 What? Yeah, we'll try to do something. You can try. There is going to be an exception. Even I don't know what's going to be an exception. Okay. Okay, great. Today is actually really going to have to be a short class. It's been 15-30 years already. It's a hard dead line. So we'll do what we can. Okay. And in the last class we were discussing perturbation theory for almond diagrams. It's a theory whose basic field for matrices. Okay. We had taken as an example a field theory whose run time was there like 12 plus IQ. And we'd studied a few diagrams that computed the vacuum free energy. Okay. And we found that all the diagrams that we found in the examples we studied. There are all diagrams that we could draw on a plane, skidded like n squared. Now I want to, I want to at least indicate to you how the systematic analysis of this, of this topic goes. Okay. So how? As you remember we drew propagators with double lines. Okay. And as you remember then our final graphs. So we had examples of final graphs for instance, with a graph like this. Which when we brought it around and brought in the double lines, in like this. And when we had another graph, like this and brought it down to the double lines, this would be like this. Okay. Now I want you to think of the formula. Once you draw this diagram final graph in double lines. I want you to imagine that, you know, the double lines are there with a very thinness. I want you to imagine that these are the shaded regions. Nine outside is also shaded on the outside. Okay. Imagine shrinking the region of the double lines to zero. So that's basically not there. So what you see is that what we are doing is a tiling of a two-dimensional surface. That these graphs are giving us a tiling of some two-dimensional surface. Is this here? Okay. Associated with any given, any given graph. The powers of n. There's one power of n for every index line. So everything we've got in the index line let's call that 6. Okay. It's one of the times. Okay. So we get an n to the power of number of faces. Every edge that we have is a property. And remember every edge came with a one over n. Because there's an n outside collection. Okay. So minus number of edges. Every vertex that we have. Yeah. It's a vertex vector that comes with the power of plus n. Okay. So plus two vertices. So the net power of n is n to the power of number of faces minus number of edges. Okay. Plus number of vertices. You see once we've worded it this way, we're looking at some quantity to do with the discretization of two-dimensional surface. Suppose you take a two-dimensional surface and tiling it. Okay. Forget about finding graphs. The tiling has some faces, some edges, some vertices. And we've written it in terms of purely geometrical points. And there is a famous identity, I think, like Euler. Okay. For this quantity. Well, before large n limits of 150, the claim that this would even do much better. Where g is the genus of the surface. We're going to explain this a little bit more. Actually Coleman's lecture on physics in a small appendix was a proof of this statement. Okay. And it's based on the proof. Well, it's reading it to you. My proof. What we're going to do, not the proof, but just to motivate you to understand it. But before I even prove this, let me explain to you what a genus is. Okay. I can explain that really with an example. A sphere that is a shape of the genus here. A sphere with some part of it eaten out. Like a dose. Like a donut. Okay. Is genus one. If you make a sphere and eat out two holes. I mean, these are holes that you can see through. Okay. So this goes like this. This goes like this. It's like a pretzel. You guys know what pretzels are. Okay. Okay. This is a genus two. And in general, the number of holes you eat out of the surface. Okay. Is the genus of the surface. Okay. Every two-dimensional metaphor oriented to dimension. That's compact. Is topologically equal to one of these sciences. Okay. This is some phase with some mathematics. So, when we do a draw of 100 graphs with this double-nitronation, we'll be tying some two-dimensional. Okay. And the claim is that the power of n that we get. The power of n that we get is determined purely by the genus. So all graphs that are spheres. Okay. All graphs that are spheres have powers by h. All graphs that are tori have powers by one. This is for the vacuum energy. Okay. This is the claim. And now we're going to try to understand it in a little more detail. But first we understand the mistake. We're going to marvel this thing because it introduces a new organization of graphs. If you think quantum field theory is summing an infinite number of graphs, then there's a very old organization of the graph then we're organizing it according to powers of the couple of graphs. That's perturbation. This is a new organization. That's sort of a problem of the revolution. Graphs are any given automain. It could have all the orders in the couple of graphs. Okay. And trying to solve the theory with a large interval is effectively trying to sum all the graphs and leading the life of the standard of automaton. Okay. Any questions or comments about what we've said before trying to understand it in detail? There is no question about that. This is a very big question. Why do you think two dimensions are necessary? Well, you see, you might even need some sort of field. You see, why do we need two dimensions? We need to get two dimensions because we were working in two dimensions. Sir, no, I'm just asking. Oh, general mathematics. I don't know. Maybe there's something. Maybe there's a way of understanding questions. Then trying to understand. You see, there's something about two dimensions about data. There's