 All right, so let me third thank Anton and Maxime for organizing this very enjoyable conference and of course, Samson for giving a rationale for it. Background independence in string theory has been an issue that Samson has been pushing for many, many years and in fact, yesterday's talk by Simeon and today in Erics Talks, this was already said and I certainly agree with this and in fact, I think it was the second issue that Samson raised after the compulsory comment about your PhD advisor when I met Samson first I think about 20 years ago in Lyon. So, but I should say that discussing with Samson and collaborating with Samson certainly affected my way of thinking about string theory and string filtering in particular and the work I'm going to present today is in some sense a reflection of it. So, so therefore I thought it would be appropriate thing to talk about here at the conference. So before going into the details, let me describe a little bit the result. A perturbative string theory, where it's usually formulated, is formulated in terms of a conformal sigma model and just by the necessity of conformal invariance put strong constraints on the background. So here background independence is immediately violated of course. And so what one would like to have is some theory which is background independent but when expanded around a classical solution would give rise to this string perturbation theory. So this is the problem I would like to solve. So I would like to go in the other direction. Now this turns out to be a very difficult problem. I'm not even sure a solution exists in the general case. So now being in France, I'd like to compare this with the menier that only obelisks of course can lift. But I consider myself more the type of a Roman soldier who sort of gets misled into the idea that lifting a rock is an achievement. So I'm going to lift a rock today. So that's sort of the equivalence. So here the equivalence of the rock here is the word line, is the point particle. So I'm going to not solve this problem, but I am actually going to solve this problem. And I think that this problem is trivial and in some sense is perhaps mechanical. But the result is not entirely trivial in the sense that what comes out is that quantum consistency of the word line theory implies gr plus an action for the Calvary-Montfields and the dilatons, which is identical with that of the Nuwe-Schwarz sector of the string. So at that level the string has nothing to add. So something that I can maybe here draw, something like commuting diagram. So done with that. So we don't know what this is in general, although I'll comment a little bit about it, but we know now this way. The upshot then is that the quantum consistency of the say n equal 4, by n equal 4 I mean four world line supersymmetries, so maybe put a curly end, spinning particle, uniquely, essentially uniquely fixes the action for the metric, the Calvary-Montfield and the dilaton to be that of the Nuwe-Schwarz sector, the muscle sector of the Nuwe-Schwarz. So uniquely I put it in bracket because there's one choice that you don't have in the string, which is you can add a cosmological constant at the price of projecting out the Calvary-Montfield. So if you give up the Calvary-Montfield then you can have a cosmological constant. So now I'd like to describe a little bit the framework of how that comes about and then how you solve this problem, so that's the rest of the talk. So the framework is the, I have to say a few words about BV quantization. The reason being that string theory or string field theory, in some respect also the point particle naturally gives rise to a BV action, so I have to say a couple of words about BV action. So the starting point of BV action is a space of fields, a graded space of fields where you also have ghosts. So let's say there's a fields, let's say maybe it's phi. Then I have an action on this space of fields, so that's some function. And then there's the BV operator. So this is not a spectral triple, but this is a different thing. So graded space of fields, an action which is a function of this space of fields, and the BV operator which is a second-order differential and the Gage invariance of this theory, you probably don't see it. So the Gage invariance implies the BV equation which I can write down here, which is e to the minus 1 over h bar s, BV operator e to the s, 1 over h bar, which we write as s comma s plus h bar delta s, and this is equal to zero. So that means Gage invariance. Today, h bar is equal to zero. So we'll only talk about the classical theory in the sense that no loops. So let's take h bar equal to zero. Quantum consistency in the same sense as the way that you quantize the worksheet in the string. So in that sense, so you have normal ordering of the operators. So the h bar is zero, but you still have to worry about quantum. No loops. Yeah, you still have to. Yeah, that's right. That's a target space. Yeah. So I say exactly an alpha prime with the analog slope in the string. So this is an odd bracket which satisfies Jacobi, or a graded Jacobi identity. And if you talk about classical theory, the consistent of Gage invariance is that the BV bracket of the s with itself is equal to zero is an odd bracket. So that's generic stuff. Now if you have in this, this defines a vector field. You can say v of ff is some function, which will simply take us this BV bracket of s with some function f. So that is a homological vector field. In the sense that v squared is zero. And then if we expand, ah, so maybe you should say one more thing. If we have some additional, so that principle, that is enough to start with a BV, get you started with a BV action. But here in the string and also in the point particle, we have an initial structure which we have a non degenerate, degenerate symplectic form. It's odd such that, let's denote it by omega, such