All right, so yeah, let me first thank Anton and Maxine

for organizing this very enjoyable conference. And of course, Samson for giving a rationale for it. Yeah, I mean, background independence in string theory has been an issue that Samson has been pushing for many, many years. And in fact, in yesterday's talk by Simeon

and today in Eric's 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.

But I should say that discussing with Samson and collaborating with Samson certainly affected my way of thinking about string theory and string filtering in particular. And the work I'm going to present today is in some sense a reflection of it. So therefore, I thought it would be an appropriate thing

to talk about here at the conference. So before going into the details, let me describe a little bit the result. A perturbative string theory, where it's usually formulated, is formulated in terms of a conformal sigma model.

And just by the necessity of conformal invariance, puts 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 Meniere, that only obelisks, of course, can lift. But I consider myself more the type of a Roman soldier who sort of gets misled into the idea

that lifting a rock is an achievement. So I'm going to lift a rock today. So that's sort of the equivalence. So the equivalence of the rock here is the word line. It's the point particle.

So I'm going to not solve this problem, but I am actually going to solve this problem. And one might 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 Kalbremont fields and the dilatons, which is identical with that of the Neuberschwarz sector of the string. So at that level, the string has nothing to add. So something 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 do know this way. So the upshot then is that the quantum consistency

of the, say, n equal four, by n equal four, I mean four word line supersymmetries. So I should maybe put a Kirlian.

Spinning particle uniquely, essentially uniquely, fixes the action for the metric, the Kalbremont field, and the dilaton to be that of the Neuberschwarz sector,

massless sector of Neuberschwarz. So uniquely, I put it in bracket because there's one choice that you don't have in the string, which is you can add the cosmological constant at the price of projecting out the Kalbremont field. So if you give up the Kalbremont field, then you can have a cosmological constant.

So now I would 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 B-V quantization. The reason being that string theory or string field theory and in some respect also the point particle naturally gives rise to a B-V action.

So I have to say a couple of words about B-V action. So the starting point B-V 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 pi. Then I have an action on this space of fields.

So that's some function. And then there's the B-V operator. So this is not the spectral triple, but this is a different thing. So graded space of fields, an action which is a function of the space of fields, and the B-V operator which is a second order differential.

And the gauge invariance of this theory,

you probably don't see it, just about. So the gauge invariance implies the B-V equation which I can write down here, which is e to the minus one over h-bar s B-V operator e to the s one over h-bar which we write as s comma s plus h-bar delta s that this is equal to zero.

So this equation. So that means gauge invariance. Today, h-bar is equal to zero. So we'll only talk about the classical theory in the sense that no loops. So let's take h-bar equal to zero.

Ah, quantum consistency in the same sense as the way that you quantize the word sheet in the string. So in that sense. So you have normal ordering of the operators. The h-bar is zero, but you still have to worry about the quantum. No loops. Yeah, you still have to, yeah, that's right. That's the target space. That's, yeah.

Yeah, I would give that. So I say exactly an alpha prime would be the analog statement of the string. So, and this is a bracket, this is an odd bracket which satisfies Jacobi. Or the greater Jacobi identity. And if you talk about classical theory, the consistent of gauge invariance is that the BV bracket of the s with itself

is equal to zero is an odd bracket. So that's generic stuff. Now, if you, having this, this defines a vector field. You can say v of f, f is some function, which I simply take as this BV bracket

of s with some function f. So that is a homological vector field. Homological vector field, in the sense that v squared is zero.

And then if we expand, ah, so maybe I should say one more thing. If we have some additional, so that, in principle, that is enough to start with a B, get you started with a B reaction. But here, in the string and also in the point particle, we have an initial structure, which we have a non-degenerate,

degenerate symplectic form, it's odd, such that, let's denote it by omega, such that delta written as omega ij

in components, d i, d j. So then we can define, we can take the interior derivative of omega with this v,

and this is the 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. Derivatives, these ones, no, they're odd, these are graded fields. So yeah, so they're not, so I should probably put a bar or something like that.

