 to do the solution of the thought model. And now I want to proceed to the study of Yankee's theories in higher dimensions. There are many preparatory things that I'd like to do to set up a study. Part of the complication of dealing with Yankee's theory analytically, for instance, in the loop calculations. As we've described, is that gauge invariance is a big thing. We need to fix gauge invariance in order to do a calculation. And then there's formulation. It's very important to deal with gauge invariance, actually, even when you're regulating the theory. But you see, as we try to emphasize, when we write at the beginning of the course, when we looked about the table and tried to quantize it, if you do a Lagrangian like d mu, a mu, all d squared, and try to set up a particular space associated with that, you would run into trouble. Because a0 would give you states that either have energy above and below, or a non-unitary. That would be some, if you try to fix up one problem another would arise. So we got along this problem by working with the gauge invariant action, read the action by F mu and a little bit squared. Now, suppose we introduce a regulation scheme that regulates gauge invariance. Then renormalization group flow. If you start with some bare action with some terms by some small amount, then it's fine. Now, gauge invariance is not there, any problem. Renormalization group flow could, in principle, generate all kinds of operators, including operators like this one, when you conduct low energies. And you could end up defining a theory. In fact, you generically will end up defining a theory. There are problems, there are sicknesses. So it's very important when dealing with gauge theories and when dealing with their renormalization to deal with gauge invariance exactly. And we're going to start moving towards understanding how to do that. Now, there are basically two different ways. One way that is used in most analytic calculations is the way by fixing gauge invariance, like we discussed by Padipa Bha. And then noticing that the gauge fixed action has a largest symmetry, so-called BRST symmetry. And adjusting your renormalization, I mean, working out, this symmetry is automatically preserved by every normalization group flows that helps you to understand various things. This is one of the things we're going to try to understand. We're going to try to, by the end of next week, we will have completed the beta function for the angle theory in four dimensions, for instance. But there's some machinery we want to set up before But there is another way of dealing with this problem or dealing with the radiation of the angle theories, which is very much used in numerical calculations. And it's also used for various conceptual ways of thinking of the theory. It's called lattice gauge theory. So I just want to tell you a little bit about lattice gauge theory today, and to do a few little calculations, with that just to get you to use the theory. Okay, let's start with the ideal theory. So the ideal theory is designed by the actual leader. Okay, let's put a 1 over 4 G squared and in my scientific mentions, I have... Now remember, of course, that this theory, that feels it, feels MU, we add an invariance under MU goes to... is to do something quite interesting. What we can try to do is to make space-time discreet. We'll try to make space-time discreet. But to do this in a way, so that we preserve a discrete version of this paper. First, we've got some lattice, I'm drawing it into dimensions, but to be mentioned, this lattice discretizes space-time. Points in the lattice are labeled back here, labeled points in space. Let's imagine that it's a cubic, a cubic lattice of size, these MU fields, these MU fields in the problem are vector fields, but they're in little arrows. Okay, so they naturally live on links that go between points in space. Okay? Of course, you can put the little arrow in any direction, but once you know basis for a particular vector, you know the vector, okay? So this is A1, this is A2, this is A3, that's all. So what we would stupidly have done is to just introduce an MU field that lives between here and here, and associate it with here. A slightly clever version of this. And that is to work the same with the object, phi, which lives on a link, so these links are labeled by N, okay? Which should be thought of, which should be thought of as approximately e to the power, okay? So let's say this phi is labeled by two points, it's a link, so it's labeled by X, and X plus NA. N is one direction vector, it's either the X hat, Y hat or Z hat, okay? A is the size of the direction in which we're going, and this we will think of as approximately e to the power RA, A of X times A, but A was a DX basically. Now, what do I mean by this as approximately that? What do we do just to try to work out what the gauge transformation property, what the gauge transformation property of this object is? Okay? And remember, you should think of A as being very small. Is that right? Okay? So under wave transformations, we know that A goes, A transforms as A goes to A plus DB lambda. So let's write this in terms of finite nature. Okay? So A goes to A plus, so suppose I define a unitary matrix, e to the power I lambda phase. So if you have charge field, the charge field of charge N would transform like e to the power I and lambda things. Okay? So this object, which is not even carrying if we promote it to a unitary matrix, we promote it, we just call it U. It's just a base. Ux. And this is the gauge transformation property. A goes to A plus DB lambda. You can be looking equivalently as U inverse. We can even transform the gauge transformations. So what was the motivation for defining finite nature? So this is sort of the blocks theorem which we do by doing the chronic programming model in some good state. Like is there a motivation for that? Motivation for what? For writing phi as e to the power. Give me two minutes. Let's see the result that we have seen. Okay? Hang on for one. Okay? So now let's look at how what? Phi transforms under the gauge. Standard gauge transformation. Phi prime is equal to phi times e to the power I tell mu lambda A simply because A goes to A plus del mu lambda. To first form the name, I can write as is equal to U inverse of X. Phi, this is remember X, X plus N. U of X plus N. Let me change my gauge transform just to make it more standard. Let me change my gauge transformation to A goes to this or A mu goes to A mu minus del mu. Then this will become U. The motivation is that it has this nice transformation under this U. Okay? Phi is our scalar fields. Phi is our scalar fields that live on each link. There's some fields that they live on links. Yeah, that's just something. Okay. Now, hang on. If you hang on two minutes you'll see where this is going. Now we have this nice object and we're going to use this observation that e to the power i a x i a x times a to lowest of an i a has this gauge transformation property to form the basis for our discretization of the gauge. That is, what our discrete theorem will be defined by is a theory whose basic variables are fars which are postulated for which we would write down an action which will explicitly contain this gauge of variance. So for our discretized theorem this gauge of variance will not be approximate as it was in the continuum with its corrections at higher orders d squared. But in exact, in the limit a goes to infinity so this will be the discrete version of the gauge of variance that we will preserve discretizing the property. So we will have a theory which will only be able to generate gauge invariant operators under the normalization of the flow because as an exact theory preserves gauge impedance and made