that delta written as omega ij in components di dj. So then we can define, we can take the interior derivative of omega with this v. And this is exterior differential of the action. So the usefulness of this equation is that it tells you that the zeros of this homological vector field, which is induced by this BV bracket together with the action, are precisely the critical points of the action. So these are the classical solutions. So whenever v vanishes, that means you have a solution of the equations of motion. These ones, no, they're all, these are graded fields. So yeah, so they're not, so I should probably put a bar or something like that. So maybe I'll write that here. v equals zero critical point, let's say phi zero of s. Now, so if you have a critical point, we can expand. We say v is equal to v zero phi zero, which is equal to zero, the vector field at the point to pay their expansion. And then you get v1 plus v2 and so on. Keep going, let's maybe put the v3 as well. No, probably enough. Now, since v squares to zero, this implies some structure on this, on these coefficients in this expansion. So first, I should say what these are. I mean, this is just, this is, let's say, this is zero at the point. So this is a map, so this is a linear map from the tangent space at the critical point to itself. And this is a map from t phi zero f into t phi zero f squared and so on. And this keeps going as you go to the higher order terms. So you can take the dual version where these are maps from v tensor v into v, where v is the vector space and so on. So you can go either way. The point is, though, there is some structure on these maps. And the structure of these maps follows simply from v squared equal zero, which is just the consequence of the way v is defined as coming from the BB bracket. So this is all, in some sense, consequences of gauging variance. And what it implies is that the v i's i bigger equal one, they satisfy what is called a l infinity algebra. Infinity algebra simply means that if you take the commutator, say this is a linear map. This is already, as we see, 2 to 1 or 1 to 2, depending on how you look at it. If you take the commutator, then this is not equal to zero. If it was zero, it would be a l algebra, but it's zero up to a v1. So this is a graded commutator with v3. So you see, if the higher order maps are zero, then that would mean it's just a l algebra. But since it's not, it's l up to a homotopy, which means that this is a linear, this is differential, and this is l up to this. So that's l infinity. So this is just a consequence again of just this being a BB action with a BB, with a gauging variance. Yes, it's l, it's Jacobi up to higher order terms, up to exact pieces, absolutely, absolutely. So furthermore, this induces, so these, this is called Si induced. It's called a minimal l infinity structure, and I'll denote it by Si on the homology of v1, which is the differential. And this is actually the physical states. These are what we call the particles, and these are just the S matrices. So this is all sort of descents straightforwardly. You start with gauging variance, you get in this way homological vector field, you expand it around the critical point, and that induces an l infinity structure on the homology, which are the S matrices. So now we can perhaps put that here, we can summarize this structure, simply one line. So I take S, I have F, that's what was my starting point, space of fields, BV action, a BV operator, that gave rise to homological vector field, that in turn gave rise upon expansion to the V i's, which is this infinity algebra, and then this, they're not really maps, and then this gives rise to the Si's on homology. So this is where string perturbation theory is defined. In string perturbation theory we compute S matrix elements. That's what we know how to do using the virtually conformal field theory. String field theory, as defined by Zweibach, would be here. So in string field theory one tries to recover the vertices. So these I should say, if you did any field theory, these V i's would be the vertices of your Lagrangian, which would give rise to the S matrix elements. So the Zweibach's construction of string filter, you try to reconstruct vertices, which give rise to the right S matrix. But these are defined away from the homology, so that's why you can compute off-shell amplitudes. But what one really would like to have is this. You would have business till background dependent, because all this is expanded around one point. So only once you get rid of this point five-zero, you can truly speak about the background independent formulation of string theory. Now, Samson gave, based on earlier work by Witten, and we've seen this in Eric's talk, Samson gave a show that there is a formal expression for the background independent version of this for the open string, which he showed to be non-ambiguous. So that was this 93, I think. And I think this is today perhaps the only background independent version of string filter we have. I mean, it has its limitations, but it's the only one really we have for the open string. It's the only one limitation which was, I could not do the Gauss' Metronautical Case, and it's still open. Still open, yeah. So that's where I go, first of all, I will not do the open string, I will do the closed string, and I don't do the closed string, I do the word line. Make my life a little bit easier, and I want to show in the word line you can get a manifestly background independent formulation, which contains quite a lot of information about string theory. So that's basically the content of this talk. Are there any comments or questions at this point? Let's see if I forgot anything important, probably. Well, maybe the comment is just that, no, I think, so what we really like to do is we'd like to go from, we'd like to go here. So that's the aim. So that's