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 is zero, phi zero, which is equal to zero, that's the vector field at the point, to pay the expansion, and then you get v one plus v two, and so on. Keep going, let's maybe put a v three as well.

No, probably enough. Now, since v squares to zero, this implies some structure on this, on these coefficients in this expansion. So first, I should say what these are. I mean, this is just, this is, let's say, this is zero at the point,

so this is a map, so this is a linear map from the tangent space at the critical point to itself, and this is a map from t phi zero f into t phi zero f squared,

and so on, and this keeps going as you go to the higher order terms. So you can take the dual version, where these are maps from v tensor v into v, where v is the vector space, and so on. So you can go either way. The point is, though,

there is some structure on these maps, and the structure on these maps follows simply from v squared equals zero, which is just a consequence of the way v is defined as coming from the bb bracket. So this is all, in some sense, consequences of gauge invariance, and what it implies is that the v i's

i bigger equal one, they satisfy what is called a L-infinity algebra. And 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. If it was zero, it would be a Lie algebra,

but it's zero up to a v one. So this is a graded commutator with v three. So you see, if the higher order maps are zero, then that would mean it's just a Lie algebra, but since it's Lie up to homotopy, which means that this is a linear,

this is differential, and it's Lie up to this. So that's given by L-infinity. So this is just a consequence, again, of just this being a bb action with a bb, with a gauge invariance. So there's, it's a, yes, it's Lie,

it's Jacobi up to higher order terms, up to exact pieces, absolutely. Absolutely. So furthermore, this induces, so these, let's say the v i, induce, it's called a minimal L-infinity structure,

and I'll denote it by s i, on the cohomology 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 descends straightforwardly.

You start with gauge invariance, you get, in this way, a homological vector field, you expand it around the critical point, and that induces an L-infinity structure on the cohomology, which are the s matrices. So now we can, perhaps, put that here, we can summarize this structure,

simply one line. So I take s, I have f, that was my starting point, space of fields, bb action, a bb operator, that gave rise to a homological vector field. That, in turn, gave rise, upon expansion, to the v i's,

which is this L-infinity algebra, and then this, they're not really maps, and then this gives rise to the s i's, on cohomology. 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 Virge conformal field theory. String field theory, as defined by Zwiebach, would be here. So in string field theory, one tries to recover the vertices. So these, I should say,

if you did any field theory, these v i's would be the vertices of your Lagrangian, which would give rise to the s matrix elements. So in Zwiebach'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 cohomology, so that's why you can compute off-shell amplitudes. But what one really would like to have is this. You would have, this is still background dependent, because all this is expanded around one point. So only once you get rid of this point, phi zero, you can truly speak about the background in the pain formulation of string theory.

Now, Samson gave, based on earlier work by Whitten, and we've seen this in Eric's talk, Samson gave 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 nonambiguous. So that was this 93, I think, word.

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 major, not decoupled case, and it's still open. Still open, yeah. So that's where I go, first of all, I will not do the open string, I will do the closed string, and I don't do the closed string, I do the word line. Make my life a little bit easier. And I want to show, in the word line,

you can get a manifestly background-independent formulation, which contains quite a lot of information about string theory. So that's basically the content of this talk. Are there any comments or questions at this point? Let's see if I forgot anything important, probably.

Well, maybe the comment is just that, no, I think I'll, so what we'd 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. This can go.

So we do the n equal four, and let me simply write down the action for it. It's p dx, so maybe e over two.

