Space Time Actions from The World Line
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Sachs, Ivo
I will describe how the full non-linear equations of motion for all massless NS-sector fields of the string arise from world line quantization of the spinning particle. Somewhat surprisingly this includes the dilation which is often thought to be stringy in nature.
 All right, so let me first thank Anton and Maxime for organizing this very enjoyable conference and of course, Samson for giving a rationale for it. Yeah, 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 Leo. So, but I should say that discussing with Samson and collaborating with Samson certainly affected my way of thinking about string theory and string filter in particular and the work I'm going to present today is in some sense a reflection of it. So, so therefore thought 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, 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, with the manier 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, so I'm going to lift the rock today. So, that's sort of the equivalence. So, here the equivalence of the rock here is the, is the word line, is the point particle. So, I'm going to not solve this problem, I'm going to, but I am actually going to solve this problem. And when 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-Montfield and the Dillatons which is identical with that of the Neuvier-Schwarz sector of the, 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 don't, we do know this way. So, the upshot then, upshot then is that the quantum consistency of the say n equal 4, by n equal 4 I mean 4 word line supersymmetries. So, I should maybe put a curlian spinning particle uniquely, essentially uniquely fixes the action for the metric, the Calvary-Montfield and the Dillaton to be that of the Neuvier-Schwarz sector, the Muscle sector Neuvier-Schwarz. So, uniquely I put it in bracket because there is, 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 PV quantization. The reason being that string to a string field to D, and some respect also the point particle naturally gives rise to a BV action. So, I have to say a couple of words about PV action. So, the starting point BV action is a space of fields, a graded space of fields where you also have ghosts. So, let's say there's a field, 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, just about. 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 that this is equal to 0. So, this equation. So, that means Gage invariance. Today, h bar is equal to 0. So, we'll only talk about the classical theory in the sense that no loops. So, let's take h bar equal to 0. Quantum consistency in the same sense as the way that you quantize the well sheet in the string. So, in that sense. So, you have normal ordering of the operators. The h bar is 0 but you still have to worry about calling. No loops. Yeah, you still have to. Yeah, that's right. That's yeah. So, I say exactly alpha prime would be the analog statement in the string. So, this is a bracket. This is an odd bracket which satisfies Jacobi or a greater Jacobi identity. And if you talk about classical theory, the consistent of Gage invariance is that the BV bracket of the s itself is equal to 0 is an odd bracket. So, that's generic stuff. Now, if you having this, we this defines a vector field. So, 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. Homolog vector field in the sense that v square is to 0. And then if we expand, 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, generic, 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. The derivatives, these ones, now they are all, these are graded fields. So, yeah, so they're not, so we should probably put a bar or something like that. So, maybe I'll write it 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, say, zero, phi zero, which is equal to zero. So, vector field at the point to pay their expansion. And then you get v1 plus v2 and so on. And keep going. Let's maybe put the v3 as well. No, probably enough. Now, since v2 is zero, this implies some structure on this, on this 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 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 equals 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 the, of Gage invariance. And what it implies is that the v i's, i bigger equal one, they satisfy what is called a L infinity algebra. So, infinity algebra simply means that if you take the commutator, say, this is a linear map. This is already, as we see, two to one or one to two, depending on how you look at it. So, if you take the commutator, then this gives, this is not equal to zero. This is not equal to the L algebra, but it's zero up to a v one. So, this is a great commutator with v three. 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, it's L up to a homotopy, which means that this is a linear, this is differential, and it's L up to this. We mean by L infinity. So, this is just a consequence again of just this being a b v action with a b v, with a gage invariance. There's a, it's a, yes, it's, it's, it's, it's, up to higher order terms, up to exact pieces. Absolutely. Absolutely. So, furthermore, this induces, so these, let's say the vi, induce, it's called a minimal L infinity structure. And I'll denote it by si on the homology of v one, 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 gage invariance, you get a, in this way, a homological vector field. You expand it around the critical point, and that induces a, an L infinity structure on the homology, which are the, are the S matrices. So, now we can, perhaps, put that here. We can summarize this structure. So, I'll do one line. So, I take S, I have F, that's what was my starting point, the space of fields, BV action, a BV operator, that gave rise to a homological vector field. That, in turn, gave rise upon expansion to the vi'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 Vergey 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, in, in, if you did any, any field theory, these vi's would be the vertices of the, of your Lagrangian, which would give rise to the S matrix elements. So, in, in, so the Zweibach's construction of string field theory, you try to reconstruct vertices, which give rise to the right S matrix. But these are defined away from the homology, so these, that's why you can compute off-shell amplitudes. But what one really would like to have is this. You would have business-still 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 formulations of string theory. Now, if Samson gave, based on earlier work by Witten, and we've seen this in Eric's talk, Samson gave a, a, showed that there is a formal expression for this, 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, work. So, and I think this is today perhaps the only background-independent version of string field theory we have. I mean, it has its limitations, but it's the only one really we have for the open string. It is only one limitation, which was I could not do the Gauss', Gauss' and Metronautical case, and it's still open. 