 Thank you, Paolo. So one thing that I noticed that you guys are really paying very close attention to everything I'm writing. So several of you pointed out some of the errors I made on the formula. So I thought I'd start in the Rata Courage section on this side of the board. So I made one mistake writing down the invariant distance in the sit there. For those of you that have written that formula, there is no square here. It's easy to see because at least has to be invariant under dilations. So only tau tau prime, not tau square. And the other thing I realized maybe since my handwriting is not as good as Lattec, that I wasn't making a clear distinction between this curly R that is the letter that I used for primordial curvature perturbations, the one that originated the whole distribution of matter in the universe and the CMB. This is what I'm going to introduce later today in a gauge invariant way, et cetera. Sometimes maybe this gets confused with the capital R, which is just appearing in the Einstein-Hilbert action as the Ritchie scalar. So most of the time, actually, this would be the R. I'm going to try to make it curlier from now on, but I hope that this isn't confusing. So every time R appears in a correlator, I'm always talking about this. In fact, most of the rest of the lecture R will always be this variable, the primordial curvature perturbation. And we will see what they are. OK, so we are going to pick up exactly from where we left the story. So what I'm going to do for the rest of the lecture is talk about zlerol. And I know that many of you already know this story, but I'm going to try to do it in a way that, hopefully, is slightly different from the way you saw it. So please don't fall asleep right away. We're going to talk about Hubble zlerol parameters and potential zlerol parameters. And then I was debating a little bit whether this should be taught at the school. But I think quite some people are working on multi-field, also given that single field inflation has been so much studied. And I think in the literature, I don't know a single paper, including my own, in which multi-field potential zlerol parameters have been written correctly. So I just thought the school is the best way to try to fix that mistake. So I'm going to discuss a little bit about what happens if we do inflation with more than one field. And then finally, I'll start kind of the sexier part of the lecture with perturbation theory. That's where you use really quantum mechanics and curse space time to predict the distribution of things in our universe. And finally, I'll tell you what this R is by the end of the lecture. Very good. So where we left was a theory with a scalar field coupled to gravity. And we computed the two relevant equations to describe the background. One is the Friedman equation. And the other is the equation of motion for the scalar field. The Friedman equation just looks very familiar. It's just 3h squared on plant square equal energy density. I'm assuming the universe is spatially flat. And this one looks also familiar, except that there is this Hubble friction term. So the whole reason why we got into this business was that we were trying to find a phase in the early universe that looks like the sitter. So what does that mean? It means that it's approximately a metric like this. And if it were exactly the sitter, this parameter here h, maybe it's also useful to rewrite this metric using cosmological time, just as a reference. This would be the metric. This parameter h is exactly constant in the sitter. Now we already said we don't want it to be exactly constant, otherwise the sitter phase is infinitely long. So it is clear that we're gonna have to have some h dot. So the time derivative of h, that is non-zero, okay? Because otherwise, if it is zero, it's always the sitter. And in some sense, we don't want this to be so big, right? If it is big, we are very far away from the sitter. Now, in physics, we try to always avoid saying that some dimensionful quantity is small or larger, because that's meaningless. So we are gonna divide this one by h squared to make it dimensionless. Very good. And then it's always fun to work with things that are mostly of the time positive. So I'm gonna put a minus sign in front of it. You can just take it as a definition and I'm gonna call this thing epsilon. And this is the first Habos-Loroll parameter. Habos-Loroll. Hopefully the reason why it's called Habos parameter is obvious that this thing doesn't know anything about what kind of matter is sourcing the expansion of the universe. This is only knowing about gravity. It's only knowing about space-time. It's only taking derivatives of geometric quantities like Habos. And first, because there is only one derivative and it's the first one I introduced. Okay, so there is some sense in which we want this one to be much smaller than one, right? To be close enough to the sitter. That's gonna be the first condition. Actually, this is a kind of a fun parameter to introduce. In fact, you can rewrite the acceleration condition that you get from Einstein equation. Usually it looks something like this. Now I shouldn't have written this out of what I remember because I probably don't get it right. But, so that's why I'm not 100% sure, but I'm sure that this thing is written as h squared one minus epsilon. Okay, this is the acceleration condition. So it's obvious that acceleration, which is what we want, right? We said we have a couple of reasons where we want an accelerating universe. It's the same as epsilon smaller than one. I have some signs, right? Epsilon smaller than one, very good. Yeah, because if it is bigger than one, then this is negative, very good. And in particular, let me just mention the obvious thing that epsilon equals zero means that we are exactly in the sitter. Okay, so this one tells me that a certain instant of time, well, we might have been close enough to the sitter, but one moment later it could be that this function changes very quickly. So I want to avoid that. So I want this epsilon to be small for some amount of time. Why is that? Because we observed more than just one scale. We observed a lot of scales. We observed at least the maximal cosmological scale that we observed divided by the minimal one is of order of maybe 10 to the three or maybe 10 to the four. So if I want to take a log of this, because usually I like to define this in terms of E-foldings. That means natural logarithm of this quantity. This is of order of eight, 80-foldings. And over this whole range of scales, from the largest scales in the CMB to the smallest one and the same in large scale structure, I see this approximate scale invariance. So scale invariance was coming from being close to the sitter, so I want to be close to the sitter, not just at one instant in time, but for some prolonged amount of time. So I'm gonna have to ask that not only epsilon is small, but also its variation in time is small. So I take the time derivative of epsilon. I would like that to be small, but that's the dimension for quantity. So I divide it by h to get the dimension less. And I want it to be small with respect to epsilon itself. Okay, so the fractional variation of epsilon is small. And I can call that one a name. I'm gonna call it seconds and all parameter. Hubbell parameter, of course. Coming from the first hubbell parameter. Let me just point out that there is another convenient way