of this discrete variable. And that is a that we will engineer to be a discretization of the model b. We wish to study. Is your taking a RG flow of a lattice that we should take? As we you stress over this very different from taking a RG flow of a material. Yeah. So actually what we were you know when we when we when we do work with a lattice we will never really take RG flows. Okay. We will just define the theory thinking of this as a variable action. Okay. Imagining that there is some way of doing the RG flow. You know using the same framework as the previous thing. But you see when we when we discussed how to define boundary theory. Okay. What we discussed was that all we have to do was to take a particular limit of a bare action. So as some correlators were heavy fixed in the eye. That we didn't know on the lattice. And that's going to give us a definition of the RG flow. Even though it will be hard to control as you say. The explicit RG flow. Okay. So we will not actually be writing down RG flow. Okay. The a is over here are the objects which live on the points where the fights are the ones which actually live on the aliens. Wait, wait, wait. Nothing lives on points. Aliens, aliens use also conceptually into things. But now when we go to the discrete theory we don't have any of them. We only have fights. So in the continuum theory we have aliens. Which were vectors. So they were always little atoms. In the discrete theories we got a scalar field per link. So those are also little atoms. The value of the gate of the vector field on that thing. Okay. Except if we are not taking the value of the vector field itself as well. Yeah. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. Now, if you try to preserve, basically what you want to do is to find is to preserve gauge and variance, not in an approximate way, but in an exact way, it is just that this object is a very nice transformation property that links you to, so see once we discretize that property, as you said there are points and there are links, but there is something that lives on points, what are those somethings, those are gauge transformations, gauge transformations are associated not with the direction, but with the point, it is u of x, the point about this phi object is that transfer, now because we discretize, we don't have u of x is in the middle, we only have u of x is at these points, and the point of this phi object is that it transforms nicely under the remaining discretization, in some exact way, you see, so now we've got a new set of variables, we've got these phi's labeled by beginning in x points, and we've got this gauge invariant, u of x that lives on each set, we want to build some sort of action that preserves the gauge invariant, okay, and a phi of x1, and suppose we put a variable which I just did not buy, this phi is just the phi that goes from here, how does it transform, it transforms as u, universe, u here and universe here are totally unrelated, how are we going to make a gauge invariant, you see, if you think about it, you see there's only one way, that is by making products of these, such that these products form some closure, okay, so that this phi, the universe of this phi, we can't be cancelled by the u of this phi, and so on, so the expressions that we write down in our action, okay, will have to be products of phi's over some loop, products of phi's over some loop, that will be gauge invariant, any polynomials of these products of phi's over some loop, very functional products of phi's over some loop, will constitute a gauge invariant action for this discretized problem, okay, now we have two options here, we can discretize space, either only space, or you can discretize space and time, and the moment I'm imagining my Euclidean, discretizing space and time, I'll very soon move over to the discussion of discretizing only space, so I'm going to have a little bit of that discussion, okay, but let's work on this one, okay, let me look at the following, let me look at what we get, okay, when we take the simplest little bit, okay, let me look at what we get when we take the simplest little bit again, there are these, oh by the way, associated with each x and x plus a, we have this phi, but of course you can ask, there's another way of going down the rounds link, that's x plus a to x, you can ask what is phi over x plus a to x, it sounds too much like two variables living on a link, and yeah, basically we divide phi over x and x plus a, it's going to minus of phi, it will require x plus nk, no, so minus inverse nk, and yeah, maybe phi is not the best, I'm going to switch notation to, I want to think of something that reminds me of a phase, let's call this w prime, okay, but w is just a, w is just a, remember the phi is object, it was just a phase, okay, so w is a phase, and then sometimes we think of w as e to the power, now, if we have this phase, it will be the other way, you know, the negative phase, the inverse is just, inverse of e to the power i a is e to the power minus a, okay, so this will be the statement that, what we define this thing, phi over x, x plus a n is equal to minus of phi of x plus a n, where phi is w z, is this clear, just evaluate this object, assuming that w was e to the power i a x, d x, so, the approximation we have, roughly speaking, you should think of this phi x, x plus a n as approximately a in the n hat direction, and then, no, w is the exponent, I'll write it, no, w was the exponent of a, and was the exponent of z, phi is this, just evaluate what we get for this little, this little basic line, in which you call, let's say, the one direction, two direction, reverse one direction, reverse two direction, this is exponential of, that's exponential of i times, and now, when you go to one direction, this was this, this phi object, which was basically a 1, and let's say, this was at, we start this loop at x, so this is x 1, x 2, this is x 1 plus a, x 2, x 1 plus a, x 2 plus a, and this is x 2 plus, x 1, so this is a 1 at x 2, and from x 1 to x 1 plus a, let's just leave it here, this is the beginning point, a 1 always moves in the x point, okay, now we get something from here, but before that, let's fill in this guy, this guy is the reverse, this is minus a 1, but at x 2 plus a, because this is reverse of this hat, this minus of this hat, similarly we will get plus, this whole thing will be modelled by mainly, so we will get plus, this is a 2 at x 1 plus a, x 2 minus a 2, x 1, x 2, so if we make the approximation that phi is equal to a times little a, then this basically will look like a, just evaluates to the subject, let me do x 1 plus a, x 2 plus a, what, the third term, x 1 plus a, x 2 plus a, because it's a, now we have lately introduced by when we start, we are leaving the a's by the starting point of it, is this clear, what should be there, let's look at each of these, this a 1 here was labeled by x 2 x, I should have said x 1, x 2 plus a, x 1, x 2, x 1, x 2 plus a, so it's labeled by the starting point x 1, x 2, this guy is labeled by the starting point, so it should be labeled by this, we will be using the factors minus of this, so it's minus of this starting point, this starting point is x 2 plus a and x 1, similarly this guy here is a 2 which starts here, so it's x 1 plus a, x 2, minus of this guy should be labeled by this, because it's minus of this, it's labeled by this, which is x 1, x 2, is this clear, what is the difference between a 1 and a 2, a 1 and a 2, are these two different functions, yeah, the two different functions, this is, you have got a vector field, it's a mu right, so a 1 and a 2 are completely unrelated functions, completely distinct, so the a's are like the wave vectors, like when we like for example a plane wave