 We started off with Li-algebra, we started off with Li-algebra that, okay, so, the actually run away, so suppose you got some vector space psi n, Li-algebra is a symmetric generator that acts on some column like u, psi prime, psi tilde, m is equal to u, m, n is psi n, okay. The groups we were talking about are the individual versions in the F-group of u, which we noted as u to the pi i epsilon a k, TAs were called the generators of the group, okay, and the associativity of the group were required, that the commutator group of the generators is itself a generator, number of TAs would be b versus the f of the C to Ci, okay. Then we noticed that if you look at infinitesimal group elements of the form e to the power i, let's say, theta a k a, and you act from these by similarity, okay, that present the infinitesimal nature of the group element, because the similarity transform acting on identity gives you identity, so something near to identity is near to identity, okay. So this action in this action of acting on a group element itself by similarity transforms a group element of the vector, right, is itself used a linear representation of the group, okay. In infinitesimal version, that representation was generated by, it was called the adjoint, but this representation is called the adjoint representation, in the infinitesimal form of that comes from the fact that this action here is essentially induced by this commutator, so the action, so if we put epsilon b Tb, an actor on this, let's say 1 T a, you would get Tb commutator T a, okay. So we concluded that the group matrices that T a the generators in the adjoint representation were simply given by f e, and the fact that these group elements of a, that they conveyed this relation was in particular, followed in particular from the jupyumbi identity of triple commutators, okay. So this stuff was purely algebra theory, nothing but physics, nothing but gauge fields, okay. Any questions or comments about this problem? Oh, we'd say one more thing, but it was algebra theory. We said that there were, that if you had an invariant quadratic form, an invariant quadratic form, which we might have left for you, that's what it's called, an invariant quadratic form, by that we meant that this object here, T a, this object was in there, so in which we meant that under, in any representation, this is some matrix, this is a matrix that transforms under, under action of the group, under action of the infinitesimal elements of the group, by commutator with those infinitesimal elements, okay. And we said that if this thing was invariant under the action of the group, okay, then we called this object here an invariant dense, okay. And in particular traces of D, A, D, D, in any representation of the group, for examples of invariant denses, the reason that it was an example of invariant denses is that it's not a commutator, trace of any commutator, we think of finite denses, okay. And we argued that if you call an invariant denser, H, A, D, this, then if you use that invariant denser to lower this index of the structure constant, then the structure constant becomes complicated. So perhaps in the last day, you may think that this is a good way to measure it, many, many invariant denses, you see because this object in any representation of the group forms an invariant denser. But you know like it's not that kind, it turns out that all of these are proportional to each other, okay. So if you work these up in different representations, you can say that invariant denser tends to different organization, okay. So typically there is only one, two index invariant denser, okay, yes. Sir, do I insert the indexing input that this is the invariant denser equivalent to one, one, two, three, four, six, seven, eight, the root thing we written as a sum of commuting evens, commuting evens. The direct sum of, sum of commuting compact, simple and evens other things. Simple algebra is different from U1, sir. What you're saying is that it's a U1's plus compacted, direct sum of U1's plus you could have these compact symbols, let me really learn. You see, the Lie algebra's direct sum of commuting compact, simple, that's the first thing. And U1, sir. Okay, no, SC2 is an example, that's not U1. Okay, good. So far though what we said works very angry. The thing that is very special for compact, for compact input, compact assembly groups is that it's always possible to find the basis in which HAP is just delta HAP. Well HAP is delta HAP lowering and raising, that's the thing. And so the fact that FAPC is anti-symmetric when once you reload means that FAPC was anti-symmetric even before you reload. So that's the second statement. Yeah, point B here is that it's possible to find the basis such as the structure constant anti-symmetric. It's always possible to find the basis such as the anti-symmetric up to reload. But for these special analysis, it's possible to find that such as anti-symmetric even before you reload. Because that's the same thing, the same as the basis in which the metric is delta HAP, at this invariant