 Thank you very much. First of all, thanks to the organisers for inviting me to this really wonderful workshop. It's been really spectacular progress on a lot of the topics we've heard about over the last few years, and it's been really stimulating, interesting set of talks. So thank you very much. What I'm going to talk about here is perhaps a little bit off topic, so I will be talking about holography. I won't be doing any localisation, at least in the physics sense, although some of the formulas that you'll see later can be localised in the mathematics sense, just using the fixed point theorem of Berlin and Verne. Anyway, this is work in progress, so there's no paper yet. We're still very much working on this topic with Chris Cousins, Jerome Gauntlett and Dario Martelli. Probably many of you here know Jerome and Dario. Chris is Dario's PhD student, and he was at the school here last week. All right, so what are we going to be interested in? Can you read this at the back, by the way? So we're interested in type 2b, super symmetric, ADS3 times Y7 solutions, and here Y7 is going to be a compact 7-manifold, and these will be holographically dual to two-dimensional 0,2 superconformal field theories. The amount of supersymmetry is such that we get 0,2 supersymmetry for the dual field theory that lives on the conformal boundary of this ADS3. So I'll make a few general remarks before getting into the details of this. On the field theory side, so typically we don't have a nice Lagrangian description of these superconformal field theories here. Usually at best what we have is some UV Lagrangian for a supersymmetric quantum field theory that we believe flows to an interesting interacting superconformal fixed point. Under those assumptions, there are then certain protected quantities, BPS quantities, that you can compute at the fixed point using that UV description. So examples of quantities in this setting would be the central charge of the 2DCFT and also the U1R symmetry. So if the 2D 0,2 theories, there's a U1R symmetry which sits in the same multiplet of the stress energy tensor. It plays an important role in the theory. For example, it determines the scaling dimensions of chiral primary operators. And you can compute these in the UV theory using C extremisation, which I'll explain a little bit later. The point I wanted to make is that whenever you can do these kind of exact calculations, of protected quantities in the field theory, I think there should also be an analogous calculation you can do on the gravity side. A priori, that's not obvious. Usually in supergravity, as we heard someone mentioned yesterday, you have to solve the Einstein equations to get anywhere. They're nonlinear coupled PDEs. It's very difficult to do that even when you've got a lot of symmetry. However, I think you should be able to compute protected quantities on the gravity side without knowing the full solution to the equations of motion. And what I'll show you today is how you can set that up. There are other examples, too, that I've worked out in the past. Of course, you've still got some PDE to solve on the gravity side. You can't get around that, but you need some existence theorem in that case. And the analogy to have in mind is Yao's theorem for Calabiao metrics. We don't know any explicit solutions on compact Calabiao manifolds, but we can still work with them and compute lots of things with them. That really does rely on Yao's theorem that tells you that they exist in the first place. That's the philosophy. For the sorts of theories that I'll be discussing, they're D3 brain backgrounds, so you'll see in a moment. They'll be D3 brains wrapped on Riemann surfaces, for example, the sorts of theories that I'm going to be talking about. Good. Maybe that's a bit low. What are the class of backgrounds? On the gravity side, I've got a 10-dimensional metric. That takes the form L squared e to the minus b over 2. L is just some length scale. In brackets, I've got the metric on ADS3. Plus some metric ds squared 7 on my 7-manifold y 7. This b here is a walk factor. It's a function on y 7. If you want, that's my ansatz or class of metrics that I want to look at. The only other field that's turned on in type 2b is going to be the 5 form. Again, there's some dimensional factor L to the 4. Then I've got the volume form on ADS3. I can only wedge that with a 2 form, which must then be a 2 form on y 7. I need to make that self-dual. It's minus the hodge-dual with respect to the 7-manifold of f. The 5 form needs to be closed in type 2b. In particular, f is also closed. Locally, I can write it as dA. This is a class of type 2b backgrounds we want to consider. They were first considered by Kim in about 2005. I've got a metric on y 7. I've got a scalar field b on y 7. I've got a gauge field f or a with curvature f on y 7. I've only got 5 form flux turned on. In some sense, these are d3-brain backgrounds, is how we should think of them physically. If you plug that into the 10d equations of motion, you'll get some equations of motion for this metric scalar and 2 form. No, but you could do. It's just my choice not to. I know. You could do something similar to what I'm going to do for those cases. This works out particularly nicely. You're absolutely right. You could do what I'm doing here in other cases too. There are equations of motion for these things, and