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 space-time metric enters in this bracket. I didn't display it, but it's there. Was 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. The e is taking care of, this is the Einstein taking care of the world-line reparameterization 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 Einstein, the superpartner, and these are the supercharges. Yes, it's not Green-Schwarz,

the world-line supersymmetry, so not space-time. So space-time have no supersymmetry. Q i, and then I have also,

since I have a n equal four, I here I take n equal four supersymmetry, which means i is equal to one or two, and so I take complex fermions here. So these are all world-line fermions. So I take complex combinations,

i goes from one to two, but it's really n equal four in terms of real fermions. And as a consequence, I have an SO4 r-symmetry, and these are the generator of the SO4 r-symmetry, which I may gauge or not. So I can gauge the SO4 or not. That is a little bit up to me. And this will play an important role,

actually how much exactly of the SO4 we gauge or at all. So that's about the only freedom you have in this theory. So that's the action. So now let me, the theta's also, yeah. So these are really theta mu i,

and so is this. So they have a space-time index. So they're also contracted with a space-time metric. So this has n equal four world-line supersymmetry, no space-time supersymmetry. So there is the metric again in the theta, d theta? As well, yeah, absolutely.

Okay. Question. If it's an open string, you can add jump atom vectors in there? Yes, you could. You could add extra factors. I mean, you could, you know, I took this to be space-time indices, but I could add extra indices in principle. This is gonna be n equals four super-generals without any case.

Ah. So yeah, I'm actually going to comment about this right now. So this is a little bit of history. So I could have taken n equal two instead, n equal two world-line supersymmetry. And then if you do BRSD quantization, until you've posted the BRSD squares to zero,

that implies the Young-Mills equation of motion. You can do it non-abelian. Okay, so this was done by Segal and Dye, I think in 2008.

So no conditions on g mu nu. But these two super-generals are actually very, I mean, do you get the full? Fully non-linear Young-Mills. So how do you couple Young-Mills?

Ah, I will, so here is the summary, but I will show it later. Actually, Berkowitz before said the initial art and proved that it is very equivalent to the initial art for particles. And then derived from it, the space-time equation 10 dimensions.

So that the kappa symmetry? Right, with the kappa. No, there's no kappa symmetry. This is not super-symmetric. But I think it's Berkowitz mapped the Grinschwarz for super-particle into his stuff. Oh, I see, yeah, yeah. Okay, this, yeah. Remember, James Ramon was teaching us about that 20 years ago.

So N equal four, again, you can impose this, and this gives you the equation of motion for NS sector of four M equals zero. So the massless NS sector of type two.

So this is type two super-gravity, if you wish. So why N equal four, why N equal two? We will see in a moment. So basically the story is that

N equal two is the particle that corresponds to N equal two is 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. And so that's sort of the hierarchy of things.

So this is, by the way, with Roberto Bonetzi and Adiel Meyer. So this turned out to be, after Siegel did this thing,

people tried to do the N equal four. And for some reason, failed. And the reason why they failed is sort of clear. These guys are asking you constantly, where the navigation structure comes? How do you put jump pattern factors? You can put jump pattern factors, absolutely.

Eric, do you think we have to put jump pattern? You put, yeah. Yeah, yeah, yeah, but how do we do it on a point? I don't have any points. So the T does, what happens is we will see in, let me maybe go a little bit, and you'll see it very quickly, very shortly.

You will see it very shortly how that works. But before I do that, I would like to say one thing is that why, I mean, what has this to do with what I said before? So why Q squared equals zero has anything to do with the equation of motion the way I described it before? So this, we need to go back here to the expansion here.

By the way, I still believe that all you can get is the equation of motion, not the action. You can only get the equation of motion, absolutely, but you can go off shell. I can quantize this theory off shell. So let me see. So actually, a comment I was going to make.

So let's look again at this V. Is 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 it perhaps here. So this is V, so this equal to zero, implies V two squared plus V one, V three

equal to zero, plus other terms. Sorry, V zero, sorry, I got that wrong, I think. V one squared plus V zero, V two. Now, on shell, on the critical point, V zero 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're off shell, V 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 squares to zero is saying that V zero has to be equal to zero. That means we are on the critical point. So that's the connection between the BRSD quantization and the BV quantization in space time.

