okay, good. So because we're interested in a lot of it, we would like to have an infinite sequence of such n equals two SCFTs. So it turns out that there is such a sequence with SO8 flavor symmetry. I think there's a big lag, but I'll try to move my slide slide slide so in these cases, the gauge group is SP. Now for n equals one, SP2n is SU2. So this doesn't de-generalize. The matter content consists of a vector i, which is in C8, tensor c to the 2n. So really i should be thought of as eight vectors and then there's two matrices. Can you guys hear me okay? Yes, yes, we can see you and hear you. So the matter content in these examples is eight vectors and two matrices with certain symmetry properties. I don't want to go into too much detail on every little thing about this example. It will be important for us. The OBEs are there's a non-zero OBE of the vector with itself and if the matrices with itself. So ii equals like one over C and x1, x2 equals like one over C. And just like in the SCT case, of course we need to introduce the BC ghosts. So because we're in the group SP in this case, B and C are symmetric. So let's if we write things a little more succinctly, let's write omega here for this 2n by 2n matrix. And the symmetry properties of B and C is expressed here. B transports omega plus omega B equals zero is already for C. An x1 and x2 satisfy a similar condition. So in sum, the field content of this model consists of B and C matrices with some symmetry properties, x1 and x2 matrices with other symmetry properties and eight vectors. I won't give the details of why this model is also anomaly free. It's a little exercise in algebra to check that twice the trace in the adjoint is equal to the trace in the model. So our goal is to study these 2D chiral systems which are pieces of the n equals 2 theories at large n. So one of these systems at large n, all we need to do is understand the single trace operators because in any holographic system, single trace operators correspond to states of the dual gravitational theory. So it turns out you can calculate the single trace operators exactly. So let me describe the answer, but maybe I should quickly pause for questions first. No questions? So I think everything we're doing so far today is really quite elementary. So the first thing to check is that this operator trace of x1 to the k is BRST closed. How do we do this? We're considering the case of n equals 4 where we have x1 and x2 adjoint value fields. The BRST current is trace BCC plus this one would just trace x1, x2 minus x2, x1 on multiply by c. So what we're asking is this BRST closed? What we need to do is place my operator point and integrate the BRST current around it. That contouring will pick up any first over poles in the OB between the BRST current and my operator. Now if we look at the diagrammatics, the operator is only x1 and the BRST current has exactly 1x2. Since x1 and x2 pair with each other when we do a Witt contraction, we can think of them in a famous diagram rotation. The propagator here connects x1 and x2. There's only one Witt contraction. So this is computing that this is BRST closed as a tree-level computation. And since there's two terms of the BRST current involving these two terms here, we'll pick up two terms in the OB. And if you calculate it, the two terms are the same, but the play appear with a different sign. So we find that indeed trace of x1 to the k is a BRST closed operator. Now from this, bootstrap this argument to build many more of the BRST closed single-trace operators. To do this, we note that x1 and x2 form a doublet under an action of SU2. This is an action which comes from or symmetry in n equals 4 sub-reminimals. In n equals 4, there's an SO6 or symmetry which rotates the six scalars. Here only two of the scalars survive, but they're rotated by an SU2. Since it's a symmetry, if I take any state, if I take trace of x1 to the k and apply an SU2 generator, that will still be the BRST closed. The things you find if you do this are the symmetrized traces where I take some or powers of x1, s powers of x2, and I symmetrize in the order. That is, I sum over all permutations of how to order x1 and x2 inside of the trace. Great. Okay, so this is an elementary discussion of how to build the BRST closed single-trace operators at large n. If you work a little bit harder, you can compute the entire BRST homology at large n, and you find the following description. There's four towers of states. One tower is the one we just described, the symmetrized trace of x1 to the order of SU2 to the s. That's bosonic. That's spin or possessive or two. And there's two fermionic towers, which look very similar, where I insert the b-ghost or the derivative of the c-ghost. And finally, there's another bosonic tower, which is a little more complicated, and I haven't written out the full expression here. It involves del x and x and then a symmetrized trace. So the result is, in equals four case, this is a complete description of all of the operators, and really of the spectrum since we know their dimensions. So all operators are obtained by taking one of elements of these four towers, differentiating them some number of times, and taking their normal order products. So our task is going to be to find a string theory, which will have the same spectrum. So let's move on to the next case. The other example of a family of chiral algebras, depending on it, which we would like to model holographically, is the one with SO8 flavor symmetry. This was a little more complicated, but if you remember, the fields of this model consisted of eight factors. I, every five minutes or so, I appear to be getting kicked out of zoom. Yes. Can you guys hear me? Yes, now we can hear you. Yeah, we lost 20, 20 