 OK, well, thank you very much. It's a real pleasure to be back in Trieste. So I thought since, in fact, we realized that I'm actually the only lecturer, I think, who's speaking about early universe, the very early universe, that I thought it makes more sense to not just focus on alternatives, but also put in the context of inflation and then some alternative. So let's just say, in general, my lectures will be entitled The Very Early Universe Cosmology. And by way of motivation, I think, as we try to decipher what happened shortly after the Big Bang or perhaps before the Big Bang, really our unique probe to figure out that question, the unique probe that we have are cosmological perturbations, primordial cosmological perturbations. So really, this unique probe is encoded in temperature fluctuations of the microwave background, or if you want curvature perturbation, the various moments there of the two-point function, the three-point function, et cetera. And, well, the sort of general theme of these lectures will be that, in fact, early universe physics is encoded in the symmetries of these correlation functions. Specifically, we will see that symmetries constrained these correlators much like they do in particle physics. So can they constrain these correlators? And correspondingly, there are ward identities we can write down associated with these symmetries. And this is what we'll derive. So specifically, the outline that I have in mind for these lectures is as follows. So in the first lecture, we're going to focus on, in some sense, the simplest case, which is what I call multi-field or spectator field. By the way, can you read in the back? Is it big enough? Yeah. Multi-field inflation. And we will see in this case, this is the story of having spectator fields around that fluctuate but are not affecting the background expansion very much. And in this case, the symmetries that govern fluctuations are, is the full SO4 comma 1 group of isometries of the sitter space. Then in lecture two, we're going to consider what seems to be an even simpler case, but in fact, from the point of view of perturbations, it's more subtle. And that's the case of single-field inflation. And we will see that single-field inflation can be thought of, at least the scalar perturbations thereof, can be thought of as the symmetry breaking pattern of SO4 comma 1 being spontaneously broken to spatial translations and rotations. And finally, in the very last lecture, we'll discuss an alternative mechanism, which in general terms can be called the conformal mechanism or conformal scenario. And this mechanism relies on the full conformal invariance on 3 plus 1 dimensional spacetime. So at the end of the day, it describes a symmetry breaking pattern, SO4 comma 2 being spontaneously broken to SO4 comma 1. And part of the reason I find this alternative appealing is that on the one hand, it relies on symmetries. So it's not just some action that we cook up out of the blue. It relies on conformal symmetry. And there are various different realizations of the scenario, one of them, as we'll see, being Galilean Genesis of Alberto Paolo and Friends. And there's also a fight of the fourth example that was proposed by Rubakov. So there are different ways of realizing this particular alternative, as we'll see. OK, so that's the outline. And by the way, if you ask me what is my, some people have asked me what's my favorite alternative. What's my favorite scenario? It's not even on this list, OK? So it's probably so provocative I shouldn't start by talking about it, otherwise you won't listen for the rest of it. But indeed, or you'll skip the rest. But OK. I'm actually serious, but I'll wait till the end. I'll keep the suspense a little bit. OK, so let us start with a multi-field or spectator field, inflation. So for our purposes, inflation in these lectures is really a phase of approximate decider expansion. And by multi-field, we simply mean a field. So we really mean a spectator field, which is a field that is light. It's present during inflation. It fluctuates, but does not affect the background much. So one example of this is something like the Curveton, where you have some massive scalar, called its sigma, some quadratic potential, whose mass during inflation is much less than Hubble. So this field has a mass much less than Hubble. And the field, it generically will be displaced from its minimum. It will be perched somewhere on its potential. And by virtue of the fact that the mass is much less than Hubble, Hubble friction sort of keeps it there. It doesn't move at all. But nevertheless, it fluctuates quantum mechanically. It acquires a scalar and spectrum. And then later on in the evolution, it can decay and convert its spectrum onto the curvature perturbation, which we later observe. So for such fields, the key point is that for such fields, to a good approximation, we can treat the background as exact the sitter. We'll see that we cannot do that for single field inflation, but for this case, it is a good approximation to do so as exact the sitter space. So for starters, let us