 So, when we ended last time, we were looking for a nonlinear theory of massive gravity. And the two questions that we wanted to answer was, first of all, is there a nonlinear theory that is free of this boulevard desert ghost? So does it propagate only the five degrees of freedom appropriate to the massive graviton? So this is a question of theoretical consistency. And the second question that we wanted to answer once we have such a theory is whether or not this theory exhibits this Weinstein mechanism. So in other words, at short distances, do we recover the predictions of general relativity? And thus, is this a phenomenologically viable theory or not? So before going on to answer these questions, I wanted to take a brief interlude to either introduce or review some mathematical formalism. This is probably stuff that you guys have seen before, however, I just wanted to go through it quickly to give some background. So this will be a formal interlude. And I want to emphasize that this formalism isn't required for writing down these theories of massive gravity for the ghostry theories of massive gravity, but I'm introducing it for two reasons. So the first reason is that it enables you to write down theories of massive gravity in a particularly simple formulation. So it's useful in that regard. But the second reason that I'm introducing it is it also helps put these theories of massive gravity into a bit of historical context. So by introducing this formalism, we'll be able to see how these massive gravity theories, in fact, are very closely related to Love-Lock theories, which are other generalizations of general relativity. All right, so let's go ahead. So the first formalism that I wanted to introduce slash review is that of differential forms. So in general relativity, we're used to working in what's known as a coordinate basis. And what this means is that we can use differential line elements, by which I mean some dx mu, to form the basis of a vector space. So these become basis vectors. So we do this when we write down, for example, a covariant vector, so sum v, which is equal to some coefficient, say v mu, times these basis vectors, d mu x. And in fact, we're so casual with using this basis that we often refer to the coefficients themselves, the v mu, as the covariant vector. But the understanding is always that these are the coefficients in this basis here as well. All right, but now I'm going to be explicit, and I'm going to hang on to these basis vectors, the dx mu, when I write anything out here. All right, so for reasons that will become clear, these dx mu's are often referred to also as one forms. All right, so we can construct a two form by introducing an exterior product, also known as the wedge product, between two one forms. So the product of two one forms will look like the following. So a two form w, we can write as u wedge v, where these guys are both one forms. So this is what we mean by the wedge product, or the exterior product. Now the key feature of the wedge product is that it's an anti-symmetric product. So we have that u wedge v is always equal to minus v wedge u, like so. And by this definition of course, it means that the wedge product of any two one forms together, sorry these are both u's, is always going to be equal to zero. Okay, so conceptually the magnitude of two one forms can be thought of as the area of a parallelogram, whose sides are the length of u and v. So if we have one vector v and another vector u, the wedge product u wedge v is going to be equal to the parallelogram formed by this guy here. All right, so we'll see that there's something related to wedge products and in general area elements, and as you go up to three forms to volume elements as well. So in particular, the differential line elements, these are differential one forms. And the two forms, so suppose I have two of them, say dx and dy, the two form, form by wedging them together, so dx wedge dy has the properties of a differential area element. And the best way to see this is the following. So suppose I start out with these two differential one forms dx and dy, and I want to perform a change of variables. So let me take x and y, and let me perform a change of variables to some x prime, that's a function of x and y, and also some y prime, that's a function of x and y as well. Okay, so we know how the differential line elements are going to transform. So dx prime should be equal to partial dx prime, dx, dx, partial dx prime, dy, dy, whereas we have dy prime is going to be equal to partial dy prime, dx, dx, plus partial dy prime, dy, dy, like so. So this means that when I write the wedge product now of the new variables, dx prime, dy prime, I'm wedging these two things together with each other. So we know that the wedge product of two of the same one forms is going to give us zero. So the only terms that we have to pay attention to are the ones that we get from wedging dy with dx and dx with dy. So this is going to be equal to dx prime, dx times dy prime, dy, dx wedge dy, and then I'm also going to get a plus dx prime, dy, dy prime, dx times dy wedge dx, but of course this is just minus the x dy. And so what I find is exactly the Jacobian of this transformation appearing in front of the two forms. So dx prime, dx, dy prime, dy, minus dx prime, dy, dy prime, dx, all times dx wedge dy. So this is the Jacobian here. All right, very good. So basic properties of the wedge product. So the wedge product in general can be used to construct forms higher than a two form. So to make a general say p plus q form, we simply wedge together some p form and some q form. So in general we're going to have say w of rank p plus q is going to be equal to say some m of rank p wedged with some form of rank q. So this is a p form. This is going to be a q form. So this is a p plus q form. And in general this is going to obey the anti-symmetry property that this is going to be equal to minus one to the p times q and q wedge mp. All right, where p, q and p plus q we call the rank of these forms. All right, so I'm going pretty quickly here, but I'm assuming that this is something that most of you have already seen before and I just want to give a bit of a refresher here. All right, so note that because of the anti-symmetry property of this wedge product, the size of the form, the rank of a form that you can construct is always restricted by the number of dimensions that you're in. So if I'm in a theory that has say four spacetime dimensions, I can only construct a four form at the most