form is delta HAP. Okay, so far so good. Now we started, then we may not try to develop some physics. So we talked about an action, some field psi, some field psi n, that was invariant under some non-ambient, non-ambient group transformation such that psi goes to u psi. It was invariant under this when u was not a function of position. But we wanted to try to make a theory which would promote the dated variance of the u1 theory to a non-ambient. So we wanted to try to make a theory in which this psi goes to u x times psi of the dated variance. Okay, and we discovered that we could do that if we invented a covariant derivative, d mu which was a mu minus i a mu provided a mu transformed in the following way, a mu times 2, u a mu. I am not going to emphasize it, but I should. And let me do that. Okay, take this transformation which is the transformation for finite, for finite symmetry action and specialize it to an infinitesimum transformation. Okay, so I don't need to suppose u is equal to e to the power i epsilon of the function of x. Okay, so what is delta? What is delta angle? Delta angle is equal to t a commutator. Let me call it t a commutator. Try to guess something on the right opposite side of the left. Okay, plus we call it both ways, epsilon a. Now you see what is the infinitesimal order? At the infinitesimal order we can't keep the identity part in u. We need the epsilon part in u. Since we want to, the first order u inverse is just identity. Okay, so that's just plus i times the, that's a minus i times plus i, so that's mu i, so plus delta mu epsilon. So this is epsilon which is epsilon a times t a. It uses the following transformation on a. What was a? Was a a generator in that representation in which the matter feels left. I'm a bit worried that look we need a new gauge field for every representation of matter that exists in there. Because we needed this gauge matrix to transform like this and these u's were built out of that, the representation in which the matter transforms. But notice, if we write a as a a t a, okay, if we choose to implement the a is equal to a a t a, then this transformation law here, in terms of a a's is simply that delta a a a. So we just equate coefficients of t. Okay, so that's right. So this is delta of a a t a. This is the i and then there was this, let's call it, epsilon a n and then f m i times f m n a. This first i was okay. The next i was okay. The next i was okay. The next i was okay. Delta a a is equal to minus epsilon m a n delta m a n plus delta m a n plus delta m a n. This relationship here uses only the structure constants, difference to which representation the matter feels appeared. So the right-hand side as a mu subscriptor is the, all the a's are a's. It's a a a. So it was a guy in my class when I was an undergraduate when I was a student. But he slowed down a class. So a very complicated equation, put up with the answers, sir, the dimensions don't matter. Okay. The point here is that if you want matrices that transform like this, without having to introduce new matrices for every representation, there's no other way to implement it. Just choose a to be a a t a. Okay. With a a's some numbers. And then if these a a's transform according to this rule, then in every representation you produce the run. The clever trick. One gauge feels, just a little. You'd have to make a field. You'd have to. And then go in a derivative. You'd have to have a a t 1 a plus a a t 2 a. Exactly. So t always refers to the p in which the matter p starts. Okay. Different covariance. Different covariance from different fields. But built up with the same gauge field. Right. You don't need to introduce new dynamical objects for every matter p you have. One dynamical object, namely the gauge field is one single dynamical object. And then you vary it with different representation matrices. Is this clear? Okay. That this transformation law for constant epsilons is the adjoint transformation. So this tells you that the gauge field transforms in the adjoint representation. At least as far as constant. It's basically dependent gauge transformation. It's a concern. Okay. I'm sorry. This is to get that point clear. Yes. As to my fields and your d u, d l u minus i a u again. Or would you have something else? Okay. So suppose we've got two manifolds. Yes. Psi 1, Psi 5. Psi 5. Suppose Psi transforms in representation r 1. Yes. And Psi transforms in representation r 2. Yes. Then every derivative that acts on Psi would be d mu. It should be d mu minus i a u a p a r 1. Okay. But every time you've got a derivative from 5, d mu on Psi would be this. d mu on 5 would be d mu minus i a mu a p a r 2. So that each covalent derivative will transform homogeneously with the unitrix appropriate to Psi of i. The single gauge field does the same. The single gauge field by itself always transforms in the adjoint. So this is a rule for infinite as much as we can. It's slightly more complicated. Basically adjointion of this rule. Yes. Former. It's a scale. One gauge field that allows you to covalentize many different