explore the properties of the sitter space. So four-dimensional the sitter space we can think of, by definition, if you want, as a four-dimensional hyperboloid in five-dimensional Minkowski space. So here's my hyperboloid. And here's Minkowski, the embedding Minkowski. So I'll denote by capital X is the coordinates of this embedding space. X0 will be time. And then I have spatial coordinates. Let's call it X1 and X2. OK? So this is Minkowski, five-dimensional Minkowski. So the line element in the radius, I should say, will denote the radius of this hyperboloid as H inverse. And of course, this H will be nothing but the Hubble radius or the Hubble constant of the sitter space at the end of the day. So we have, as our embedding space, five-dimensional Minkowski space. So five-dimensional line element is just eta eb dxa dxb, where a and b run from 0 all the way to 4. And this being a hyperboloid, this surface is described by the following equation. It's simply eta eb xaxb is equal to 1 over h squared. Now we can choose coordinates on the hyperboloid in various different ways. We will choose coordinates. So the coordinates relevant for inflation are the spatially flat slices, I should say. OK, so if you take your hyperboloid, you can slice it in different ways. Yes? The only motivation from my point of view is that the symmetries are easier to describe. Because here you have the full SO4 comma 1 group. So that's the logic I'm following. We'll see that in this case, correlation functions are quite constrained by virtue of the fact that they have the full SO4 1 group. While when we get the single field, then if you want the fact that the inflaton is both driving the background and generating the fluctuations, its fluctuations are directly sensitive to the fact that you're not in exact the sitter space, that you have some finite epsilon, some slow-wall parameter. And so as a result, the fluctuations of the inflaton, they're actually described by a spontaneous breaking pattern of this kind. So it's just for simplicity. We lack scalar fields. Yeah, we have 1 and 1. It depends. So we only know we only have one at the TV scale, but who knows at very high energy scales? Yeah, so in general, in string theory, string theory is infested with scalar fields. They have more than they need. So at high scale, why not? And the fact that you would have some that are light, well, in my opinion, if you have one that's light to drive single field inflation, why not have many? OK. Anyways, if you don't like multi-field, this is just a warm-up if you want. OK. So as I said, you can slice this guy in many different ways. In particular, if you slice it with space-like planes, you'll get the closed slicing of the sitter space, the one where you have positive spatial curvature. If on the other hand you slice it with time-like planes, then you will get the open slicing. So the one that corresponds to spatially flat slicing is the one where you slice it with null planes. So let's draw again. So here I will use X4 for concreteness. This will be X1, let's say, X0. So if you pick a null plane, so here's null surface, intersecting this guy. You draw a bunch of these planes in this direction. They will intersect the hyperboloid into these kinds of surfaces, going on the other side, and so on and so forth. OK. Each of these slices obtained by intersection with a null plane corresponds to a surface of constant time in the inflationary universe. Does that make sense? Yes. So in fact, we can write this using equations. One way to do this is to write the constraint equation for the hyperboloid in this way. So we're going to introduce a tau parameter, introduce a tau parameter which splits the constraints, which splits the constraints into two in the following way. So that constraint that describes the hyperboloid, we can introduce a parameter to break it into two constraints as follows. So I'm going to write it as X0 squared minus X0 squared plus X4 squared, will be equal to 1 over H squared, 1 minus little x squared over tau squared. And then the other constraint is just the sum of the spatial Xs, X1 squared, X2 squared, plus X3 squared is equal to 1 over H squared, little x squared over tau squared. So clearly, if you sum these two, you recover the constraints that we had. And now we have this parameter tau, which is an arbitrary parameter. Where tau here, I'm going to take to run from minus infinity to 0. So one way you can implement these two constraints is as follows. We can make the full, exactly, you'll see that in a minute. So now I'm going to write it explicitly. That's exactly right. So the tau parameter is parameterizing which surface or in which null plane is intersecting the hyperboloid. I will make this clear now. Thank you. So indeed, a nice coordinate system consistent with this parameterization is the following one. So let's write the embedding coordinates as follows. So X0, I'm going to write as 1 over 2 minus tau, because remember tau is negative, 1 over 2 minus tau, times 1 over h squared minus tau squared plus little x squared. Let's write capital XI as 1 over