because I only can wedge together four different line elements at a time. So in particular we could consider say a space that has three spatial dimensions. In this case there's only a limited number of types of forms that I can write down. So we can categorize the most general forms that I can write down in a space that has three spatial dimensions. So a zero form is just going to be some function that depends on my three coordinates. That's going to be some function f, y, and z, and that's the most general zero form that I can write down. The most general one form that I can write down is something with arbitrary coefficients that multiplies the dx, the dy, and dz. So I can write down some g1 of x, y, z times dx plus some g2 of x, y, z, dy plus g3 of x, y, z of dz, like so. And to simplify this I can write this notationally as say some gi of x, y, and z dx i, like so. Let me do the other ones over here. All right, I'm still considering a three dimensional space. The most general two form that I can write down is going to take the following form. So I can have three functions, let me call them h12, x, y, z, that's going to multiply a dx wedge dy plus h23 of x, y, z that multiplies dy wedge dz, and an h31 x, y, z that multiplies a dz wedge dx. And of course I can write this more compactly as some hij dx i dx j. And finally I have a three form which is going to be characterized by only one coefficient. So let's call it some m123 x, y, z that's going to multiply dx wedge dy wedge dz. And I can express this as some mij k dx i wedge dx j wedge dx k, like so. And I'm not being very careful with overall factors in front here. All right, but that's it for a three dimensional space, because there's no four form that I can write down, because that would involve wedging together say two dx's or two dy's. So this summarizes all the possible kinds of forms that we can write down in three spatial dimensions. All right, so the next important operation besides the wedge product is that of the exterior derivative. So the exterior derivative is an operation that takes a p form to a p plus one form in the following way. So if I start out with some function, so some zero form f, then the exterior derivative d of f x, y, z is going to be equal to df dx i dx i, like so. So this is a zero form going to a one form. For a one form going to a two form, we can write say d, and now we're going to have d of say g i dx i, actually let me make these j's for simplicity. So this is going to be equal to d g j dx i dx i wedge g x j, like so. And you can see how the pattern is going to continue for higher forms here. So I can let me do one more. So this was the exterior derivative taking a one form to a two form. And let's see what it looks like to take a two form to a three form. So there we can write d. And I'm going to use the same variables that I introduced here. So d hj k dx j wedge dx k. This is going to be equal to partial. d hj k dx i, and now I'm going to have dx i wedge dx j wedge dx k. All right, but these should look a little bit familiar, right? So if we consider this three dimensional space, this process here, this is simply a vector that I've ended up on the side whose coefficients depend on the derivatives of the function x. So in fact, this is the same thing basically is just taking the gradient of f. Whereas this here, so now I can also think of this as a vector in three space that points in the k direction if this is i and j. But because this is anti-symmetric and i and j, this is like taking the curl of g here. So this is curl of this function g. Whereas here, in three space, I can think of this as a scalar, basically, because I have only one three form in three space. So this is a scalar, and in fact, this is like taking the divergence of h, like so. All right, so an important property of the exterior derivative is that taking the exterior derivative twice of any form always gives you 0. So if some Wp is always equal to 0. And we see how this plays out in terms of these functions here. So this statement here is just equivalent to the usual vector identities that curl of a gradient is equal to 0. And also that divergence of a curl is equal to 0, is equal to 0, like so. All right, so those are the main properties that I wanted to introduce in relation to these differential forms. Are there any questions about this? OK, very good. So the next bit of formalism I wanted to introduce was the notion of tetrads here. So we started out by introducing this coordinate basis in which the dx mu were the basis vectors of our tangent space. But we can introduce a different basis instead. In particular, we can introduce an orthonormal basis and denote the basis vectors by some EA. And what we mean here by orthonormal is the usual definition up to the appropriate signature of the metric. So we want to be taking inner products of these guys that reflect the correct signature of the metric. So in other words, what we mean by orthonormal here is that if I take the inner product of two of these basis vectors, say an EA and an EB, and I use the metric to define the inner product, this should be equal to some eta AB, like so. OK, so this is the definition that we're taking. So we can express the new basis vectors in terms of the old basis vectors simply by writing EA is equal to some E mu A dx mu, like so. So express new basis vectors in terms of old. And now this matrix E mu A, this is going to be a 4 by 4 invertible matrix. And it's often referred to as the tetrad. So you'll also hear other words as well. It's occasionally referred to as the veal bind. In four dimensions, it's also referred to as a veer bind. In three dimensions, it's a dry bind. And I think in two dimensions, it's a svi bind. I'm going to refer to them as tetrads throughout here. All right, so we have this 4 by 4 invertible matrix. We can define the inverse of the matrix then. So the inverse is going to be denoted by E mu with upper indices, A lower index. And the property of the inverse should be such that both E mu A, E mu B is equal to Kronecker delta AB. And likewise, E mu A, E nu B should be equal to Kronecker delta mu nu, like so. So written in terms of these inverse matrices, sorry, written in terms of these 4 by 4 matrices, this requirement here, so the orthogonality requirement, can be expressed in the following way, A, thank you. So we can re-express this orthogonality requirement in terms of these 4 by 4 matrices. So this is basically just saying that E mu A, E mu B, G mu nu should be equal to eta AB, like so. That's