So now coming to Samson's question, or comment rather. Samson doesn't ask questions, he makes comments. Rightly so. I can't do that anymore either. Yeah. Let me put that here, perhaps.

So what does that have to do with the action? And I agree, that doesn't give you the action. So what we have is we have V1 we can compute from the point particle, which is the first derivative of the vector field. We can compute it at any point. It just doesn't square to zero, the generic point.

So in principle, you can integrate from phi zero to phi. You can get the V, but of course, where this is integrating over the field. Of course, there are questions here. This is formal because you have to choose a path. You have to make sure the thing is perhaps path independent and so on. But once you have this, if you have V,

you know then we have this IV omega is equal to ds. Since the symplectic form in this case is independent of the background, this is a constant thing. So you can integrate once more and you can in principle get s. But of course, in principle. So you get the equation of motion for s.

Yeah, I mean, this equals zero. Yeah, I mean, V equals zero gives you the equation. The on-shell data, it's not the same as the s. That's the on-shell data. Point is the motion of the on-shell data. I get the first derivative of s. Would you agree? I mean, IV with omega is the first derivative of s,

which I can integrate. Of course, formal, but can be integrated. So this is just to address this comment by Samson. Okay, but now let me describe a little bit what has this to do with John Paton. 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. And this is gonna be familiar to most of you. So Q BRSD is equal to C times the Hamiltonian

plus gamma i Q bar i plus gamma bar i Q i plus gamma, gamma bar times B. So this is the structure. Gamma is the word line ghost now. 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 Dirac constraints and we BRSD quantize the system. So what is the Hilbert space? Hilbert space is functions, let's say square integral or perhaps a little bit less

if you want scattering states on space-time times while algebra to some power n times Clifford tensor n.

So these are the functions in space-time. These are generated by the Tdas, sorry, the Cliffords are generated by the Fidas. These are, and the while are generated by the ghosts. So in general, we have a state H. In principle, you have to sum, yeah, sum over n.

So H is some state psi, which is some polynomial in these theta i's and the betas and the gamma i's with coefficients in L2 of Rd.

These are the functions, so the wave functions. 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, 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 it's never gonna square to zero. Not even in the n equal two case it's going to square to zero. You have to restrict the Hilbert space, otherwise it only squares to zero in zero background field. So the only solution is zero background field. Please remind, when is the 10 dimension coming to this?

An equal four requires 10? Dimension is free. I can do for any dimension. I mean, up to one, two, three, four, one maybe not. So dimension is not fixed because I have no conformal symmetry here. You see, I don't need a central charge to be zero or something like that. So I don't have to. Space time supersymmetry. I have no space time supersymmetry.

We know. No space time supersymmetry. Size of the multiplex depends on the dimension. That is true, yeah. So eventually the spins that you end up with. No, the spins that you end up with actually depends on the number of supersymmetries.

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 respect the Hilbert space is by gauging some of the SO4 R 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. Because say you take a state which is theta one mu, theta two mu, let's say mu one,

mu one, theta one mu two, theta two mu one, theta two mu two, and so on. So it has to be anti-symmetric in mu's, has to be anti-symmetric in the nu's. I have a naive question. Yeah. If I think about this as the zero slope limit of strings,

point particle limit, then am I not in the Rommel sector? It's a good question. So let me, you can relate it to the zero slope limit. That is something you can definitely do.

Theta and mu's, let's write that down here. So there's two things you can do. You could do that, what do you suggest? Or what you can do is you can do a twisted reduction on the word sheet in this sort of space direction. And then this theta, they correspond to the psi,

one half, and psi minus a half, and bar in the Neuville-Schwarz sector. So they just count the first. It's like a level truncation. But you could also do Ramond. The question you write it like, here it's. You could also do Ramond, and then you get fermions in space-time. If you put Ramond together with the Neuville-Schwarz,

you get space-time supersymmetry if you want, but there's no need to do it because you don't have, not very 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 thetas you can have. So it tells you about the maximal spin you can have. So it's totally anti-symmetric in the mu's. Totally anti-symmetric in the mu'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 SO4 R 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. So if you fix this to be one, which we say we take the charge to be one, the U1 charge equals 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 theta ones.