a great variety of three dimensions. If you're trying to make three-dimensional surface using three-dimensional sub-values, there's a great variety of the kinds of things you could do. I don't get it. Sir, here we are talking about two-dimensional tidying. There are things genuously of poles. But the poles are... Imagine a pole. There is locally no pole. It's a topological feature. You understand this? There's no special point on the surface. It's not like a sheet of paper where you cut out. It's like a copper and domino analogy. Like a cup has a hole in it. A cup has a hole. I see. Yeah, I suppose. But if you had a sheet of paper, you cut out a hole in it. That's a hole in it. There are special points on the sheet of paper. There are points on the edge of the hole. They're different from points anywhere else. On a donut, every point locally... I didn't mean every other point. There's no local fun in it. I think these local things are together. Make your point. Do you understand this? The two ways in which you can work with a pole... I just want to make sure you understand my idea. It's not like I've torn the surface, anyway. You know, you could have a balloon. This genus-1 surface, if it was a balloon, would keep its air inside it. Do you understand? So what's the genus-2 surface? Is this clear? Consider that diagram. Consider that diagram also. We can form a sort of a balloon. Basically, this edge and this edge, we can make a face out of it. On the back side. So it is two-dimensional. It's a three-dimensional embedded edge. The embedded edge is not important. This looks like two poles coming together and merging up as one. Where are the poles? These are not the poles. In fact, it's an artifact of the tiling of the stairs. You should imagine this joint. Imagine tiling in a bathroom. I've drawn that segment so that we can draw that. But there's no gap between it. Building the surface. There are no poles at the tiling. It's just the question of what surface it's tiling. So it's just a diagram. Which is the topology sphere. A particular tiling of an object that's topologically tiling. Do you understand? So is it like cutting a sphere and that gives us a surface because they're tiling on that surface? No, it's like... It's really just drawing tiles on a building. So this is genus zero. Genus zero? This brings me to the genus zero. Can you show it on the other side? Hang on. Let's first understand the genus zero. Well, we'll see. For the genus zero surface, we're getting a sphere by taking the last edge and closing it up. If you don't know the closing, it's convenient to say that these are planar. Because we don't close the last edge, we draw them on a plane. And for genus zero graphs, you don't even need the double line. Because you see, if you draw them on a plane, in double line, you can also draw them on a plane with single line without having propagators jump over each other. So genus zero graphs can more simply be thought of as those graphs either in double line or in single line rotation. Which you can draw on the sheet of paper without having propagators cross over each other. So the leading order are often referred to as planar graphs. Let's understand these systematics in a little more detail. This is a geometry. Suppose I had some tiling of a surface. And this was one of the tiling. How do I change the tiling of a surface? One simple way, more or less the only way is to break one tile up into two, or one tile up into eight. So I just want to convince you that this breaking up operation does not change the end to the power verticals. It faces minus angles. It does not change this quantity. This is just a local thing. So once I convince you of that, we check that this quantity is a topological ingredient. The fancy way is proving it. But we don't need that. We're just working in physicists. Follow your notes. So there are many ways of breaking up a tile up. Let's check a few. Just to convince you of that. Let's break it up like this. Let's draw a little graph. I'm drawing a single line relation now. Choose this guy. So each of these would be a facelift with dot and dot. Let me choose this guy and break up this. What is the change in the number of faces? Well, by definition, I broke down one facelift. So delta is... What is the change in the number of ages? This is the only thing that has to be a bit different. No, one is a bit different. Let's look how many propagators have been affected. This propagator, which was one, has become two. This propagator, which was one, has become two. And this propagator, which was not there, has appeared. So delta is the propagator. What is the change in the number of vertices? Delta v is equal to... Not changed. Let's try another way of doing it. Because it does all of these ways. But it's basically... We're not going to try these ways. Let's try another way of doing it. Let me try this one. Let me... You know, however I think, let me do this guy. Now, there are many differences. Let's start with another shape. I started with a line that had an edge. Let's check. Number of faces, the increase of delta is equal to one. What about the number of propagators? Now, be careful. One, two. Delta ages is equal to two. Whatever changes the number of vertices is one. So because they're used as an existing vertex, we've got one less additional vertex. So we also have one less additional propagator. One. This has changed. And so on. You can easily play around to say that any way you have of tiling the surface, you break up one tile into two. It will not change this offset. Everything is done as local. It doesn't care whether this lies on a spherical topology. It's the local part of the task. So what we have done is