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 wordline. Make my life a little bit easier. And I want to show, in the wordline, 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 I'll... 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. Okay, so then, let me introduce the spinning particle. I think this can go. So, we do the n equal 4. And let me simply write down the action for it. It's Pdx, so maybe E over 2. 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? It'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, from consistency. 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 imbine taking care of the word line repormeterization invariance. And then this, since I'm doing supersymmetry, I need an analog for 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 imbine, the superpartner, and these are the supercharges. But you take green schwarz and such a green schwarz. Yes, it's no 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 4, here I take n equal 4 supersymmetry, which means I is equal to 1 or 2. So, I take complex fermions here, so these are all word line fermions. So, I take complex combinations of I goes from 1 to 2, but this really n equal 4 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... Q I, and also the spacetime index. The t does also, yeah. So, these are really tmui, and so is this. So, they have a spacetime index. So, they also contract with a spacetime metric. So, this has n equal 4 word line supersymmetry, no spacetime supersymmetry. Q So, there is the metric again in the theta vteta. A as well, yeah, absolutely. Okay. Q One question. If it's an open string, you can add gem pattern vectors in there. A Yes, you could. You could add extra factors. I mean, I took this to be spacetime indices, but I could add extra indices, in principle. Q This is going to be n equals 4 super-n-miles without n in case of... A 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 2 instead. n equal 2 word line supersymmetry. And then, if you do a BRSD quantization, you can suppose that the BRSD squares to zero, that implies the Young-Mills equation of motion. You can do a non-habilian. So, this was done by Siegel and Dye, I think in, I think, 2008. So, no conditions on G-Menu. Q These are the linear signs, actually. I mean, do you get the full... A Fully non-linear Young-Mills. Q How do you... A I will... So, here is the summary, but I will show it later. Q Actually, Berkovic, before said, did visual arts improve the de-timidary equivalent to visual arts for particle? A And then derived from it the space-time equation, 10 dimensions. Q So, that is the kappa symmetry. A Right. Music kappa symmetry. Q No, there is no kappa symmetry. This is not super symmetry. A But I think Berkovic mapped the green-schwarz for super particle into his stuff. A Oh, I see. Yeah, yeah. Okay, this... Remember the James Ruhmund 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, 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 will see in a moment. So, basically the story is that N equal 2 is the particle that corresponds to a photon or a gluon, 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 Nabilian structure comes, how do you put jump pattern factors? Yes, you can put jump pattern factors, absolutely. Eric, do you think you have to put jump pattern factors? You put, you know, yeah. Yeah, yeah, yeah, but how do you do them on a point? I don't know. 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. 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 a question 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. 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 is 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. So, here 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, a comment rather. Samson doesn't ask questions, he makes comments. Rightly so. I don't do the thing more easily. 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 at 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. So, 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, the V equals zero gives the equation. Yeah, gives you the equation. The on-shell data, it's not the same as the S. 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, 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. Now, 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 bar times B. So, this is the super ghost gamma. It's the wordline ghost now. It's the C is the reparameterization 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 quantized 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 while algebra do some power n times Clifford tensor n. So, these are the functions in spacetime. These are generated by the T does, sorry the Clifford is generated by the feed-off. These are and the while are generated by the ghosts. So, in general we have a state H. 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 and the wave functions. So, but they do not depend on theta bar, the beta bar and the gamma bar. These are not the states, they annihilate the vacuum on the highest rate. 