this question mark up there. Okay, so then let me introduce the spinning particle. This can go. So we do the n equal 4, and let me simply write down the action for it. So this would be just a bosonic particle. So you have the coordinate, the momentum, and then the Hamiltonian. There are metric enters here, indeed. So that's a spacetime metric enters in this bracket. I didn't display it, but it's there. Is it the blackboard? That's the background metric. Yeah, absolutely. So this is an external field, if you wish. For which we want to derive equations in the end. So then we have, since it's supersymmetric, I say the I, and then I have to gauge it, and make local supersymmetry. This is, the E is taking care of, this is the im-bind taking care of the word line reparameterization invariance. And then this, since I'm doing supersymmetry, I need an analog for the super translations, which are supersymmetric. The super translations, which are super, so it's one-dimensional supergravity, and then plus R, I, P. So this you can think of, so this is the superversion of the im-bind, the superpartner, and these are the supercharges. Yes, it's not green-schwarz. The word line supersymmetry, so not spacetime. So spacetime have no supersymmetry here. Q, I, and then I have also, since I have a n equal four, here I take n equal four supersymmetry, which means I is equal to one or two, so I take complex fermions here, so these are old word line fermions. So I take complex combinations, I go from one to two, but this is really n equal four in terms of real fermions. And as a consequence, I have an SO4, R-symmetry, and these are the generator of the SO4, R-symmetry, which I may gauge or not. So I can gauge the SO4 or not. That is a little bit up to me. And this will play an important role, actually how much exactly of the SO4 we gauge or at all. So that is, that's about the only freedom you have in this theory. So that's the action. So now let me, the T does also, yeah, so the T, these are really TmUi, and so is this. So they have a spacetime index. So they also contract with the spacetime metric. So this has n equal four, word line supersymmetry, no spacetime supersymmetry. So there is the metric again in the theta d theta. As well, yeah, absolutely. Okay. If it's an open string, you can add chain pattern vectors in there. Yes, you could. You could add extra factors. You mean that you could, you know, I took this to be spacetime indices, but I could add extra indices, in principle. This is going to be n equals four, so without any case. Ah, so yeah, I'm actually going to comment about this right now. So this is a little bit of history. So I could have taken n equal two instead. n equal two, word line supersymmetry. And then if you do a BRSD quantization, and then in both of the BRSD squares to zero, that implies the Young-Mills equation of motion. You can do a non-Abelian. Okay. So this was done by Siegel and Die, I think in, I think, 2008. So no conditions on G-Menu. These are the Young-Mills. I mean, do you get the full? Fully non-linear Young-Mills. So how do you talk about Young-Mills? Ah, I will, I will, so here is the summary, but I will show it later. Actually, Berkovitz, before that, did Grinschwarz and prove that it is unitary equivalent to Grinschwarz for particle. And then derived from it the space-time equation ten dimensions. So that the capacimetry. Right, musical. No, there's no capacimetry. This is not supersymmetry. I think it's Berkovitz mapped the Grinschwarz for super particle into his stuff. Oh, I see, yeah, yeah. Okay, this, yeah. Remember, the James Ruhmann was teaching us about that 20 years ago. So N equal 4. Again, you can impose this, and this gives you the equation of motion for NS-sector, or for M equals 0. So the massless NS-sector of type 2. So this is type 2 supergravity, if you wish. So why N equal 4, why N equal 2? We'll see in a moment. So basically the story is that N equal 2 is the particle that corresponds to anything that is a photon or a gluon. It cannot couple to itself in a gauge invariant way unless it is around the classical solution, but it can couple to gravity. Graviton cannot couple to itself unless the gravity equation of motion is satisfied, linearized. So that's sort of the hierarchy of things. So this is, by the way, with Roberto Bonetti and Adil Mayer. So this turned out to be, after Segel did this thing, people tried to do the N equal 4 and for some reason failed. And the reason why they failed is sort of clear. These guys are asking you constantly where the nabilia structure comes, how do you put on pattern factors? You can put on pattern factors, absolutely. Eric, do you think you have to put on pattern factors? You put, you know, yeah. Yeah, yeah, yeah, but how do you do on a point at all the angles? The T does, what happens is the, we will see in, let me maybe go a little bit and you'll see it very quickly, very shortly. You'll see it very shortly. How that works. But before I do that, I would like to say one thing, is that why, I mean, what has this to do with what I said before? So why Q squared equals zero has anything to do with the equation of motion, the way I described it before. For this, we need to go back here to the expansion here. Basically, I still believe that only you can get is the equation of motion, not the action. You can only get the equation of motion. Absolutely, but you can go off-shell. I can quantize this theory off-shell. So, let me see. So, actually a comment I was going to make. So let's look again at this V. This V squared to zero anywhere. You don't have to be on shell for this to square to zero. But you see, if you want V to be equal to zero, I'll write that perhaps here. So this is V, so this equal to zero implies V2 squared plus V1 V3 equal to zero, plus other terms. Sorry, V0, sorry, I got that wrong, I think. V1 squared plus V0 V2. Now, on shell, on