solution into the itx, so the a's are analogous to the k's, see there's no wave functions or anything, yeah, this is just parameterization of the very good problem, limit, in the limit that this guy, that the little a is small, we can approximate this for the first term in the Taylor series, okay, so this guy here is del 2, so this object is approximate, e to the power i, okay, del minus del 2, minus del 2, a 1, okay, and this guy here is approximately e to the power i plus del 1, a 2, and then the whole thing is all the way square, so it's already an a here, this is the a to the a, if you look at, so this, this is again, is approximately e to the power minus i, that's 1, 2, where 1, 2 is equal to del 1, a 2, and del 2, we have product of these four files, product of these four w's, so this w's exist of it, exist of it, exist of it, exist of it, because you see I'm going to build my action from products of w's over placents, over, over parts, those are the most general gaussian variation, now I want to understand what this action that I get will reduce to in the continuity, this is the prepper traces after that, and in the simplest little product, that is a product over an elementary product, and see what that reduces to in the continuity, so what has it reduced to, it's reduced to the exponential of, it's reduced to the exponential of, it's reduced to the exponential of the field strength times the area, so basically the integral of the field strength over the little area level back up again, okay, what's going on here is of course very obvious, what you're seeing here is a discrete version of, I mean, is stock stock, because this thing was e to the power i a dot x, and stock's theorem tells us that a dot x is equal to f, f integrated, so that's okay, that's okay, that's what we're seeing, what I'm trying to do is to build a placent action, that approximately in the little a goes to zero limit, okay, reduces to f means, how do I do that, well, let's look at this object, suppose I take this object e to the power i f 1 2 a square, plus e to the power minus i f 1 2 a square, okay, minus 2, so this becomes the lowest order in the data expansion, okay, so to lowest order in the data expansion, this becomes 1 plus i times this plus the square, now the 1 and the 1 from there cancel the 2, the linear terms cancel the 2, but the quadratic terms don't, they add up, okay, so this becomes i squared and f 1 2 squared times a square e to the power 4, the half of the data expansion cancels the fact that there are 2 of these, okay, so now we're slowly getting into business, okay, we're slowly getting into business, so look, so now suppose I take the following action, I take the action that is this placent, what do I mean by that, that means the product of these four values, plus the inverse placent, that's the product of these four values, that's just a minus of this, I mean the inverse of this, this kind, okay, minus 2, sum of this over all the placents of the, all the basic placents of the problem, do you see that that will approximate minus f mu nu squared e squared, okay, what internal cancel, this internal, I mean if you sum up all those placents, it seems like ending up with the boundary, no, no, no, no, there's nothing left, see we just come to this little placent, this little placent is giving us f 1 2 squared, now we sum up the other little placent, we're only summing basic the answer, now we give us f 1 3 squared, you sum up now the other placent it will give us f 2 3 squared, so that will give you this f mu nu and then f, it's exact, in fact, a 1 is just, yes, is the log of this w, a 1 is just this phi of this factor, okay, in the one direction, in the other direction it will be a thing, however many dimensions you have, you see that if you do this sum over all placents, including all orientations, you simply get this f, minus f mu nu squared, is this clear, sorry, 8, 8, thank you, and actually this is probably f mu nu squared by 2, because f mu nu squared has f 1 2 squared and f 2 1 squared, it's the same thing if we just write it down, is this clear? Sir, may be sure that we do this for 1, 2, 3, would be redundant orientations for 1, 2, 3, may be say that for some of you, the one or the problem is that you can get some more similar results, sir. No, no, all basic little, so let me, be clear, probably to sum over anything but the basic square in it, so let me suppose, suppose this is minus, that was 2 dimensions, just because this is the sum over this placent, and this placent, and this placent, only the basic square in it, f 1 2 here, actually is integral of f 1, suppose we have 2 dimensions, the full action is integral of f 1 2 squared, so we need to get f 1 2 squared each point, so we have to generate that integral, now we have 3 dimensions, we have to get that integral to be a qubit lattice, and we are going to have 3 types of placents to sum over, we have these placents, then these placents, and these placents, that will give us f 1 2 squared, f 1 3 squared, and f 2 3 squared, but that's what we need, right? The geometry of our lattice, yes, yes, it's true that we would have to think hard that we were using not a qubit lattice. Sir, you see what we are interested in is a discretization of the continuum, we are not that interested in the model by itself, except in the fact that it's a discretization of the continuum. So, of course, when you discretize something, you can always find complicated ways to do the discretization, but you look as you said, because that's simple. Sir, are there any associated boundary conditions if you consider a sum like lattice, it isn't true that the ends will be full placents, it might be partial. It's a good point, it's a good point. So, yes, this is of course, if you have boundary conditions, say if you, the way that's normally done is to choose a torus, period, period, period, period, period, period. That's very well suited to dealing with a qubit lattice. Now, if you actually end somewhere, then you have to specify what happens to the gauge transformations at these end, that becomes a complicated issue. Sir, the interesting physics is, let's imagine for the moment that's a torus. Good question. Other question. So, this is the sum of all elements. So, in two dimensions, this is the orientation sum as well as the locations. So, in two dimensions that is where there is no orientation sum is this again plus this again plus this again and so on. That is the x in this. So, that sum goes over to even h minus square a square upon 2. Is there a sum? Yeah. So, this sum goes over to the 8 into the 4 plus 1 in fact. Is that equal to the number of relations still? Ah, ok. You are right, you are right, you are right. I will have to put the right power here. Thank you. Now, what we have got? So, this thing here would really have had a to the power d. We have a to the 4. So, we will have to. So, this divided by e to the power b minus will be the, will be the, will be the yeah. And this of course is reflective of the fact that the angles coupling before I mentioned is a dimensionless, but the angles coupling in every higher dimension carries a dimension. So, that is it, that is it. So, now let us say it is better to say it more concretely. That is d x. What? d d x. Why? boundary is so complicated visually. I would assume just if there are 3 rings on the boundary, you just put gage transformation under the 3 rings. No, it is a question of what, whether it is sum of a gage transformation. Suppose we end our lattice here, it is, I think it is an ending point of the lattice. It is a question of whether you are going to impose gage in there is over these points or it is easy to. See, in a normal, it is going to be the question of whether you impose a Gauss law