of writing these things in terms of another variable that I call the number of E-foldings. This is equal in differential form, this is just the differential of the logarithm of the scale factor. Maybe more evocatively, this is just the logarithm if you integrate this quantity of the scale factor at the end over the scale factor at the beginning. Okay, we call this one number of E-folding. Or maybe for short E-folds. Okay, by its own definition, which is dA over A, I can use all the fun chain rules that I get to using cosmology. So this is dA over A, which is dA over dt, dt over A. And this is just h dt. So in particular, I discovered that this quantity is the same as the n of epsilon over epsilon. Okay, so sometimes maybe it's easier to write it like that. In fact, because of that analogy, I could define an infinite series. So another way is that eta is just the derivative with respect to the number of E-foldings of the log of epsilon. So it's a very compact notation. And since this looks so compact, I might just keep going. Let me define another slow-roll parameter, which you could say maybe it's the third one, to be, again, the logarithmic, the n derivative of the log of the previous one. And so I can go on infinitely. Okay, this is just the definition, not so important. Probably the first two are the only one that I mentioned, but I just wanna show you the structure. And so I'm gonna wanna ask that these are all small. And maybe the intuitive reason is because this guarantees that the spacetime looks like the sitter for a long time, but maybe a little bit more in formula. Let's write this in another way that is maybe, well, for me it's evocative. At least when I think about this, that's how I think about it. So let me take epsilon as a function of n, just because the algebra is simpler than as a function of t, and Taylor expanded around some fixed number. So the number of E-folding now is parameterizing time. So I can Taylor expand it. And we all know how to do a Taylor expansion, et cetera, et cetera. So I'm just pointing out, so let's collect this epsilon of evaluated as some fixed value. And notice here what appears is dn of epsilon divided by epsilon. That was my definition of eta, right? And here what's appearing, the second derivative. So I'm gonna take this one and divide it by the first derivative and multiply by the first derivative. And then divided by epsilon, multiplied by epsilon, so this becomes eta psi three, n minus n star square, plus dot, dot, dot. So in some sense, all that infinite series of slow roll parameter, Hubble's slow roll parameter is really what is telling me how epsilon evolves with time, or if you want with E-foldings. Now imagine that all of these parameters, they are so small that after some amount of E-foldings delta n, this is still much smaller than one. Then this quantity is gonna be smaller than the first one and so will be the second and so on and so forth. Then from this, it's easy to see that I can think of eta to be one over delta n thought. So eta is in some sense is what's telling me how long inflation goes. This is very rough, of course, and I got it from Taylor expanded. When I weigh the certain number of E-foldings of order one over eta, this second term will be of order one and this expansion is breaking down. So probably epsilon is growing, is growing large, okay? So that is gonna be close to the end of inflation. This is just for intuition. You don't have to think about it like this. Intuition. We will see in a second that you can be more formal, but this helps me. So in general, if you wanna do calculations where epsilon or eta appear inside time integrals, is always useful to Taylor expand them. And so at least you know the things you're dropping at what order in slow roll they appear. Some questions about this? Two questions, yeah. It shouldn't diverge because epsilon is positive definite and we're not gonna have bounces in which, yeah, which is going through, is it going through zero? No, well, this is never gonna be zero. Actually, typically what happens in a typical inflationary model, but I'll show you this a little bit later, is that epsilon is gonna grow with time. So typically it starts small, eventually it's of order one and usually that's where inflation ends. Yeah, so it's not a problem. You might have cases in which some of these diverge and maybe you have to be careful, but. Do you have a question? I'm sorry, go ahead. Now what I said is that after a certain number of refoldings of order one over eta, then the corrections to epsilon starts being bigger than epsilon and so you're probably close to the end of inflation. So very roughly you can take that as to be the zero-thorther approximation of how long inflation is gonna be. In practice you literally have to take this function and see when it's one. But I was trying to give you some intuition on the meaning of these slow roll parameters. If any one of them is anomalously large, that term eventually is gonna overcome the initial value and that is gonna make epsilon go to one and that's when inflation will end. So in some sense, the largest one of these slow roll parameters is gonna make inflation end eventually, assuming that this is an analytic function, et cetera. But this is just intuition, okay? So there is the way I think about it. I don't know if it is useful for everyone otherwise you can just take the definition. Okay, potential slow roll parameters. In some sense, these how long parameters what I want to stress is that one is they are only written in terms of geometric quantities, only how long. As long as you know how long is the function of time you can compute them all. And I didn't assume anything about how many scalar fields or even if they were scalar fields. They just know about the geometry and they don't know about what is sourcing this expansion or this accelerated expansion. So they have a very universal character. There are other slow roll parameters that you can define that do not have a universal character. They are much more related to this specific example. And those are the potential slow roll parameters. Since they are widely used and useful in certain situations I'm gonna introduce them as well. So suppose that you have this intuition that the given potential V of five is gonna resemble a constant. Well if it is very flat, if it is in fact if it is exactly flat that the thing is exactly a constant. So you might very tentatively try to define quantities that have to do with the derivatives of this phi. And now you might not find any reason to define these but at least these are dimensionless quantity. And then you could try to define these quantities as they are often defined. At this point there is no reason why I'm defining them but it will be clear in a second why you should define it. And then you can define, you can go crazy. Triple derivatives. These parameters, the first thing that I want to tell you is that they do not know about the dynamics. They only know about the property of V. But the property of V are of course not sufficient to find the solution of this differential equation because this is a second order differential equation so it takes two initial condition. So you still need to specify those two initial condition to know the solution. Okay, the only thing that those parameters know about is V, the functional dependence of V. The reason why they're very useful is because there is a regime