h little xi over minus tau. And finally, X4 is equal to over 2 minus tau 1 over h squared plus tau squared minus little x squared. And you can convince yourself in the privacy of your room tonight that indeed this coordinate system obeys these two constraints. These sort of exercises you don't do in public, but you can do it alone tonight. Very nice. Now, coming back to the question that we just asked, indeed, you can show by construction it's manifest from what I just wrote in particular that if I add x0 plus x4, and this will get right to the question I was asked, so if you take x0 plus x4, then clearly the tau squared x squared bits cancel. And I'm just left over with 1 over h squared tau. So this sum is 1 over h squared minus tau. And given that tau runs from minus infinity to 0, this quantity is positive and describes, indeed, the fact that this coordinate system only covers the upper half, only half of the hyperboloid, you see? Because this null coordinate, if you want, has to be positive. So this doesn't cover the entire hyperboloid, only half of it. You can also check tonight that using this coordination of the hyperboloid, that the induced metric under hyperboloid ds squared. So all I'm doing is I take ds squared, which was the embedding metric. And I evaluate this guy using these various coordinates, so x0 equals dada dam, xi dada dam. I do that, plug it in, use a chain rule, and at the end of the day, this gives, I should write this as ds4 squared, you get, at the end of the day, the familiar decytermetric, 1 over h squared tau squared minus d tau squared plus dx squared, written in terms of conformal time. So if you've seen this before in terms of exponential, that's just when you transform to proper time. This is written in conformal time. So in particular, the scale factor in this coordination is just 1 over h minus tau. OK, question so far? Crystal clear? OK, very nice. Now, I'm sure you've seen this before. The reason I take this review route is to get to the symmetries. So if you want this hyperboloid in Minkowski space, is exactly the analog of a sphere in 3D Euclidean space. So it's like a sphere. And as you know, the sphere gets induced on it the rotational symmetry of the ambient space, of Euclidean space. So similarly here, the ambient Lorentz transformations, the Lorentz transformations of the 5D space will map points of the hyperboloid to other points on the hyperboloid. So they correspond in our coordinate system on the hyperboloid to an isometry of the metric. So the hyperboloid, in other words, the hyperboloid is preserved by 5D Lorentz transformation, 5-dimensional Lorentz transformations, with algebra SO4 comma 1, since we're in 4 plus 1 dimension. And the generators of this algebra we're going to denote as jAB, as usual, xA dxB minus xB ddxA, just like the angular momentum generators in quantum mechanics. And these guys obey the SO4 comma 1 algebra. So the two commutators, jAB, so the commutator, jAB, jCD, is just equal to a combinations of eta's and j's, AD, jBC. It obeys this algebra. This is just, of course, a generalization to 4D of the usual angular momentum, commutation relation in quantum mechanics, the SO3 commutation relations. That's a generalization to 4 plus 1. Now if we restrict ourselves, so now we ask how do these act on the hyperboloid, we can just use the chain rule. So in these tau x coordinates, it's another nice exercise you can do tonight, in these tau x coordinates, you can derive fairly straightforwardly that these guys will induce particular symmetries on the hyperboloid. And they're given as follows. So let's write them down. So first of all, there's going to be jij, where I take, so i and j, of course, run from 1 to 3. So these are like spatial coordinates on my hyperboloid. You can show that these components simply reduce to xi d dxj minus xj d dxi. So they just generate good old spatial rotations on a hyperboloid. I can take a linear combination of a rotation involving the fourth spatial dimension and i with a boost, j0i. And that you can show, again, just using chain rule, that this is 1 over h times partial partial xi, which therefore generates a spatial translation. By the way, as a trivial point, but it's worth emphasizing, it's not translation in the ambient space that does this, because translation of the ambient space doesn't preserve the hyperboloid. It's a boost combined with a rotation that does it. And you can visualize it this way. So here is the hyperboloid. Remember, and let's take this is the x4, x0 coordinates. And here are my null slices. So to generate a translation from one point to another, sorry, let's make it dramatic to this guy, what I'm doing is I'm doing first a boost along this guy and then a rotation. So this is the xi direction. So a boost along xi x0 followed by a rotation from in the i4 direction. And that's equivalent to translating infinitesimally. Now these are manifest symmetries of our 4D decyterometric, because clearly the spatial part is both rotationally and translationally invariant. But of course, there's more. So I can take the boost along