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 cases going to square to zero. You have to restrict the Hilbert space otherwise it's only squares to zero in zero background field so the only solution is zero. Please remind when is the 10 dimension coming to business, n equal 4 requires 10? Dimension is free. I can do for any dimension, up to 1, 2, 3, 1 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 space time super symmetry. No space time super symmetry. Size of the multiplat that 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 super symmetries. 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 or symmetry 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 a theta 1 mu theta 2 nu. Let's say mu 1 mu 1 theta 1 mu 2 theta 2 mu 1 theta 2 mu 2 and so on. So, it has to be anti-symmetric in mu's, has to be anti-symmetric in the nu's. If I think about this as the zero slope limit of strings, point particle limit, then am I not in the Rommel sector? So, it is a good question. So, let me, you can relate it to the zero slope limit. That is something you can definitely do. So, there are 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 1 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. Because you write like here. You could also do Ramon and then you get fermions in space time. If you put Ramon together with the Nuwe-Schwarz, you get space time supersymmetry if you want, but there is 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 space time 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's, totally anti-symmetric in the Nuwe's. 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 you want 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 symmetriser 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. 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 pondons in string to read up the level matching and level truncation. So, this is the sheet parity. It gets rid of the B 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 and you cannot for... Actually, no, it's not true. This is 3. Sorry. I take this back. So, we have a little bit more freedom. Do you want the symmetrize to get a more gravitation? Do you mean to not insist on this one? I think not. So, you want to keep the B 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. 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 have a little bit more freedom because we can get rid of the graviton. It's about the only thing which we can do. So, how do you write on the symplectic structure out of all this? Usually, you take a scalar product in Rd times usual... Are you using it or you will derive the equation 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... If you do maximal gauging, then the generic state that's left is H mu nu theta 1 mu theta 2 mu plus there is some psi mu theta 1 mu beta 2 minus theta 2 mu beta 1 plus antifields plus auxiliaries. So, this is what I was saying at the very beginning. Whenever you beeristic quantize and you look at the homology of the point particle and the same thing in the string, you automatically end up with a whole bv-multiplet. 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 beeristic. So, 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 beeristic 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 that the Hamiltonian commutes with the supercharges. Now, you might say this is automatic because Hamiltonian is q with q bar and then Jacobian identity gives you that q with h is automatically zero. But it's not true because you're working on a 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. So, let me say couple into backgrounds. So, what we do is we start with the supercharge, qi. Remember the supercharge up there, qi, which I write as t i mu. So, you take usual derivative plus spin connection and then t the bar B, slightly anti-symmetric, and say let's leave it there. 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. So, you square the B of the supercharge. 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 the B of the d-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 has a c-ghost, 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, you know, 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 these, both of these terms vanish separately. Now, let me first do this one. That gives you gamma, gamma bar with Ritchie, t the mu, t the bar mu. And that's it. So, now you see, that gives you a Ritchie flat. So, this is zero only if the background is Ritchie flat. So, you get a Ritchie flat. Now, there is a little bit up, so that's part of it. Then you have to check if this is equal to zero. And if I am not zero, you have to add some non-minimal coupling in order to get this to zero. So, even on Ritchie 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, R mu nu lambda rho t the mu, t the bar rho, nu, t the nu, t the bar rho. So, this is a non-minimal coupling, which is, you know, imposed on you by imposing that this is equal to zero. But, turns out there is a little bit of freedom. One thing you can do, you can add a non-minimal coupling to gravity. It's a free parameter, so I put it here. And then if you restrict that on the Hilbert space, you'll find that the condition you get is R mu nu is equal to lambda g mu nu, and g mu is free parameter. So, you get the sitter until the sitter, Einstein space. So, this is the most general solution you can 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 B 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-Mont field, 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 only surprise, we'll only get a solution if, see if I can put it perhaps here. If R 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, nubla 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. You know, 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 to them, here I should say it's T to the 1a, T to the 1 bar ab minus T to the 1a, T to the 2a, T to the 2 bar ab. So, you know, you violate the, that's where the symmetriser comes in. You violate the 1, 2 symmetry, which from string theory is also understandable because your left-movers and right-movers couple differently to the B mu nu field. So you could sort of, from string theory, you can sort of infer that this should be the right coupling. What about the dilaton? So you might say, well, does the dilaton, the dilaton appears in the spectrum, so somehow it should be able to couple, but how does it couple? Now, first thing you might think, string, in string