You have only one of the theta twos. You have an equal number of them, so you match the level and you fix it, you make level truncation. So that's the minimal gauging you can do. If you don't insist on this, you will not get any solution. But you can do more. You can symmetrize this thing.

So there's another, say, if you can take the young symmetrizer in SO4, then you get symmetric. So that projects out the B mu nu field. Then the kalbramond field is gone because of the insistent symmetry between mu and nu. Kalbramond is out.

And you can do one more thing, which I think you cannot do in string. There's a trace, what I call the trace in the SO4 generators, which removes the trace. And that means that, so here, let's say, here the B mu nu is out. So this, and here the dilaton is out.

So these things, as I said, have pondons in string theory, that be level matching and level truncation. This would be worldsheet parity, gets rid of the B mu nu like in type one. This, I don't think, has an analogue 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 true, sorry. I take this back. So we have a little bit more freedom. 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.

So my memory is that we haven't succeeded in it. Maybe we didn't try hard enough, but I don't think we were able to do that. I think the graviton, no, actually, the thing is, if we gauge completely SO4, the graviton is left. Yeah, I don't think you can get rid of the graviton. It's impossible.

So that's the structure. You see, we have a little bit, we don't have much more freedom than in string theory, but a little bit more freedom, because we can get rid of the dilaton. It's about the only thing which we can do. So how do you write on symmetric structure out of all this stuff? Usually you take scalar product in rd times usual...

Are you using it, or you will be deriving rather the 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 take this maximal gauging, then the generic state that's left is h mu nu, theta one mu, theta two mu,

plus there is some xi mu, theta one mu, beta two, minus theta two mu, beta one, plus anti-fields, plus auxiliaries. So this is what I was saying at the very beginning. Whenever you barystique quantize,

and you look at the cohomology of the point particle, and the same thing in the string, you automatically end up with a whole b-b multiplet. It's given to you right away. So I don't see why. I mean, if I act now with q on these states, don't I get gamma one, gamma two? Because each... Yeah, okay, I will say a word about this, yeah.

So this is the graviton, and this is the diffeomorphism ghost, space-time ghost, in the b-b language, or b-r-st. So 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... So you have specified the Hilbert space, it is maximal gauging, and then now you want to see what happens to the b-r-st charge. So how do we couple the system to these background fields? And for this,

I just tell you how that... I mean, 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 Jacobi identity gives you the 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 coupling to backgrounds. So what we do is, we start with the supercharge q-i. Remember the supercharge up there? q-i, which I write as theta i mu. Take usual derivative, plus spin connection,

and then theta a theta-bar b, which is 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. Then 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, q squared would be the speedcharge.

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 is s. And then I'll find that q squared, so this is the BLS-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 goes, this doesn't, so they have to separately vanish.

Now Giuseppe, you see, now you get this structure. So the s has a gamma gamma bar, but so does the h, because if you commute the B with the c here, you get the gamma gamma Hamiltonian. So for this to vanish, so then you have to make sure that both of these terms vanish separately.

Now let me first do this one. That gives you gamma gamma bar with Ricci, theta mu, theta bar mu, and that's it. So now you see, that gives you a Ricci flat.