they come from an action. You can write it down. I'll write the action as s. The function of this 7-dimensional metric, b, which is a function, it's not the b field. It's a function. And a, which is a gauge field on y y y y y y y y y y y y y y y y y y y. The precise form is not so important. It is, if you want to calculate with it. E to the minus 2 b, you've got a volume form. There's a Reachy scalar on y7 minus 6, o'r costumlogiwcod yma i fynd yn ddefnyddio'r ddweudwch yn bach ac ydym yng nghylch yn ffwrdd A o'r cyfrifed yn F. Mae'r ffordd yw'r gweithio'r ddechrau i ei ddweudio yma o'r ddweud o'r ddweud o ffwrdd yma o'r ddweud o'r ddweud o'r ddweud o'r ddweud. Felly, mae'n gweithio ychydig o bosionau mewn gwahodau supersymmetri. Mae'r gwahodau cydweithio ddechrau. Mae'r gwahodau cydweithio ddechrau sy'n gweithio gyda gyfyrdd mewn gwahodau gyllideb yn y cerddau gyfyrdd gyda'r Y7. Felly, dwi'n gweithio, mae'n gweithio'r ddechrau, mae'n gweithio'r ddechrau. Felly, mae'r norma unig, y ddau'r gilydd. Mae'n ystafell y Y7, ac mae'n bilyniad. Mae'n rhaid i'r ddau'r gynhyrch sy'n gwybod i si, mae'r ddweud i'r ddweud i si. Mae'r gwybod i'r ddweud i'r ddweud i si, mae'r ddweud i si sy'n gwybod i si, plus p squared plus e to the b, ds squared. And this metric here is caler. So it's a 6D transverse caler metric. And it's transverse to this vector field. So you've got some vector field on your 7-manifold. And transverse to that, there's some 6D metric. And that metric has to be caler. Super symmetry tells you that. Everything is now determined by what I've just told you. So e to the b, this scalar, is just 1 eighth times the Ricci scalar of the caler metric, the 6D Ricci scalar. There's this connection term, capital P, here, appearing in the metric. So dp is rho, which is the Ricci form of the caler metric. This Abelian gauge field, flux on Y7, is minus twice the caler form, so j is the caler form. And then there's an exact piece, plus a half d e to the minus b e psi plus p. So hopefully you can see once you've fixed this choice of vector field and the choice of caler metric, everything else down here is now determined in terms of that data, the other field. And I want to label that set of conditions as star, because I'm going to use it later. And it's equivalent to supersymmetry. So imposing supersymmetry is equivalent to these geometric conditions on Y7 and these other fields. Finally, there's one more equation to impose, and that's the Bianchi identity for F5, which is equivalent to the equation of motion for F. And that gives you this funny looking equation. So box R is a half R squared minus Ricci tensor contracted into itself. And everything here is for the caler metric. So that's box is for this caler metric. Everything is caler metric. And I'll call that dagger. And that's equivalent to the Bianchi identity for F5. It's a fact that supersymmetry plus the Bianchi identity implies the equations of motion in this case. So these solutions are necessarily critical points of this action. And you should compare this. So CF, some of you at least, have worked on Suzuki Einstein geometry. If you haven't, then just ignore the comment. But this looks very similar to Suzuki Einstein geometry, which arises in a very similar way for ADS5 times Y5 solution. There's a very similar structure. So this is, again, a unit killing vector field. There's a transverse caler metric and so on. OK. What do I want to do next? So here's a key remark. If we restrict this action functional, S, over there, to a subspace of the field configuration space. So you start off with a space which is metric, the scalar B, and this U on gauge field A. And solutions to the equations of motion are critical points of this action on this space. But I can restrict to a subspace. And if I do that, solutions are also critical points of the action restricted to the subspace. It's kind of a trivial statement, but it's important. So all I'm saying here is that, so here you're setting the derivatives of your action functional to zero in every direction for a critical point. Here I'm just requiring derivatives to be zero along some subspace, but that's a necessary condition to have a solution. That's all that I'm saying. And then the trick is to pick a smart subspace in this case, but there's a completely natural one, namely the solutions to the supersymmetry equations, which is star. So there's now a computation to do, so you compute. So in a horrible abuse of notation, I'm going to take my original functional, S, my action functional, restrict it to star, those are the solutions to the supersymmetry equations, and I'm going to keep calling that S. And if you do that, you get the following formula. So, of course, there's an integral over Y7, but it's extremely simple. So that's it. Eta, I have defined as this one form that's dual to xi, so that's just the one form dual to the killing vector. And so in particular, d eta is a half times the Ricci form rho. So you get this very simple formula for the action functional, eta wedge rho, rho's the Ricci form, and j is the Caillir form of this Caillir metric. No, I'm off-shell still, right? So all I've imposed at this point is supersymmetry. I have not imposed this equation of motion up here. So I'm off-shell. So far, I can pick whatever vector field I like still, and I can pick any Caillir metric. Of course, for solutions, you