all that we mean here. And we can rearrange this just by multiplying by inverse matrices and inverse verbines to express this as G mu nu is equal to E mu A, E nu B, eta AB. So these are the same things here. So the main points about this tetrad matrix. So the first thing is that even if you don't care about any of this, the mathematical formalism, what we're basically doing is we're defining some matrix that's essentially like a square root of a metric. So you can think of the tetrad is like a square root. But there's another way to think about it as well. We can also think of it as a transformation that takes us from a locally inertial frame, so some frame where I can define a flat matrix eta AB to some non-inertial coordinates, G mu nu as well. So in particular, if at every point denoted by some capital X, say, we construct a set of coordinates, and let's call those coordinates psi of capital X with index alpha that are locally inertial at X, capital X. Then the metric, can you guys read if I write down here? So then the metric is given by, again, G mu nu equals E mu A of little x E nu B of little x eta AB, where now we can define E mu A at the point capital X is going to be equal to the coordinate transformation d psi x alpha of little x dx dx mu evaluated at x is equal to capital X, like so. OK, which is all to say, again, that we can treat these tetrads as the transformation taking us from a locally inertial frame to the full curve metric as well. Important properties of this tetrad. OK, so note that the tetrad is really four covariant vectors rather than a tensor. And what I mean by this is the following. So if I do a general coordinate transformation, so if I take X mu to some X prime mu equal to X mu, well, just let's leave it at X prime mu, like so, then only the mu index of the tetrad is going to transform. So in other words, I treat the A index as labeling four vectors each that have mu index. So the tetrad is going to transform, like E mu A goes to E mu A prime, which is equal to dx mu dx mu prime E mu, like so. OK, so we're only transforming this index here. So in general, I'm going to refer to the mu index as the spacetime index. But notice on the other hand that the definition of the tetrad is unique only up to an overall local Lorentz invariance. In other words, I can transform the A index of the tetrad by a local Lorentz transformation and end up with the same metric, g mu nu. So in other words, if I define now an E mu prime A that's equal to lambda AB, E mu B, like so, and I stick it into this definition where I write g mu nu in terms of the tetrad. Then I have that g mu nu prime is going to be equal to, I'm going to have two Lorentz transformations. So A, C, B, D, E mu C, E nu, D, eta AB. But of course, the two lambdas combined with the adas just gives me an eta CD. So this is E mu C, E nu D, eta CD, which is of course just equal to my original g mu nu, like so. So I also have this Lorentz invariance as well. In fact, this invariance is really important and makes sense if we want to count degrees of freedom or count components in these theories. So we're starting from a theory that has a metric, which we said before because the metric is a 4 by 4 symmetric rank 2 tensor. This has 10 components contained in it. But now we're saying, OK, but we can also express it in terms of these tetrad variables, which are also 4 by 4 matrices but are not constrained to be symmetric. And so in fact, the tetrad has 16 components in it. So in order to get this equation to balance in terms of components, we see that it's precisely the six components contained in the Lorentz transformation that I can use to get rid of the six extra components of the tetrad here as well. So the Lorentz transformation here contains, of course, three rotations and three boosts. And I can use these always to kill six components of the tetrad and end up with the same 10 components of the metric that I want as well. Very good. All right, so this index A, so mu is the spacetime index. I'm going to refer to A as the Lorentz index of the tetrad. So finally, I just want to note that tetrads are more than just a mathematical formalism, a mathematical curiosity. If we want to talk about spinners in curved space time, we're in fact forced to introduce tetrads rather than just the metric instead. And this is because in order to describe a spinner, spinners are really only defined as representations of the Poincare group. If you have a curved spacetime, you no longer have Poincare invariance. And so you need to set up locally inertial coordinate systems at every point in space in order to define spinners on a curved background. And so formally, what this looks like is that if we want to go from, say, the Dirac equation in flat spacetime to in curved spacetime, that means that we have to take, say, the Dirac gamma matrices, which are defined with a Lorentz index. And we can use the tetrad to promote this Lorentz index to a true spacetime index. So we can define a gamma matrix, gamma mu, which is just equal to e mu a, gamma a, like so. Which means that now we can write the Dirac equation in terms of these new gamma mu matrices. So we could have some i, now gamma mu, covariant derivative, mu minus m psi equals to 0. And the point is that this covariant derivative is, of course, already a tensor. And so we need to introduce the gamma mu in such a way that this also transforms as a spacetime tensor as well, a spacetime vector. OK, I'm going very fast. Are there any questions about this? Yeah, so in fact. OK, so the question was, what we've described about massive gravity so far, you don't have a general covariance. So what's the relationship between these massive gravity theories and geometry? And so yeah, I guess I would argue that you're right, there's no longer this beautiful different variance of the theory. Nevertheless, we're going to see that they can still be elegantly constructed in terms of these forms. And so there's some sort of topological inspiration behind these theories, without these mass terms really being topological terms. But hopefully it'll become more clear. Are there other questions about this? OK, so before again moving on to massive gravity, what I want to do now is I just want to, in fact, go through general relativity in terms of this new formalism here and talk about how the structures that we've introduced here can in fact be used to write GR in a very compact and elegant notation. All right, so in order to see what GR looks like in this new formalism, I'm going to