the fourth dimension, j04. And that corresponds, that will define as d, capital D. And it corresponds to the following transformation. It's tau partial partial tau plus xi partial partial xi. And you recognize this as a dilation, spacetime dilation in which, so this is a spacetime dilation in which tau gets rescaled by lambda tau and xi gets rescaled by lambda xi. That's clearly a symmetry also of our decyterometric. Simply because, yeah, you just see it explicitly. If you rescale tau and x, similarly, it gets absorbed by the 1 over tau at front. So that dilation is an isometry. And finally, the one that's really non-trivial is this one. So it's when I take the other linear combination, so j4i plus j0i. And that guy will define as k sub i. There are three of these. And this, you can show, is given by the following expression. So it's h time 2xi tau d tau plus 2xi xj. Sorry, should be bottom partial xj minus minus tau squared plus x squared partial partial xi. Now that is less trivial. And that's something you can do tonight in the privacy of your room to check that this transformation is indeed an isometry of this metric. So that's going to be an exercise for you. Check that Ki is an isometry. The other ones are clearly isometries. So now I want to consider, so now we're going to move on to perturbation. So let's consider a massive scalar field in the sitter for simplicity. Let's consider a massive scalar on this sitter background. So the action for my scalar d4x minus g minus 1 half d phi squared minus 1 half m squared phi squared. So now when I specialize to this fixed metric, which is this sitter metric over there, you get the following. So you're going to get, so you just get powers of the scale factor. So you get 8 of the fourth from this, 1 over a squared from the inverse metric here. And so this guy together gives me a, so you get a squared, sorry, 8 of the fourth from here, 1 over a squared from there. So I'm left over with 1 over h squared tau squared. And then from this guy, I get 1 half phi prime squared minus a half grad phi squared. And now for the master, I get an extra power of a squared with this guy. So I have 1 half m squared h squared tau squared phi squared. Now this theory, by construction, better be invariant under our symmetries. So we have the following. So since phi is just a scalar, it must transform as a scalar under this coordinate transformation. So this s is invariant under the following field transformation. Must be invariant under the following. So if I change phi via dilation, so delta phi, so delta sub d, is lambda, where lambda is a small parameter, tau dd tau plus xi dd xi on phi. This is a symmetry of the action. And similarly, if I transform phi by this other transformation, there are three of them. So I have three parameters, which is a vector bi. And here is the transformation. It's just what we wrote down. So 2 xi tau dd tau plus 2 xi xj. And finally, minus tau squared plus x squared partial partial xi on phi. So by construction, these are symmetries of the action. And this will be relevant for us in a minute. Absolutely. That's right. Yeah, that's right. So if you want, these transformations are nothing but global difumorphisms. They're just global transformations, as you said, isometries. This action is a scalar by construction. So it's invariant under any diff. In particular, it's going to be invariant under a Killing symmetry. So the only thing I've used here is the fact that phi is a scalar under diffs. And so I can ascertain how it transforms. So all I'm using, if you want to get to this point, I'm just using the fact that phi in the new coordinate system is just phi in the old coordinate system. And then I expand in small parameter away from that. So for small diffs, and then you get. So now in other words, yeah, so thanks for your question. So in other words, in this language, I don't think of it. This is maybe the passive or active form. I always forget. But this is the form in which, if you want, coordinates are staying fixed. But instead, I'm thinking of just transforming my fields. Yeah, thank you. OK, so now we can use as well this action to study the equation of motion. I want to emphasize, by the way, that although this looks classical, I'm going to think of this completely quantum mechanically, meaning I'm going to think of phi as a Heisenberg picture operator, in which case it obeys its classical equation of motion. So if you want, thinking of phi as a Heisenberg operator, I can just write down its classical equation of motion, but really thinking of it as a quantum operator. And then it obeys. If I just vary this with respect to phi, I get the following equation of motion. So phi double prime minus 2 over tau phi prime minus grad squared phi plus the mass term equals 0. And we will be interested in the limit where the gradients are small. So namely, this is to be thought either as late times when modes have been stretched well outside the horizon. So namely, tau goes to 0. So you can see this guy is suppressed by spatial derivatives compared to 1 over tau. So we're going to ignore