expect some restriction on this vector field. I should have said this earlier, perhaps. This vector field is dual to the r symmetry, and we expect that to be uniquely fixed for the solution. So you can probably see where this is going now. This is the function I need to extremise. There are a few more details, though. So a few remarks. So this S, it depends on the Caillir metric only via the Caillir class. And to be precise, it's the transverse Caillir class. So if you shift j, choose a different Caillir metric, but keep the same Caillir class, this functional doesn't change. I don't know whether to write this as perhaps too technical. So this thing lives in some basic carmology of the filiation for the vector field, but anyway. In the nice case where the orbits of the vector field all close up, so you get a circle action on Y7, this is just the Caillir class of the base, Caillir manifold. But I want to move the vector field, and in general, its orbits won't close. Anyway, that looks like a good start. There's also this equation of motion to impose over here, this rather horrible quadratic thing involving the curvature. So F5 is closed, if and only if dagger holds over there. All right, hand side. And so a necessary condition for solving that equation is that its integral of the left-hand side is the integral of the right-hand side. So if you integrate over Y7, you also get something very simple. Eta wedge rho wedge rho wedge j is 0. So it looks almost like this one. It's just you swap one of the j's for a rho. That's what you get if you integrate that equation over Y7. You get this. So we also certainly need to impose this condition if we want a solution, and I'll call that double dagger. So another remark is there's also an eight-dimensional geometry that's naturally hanging around in this problem. So I'll call that capital X. And it's a product of the real line, the positive real line with Y7. And I'll introduce a coordinate little r is positive on the positive real line. And this eight-dimensional geometry is complex. In fact, in some sense, it's calabiaw. So it's a complex manifold. And it has a holomorphic volume form. In that sense, it's calabiaw. So with a holomorphic volume form. I can write that as capital omega. So 4 comma 0. It's got four dz indices and no dz bar indices in complex coordinates. And so there's some expression for that. So I'll write it down. It's not terribly important, but here it is. So there's dr plus i r eta here. And this omega is a global transverse 3 comma 0 form. So on the scalar manifold, that scalar manifold will have a canonical bundle. And that canonical bundle will have some section. It won't be globally defined section. However, if you look at the metric that I've just erased, you see that this coordinate psi is a connection on the canonical bundle. So if you take the 3 0 form on the scalar manifold and you multiply it by e to the i psi, that's a global 3 0 form on y 7. And that also tells you it must have fixed charge 2 under this vector field. So this is a remnant of the r symmetry that you'd see in field theory. This vector field will be due to the u and r symmetry. And this is telling you that the spinners have got some particular charge in some normalization. And here this has got charge 2. Follows from this fact down here. However, so although it's complex and it's got a holomorphic volume form, it's not scalar. That's what makes this slightly more exotic and interesting geometry than the Suzuki Einstein case that I mentioned over there, where this cone is scalar. So there's one more thing to do. I have a 5-form flux, and I need to impose flux quantisation for this background. And that's essentially the last step, flux quantisation. For all five cycles, certainly call them sigma a in y 7. So this is a five-dimensional manifold, and there's some number of them. I label them by capital A. Then the period of the 5-form flux over these should be integer with appropriate factors. So here are the factors 1 over 2 pi L string to the 4 G string. And you integrate the 5-form over sigma a. So the left-hand side here should be an integer. Back flux quantisation in this case. And we know the 5-form flux explicitly for the class of background star, the supersymmetric backgrounds. And I'm looking at, and you evaluate it, so you get these annoying factors at the front. But here's the key point. The integral is eta wedge rho wedge J. This is exactly the same factors again appearing. So there are sort of three key ingredients here. There's the action, which has got eta rho J J. There's this constraint. That's from solving this equation of motion over here, which is very similar. It just swaps a J with a rho. And finally, the flux quantisation also depends just on these quantities. So I should say, remember that this came from closure of F, this condition. And in fact, this isn't a topological invariant unless that holds because of that reason. So if you don't impose this, this form isn't closed. And so this makes no sense as a cycle. It would depend on the specific submanifold. So to make sense of that as a relation in homology, you need this double dagger to hold. It's a technical comment, but it's important when you come to calculate things. So here's a summary that you fix a complex cone. There's this eight-dimensional complex geometry with a holomorphic volume form. Fix one of