the critical point, V0 is equal to zero. So this has to be equal to zero. This is the BRSD charge. So the linear term in the expansion of the vector field is the BRSD operator of the perturbative calculation. If you off-shell, V is still squares to zero, but the BRSD doesn't square to zero, but it's compensated by another term coming from here. So this is a well-defined object, just doesn't square to zero, but on shell has to square to zero. So BRSD square to zero is saying that V0 has to be equal to zero. That means we are on the critical point. So that's the connection between the BRSD quantization and the BV quantization in space-time. So now coming to Samson's question, or comment rather, Samson doesn't ask questions, he makes comments. Rightly so. Let me put that here perhaps. So what does that have to do with the action? And I agree, this doesn't give you the action. So what we have is we have V1 we can compute from the point particle, which is the first derivative of the vector field. We can compute it at any point. It just doesn't square to zero, the generic point. So in principle, you can integrate from phi0 to phi. You can get the V, but of course, where this is integrating over the field, of course there are questions here. This is formal because you have to choose a path, you have to make sure the thing is perhaps path-independent and so on. But once you have this, if you have V, you know then we have this IV omega is equal to dS. Since the symplectic form in this case is independent of the background, this is a constant thing. So you can integrate once more, and you can, in principle, get S. But of course, in principle. So you get the equation of motion for S. Yeah, I mean, this equals zero. Yeah, I mean, V equals zero gives the equation. Yeah, gives you the equation. That's the on-shell data, it's not the same as the F. That's the on-shell data. Which is the motion or the on-shell data? I get the first derivative of S, would you agree? I mean, IV with omega is the first derivative of S, which I can integrate. Of course, formal, but can be integrated. So this is just to address this comment by something. Okay, but now let me describe a little bit. What has this to do with John Payton? How does all this stuff enter? So let me describe this. So first thing I need to do is I need to say, what is the BRSD charge? This is going to be familiar to most of you. So QBRSD is equal to C times the Hamiltonian plus gamma i Q bar i plus gamma bar i Qi plus gamma, gamma bar times B. So this is the structure. So this is the super ghost, gamma. It's the wordline ghost now, which is the... C is the repermeterization ghost, and B is the conjugate to C. So this is the usual structure of BRSD charge we have whenever we have DRAC constraints and we BRSD quantize the system. So what is the Hilbert space? Hilbert space is functions, let's say square integral, or perhaps a little bit less. If you want scattering states on spacetime, times y, algebra to some power n, times Clifford, hence our n. And so these are the functions on spacetime. These are generated by the T-dust. Sorry, the Clifford is generated by the T-dust. These are... And the y are generated by the ghosts. So we have a state, h is psi. In principle, we have to sum over n. So h is some state psi, which is some polynomial in these theta i's and the betas and the gamma i's with coefficients in L2 of Rd, these are the functions, so the wave functions. But they do not depend on T the bar, the beta bar, and the gamma bar. These are not the states, they annihilate the vacuum and the highest weight. So the thing is now you could say, okay, this is my Hilbert space and I just want q to square to zero acting on this Hilbert space and the result is never going to square to zero. Not even in the n-eagle-2 case is going to square to zero. You have to restrict the Hilbert space, otherwise it only squares to zero in zero background field, so the only solution is zero background. When is the 10-dimension coming to business? n-eagle-4 requires 10? Dimension is free. I can do for any dimension. I mean, up to one, two, three, four. One maybe not. So dimension is not fixed because I have no conformal symmetry here. You see, I don't need a central charge to be zero. I have no spacetime supersymmetry. No spacetime supersymmetry. Size of the multiplex depends on the dimension. That is true, yeah. So eventually the spins that you end up with... No, the spins that you end up with actually depends on the number of supersymmetries. Okay, yeah. Okay, let me explain it right now, actually. So the thing is, so now we have to restrict this Hilbert space, otherwise you get exactly zero background fields and the only way you can restrict the Hilbert space is by gauging some of the SO4 asymmetry which you have. So let us explain a little bit how that would work. So a generic state in H will be of this form. Right, because say you take a state which is, say, theta one mu, theta two nu. Let's say mu one, mu one, theta one mu two, theta two nu one, theta two nu two, and so on. So it has to be anti-symmetric in mu's. It has to be anti-symmetric in the nu's. I have a naive question. If I think about this as the zero slope limit of strength, the point particle limit, then am I not in the normal sector? So it is a good question. So you can relate it to the zero slope limit. That is something you can definitely do. So there's two things you can do. You could do that, what you suggest. Or what you can do is you can do a twisted reduction on the word sheet in this sort of space direction and then this theta, they correspond to the psi one half and psi minus a half and bar in the Nuwe-Schwarz sector. So they just count the first. It's like a level truncation. But you could also do Ramon's. Because you write like, here it's... You could also do Ramon's and then you get fermions in spacetime. If you put Ramon