here or not. See, if we are going to look at discretization, suppose we are going to end out. We will continue with some. Basically what we need is boundary conditions. Basically, suppose you have a classical action. You take any classical action and you end it at some boundary. Just even as a classical problem, that is an ill-poor problem unless you have boundary. So, if you define this problem as well, you will need some sort of boundary conditions. You should know what your boundary conditions are. We are going to be definition of it. That is why it is complicated and that is a, that is some choices involved. That is a fact. Sir, in that function of what was your fault? Yes. It is good for left and right. It does not have to be same here. Meaning, the Gauss matrix is different. There is u of x. It is not the same as u of x plus a. But it is of course an element of the same algebra. What is u of x? What can I do? It could be. It could be. It could be that. That is an interesting question. You know, the way we can think of that. The way we can think of that is as a bifundament if we did that. The kind of thing that you are thinking of is that there are two kinds of, there are two kinds of lattice sites. You could have this, this, you could have a lattice like this. Such that on a cross, there is one Gauss matrix, on a dot, there is another. So now, this whole thing can be thought of as follows. You can think of this whole thing as a lattice. What do you have here? I am just wondering whether this would be basically a five of a theory. No, but then you have these gauge fields that are going between the two. What is this? Interesting question. It is like some sort of gauge theory that has basic vector fields associated with that. Now, the adjoint of a particular group or a fundamental of one group and the anti-fundamental of the other group. Sir, but since you are taking square pairs, it is possible if you take some other structure like this. Sir, since you are a different lattice, why is there a bit of kind of transformation in space? That is what he said. He said that there are two gauge groups basically. But he said that both gauge groups do not deliver each point. As a lattice model, it makes sense. The question is whether it approximates some interesting continuation. He said interesting. Let me think about this. That is an interesting question. Now, so what was our net conclusion? Our net conclusion was following. Our net conclusion was that the action d, so the action because 1 by 4g squared, d dx s mu nu squared. Let's say that we were looking at this hdn. So this thing can be written as sum of over placets. 1 by 4g squared and then 2a to the power d minus. So a to the power d minus 4 by 2g squared. p plus p inverse. Did I miss a minus sign? It was this 2 minus. Let's go 2 minus p minus p inverse. And then sum over all the placets. Where p8 represents the product of these. The product of these things are all basic options. Is this clear? And so the euclidean angle is positive. In this case, so far just u1. So far just u1. The euclidean angles are positive different. Can be rewritten as z is equal to product over every link. So x and n hat. d pi x n hat. Exponential of minus a to the power d minus 4 by g squared. 2 minus p minus p inverse by 2. Sum over all the placets. My action should be. Now I move to the euclidean space. So let's conduct. This is the right sign because this is a positive definite. This is now very beautiful and popular. Because it's very concrete. What we've got is the integral over a phase. This integral for each of these guys runs from 0 to 2 pi. It's a phase at each link. It's related to action. It's related with build up of these phases. So the action builds up in each of these phases. And that object gives us a positive definite. Notice that g squared, as we mentioned before, g squared in d dimensions has dimensions of what? Well, d lengthen. Suppose we have the length dimensions. d minus 4, the length dimension of g squared, is equal to 0. So g squared has length dimension 4 minus b. I messed up. Firstly, I get the length dimension. So it's g squared f nu nu. So if I count length dimensions, this is okay. So it's d minus 4 minus the length squared and length dimension of g. Ah, I just broke this one. So this is g. So g squared has length dimension d minus 4 and then everything's okay. The action that goes in the exponential is always a dimension in this. Here, we had a 1 by g squared which counted some dimension. In the continuum theory, the dimensions of the integral make it up the length. There's no integral in it, right? These are all numbers, phases, no dimensions. So the dimension factor has to come out explicitly. Then we've seen that. Is this clear? Okay. Very clearly useful discretization of the angle's path in depth. As you can see, this is very useful to put out an auto-domain. While we were writing this p, 2 minus p minus b inverse, all the higher order terms should drop out only in the continuum level. But will they have an effect when we keep them in the discrete version of this? Right. So what we need to do is to look at this action in the limit a goes to 0. Is that limit? They should have any effect. It's like discretizing an error. Okay. Now, of course, you could ask, what does it mean from the point of view? Effectively, then there is this coupling. There's this coupling here, which has, which is wherever your angle's coupling was, your bare angle's coupling was, times the a to the power d minus 4. This combination, if you remember, was a kind of thing we very, we usually kept fixed in our discussion of our jiffles. You remember what we did? You imagine that there was some bare theory at some scale, now, no, not. We imagine that all parameters at bare theory had dimensionless numbers of order. This combination is very, very natural. Okay. Now, the fact that this a is going to infinity is going to 0. What does it mean? It means that we should be calculating correlation function in very large numbers of lattice units. Because the size of the lattice is a, and we are interested in physics at some fixed scale. So, taking a to 0 is basically the same thing as working in the limit, where you're calculating correlators on very large number of lattice units. That is to size a. You want to do some calculation to continue, let's say it's size at distance 1 meter. Take a to 0, the number of lattice units between 1 meter is 1 by a, and so it goes to infinity. Is that clear? Now, let me first say the power, let me first just tell you, I'm not going to go through the algebra, but I'll just tell you. And there is a non-Amelian generalization of this statement, okay, which actually is very similar to this. Okay. So, in the Amelian theory, in the Amelian theory, we had this W variable. In the Amelian theory, we have this W variable per link. In the Amelian theory, W was just a phase, just unit-norm complex number. So, it was in the exponential of a phase. So, we parameterize things in terms of the phase. The non-Amelian theory, it's not convenient. So, we continue to work with the W's this. Okay. So, what are the W's of the non-Amelian theory? The W's were exponentials of i times a. So, they were unitary matrices. Because the exponentials of the algebraic elements were unitary matrices. And group elements. Okay. So, what we have in the non-Amelian theory is a group element. So, our gauge transformations, which are group elements, at each point, the gauge transformation act of the group element that goes from x to x plus a. It acts as u of x, u inverse of x plus a. Exactly as a dig when we looked at it in terms of W's. That's why I introduced the W. Again, exactly as a dig when we looked at it in terms of W's, what did we look? Here, we