theory, the dilaton couples to the vergy curvature. There is no vergy curvature here, so, cannot couple. Because, 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 divergence, there's no anomaly in the ghost current here. So you might think, you're not quite clear how to couple. So, for the dilaton, it 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 do electromagnetism with, in gravity. So this is a non-hermition. You see, it doesn't come with an eye here. It's not a U1 gauge field. It's a sort of 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 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 the Neville-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 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, you 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 beam and nu field if you want. And then you say, well, this squares to zero, so then q squared is 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 got a state. Unfortunately, I deleted it, but there was such a state. Remember, there was the diffeomorphism ghost, which was denoted by xi, which I wrote as xi mu, tmu1, beta2 minus tmu2, beta1. 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 a q closed, but it will not be exact. And that's it. In fact, this is the graviton, the beam and 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 BRSD quantization, which I claim again is really background independent. The only condition that q squared to zero will give you that 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 it has, as far as the master sector is concerned, has the same information. That's probably where I should stop. Thank you very much. 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 and etc. in the 70s. They did, well, the one thing, 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 when you come from string theory, it's somehow the obvious thing to do because in string theory we also restrict the Hilbert space. 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 S04. You only have a U2 times U1 times U1, really. When you could think of perhaps the Vergi parity which becomes here part of S04, I think in string theory usually we don't think of it that way. But it's really this sketching which for some reason, I mean, I, 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 some papers, a series of papers from the 80s by Paul Howell, Paul Paxenden 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 expect they would be, you would get spin 4. And they went through this, they restricted the Hilbert, essentially they put at a constraint 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 arbitrary 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? I think in 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 Q0 will always square to 0. But the moment you try to couple it to a background field, you will find that the Q only squares to 0 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 graviton. 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? Yeah, free particle will not be a problem. One like symmetric space, I mean nice spaces without... Yeah, 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 sometimes. Yeah, maybe you're in a set of space, you might expect there to be a solution. 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 nu equals zero, 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......normal code? 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 net equals zero, 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... This, when I couple the covariant derivatives, right? So 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. So this is where nonlinearity, how did nonlinearity start? 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, you know, by this... I mean this has, of course, a linear contribution, but in, you know, it's made out of the inverse fear band. So it has no exp... I mean, it's fully nonlinear. But it's true that the minimal coupling is almost sufficient up to these nonlinear terms, non-minimal couplings, which I guess, you know, have anymore here, but they had to add this non-minimal coupling to the Hamiltonian, which was a had to add H for some of the box, plus R mu nu lambda rho, theta, theta bar, theta, theta bar, things like this 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 metric and use some Riemann normal coordinates, such terms will probably show up. But for instance, do you prove that in an amplitude, if you make a gauge transformation from the external state, this is the same as part of the BV? Because in string theory, that's the way you prove your full-on. How sorry is this last part? That you are gauging invariant. I mean, different morphisms invariant for the graviton state. Yeah, that's fine. Is it part of the BV? 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, right? 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 gauging 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're not expanding around the solution of the equation of motion. There is no such thing as a graviton. No, there's a deframorphism. No, I have a comment. Can I make comment? You can make comment. I have not the first. The ground dependent. Why? Because notion of metric is a choice of the ground. Do you think so? No, in string theory, of course, we'll be back there. But in point particle notion of metric is... No, in here, no, no. No, no, in string theory, of course, there may not be a metric in the first place. In engineering background, I agree. In the initial thing, there is 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. The string model is a larger... 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. Closed. Closed string. No, open. For open, maybe, possible. For open. So you wanted 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... I'd have said the 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. But that is the work. Maybe that's what Eric is trying to do, starting from this trans-simon thing to starting three. We discussed this at the time. And we give up. We give up. So many. Okay. Never late. Good. Well, excellent. I think it's an opportunity to thank Ivo again and also the speakers of the day.
