 Right. So this only needs to exist in the vicinity of the conformal boundary. And there are theorems that in asymptotic, at least in the Euclidean case, you can always do that. So now what we do is we want to rephrase this action in terms of these new variables. So this is like an exercise if you want. So you can decompose the bulk Ritz scalar in the following form, where this vector n nu can be expressed. So it has like the radial component is 1 over n. And the transverse component is minus ni over n. And it's normalized. So it's a unit. So you can compute the normal using this metric. And you see that it's normalized. And the orientation is such that it is outward. So it's like a long-increasing radial coordinate. And here also, so k, remember it was the term that appears here. This is the trace of this tensor. So this is the extrinsic curvature. And it takes the form gamma dot ij minus ni, where dot is defined to be the radial derivative here. Right. So now using this expression here, and if we just plug it in here, you can easily verify that we get, you can write the action. So as a radial integral over a Lagrangian density, and this Lagrangian density is the following. And notice here that it's crucial, in order to formulate the problem in this Hamiltonian language, it's crucial to have this Gibbons-Hawkin term. So what this Gibbons-Hawkin term does is it cancels precisely this total derivative term here. So in particular, if you do the calculation, you will see actually that this term is, once you pull it back like on the hypersurface, actually this term is identical to zero. And it only cancels this. So this term here cancels the Gibbons-Hawkin term. So that's why you can write as an integral over this Lagrangian density. Right. So this is the Lagrangian. So we can do LSN, OK. Before the LSN transform, let's compute the canonical momenta. So the canonical momenta, if you recall, from classical mechanics is defined as the derivative of the Lagrangian with respect to the velocities. So in this case, so the momentum of the metric is this. And then there is also an important observation that you see here in this Lagrangian. So remember, k, like from this expression here, has a first of the derivative in the radial coordinate. So in this kind of Hamiltonian time. So when we, like, but this Lagrangian here, it does not have any, it does not depend on n. Actually, sorry, it depends only on n here. So there is an n here and there. So basically, now if you compute the canonical momenta, so pi of n is, well, it's the derivative with respect to n dot, and its identical is zero. And the same for pi n i. And this is also identically zero. So this implies that n and n i are Lagrangian multipliers. So they are non-dynamical fields. So they are kind of auxiliary fields. Sorry? At the monitor. At the monitor. Ah, sorry, yes. It's just on it. OK, sure. Thank you. Ah, it removed that. OK. Oh, sorry, do you need that? Well, for now it's OK. Thanks. Right, OK. So now that we have, like, the Lagrangian density and the corresponding momenta, we can do the Legendre transform to obtain the Hamiltonian. So the Hamiltonian is given, we have to sum over all the canonical momenta. But since only one of them is non-trivial, so we just have this expression here minus l. And then if you plug explicitly this expression here for the Lagrange density and you invert this expression here, so, right. So now you can combine this, right? I mean, you can, this expression here. You can invert this expression for the extrinsic curvature to write in terms of the canonical momentum. And then we plug it back there. So if you do that, then you can express the Hamiltonian in this form plus n i is i, where this h and h i are the following, right. So this is like what you get from this calculation. But now you see that, remember that this n and n i, they are Lagrange multipliers. So the corresponding Hamilton equations imply that this f2-1 is identically, right? For n and n i, apply the constraints h, h i is equal to 0. So this constraint, the fact that h has to vanish is the Hamiltonian constraint. And this is sometimes referred to as the momentum constraint. Sorry, here this is like an overall n. So n i appears only inside here, the extrinsic curvature. But there is no, so these derivatives are, I didn't mention, sorry. So this covariant derivative here is d i is covariant with respect to the induced metric gamma ij. OK, so now, so here before I continue like this Hamiltonian description, I'll just make a side comment because it will be useful later on. And this is the Hamilton Jacobi. So the Hamilton Jacobi formalism, like probably you're familiar again like from classical mechanics. But now you can, like essentially, very briefly, I mean what it amounts to is expressed in the canonical momentum, not as a derivative of the off-cell Lagrangian with respect to the corresponding velocity, but rather as a derivative of this functional s of gamma with respect to gamma ij. So notice there is no radial derivative here. So it's not the velocity, it's just like the field itself. So this s is related, like if you remember from classical mechanics, this s is called like Hamilton's principal function. And it is up to some constant, this equivalent is equal to the on-cell action. So I don't have time to derive this like from scratch. So it's the same derivation that you have like in