take products of links that close. Okay. Once again, if we take products of links that close. So, suppose I take a product of a link and I open it like this. Now, at the moment this is the gauge element. Okay. It's... Let me do the way I done before. u, u, u, u, u, u, u, u, u, u, u, u, u, u, u, u, u, u, u, u, u, u, u, u, u, u, u, u, u, u, u, u, u, u, u, u, u, u, u, u, u, u, u, u, u, u, u, u. How do we transform under gauge transformations? It will transform like u of x if I begin an n matrix times whatever the circuit is, times u inverse of x. Is this clear? because everything in the middle cancels it. This is not quite setting A to B. But it will be if we take the trace, because the trace of the property that we take in the universe, this side, the new density of the universe, is 1, it goes away. So the non-Abelian theory, very similar lattice discretization. The non-Abelian theory is a very similar lattice discretization and it is given simply by the ordinary formula, product over x and hat. Now, this is dUxx plus n hat. A product over the integrals over the unity group for each link. Exponential of minus 8 above d minus 4 by d squared into trace of 2 minus 3 minus 3 in us. It works almost without modification. It works almost without modification, yes? Use of the ones we are living at the points. No, no, no, no. Now there are two different views. There are the views that later gauge transformations. Which are the ones we are living at. But there are the views that are replacing the double views. But these are different views. These views are the variables of the action. The earlier views you talked about, the views of, I should have called these views. Let me go over these. Okay. These views are the variables that enter the pattern. The views we put in the equations of the, of the symmetries of the action. The action here has invariance under v goes to u of x because of x plus a and hat. So use parametrized symmetries of the action. These are the variables of the action. These views we don't have. Ah, that's a great thing. Exactly right. And so where does all this business about gauge fixing and uniquely specified one point and each gauge output going this way? These you didn't have to. You see, what we were doing was, you see, if you remember what we did for gauge fixing is that we said that the original path integral was correct except that it was infinity because for every configuration, there's an infinite number of configurations that are comparable to the same action, of infinity, gave us trouble if we were actually trying to do calculations. But since we only cared about the path integral up to overall normalization, if you would regulate that infinity somehow, then you just don't have to gauge this. Let's say how we regulated the infinity. Then infinity came from the fact that it was the infinite volume of the group of gauge transformations. Why was the volume of the group of gauge transformations infinite? For the simple mundane reason, the number of spacetime points was infinite. Because at each spacetime point, what is the span of gauge transformations? It's the space of unitary matrices. Set of unitary matrices is a finite dimension, finite volume. Integrate them with higher dimensions, they should be unitary dimensions. If you just regulate the number of spacetime points, then the fuller volume of the gauge group is finite dimension. What would be done here? So what we are going to get by this gauge is differs from what you would get by the gauge reaction procedure by a factor of the volume of this space of unitary matrices times the number of spacetime points in here, to the power number of spacetime points in here. But if we only care about path integral up to overall normalization, we don't care. So if you think about our path integral procedure, the starting point was that this is the path integral we wanted to compute. Divide it by some overall number. And then we process that. So the volume of the gauge group is finite. Is finite? Because it's a compact group. Because it's a compact group? And? But this infinity is arising because the number of spacetime points was infinite. And here, it's just everything is finite. We're never going to get an infinity in doing an integral. The problem was that then we actually tried to do the continuum element. For instance, a propagator was 1 over 0. We were getting infinities and we had to regularize those infinities. And that basically was what we were getting with this. With this finite. But here, because we made the number of spacetime points by its finite, there's no infinity. So we don't have to bother with all that nonsense. It's a great thing to be able to think of the gauge theory without having to think of course and so on. Even in your mind. Because that's all. Don't be in this ancient. But it's needed for calculation. You'll see when we actually compute the beta function of... of... of Yangtze's theory, what a pain it is to keep crack of all these things. On the other hand, there's actually the lattice way of computing this beta function which is... which we were discussing last week. It's up. I mean, it gets rid of all those things. And, you know, there are some things that are much better to do with pen and paper rather than, you know, in terms of momentum, dimensional regularization, you know. So you should be able to think of both of these. You should not be committed to one other issue. Yeah. Excellent. Any questions or comments about this whole issue? You will be able to think of both. What? You will be able to think of both. Any questions or comments? Naturally, isn't it? Yes, we are interested in you and me looking at it like this. No, the only reason... Oh, I see. No, there isn't. There isn't. There isn't. Yeah, it's it. In this way of looking at it, the S-U-N and U-N are not truly equal. It's not... it's not... there's no factorization. On the other hand, Senghar is that even if you want problem, if you think of this as an exact lattice model, it's not a free problem. Again, this is quite a complicated problem. No problem. Okay, sir. In the non-ambiguous case, pen and paper are simply as they... In the evident case, I actually know phi of x, x plus a is minus of phi of x plus x. In the non-ambiguous case, this is... Yeah, it's a big change. The way to think of it is in terms of what was the exact relationship for w? For w, you see, there's no analog of phi. But there is an analog of w. So the analog of w was w of x, x plus a. It is w of x plus a, x inverse. That's that same thing for you. So v of x, x plus a n is equal to v inverse of x plus a n. We should obviously think of this in terms of group value. Group value can be merciless make a lot of sense. There are some things in the adjoint. Which? Which? v. v is the unit matrix. And the transform in the adjoint, that's this thing. Yes. It's in the adjoint. I mean, yeah, this is this thing. This is the discrete version of the statement that we... Yeah, this is the precise statement. We normally talk about, when we say something transforms in the adjoint, we normally talk about the infinitesimal things. This is the finite guy. So the statement is that the transform in the adjoint is not precisely precise. But this statement is completely precise. And that takes all the integration. In the abelian case, the minus was essentially giving me the current. So how do I get it out? Do I say that v inverse is the minus a? It's the same thing. Because you see, in the abelian thing also, it's been worked in w's. The transformation was w of x is equal to w of x plus a in the x inverse, plus or minus. Yes, but the phi would get a minus, which from this I could get a del 2a minus del 1. That's it. Yes. So once again, the way you would show that you get