in which these Hubble's Lorentz parameters that will compute you should solve the whole dynamics are approximated by these potentials of Lorentz parameters. So they're very useful as an approximate way to compute the Hubble's Lorentz parameters. And in fact that's very easy to see. Let me take the Friedman equation and then I claim that that equation can be written in this form. And maybe you can try to prove this and to prove it we'll have to use the fact that there is an exact relation like this. And easily derive by taking the time derivative of the Friedman equation and using the equation of motion you can prove that this is an exact relation. And then this h dot is epsilon and h dot is also phi dot square. So this is the same as the Friedman equation. Okay, so this is a convenient form because of course from this I can take time derivatives on both sides and on this side every time I take a time derivative I'm gonna use the chain rule and transform this as a d and d phi phi dot. And on the other hand, I'm really gonna take the time derivative. And so I'm gonna get time derivative of epsilon and of eta. So it is obvious that like this I can find the probably very long and complicated expression with all the possible derivatives of v on the one side and everything else on the other. Okay, just a method of take another time derivative and this becomes the second derivative, the third derivative and I can keep going. And just to give you an idea of what happens if you do that suppose that I use this equation I solve for d phi of v by putting phi dot on the other side and I just compute it all. This is what I would find and then I can even more complicated. In fact it's so complicated that I don't even wanna write it all down. So I get this enormous equation that grow larger and larger as I go on. These are all exact equation. So there is an exact relation between Hubble's Lorentz parameters and potential's Lorentz parameter. So at this point it seems that I haven't gained anything. This expression says much more complicated than the one I had before. But notice that as long as all of the Hubble's Lorentz parameters, so if I assume now that all of these Hubble's Lorentz parameters are much smaller than one, then these expressions simplify dramatically. And in fact they give me these nice expressions. So under the assumption that those Lorentz parameters are small, called Hubble's Lorentz parameters are small, then the Hubble's Lorentz parameters, sorry, are given in terms of the potential one. And so this gives me a shortcut to compute them. So this is why they're useful because now these one are easy, I just take the derivative of the potential and then I can compute them. This I think is what the purpose in life of the potential's Lorentz parameter is. And there is nothing magic, there is an exact expression and we are just approximating it. It was just a little bit of algebra and now I'm gonna tell you how this is gonna help us to find the solution of those equation of motion that I told you at the beginning in general, very hard to solve, this equation of motion. But I'm gonna argue in a second that under the assumption that all of these Hubble's Lorentz parameters that I've introduced as long as they are small, I will find some simple approximated solution of these equations. Questions? Okay, if not, yes. This is dimensionless because V is the same dimension as V and here there is a one over five and then Planck takes care of that. So this is dimensionless. So this is the first derivative of V with respect to phi. So it's dimension three. And I square it is six plus two eight and downstairs I have V squared which is dimension four times two eight. So it's dimensionless. Is that the question? And this is also the, all the factors of Planck, in fact some people leave them as an exercise. I'm gonna try to write down all the factors of Planck but you can just fix them doing dimensional analysis. And so all of these things are dimensionless and that's why I put them Planck to the fourth here. Pretty much they count the number of derivatives. Okay, so now I'm gonna talk about some dynamics. And to do that I'm going to introduce yet another little piece of notation which I know is a little bit annoying but it's gonna make it so straightforward to generalize all of these calculations to the multi-field case that I thought I should introduce it. So the other piece of notation is that I'm gonna call the kinetic term X. This is just a definition. On the background, this quantity is simply one half phi dot square. Because that's kind of, it's a nice function to work with. Why is that? Well, it's the most natural function to write down. Probably remember that the energy density was one half phi dot square plus V. So this is just X plus V. And the pressure was the same thing. Minus V, so this is X minus V. And therefore the conservation of energy momentum. This is the conservation of the energy momentum tensor. You nicely see that the V cancel here. This is exactly what you expect from the cosmological constant that doesn't get dilute with the expansion. And of course this gives you this equation. This equation of course has to be equivalent to the equation of motion of the field phi because we know that the equation that we get from the conservation of the energy momentum tensor, they are not new things. They should be equivalent to the equations of motion. They are a consequence of general dephiomorphism invariance of the theory. And in fact, you can easily check this by noticing that phi dot is just phi double dot times phi dot. And so this equation is nothing but phi dot times the equation of motion. So this one is the same as that, just multiply with a phi dot. It will become clear in a second that the notation is much easier and all the algebra is if I use these variables. Okay, so what we define epsilon to be minus h dot over h squared. And that's easily written as x over h squared. And therefore, three x over v plus x. So x is kinetic term, v is potential term. Since I want this one to be much smaller than one, the first thing that I discover is that the potential has to be much bigger than the kinetic term. That's kind of obvious from the Friedman equation. I want to be dominated by the potential part so that it looks like a cosmological constant. That's the first assumption. So I can drop this kinetic part here and only keep the potential at first order in this approximation. So in some sense, if you want, the usual solution for a of t is just e to the integral in dt of h. And if I can drop this phi dot, this is just given by the integral in dt of the square root of v over three and Planck square. In some sense, this is telling me what the scale factor is. Very good. So this is actually the trivial part. The more interesting one is the second slow roll equation. The definition is written up here. And you can show, I leave it as an exercise, how this is written in terms of the kinetic term. This is the formula. eta in some sense knows about the smallness of epsilon, but it also knows that the kinetic term must not change fast as a function of time. So asking that eta is much smaller than one, automatically tells me that x dot, since I already knew that epsilon is smaller than one, it really tells me that the second term is smaller and therefore x dot is more than x h. This is very important because it tells me that one of these three terms appearing