together with the Nuwe-Schwarz, you get spacetime supersymmetry if you want. But there's no need to do it because you don't learn invariant. But that is true. So this is in some sense related to level truncation. So now you see, if you build states like this, and now as you say, the dimension of spacetime tells you how many of these t-dots you can have. So it tells you about the maximum spin you can have. So it's totally anti-symmetric in the Muwe. Totally anti-symmetric in the Nuwe. So you have this structure. So that would be a generic state in the Hilbert space. Now, what you can do is you can gauge a U1 times U1 subgroup of the SO4R symmetry. And in fact, turns out you have to. That's the minimal gauging you can do. If you don't do that, you get zero background field. So this actually has an interpretation in string theory as well. The first U1, you can think of as level matching. And the second U1, you can think of as level truncation. So what happens is you get this instead. Say if you fix this to be one. We say we take the charge to be one. Do U1 charge equal one-one? So then you go from this general state to go down to this state. So now, but that means exactly, you have exactly only one of the t-dots. You have only one of the t-dots. You have an equal number of them, so you match the level and you fix the level truncation. So that's the minimal gauging you can do. If you don't insist on this, you will not get any solution. But you can do more. You can symmetrize this thing. So there's another, say, if you can take the young symmetrizer in SO4, then you get symmetric. So that projects out to be mu nu field. Then the Kalbremont field is gone because of insistence symmetry between mu and nu. Kalbremont is out. And you can do one more thing, which I think you cannot do in string. There's a trace, what I call the trace in the SO4 generators, which removes the trace. And that means that, so here, let's say, here the mu nu is out. So this, and here the dilaton is out. So these things, as I said, have bondons in string theory. That would be level matching and level truncation. This would be world sheet parity. It gets rid of the mu nu like in type 1. This, I don't think, has an analog in string theory. But we can do it. And that's, by the way, the reason why you can get the cosmological constant for the point particle. Actually, no, it's not true. This is three. Sorry, I take this back. So we have a little bit more freedom. Why don't you symmetrize to get a more gravitation? Do you mean to not insist in this one? I think not. So you want to keep the mu nu, but get rid of the graviton. My memory is that we haven't succeeded in it. Maybe we didn't try hard enough. But I don't think we were able to do that. I think the graviton, no, actually the thing is, if we gauge completely SO4, the graviton is left. Yeah, I don't think you can get rid of the graviton. It's impossible. So that's the structure. You see, we have a little bit, we don't have much more freedom than in string theory, but a little bit more freedom because we can get rid of the dilaton. It's about the only thing which we can do. So how do you write unsympathetic structure out of all this stuff? Usually you take scalar product in Rd times usual... Are you using it or you will be deriving rather the equations of motion? Equation of motion. We come straight to equation of motion. From q squared equals zero, we give you straight equation. So if you do all this, let's say, let's take this maximal gauging. Then the generic state that's left is a h mu nu, theta 1 mu, theta 2 nu, plus there is some psi mu, theta 1 mu, beta 2, minus theta 2 mu, beta 1, plus anti-fields plus auxiliaries. So this is what I was saying at the very beginning. Whenever you characteristic quantize and you look at the comology of the point particle and the same thing in the string, you automatically end up with a whole BV multiplied. It just comes, it's given to you right away. I don't see why. I mean if I act now with q on this state, don't I get gamma 1, gamma 2? Because it's... Yeah, okay. I will say a word about this. So this is the graviton and this is the diffeomorphism ghost, space-time ghost in the BV language or beer steel. This has ghost number, this has the wrong ghost number to be a physical state and this has ghost number zero, this has ghost number one. And now you have specified the Hilbert space, it is maximal gauging, and then now you want to see what happens to the BRSD charge. So how do we couple the system to these background fields? And for this, I just tell you how that works. The generality is a lot of formulas, but let me just describe a little bit how this works. So the generic procedure is you take the supercharges, you always start with the supercharges and you couple them to the background fields by changing derivatives to covariant derivatives. It's as simple as that. And the first thing you have to check that the supercharges commute into the Hamiltonian, like in Dirac, and the next thing you have to check is how the Hamiltonian commutes with the supercharges. Now you might say this is automatic because Hamiltonian is Q with Q bar and then Jacobi identity gives you that Q with H is automatically zero. But it's not true because you're working on the constraint system, on the constraint Hilbert space. So just because Q, Q bar on the constraint Hilbert space gives you H doesn't necessarily mean that Q with H on the same constraint Hilbert space gives zero, so you have to check it separately. But you do it. Let me say coupling to backgrounds. So what we do is we start with the supercharge, Q i, remember the supercharge up there, Q i, which I write as T the i mu. Take usual derivative plus spin connection and then T the bar B, slightly anti-symmetric, and say let's leave it there for time being. And then you have the same for the Hamiltonian. You take it, let's say this