the spatial gradients for simplicity just to see how the mode behaves as tau goes to 0 at late times. So we take, in other words, we can either think of this as taking tau goes to 0 or equivalently, since you see that this, I can just multiply the whole equation by tau squared. So clearly that the parameter of interest is k times tau. Of course, so this is equivalently just k goes to 0, ignoring those spatial gradients. And so now we can figure out the time dependence of this growing mode. And it's easy. Clearly it's going to be a power law. So we can let tau, sorry, let phi goes as tau to a power delta. And again, it's straightforward. You just plug it in. You get a quadratic equation for delta. And the solution, you get two roots that delta is equal to 3 halves, 1 plus minus. This is the famous answer. 1 minus 4m squared over 9h squared. So these roots are real if the mass is less than whatever, 2 thirds h. That's the usual bound. Otherwise you get oscillatory solutions. But for us, let's just focus on this case where m is less than 4m squared is less than 9h squared. That's what we'll focus on. So these roots are both real. So as a result, we get phi at late times growing at some power of tau. We want to pick out the growing mode. And the growing mode, since tau is negative and going towards 0, the growing mode is actually the power of lambda, which is smallest. So it's for the negative sign. So in fact, the growing mode solution, actually maybe I go here so everybody can see. So the growing mode solution will simply be delta minus, sorry, this was to be called delta plus minus. The growing mode is delta minus. So it's tau, sorry, phi going as tau to the delta. We're now delta. I'm simply referring it to delta minus. This is the growing mode solution. And what this tells us is that, in fact, in these symmetry transformations for phi, so these symmetry transformations, I can replace at late times. I can simply replace this tau d d tau. This logarithmic derivative with respect to tau by delta, you see, because I know the mode behaves in this way, quantum mechanically. So in other words, in the symmetry transformations, I can replace tau d d tau with delta at late times. And moreover, if you look here, you see that there's a term here which is minus tau squared plus x squared. I can also neglect tau squared relative to x squared, since tau is going to 0 relative to the special scale of interest. This is the same approximation we made here. And so similarly, I can drop this tau squared x squared. And so at the end of the day, these transformations at late times on the field boil down to the following. So I replace tau d d tau by delta plus, let me write it like this, x dot grad phi and delta k is bi 2 xi delta plus 2 xi x dot grad minus x squared partial xi on phi. So this is how symmetry transformations should act on our field at late times in the inflationary universe. And these you recognize, they're nothing but ordinary spatial dilation and special conformal transformation on the three plane. So these transformations, they're recognized. So this is spatial dilation. And this is just the spatial SCT. So special conformal transformation on the plane. And so this is the well-known result. So asymptotically, the symmetries of the sitter space. So SO4 comma 1, the isometries of the sitter space, all that we've used essentially is that its equivalent has an algebra to the conformal group R3. And this, of course, is an observation that's at the basis of the DS-CFT correspondence. So it acts as conformal transformations at late times. Now, in cosmology, we are precisely interested in computing correlation functions at late times in the inflationary universe. So we're precisely interested in this limit where k tau goes to 0. So this is particularly convenient. So we're precisely interested in this regime where k tau goes to 0. OK. And so in fact, we can capitalize on this to realize that our correlation functions, our late-time correlation functions, should be invariant. So we can use two facts. We can use the fact that, first of all, our correlation functions that involve various powers of phi, they must scale as tau to the particular power delta, where delta, remember, is related to the mass of the field in question. And secondly, we can use the fact that the correlation functions, the x dependence, must be invariant under the full conformal group in three dimensions. And by the way, yeah, so let's just get started. So we can immediately, and by the way, this is an observation that's very old. I think the first people who talked about this were Antoniades, Motola, and somebody else that escapes me in the 90s. And then, of course, Schrodinger with a DFCFT. But one nice paper in particular that discusses this is by Paolo. So if you look at the paper criminally, here's the phone number 1108.0874. He nicely describes the consequences of these symmetries. So let us indeed start with a two-point function. So with a two-point function already, you know that the fact that the spacetime is translationally invariant and rotationally invariant in space, so