in the equation is negligible, namely the first one. That term is the acceleration term. So in single field, this statement of eta being small tells me that I can neglect the acceleration term. We will see in a second that's the peculiarity of single field inflation and it's not a general statement. Okay, then the equation simplifies to six h equals minus v prime phi dot, or phi dot equals minus v phi over three h. So there is more about this equation than meets the i. Okay, so this is the final equation that we got by neglecting the acceleration term. And we got this remarkable equation, okay? There are at least three things that are remarkable about this. Perhaps the first thing is that it's a first order differential equation, where we started from a second order one. So this approximated equation, the first thing is that it is first order. So in some sense the assumption of being in slow roll, it means that we lose the memory of one of the two initial condition. If you have a second order differential equation, two initial condition, assuming that in slow roll, one of the two initial condition disappeared. So that means that there is some kind of a tractor trajectory in phase space that irrespective of where you start, you're gonna end on the same trajectory. So that's the first remarkable fact. Second remarkable fact is that the left-hand side knows about dynamics. Knows about if you want the solution, because it's phi dot. No, I need to have a solution to compute phi dot. While the right-hand side, no, it only knows about V of phi. That we said that priori doesn't know about the solution. Yes. Yes, completely independent. In fact, here I haven't mentioned potential slow roll parameters ever. In fact, those are not useful concepts in general deriving this slow roll expansion. Yeah, I don't find them as useful. In some single field regimes, they are useful because they're so quick to compute, but I think they put you on the wrong mindset, especially when you generalize to two fields. And I'm gonna show that specifically. So, nowhere here I've ever used this potential things. Yeah, that's a very good point. Okay, so this is the magic relation that happens in single field. Because of this slow roll expansion, things that know about the specific solution are somehow determined only by the properties of the potential. That was the magic that we were seeing here. This is just another way of saying the same thing. This does not happen when you have more than one field. So it's a very remarkable property of single field inflation. Okay, finally. So, of course, one could now spend the next three lectures writing down different V of phi and computing all of these slow roll parameters. But I'm not gonna do that. On the other hand, I encourage you, if you're serious about learning about this, is just to play around with all kind of potential that you can come up with. Start with simpler ones, like this, and then maybe you can go crazy. Basically, what you discover is that there are potential like this, or potential like this, or some other ones. And all of these are V as a function of phi. And more or less where they are either very flat or very high, those potential support inflation as described by these potential slow roll parameters. I don't think I have too many interesting things to say about this. So just play around with it and you convince yourself that this model actually can give you inflation. This one as well, actually, pretty much any curve you write down. So I'm gonna let you play with it. I'm gonna deal with one more conceptual issue, which is how long does inflation have to go? We probably all have heard the number 60, usually coming up in this discussion. So it would be nice to know just at least a quick way to get the number 60. So that's what I'm gonna give you a long discussion to get 60 as a result. And inflation. So when does inflation end? By definition, inflation ends while when we stop the accelerated expansion and that's the same as asking when epsilon becomes one. Notice that this is Hubble epsilon, not potential. Only using Hubble's slow roll parameters. Okay, this is when it ends. So how can we count how long it is? Well, we've used the same number before. It's usually much more convenient to use n as a measure of time rather than t. Even when you're solving this numerically on a computer, for example. For one thing, n is dimensionless, but also it doesn't change as crazily as t does. So it's much more easy numerically and I think it's conceptually simple. Well, we know what n is. We just said it's the integral in d log a and we said this is the same as the integral in h dt. But we don't stop with this chain rule things. We're gonna write d phi over phi dot. So d phi in d, sorry, d t in d phi. d phi, which is equal to h over phi dot in d phi. Okay, with chain rules, as long as you do background inflation, you can pretty much get anywhere. So in principle, if you do this integral from some initial condition up to the phi for which epsilon is equal one, that's how long inflation lasted. So given a solution, you know what h is, what phi dot is, you can perform this integral and approximate the result. Well, we can, now we go to the slow roll parameters. Is by approximating this h over phi dot with v over v prime, which you can just using the this equation and the fact that v is much smaller than much bigger than h at d phi. Okay, so if you wanna have an approximation of how long inflation goes, you can use this formula. How long do we need it to go? So for that I need to do a plot. The vertical axis will be commuting distances, commuting scales. One important commuting scale that we discuss in this class is the Hubble radius. And we said that the commuting Hubble radius is one over a h. And this is this line here. Another interesting scale, it might be a certain k that you're looking at. And this is, let's say logarithm of a scale factor. Horizontal is some measure of time. As the Hubble radius is decreasing with time, that's where inflation is. And then later on is either radiation domination or matter domination. Eventually these things comes back down and you have dark energy, but we don't worry too much about that. This is the, actually this is the whole history of the universe somehow in one plot. So it's good to keep in mind how to do this picture. Okay, so let's consider a single mode. Let's suppose that I go in the sky or in the last case structure and I measure a mode of a certain physical length today, k, but that is the same as the commuting value of that field. So you can see that that was the same as the Hubble radius twice during the history of the universe, one recently. And that's when after it re-entered the Hubble radius, we measured it. But then it had been outside of the Hubble radius for a long time until it re-entered during inflation. Good, so when it re-entered by definition, it was equal to one over a h. So here it was equal to one over a h of k. And here when it re-entered, let's say it re-enters today. So it's one over a h today, zero. And here I'm gonna call it this the end of inflation. So the Hubble radius there is one over a h and. Okay, now to tell you how long I need inflation to be, I'm gonna compute the ratio of these two scales in two different ways. I'm gonna compute it from the left side of the figure and from the right side of the figure and the left side of the figure. I'm gonna compute