 All right, so let me first thank Anton and Maxime for organizing this very enjoyable  conference and of course, Samson for giving a rationale for it. Yeah, 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 Leo. So,  but I should say that discussing with Samson and collaborating with Samson certainly affected my  way of thinking about string theory and string filter in particular and the work I'm going to  present today is in some sense a reflection of it. So, so therefore thought 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,  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,  with the manier 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,  so I'm going to lift the rock today. So, that's sort of the equivalence. So, here the equivalence  of the rock here is the, is the word line, is the point particle. So, I'm going to not solve  this problem, I'm going to, but I am actually going to solve this problem. And when 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-Montfield and the Dillatons which is identical with that  of the Neuvier-Schwarz sector of the, 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 don't,  we do know this way. So, the upshot then, upshot then is that the quantum consistency  of the say n equal 4, by n equal 4 I mean 4 word line supersymmetries. So, I should maybe put a  curlian spinning particle uniquely, essentially uniquely fixes the action for the metric,  the Calvary-Montfield and the Dillaton to be that of the Neuvier-Schwarz sector,  the Muscle sector Neuvier-Schwarz. So, uniquely I put it in bracket because there is, 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 PV quantization. The reason being that string  to a string field to D, and some respect also the point particle naturally gives rise to a BV  action. So, I have to say a couple of words about PV action. So, the starting point BV action is a  space of fields, a graded space of fields where you also have ghosts. So, let's say there's a  field, 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, just about. 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 that this is equal to 0. So, this equation. So,  that means Gage invariance. Today, h bar is equal to 0. So, we'll only talk about the classical  theory in the sense that no loops. So, let's take h bar equal to 0. Quantum  consistency in the same sense as the way that you quantize the well sheet in the string. So,  in that sense. So, you have normal ordering of the operators. The h bar is 0 but you still have to  worry about calling. No loops. Yeah, you still have to. Yeah, that's right. That's yeah. So, I say  exactly alpha prime would be the analog statement in the string. So, this is a bracket. This is  an odd bracket which satisfies Jacobi or a greater Jacobi identity. And if you talk about classical  theory, the consistent of Gage invariance is that the BV bracket of the s itself is equal to 0  is an odd bracket. So, that's generic stuff. Now, if you having this, we this defines a vector field.  So, 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. Homolog vector field in the sense that v square is to 0.  And then if we expand, 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,  generic, 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.  The derivatives, these ones, now they are all, these are graded fields. So, yeah, so they're not, so we should probably put a bar or something like that.  So, maybe I'll write it 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, say, zero, phi zero, which is equal to zero.  So, vector field at the point to pay their expansion. And then you get v1 plus v2 and so on.  And keep going. Let's maybe put the v3 as well. No, probably enough.  Now, since v2 is zero, this implies some structure on this, on this 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 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 equals 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 the, of Gage invariance.  And what it implies is that the v i's, i bigger equal one, they satisfy what is called a L infinity algebra.  So, infinity algebra simply means that if you take the commutator, say, this is a linear map.  This is already, as we see, two to one or one to two, depending on how you look at it.  So, if you take the commutator, then this gives, this is not equal to zero.  This is not equal to the L algebra, but it's zero up to a v one. So, this is a great commutator with v three.  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, it's L up to a homotopy, which means that this is a linear, this is differential, and it's L up to this.  We mean by L infinity. So, this is just a consequence again of just this being a b v action with a b v, with a gage invariance.  There's a, it's a, yes, it's, it's, it's, it's, up to higher order terms, up to exact pieces. Absolutely. Absolutely.  So, furthermore, this induces, so these, let's say the vi, induce, it's called a minimal L infinity structure.  And I'll denote it by si on the homology of v one, 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 gage invariance, you get a, in this way, a homological vector field.  You expand it around the critical point, and that induces a, an L infinity structure on the homology, which are the, are the S matrices.  So, now we can, perhaps, put that here. We can summarize this structure.  So, I'll do one line. So, I take S, I have F, that's what was my starting point, the space of fields, BV action, a BV operator, that gave rise to a homological vector field.  That, in turn, gave rise upon expansion to the vi'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 Vergey 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, in, in, if you did any, any field theory, these vi's would be the vertices of the, of your Lagrangian, which would give rise to the S matrix elements.  So, in, in, so the Zweibach's construction of string field theory, you try to reconstruct vertices, which give rise to the right S matrix.  But these are defined away from the homology, so these, that's why you can compute off-shell amplitudes.  But what one really would like to have is this. You would have business-still 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 formulations of string theory.  