seconds or something like that. Is the audio okay? Now it's good. I'm so sorry for all these troubles. Sure, no problem, but now we can hear you so you can go on. Okay. So in this case, and the other example with the SO8 flavor symmetry, then, if we recall, there is eight factors, two matrices, as well as the B and C coasts. So in this case, what has other open strings you can build, which looks like a rock. This means I in the middle of a bunch of X's. So it's a vector matrix. Take your vector, then you hit it with a bunch of matrices, and then you end with a vector again. So the large and single, these, the terminologies usually that expressions like this are also called single traits or single string really. So in this case, we can also enumerate all of the states and we find these extra open string states together with the bosonic states that we saw on the n equals four case. The fermionic states of the n equals four case turn out not to contribute because of some symmetry reason for matrices. So in this case, all the operators are bosonic. And if you look at the lowest or order one, TRS 00, this is just the SOA current around for the SOA flavor symmetry. Whereas these other ones, that's some kind of infinite dimensional extension of these SOA current articles. So as I said, our goal is going to be to find a string theory or gravitational theory, which is holographic in dual to these models. I have a question. So that is, we want it. Yes. So for the case of n equals four superior meals, we know that there is a dual is a string theory on a DS5 versus S5. So for the other class of example, SP2n with these eight flavors, do we know about standard holographic pair before the twisting? I think so. I'm not an expert in this, but I mean, there's a string theory embedding of this system as a D3 ring lying on a D7, four D7s and an O7 minus plane. So I would guess that it's just going to be ADS5 times S5, or you've also inserted certain seven brains into the geometry. Okay. Thanks. So in this twisted set up, what we're going to find is that the topological string will capture the holographic dual or will capture these parallel numbers at large. And what I'm hoping to do is there's some really fun computations you can do with these parallel numbers from which you can see the aspects of the geometry emerging. The origin of this is in a conjectural relationship between the physical and topological experiments. So if you remember, we got these chiral algorithms by thinking about placing a D3 brain on a cigar in some supersymmetric setting where the cigar was forced to rotate on the Q-square and was location. The idea is that we should do the same thing for type to be symmrographic. And we have a product of two cigars times C3. And the conjecture is that after an appropriate twisting, this becomes a topological B-model in C3. So this is not a new conjecture, or at least not completely new, because it's these kinds of connections between physical and topological strings go back to the early days of topological strings in the 90s. So it's discussed in the same paper of Bershadsky-Cigar-Huy growing in Vafa, where they talked about the gravity float on the background. I think we lost you. Can you guys hear me? We can hear you, but we cannot see your screen. Yeah, it appears that your screen is frozen. Maybe using the PDF was using your tablet is too complicated. This is what a disaster. I don't know what's going on today. So conjectures relating the B-model on C3 to type to B-spring have been made since the early days of spring here. We cannot, we can still, we cannot see. Yeah, I think there's nothing I can, I think maybe we'll just have to go without seeing. Is that okay? Can you see my screen, my slides? No, we cannot see anything. Yes, now we can. I really wish I knew where the problem was. Anyway, so these kinds of conjectures have been made since the early days of spring theory, but you know, things were really exciting back then. And I think people didn't want to take the time to stop and really derive everything from first principles. I would say the first principle is proof of these relationships is still lacking. But there's really nice work by Severian Williams and Agar and Hunter, where they studied, they take the super gravity multipliers and take its q-chromology and they identify that with the fields of topological strength. So I think there's really solid evidence that these relationships hold. And I'm going to accept that it should work for the rest of the lecture. So our approach is are they type to B on the product of these two cigars with the D3 brain wrapping one cigar times a plane. This is going to become the topological string on flat space. But now with the brain in the topological spring, called the D1 brain, wrapping a copy of C inside of C2, the theory on the D1 brain will be the chiral algebra that we've been discussing when it was born. When we back-react the D1 brain in the topological spring, we'll find we'll get a new geometry, which is the sixth manifold SLTC, which you can think of as ADS3 times S3. And the statement of twisted holography that the topological spring in SLTC is equal to equivalent to the large N chiral algebra. The connection with ordinary holography is a conjecture, which is kind of, I think I'd read at the moment, that the topological string arises as a twist of type to B cybergravity on ADS5 times S. So I'd like to move on to discuss what is the topological string. But before I do that, are there any questions? Yes. Yes, one can think of SLTC as ADS3 times S3, but one can also think of it as the cotangent bundle of the threesphere. And that has a well-behaved calabiow metric. So I was wondering if it's helpful to think of it in that way. If the scalar structure