twice the same quantity and then of course I impose them to be equal. Okay, so let me go how I do that. So the quantity I wanna compute is k over a h and. The e stands for end of inflation. So I wanna compute the ratio between the wavelength that I'm looking at and the Hubble radius at the end of inflation. I'm gonna compute this first on the left-hand side. How would you compute it if you were on the left-hand side? Well, I know that on the left-hand side, k is the same as one over a h k because at some point it had to re-enter the horizon. Sorry, this is one over k, of course, because we're measuring distances, so it's one over a wavelength. So this thing is a h at some moment k divided by a h end of inflation. But during inflation, h is approximately constant so I can simplify it out. So it's a of k over a end and I'm gonna say that this is exactly the end we want to compute. Sorry, it's e to the n, right? Because we said before that n is just the logarithm of a final divided by a initial. So I guess this is e to the minus n. That's nice, the first way of computing this gave me an expression that depends on n. So if I can evaluate the same quantity with an expression that doesn't depend on n, I know how long n has to be. So this was computation a. Then I do computation b. It's pretty much the same way, but I have to be a little bit clever. So again, I wanna compute the same quantity. And again, I'm gonna use the same trick that k, well, the denominator is the same, but upstairs k, I'm gonna evaluate it now when it re-enter. When it re-enter the Hubble radius, well, it's equal to the Hubble radius and so k is equal a h today. So that's what I put upstairs. Now it's easy to know how a h evolves in time because you get it from the Friedman equation. Friedman equation just says that a h goes like the square root of a square times rho. So for really simplicity, let's imagine that the universe was radiation-dominated for the whole time, then this is just one over a because rho radiation goes like one over a to the fourth. This is the simplest possible thing you can do. If you wanna do something more elaborate, just put it on Mathematica and you can. It just becomes cumbersome. Okay, well, we found that a h goes like one over a so this is simply a n over a zero. We need some way to compute the, this basically tells me how many e-folding there has been of hot big bang universe and I want those e-foldings to be the same as the e-foldings of inflation and then I equate them. Well, this is simple to equate because it's the energy density today divided by the energy density at the end of inflation to the power one quarter because rho goes like one over a to the fourth. Okay, so that's pretty much done. I'm gonna, now I just do a equals b. I know what is the energy density today. I don't know what is the one at the end of inflation. So I'm gonna write the result in which n depends on the energy density of the end of inflation and that formula is this. n is equal 63 plus the log of the energy density at the end of inflation divided by some typical number. Actually, typically, rho at the end of inflation is to be lower than this. So the number of e-folding will be a little bit lower than 63, but that's where you get the number 60 from. Okay, so I know that there was a little bit of algebra in this, but I think it's good that you have one time model in which you're able to compute to where the number 60 comes from. So just forget about that kind of energy, method domination, just do it as simple as possible. And this is relatively simple algebra. So this is why we need the 60-foldings of inflation. Perhaps a little bit less if inflation is low-scale, so it reheats at some lower energies. So maybe as low as when people do this in a more precise way, people find that maybe you can go as low as 25 and probably as high as 65, but not very different. It is customary to take n equals 60 when people compare results, but probably 50 would be a more appropriate number. Okay, any questions about, yes, yes. I mean, fixed as much as that range you would call it fixed. It varies by a factor of two between 30 and 60, but yeah, that doesn't matter so much. It mostly depends on, because on the left-hand side of this plot, for the calculation of A, the only thing I use was that inflation is quasi the sitter, and so H is counter. So any dynamics that you have which is quasi the sitter is gonna give the same result on this side. And on this side, it's just gonna depend on what you assume about the big band history. In that sense, I think this is a neat way of separating the physical dependencies. Okay, maybe some other questions. I know that was a little, a lot of algebra, maybe a bit fast, but yes, upper limit. Well, yeah, I showed you that one in this simple minded universe in which is all radiation after some time. Oh, this is, I'm sorry, yeah, you're absolutely right. Yeah, I should have made this clear. This is the number of e-foldings corresponding to a mode that we observed today. It could be that inflation in the past was going on for another billion e-foldings. Absolutely, yeah. We don't see that, but the number of e-foldings that we get to see, yeah, that was the number, very good point, yeah. In principle, inflation actually could go down forever in the past. We just never get to see that. But this is the one, but infinitely, I guess, in number of e-foldings. Very good, yeah, some other questions. So then I'm gonna spend 15 minutes discussing multi-fields. So this thing is already complicated enough. Why do I need to make it more complicated? Why should I care about multi-fields? Well, things should be as simple as possible, but not any simpler. So I'm gonna give you some reason why you might care about multi-field, and then you make up your mind. On the very general side, just consider a normalization group that tells you that there are more degrees of freedom at high energies. So it's unclear why there should be only one scalar field during inflation. Yeah, well, it's a very general argument. This slightly more detailed argument is that try to build inflation in any UV-complete theory of gravity theory, for example, and in fact, pretty much the only example, in string theory, then you get the number of fields is very, very large. Typically of order 100 to 1,000, and that is in fact associated to the moduli. In fact, in any extra-dimensional theory, extra-dimensional theory, you have a lot of scalar fields that corresponds to the deformation of the compact dimensions. So even if you have a five-dimensional theory, for example, if you take a theory which is Minkowski four times a circle, then you will have in four-dimension effective action, you will have a scalar field which is represented by the zero KK mode of the metric in the fifth dimension. Inflation and string, sorry, string compactification are just a glorified, complicated example of this simple behavior. So that's another reason why you do expect more than one field. Perhaps slightly more to the point, there are two things in which single field inflation is very unique. And so it seems that you're really looking under a lamppost. There are some accidental symmetries about one-dimensional manifolds that you don't wanna be dominated by. First thing is that a one-dimensional manifold is always flat. That's a kind of a fun statement. Basically, it comes about because because of the symmetries of the Riemann tensor in