derivatives for matter fields in many different representations. Then you make a term to talk about the construction of the field structure. The ninth generalization of d mu a mu minus d mu a mu in work. It didn't turn out to transform homogeneously. Well, we generalize the following observation. We generalize the observation in d mu 1 theory. d mu minus i e a mu. Homogeneously d mu minus i a mu on 5 turned out to have no terms with any derivative acting. This is true also with an homogeneity. Okay. And since it's an object that has no derivative acting on 5, it's just an algebraic object multiplying 5. Since it's an algebraic object multiplying 5, that is guaranteed to transform correctly. Because each covalent derivative to transform correctly. This gives us a two derivative algebraic object. That is anti-symmetric. And in the Abelian case, it reduces to the usual field strength. So clearly it's a good candle for the name field strength. Okay. And we deduce the formula for it. So this was equal to i d z mu minus i mu minus i mu. So I want to do that. It is equal to a mu. And that was a mu. d mu a mu. I want to do it first. So that's not like that. It's good. So d mu a and then i minus i minus i. You might think, wow, we've got a new field strength for every representation we can deduce. But if we work out what this is in components, once again we write f mu mu a. f is equal to f mu mu a. That's the expression for this. It's easy to use. Basically it's d mu a mu a minus d mu a mu a plus minus i f abc a mu. Yeah, that's good. Okay. So then now which representation you choose? It's a geometric object. That's only a negation. Okay. And we decided to try to do some stuff with some physics. So we tried to write down some sort of Lagrangian, first involving justifications. And the Lagrangian we don't know is the so-called Gengler's Lagrangian. Okay. So after some number, which we'll come to in a moment, we had trace of that. This we could write as this here of TAPP is an invariant tensor. Okay. It's this h a. We come to the important point. So this field is up to some number proportional to h a. Basically the unique in that. And now we come to the important point. The important point was that look, as we discussed for compact series of building groups, to make h, choose a basis such that h a b was one. More generally, h a b, compact series of building groups is a positive definite quadratic form. But for these non-compactly groups, that's never true. s u 2 comma 1. I said 3 comma 1. Let's look at an example. Okay. Well, this h a b is like an integral space. Okay. To look for the minus one. And what is that? Now, if we try to build a quantum theory based on this, we have the same problems that we had for that particular, those particular years for which the h a b was minus one. We would have the same problems that we had when we tried to quantize the scalar field with the wrong side. Then we would get unbounded disparity with the problem of unitarity. Okay. And we don't yet know how to make sense of these theories. So these theories by themselves, unless you accompany them by, you add to them some other principle which cuts out some of their states or something like that. Sometimes it works. But by themselves, just the theories by themselves, they do not, we don't know now how to make sense of them. We will not consider them by and large in this class. Yeah. So we will study basically theories in which this h a b is positive definite. It's the same thing as studying these positive states as a compact set is a positive. Yeah. Okay. By the way, by the way, there is a classification of all such groups. Okay. Or there's a classification of all the compact set is simple. Yeah. Okay. You know, of course, you can make a compact, the algebra of adding two. So we're interested in classifications, irreducible classifications. You know, things that you can add to build more. And it's a very simple classification. A simple classification basically says that there are three, sometimes called classical groups. U n, the space of unitary n cross n matrices with the term trace, trace 0. No, in the term 1. Fine. Okay. There's s o n. There's x p n. These are the obvious examples. You know, anyone fooling around with matrices will discover these examples. Okay. In addition to these obvious examples, there are a few exceptional guys that are not parametrized by a number like n. You know, un is an infinite series of such, sequence of such. S u n is an infinite sequence of such groups. Un is not a character. That's s u n, s u n, s u n. S u n, s u n, s u n, s u n, s u n, s u n, s u 3, s u 4. All the way up to infinity. Same for s o n, same for s p n. Okay. But these exceptional guys are really discrete exceptions. The biggest of these exceptional guys is a group that plays a big role in string theory. It's called E8. And it's sort of a big group, but it's not that big, you know. It's one that you can get to know and love. Unlike actually in the classification of discrete groups, have you ever seen them in the classification of discrete