the right thing is to say that approximately these u of x's are into the i, a, x, x. Okay? So that this inverse thing will become the same thing except you have to be the only thing that's written is you have to be careful of taking things through each other. Again, because it's not written. So when we look at this placet, the one thing I did not show you is this. That when you take this placet plus its inverse, that you wear f mu nu squared and take the praise, and then you approximate the placet by each link as in i, a mu, x, a. The only thing I'm not sure of here is that when you carefully discretize that, you get the non-immediate f squared trace. Okay? This is an exercise for you. It's obvious that it's to work because whatever you get has to be gated at it because you start with something gated at it. Whatever you get has to have the right limit. That is u1, but it's great. It's actually irreversible. Okay. So now there are two or three things that I wanted to... Okay. So far we've touched on formalism. It's nice to have this formalism, but it's even nicer to have physics. So what we can now do is to study these lattice gated theories in a particular limit in which they're very easy to study. This limit is for the strong component. Okay? So don't let me... I'm going to do this study. I'm going to do this study. I'm going to do this study. First, for the u1 theory, I'm not going to discretize that for me. Write this action once again where g out squared, g squared, right, a to the power of 2. This g out squared is some number in some dimension of this number. And what we're going to try to do is to see what we can say about this model as a function of this path integral as a function of g out squared. Okay? Can this... this guy accurately... repent that this guy accurately reduced to the continued theory? Okay? You can see when it accurately reduces to the stronger... to the theory we're studying. All of these guys here were exponential phases. In order to get the continuum limit, we needed that the phases... I'm going to say one inaccurate thing here. We'll correct that next class. But we needed that the phases were not very large. The inaccurate thing is that what we actually needed is the phases are not very large when you do 2 pi. And this is the important thing that we encountered. It's a very important thing. We're taking only 0 to 2 pi. We're taking only 0 to 2 pi. Right, that's right. That's right, that's right. So we're within 0 to 2 pi. It's fine. Yes. But you'll see something that can wind. You'll hang it. But... now... if that's true, we're doing the path of death. So dynamics should impose that. When will dynamics impose this? You see, when this phase is very near to 0, 2 minus this minus this is near to 0. So the deviations of the phase are, you know, the action. Contributions where this p deviates significantly from the phase, from unity. Are suppressed by 1 over gr squared. So if gr squared is very small, that will tend to suppress contributions in which this phase goes all over the place and will try to keep it near phase equals, near the complex number 1. So the continuum limit of this model, the continuum limit of this model can reasonably be expected to be achieved. Only when this gr is smaller. Okay? And that, especially for the non-linear thing, it's a very difficult problem. It's a problem-solving problem. Opposite limit. Even just mathematically, this model is easy. That opposite limit is when gr squared is very, very large. In this limit, the coefficient of the action is small. Okay? So gr squared is a large, and let's say 1 over gr squared, we call it, just to remind us of that. Okay? So then, in this limit, what we've got is product over x and a hat, z of sigma y, d of y, x and a hat, exponentials of minus epsilon, sigma, inverse by 2. Okay? Now, in this limit, these quantities here are not at all forced to be near unity. Okay? It's very far from the continuum limit, but it's a very easy limit in which to study the problem. Why is this easy? Because in this limit, we can effectively, to tailor expand in epsilon. Okay? Now, I want to do a very small little calculation. I want to do a small little calculation. Okay? To compute some quantity into leading order in epsilon. Okay? What is the quantity I'm going to calculate? The quantity I'm going to calculate is the expectation value of obey this. Whatever I mean by that. I've got these, I've got, on my lattice case study, this is not one of these basic small flow spots. It's an arbitrary, arbitrary closed path. Make it closed. So, these observable, as an agent management observable, just like the basic in the, in the, I would take this closed path and then also take the place. Then just like the basic plaquettes for the agent management, this guy would be getting there. And I want to compute the expectation value of this object in this lattice point. Okay? That's what I want to do. Is this clear? Let me be a little more specific. If I want to calculate, I want to calculate the expectation value of product of e to the power i pi where li's are the links along some closed path. That is basic. These links are on this little loop. What it's inside the loop. The links are the links. So, I, so let me show you in two dimensions so it's a general model. I specify some path. By the, the statement that I want to compute the expectation value of the Wilson loop, is the statement that I want to compute the expectation value of the products of these links along this path. Is this clear? So let's go on this one, two, three, four. Just lay it in the links. Up to wherever. And this is product of e to the power i pi, the product of the w's on each of these links in that order. If we, if we use here the order as a matter, not to be related. Is this clear what I want to compute? So what I have to do is to do the calculation of this object pi x n hat 0 to pi, that is the expectation value. This is the expectation value in del x function. The expectation value is just 1. What is the answer if I do this integral when I take 1? I get 0 because each of these, the integral to each of the 5s along in one of the links that appears in the Wilson line just matches. Because exponential of e to the power ix from 0 to 2 pi is 0. How could I have to bring down these factors from the from the path integral? But what are the factors from the path integral? Factors from the path integral are basically plug-ins. So suppose I was interested in having this guy not vanish. The integral of this loop and this loop now. I could achieve that by bringing down this length. I could continue to have the Wilson loops over these these guys as well as these guys. So the integral over these guys will vanish. Are you understanding me? I get the inverse 5 from this. So that cancels these two. But these integrals will vanish as well as these will continue to vanish. I make it. So that the whole thing is not there. Will you see what you have to do is to take exact same. You have to take a clucket. We have to take as many cluckets as fill out a surface bounded by the Wilson. So for instance we do all the cluckets drawn on the wall. If we do this, this, this, this, this, this, this. That will do. And we chose the guy with the right orientation. So it cancels. We bring down the fellow with the right orientation. Okay. Now for every e to the power 5 there is an e to the power 5. Is this clear? What is the weightage that we would get? Now when you have the e to the power 5 and e to the power minus 5, now the integral is just 1. Okay. So what is the answer then? In the strong clucket element. Okay. So firstly I have moralized an exact statement. Now I have done this in two dimensions. Where the only outer surface is basically four. But let's say we have