those. It needs to have a holomorphic U1 to the S action for some S that's at least one because I need to have a holomorphic killing vector field. So this is killing, and this equation here tells you it's also holomorphic, because that's the complex structure. So I need at least a holomorphic U1 action to get going. In general, I'll have some tourist action, especially if I want to move the vector field around inside this tourist. Then S, the action that I've defined, depends only on the choice of vector fields. And you can think of that as just sitting inside the lee algebra of this tourist. That's S real choices. And the transverse scalar class. So I'll just call it J. And that's finite dimensional. That's the key point. So we started off with an action that was a function on infinite dimensional space. This S is still a function on infinite dimensional space, but it only depends on a finite number of quantities in that space. Just this scalar class and the choice of vector field. You pick that setup. You need your holomorphic volume form to have charge 2, just saying the superpotentials got charge 2 is that statement in field theory. We also need this constraint equation, double dagger, this one to hold. And that's also necessary to impose flux quantisation, which is the last step. I.e. you fix a choice of these flux quantum numbers. Then a solution will extremise the function. Once you've imposed those things, you extremise S, which I'm about to erase. And that will determine whatever else remains for a critical point, it's a necessary condition. OK, so what's this got to do? I said nothing about C extremisation so far, but in some sense this is precisely the analogue of that in geometry, so central charge. They recall in supergravity there's a formula for the central charge, dual to ADS-3 solutions by Brown and Heno, and it's just 3L over 2 times the three-dimensional Newton constant. Capital L is this length scale that I referred to earlier. It's still hanging around in some formulas. So for example, it's still sitting here. Well, that can be computed by dimensional reduction from 10 dimensions. So it's equal to 3L to the 8 over 2 times the 10 dimensional Newton constant, integral over Y7. Then you've got this warped volume. So I pulled this formula from the paper of Benini, Bobbev, and Cricino. They're all here. Anyway, if you do dimensional reduction from 10 dimensions, you get this warped volume formula for the central charge. And again, if I restrict this to supersymmetric solutions to the supersymmetry equations but not imposing the equations of motion, it's precisely proportional to this action. Again, there are some various factors of pi and string lengths and so on to make everything dimensionally correct. It's proportional to S. That's the key point. So they started life very differently. S was the action originally. And when I restricted it to S, I got my functional. This is some warped volume. It's definitely different of shell. But when I restrict to supersymmetric solutions to the supersymmetry equations, it's precisely this functional. So extremising the action is the same as extremising the central charge. So perhaps here I can just say a sentence about the extremisation. And then I'll conclude just with an example. See how you actually, you can do this in practice, carry out this program, and actually compute central charges of solutions without knowing what the solution is. So C, extremisation. So this is Francesco and Nicolai. So they told us that the superconformal U1R symmetry in 2D0,2 theories extremises C trial. And it's a quadratic tuft anomaly. So I won't write down what it is. It's a quadratic combination of tuft anomalies. And it's because of that that you can compute them in the UV theory and know that you're computing correctly in the IR theory for a tuft anomaly matching. And you extremise that over all possible U1R symmetries. So that's just symmetries under which the spinners have the appropriate charge, or the superpotentials got charge too. And you extremise it though over possible mixing with other abelian flavour symmetries. So to go to the gravity side, this torus action that I've got acting holomorphically, there'll be dual to some flavour symmetries. So there'll be an R symmetry under which the spinners are charged. But the other combinations, the spinner will be uncharged. Those will be dual to precisely to these sorts of flavour symmetries. And C extremisation resolves that mixing by extremising C. And hopefully it's clear that I've set up a problem that will do that on the gravity side. Do you want to ask any questions before I give a quick example? I've got 10 minutes. So let me actually, all that was just general formalism. Let me actually compute it. See, someone mentioned earlier there are solutions with three form flux. Actually there are large classes of solutions just within this class that I've been talking about. There's a whole zoo of different examples with different topologies for Y7. Of course I can't possibly go through any kind of general analysis of that here on the board. So I'm just going to look at a simple case where I've got a two torus times some five manifold. And you should think of these physically as D3 brains wrapped on a T2. To give you some 2D CFT. Well firstly, you can prove that there aren't solutions H2 of Y50. So you need two cycles or three