motivate the mind. The day's quantity, face of mind's edges, is a topological entity. Clearly, if you take some tiling of a surface, and you break one tile into two, your tiling will set surface. It's not changing the topology of the surface. This quantity cannot, that's the aim of the inversion. And this quantity does not depend on the topology of the surface. It does not depend on the, or which drop within the topology. But it might depend on the topology of the surface. Now let's try to understand the way here. Now let's try to make a way of increasing the topology. So let me first take an example of a graph. That is not a graph like this. So that this propagator is jumping over this one. That is not a graph. It cannot be written, cannot be drawn. Okay. On this sheet of paper, without having one line jump over it. Now let me do something here that I want to emphasize. If we weren't dealing with a matrix theory, if we weren't drawing topology like this, we should have drawn the matrix theory. In the theory of matrices, the vertices have a given order, lines coming out of given vertices given by the matrix structure. This is I, G, this is G, G, this is G. So this follows this, follows this in this order. That's not equivalent. You can't just take this and put it at the outside. Are you understanding what I'm saying? Okay. So the ordering of lines coming out of a vertex is meaningful because it keeps track of the index structure. So this graph and this graph are not equivalent. With the understanding that what these graphs are, or they're in the double line notation, keeping track of the index structure. Okay. So what this graph would have been like? Is written. This graph here is not written. Understand this graph in my case. Firstly, I just do a simple mental counting of the powers of n. I'll just do a simple mental counting of the powers of n. Let me try to answer it. Simple mental counting means draw it in double line notation. So what do we have? We have this, and we have this. We have this, that's this vertex in double line notation. Okay. Then come here, we've got something very similar. This vertex. Now here we've got something very similar. Now here we've got something very similar. This guy just joins up to this guy. But this guy jumps over this guy. Two, three, four, five, six. So you've got n to the power minus six. Number of vertices. One, two, three, four. Last four. Number of index lines. Now this is a bit harder to do, right? To do that. So let me do one thing. One thing is easy to see with your eyes. If you do a piece of chalk, the whole internal line is one index line. And the external line, of course, is one index. Okay. So this graph here has number of vertices equal to two. So then it does it. It's a good agreement with the formula. Oh, if you believe this, you can see what I'm expecting. Moment. It's a good agreement with the formula. That n to the power two g minus two. Where g is one, g is two. That's the end power set. n to the power two minus two. And g is one. It's a good agreement with the formula. Okay. You see that this is an example of a non-player graph. And you see that it has a lower power domain than its player graph. But let's actually look at its player graph. Remember, we had this graph that is only five to the fourth theory or five to the third theory. Just in the same graph. That was later. Let's broaden this guy out into the next. And look at how it looks. So this guy, a couple lines of this. And then we come here. This guy goes this way. This guy goes here. Go like this. Cook up. Pick up this. And then this guy. Number of propagators not changed. n to the power minus six. Number of vertices not changed plus four. Same graph except for the indexed. What were indexed? Let's look. Two of these things. B, minus zero. If they scare. Now you see. The king of a genus. One guy. In the book. You take a genus. You draw two holes. The hole, when you're actually cutting the paper. And you're drawing these two holes. That will make a genus. You understand? I have a balloon. I cut out two pieces. I have a tube. That makes a two of us. We will like, what we will try to do. The problem is try to implement this cutting procedure, cutting making a handle and try to see how it changes edge minus the input speed. You see the problem is I am using the word hole in the photograph. Actually the word hole is used in the discussion problem. I need a better word for that, the hole of the dots, the hole of the donut. Handle this thing, it's the hole I want to do it. The thing you can put your hand through. It's the face piece. See though, I call it a donut. Again, it's the thing you can look through. Whereas this other hole is cutting the paper. You understand? These are two different things. I make two cutting holes. Join the by-handle that will be between one and two hundred. Have you understood this? Yes sir. That's if we cut two holes and then tear whatever is connected to the little weird donut. Join with the by-handle. That will be one. What is your name again? Nisha. Nisha will do this. I know that it's something I also do when I'm a dream. I'll give you a one minute story. I turned my PhD. The first few meetings I had with my advisor. Somebody asked me, look, are you doing this? There's that thing you say no, don't you? I don't know what I was doing the other day. I can say I was forewarned. My colleague this guy, he's my colleague. My colleague this guy, Curtis Callan. They were with me. Once he got in India, he was given an lecture. And all of a sudden... He was getting a counterfeit. He farted again or? He farted again. He farted again. He farted again. He farted again. He farted again. Excellent. So, it's fine. So, relax