So this is zero, only if the background is Ricci flat, so you get Ricci flat. Now there is a little bit, so that's part of it. Then you have to check, is this is equal to zero? And if it's not zero, you have to add some non-minimal coupling in order to get this to zero. So even on Ricci flat, in order for this to vanish, you will have to add a non-minimal coupling, which is like this,

somewhere I have it, plus R mu nu lambda rho theta mu theta bar nu theta nu theta bar rho. So this is a non-minimal coupling,

which is imposed on you by imposing that this is equal to zero. But turns out there's a little bit of freedom. One thing you can do, you can add a non-minimal coupling to gravity, so three parameters that it puts 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 three parameters. You get the sitter onto the sitter Einstein spaces. 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 Kalbramond 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 you will only get the solution if, see where can I 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 motions. You also get the equation of motion, lambda mu equal to zero. So if you allow, because now the restriction becomes a little bit stronger,

because you project on a bigger space, so you get more equations. But this thing, you can more or less infer from string theory what you should do. This is sort of a 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 theta one a,

theta one bar a b, minus theta one a, theta two a, theta two bar a b. So you violate the, that's where the symmetrizer comes in. You violate the one-two symmetry, which from string theory is also understandable,

because your left movers and right movers couple differently to the B-mu new field. So from string theory you can sort of infer that this should be the right coupling. But 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,

in string theory the dilaton couples to divergent curvature. There is no divergent curvature here, so it cannot couple. And you can say, well, you know, that's sort of derived. Really the dilaton couples to the divergence of the ghost current, that was the initial coupling for the dilaton. There is no divergent, there's no anomaly in the ghost current here. So you might think, not quite clear how to couple.

So for the dilaton it took us a long time 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 Weyl's idea of how to remember to electro magnetism

with gravity. So this is a non-hermission, you see, it doesn't come with an eye here. It's not a gauge, it's not a U1 gauge field, it's a sort of R1 gauge field. But this works, and again gives you a modification. Of course, so 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. Once you've done that, you can project out again the B field and you can add the cosmological constant. So 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 Neuberschwarz 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 h-mu-nu satisfies the linearize Einstein equations from the... 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 a deformation of the background.

You make an infinitesimal deformation of the metric, for instance, or of the dilaton, or of the b-mu-nu field, if you want. And then you say, well, this squares to zero, so then q squares equal to zero, since it implies that q0, delta q is equal to zero to this order,

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 a diffeomorphism ghost, which was, I denote it by psi, which I wrote as psi mu theta mu one

beta two minus theta mu two beta one. This is q closed. It's in the cohomology of q0, but it's the wrong ghost number. It's ghost number one, zero minus one. So then I can say, well, but what about if I act with 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 b-mu-nu field, and the dilaton. So you get the vertex operator simply by acting with a variation of q with respect to background field

on the diffeomorphism ghost state. So that's the analogue 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. 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 BRS dequantization, which I claim again is really background independent. The only condition that q squares 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 massless sector is concerned, has the same information. And that's probably where I should stop. There's a comment.

In this time it's a question. Oh. What was the actual 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 et cetera in the 70s.

Well, I think the one thing where people got stuck is that you have to properly restrict the Hilbert space to do this. And that is, I think, thanks to Warren Siegel, who had this first suggestion, the N equal two case, to introduce this restriction.

And when you come from string theory, somehow the obvious thing to do, because in string theory we also restrict the Hilbert space. When we do level matching, we immediately restrict the Hilbert space. The only thing is, here we have a little bit more freedom. In the string theory, you don't have an SO4. You only have an U2 times U, you only have a U1 times U1, really. When you could think of, perhaps the world sheet parity,

which becomes here part of SO4, I think in string theory usually we don't think of it that way. But it's really the sketching, for some reason, I, including myself, didn't think about restricting the Hilbert space in the first place.

Can you go beyond N equal four? Yeah, this would be an interesting... So you could, in principle, do N equal six. We haven't done it, because this was already... 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 tree theory.