introduce the following sort of dictionary between the two theories. OK, so our starting place, so this is going to be GR and this is going to be tetrad one-forms. So I should say this is GR with metrics and index notation, so compared to GR with tetrads in one form. OK, so the idea is that instead of writing down the metric g mu nu, we replace it like we said with a tetrad one-form ea, which is equal to e mu a dx mu, like so. In GR, we also usually introduce the Christoffel connection, so gamma mu nu lambda. In the language of these one-forms, instead we introduce what's known as the spin connection one-form and it's denoted by some omega ab, which is equal to omega mu ab dx mu. So remember that the rank of the form always counts the number of spacetime indices and it's independent of the number of Lorentz indices. So both e and omega are one-forms because they have one spacetime index, even though they have different numbers of Lorentz indices here as well. All right, in place of the Riemann curvature tensor, so r rho sigma mu nu, we use instead a curvature two-form. So the curvature two-form carries two Lorentz indices, rab, and it's equal to an index notation, rab mu nu dx mu, wedge dx nu, like so. And it's expressed always in terms of the spin connection omega so that rab is equal to d. So this is our exterior derivative, omega ab plus omega ac, wedge omega cb, like so. And just to be sure that our counting is right, just to check, so we're saying this is a curvature two-form. It has two spacetime indices. Omega is a one-form. If I act with a d operation, that gives me a two-form here. If I wedge together two one-forms, that also gives me a two-form here. So in fact, quick check just to make sure that this is indeed a two-form. All right, the covariant derivative, so nabla mu, we're going to replace by an operation capital D that carries no index, but is defined in terms of the exterior derivative and also the spin connection one-forms. So when taking derivatives with respect to this d, the d carries only about the number of Lorentz indices on your form. It doesn't care about the rank of the form at all. So in other words, if I take d of some zero-form phi, this is just going to be equal to d phi, the usual exterior derivative. If I take d of some form, which only carries one Lorentz index, tau a, this is going to be d tau a plus omega ab wedge tau b, and so forth. If I take the covariant derivative of a form that has two Lorentz indices, then I'm going to get two factors of the spin connection. And in fact, just like in GR, the sign is going to change depending on whether these are raised or lowered indices. So d of some sigma ab is going to be equal to d sigma ab plus omega ac wedge sigma cb minus, let me get this right, omega cb wedge sigma ac. And so again, the number of omegas appearing depend only on the number of Lorentz indices and not on the rank of the form here. So this is what we mean by a covariant derivative here as well. All right, we can relate the covariant derivatives to the curvature in the following way. So we're used to the expression that the commutator of two covariant derivatives, so say d mu d nu acting on some vector a lambda. We can write in terms of Riemann r kappa lambda mu nu a kappa. This has the analogous expression in terms of these forms here. So we can write, say, two derivatives acting on a form with one Lorentz index tau a. This is equal to R ab wedge tau b, like so. OK, just a couple more. So let me do it here. Yeah, in order to do that, you would have to restore the Christoffel connection as well. Yeah, first I think you would have to go back to index notation, and then you would have to restore the Christoffel. So in GR, we always assume the metricity condition, which is that the covariant derivative of the metric should vanish. In the language of forms, this becomes a very different looking condition, which is that the Lorentz indices of the spin connection should be anti-symmetric with respect to one another. We also often assume the torsion free condition in GR, which we can express as the lower indices of the Christoffel connection being symmetric with each other. So the anti-symmetric combination of these two indices should vanish. In the language of forms, this becomes the expression that the covariant derivative of the tetrad, which is equal to d ea plus omega eb wedge eb, this should be equal to 0. So it's a little bit curious, because it seems like these expressions almost flip in terms of form. So here we're saying that the Christoffel has to be symmetric in these two indices. Here we're saying that the spin connection has to be anti-symmetric. Here we're saying that the covariant derivative of the metric vanishes. Here we're saying that the covariant derivative of the tetrad vanishes. But the implications of each one becomes flipped in these two languages. So this I'm always going to assume to be true, whereas this condition here I'll be relaxing in various places. So I'm not automatically going to assume torsion free condition in everything that follows, so relax. All right, let's see. We also have the Bianchi identity. So in usual notation, this can be expressed as a covariant derivative of Riemann vanishing. So anti-symmetrized in these indices. In terms of this language, this is just the statement that dRAB is equal to 0. And in fact, this is indeed an identity. So you can see this simply from the definition of RAB. So if you take RAB, you take the covariant derivative according to how we've defined it. And you just massage a little bit. You find that by this definition, this is always automatically satisfied here. OK, finally, we can get around to writing the Einstein-Hilbert term in this language, because what we want at the end of the day is an action in terms of these variables. So the Einstein-Hilbert term in our usual notation is d4x root g scalar, root g scalar, as a function of g. In terms of these tetrad-1 forms, this expression is simply the wedge product of the curvature 2 form with two of the tetrad-1 forms with all the Lorentz indices contracted by an epsilon tensor, epsilon a, b, c, d. And that's it. And finally, in GR, we also can write down a cosmological constant, so the cc d4x root g. And in this language, in fact, the determinant of g is just equal to the wedge product of four of the tetrad-1 forms, all contracted with an epsilon tensor, so a, b, c, d, like so. And in fact, let me go through the derivation of getting between this one and this one, because that's going to be useful for