is equal to D, covariant derivative. And the Hamiltonian is equal to D mu, D mu. And you might say that's perhaps good enough. Now you compute, let's see, now you compute simply Q squared through squared B at a state charge. And what you get is, let's say, let me combine these two things, these two together, just to simplify notation. So Q, I denote them as the combination of the two supercharges as S. And then I'll find that Q squared, so this is to be a state charge squared, is equal to S squared plus gamma, gamma bar Hamiltonian plus C times commutator of Hamiltonian with S. Now the point is, you get two terms, this as a C goes, this doesn't. So they have to separately vanish. Now you see, now you get this structure. So the S has a gamma, gamma bar, but so does the H, because if you commute the B with the C here, you get a gamma, gamma Hamiltonian. So for this to vanish, so then you have to make sure that both of these terms vanish separately. Now let me first do this one. That gives you gamma, gamma bar with Ricci, T, the mu, T, the bar mu, and that's it. So now you see that gives you Ricci flat. So this is zero only if the background is Ricci flat. So you get Ricci flat. Now there is a little bit up, so that's part of it. Then you have to check this is equal to zero. And if I'm not zero, you have to add some non-minimal coupling in order to get this to zero. So even on Ricci flat, in order for this to vanish, you will have to add a non-minimal coupling, which is like this. So I'm going to have it. Plus, arm mu nu, lambda rho, T the mu, T the bar rho, and mu, T the mu, T the bar rho. So this is a non-minimal coupling, which is imposed on mu by imposing that this is equal to zero. But turns out there's a little bit of freedom. One thing you can do, you can add a non-minimal coupling to gravity. It's a free parameter that I've put here. And then if you restrict that on the Hilbert space, you'll find that the condition you get is arm mu nu is equal to lambda g mu nu. And g mu is free parameter. So you get the sitter until the sitter Einstein spaces. So this is the most general solution you get in this maximally gauged SO4 Hilbert space. So you get just Einstein is the only solution. No, with cosmological constant. So now you can say, okay, now let's be a little bit less restrictive. Let's perhaps bring in back the beam mu nu field. Let's not impose this strong gauging. Let's stay here. Just do the u1 times u1 gauging. Then you have to modify, you add a coupling to the Calvary-Montfield, minimal coupling. And again, you will have to add some stuff here, which I'm not going to display because it's a little bit cumbersome. But again, you get it to zero, but we'll only get a solution if, let's see if I can put it, perhaps here, if arm mu nu plus h mu, what is it? lambda rho h nu lambda rho equal to zero. So again, you get the string equation of motion. You also get the equation of motion of lambda mu equal to zero. So if you allow, because now the restriction becomes a little bit stronger because you project on a bigger space, so you get more equations. So this is, but this thing you can more or less infer from string theory what you should do. So this is sort of natural thing to do. You couple it non-minimally. You just introduce a spin connection. This is also more or less, oh, actually here it's not correct what I'm saying. Here I should say it's T to the one a, T to the one bar a b, minus T to the one a, T to the two a, T to the two bar a b. So you violate the, that's where the symmetriser comes in. You violate the one to symmetry, which from string theory is also understandable because you're left-movers and right-movers couple differently to the beam and new-field. So you could sort of, from string theory, you can sort of infer that this should be the right coupling. So what about the dilaton? So you might say, well, does the dilaton, I mean dilaton appears in the spectrum, so somehow you should be able to couple, but how does it couple? Now, first thing you might think, in string theory the dilaton couples to the virtual curvature. There is no virtual curvature here, so cannot couple. And you could say, well, you know, that's sort of derived. Really the dilaton couples to the divergence of the ghost current that was the initial coupling for the dilaton. There is no divergent, there's no anomaly in the ghost current here. So you might think, it's not quite clear how to couple. So for the dilaton took us a long time to, just by trial and error, to see how can we couple the dilaton to this point particle. It turns out there's a way to do it, simply by d mu phi. So this is like akin of the vile, vile's idea of how to remember to electrical magnetism with, in gravity. So this is a non, non Hermitian. You see it comes, it doesn't come with an eye here. It's not a gauge, it's not a U1 gauge field, it's a sort of R1, R1 gauge field. But this works and again gives you a modification. Of course, then you get here, you get on top of it, you get the d mu, d mu phi. So you get again the string equation of motion. When you've done that, you can project out again the B field and you can add the cosmological constant. So there's, the moment the B field is gone, you have a little bit more freedom than in string theory, which is you can add the cosmological constant. But that's it, no more. The moment you have the B field, you get exactly the low energy equation of motion from, from the Neva-Schwarz sector of type 2. So, I should probably wrap up. So we see really the point particle contains all the information about the massless sector of the string. So you don't really need the string to derive the Einstein equation, you can just quantize the point particle. But you also get that each mu nu satisfies the linearized Einstein equations probably. Ah, yeah, that comes from, yeah, I will say something about that right now. So the next