classical mechanics. So now it's just like you applied just to this Hamiltonian field theory. But now once you write the canonical momentum in this form, you can go back to these expressions here. And then you can use the constraints to obtain some equations. So the Hamiltonian constraint, it tells us that we get, so you can plug it in. So maybe I just write the first terms here. So you get gamma ij, and then you get the other term. And then you get one more term. So this one more equation, which is basically the conservation of the derivative of this s with respect to gamma. So and these two are the so-called Hamilton Jacobi equations for gravity. So in just a side comment again, and a second side comment that is related, so notice that now we have two expressions here for the canonical momentum. So one is this Hamilton Jacobi expression, and the other one was the standard Hamiltonian expression in terms of the derivative of the Lagrangian, which we found is equal to the exchange in curvature. So if we combine the two, so we can write something that is of this form. So it's like 2k squared, and then gamma ik, gamma gl. So I might have missed some factors here. So this flow equation looks like that. Now why is this important in what we want to do? So this connects precisely to some of the points that I mentioned earlier. So here, if you actually solve this Hamilton Jacobi equation to determine s, now why we want to do this, we'll see later. But suppose you have a solution here of this equation. Then if you plug it here, you get the first order equation for the induced field gamma. So in other words, essentially, what you have done is you have integrated once the equation of motion. So once you find the solution of the Hamilton Jacobi equation, you've integrated once the equations of motion. Now the statement about this superpotential that I made is that now if you consider, for example, black hole solutions that don't depend on the transverse coordinates, some kind of homogeneous solutions, then you can find the exact solutions of this Hamilton Jacobi equation. And one of the solutions will be precisely the supersymmetric superpotential if the theory is supersymmetric and that meets as a superpotential. So a very special solution of the Hamilton Jacobi equation is supersymmetric theories is the supersymmetric super potential. And if you plug this particular solution back into this flow equation, you will get the BPS equations of the theory. But of course, this applies more generally. So it applies to all solutions of the theory. It doesn't apply only to the supersymmetric ones. But in the case of non-supersymmetric solutions, solving this Hamilton Jacobi equation is much harder. And usually, the solution does not exist globally, so it exists like only locally. So in supersymmetric theories, then usually there is always a solution of this Hamilton Jacobi equation which corresponds to the supersymmetric superpotential. So if you plug this supersymmetric superpotential, so the solution here that you get, let's call it like SBPS. So let's take the black hole solutions that Stefan was talking about earlier, like in n equal to gate supergravity. And so there is a superpotential that describes these solutions, right? The pre-potential, OK, a superpotential. He was talking more about the pre-potential. Right, that's right, that's right, yes, OK, sure. But you can also, like in n equal to 1, let's say, you can have a superpotential. So then the superpotential, it will be a solution of this equation. And then if you use the superpotential, so let's call it SBPS. So if you use this superpotential as a special solution of the Hamilton Jacobi equation into these general first order equations, so what you will get is that these flow equations will reduce to the BPS equations. But of course, both the Hamilton Jacobi equation and these first order equations are more general than supersymmetric solutions. So they apply to all solutions. However, for non-supersymmetric solutions, in that case, it is generically very hard to solve the Hamilton Jacobi equation, although you can do it. So there's something called fake supergravity that essentially boils down to this, solving this equation for non-supersymmetric solutions. But the last point I made was, usually, for non-supersymmetric solutions, so the Hamilton's principal function will have branch cuts in field space. So it will not be globally well defined. So that's why supersymmetry ensures that the solution is globally well defined. Action obtained by just a time term or something like that. So by SBPS here, I just mean the solution of this Hamilton Jacobi equation that corresponds to. Yeah, so just look at the Gravitino, for example, or the Gravitino equation, right? And then if you look like this will look like, so it will give you some equation that is like gamma dot minus like w epsilon equal to 0, right? I mean, if you demand very schematically, right? I mean, something like that, right? And then from the Latino, let's say, I mean, you will get something like that is like phi dot minus di phi w epsilon equal to 0. Very schematically again. So this w is a solution of this Hamilton Jacobi equation, but it's a very specific solution. And this first order equation, then you interpret as BPS. Yeah, so if you plug this very special solution into these equations, these equations reduce to this. I mean, of