one dimension, it's, well, you can convince yourself that it has to be anti-symmetric and you do all the games, so you cannot write any component that is non-zero. So one dimension is always flat. That means that the kinetic term is always canonical. I can always do a transformation of what I called phi. I can think of phi as parametrizing a one-dimensional field space, the target space. And up to changing phi into some phi prime, I can always take this kinetic term to be canonical. So here I didn't multiply times some complicated function of phi. I could, in principle, but I can always go from phi to phi prime and get rid of this function, always in one dimension. Because one-dimensional manifolds are always flat. Yes, here I'm assuming a smooth manifold and for smooth manifolds, you can do it. Yeah, if you have some singularity somewhere in the target space, it gets complicated. But here I'm assuming one to one, exactly. So the transformation, you should be able to do it. I mean, this is a covariant tensor. So if it is zero in one set of coordinates, it's zero in every set of coordinates, so. Okay, so this is one fact that is very peculiar of single field. So you might wonder what happens to multi-field and we will see it in a second. And the last thing is that there is no sense in which potential slow-roll parameters, they don't exist. No potential slow-roll parameter, generally. And maybe this is the last reason why I wanna make this point. Maybe a couple of you in the rest of your life will write a paper about multi-field, so at least you know what is the story with the slow-roll parameters. Okay, so we're gonna do something very simple to what we did before. That's why I left all of these equations on the board. Now we're gonna assume that phi instead of being a single field, we're gonna upgrade it to a vector in some n-dimensional target manifold. For example, just our n, or for example, something more complicated. So then this kinetic term should be improved to include for the metric of this manifold. So this is a generic function of phi vector. This is the one thing that we didn't have in single field. So this formula changes slightly, where now the i's are contracted using this capital G. So besides using this trick, every other formula that I've written on the board pretty much is gonna remain true. That's why I think it's very easy to do this, and I'm gonna try to do it, and I think it's interesting to see what the result is. So up to putting these indices, most formula remain true, and it is again true. I have to tell you what the equation of motion is also very similar to this, with just two exceptions. One exception is that the second derivative is substituted with this directional covariant derivative, and then I have to put little indices i everywhere. And then again here I can just call it x, so I'm done. So these are the equations of motion in the general case. I haven't told you what this d is. All of this remains true. So I have to tell you what d is. dT of phi dot i is equal phi double dot i. So now the second derivative is a little bit more complicated because I wanna make it covariant under change of coordinates in the target manifold. So we get this capital gamma, where this gamma now is the Christopher symbol of this metric. So upon using these tricks, the equation of motion look formally the same as before. But what I want you to appreciate is that there is an interesting new player that before was forbidden. Is that there is a new force that plays a role in this equation, which is not the Hubble friction, is not the force coming from the potential, is the force coming from the fact that the manifold is curved. That's usually what we call gravity. Not the fact that the manifold is curved and you move. In this case it is the target manifold of being curved. So this is in some sense the target manifold gravity, but it's just a new element that was forbidden before. So you might find interesting dynamics by looking at this system. And notice that all of this formula are both general covariant under space time changes, but also under five changes in the target space. So that's why they're cute. All of these equations are the same as before. X plus V, X minus V, where X now is the upgraded X. And again, I wanna consider inflation, so epsilon has to be much smaller than one. Notice the beauty of using Hubble's Laurel parameters. Doesn't care if I have 10 million fields. It's the same equation because it only knows about geometry. And the same for epsilon. This also doesn't care. So it's still much smaller than one. Same, same. Okay, so all of these equations are the same. Okay, so sometimes people write down, I just wanna say, sometimes people write down these incorrect equations and they say this is as long as this is small, you can do inflation. So this would be pretty much the analogous of what we were doing before by just putting indices. This is not a necessary condition. Sometimes people define the Hubble, the second one, to be the minimal eigenvalue of Vij divided by V, we have a Planck square. This is neither necessary nor sufficient. Both of them are not great things to use. So I just thought they're not good things. Okay, let's derive the right one. Well, pretty much is all written here. Okay, this is still true. Axiom is still this quantity, which is still h over x square. So this is still true. So again, V is much bigger than x. Again, we need to be dominated by the potential term. Yes, very good. Yeah, you're right. Yeah, thank you. Yeah, that's important. So now I have to contract them. In fact, all of this formula, you can just guess how they become covariant just by putting indices and asking that they look okay. Okay, so very good. Theta, same formula also doesn't change. This is the same formula as before. And again, I can say that x dot is much smaller than xh because it's the same formula. Now, what happens is that before we could conclude we could find the solution for phi dot. This was the big step and I kind of spent some time saying how remarkable was this relation. As you imagine, this relation now doesn't hold anymore. And this relation does. So I can neglect this x dot term. Let's write what x dot is. Is equal phi dot i. This is not just the acceleration that is small, but is the projection of the acceleration along the direction we are moving. That is the quantity that has to be small. You can move as fast as you want and as accelerated as you want as long as you're accelerating the direction perpendicular to the direction in which you're moving. For example, if you move on a circle, this term is always zero because the acceleration is always perpendicular to the velocity. Okay, well, that's pretty much the final. There are a couple of algebraic steps. You can use this equation to find the solution which is valid in slow roll for x. x is just the other thing. And that gives you the correct, the correct slow roll approximation which now I'm assuming I'm neglecting some higher order things which is very similar to that one, but differently in an interesting way. Cosine theta, where cosine theta is defined to be the angle between the gradient and the direction in which you're moving. Very intuitively, if you're going down a ski slope and you're skiing perpendicular to the gradient, your altitude is not changing. So the cosmological constant is not changing and so you're really inflating no matter how fast