groups? This thing called the monster group. It's really big. Nobody gets to know and love the monster group. Okay. So, you know, the space of all possible re-algebra that are relevant for building in the simplest context, the geisha space, is a very civilized space that's listed, that's listed, you know, in a page. So, anyone who wants to learn more about re-algebra as a physicsy context, I recommend this book by our George I, called Lea Algebra as a particle physics. It has a lot of very nice stuff. He will in particular take you through the classification of these re-algebra. Okay, great. So, in order to specify one of these non-linear geisha, you need to specify the group, but this is a classifying space. Okay, and fine. So, this was the action. But the last thing that we did, that we studied in the last class, was that we studied what the Hitler space interpretation of, what we studied with the Hitler space interpretation of this, of the pathological basis of such an action. And if you remember, what we found was that it was very much like the human theory. We found that the Hitler space was a framework like quantizing the spatial oscillators in one, two, three, four dimensions, subject to constraint. And the constraint was that all wave functions had to be invariant under gauge transformation law. And the gauge transformation law was simply that the gauge transformation law was simply this action, or the finite version of this action. Okay, that any wave function that's a function of A is had to be invariant under finite version of this action. The generator, you remember, what we're trying exactly is. This, by the way, is covariant derivative. So, I think it's the covariant derivative of this action. Any questions or comments about that, we'll probably move on to... We're going to study this, the path integral that defines an unabated gauge theory in a little more detail. Brush, and try to understand this path integral. And that formal brush was, as we just discussed, the path integral to the path integral of A is and the path integral of A is Europe, giving a reverse phase interpretation of the path integral of A is, and then a projector interpretation of the path integral of A is. But there was one issue there, this projector interpretation, which let me bring up as an analogy. You know, suppose we've got... Suppose we've got a theory that lives our self. It's suppose we've got function that represents. And I've got to project only to those functions that are constants. One way, one crude way of bringing it is to say, well, the projection that does it is the integral, because you've picked out the zero. It's not in the projection. But as you know, if I really wanted to make it a projection, I should actually divide the space by A. Because if it's a projector, then when acting on a constant, it should give back the constant. R is the radius of the set. The volume of the space. So when you do this integral, normalize the norm, to include in the denominator the volume of the space over which you're doing the integral. That you take, you take greater projectors. Now, if you remember our discussion that gave us our projector, we were doing this integral dA0. Exponential of I times cI times dI dA0. This was a generator of gauge transformations generated by this integral, dA0. And we were claiming that this was a projector. And that's true up to a normalization. Okay? This sound over, you know, it takes your gauge transforms. It sums over all gauge transforms. But you're far from guaranteed that if you do this in any kind of a way, you will have generated a group projector that has the idea of generating the normalization. But we're integrating over, we're integrating over all gauge transformations. All space-dependent gauge transformations. Now, if you remember, in our interpretation with the path integral, we got one such integral at each point each slice of time. Now, if the normalization was that of a projector, it didn't matter. The space-quad was clear. But if our normalization is not right, then we get extra normalization factors at each point of space-time. So, what is our extra normalization? Our extra normalization is essentially, as you might believe, the volume of gauge transformations that depend on space and time. That's my spend on that operation. That's that. So, the volume of space-time with any gauge transformations. So, you might think that the correct path integral, the path integral that we want to compute that will have a good Hilbert space interpretation is dA mu times d psi. There's a matter of means to be just to give you a case name for one moment. Divided by integral d omega, where omega is the space of all gauge transformations and depends on space and time. Yes, the volume of gauge transformations. But there was another point of view that also suggests that this is the case. Because, as all of you know, by construction, animal's action here was invariant under A mu was to view a obituary group element to you. Space-time dependent group element. Part of that group by itself, if you try to