three dimensions. We have this surface drawn 3 and 2. The lattice with three dimensions. We could have had surfaces that came out. It is like a Lego building. Think of it. What this would turn into into some over all surfaces that are bounded by our Wilson life. Okay. Now with what number would each surface be weighted? Exactly. Number of cluckets to the power epsilon. So it would be weighted by epsilon. It would be the number of cluckets. Suppose I consider this example of two dimensions on the surface. Then the number of basic cluckets is this one. Yes. So there is no effective sum in the exponential. Because that is the sum of all basic cluckets. No. And what is the sum of all space segments? Apart from the space segments. No. The space segments are very important. Because you just bring out this. One space segment plus this. But that would give you how many cluckets you need. How many cluckets you need? The number of cluckets you need is the area. Right. So that is one basic clucket. And times number of cluckets which comes from the sum over all points. The number that you need to bring down is the area. Okay. Each time you bring something down you get down the power of epsilon. Okay. And so what you will get is epsilon to the power number of cluckets. Okay. And therefore, we have to work out all the details in order. This can be done with great precision. Sir, this is the space segment. Space should be called space. Very much. Exactly. So this would be epsilon to the number of cluckets which would basically be some number to the power area of the surface. Surface is a greater area than this minimal area. Okay. Would be some number in this limit. So this wills so what we conclude from here is that in this strong coupling expansion what we find is that expectation value of this Wilson line goes like e to the power minus some number and the area of this minimal area goes by this. It doesn't matter. It's just a log of epsilon. Yeah. That's the sum. The thing that I said actually goes through almost without modification for the number. The only thing that we used is the property that if you had one e to the power i5 it's integral over all 5 to the power 0. But it also could not be the theory that integral the u of u is equal to 0. So that continues to be true. Okay. And then we use that if we had an e to the power i5 and e to the power minus i5 we know now that there is 0. Okay. There is the following statement that the u of u alpha u u inverse So why is integral du equal to 0? Why is integral du equal to 0? This is immediately obvious from rotational variance. Okay. Let us suppose when integral du equal to 0 and it was something. Yes. Now let me do a change of variance. Let me do an variable change. u is equal to v times u prime. v is some fixed unit matrix. Okay. So what we will get is that what we will get is that this is v u prime times v times u prime. Okay. But the integration measure when we use the higher measure which is what we are doing integration of unitary matrix. See. The higher measure has this following beautiful index. It is invariant of both left multiplication and right multiplication of loop elements by constant loop. Okay. So this. So this is the same to the other. Okay. So this f whatever it is is some object that is invariant and left multiplication as well as right multiplication. Now there is no matrix that is invariant on the both left and right multiplication of an arbitrary loop. Just like there is no number that is invariant on the left multiplication of a phase and right. The after matrices are invariant on the left number left inverse identity. Okay. So that's the way. That's the way. Okay. But there are similar statements if you take d u of u times u inverse then using very similar logic you will get clues of those one things. Okay. By demanding group invariance of this thing you can deduce what this is. Okay. And using the fact that it gives you something else here is the analog phase tensor anti-phase. And using that once again you basically conclude and leave your work out of the details or maybe I'll enter your assignment. What happened to the assignment? Okay. I'll tell you what. We'll give you another week for the anniversary problems. This will be one of them. Okay. So we're done. Okay. So using that once again you see the only way to get something non-serious to fill out these placets. Because every time you have u here you need a u inverse from the placets. And once again you conclude that the last thing I want to say is the physics. Now physics of areas was very important. You see what is the physical interrupts. So now we've been talking about that's already stopped. It will look for 3-4 minutes. 3-4 minutes. So I'll finish. Famous last words. Okay. So see suppose we cut a gauge theory and we take a Wilson loop like this. One in which we make a rectangle. Which is very long and the width is smaller than that. Okay. So we're going to ignore all the effects. And we're working in previous phase but now we'll do this direction. Go back to the next stage. What is the interpretation of the path in the integral with the insertion of this Wilson? Okay. The Avivian theory it's very clear. Say it's actually similar in the non-villain theory. Let me just tell you when I'm going to discuss it. You see the presence of this Wilson see what is let's first work on the Avivian theory. Integral e to the power i a dot dx. So it's an addition of a dot dx to the action. Now when you add something to the action then in any field theory there's the current of it. The current of it is derivative of the action with respect to a. So what do you do if you're adding the derivative here it's very clear. You're adding what a current that is delta function localized. If this is going in the time direction delta is going in time. So it's just is J0. So it's J0 of something that's since our point. Okay. But when you come back this is this way and this is that way. That way is minus of that way. You put J0 here and minus of that J0 components. So do you see what you're doing by adding the Wilson loop? It's working on the path integral in the presence of a charged particle at one point and a negatively charged particle at the opposite. The analog statement of that in an ordinary theory is that the Wilson loop gives you the path integral of the problem in the presence of a classical quark. By classical I just mean you're not doing the path integral over the position of these ends. They're just fixed. So presence of an external quark here and an external antenna are integral in the presence of these fields. Okay. What's the interpretation of the value of the path integral in the presence of these fields? The interpretation of the value of a path integral is always vacuum energy times time. This year too, right? Because the path integral of the Euclidean path integral is e to the power minus beta H over very long times you get contributions essentially only over the lowest energy state. So it's e to the power minus beta which beta is the time direction times Z0. Okay. Now we're completing the vacuum energy because we've been denominators reflecting the vacuum energy of the problem without the concealment. So what this expectation value gives you is the change in the vacuum energy, change in the energy of the problem, the lowest energy of the problem in these two times in the picture. The delta e is proportional to length. Proportional length tells you that the lowest energy configuration in the presence of a quark and an anti-quark is proportional to the distance between the quarks. This is the statement of confinement that a fundamental charged object cannot exist by itself at finite energy without having an empty object sometimes. If you separate them very far