Now, if Samson gave, based on earlier work by Witten, and we've seen this in Eric's talk, Samson gave a, a, showed that there is a formal expression for this, 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, work. So, and I think this is today perhaps the only background-independent version of string field theory we have.  I mean, it has its limitations, but it's the only one really we have for the open string.  It is only one limitation, which was I could not do the Gauss', Gauss' and Metronautical case, and it's still open.  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 wordline.  Make my life a little bit easier. And I want to show, in the wordline, 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 I'll...  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.  Okay, so then, let me introduce the spinning particle.  I think this can go. So, we do the n equal 4.  And let me simply write down the action for it. It's Pdx, so maybe E over 2.  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?  It'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, from consistency.  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 imbine taking care of the word line repormeterization invariance.  And then this, since I'm doing supersymmetry, I need an analog for 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 imbine, the superpartner, and these are the supercharges.  But you take green schwarz and such a green schwarz.  Yes, it's no 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 4, here I take n equal 4 supersymmetry, which means I is equal to 1 or 2.  So, I take complex fermions here, so these are all word line fermions.  So, I take complex combinations of I goes from 1 to 2, but this really n equal 4 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...  Q I, and also the spacetime index.  The t does also, yeah. So, these are really tmui, and so is this.  So, they have a spacetime index. So, they also contract with a spacetime metric.  So, this has n equal 4 word line supersymmetry, no spacetime supersymmetry.  Q So, there is the metric again in the theta vteta.  A as well, yeah, absolutely.  Okay.  Q One question. If it's an open string, you can add gem pattern vectors in there.  A Yes, you could. You could add extra factors.  I mean, I took this to be spacetime indices, but I could add extra indices, in principle.  Q This is going to be n equals 4 super-n-miles without n in case of...  A 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 2 instead.  n equal 2 word line supersymmetry.  And then, if you do a BRSD quantization,  you can suppose that the BRSD squares to zero, that implies the Young-Mills equation of motion.  You can do a non-habilian.  So, this was done by Siegel and Dye, I think in, I think, 2008.  So, no conditions on G-Menu.  Q These are the linear signs, actually.  I mean, do you get the full...  A Fully non-linear Young-Mills.  Q How do you...  A I will... So, here is the summary, but I will show it later.  Q Actually, Berkovic, before said, did visual arts improve the de-timidary equivalent to visual arts for particle?  A And then derived from it the space-time equation,  10 dimensions.  Q So, that is the kappa symmetry.  A Right.  Music kappa symmetry.  Q No, there is no kappa symmetry.  This is not super symmetry.  A But I think Berkovic mapped the green-schwarz for super particle into his stuff.  A Oh, I see.  Yeah, yeah.  Okay, this...  Remember the James Ruhmund 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,  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 will see in a moment.  So, basically the story is that N equal 2 is the particle that corresponds to a photon or a gluon,  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 Nabilian structure comes, how do you put  jump pattern factors?  Yes, you can put jump pattern factors, absolutely.  Eric, do you think you have to put jump pattern factors?  You put, you know, yeah.  Yeah, yeah, yeah, but how do you do them on a point?  I don't know.  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.  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 a question 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.  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 is  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.  So, here 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, a comment rather.  Samson doesn't ask questions, he makes comments.  Rightly so.  I don't do the thing more easily.  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 at 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.  So, 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, the V equals zero gives the equation.  Yeah, gives you the equation.  The on-shell data, it's not the same as the S. 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, 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.  Now, 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 bar times B.  So, this is the super ghost gamma.  It's the wordline ghost now.  It's the C is the reparameterization 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 quantized 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 while algebra do some power n times Clifford  tensor n.  So, these are the functions in spacetime.  These are generated by the T does, sorry the Clifford is generated by the feed-off.  These are and the while are generated by the ghosts.  So, in general we have a state H.  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 and the wave functions.  So, but they do not depend on theta bar, the beta bar and the gamma bar.  These are not the states, they annihilate the vacuum on the highest rate.  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 cases going to square to zero.  You have to restrict the Hilbert space otherwise it's only squares to zero in zero background  field so the only solution is zero.  Please remind when is the 10 dimension coming to business, n equal 4 requires 10?  Dimension is free.  I can do for any dimension, up to 1, 2, 3, 1 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 space time super symmetry.  No space time super symmetry.  