the fact that you have 3D spatial rotations and translations. So just that fact, of course, tells you that the two-point function, so we're only going to consider equal time correlation functions, so phi of x1 at tau, phi of x2 at tau, already you know that this quantity is simply, just by these symmetries, must be in general terms, it must be a function of the distance x1 minus x2 and time, tau. But of course, we have more than this. We have the full conformal invariance plus the time dependence. So in fact, the spatial dilation is enough to fix everything. So just the fact that it must be spatial. So what you know is the following. So their correlation function, x1 tau, x2 tau, by conformal invariance, by dilation invariance, I'm sorry, it's enough. It must be a function of the distance to the 2 delta. And the rest you fix by knowledge of the time dependence of the fields. So this must go as tau to the 2 delta. And that's it. So just based on symmetries, the two-point function is fixed up to an overall constant, which of course is just a normalization of the terms. Ah, it doesn't matter. This is completely fixed. As long as it's preserving the symmetries, as long as it's preserving the symmetries of the background, then it's fine, yes. Otherwise, it's completely independent. Yeah. Oh, good question. No, so you just have to be careful in taking this limit. So let's just put the minus here. So in the limit where m goes to 0, you see delta goes to 0. And what you get at the end of the day, if you start expanding this, so you see, first of all, the answer becomes scaling variance. And there's a correction in delta, which goes as a log. As a log of x1 minus x2, which is what you expect. Yeah. Yeah, but this is the assumption we make all the time when we do inflationary computations. Now, of course, you can also have people have considered departures from what we call the Bunch-Davies state, the adiabatic vacuum. As long as these are not too violent departures from Bunch-Davies, these departures will be erased as tau goes to 0. In fact, most of the time when people start fiddling with these states, they find huge departures, which in fact lead to a breakdown of inflation. So you have to be quite careful about the kind of departures you're allowed. So if you want, that's just for this discussion, assume Bunch-Davies, just to fix ideas. The point I want to make is just that even with this assumption, irrespective of the Lagrange and I wrote down, I could deduce the two-point function without calculating Hankel functions or anything. And it becomes powerful now when we look at three-point functions and higher-point functions. So for two-point function, it's kind of trivial. So let's indeed move to, ah, so I want to stress, yeah. So at the level of two-point function, there are no further constraints. So if you want the special conformal transformation acts trivially, okay? So everything is fixed by dilation. At the three-point level, on the other hand, I have the following. So let's consider again three fields at equal time. Now again, by virtue of the fact that I have three powers of the field, the answer will go as tau to the three delta. And now I want to first write down something that's dilation invariant. Well, there are many things I can write down now because I have three points, so I can construct two distances. Okay, so in particular, so I can write three distances. So delta, I'll write first the answer. So I can write this, x2 minus x3. But if it were just about dilation invariance, there are other things. I could multiply this by an arbitrary function of ratios of these distances, right? So if it were just about dilation invariance, you could consider ratios like these, okay? Et cetera, et cetera. All the various ratios. So by dilation invariance, it's not enough, but in fact, once you impose special conformal transformations, you show that this function must be a constant. So that in fact, this is the unique answer. So even the three point function is fixed completely by symmetries up to a normalization. And so on and so forth. Of course, as you go on to higher and higher point functions, there are more freedom, but they're fixed by conformal invariance. At the four point level, you're forced to have these cross ratios function of the cross ratios, and so on and so forth. Okay, so the last thing I wanna do is to actually, because this will be helpful for the next lecture, I wanna describe these transformations in momentum space. So let's write down the momentum space version of these symmetry transformations since we're gonna make use of this next time. So we have in real space, as we've seen, let's just do dilation. And I'm gonna let you do the SET as an exercise. So in dilation, we saw that delta D of phi in real space is just lambda delta plus x dot grad on phi. So let's see how that acts in momentum space in Fourier space. So let's write phi. So I'm gonna write phi in Fourier space. So let's do it directly here. Well, in fact, let's do it here. So I have that phi. So if I write it in Fourier space, so phi of x is D3k, e to the