evaluate it in any naive way, it's just going to give you infinity. Why is it going to give you infinity? It's going to give you infinity because given any gauge configuration here, they have the same action. These are gauge configurations related to A mu by gauge transformation. The integral over this gauge transformation is going to give you a factor of the volume of the gauge group. Dividing by this volume of the gauge group is precisely removing integral over the gauge transformations so that you can get some finite elements. As you know, roughly speaking, what we want to compute really is about the integral which is the integral over A mu divided by integral over the volume of the gauge transformation. Please. We're only going to look at fractions and you're only going to cancel them. What? What? What? Yes. You see, but you see, it's true. If you cancel them, that's what you mean. If you mean lattice-gauge variables, you wouldn't worry about this. But, you know the fact that there is a zero-mode direction. It shows up at every order, including quadratic. What this means is that the quadratic part of your gauge theory action is not inverted, okay? Which means that if you just try to do perturbation theory first, without dealing with this issue, it will be ill-defined, have all kinds of problems. So to make sense of analytic computations in this theory, it's actually necessary to get rid of this, this zero-mode, okay? What we're going to now do is a trick that we get rid of this. It's called the funny pop-off trick and it's very simple if you're going to do a surprisingly long time for people to understand. Roughly what to do is first understood by finding them. Just from consideration of unitary. Just make things unitary. You roughly understand what should be done. Then the simple logic behind this could sometimes be... Is this somewhat similar to what we do when we divide the vacuum bubble diagrams and we do the ethnic tricks? No, this is more serious. You see, that as was asked, we don't have to worry about it because we can just do the calculation without doing some division. And then divide by zero insertions, okay? But if we didn't do this in the first place, we wouldn't be able to do any calculation. The reason is that these zero-modes, you see here, we wouldn't be able to do any good development. Why is that? Let's understand what a perturbation theory is in terms of a path integral. See, a path integral. A path integral. So let's take one minute of deviation down to the major perturbation. Okay, so suppose you have a path integral with some parameter h bar. Okay? Now, the rough idea is the following. Suppose I want to compute some path integral as if I were going from 0.8 to 0.3 in some time. Now, there is a classical trajectory. Now, as the property with small deviations from this classical trajectory would not change the action. Every other trajectory almost cancels out if this h bar is very small because of very rapid oscillations. This is the steepest descent over this stationary phase of optimization. Okay? So perturbation theory is the attempt to make that system active. Is to start with assuming you're on the taciturn trajectory, which in very simple case, which is same. Okay? And look at small fluctuations around the trajectory systematically. Okay? Now, if you're running exact zero-mode in this problem, exact zero-mode, of course, those modes are not suppressed. There are no fluctuations. Okay? So perturbation theory is going to break down. As we've discussed, if you regulate this issue, it's not a serious issue. It's not an issue of principle, but it's an issue of practicality. Practicality. Okay? So practically, it means this is more important. This is then natural. Okay? Now, other questions. Okay. So, now, we're going to try to look at a trick that has to deal with this thing practically. Okay? So, the trick is the following. Look. Let's draw a picture first. What do we incur? Will any gauge be laid new? There are all the other gauge fixings that are related to it by gauge observations. Let's call all these gauge, other gauge fields gauge-orbitancy by gauge-orbitancy. Okay? So, what we've seen is that physical states are states that are independent of where you are on the gauge-orbit. The wave functions don't care. They're not functions of positions on the gauge-orbit. They're only functions of which orbit you're on. Okay? Which is a physically state, a characterised by which gauge-orbit you're on. And not aware of the gauge-orbit. That is the reason. That's the underlying reason why we have to decide by these gauge-orbits. That's why we've got these gauge-orbits. They parametrize on physical image. Okay? So, imagine we've got this set of gauge-orbit. What we want to do is to do a positive integral really only over individual gauge-orbits. Okay? And one way of doing that practically is to draw a line, some line, that goes through this place of gauge-orbitancy and cuts each orbit exact truth. Even manage to draw such lines. Then what we need to do is only to do the integral over all these