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. Yeah, so my question had some overlap with that. But there was some papers, a series of papers from the 80s, from Paul Howe and collaborators,

where they looked at general numbers of supersymmetries. So for N supersymmetries, they found they got a maximum spin of N over four, N over two, so that if you went to N equals eight, you would expect you would get spin four. And they went through this, they restricted the Hilbert... Essentially, they put added constraints, which I think corresponded to your restriction

on the Hilbert space. Maybe. And they got the free wave equation for spin four 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? Yeah, I think the free level, indeed, you can do in principle for any N. And exactly as you say, the spin is N over two. For N equals four, you get spin two, and accordingly for higher. By the way, for spin one, you get fermions. For spin three, you get the gravitino.

For N equals three... But indeed, as you say, the thing is that the Q zero will always square to zero. But the moment you try to couple it to a background field, you will find that the Q only squares to zero when the background field identically vanishes. And here we just found a little bit of a way

by suitably restricting the Hilbert space, that you can do it. But that's really where, I think that's where you will fail. The moment you try to couple it to background fields, it will just not close anymore. 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 the... Yeah, free particle will not be a problem. 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 good coupling of spin 4 to an arbitrary metric, which you know those issues. The metric is not arbitrary. It's just a constant curvature background of some type. Yeah, maybe in actually a set of space you might expect there to be a solution.

It's still the expansion. You don't get the background. Yes, I do. I never put in an expansion. This coupling is, you know, this coupling... Where is it? I've deleted it now. But the spin connection is not linearly coupled.

This is fully non-linear coupling. This is the full non-linear 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. Non-linear. So it's not like expansion or... No, no, no. So that is really fully background independent.

But you work in some normal code. No, no, no. Just completely coordinate invariant. We didn't use coordinates anywhere. He says that he just writes H mu nu and then he pops up the R mu nu R net equals zero top side. 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 AB theta A theta bar B. So that's where the coupling comes in. It's not super non-linearity. How did non-linearity...

Yes. This is a non-linear 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 is... I mean, this has, of course, a linear contribution, but, you know, it's made out of the inverse Wirbein. So it has no expand... It's fully non-linear. But it's true that the minimal coupling is almost sufficient up to these non-linear terms,

these non-minimal couplings, which I guess you don't have anymore here. But I had to add this non-minimal coupling to the Hamiltonian, which was I had to add H was somehow box plus R mu nu lambda rho T theta T theta T theta bar. Things like this I have to add to the Hamiltonian for the system to close.

So it's non-minimally coupled. Such terms pop up. Otherwise, just q will not square to zero if I don't have them, 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 in generic background metrical use from Riemann normal coordinates, such terms will probably show up, I think so. But for instance, do you prove that in an amplitude, if you make a gauge transformation from the external states, this is the same? That's part of the bb thing. Because in string theory, that's the way you prove it.

Sorry, this last part? That you are gauging invariant. I mean, if you are an orpheism invariant for the graviton state, is it part of the bb format? That is part, yeah. Basically, the statement is that if you take Einstein-Hilbert action and expand it around R of g0 plus terms,

and then you linearize. So let's write it as s0 plus s1 plus s2 in expansion. So this is the quadratic term in the expansion. And if you want this part to be gauge invariant, diffeomorphism invariant, this linearized part,

you need this one to be equal to 0. So you need to be on shell. So basically, the statement is a 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. Not as a diffeomorphism.

You can make comment. I've not the first. The ground dependent. Why? Because notion of metric is a choice of background in string theory. No, in string theory, of course, would be backward. 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 condition of string theory, 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. That's why I was saying, probably, in the string theory setting, it's

perhaps even not possible to write down a background independent action in the full string theory. Closed, for closed. For open, maybe positive, actually, for open. I really want you to consider the streamline pole, where the center, whatever. The reason why it works for open is that in the open string, whatever you put in 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 set of 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 manifolds, where the two manifolds is a boundary, and then somebody, but that doesn't work.

Maybe that's what Eric is trying to do, starting from this John Simons thing, to start in three. We discussed this at that time. And we give up, to be careful. 20. OK. Never late. Good. Well, excellent. I think it's an opportunity to thank Ivo again, and also the speakers of the day.