the other terms that follow. So if I want to go, say, to check that this form is equal to this form here, let me first do that by restoring the index notation. So this is saying e mu a, e nu b, e, let's say, alpha c, e beta d, epsilon a, b, c, d times dx mu wedge dx nu wedge dx alpha wedge dx beta, like so, is equal to this. So all I've done is written these guys in terms of their components in the one form, because these are just coefficients and not vectors. I can pull them out in front of the one forms that just appear here. All right, but now, this quantity here, the wedge product of the four dx's, because this is a four-dimensional space time, this is always just going to be a product of dx0 times dx1 times dx2 times dx3 with some anti-symmetrization of the indices in front. So in other words, this thing here, in fact, I can write this just as epsilon mu nu alpha beta times dx0 wedge dx1, wedge dx2, wedge dx3, like so. All right, so this, in fact, this is just my usual volume element, so I'm going to call this d4x, whereas the combination of the two epsilon's contracted with the four identical matrices. So these guys here, this just gives me a determinant of E, like so. So I see this whole thing is just equal to determinant of E, d4x. And because determinant of E is equal to the square root of the matrix G, in fact, this is just equal to root minus G, d4x, like so. All right, so that's just to give you some feel of how to get back and forth between the two. All right, so let's actually look at GR in this language. Are there any questions? Yeah, that's right. So what's going to happen? So there are two ways that you can go about it. So either from the beginning, you can impose this torsion free condition by hand, in which case that lets you solve for the spin connection in terms of the tetrad. And then you can write the curvature just in terms of the tetrad without the spin connection. Or the way that we're going to do it is we're actually going to treat the spin connection as an independent variable. And then you find the torsion free condition as an equation of motion. And then you get exactly the same thing you solve for the spin connection in terms of the tetrad. And you can express the curvature just in terms of the tetrad as well. Yes? I don't know. So OK, I think it's a beautiful formalism. I think it's a very compact way of writing GR. Unfortunately, I think if you want to calculate anything, like cosmological perturbations, this is not the most efficient formalism for doing this. So I think most of the time it's not a very practical formulation for calculating things in GR. But it sort of gives you insight into the geometrical background of it. But I think more textbooks should have this. OK, so let's start out with our action in this language. So we're saying that the Einstein-Hilbert action, S-E-H, we can write roughly. And here I'm just going to drop all factors in front. So all m-planks, all factor of 2, just to give you a schematic idea of what's going on. So as Einstein-Hilbert, we said we could write as the integral RAB, wedge EC, wedge ED, epsilon AB, CB, like so. OK, let's note the following things about this action. So first of all, we have an overall Lorentz invariance. So in other words, I can take EA to lambda AB, EB. And similarly, I can take RAB to lambda AC, lambda BD, RCD. And this action is invariant under these transformations, simply because I get four lambdas that appear in front here. When contracted with the epsilon tensor, this gives me a determinant of lambda. And determinant of lambda is 1. And so in this way, this action is invariant under these Lorentz transformations. And again, that's important because it's that Lorentz transformation that's responsible for killing the six extra components of the tetrad here. So giving me the right number. All right, so in addition to the overall Lorentz invariance, I have a diffeomorphism invariance, which I'm going to put in quotes because it's not exactly the same as the usual diffeomorphism invariance. And what I mean by this is the following. So I can take the tetrad EA, and I can send it to EA plus covariant derivative D of some scalar, some four scalars, phi A. And sorry, what I mean by scalar here is, in fact, these are four zero forms, phi A, like so. OK, and also in this language, if I'm treating the spin connection as independent, then the curvature tensor is not going to transform. So this is just a transformation of the tetrad EA. OK, so now if I do this, I get the following. So the variation delta EH under this transformation is going to look like the following. So it's going to go like RAB wedge EC wedge D phi D, like so, epsilon ABCD. And this is going to be equal to 0 via integration by parts. And that's simply because when the D hits the R, it's going to give me 0 by the Bianchi identity. And when the D hits the E, it's going to give me 0 by these torsion-free conditions as well. So the action is invariant under this transformation here. So this is 0 by integration by parts. That's right. Yeah, yeah, so it's actually straightforward to show because so D only ever hits Lorentz indices. And so when you have, say, D of some tensor, say, TABCD, as long as all the indices are contracted, this is just equal to a little D of the same thing. So you can always integrate by parts. That's a good question. OK, but let's see, in fact, so I said this wasn't exactly a diff. So let's see, in fact, what a general coordinate transformation looks like. In this language, so if you perform a general coordinate transformation, it has the following effect on the tetrad. So if I let x mu go to x mu plus some psi mu, the transformation of the tetrad is the following. I get the delta EA is equal to D phi A minus lambda AB where I've defined phi A is equal to psi mu contracted with a tetrad, E mu A, and lambda AB is equal to psi mu omega mu AB, like so. So in fact, a general coordinate transformation acts as both this kind of different variance in addition to a Lorentz transformation on the tetrad. But because our action is invariant under both of these, it means that the action is also invariant under a general coordinate transformation as well. OK, let's look at the equations of motion. So again, we're starting from this action here. And the simplest way to derive the equations of motion is, like I said before, to treat the spin connection as an independent variable. So this is known as first order form. And we'll see that this is equivalent to second order form, which means setting the