a little. So what do you want to do is to join in strictly. Do the procedure of cutting out two paper cut holes and joining with the... Let's implement it. This takes a little visualisation. But because of some surface, let's make it as simple as possible to avoid complications. There's... It doesn't matter. It's to... Let's make it all a triangle. It's called a triangle X. By which I mean now there's nothing here. So this is tidal. This is tidal. And this is tidal. And then I've got something similar. I've got a similar triangle. Drawing one tile from here to here. From here to here. This takes a little visualisation. Can you imagine? It's like one strip of paper like this. One strip of paper like this. One strip of paper like that. That makes the tube. These are the two cut holes. And then I'm making the tube that joins them to give the little handle between these two cut holes. Can you understand this? If I'm trying to do clear? This is... No. What am I... The side is from the faces of the tube, right? There's some new surface. Let's go into the sphere. Again. But this is some local part of the surface. This is some other local part. They're joining somehow. Some tidying. I don't care. What I'm... You see, what is the strategy of what I'm trying to do? What I'm trying to do is that if I had a surface and I just changed its tidying by breaking up one tidier to two, that didn't change by that. Now what I want to do is a more drastic operation. I've got a tidal surface. I'm going to change the tidal surface by, in a way, that I know we increase the genus of the surface. Okay? So I took a tidal and ate out the hole in it. A paper. Took another tidal and ate out the paper hole. That now is not a closed surface. I want to continue looking at closed surfaces. The one I do is try... I have one tidal going from here. One tidal going from here to here. One tidal going from here to there. Isn't this clear? So the index structure on that tidal on which the whole structure is now added to three different index structures. That's right. But now since we've reduced the problem to two much, it's a forever index structure. Just think about edges, vertices, and faces. So that puts the constraint on the words, right? That is a triangle. That's right. That is a triangle in the south side. What we have already shown is that once you've got the surface of a particular genus, you can take it up to Hawaii. We just want to see what happens when you change the Venus. So if I just show you what happens when I add an angle, then that's sufficient to prove this graph. Prove this here. I can say I'm not doing it like a regular proof. Coleman has beautiful proof, but you're actually shorter, less time to prove than what I'm trying to do. But you know the reason I'm trying to do the way I am is that until you... you know, in a new problem, you never think of the elegant proof. At least not till you're... until you approach... at least people are there. The way I actually approach... is by hands of... so it's valuable to think this way. After you've got the answer, you're trying to make it elegant for the people. But that's the way. Very rarely I'll actually think of it. So it's valuable in teaching to do things the way you actually do it. No, just from here. So then also there... No, she... No, the holes have been... yes, other than that. Suppose it was shut here. Suppose it was shut here. Then by itself, without the handle, this thing would be... a set of holes. You cut out two holes. The amazing thing is the handle is solid. You understand, right? You cut out two holes from here and then you join it back together. This is the... don't matter. These little things were the paper holes. Is this here? It should be connected to outside holes. Yeah, yeah, yeah, yeah, yeah. You have to make your full service. Steady, yeah. Now let's analyze this problem. This requires visualizations easier here. Let's do it. What is the change in the number of faces? Okay. Let's see. In the way that I've done it here, this was one face. We get one, two, three. This was one face between one, two, three. Is that the only change in the number of faces? Yeah. We'll introduce three new faces. This one here. This one and this one here. See. We've got this face that goes from here to here. This face that goes from here to here. Let me draw that part of it out. It's basically a text. It's being joined on the nose. These were paper holes. Those are not faces. One, two, and we added three faces. Is this clear? Delta FZ is to connect the upward face on that delinear tube. Can we take one as a one surface? No, but in the case of double line rotation. No, it's a different dilute. We'll start adding that. Come here. Why that is not like a face? If I take this triangle. Why that is the three surface? Basically this. This triangle. And now this face connecting these two triangles. I can take this as the whole as one surface. Like this triangle here. This triangle here. And this edge connected to itself. It was this surface. So I can take it as a one surface. Like this. What I was doing was breaking this one tile up. One tile, one tile, one tile. And then new tile. New tile. And also because there are other ways of doing it. But you understand what I'm saying? I had a tile here. I break this up into these three tiles. And then I attach one more tile here. One more tile here. This is for wide construction. It's possible that there's some way of doing it. Without breaking this tile. Extending this tile. But that's not what I'm doing. Always we shouldn't give you the same answer. You understand? I've got a tile in it. I'm giving you a certain set of operations of breaking this tile. Is this clear, Bangladesh? Okay. So a one more double line is introduced. It won't make any change actually. Yeah. Because that will make that kind of difference. Exactly. Exactly. All things will change. Okay. So. Fine. The number of change. Number of change in the vertices. This is very clear. 