i k dot x, phi of k, i sub k of eta. And so now when I apply this operator, this operator x dot grad, so if you want x dot grad on this object, the x dot grad will only act on the phase. So that's the only one that carries x dependence. So I'm gonna have D3k. So phi sub k of eta comes out. And then I have here x dot grad on the phase. So it's a simple exercise to convince yourself that the grad, of course, brings down an i k. Meanwhile, x is equivalent to idk. So at the end of the day, or one over idk, this guy is the same as k ddk on this. So this is the same as k dot partial partial k. And now I can integrate by parts the ddk, and you see that the ddk is gonna act both on the phi, but also on the k. And when it does, it gives me a factor of three because it's the divergence of k. Okay, so I get, at the end of the day, this is equivalent to minus, by integration by parts, minus d3k of three plus k dot ddk on phi sub k. Eta the i, little nap. Okay, and so at the end of the day, in momentum space, the transformation amounts to the following. So therefore, in momentum space, delta D of phi of k, oh sorry, I switched to eta, I meant it's tau. Okay, sorry about that. Hopefully that's clear. I kept switching in my notes as a style for conformal time. And so the variation in momentum space is simply the following. It's lambda times delta minus three minus k dot ddk on the four year mode. Now y minus three, it's obvious, minus three, why? Because you just lower the dimension, right? So if you want to really, if phi has a dimension delta, then its Fourier transform must have mass dimension delta minus three because of the dk. So this is just telling us that in Fourier space, this field has a mass dimension delta minus three, as you'd expect, the weight of delta minus three, and then the k to dk scaling as usual. Okay, so now we wanna use this fact, what we're getting at, again, which will be useful for next time, is to now consider how a correlation function transforms under this dilation. So let's consider a general correlation function, which is O of k one k sub n, so a general endpoint function. And the various k's here, they correspond to different fields in the correlation function, and each of these guys can have different weight, so different deltas, okay? And we wanna see how does this transform under dilation. So first of all, since what we know is for sure this correlation function is gonna be invariant under spatial translations, so by spatial translations, we know that this will be proportional to a momentum conserving delta function, so by spatial translation, we know that this object O must be proportional to a delta function conserving the sum of the momenta, okay? And we don't really care about this delta function aside from the fact that it comes along for the ride, so we're gonna define, to make our life easier, we're gonna focus on the amplitude, the pre-factor that multiplies this delta function. So I'm gonna write it as follows, we're gonna define, so O, so this correlation function, k one through k n, oops, is gonna be by definition a convenient two pi cubed, a momentum conserving delta function delta, I'm gonna call it capital P, capital P will be the sum of the ks, and now the correlation function by definition O prime, okay, this is kind of a trivial thing, okay? But it just means the correlation with a prime is a correlation function without the delta function quite simply, okay? And we wanna ask how does the prime correlation function transforms? And so P again, just to write it, so where P, capital P is the sum of the ks, chock, chock, chock, chock, chock, chock, chock. So let's consider how this object transforms under dilation, so delta D of this guy. So delta D of O is delta D of this delta function P, times O prime, okay? And now we just have to be careful about how these different pieces transform. So by definition, this object, which is an object that depends on nk's, will just be summing over the variations of each field participating in the correlation function. And so it's simply gonna be a sum from a equals one to n of this operator, of this particular field variations, but sum for each delta in each k. So this will be sum over A of delta A minus three minus k sub A dot DDK sub A, okay? Acting on the whole thing. So acting on the delta, delta three P times O prime. And the reason I do this is just to focus on how this operator will act on the delta, because at the end of the day, what my goal is is to write this as delta times the variation of O prime. So we just wanna fish out how delta gets transformed, okay? So let's do it. So this is a linear operator, so I can just do the chain rule. So I get two terms, so I get minus, and let's see how I wanna do it. Ah yeah, so the delta minus three, of course, doesn't care, right? It's just a number, so let's just focus mainly on this guy. So I have minus sum over A one through n of k A dot DDK A, sorry, like this, acting on the delta times O prime, okay? And then I have the same thing now, but acting on the O prime. So I have sum, so I have the delta function out front