torsion free condition equal to 0 from the beginning. So let's treat the spin connection as independent. And let's first look at the variation with respect to the spin connection itself. So let's look at variation with respect to omega AB. So the only thing in this action that depends on the spin connection is the curvature to form R. So we want to see how this varies with omega. So delta of R AB, this is equal to D of delta omega AB. And then using the definition, the rest of the definition of the curvature, this is equal to delta omega AC wedge omega CB plus omega CB, sorry, AC wedge delta omega CB, like so. But in fact, because I have these two wedge products with the curvature, sorry, with the spin connection, this is just equal to the covariant derivative of D omega AB. So now if I look at the variation of the action with respect to the spin connection, I have delta S goes like D delta omega AB wedge EC wedge ED epsilon ABCD. And so again, using integration by parts, I see that setting this equal to 0 corresponds to the expression that DEC wedge ED epsilon ABCD is equal to 0. And it's somewhat non-trivial, but you can massage this equation enough, so massage the indices, such to show that this is uniquely solved by the torsion free condition, DEA is equal to 0. Which is to say that we could have imposed this by hand from the beginning, and we would have found the same equations of motion. But now we should also vary with respect to the spin connection. So if you vary with respect to EA, now we have delta S. EH is going to go like RAB wedge EC wedge delta ED epsilon ABCD. So now our equation of motion is just going to be RAB wedge EC epsilon ABCD. Is equal to 0, like so. And after massaging, you can show that this is equivalent to the Einstein's equation as well. And I think I'm going to skip showing that explicitly. But let's see, just to make a little bit of connection with the usual results. So here I've suppressed all spacetime indices. So I can restore them in this expression here. So this expression here, if I restore my spacetime indices, this is telling me that d mu E nu A dx mu wedge d x nu is equal to 0. But the only thing that's important here is just the fact that the wedge product enforces the anti-symmetry of the mu and nu indices. So in fact, this is just equivalent to saying that the anti-symmetric combination d mu E nu A has to be equal to 0. All right. So that's just what it looks like in terms of usual indices. Are there any questions about this? Yeah. Yeah, so if you write down non-Einstein-Hilbert terms and use this first-order formalism, you'll have torsion. But generically, torsion is going to come with extra propagating degrees of freedom. And so it's not going to describe the representation that you want necessarily. So yes, if you switch to nabla, then you also have to introduce Christophels as well. But yeah, that's right. Sorry, let me think a second. Yeah, no, sorry. I think that's correct. You can just interpret it as a nabla. OK, so far, this is just sort of a rewriting of the usual equations of general relativity. But in fact, this formalism has already found its place in a non-trivial extension of general relativity that goes by the name of love log gravity. So in 1971 and 1972, David Lovelock looked for extensions of general relativity that are higher-order polynomials in the curvature tensor, but constrained so that the equations of motion remain second-order in the metric. So the reason why you want equations that remain second-order in derivatives is that generically higher-order equations of motion, basically you can think of it means that you're going to have to specify more Cauchy data. So in other words, you have extra degrees of freedom in your theory. By demanding that the equations of motion remain second-order, it means that you propagate the same number of degrees of freedom as the original theory. And in particular, you expect a theory with no ghosts or no extra degrees of freedom. So we'll talk about the relationship between these two more carefully later. But let's just consider this requirement here. So basically, he was looking at terms of the form. So S is equal to MP squared over 2, d4x root gr. But then also wanted to add terms of the form, say root gr squared or root gr mu nu alpha beta, and so forth, like so, by enforcing this requirement. All right, and what he found was the following. So in d equals 4 spacetime dimensions, there are only three such terms that you can write down. And these can be expressed very nicely in terms of this formalism that we just introduced. So all the terms with second-order equations of motion are of the form dA, wedgie B, wedgie C, wedgie D, epsilon ABCD. So this was just our cosmological constant that we introduced before. You can write down, like we said, RAB, wedgie C, wedgie D, epsilon ABCD. So this was just our Einstein-Hilbert term. Or in fact, there's a third term that you can write down, which looks like RAB, wedgie RCD, epsilon ABCD, like so. And this term in index notation is what goes by the name of the Gauss-Benet term. So this looks like d4x root minus g. And then you have R squared minus 4 R mu nu R mu nu plus R mu nu rho sigma R mu nu rho sigma, like so. So this term satisfies this criteria that the equations of motion remain second-order in derivatives. However, in four dimensions, this term is purely a topological surface term. So it doesn't actually contribute to the physics. And in fact, in this notation, it's really easy to see why that would be the case. So if we vary this action here, so if we look at delta of R wedge R, this is going to look like R wedge D of delta omega. And then by integration by parts, this D is going to always hit an R, which is going to give me 0 up to a surface term as well. So in D equals 4, this is just a surface term. All right, but something interesting happens when you go to higher dimensions. So in dimensions higher than D equals 4, this turns out to be a legitimately new term that contributes to the physics, the equations of motion of your theory. Sorry, I'm sorry, I'm having trouble hearing your question. I've written down terms like R A B wedge E A wedge E B. So because here, the harsh dual is involved, right? Yeah, so you're talking about writing down terms like R A B wedge E A wedge. Oh, yeah, the A wedge B. And also for the other one, the third one. Yeah, so let's see. I think in general, these are going to be equal to 0, right? Because of the anti-symmetric property. So you're basically taking the, sorry, let me think