thing you might say, but what about the states? How do I get states? I mean I want to perhaps do some scattering. Well, in string theory we know how to do scattering, we simply use the operator state correspondence and insert an operator at the origin of the complex plane and then that's it. You say, well, this is a point particle, there is no such thing. Turns out there is such a thing. And the way you do it is simply in order to get a state, so states, so let's say one particle state, what you do is you take q, q, write q is equal to q0 plus delta q and this is the deformation of the background. You make an infinitesimal deformation of the metric, for instance, or of the dilaton or of the b mu nu field if you want. And then you say, well, this squares to zero, so then q squares equal to zero since implies that q0, delta q is equal to zero to disorder because q0 squares to zero. So if the delta q acts on something that is q0 closed, you've got a state. Unfortunately, I deleted it, but there was such a state. Remember, there was the diffeomorphism ghost, which was, I denote it by xi, which I write as xi mu t mu one beta two minus t mu two beta one. This is q closed. It's in the comology of q0, but it's the wrong ghost number. It's ghost number one, zero minus one. So then I can say, well, what about if I active delta q on it? q, and if this commutes with this, then that will be q closed, but it will not be exact. And that's it. In fact, this is the graviton, the b mu nu field, and the dilaton. So you get the vertex operator simply by acting with a variation of q with respect to background field on the diffeomorphism ghost state. So that's the analog of the operator state correspondence in the word line. Now you can write down n-particle amplitudes in principle. You can write down these things. Simply what you have to do is you put the diffeomorphism ghost here. You put the diffeomorphism ghost here. This involves a little bit of information. You have to put the momentum of that ghost state. And then you put vertex operators, delta q, delta q, delta q, delta q. And then you have to integrate over the propagators, of course. And that gives you the right amplitudes. So all of this with the point-particle sort of is very naturally just simply doing BSD quantization, which I claim again is really background-independent. The only condition that q squared to zero will give you that you are you reproduce the equation of motion. So that is basically what I wanted to say, is that in the point-particle all these things which in string theory we cannot do, we can do. And as far as the master sector is concerned, it has the same information. That's probably where I should stop. There's a comment. This time it's a question. What was the actually main idea that made it that it was not done before? I think all this kind of thing. Yeah, this is a good question. So why didn't people do this in the 70s? You know this spinning point-particle was introduced by Brink, et cetera, in the 70s. They did... Well, I think the one thing where people got stuck is that you have to properly restrict the Hilbert space to do this. And that is, I think, thanks to Warren Siegel, who had this first suggestion in the Nical 2 case to introduce this restriction. And if you come from string theory, it's somehow the obvious thing to do, is to restrict the Hilbert space. And we do level matching. We immediately restrict the Hilbert space. The only thing is here we have a little bit more freedom. In the string theory, you don't have an SO4. You only have a U1 times U1, really. When you could think of perhaps the world-sheet parity, which becomes here part of SO4, I think in string theory usually we don't think of it that way. But it's really the sketching, which for some reason, including myself, didn't think about restricting the Hilbert space in the first place. Can you go beyond Nical 4? Ah, yeah, this would be interesting. So you could, in principle, do Nical 6. We haven't done it because this was already... You know, working at all these commutators is quite messy, but there's no in principle obstacle. But I suspect you'll get a contradiction that there will be no solution because otherwise it means we have a consistent spin-3 theory. So, yeah, so I think it's not going to work. But it would be an interesting exercise to do, but I suspect it's not going to work. So my question had some overlap with that, but there was a series of papers from the 80s by Paul Howell, Paul Townsend and collaborators where they looked at general numbers of supersymmetries. So for N supersymmetries they found they got a maximum spin of N over 4, N over 2, so that if you went to N equals 8, you would get spin 4. And they went through this, they restricted the Hilbert... Essentially, they put addic and strange, which I think corresponded to your restriction on the Hilbert space. And they got the free wave equation for spin 4 coming out, or for monetary spin. They looked at various generalizations including masses and various other representations. And so the question would be then, which was very close to what was just asked, that given that it works at least at the free level, what would happen if you tried to implement your program in that? Yeah, I think with the free level indeed you can do, in principle, for any N. And exactly as you say, the spin is N over 2, for N equals 4 you get spin 2 and accordingly for higher. By the way, for spin 1 you get fermions, for spin 3 you get the gravitino for N equals 3. But indeed as you say, the thing is the Q zero will always square to zero. But the moment you try to couple it to a background field, you will find that the Q only squares to zero when the background field identically vanishes. And here we just found a little bit of a way by suitably restricting the Hilbert space that you can do it. But that's really where I think that's where