course, you will have another one for the scalars. You'll have another of these flow equations for the scalars from the Hamilton Jacobi construction. Any other questions about this point? OK, so right. So now let me, my comment is that in principle, in general, this is actually what is called a Hamilton's principal function. So it is any solution of the Hamilton Jacobi equation. Now it actually corresponds to, is equal to the own selection. You can show this, but there are some caveats. So there are some constants. So when you integrate the Hamilton Jacobi equation, there are some, so of course, the general solution of the Hamilton Jacobi equation is a PDE. Generically, we have many fields. So you will get like functions, integration functions. However, the only, the general solution that you need, the most general solution that you need is what is called a principal function, which contains like a number of integration functions, constants, not functions. And you need as many integration constants as the number of fields. So however, the general solution, this principal function, will contain this integration constants. And then the point is that like, you have to, for a very specific values of this integration constants, this will agree with the own selection. OK, so let me see here, right. So now let's see another aspect of this formulation, which is like the symplectic structure. And then, so here the point is that now this space, like of momenta and fields, right, I mean, so like in classical mechanics, admit like a symplectic form. And in this case, the symplectic form is expressed in terms of these fields and the canonical momentum. So here, this is like a wedge product in the tangent space, right, I mean, in the field space, the tangent space of phase space. So this is the symplectic form. You can derive it in different ways. You can also use something called like the covariant phase space formulation. So you get the same result, like if you pull it back like on a radial slice. But in any case, this is like the symplectic form of the theory. And an important property is that the radial derivative of this symplectic form is 0. It's constant. So this is crucial. And that is one of the reasons why I mentioned this caveat about the manifold here. So this is like on the radial slice. And why this radial slice has to be compact. If it's not compact, then I mean you get, this is not so trivial to show. So it might not be the case without additional boundary terms here. So let me say sigma r compact. So now this, so we'll see in a moment that this is quite important. So now we can invert this, and then we can construct the corresponding Poisson bracket. So this is like the equal time Poisson bracket, where time is like this radial coordinate now. And then this Poisson bracket can be realized in the following way. So if you have like functionals on phase space a and b, so you can represent it in this form. So it's delta a gamma ij delta minus a goes to b. OK, so then you can check that this radial derivative, like so Hamilton's equations now, they can be written in this form. So we have that it's delta h by delta pi ij. And this is the Hamilton that I wrote earlier. But now we can check that it can be written in terms of this Poisson bracket with gamma ij. And then similar pi ij dot is minus delta h delta gamma ij. And this is minus h pi ij. So now what we can do now is, well, for no good reason, let me just define for now some functional on phase space. So this is a vector that has a vector on the manifold m. And then it's also a functional on this phase space. And I split it in this way. So it's some arbitrary function. It's like the radial component of the vector plus psi i h i. So and then what I can do is, so here is an exercise. So the claim is that this psi x then generates bulk diffeomorphisms with parameter psi tilde mu, which is the following. Psi r over n. And then here is psi i minus psi r ni over n. So that is, if you compute the Poisson bracket with gamma ij, it should give you that you get gamma ij. And then it's for pi ij. So you know these are like tensor fields. Actually, you can compute how they transform diffeomorphisms independently. But then you can also check as an exercise like that if you compute the Poisson bracket of this constraint here with the corresponding field, you will generate the corresponding diffeomorphism where the parameters now are these expressions here. Delta is the diffeomorphism of gamma ij. What is on the right hand side of this? Yes, so this is like psi tilde. So psi tilde is this. So the point is that, so here if you just introduce this psi, this vector psi in the constraint, the diffeomorphism is slightly modified. The parameter of the diffeomorphism is slightly modified. And the parameter is this psi tilde. So it's psi r over n and psi i here minus psi r ni over n. So if you go to the Gaussian coordinates, then they coincide. But for non-Gaussian coordinates, they don't. Yes, so here it's not direct brackets. So I will probably mention a bit later like something about direct brackets. So this is like just the Poisson bracket. It's not the direct bracket. So and then you can also compute the algebra. So namely take these two constraints of this form. Then you will find that indeed it closes. So you can also write it. So if you take like two constraints of this form, you take the