about this a sec. I mean, so actually, no, no, sorry, R A B is symmetric. No, sorry, R A B is anti-symmetric in these two indices. So if you anti-symmetrize, this will give you a minus sign and then redefine, no, no, this is OK. I think the issue is that these might be parity violating terms if you work them out. I'm not sure what the physics of these are exactly. Yeah, but I think it's possible that. What would it be in the metric formalism, in the second order formalism? In second order form. So yeah, I mean, I guess you could just naively restore the indices, right? So you would get something that looks like this. So you would have R A B mu nu, say some E alpha A E beta B epsilon mu nu alpha beta, something like this. And you could probably pull out some determinant of E in front. And so it would just be some new way of contracting the indices. But let me, yeah. And we had similar thing for the second option, R A B which R A B down. That's right. No, that's right. Yeah, I mean, I guess if you want the requirements of different variants, second order equations of motion, Lorentz invariance, and probably parity conservation as an additional requirement, you're really forced to write down these three terms. But yeah, but there are other possibilities. Yeah, I have to think about this more. OK, so if we extend this story to higher dimensions, what we find is that this Gaussian term becomes a legitimate new term. So if we look at D equals 5, the answer to the question that Love-Luck posed is, again, you have the cosmological constant. So E A B, sorry, E A, wedge E B, wedge E C, wedge E D, wedge E E all contracted with an epsilon. You have the Einstein-Hilbert term, R A B, wedge E C, wedge E D. Now with the third E E, epsilon A B C D E, A B C D E. But now the Gauss-Penet term takes the following form. You have R A B wedge R C D wedge E E, epsilon A B C D E, like so. All right, and now this term is no longer purely a surface term. And yet it still has this requirement that the equations of motion are second order in derivatives. So in higher dimensions, in other words, there are these extensions of general relativity that preserve the number of degrees of freedom of the theory. And in fact, you can extend this story to even higher dimensions. So in D equals 6, you can also introduce a term that looks like the wedge product of 3Rs. But in D equals 6, this is also going to be purely a surface term. However, in D equals 7, there's a natural extension of this term, which is the wedge product of three curvatures with one tetrad. And now this is, in fact, a new extension. So this is a new dynamical term. And so forth, in higher dimensions. All right, so the story is the following, that these extra terms exist, these natural generalizations of general relativity in higher dimensions. And they can be formulated in a particularly simple way using this wedge product formulation. So in particular, we see that in the example of the Gauss-Benet term, this looks like you have somewhat arbitrary coefficients appearing in front of the combination of these three terms, whereas in terms of the wedge product, it's very naturally expressed. It's just a wedge product of two curvatures. And so this simplifies our notation a bit. All right, questions about this? Yes? Sorry. Yeah, so that was the dot, dot, dot. That includes everything else. But yeah, you can have the arm you knew, arm you knew as well. Sorry. How do you transform it? Under the, I know, I'm sorry, could you say? Oh, it's part of the Gauss-Benet term. Yeah, so it belongs as part of the Gauss-Benet term. I mean, all these terms are still diff invariant. And maybe you can write how you can write this term in your notation. Just, you mean just the arm you knew, arm you knew one? Yes. So the answer is, I mean, in this notation, this is sort of the unique term that you can write down that involves the wedge product of two curvatures, two forms. And this gives you only this term. So in this notation, you can't really extract just the arm you knew, arm you knew part. If you go to index notation, you have more flexibility. And you can write these things down. But the statement is that if you're just stick to wedge products of this form, this is the unique quadratic term that you can write down. Are you saying that it's obvious that once written in this form, the equations of motion are second order? Or are you saying that it turns out to be? It turns out to be. It turns out to be. It's somewhat non-trivial to show other questions. OK. So in the last 15 minutes, I finally want to get back around to massive gravity. So in a sense, we're asking a question that's very similar to the question that Lovelock posed. So can we write down interacting nonlinear theories of a massive graviton that propagate no extra degrees of freedom? So no ghost-like terms. And the answer to this question is yes. So we call this section 5. And this brings us around to the RGT, massive gravity. And we want to answer the question of, is there a nonlinear theory of massive gravity that only propagates five degrees of freedom? Read the question. All right. So in order to do this, we're going to again use this tetrad notation. We're also going to introduce, so we have a dynamical metric, g mu nu, which we're saying we can write as e mu a, e nu b, eta a b. And we define the equivalent tetrad one form, e a is equal to e mu a dx mu. We're also, like we said before, we need to introduce a non-dynamical reference metric in this theory in order to write down mass terms. So for now, I'm going to take this just to be a flat Minkowski reference metric, simply because I want to Lorentz invariant theory at the end of the day. So my Minkowski reference metric, I'm going to take to be eta mu nu. So this is just a fixed metric that appears in the theory. I'm going to denote this by a tetrad that's basically delta mu a, delta nu b, eta a b. And I'm going to denote the corresponding one form, simply by a matrix 1 a, which is equal to delta mu a dx mu, like so. All right, so now the non-linear theories of massive gravity that are free of the Bulwar Desert ghost take the following form. So you have your action is going to contain the usual Einstein-Hilbert term. So in this language, R, A, B, wedge E, C, wedge E, D, epsilon A, B, C, D, like so. And in addition, you're going to have terms that mix, that now contain no derivatives, but mix the dynamical tetrad one form with the non-dynamical tetrad one form, and that are simply going to be every possible way of wedging together