configurations. On this line, do it with the right weightage. With the right weightage of each point in life. Because as you see, different gauge-orbits could amount to different volumes of field space. As you see, visually it's actually. Let me give you an analogy. Suppose you were interested. You had a function in two-dimensional space. That was function only of radius. You want to do the integral over the whole space divided by 2 pi, which is the volume of rotations. Okay, I'm doing this to say, well, let me just do the integral of the function only over x. But if you did the integral f of x divided by 2 pi, you'd be doing it totally wrong. Or maybe not divided by 2. You'd be doing it totally wrong. Because you've come to the major factor. The right way is, of course, as you know, r. The r, r is r. Let me say that. That's a useful analogy. Suppose I want to do, that functions only for r. This is the analog of gauge invariant. Gauge invariant action. Okay? The analog of the gauge group is the group of rotations. The volume of the gauge group in this case is 2 pi. It's a symbolizer. W. What you want to do is to evaluate d2x f of r divided by integral p of a r. This, as we know, is that you'll need 2x divided by 2 pi. We're going to do this in a row. We change from cartesian to polar coordinates. Okay? Then we do the integral over over theta. That's 2 pi. That's r d r. That's 2 pi. So the right answer is r d r. Now, you might crudely have thought, well, you might crudely have argued, well, let's look at this. My function is a function only of r. So I have one slice. Let me call it the x-axis. So I'll integrate from z over to infinity and on the x, and the answer should just be vx f of x. That's wrong, because we put on this r. Why we put on the r? It's one that different gauge organs, different rotation organs, and different sizes. As you rotate over, a particular gauge organ, in the rotation organ, you've got a larger gauge organ in physical measure space. As you go larger and larger. Okay? So you have to do things correctly. But if you do things correctly, you can get the right, somehow you get the right x factor, then you have to hit right. Just choose any. Any slice that cuts the gauge organ once into the integral. That's what we're able to do in a more complicated context. Is this clear? So this is actually very easy to achieve. By the following, by the following device, suppose you invented some function whose zeroes, whose set of zeroes are this line. In this case, the function will be y. Okay? And then somehow we have to block out this one. It's s. The y is equal to zero and something like that. Okay? Suppose you invented some such function. Then one way of restricting you from just this line is to put a delta function inside your bottleneck. But you don't want to change the bottleneck. Okay? So suppose you invented some such function. This function is some function which usually chosen to be a local function. And we need this function to be one for each point in space, space time, because the amount of gauge invariance we want to fix, the amount of gauge invariance we want to fix is one adjoint value field point in space time. Because epsilon a of x parameterizes your gauge invariance. So if you want to fix that gauge transformation of big duty, you need as many conditions as there are in big duty. Okay? So suppose you got some such, some such function. Can we even discuss examples a bit? One example if you can mind us because all you need is the example of such. Given this f of a, what you do is a following. Do an integral over the gauge rule. Okay? With the delta function of f of a x, gauge transform with the zeroes of this f of a, defined once a slice in gauge space. By definition in this integral over the gauge orbits, this delta function will click otherwise you're not going to get a good slice. Okay? So you're going to get this delta function clicking once at a difference. Because it clicks, you won't get zero. But this integral is not necessarily one. Because you can integrate an delta function dx, delta of f of x. If you don't get one, you get this is one over mod x prime of x. This is only one. Whatever stops it from being one, we will define the delta of delta called the Paddy Pope. That's all. So let us define this Paddy Pope of determinant by the identity, by the requirement of all the things. Then this is the integral. So the Paddy Pope of determinant, one over the Paddy Pope of determinant is this, the Paddy Pope of determinant is this wood in this example will be mod f prime of x. Because this integral is one over. It's saying something that's utterly obvious but I'll prove it in an algebraic way if I'm actually confused. Suppose we now are as k. Suppose we, on this equation, this equation which we think of as a definition of the Paddy Pope of determinant which by the way of course depends on n. In this equation we make the following change of variables. We say that a be replaced by a twiddle gauge transform. Let's just make that change of variables. Then inside my definition that the omega