you will fail. The moment you try to couple it to a background field it will just not close anymore. I think that's what will happen. But mass, by the way, adding a mass should not be a problem. So I think we can add, we didn't try that. You could probably add a mass to the gravitone. And I don't know what's going to happen. Maybe again we end up with an inconsistency. Maybe not, we haven't tried it. Free level will always work. It's only once you couple it to background fields that you will get punished. So it seems that you can get pretty much any free particle in there. Free particle will not be a problem. One like symmetry space, I mean nice spaces without... Yeah, it has to be flat probably or something similar. Something simple. Otherwise you'd be constructing a coupling of spin 4 to an arbitrary metric which you know those issues. But if the metric is not arbitrary it's just a constant curvature background of some time. Yeah, maybe in that sort of space you might expect there to be a solution. Yeah, it's still the extension. You don't get the background. Yes I do. I never put in an expansion. I just, this coupling is, you know, this coupling. Where is it? I've deleted it now. But the spin connection is not linearly coupled. It's fully nonlinear coupling. This is the full nonlinear coupling. I mean there's nothing to be added. And I don't get linearized Einstein equation. I get the full Einstein equation. I mean I get, you know, R mu equals 0 but not linearized. Nonlinear. So it's not like an expansion of... No, no, no. So that is really fully background independent. Did you work in some... on normal code? No, no, no. Just completely coordinate invariant. We didn't use coordinates anywhere. He just writes H mu nu and then he pops up the R mu, nu, R mu equals 0. So he just puts H mu nu as a coefficient of that... But you have to prove the cubic vertex, for instance, is the one of... When I couple the covariant derivatives, right? The coupling comes through this d mu plus omega mu a, b, theta a, theta bar b. So that's where the coupling comes in. That's not the word nonlinear. How did nonlinear... Yes. This is a nonlinear object. Omega is not linear in H. Yeah, but the cubic... The fact that the cubic term in Einstein's is given by this is... I mean this has, of course, a linear contribution but it's made out of the inverse fear band. So it has no... I mean it's fully nonlinear. But it's true that the minimal coupling is almost sufficient up to these nonlinear terms which are these non-minimal couplings which I guess we don't have anymore here but they had to add this non-minimal coupling to the Hamiltonian which was... I had to add H for somehow box plus R mu, nu, lambda rho, theta, theta bar, theta, theta bar. Things like this I have to add to the Hamiltonian for the system to close. So it's non-minimally coupled. Such terms pop up. Otherwise just Q will not square to zero if I don't have it. Or even on shell. But this is perhaps... I mean I think you can even understand this from string theory probably because indeed if you were to write down the sigma model engineering background matrix and use some Riemann normal coordinates such terms will probably show up. I think so. But for instance, do you prove that in amplitude if you make a gauge transformation from the external states this is the same, that's part of the building. Because in string theory that's the way you prove the whole answer. How sorry is this last part? That you are a gauge invariant. I mean if you are a morphism invariant for the gravity of the state. Yeah, that's fine. That is part, yeah. So basically the statement is that if you take Einstein Hilbert action and expand it around R of G zero plus terms and then you linearize. So you have, let's write it as S zero plus S one plus S two in expansion. So this is the quadratic term in the expansion. And if you want this part to be gauge invariant the morphism invariant this linearized part you need this one to be equal to zero so you need to be on shell. So basically the statement is that graviton only makes sense. I mean it doesn't make much sense to talk about the graviton if you are not expanding around the equation of motion. There is no such thing as a graviton. No, there's a diffeomorphism. You can make comment. I'm not the first. Why? Because notion of metric is a choice of the ground in string theory. No, in string theory of course we'll be back to that but in point particle notion of metric is... No, no, in string theory of course there may not be a metric in the first place. In engineering background I agree. No word metric or no word gauge field or something. When you say word metric you pick the background. When I say spacetime metric I choose a background. In string theory, yeah. So I agree that's... Yeah, yeah. I mean that's why I was saying probably in the string theory setting it's perhaps even not possible to write down a back-end independent action in the full string theory. Closed, for closed. No open. For open maybe possible, for open. I already want you to consider three manifold where... I mean the reason why it works for open is that in the open string whatever you put on the boundary in your thing is induced from the bulk and the bulk is solid. The bulk doesn't change. So you know at least what is the what I would say to vertex operators what is the Hilbert space. In the closed string the moment you go away from background you don't even know what is the vertex operator what is the Hilbert space you lose everything. Well if you would have three manifold where the two manifold is a boundary it would work. Maybe that's what Eric is trying to do starting from this John Simon thing to starting three. We discuss this at the time. And we give up. We give up. Too many. Never late. Good. Well, excellent. I think it's an opportunity to thank Ivo again and also the speakers of the day.