1, 2, 3, 4, 5, 6. That's amazing. Not a simplistic number. Because I should now get 15. It sounds like an option. Okay. The number of changes here is very clear. 1, 2, 3. At each of those edges have been divided into two. So 6 new edges here. 6 new edges here. 3 new edges. And 3 more. Yeah. 15. So delta E. I'm sure I could do some economically right. Okay. But what we've seen is delta of F minus 3 plus 3 is changed by minus 2. What we've seen is that adding a handle changes the quantity by minus 2. Okay. What I was doing was introducing 6 new edges and 6 new edges. Exactly. Exactly. Exactly. There will be many other ways of introducing new edges. Every way. Every consistent way of interval. Changing the time to introduce the handles will decrease this by 2. And even to you to fool around. And even to you to fool around and to check that any operation you do again is consistent. If you convince yourself that every way of introducing handles decreases the quantity by 2. And every way of introducing new tiles that does not change the genus, does not change the power of this. So why we haven't ready proved any theorem so we have motivated the answer. Okay. I'll leave to you also as an exercise to begin this appendix of the coding of the aspects of symmetry. And look at this very simple and very beautiful proof of the statement. But we will promise. Okay. Excellent. Now, there's still one more. One more here for a formal bit of reflection. I hope you may not even get the first paper. Okay. There's still one more formal bit of manipulation that I need to do. So far we were working with theorem. You see, by the way, in what I did here I made no assumption about whether my field was a scalar or a vector or fermion. The only thing we cared about was that it was in the Adlern representation. That was an Inconvenient matrix. I did not assume that I had only cubic vertex. This is popular. It's true no matter whether a vertex is cubic or sextic or rectangle. It's a completely generalism. But there is one T assumption in it. That is that every field in the theorem is an n-cross-end matrix. n-cross-end matrix is the U.S. theorem. It's an adjoin matrix. And because the gauge fields themselves are adjoin matrix. That is, of course, very useful to some extent. You try to apply a pure gauge to it. Or pure gauge theory coupled to other adjoin matrix. This is the way to try to approach the large end limit of such theories. So it's the appropriate analysis, by the way, to the theory we studied in two or three lectures ago. Two lectures ago. The zero-to-zero dimension limit. If you try to approach that theory in perturbation we have much more alien way of solving it. We summed all the graphs. But if you try to analyze that integral in perturbation theory, you could do that. And what we have done is sum all the play-up graphs. How many play-up graphs are there, by the way? From the free entry. Zero-to-zero dimension matrix model. With an arbitrary view. Unless every of five is 524. It's 5 squared plus 524. How many play-up graphs are there? The number between the free entry and the interactivity. It's a simple question. The answer is zero. Maybe one. The answer is zero. The possible dynamics. This one. There isn't even a number of ways of writing this. So by the point of view of this, from this point of view, our very elegant solution of the zero-to-zero dimension matrix model was, in fact, summing an infinite number of five. Number of why five? Only those that will be drawn in a sphere. But, indeed, they're infinite. Isn't that interesting? That is the whole point of the large-edged interface. It's a very orthogonal organization of graphs from perturbation theory. If you solve all orders in the large-edged interface, you're doing an infinite amount of work from the point of view of ordinary coordination theory. But it's orthogonal. You're doing an infinite amount of work, but you're missing even free diagrams, or one-loop diagrams. That happened not to be. Excellent. Let's continue. The reason I said this is that the next theory we want to study is the theory with that. It's the theory of two-dimensional utility coupled with two box and the fundamental. Okay? And, unfortunately, analysis is not allowed to study box and the fundamental. A prototypical example is something like this, when you've got... This can't be a given answer. See? I'll say it's a given answer. Next one is no. D mu phi d mu phi plus x squared. D mu phi now is not a commutator. A commutator. This is one. This is d mu plus i d mu times phi. What do we need to study? I'll tell you the left class in the article. Phi star in the article. What we have here will be some sort of... A new is the article. A new is the matrix. Because an adjoint is a matrix. The fundamental is a column. And matrix is multiplied by all these two. Okay? So, the kind of interactions that we will get from this, for instance, there will be a delta that is A phi star phi. So, what is the adjoint? Index root. What is the adjoint? A, you know, is like this. And then these are the fundamental fields. They didn't only buy a single line. They didn't only buy a single line. Because they only bought a single line. Look at some graphs. An example of some graphs. Which have both adjoint as a matrix. So, suppose the graph that we looked at before. This graph is known in the adjoint. What are the matrices? And in this graph, I do the following. That there is a self-energy