delta of f of a of a twiddle gauge transform. Okay, now we use the difference. So this is gauge transform by gauge transformation. This denotes a gauge transformation. Times omega delta f of p of a middle gauge transformation. And then for the twiddle. Mineral gauge transformation. Because there's an integration variable. Gauge transformations. You see, what are you doing here? First, here you're doing the gauge transformation by the twiddle gauge transformation. And then the gauge transformation by the gauge transformation will be. But that's the same thing you want times you do. Gauge transformations from the group. It can be multiplied. And if you've chosen this integration measure to have the property then it is both left and right invariable. That is a change of variables which is a change of variables by multiplying from the left or multiplying from the right by another group transformation. Because this group is now a gauge transformation. It's an invariance of your measure. And this we will always say it was always possible and always choose such a measure. Then you can just make a change of variables to the compounded gauge transformation. As your new gauge transformation. So then having made that change of variables and changing the dummy index and stuff like that. The gauge transformation I think on a, right, not under function. You see the function is a function only of here. That's fine. But what I mean by gauge transformation of a function of a is act on the a with a gauge transformation. There's no other notion of acting on only fields transformation not functions. Become as many derivatives as you want. Doesn't act on them. Is this clear? This was a dummy variable. We can get rid of them. What we've got now is the identity of this equation. You see they're actually the same. Only if this and this are the same. The proof is that this Fadi Popov gauge and Fadi Popov determinant is a gauge invariant function of the gauge. A is a variable. Yes? What? Are you going to function the same? No. But now the integration variable is just an a. Let's go through it again. What's with the the definition? The definition isn't a definition. Yes, it's a definition. No. This is not a function of a omega. You can take that out. Let's just go through this one. See, this is our definition. Now on this definition I'll take a change of variables on a. On a, I call that gauge transform of the Monday. That by definition gives us the identity of the gauge transform of that Monday. And then I show that this part here is unchanged. And therefore the Fadi Popov determinant doesn't care about gauge transform. So the delta FD of a is equal to delta FD of a to the omega delta. I just proved that delta FD of a to the omega delta. Okay, let's do it. No, because a and a to the omega delta are the same thing. Okay, let me say this part here by definition now See, I've dropped an a to the somewhere. This was an a to the delta. This was an a to the delta. This part here times delta FD of just a to the is by definition one. Okay, but we also see that it is one with delta FD of omega a to the omega. Therefore this is actually think about what this is saying dramatically. In terms of orbits, it's totally obvious. It's totally obvious. What it's telling you is that the Fadi Popov determinant is the same at different points of the age group. That's clear because you see in the integral you only click once. You only click when you hit that slice. That's not a very common problem. That's an utterly obvious statement but this is just an algebraic proof over the notion of this Fadi Popov determinant whatever it is. We're going to proceed to have an evaluation of the angle of the particle there. Let us proceed. So, this is the quantity we wanted to evaluate. Now inside this quantity let me insert dA mu d omega and then let me just insert one. So, I'll insert it. Integral d omega delta of F dA gauge charge from respect to omega. Let me take this outside the integral so that's I s of A divided by integral d. There is no particle that we want to compute. So, all I've done is multiply the particle I want to compute by one. Now, I make a change of variables. Let A omega equals take the variables as such that it's an invariance of the equation. If you regulate the probability. Gauge transformations are symmetry not just of the action but also the measure of the quantity. Now, if that's the case then the measure is unchanged. So, this becomes dA delta is integral d omega this is delta of F of A with because A delta was A omega. Okay? And then we get delta of A d of omega inverse and e to the power I s of A d of omega inverse. Because the value of the determinant and the action of the matrix of A not A delta. And A delta is A delta times omega. But this would be too careful. Because we've seen that the value of the determinant is gauge invariant. And we know the action is gauge invariant. We also have this integral division by integral of d omega. And now suddenly we see that nothing actually in the particle temperature depends on that expression tells