you go. Well, with some limits. But so in this sense, in multi-field, the right concept is really slow descent. It's asking how much do you descend rather than slow roll. Okay, you can be moving fast as long as you go perpendicular. So that means that even if the potential is steep, as long as you're moving perpendicular to the potential, you can still be inflating. Of course, if you impose that cos theta is one, then you recover, for example, single field equation. But this is not mandatory. And just for completeness, let me mention what is the other approximation that you could try to call eta v. It's unfortunately slightly more complicated equation. Okay, so these are the correct approximation for the Hubble potentials in multi-field. So I think that there is a large chunk of multi-field inflation that no one has ever really explored. Many Nogo theorem in string theory are just based on epsilon without this cosine being small. So perhaps should be at least revised. Okay, so I leave that for the future generations to think about. And in the next, yes. Very good, yes. Well, I mean, this would be got some bosons if I didn't have the potential term, right? The potential is breaking all of those shift symmetries. Yeah, I mean, the current approach is two ways. Either you really try to construct it in a specific theory like string theory, and then let's say you are describing a D3 brain, which is a point like in a compact space like a Calabi Al, which is not flat, it's a rich flat, but not flat. Then it's kinetic term, we'll have those kind of geometric factors. We will feel about the target space geometry. So that's a simple case. Whether you ask me whether it generically appears one way or another, yeah, it's hard to tell. I agree with you that you cannot get it from an exact ghost on boson because you need a potential, otherwise nothing happens. In fact, it's a really cool exercise. Take an exact non-linear sigma model, I compute epsilon is three, irrespectively of the geometry. For any non-sigma model in any number of target space dimension, epsilon is three. I was actually shocked by this fact, but I think it's true. But if you put the potential, no. Then if you put the potential, things change. So that's a good comment. Okay, so the next 15 minutes, we're gonna introduce the notation that we will use to finally compute the nicest story of all of this is how quantum perturbations in the early universe generate what we observe today. How did that come about? Well, you kind of know the story. They got expanded and blown to cosmological scales and the amplitude is set by the uncertainty principle. So that's the calculation we wanna do. To do the calculation, we need to do a few fun things. What is that? We will consider QFT incurred space, time. And we're gonna do it in the simplest possible, algebraically simplest possible context. So GR plus a scalar field. So we're really after kind of the idea rather than the complicated calculation. So to do this, we clearly have to go beyond the homogeneous background. We're gonna go beyond an isotropic background. Was all of this discussion of the background useless? No, it was not because the background determines the symmetries obeyed by the action for the perturbation. That's kind of a general lesson in physics. Once you've solved the background and that has some symmetries, for example, homogeneity and isotropy and maybe some more. Those are the symmetries of the action for the perturbations. And it's gonna be much easier to compute correlators of perturbations if we have some symmetries lying around. Okay, symmetry number one, homogeneity. What does this tell us? Tell us that it is a good idea to work in Fourier space. Because two different nodes, well, let's say five for example, five of K. Oh, sorry, let me introduce this notation. I'm gonna call perturbation to phi. I'm gonna call them curly phi. I don't have to write delta phi all the time, which takes me twice as long. So I'm gonna use this curly phi for perturbations and this other phi here for the background. So perturbations with one Fourier mode are completely independent from perturbations. We have a different one. Okay, different from K prime at linear order. Perhaps this is the reason why we're all here today. No, it's very easy to do cosmological perturbation theory because different K is the couple and all of the transfer function for C and B in large scale structure are just one function of K, pretty easy. Otherwise it would have been very hard to solve this problem. And in fact, people in large scale structure, that's what they are trying to do, push beyond the linear order and that's tough. Okay, so that's simplification number one. So you can always think of superposition principle. You take the universe, homogeneous with one wave, find the result and then sum up over all possible waves. And that's the correct result. That is much, much simpler. Point number two, isotropy. But what does that tell us? It tells us the very useful factor of scalar vector tensor, the composition. Scalar vector tensor, what? You can decompose it and after you decompose it, the scalar vector and tensor are independent of each other at linear order. S, V, and T independent again at linear order. Makes it even easier. And I'll discuss in a second why this is easier. Both of these properties, well, you can figure them out in a little bit more formal way. The way I think about them, which is clearly very sketchy, but is try to write down a way because there is no way that you can make two different case couple or a scalar couple to a vector or a couple to a tensor. So my proof would be by exhaustion. So try to write down an equation with K. You can try to put derivatives in Fourier space. Those are just multiplications. Well, let's write it in real space. But you're not allowed to write a random function of X. That you're not allowed because the background is isotropic. So this would break, sorry, homogeneity. And the same thing, you're not allowed to write down something squared. So this would be no linear. This could be a very sketchy form of an equation. When you do the Fourier transform, clearly this is phi of K. And this is K times phi of K, with the same K appearing in both of them. And the only case in which you get something interesting is when either you break the homogeneity of the background or you're going on linear. Probably obvious to all of you, but I think it's good to keep in mind what's the basic reason. It just, there is nothing I can write down to make K and K prime couple, except for these two things that I explicitly said are forbidden. Perhaps the most intuitive way. Okay, so let's do this scalar vector tensor, the composition in this monochromatic universe. Monochromatic because I can only, I can consider ways one at a time. As an example, I'm gonna give you a scalar vector tensor, the composition of the metric. So g mu nu will be some background g mu nu. This will be some FLRW metric. And I think you did a little bit of cosmological perturbation theory. When you studied arc energy with Koyama, so I'm gonna do it a little bit faster, perhaps. But I'm gonna use the notation of scalar fields. So if you cannot follow, please stop. So this is, I'm gonna call h the perturbations to the metric. So how many degrees of freedom there are in this h in principle? 