these two one forms. So the mass terms will look like the following. So let me write this as m p squared, m squared integral, sum from n equals 0 to 4 beta n times these functions s n that depend on E, A, and 1, like so. That are defined the following way. So s 0, maybe I shouldn't use s. Let me call them m. m 0 is going to be the wedge product of every dynamical one form, which as we said, this is just a cosmological constant. So this is dead E. But now m 1, we're going to wedge together three dynamical guys with one non-dynamical guy, D, epsilon A, B, C, D. m 2 is going to be the wedge product of two dynamical ones with two non-dynamical reference metrics, like so epsilon A, B, C, D, and so forth. So m 3 is going to be equal to E, A, wedge 1, B, wedge 1, C, wedge 1, D, epsilon A, B, C, D. And m 4 is equal to, it's a non-dynamical term. But I'm just putting it in for completeness, the wedge product of all four of the non-dynamical guys, like so. And so this is what ghost-free mass of gravity looks like in a very similar form to the love-lock terms that we derived. So in this notation, it's not obvious to show why this propagates the right number of degrees of freedom. But in fact, we can go through and I'll show you. I'll give you a general outline of the proof of why this is the case. But this is just to show you that the structure that leads to the ghost-free theory for massive gravity is a very close parallel with a structure that led to a ghost-free theory for these non-trivial extensions of general relativity as well. All right, so just to say a few words about this theory. Sorry, I have till 1045. Is that correct? So just to count free parameters in this theory. So I'm starting with a theory that contains a Plink mass, which I have to specify. I'm allowed to include also this cosmological constant here. So a coefficient that appears in front of this M0 term. So lambda, like so. As we said this last term, the M4 is non-dynamical. And so it doesn't contribute to the equations of motion. So I'm left with three terms here that represent genuine mass terms for this theory. And they appear with independent coefficients. So beta 1, beta 2, and beta 3. I can take one of these coefficients and I can absorb it into the mass of the graviton. So I can define some little m. But then I'm left with two remaining dimensionless coefficients, so say beta 2 and beta 3 as well. So this is why you hear people sometimes refer to as these DRGT theories as being a two-parameter family of massive gravity theories. Because on top of the Plink mass, the cosmological constant, and the graviton mass, you have these two independent parameters as well. All right, and since I just have a few minutes left, maybe the last thing I'll do is just take these terms and switch back to index notation just to give you a sense of what these look like in a less formal language. All right, so we've said that M0 looks like just determinant of E. I can restore indices now on the rest of these terms. So write E mu A dx mu, take the wedge product, massage things to pull out a determinant in front. And I'm going to introduce the following notation. So I'm going to use square brackets to denote the trace of a matrix. So in particular, say square bracket of E inverse 1, this would be equal to E mu A delta mu A, something like this. All right, so M0 goes as determinant of E. M1 looks like determinant of E times trace E inverse 1. M2 looks like 1 half determinant of E times trace E inverse 1 squared minus E inverse 1 quantity squared trace, like so. M3 goes like 1 sixth determinant of E. So now we're going to have E inverse 1 cubed minus 3 trace E inverse 1 times trace E inverse 1 squared plus twice E inverse 1 cubed, like so. And for M4, this is going to go like 1 over 24 determinant of E. And now some combination of four powers of the E inverse 1 traces. So there's going to be an E inverse 1 to the fourth. Let me see if I have this in my notes. OK, I only got, no, I think I have it. Here we go. Minus, OK, so there's a 6 E inverse E squared trace E inverse 1 quantity squared. There's a 3 E inverse 1 squared squared. And then there's an 8. So trace of 1, trace cubed. And then minus 6 trace to the fourth, like so. All right, and in fact, so this structure is something that you can keep continuing to higher dimensions. But the idea is that these terms are really related to the determinants of matrices in various dimensions. So in fact, in one dimension, this is just the determinant of the matrix E inverse 1, this piece here. In two dimensions, this would be the determinant of a 2 by 2 matrix. In three dimensions, this is the determinant of a 3 by 3 matrix and so forth. And in fact, these expressions vanish. They're identically 0. If you try and take, if you plug in, say, a 4 by 4 matrix into the expression for the determinant of a 5 by 5 matrix. And so that's why this series terminates at this order. Yeah. Oh, yeah, let me go through it. I completely skipped the connection between the two. All right, so if I look at M1 and I restore indices, so this is going to give me, let's do it this way. So I can write it as epsilon mu nu alpha beta epsilon A B C D. And then I'm going to have an E mu A E nu B E alpha C 1 beta D, like so. All right, and I can use the fact that I can write, say, determinant of E times the inverse matrices. So some E, let's do, what's the best way to do this? Rho F E sigma G E lambda H E kappa I, something like this. I can write this. This is simply epsilon rho sigma lambda kappa, like so. Epsilon G H I. So I can use this expression to plug in here. And I'm going to get that so every time an inverse tetrad hits a usual tetrad, it's going to give me a delta, basically, except for the last one where I'm going to have an inverse tetrad hitting the identity matrix. And that's going to give me the trace. And so it's just a matter of massaging this, pulling out the determinant in front, and you end up with these different powers here. But the interesting thing about this expression, so the right-hand side doesn't depend at all on what matrix I'm using. So I could have pulled out a debt E in front, or I could have pulled out a debt 1 in front instead. And these expressions would have had a different form, but they would have been entirely equivalent to each other. But I'm just doing it like this to put it in a more familiar form. All right, so next time we'll go through and we'll talk about how to show that these are, in fact, ghost-free theories of massive gravity. But why don't I stop here for now?