Size of the multiplat that 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 super symmetries.  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  or symmetry 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 a theta 1 mu theta 2 nu.  Let's say mu 1 mu 1 theta 1 mu 2 theta 2 mu 1 theta 2 mu 2 and so on.  So, it has to be anti-symmetric in mu's, has to be anti-symmetric in the nu's.  If I think about this as the zero slope limit of strings, point particle limit, then am I not in the  Rommel sector?  So, it is a good question.  So, let me, you can relate it to the zero slope limit.  That is something you can definitely do.  So, there are 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  1 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.  Because you write like here.  You could also do Ramon and then you get fermions in space time.  If you put Ramon together with the Nuwe-Schwarz, you get space time supersymmetry if you want,  but there is 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 space time 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's, totally anti-symmetric in the Nuwe's.  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 you want 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 symmetriser 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.  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 pondons in string to read up  the level matching and level truncation.  So, this is the sheet parity.  It gets rid of the B 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 and you cannot for...  Actually, no, it's not true. This is 3.  Sorry. I take this back.  So, we have a little bit more freedom.  Do you want the symmetrize to get a more gravitation?  Do you mean to not insist on this one?  I think not.  So, you want to keep the B 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.  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 have a little bit more freedom because we can get rid of the graviton.  It's about the only thing which we can do.  So, how do you write on the symplectic structure out of all this?  Usually, you take a scalar product in Rd times usual...  Are you using it or you will derive the equation 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...  If you do maximal gauging, then the generic state that's left is  H mu nu theta 1 mu theta 2 mu  plus there is some psi mu theta 1 mu beta 2  minus theta 2 mu beta 1  plus antifields plus auxiliaries.  So, this is what I was saying at the very beginning.  Whenever you beeristic quantize  and you look at the homology of the point particle  and the same thing in the string,  you automatically end up with a whole bv-multiplet.  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 beeristic.  So, 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 beeristic 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 that the Hamiltonian commutes with the supercharges.  Now, you might say this is automatic  because Hamiltonian is q with q bar  and then Jacobian identity gives you that q with h is automatically zero.  But it's not true because you're working on a 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.  So, let me say couple into backgrounds.  So, what we do is we start with the supercharge, qi.  Remember the supercharge up there, qi,  which I write as t i mu.  So, you take usual derivative plus spin connection  and then t the bar B,  slightly anti-symmetric,  and say let's leave it there.  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.  So, you square the B of the supercharge.  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 the B of the d-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 has a c-ghost, 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, you know,  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 these, both of these terms vanish separately.  Now, let me first do this one.  That gives you gamma, gamma bar with Ritchie,  t the mu, t the bar mu.  And that's it.  So, now you see, that gives you a Ritchie flat.  So, this is zero only if the background is Ritchie flat.  So, you get a Ritchie flat.  Now, there is a little bit up, so that's part of it.  Then you have to check if this is equal to zero.  And if I am not zero, you have to add some non-minimal  coupling in order to get this to zero.  So, even on Ritchie 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, R mu nu lambda rho t the mu, t the bar rho,  nu, t the nu, t the bar rho.  So, this is a non-minimal coupling, which is, you know,  imposed on you by imposing that this is equal to zero.  But, turns out there is a little bit of freedom.  One thing you can do, you can add a non-minimal coupling  to gravity.  It's a free parameter, so I put it here.  And then if you restrict that on the Hilbert space,  you'll find that the condition you get is R mu nu is equal  to lambda g mu nu, and g mu is free parameter.  So, you get the sitter until the sitter, Einstein space.  So, this is the most general solution you can 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 B 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-Mont  field, 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 only surprise,  we'll only get a solution if, see if I can put it perhaps here.  If R 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,  nubla 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.  You know, 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 to them, here I should say it's T to the 1a,  T to the 1 bar ab minus T to the 1a, T to the 2a, T to the 2 bar ab.  So, you know, you violate the, that's where the symmetriser comes in.  You violate the 1, 2 symmetry, which from string theory is also understandable  because your left-movers and right-movers couple differently to the B mu nu field.  So you could sort of, from string theory,  you can sort of infer that this should be the right coupling.  What about the dilaton?  So you might say, well, does the dilaton,  the dilaton appears in the spectrum, so somehow it should be able to couple,  but how does it couple?  Now, first thing you might think, string, in string theory,  the dilaton couples to the vergy curvature.  There is no vergy curvature here, so, cannot couple.  