correction for the flow. So, I'll draw this in double line. So, the graph that we had originally was this. This blue one, which was propagating along. At some point, those have become the fundamental adjoint. This first blue one, blue one, blue one. Then this blue one. That's true. Yes. So, this graph brought along with double line notation. So, let's first count the powers of n. And then we will try to understand better the system. Let's do this. So, let's just count the powers of n. What are the powers of n? The rules, of course, are the number of powers of n because of an index line. The number of powers of n because of an index line are changed. So, the number of index lines is 1, 2, 3. So, we get n to the power 3. We propagate this to n. Okay? Once again, this whole Lagrangian area. This whole Lagrangian area has an n inside it. So, once again, if the propagator comes to the power of n, how do we propagate this to n? In this graph, 1, 2, 3, 4, 5, 6. 1, 3. There are 4. See, we got something that was never possible with just. Just to enjoy that. Because there you have a square for n to the 0, n to the minus 2. We have an n. What we have seen here is an example of a rule. First, I will do one more graph and then I will tell you the general rule in the very first time. Let me do another graph. Let me take the same guy, this guy. Now, it emanates out of somewhere else. So, the general rule has 3 blue ones coming out of it. Let's do this guy in double language. So, this, this, this, this, this, this, this, this, this. Let's do the counting again. 1, 2, 3, 4. n to the power of 4. How many propagators? These are the propagators here. 1, 2, 3, 4, 5, 6, 7, 8, 9. How many vertices? 1, 2, 3, 4, 5, 6. Let me tell you the general rule. Okay. The general rule goes like this. See, whether you got a fundamental, it must meet an empty fundamental. In order for it to be able to talk to an attribute. Uh, you can never have a vertex which there is a single fundamental line. The fundamental always has to continue to reactivate. So, there's some conservation of charge. It's like when you've got electrons always a positive. So, conservation of charge. Any graph here can be drawn with a closed fundamental loop. We've seen that here. We've seen that here. Not about the in the double double line. It becomes quite clear here. You see, what happens to the closed fundamental loop as far as double lines go? As far as double lines go, if we had added an extra line here, this would have been just another face. Yeah, it would have been an square. It's then been another line. But there isn't another line because it's fundamental. Now, this is in the sense of the tile. There's this tile here, there's a tile here, there's a tile here. There's no tile here. So, this is the paper hole. Of this fundamental matter, removes a tile from the graph. Tile back in. How many extra propagators would have been introduced? Because this was already important. It just happened to be a fundamental. How many extra vertices would have been introduced? None. How many extra vertices would have been introduced? Which is, removing a tile gives you a factor of 1 over h. So, the general formula goes as follows. The general formula goes as n to the power 2 2 minus 2g minus h. h is the number of paper holes. Every graph with fundamental interest can be thought of as some genius surface with some number of paper holes cut out by fundamental loops. Fundamental index loops. Okay? Yeah, but fundamental index loops. Okay? Every time you have a fundamental index loop, that gives you an additional suppression by one version. When the same fundamental index loops come in, it will fill in and it will break the tiling as you are. It doesn't take the power. The final formula. In theory as well, adjoint as well is fundamental index. The powers of n go like n to the power 2 minus 2g minus h. But g is the genus of the surface and h is the number of paper holes. Let me give you a slight aside. Let me give you a slight aside about the connection between stream. You know, in stream with KLE, what we do is that paternal expansion is not in terms of finding graphs. But in terms of x, which are basically lines that have a particle. Trajectories for particles. But in terms of trajectories for strings, then Euclidean space of trajectories for a string extending Euclidean space makes some two-dimensional surface. And it turns out that according to the rules for strings, surfaces with topology with a topology of gsg h number of holes in it now are weighted by the string constant, the square the string constant to the power 2 minus 2g minus h. So the fact that the same extension appears with the string coupling constant replaced by 1 over n. I said, string coupling related by 2g plus h minus 2g. So the string coupling replaced by 1 over n. Looks like a striking observation. And of course I've been noticed by the people who first discovered this, by thought. In the 1970s and 80s and maybe in the annuals there is somehow building up a string in the world sheet. And this related to entilising but unfulfilled suggestion until we have been a various year to our students. Well, in some way now in many other situations they are very precise. This is actually good fun. So this larger expansion here the larger expansion of Ironman gloves is one of the links between the annuals theory and string theory. But I will say more about that since this is not the most important topic in this class. So apart from giving you a way of organising suppose that organising is going to some interest in geometry in the class suggesting the role for some new geometrical objects in this class. But we will take that as an example. Any questions about larger