10, very good. So this has 10 things. So that would be a hell of a job to work with all of those 10 coupled differential equation. Partially even. Maybe I should have said, this means that the partial differential equation becomes already a differential equation. That's why it's cool. Okay, 10, but we're gonna see what those 10 are. So in particular, h00 doesn't transform under rotations. Well, it transform like a scalar. So it's argument transform. So when I say scalar vector tensor, the composition, I mean with respect to spatial rotations. It should not be confused to scalar vector tensor with respect to general defilmorphism, which is not what I'm talking about in this specific discussion. So the 00 component is clearly a scalar under spatial rotation because, well, it doesn't have an I index. The h0i component, I can decompose it into two things up to a conventional factor of A in front. And here I'm using the notation from Weinberg's cosmology group book. I decompose it into two things. One is a vector and one is the derivative of a scalar. And this makes sense as long as that vector is transverse. Finally, the ij part, that's the most complicated one. Can you see here? The ij part has another scalar A, another scalar B. So let's summarize all the scalars. They are A, B, E and F. So that's four of them. There is another transverse vector here. So di, ci equals zero. So there are two vectors, ci and gi. Each one have two polarizations plus and minus one. So this gives me, oh, I'm sorry, two times two, so four components. And finally, there is a tensor one. This one you know very well because you know about gravitational waves. This is the graviton. This is transverse traceless. And so dii is equal di, dij equals zero. And so we have one tensor. Transverse traceless, only polarization plus two and minus two. So this is two degrees of freedom, total 10. Good, so we got the right counting. The isotropy of the background, what it allows us to do is to throw away these. Well, we would have been able to throw them away anyways because they decay at linear order. And we can also disregard the tensor as long as we talk about the scalars and vice versa. This regard the scalars as long as we talk about the tensor because these things do not talk to each other. Okay, the last piece of thing that you need to actually do a calculation is a trick to get rid of yet two more of these two scalars. And you probably know that the way you do that, in fact, that's really necessary is gauge transformations. So coordinate transformations. What does a gauge transformation look like when you go from one variable to another one and they can differ by an arbitrary function of this variable. This is the most, sorry, this is the most generic change of coordinates that I can do. And I'm gonna work instead of using change of coordinates which is a passive way of thinking about this. I'm gonna use an active way and think that every change of coordinates rather is a gauge transformation of the field. So I'm gonna define the gauge transformation of a scalar field, for example, to be its value with the new coordinate, sorry, it's new value minus its old value evaluated on the same point. That's very important. Okay, well, we know very well how a scalar field transform is x, phi prime x prime equals phi of x. So I can just use that into this equation and you will see that this is equal minus the lead derivative in the psi direction of the field phi. If you're not so used to lead derivatives, this is nothing by psi mu d mu phi. Ah, sorry, d mu phi. And since we're gonna treat this transformation as a linear quantity in perturbations, at linear order, the thing that will appear here is psi mu d mu phi, where this is the background. Since the background only depends on time, this is just minus psi zero phi dot. This is how a scalar field transform under a general change of coordinates. And this formalism is powerful because we've lead derivatives because then it's very easy to know how the metric transforms, what is minus the lead derivative of the metric. And the lead derivative of the metric is probably familiar to you from discussing isometries and you know that it's just the killing equation, right? So this is just minus, so this is the transformation of the metric. Okay, I just need one equation and then I'm done. Sorry about going a little bit long. This psi mu we can write it under scalar vector tens of the composition as a zero component, a di component of a scalar part plus a transverse component. Okay, there are four of these four functions, one function here, another here and two here. Notice that two of these functions, the scalar part and the zero part, they all transform as a scalar. So I can use them since they are arbitrary to get rid of two of the scalars that I had down there. So minus two scalars. So this has made my life so much easier without having to do a single calculation and down to just three scalars. And next time actually we will discuss is only one that we really need to take care of and quantize. So this is probably a good time to stop and ask you if you have any questions. I know it was a little bit of a long lecture and it's getting late so we're all tired but yeah. Could you please repeat why you have two degrees of freedom in the tensor part? Yes, because this is a three times three matrix, symmetric so it's six. One, two, three, four, five, six. And then I lose one here because it's traceless. And three here because it's transverse. So it's six minus one minus three and that's the two. And those are really the two polarization, the cross and the plus polarization of the graviton. And I think Michele tomorrow probably will do a little bit of theory since he will do gravitational waves about these two things. Very good, yeah. Very good, yeah, yeah, sorry. All right, do you need some special properties about the GUIA and multifield inflation? Yes, good question. It has to be positive definite because I want to avoid ghosts. If there was one negative definite, I could diagonalize it there and I would have a ghost. So I would like it to be positive definite and for the rest, smooth, yeah. Symmetric, well it has to be symmetric, yes. Well any anti-symmetric part is gonna drop because I'm contracting it with something which is symmetric. So yeah, it's only the symmetric part that is relevant. Is there any use in proposing instead of a multiplet of scalar fields like something else like spinors or something a little bit more elaborate or? Yes, so typically things that are not scalar when they, if they're allowed to get an expectation value and they get it, they usually break isotropy. Think about the vector, it's gonna point somewhere. That's annoying because the universe we see is isotropic. So then you can either do some smart tricks to get rid of that anisotropy by gauging it somehow and introducing some internal symmetries but that's maybe more advanced. The other thing is that you can put a bunch of them and make them point in random direction. And if you put, I don't know, 10 to the 10 of them, the square root of that 10 to the five is maybe the level at which isotropy. So those are, people have investigated those possibilities. Doesn't feel as natural. And especially it doesn't feel natural from this effective theory point of view. You're really just giving a clock to the cosmological constant. So it's unclear why a clock gives you a different time when you look at it from another rotation. Other questions? Okay, if not, let's thank Rico.