Poisson bracket, you will find that you can express the right-hand side as a constraint of this form again, but with different parameters. We are now psi, this psi double prime. It has some complicated expression, but it's important to write down for a reason that I will mention, minus. So this is the radial component. And the other components are. And then there is this other term. So this is like the radial component of this new vector. And this is the transverse component. But now you see here, we have an issue. Namely, this covariant derivative appears. And the covariant derivative depends on the Christoffel symbol. The Christoffel symbol depends on gamma. So the structure constraints are field dependent. So that's why the algebra of the morphisms is not a standard like Gaetz algebra. So there is this issue here. I'm not going to discuss it more now, but we will see that we can bypass it in a way that I will mention. OK, so that's all I had to say about the generic kind of Hamiltonian formulation of gravity. So next, I'm going to go like to asymptotically ADS, like supergravity, in a asymptotically ADS space. Just pure gravity for now. Are there any questions? This line? Second line. Second line here. This is i, i, j, prime j. So the free parameter, I just wrote it j here. So if you want this xi, it's xi double prime r. And yes? And that's what it is. Yeah, j, sorry. It's just like a parameter in terms of j. Right, OK. So now let's discuss this reduced. Well, I will go to the gravity part here, but you can extend the whole thing. Maybe I can do it a little bit like in the next lecture. So now what I want to discuss here, first of all, is the structure of the general asymptotic solutions. So asymptotically ADS spaces, maybe I can also mention this later more systematically. So they have a conformal boundary. And you can construct the general asymptotic solutions of the equation of motion in the vicinity of this boundary. So there is a huge literature on that. So one way to do it is we can pick a gauge. So I'm going to put n to 1 and ni equal to 0. So remember, these are Lagrange's multipliers. They're non-dynamical, so I can decide to fix them as I want. And then typically, this is called the Phefermann Gram Gauge. So in these gates, and I'm going to set also like the ADS radius to 1. So in these gates, we can write a metric in this form where this gamma ij takes the following form. So it's e to the 2r. So this is the general structure. So here, now let's see. So this constant, well, not constant. Like this leading term here is independent of the radial coordinate. So all the coefficients here are independent of the radial coordinate. So it's an asymptotic expansion in the radial coordinate. So this here is the boundary metric and this arbitrary. So if you solve the equation of motion, so this can be one way to determine this. It's just like an expansion of this form. Just plug it into the equation of motion and then you will determine this coefficient. But there is no constraint on this leading coefficient here. So it's completely general. It's arbitrary. So now these other coefficients, so this g2 ij up to this h4, sorry, hd ij are local curvatures of g0. So in particular, g2 here is the rigid curvature. And then the higher you go, you get higher order curvatures, like of the boundary metric. And finally, so also this gd here. So I should also mention that I'll just write here. So hd, so I'll just finish this. So hd is non-zero if and only if d is even. Remember, the bulk dimension is d plus 1. So that means d even means odd dimensional adi space. So otherwise, this h is 0. OK, so was there a question? Yes. Can you screen the last three lines? Yeah, can you screen it? OK, I see. This line is here. Yeah, you went the one before, yeah. So e to the 2r? Yeah, so this one, I'm going to write this again. So this e to the 2r, g0 ij. And then it's like g2 ij, et cetera. And then the bracket closed, right? So this closed is here. Yes, indeed, there is another bracket. Thank you. Right. So Samir, is this line clear? OK, so I will rewrite the other two. So g2 up to hd ij are local curvatures. Yes, but curvatures, I mean also like the filmorphism covariant quantities. So I will explain what they mean. So eg is proportional to r of g0, so rij of g0. Then g4 goes like to rij, let's say, rik g0 rkj g0, plus other terms of this kind of line. Sorry, it's OK. Right, and then finally for gd, so the other thing that I said is that hd ij is non-zero if and only if d is even. This applies in the case of pure gravity. So if there is matter fields, this is not always true, although like sometimes it's stated like that. So sorry for my handwriting. So this is hd ij. Sorry, that was a mistake. So local curvatures of g0. Sorry. Right, and finally gd is arbitrary up to two constraints. So now the constraints, I'm going to write them in a compact and implicit form, so they take this form. So this is the covariant derivative with respect to the metric g0. And then the trace of this, for the stress tensor, well, of this quantity, is some local function, where tij is defined to be ij minus, right, and this. So these terms here, so let me just write this in red. So again, these constraints. So gd satisfies, like it's arbitrary, but it has to satisfy these two constraints. But now this form of this a and x, so a is some local function that depends locally on the