expansions of matrix theories the plagiar expansion the genius expansion. What is that? The specific case is a case replaced by 1 by n 1 by g. g is not g. It is string theory. So larger dimensions begins to work Yes, that's the point. In a string like a limit string theory becomes classical. You see like I told you in the last class the two ways of getting classical limits do not ways of getting classical limits. We are coupling limits now. And what we are seeing is that we are coupling a bit of string theory. It is like the large end of the world. In the context of the area it is not just like it is the same case. However listen it is something that we understand very easily. What type of thing? Ok Other questions about is there a way that we could get another variable surface where we could not we could not we could not we could not you see the orientation comes about because the adjoint is split into a fundamental key and then the orientation is given by the sense of circulation of the anus. You understand what I mean? All of these graphs although I haven't dropped them have atoms. And the orientation of the convention so this is automatically an oriented you understand? However in an oval theory there is no fundamental in an oval theory you get the same you get the same tight link of the anus but without atoms. Because there is no atoms there is no intrinsic sense of orientation and that it does allow you to build up obvious structure. So the answer is UN theories must be oriented UN and SPN theories Of course you could ask why you fix it with UN, SPN and UN What is the answer? There are other groups of the world There is E6 What is the answer? These are the only ones which have a larger You see What is the answer? Leave Leave And these groups Who ever classifies the groups? I don't know Some famous mathematics Some famous mathematics Some famous mathematics Some famous mathematics Some famous mathematics Some time classified the groups What was the hard part of the analysis? The easy part was He knew what UN was He knew about UN He knew about SPN This is a bit from the classical books And these are these parameters We have matrices, matrices can be made as big as you want What was the non-trivial part of the classification is that apart from these obvious series from some exceptional groups there is infinitely There is E6, E78 There is three exceptions There is no such thing as an EN Where you can only do large and limit And there is a larger group There is a parameter you can make You can only do that to the big class groups What about SLS? No, we are doing compact Legos Why? Because we are worrying about what our gaze there is As we discussed in the beginning of our course unless you have a compact group you can do large and limit There is no problem with that But we will You see, because in the over-end and below the fundamentals what is important is the fundamental idea What is important is that in the UN product of fundamental and fundamental is the adjoint plus a significant In the over-end the corresponding statement of that is product of the vector plus a significant Let's look at SU2 Let's look at SU2 Let's look at SU2 let's look at SU2 we would have SU2 So in that case now as I said the product vector is symmetric but it contains that is the principle Let's look at S O N, VUED, S O 3 VUED as VUED as, what we have D VU 5 and D VU 5? All these 5 are just ordinary vector fields. There's no saying that this guy is a fundamental and this guy is an anti-fundamental. Both are the same representation. You understand what I mean? In S U N, U N, there are two different representations of dimension A. There's a fundamental and the anti-fundamental. In O N, there's a single representation of dimension A. Now, the sense of orientation, the sense of orientation goes along these index lines. Actually, I didn't even intercept it. Even in the matrices, in the matrices that describe O N matrices, the first and the last index are basically symmetric. Because O N matrices are, if you look at the generators, they're just anti-symmetric. U N are anti-symmetric of the complex combination. So, the first index and the last index are not an equal footing. So, what is the fundamental index? The other is the idea. So, in S O N, you have two indices in the matrices. There's no way. Now, what I'm saying is that there's no up and down in S O N. You see? But a fundamental index can only contract with an R. In a U N there, you've got a matrix like this. I J. The only thing consistent with gauges there is, you've got another matrix like this, which is M N. This contraction is an R. But this contraction is not R. Why? Because that's time N. That's contracted with this index. Why is this gauges there? Because M transforms like U M U inverse. N transforms like U N U inverse. And these can. Now, suppose they try to do contraction M. That's N transpose. So, they try to contract the other way. N transforms like U M U inverse. N transforms like this. So, the transpose transforms like U inverse transpose. N transpose U transpose. And for unity of matrices, it's not true that U inverse transpose is U inverse. Of all matrices, is it? There's no distinction between the two indices for all matrices. If a unity of matrices is this, that's the key. Okay? We're going to have to stop the issue. The other question. Let me introduce the problem that we're going to be studying. Okay? I'll introduce the problem. And I will maybe give you some details next time. I'll give you a few details. Next class, leave the rest for you. Work out. Here we go. So, in the top model, we're going to be studying. Now, I think you'll see the interaction with something like that. We're going to try