cycles. Otherwise there aren't solutions. For example, there's no T2 times S5 solution. It's very easy to see that from these extremal functions that there's no critical point in that case. So I'm going to assume that H2 of Y5 is R. In other words, I've got a single three cycle in Y5. And I'll call that little sigma. So sigma is a three cycle in Y5. I can write the caliform. This is the transverse caliform on Y7. In general, it will be proportional to the volume form on T2. So A is just some constant times the volume form on T2. So that's the scalar class along the T2. And then there will be some scalar class on this Y5. Transverse to this vector field. Pretty much the most general thing I can write down. So it's a piece on the T2. And it's got a piece on Y5. I need to enclose flux quantisation. There are two five cycles in this case. So I'm going to impose capital N units of flux over Y5. That's at a point on T2. So if I fix a point on T2, I get a copy of Y5. That's a five cycle. And the integral of F5 over that, I call that N. But there's another cycle because I've got a three cycle on Y5, which I can take the product of that with the T2. And that's a five cycle as well. And that will be the only five cycles if Y5 doesn't have any four cycles, which will be the case in my example. So there's two integers then, N and M. And you can plug those into these formulas that I've given to you and simplify and rearrange a bit. And you get the following nice expression for the supergravity central charge. So it's some numerical factors times M times N over the integral of eta wedge rho over sigma. And that's it. So that's the central charge. We're still off-shell though, so I've not fixed this vector field yet. So and this thing in the denominator is a function just of the vector field. There's also the constraint to impose. So I've quantised the flux already. There's also the constraint. And the constraint in this case reads integral over Y5. Eta wedge rho wedge rho is 0. So that's it. That's the problem in this case. Very simple now. And both of those are just functions of the vector field. Its smoothness, at least at the level I've described the solution so far, is automatic. So I'm assuming my Y5 is smooth, et cetera. The complex structure. Well, I'm going to write down a case where I fixed complex structure on the cone in a moment. And that will all be smooth. Whether or not there exists a solution to the PDE, that's an open problem. But you can give that to Yao or someone like that and say, I have this horrible PDE, but I suspect that there always exist solutions to this whenever I'm at a critical point of this function. Can you please help to prove that? I think that's the strategy. These depend only on the vector field. There is an example of an example just to make things completely explicit. So I've got this five manifold. I've got this eight manifold. And then this x hat six is going to be the cone over Y5. And I'll take that to be the total space of complex line bundle. So F2, the second Hertzabruck surface. It's just some particular complex surface that's different morphic to product of two two spheres with churn numbers p and q. And x6, which is this cone over Y5, will just be this x hat six minus the zero section. That is then topologically a product of r times a five manifold. So that's the original eight dimensional complex manifold I was talking about. It's just a product of this with t2. These are toric. That means that there's a holomorphic u1 cubed action. So s that I had earlier is three in this case. And so you can write your vector field in a basis for the Lie algebra of this torus, say b1, b2, b3. These are called the YPQ singularities that some people know what they are. There's a standard basis to use in this case. Anyway, that's my vector field in this basis, almost done. So the constraint, double dagger, that's down there. If you impose that, that fixes b3 to something, p minus q, p1p plus b2q, p minus q. See, it now starts to get very explicit. And you can evaluate these, so I definitely don't have time to do this. You can evaluate this the simplest way is using localisation, actually. So I've got some nice resolve space here. You can promote these to equivariant forms on this space. And they localize to the fixed points of the vector fields, which are on the zero section here. So I localize something. I cover that. And finally, the important thing is this object here that appears in the central charge. If you compute it, it's a quadratic function of b1 and b2. The details don't matter, but just to show you, it's completely explicit. You can use your favorite Toric methods to compute it, some denominator. The formula is not important. Just notice it's quadratic in b1 and b2. And so that's what you need to extremise. You don't even need Mathematica to extremise this particular function. It's just quadratic, so you can do it by hand. And if you extremise it and plug back in, you get the central charge for a solution. So if there exists a solution for this particular geometry that I've described to you, that is definitely its central charge. And of course, I picked an example where I know there is a solution. So there is an explicit supergravity solution, in this case, that solves the horrible PDE. And this correctly reproduces the central charge. And I think I'm out of time, so I'll stop there.