away you pay an infinite energy price and moreover it's the statement of confinement by very particular mechanism by a linear potential between the two between the two between the two objects. Okay. I think as we go along this linear potential may be thought of as producing some sort of stream a flux stream where you stretch out more to avoid it. Sir, is it necessary to think of these terms of just quark or are you thinking of the red problem for the quark? In the U.R. problem it's just a eucalyptus problem. Where in the non-eucalyptic theory what we call is something charged in the fundamental energy. So confinement is very important for the non-eucalyptic theory. So we wouldn't see confinement as a primary possibility. So this analysis wouldn't be worth it. You see all we have done that's a good question is to show that in this strong coupling limit we get the carrier down and now we interpret it. So in the strong coupling level you have seen this conflict both for the eucalyptic problem as well as for the non-eucalyptic. Ok? Now the key question is the problem. The key question is to get to the quantum field theory we need to take this lattice model and take its coupling from D-merits from where there is conflict to very weak where as we have seen you reproduce the actual dynamics to continue there. The question is in this process is there a phase transition? This object this area of behavior of the Winston loop is an order parameter that characterizes the spaces. Ok? In strong coupling for this lattice model we have a question as we take the transition from to the end where we are very interested in is there a phase transition? Conjecture for not a big angle theory there is no phase transition that there continues to be area of defendants even in the continuum Ok? This is one statement of confinement of pure agents. You are not telling we know that there is a phase transition similar to the lattice model Ok? And when you are weak enough I will show you that this is not our last analysis we will study these lattice theories a little bit more in the next class Ok? Ok, one of the things we will do probably in the next class is demonstrate that continue that the continue 2 plus 1 dimension undergoes conflict as area of further Winston loop this is a beautiful proof of confinement I will explain that We will understand these lattice theories a little more it will be quite fun but then we will assume how to continue Ok? I am hoping that by the end of the next class we will continue our discussion of lattice theories and we will discuss BRSTs of n-trade and renormalization of pianos area and computer one computer function that will be the answer This is the real law Do you remember about number of brackets about number of brackets about number of brackets Yes Yes Now, how good is that for it Probably not Probably not Because there is Susie What we had was e to the power I will let us give the section of this epsilon times b1 plus b2 bn But these are the various brackets we need Ok? So what we want is epsilon to the n as we said So So epsilon to the n by n factorial times b1 plus b2 bn But we also have the number of ways of choosing b1, b2 and then it says What? No, you will have because you just have these places It's the same So this is just the number of ways You understand, right? Excellent was the exponential So we have e to the power epsilon and we have to bring it down as many factors of this object as we need it to get We have to bring down epsilon to pi by n factorial times this p1 plus the exponential Sir, if you have even that places and then you have a matrix that you have it's not going to cancel it It's going to cancel because of the intent to lessen that Let me explain this So suppose, let me take an example Let me take a two basic brackets with two basic brackets So suppose this was a matrix Ok? So this is the thing I was planning to do and you have to do all the exercises and you need to get help Ok? So this guy is u1, u2, u3 u4, u5 u6 and let's call this guy So what do we want to compute? We want to compute the trace of u1, u2, u3 u5 What we bring down is trace of u1, v, u5, u6 u3 inverse u6 inverse u5 inverse v2 inverse and bring down another place These guys are also inverse Ok? This will actually be So u3 So this is actually v times u4 inverse u3 inverse and now I want to show you that this integral this thing integrated over u1, u2, u3, u4, u5, u6 and v is not What is the integral? Ok? Basically what this is the product of u's and u inverse and then various index contractions So you need to know the basic answer to the problem what you get by integrating u tends u inverse u2 matrices u inverse u2 matrices You know what I mean? Let's quickly try to understand how to figure that out I am sure you will understand since it is that way umn, u inverse suppose I am going to do this What could we get? What could we get? Something to be left and right in there There are only two possible structures Delta NP Delta NQ upper and lower indices That is one structure The other structure is Delta MN Delta NP So this object has to be equal to A times this, C times this Ok? We are normalizing the heart measures so we get little of 1 is 1 So this is this is true and now we have to figure out what A is Ok? So let's do this The second case is just the fact that they are right there Yeah, exactly They are anti-partisanal fundraisers in this case They fit in some many different ways Another way of saying it is that we want this to be invariant under You know Ok, let's figure it out and then we will check So one way of doing that is now to take N and P and contract If I take N and P and contract then this thing is just Delta NQ So Delta NQ is du in general 1 is over du is 1 and AQ So if I take N and P and contract N N and P and I contract here I get B, Delta NQ But here I will get A, N Delta NQ So one equation that we get from here is that A, N plus B Then the other thing that we could do is for N Suppose we contracted each of these with themselves Now it depends on whether we are doing the SUM theory or the UN theory Suppose we are doing the SUM theory Contracting these with themselves is confusing So if we contract M with we should just get 0 M with any arguments A is equal to 0 So these two equations simultaneously that gives you A and B See So what this is Okay, I will give you something Plug that into here You will get something Hard measure Okay Given the group manifold The group manifold is labeled by any manifold you can It's a quantum science And then there is a useful integration measure for this The hard measure for any compact group which is invariant under left and right multiplications of the group Okay Let's take Let's take Let's take a very simple example Sorry one minute Let's take a very simple example Suppose the group was just an abelian group Okay So the group was just The hard measure is just defeat Because left multiplication is translation Right multiplication is also translation Just translation invariant Suppose the group was S3 In that case Group manifold is S3 And there the metric The hard measure It's just the measure The metric that you know The metric you will write for an S3 Okay Reserves SO4 So in particular it preserves Multiplication from the left by SC2 And the right by SC2 That's the SO4 Suppose you have some parameters Okay That parameter is where you are For coordinates Now Do change variables According to Multiplication of the group time So the group Basically the symmetry of this problem And you want your measure to respect the symmetry And the problem Point is that you can always do it Many properties about integration Over this symmetric measure Can be deduced to the actually ever doing an integral Just by using symmetry Very good