is not going to play any role in our analysis, though it's only as a common manifold we're going to see it. So I don't think one can really see the distinction between them. That answers my question. Thanks. Okay, so what is the topological string? So the most important field of the string is the Boltrami differential, which is a tensor that looks like this. And it is constrained. This is not going to be super important for us. Constraint is how to satisfy an equation, which really just says that it's diversions free. So the geometric meaning of this, the Boltrami differential plays a geometric role very similar to the graviton in an ordinary gravitational theory. So if we have a manifold, it has S3 on it, and it implies that it would be too flat, and so satisfies the vacuum inside equation, then the data of the metric parameterized by two pieces of data, omega i j bar, scalar form, and mu j bar i, the Boltrami differential is very complex. So from this perspective, the Boltrami differential is like literally some of the components of the metric. So the conjectural picture, the omega i j bar components, they are q exact, but q is the super charge. In other words, the topological string captures a piece of some version of instant gravity, or we've discarded the scalar form, which we changed the Boltrami differential. So the equations of motion of the topological string and differential find an integral complex structure. So which means in which way is the equation of the usual db, dz bar, d bar, if it looks like this, plus mu i j bar, dz bar j, this whole thing squares to zero. So in addition to the gravitational field, we have, we had two classes of model of parallel profiles, one with S and weight, favor symmetry, and one which is just n equals four. So there's the type two, ordinary topological string. In this case, the other fields are a pair of harmonic fields, a little Gaussian, but these fields is this very simple expression where we've coherently coupled the Boltrami differential with the type one topological string, which is dual to the n equals two series with SOA flavors of three. In the bulk geometry, we also have an SOA gauge field with the homomorphic and sine zones of all of them. In this case, the equations of motion for a defines SOA Okay, so now we have these two models, each of which has a field which is a Boltrami differential together with some other fields from Rheomic in the case of type two, and Bosonic in the case of type one. We're going to be able to match these theories on SO2C with the chiral algebra. I have a question. Yeah. Is there some constraint on the, on this SO8 bundle that it has to be SO8 or any bundle would give a good topological string, type one topological string? Yeah. As far as I know, it has to be SOA. There are suggestions from F3. This is a bit more complicated. If one tries to build a holographic joules of certain non-ragrangian n equals two theories, they will have flavor symmetry, which might be E6 or something like that. So there's a possibility there's an exotic topological string, the E6. But for this kind of fairly simple things we're discussing, the constraint comes from, if we were studying the ordinary type two, type one string, we find that SO32 comes by thinking about the hexagon anomaly. In this dimension, there is a box anomaly. This diagram fails to be gauge invariant. This is a diagram in the gauge theory. Gauge variation if I vary on A goes to A plus D bar C. This diagram varies like this. So this is an expression that descends from an A form, as you would expect for an anomaly in six dimensions. And the A form is traced after the fourth. But this is cancelled in the fourth mechanism, which involves an exchange of closed string fields, that's new. And this requires trace in the a joint, back to the fourth, is traced in the fundamental x squared, or the proportionality. So just like in the familiar case in 10 dimensions, there are certain trace identities, and also a constraint on the dimension, which forces you to tightly constrain the group. Let's see. The next thing I want to talk about was the back reaction. I think I should postpone that next time, about how we back-reacted a D1 brain to get us to SLTC. Maybe I can stop now and we can have questions. I mean, if you want, since you lost a lot of time, you can take, I mean, you still have three minutes, but you can take five or one. One brain, C inside of C3, I got to take my coordinates, BZ1, Z2, Z3, brain, W1, W2, the brain is in the look as WI0, and this sources, I'm not going to derive the equation for the fielded sources, I'll just write the equation down. The equation is that the Z component would be delta function, W1, W2. So what we'll find is if we solve this equation, the three with the brain removed, let's try to solve the equation. This is a pretty standard PDE, and the solution is given by the Bocke-Mortenelle-Kerl, where the Z component of mu is given by 10, and then there's some kind of factors of 4 pi or something like that. And so it's kind of an amusing exercise to try to solve, to try to check that this Bocke-Mortenelle-Kerl needs satisfies that equation. So how does this deform the geometry? Well, it deforms the geometry by changing the Cauchy-Wohlman equation of F, which is a function of all the variables, and the equation is going to be d bar plus mu Z, d by d Z of F is equal to 0. Now if we collect the components, d F, d W bar 1 minus 10 over w to the 4, let's come from up here, W bar 2, d F, d Z, 0, last equation is that d F, d Z bar. So what we will do tomorrow is solve this equation, and we'll find that the solution to this equation is your coordinates on the manifold SLTC. Okay, so I think that's the reasonable point to stop. Okay, very good. Thank you.