metric. So it's kind of like derivatives, but in a covariant way. And then this xij, it's a symmetric tensor, again local in g0. But the precise expression of this depends on the particular dimension. So if you do it like in two dimensions, you will get something. So in two dimensions, this will be the Ricci k, for sure. So in higher dimensions, it's something more complicated. And then, again, here, you get some particular tensor that depends on g0. And again, it's dimension dependent. So maybe you can summarize it in a sentence. So what is the original, simply, space? What is the reduction, the symmetry that you're using? It's just the radial. And what is your reduce? So I didn't space by the reduced phase period. So before, I was just doing some preliminary setup for the Hamiltonian formulation. So now I'm going to use, so here I'm describing just the general syntotic solutions of ADS gravity. So this is just by solving the equation of motion. So there are more fancy ways of doing it, but you can just get what I said here in this board now just by solving a syntotically the equation of motion. There is nothing to it. I mean, just brute force. Yeah. Just brute form solving the equation of motion. That's all, right? I mean, without knowing anything. So of course, it's not very efficient doing it that way, but you can do it. So then you get these constraints as well. And then I'm just saying that, like here, I just don't write explicitly because it depends on the dimension. So this a and the xij. So any some curvatures, like, I mean, of g0. So now, now comes the first important thing, which is like this, towards like this, what I call this reduced phase space. So now what I want to show is that, like, if you see now, this is like a space of asymptotic solutions. And then my first claim is that it inherits a non-trivial symplectic structure. So that's why I call it like a phase space. So now recall that the omega that we determined before, and it was like this expression. So it's satisfied omega dot is equal to 0, right? So this was like independent of the particular radial slice, assuming the slice is compact. So now this means that we can just plug in the asymptotic expansions that I told you, right? And that I just erased, just by brute force, use the expressions that we know from the Hamiltonian formulation to just, like, get something, like, just plug it into this expression and see what we get. And of course, you need all the coefficients up to this order gd that I mentioned. So if you do this, what you find is that on this reduced phase space, omega becomes, so I will explain this notation in a moment. So this is defined to be minus 1,5 Tij. And then this Tij, remember, is this expression here. Like, so it's related to gd. It's this particular combination. So you see, first of all, that indeed is independent of the radial coordinate, so as expected. And secondly, we see that t and g0 are symplectic conjugate variables. So now we have, like, this space of asymptotic solutions with a non-degenerate symplectic form on them. And from here, you can go on and define, like, again, Poisson bracket. So I'm going to define this and then stop. So we have, like, g0ij of x. And then by the kl of x. The variables that were in the expansion, there would be an appropriate bracket because it follows from g0. Which are the terms? So you have now g2, g4, up to gd. Ah, yes. But this is crucial. So let me just write down the symplectic form and I will come back to this point. So right. So this is now the Poisson bracket that follows from this symplectic form. And then we can realize it again in a similar way as before. So now this is, like, the space, the symplectic space we're working on is this reduced phase space of solutions. OK. So next time I'm going to discuss the symmetries, like, of this reduced phase space and then, like, how all this is connected to field theory. But for now, let me just, like, answer Leo's comment and then stop. So what happens is, like, if you plug these expansions that I mentioned, right, I mean, with the g2, et cetera, so it is crucial that these are, like, local expressions in terms of g0, right? So because they are local expressions in terms of the g0, you can write them. So you see that here it will cancel with the corresponding terms from here. So it's a non-trivial exercise, so you can just do it. But I mean, you can see that the fact that the higher order coefficients, like g2, for example, g4, et cetera, are local functions of g0. If you put them in the symplectic form and you take the corresponding product here, like, I mean, with the variation of the metric g0, for example, the symplectic product, the variation, I mean, is 0. So the only non-zero terms you get is, like, when you hit, like, from one of the expansions of, so you get terms here, right? I mean, both of these get expanded. So and from here, let's say you get, like, delta gd. And then here you get, like, delta g0. You also get the cross term, so it's, like, delta g0 from here and delta gd from here. So but these are the only cross terms that survive in this expansion. And the crucial ingredient in this, in showing this, is that the intermediate terms are all local as functions of g0. And that's why the symplectic form does not get contributions from those, which is also why it is actually finite. It's independent of the radial cutoff. OK, thank you.