Because, 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 divergence, there's no anomaly in the ghost current here.  So you might think, you're not quite clear how to couple.  So, for the dilaton, it 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 do electromagnetism with, in gravity.  So this is a non-hermition.  You see, it doesn't come with an eye here.  It's not a U1 gauge field.  It's a sort of 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 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 the Neville-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 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, you 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 beam and nu field if you want.  And then you say, well, this squares to zero,  so then q squared is 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 got a state.  Unfortunately, I deleted it, but there was such a state.  Remember, there was the diffeomorphism ghost, which was denoted by xi,  which I wrote as xi mu, tmu1, beta2 minus tmu2, beta1.  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 a q closed, but it will not be exact.  And that's it.  In fact, this is the graviton, the beam and 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 BRSD quantization,  which I claim again is really background independent.  The only condition that q squared to zero will give you that 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 it has, as far as the master sector is concerned, has the same information.  That's probably where I should stop.  Thank you very much.  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 and etc. in the 70s.  They did, well, the one thing, 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 when you come from string theory, it's somehow the obvious thing to do because in string theory we also restrict the Hilbert space.  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 S04.  You only have a U2 times U1 times U1, really.  When you could think of perhaps the Vergi parity which becomes here part of S04, I think in string theory usually we don't think of it that way.  But it's really this sketching which for some reason, I mean, I, 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 some papers, a series of papers from the 80s by Paul Howell, Paul Paxenden 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 expect they would be, you would get spin 4.  And they went through this, they restricted the Hilbert, essentially they put at a constraint 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 arbitrary 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?  I think in 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 Q0 will always square to 0.  But the moment you try to couple it to a background field, you will find that the Q only squares to 0 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 graviton.  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?  Yeah, free particle will not be a problem.  One like symmetric space, I mean nice spaces without...  Yeah, 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 sometimes.  Yeah, maybe you're in a set of space, you might expect there to be a solution.  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 nu equals zero, 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... ...normal code?  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 net equals zero, 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...  This, when I couple the covariant derivatives, right?  So 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.  So this is where nonlinearity, how did nonlinearity start?  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, you know, by this...  I mean this has, of course, a linear contribution, but in, you know, it's made out of the inverse fear band.  So it has no exp... I mean, it's fully nonlinear.  But it's true that the minimal coupling is almost sufficient up to these nonlinear terms,  non-minimal couplings, which I guess, you know, have anymore here,  but they had to add this non-minimal coupling to the Hamiltonian, which was a had to add H for some of the box,  plus R mu nu lambda rho, theta, theta bar, theta, theta bar, things like this 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 metric  and use some Riemann normal coordinates, such terms will probably show up.  But for instance, do you prove that in an amplitude, if you make a gauge transformation from the external state,  this is the same as part of the BV?  Because in string theory, that's the way you prove your full-on.  How sorry is this last part?  That you are gauging invariant.  I mean, different morphisms invariant for the graviton state.  Yeah, that's fine.  Is it part of the BV?  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, right?  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 gauging 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're not expanding around the solution of the equation of motion.  There is no such thing as a graviton.  No, there's a deframorphism.  No, I have a comment.  Can I make comment?  You can make comment.  I have not the first.  The ground dependent.  Why?  Because notion of metric is a choice of the ground.  Do you think so?  No, in string theory, of course, we'll be back there.  But in point particle notion of metric is...  No, in here, no, no.  No, no, in string theory, of course, there may not be a metric in the first place.  In engineering background, I agree.  In the initial thing, there is 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.  The string model is a larger...  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.  Closed.  Closed string.  No, open.  For open, maybe, possible.  For open.  So you wanted 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...  I'd have said the 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.  But that is the work.  Maybe that's what Eric is trying to do, starting from this trans-simon thing to starting three.  We discussed this at the time.  And we give up.  We give up.  So many.  Okay.  Never late.  Good.  Well, excellent.  I think it's an opportunity to thank Ivo again and also the speakers of the day.
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