times sum times delta A minus three minus k A minus k A dot DDK A, acting on O prime. Okay, so this last term is of the form that we want. This last term is a delta function out front and then some crap multiplying it. And so we just have to massage this first term, and it's very easy because of the fact that delta only depends on the sum of the ks, okay? So first of all, okay, so since it depends only on the sum of the ks, each of these derivatives is equivalent to differentiating with respect to P, okay? So this is just DDP. And now the sum of these derivatives acting on this guy, right? It's just gonna be the sum over k because DDP on delta P is just independent of N, of A, sorry. Okay, so this will just give me P. Does that make sense? Yes, just sums up. So I have minus, as a result, minus P. Let me see where am I? Yes, so minus P dot DDP, acting on the delta function and this whole lot times O prime and then plus the rest. So minus this stuff. The delta alpha was just multiplying the whole lot. So I just shoved it into this term. No, no, no, delta is not the Laplacian. No, no, no, it's just the weight. It's just the mass dimension. Yes, so it's just each, if you want, I think of these as a collection of massive fields, each with different mass. The delta for each of them is what we wrote down. So it's three halves. So delta sub A will be three halves, one minus square root one and four, A squared over nine, H squared. It's just the weight, the mass, if you want, of each of the fields participating in the correlator. Yes, I know in Russia, because my post doc writes Laplacian as triangle and we have endless fights about this, but okay. Now here it just means weight, okay? Now, okay, so we're basically done. So of course this correlation function, at the end of the day, we think of it as going back to real space. We integrate over momentum, so I can integrate by parts, this DDP, and then I'll get two terms. The DDP can act on the P, or it can act on the O prime. But notice that if it acts on the O prime, it will be what I'm gonna be left with will be P times delta, which is zero, you see? So the only place where I can get something on zero, if it's this DDP after integration by parts, hits on the P, which will give me, of course, a factor of three. So this gives me plus three, okay? All this worked for just a three, but okay. So is life. O prime plus the rest, and the rest is this. So it's minus delta. Yeah, let's do it. So it's just sum over equals one to N. Let's see, how do I wanna do it? That's right. Yeah, so let's write it as follows. I can write it as follows. So I can write as this guy times delta A minus KA dot DDKA. Cause that's the term that depends on A gets summed over. And then the factors of three, they just summed over so I get minus three N from this. And all this multiplies O prime. Okay, so after this long song and dance, the end result is the following. It's not very profound at the end of the day, but it's worth going through it. At the end of the day, now I can extract how O prime varies. So at the end of the day, the variation of O prime, which I ascertain from this relation. So now I just drop the delta functions everywhere and I ascertain that O prime transforms as follows. It simply transforms as sum from A equals one to N of delta A minus KA dot DDKA minus three N minus one. And this is multiplying O prime. Okay, now just to philosophize on the result, it's very easy to see. So remember, since we're working in momentum space, every guy in here has a minus three because of the D3K. So since I have N fields, I get minus three N. All right, that's the minus three N. But then I stripped off a delta function, which would have given me another minus three, so plus three. Okay, so at the end of the day, this guy must have this effective weight dimension. And I'm gonna leave it to you as an exercise to show the following. As an exercise, you can repeat this for the SET. So repeat for delta KI. So namely, you can work out how the SET acts on momentum space first on the field, like we did here. And then you work out how it applies to a general correlation function. And then on the primed version thereof, and the answer is the following. So it's delta K sub I on O prime. By the way, can you see? Is it two? Is it okay? Feel in the back? Yeah? No? Okay, I write the answer there. So the answer is the following. It's very nice. Is it on your balcony, watching the sea, compute. And you find the following. Delta KI is O prime is I, sum from A equals one through N of two delta A minus three, D, DK sub I, D, DK, A, I, plus K, A, I, two derivative D, K, A squared. So this is actually Laplacian in momentum space. Minus two K, A dot DK, DK, I, and this whole thing acting on O prime. So please do this tonight. The copies are due on my desk tomorrow. Just kidding. I'm very tempted to talk about my alternative to inflation from Atheist, but I'm gonna restrain myself for tonight. Okay, I mean, he talked about aliens, I mean, for Christ's sake. I'm allowed to talk about whatever I want after that. Okay, thank you. That's it. Question?