us that this new variable A tells us that. Nothing in the particle temperature depends on omega. So, we can just cancel the scratch. So, in the particle temperature that we set out trying to compute we have shown is the same thing as integral of the A delta of rewriting once again the dummy variable is re-naming the dummy variable A from A. The dummy variable times delta but what have we achieved? We have achieved what we set out to do. This delta function will ensure that we only integrate over the slices. And this value of whatever it is accounts for the volume of field space contained in one gauge of it. The change in this function is the angle of the r and r v r not too much. This is pretty formal. This is pretty formal because it looks pretty formal but it is not so. You see the only thing that is formal about this object is this delta. This is useful unless we can write an expression for delta p. But there is a beautiful expression for delta p. Let us suppose that as imagined that we have chosen our gauge so that delta fp of a is z Now let's make small infinitesimal gauge transformations out of that. Suppose the variation of delta fp because of an infinitesimal gauge transformation labeled by epsilon b is delta fp is equal to let's say some and this is a a of x comma y a p of x comma y epsilon b of y. So what else can it be? It has to take this form. So we don't need it for what I am saying in all practical applications of business. We will choose an a15 condition so that this a is a local function. Local. That may be derivative. This is the expression. What is the expression for this fp of the derivative? This is just the generalization of the one over more f prime that we discussed. It is simply one over fp of the derivative of this operator. Actually the space of adjoint value functions. This thing is sometimes written as this aab may be thought of as a functional derivative with the respect of x with respect to epsilon b of y. By definition of functional derivative the first order change, epsilon is this functional derivative and the derivative of the derivative simply the measure factor for the delta function is one over the determinant of this object. Remember the determinant of both aab space as well as an xy space. It is an operator ideal in the space of functions. Capital A is the derivative. No, no, no, the capital A is not the derivative. Capital A is whatever it is. Capital A is this. Against me the factor is only one aab. This is an operator. It is two aab. So, are we happy? This part of the determinant is this object. Now, okay, so you have got an expression as a determinant. Why is it a determinant? Why is it a determinant? It is because of the formula of delta function. You have got a multivariable delta function. Delta of let us suppose fm delta of fm of some number of variables that is called yj. Okay. And you want to do the integral pi of dyj. So, it is very important. Now, as many fms as yj is because otherwise you do not have a determinant. Okay. Then the formula of this delta function is the generalization of 1 over f prime. So, this is equal to 1 over the Jacobian delta. From Jacobian prime. Exactly. So, 1 over the Jacobian that gives you del fm by del prime. This is the formula that you can easily derive by as long as you are changing that. Okay. I have done exactly the same thing here. Except that it is an infinite dimension. It is this functional determinant. Okay. We have to actually go ahead. Our postdocs are leaving. But, this formula by itself would not yet be too useful. However, it actually becomes useful when married with something else so, that would be a very interesting formula for Krasmanian. So, this is the formula we are asking to prove. Prove the following. Prove that if you got di, dcj product over i product over d e to the power di in ijc. Okay. Then this up to some plus minus sign d is and c is a Krasmanian. Otherwise, it would have been doubtless. Exactly. The same formula would be true for Krasmanian. There are some signs. The same formula would be true for Krasmanian. So, Krasmanian quadratic integrals. As simple as Krasmanian quadratic integrals. You get determinants only up rather than determinants. Now, what we want for the Fadikopov determinant and so, this formula, this determinant can be re-recognized as exponential of we have to go down, right? I am coming. Yes, yes. The determinant of P a of x delta x a by delta x on B x y c B of y. B B Now, instead of writing this Fadikopov determinant, we replace it by a passing table where the passing table runs over the gauge fields. As well as two new auxiliary adjoint valued B and C of at least standard passing table. You can use the delta function sometimes we just do the delta function. Let's go and initiate it. Okay. And do the passing table. So, then you have to slice. And then you do the passing table over A as well as these B's and C's. These B's and C's are called constants. Okay, we better stop now. Okay. But we continue one more time. Okay. Absolutely. Very important one over here. You have turned it on.