to study this in the larger part. Now, in the larger end limit, we still have to sum an input number in the graphs. So, you might think that I'm just going to have some clever trick like we did for the zero plus zero dimension input. But, though, it didn't make a clever trick. This clever trick was choosing the right gauge. Yeah? What was the point? You see, two dimensions. All of the nonlinearity, the self-nonlinearity of the endless theory comes about because the commutator between A1 and A2. So, if we choose a gauge in which one of these two A's is just set to zero, then the angles here becomes quadratic. It doesn't become empty. It would be empty without the A. It's not empty. It's not non-interactive theory because the interaction of the matter is one of the qubit vertex. A, A5. The really complicated part about it is that it's the matrix-self-interactions. It becomes three. So, now, if there's theory, suppose you have no equation. That's a phi squared. Phi-phi. Phi-phi. Phi-phi. Some operator itself. Some other operator itself. Point function of this operator. What I have to do is to sum, and I go to the large element, I have to sum all the graphs with the least number of fundamentals. Now, because we've got a psi, the psi, the psi, the psi, the psi, the psi, the psi. We must have at least one fundamental. It's impossible that we don't get that. But we can look at those graphs with a minimum number of fundamentals. Not that requires filling in the stuff. We'll play the stuff. Now, in general, the stuff is very complicated. But, hold on. We have no glance at it. Enough, I need to tell you. Now, the way it comes is how simple does it become? Well, it's essential to this. In fact, we can also take these property elements and correct them. In this graph, if you use the convention that each of these propagators is this exact corrected property. Then this is the first sum of all the inner graphs. Now, if you solve two-dimensional angles here, we have to sum play the graphs. And play the graphs are pretty simple because there are no glance at the directions. The property it has, I'll give it by summing all of these so-called rainbows. We have the exact property that I have to use to build all of these inner graphs. If we had glance, a further correction, we had glance which connected two different propagators, then that is counted as the property. Yes. You see, suppose you had this. If you try to do it on this side, it's clearly going over something. And you try to do it on this side, that's counted as nothing. You see, the reason it's counted as nothing is that once we have this fundamental line, where is the hole in the graph? The hole in the graph of that side. This is not there. If we modify the graph in a way that will introduce new steps. Exactly. So, this will be an initial factor. Sir, if we had this sum of graphs, does it become a GP node? Yeah, it becomes all, we will submit some of this actually. Okay? Not a GP node. GP is something plus itself, plus itself, plus itself. This is more than that. Very similar to something we did for vector number of years. Can somebody point out this similar? Suppose, sigma is the self-energy. Yeah, sigma is the self-energy for this fundamental line. Then what we are getting is that and I know this and the exact quantity. Can you see that what we are getting is? Can you feel this? It's such a beautiful line. Such a beautiful signal. That sigma is obtained by taking the exact problem data between one and four. Because you can see that every one of the graphs are generated by this signal. So, it is like a gap equation even in this case. What? So, it is like a gap equation even in this case. Exactly. You obtain a gap equation which you can solve even in this case. Okay? Once you have this, then this is like a gap equation. So, that is the fundamental equation. Exactly. This is the basic problem data. Everything can be expressed here. It tells us this kind of thing. Yes. Okay? So, we are going to do this. Okay? We are going to do this just because it's fun and because it's... I mean, it illustrates techniques if nothing else. It's not a baby problem, but still it's solved with a baby. We knew this, we solved this problem. But this will happen in the next class. Because we are looking at the largest problem, the leading problem of it. And again, first let's think about it. First we have this thing. Let's work this in the other light. First we do a counting and then we'll think about it. How to think about it. So, if I had a revolving piece, how would it look? So, this guy becomes this. This guy becomes this. Oh, sorry. Let's say I'm not moving. Right? So, let us now look to see first we're counting and then we'll do this. Okay? How many new propagators are there? One, two, three. So, ten to the power minus three. How many new vertices? One, two, two, two. Okay? How many new index lines? Exactly. So, initially, there were two at the time. Yes. Two minus two. Something. Let me- It's the time that you- One minus two. What? This is- This is equal to one minus two. This is the time. This is- This is equal to- One minus two? One minus two? It is one minus- One minus two? Yeah. Okay. And to the power of this two. Okay. And to the power of this two. Now, the way of thinking in this problem I was thinking of this graph like I was trying to say to Machiavelli. You see, suppose I regarded this sphere as a graph that goes on to the sphere. Then what was the hole that was in it? The hole was the back of the sphere. And now I understand, how did I make a planar graph sphere? I took the outside edge, which is also a double line. And joined the outside part. Now because the outside is a single line, there is no this kind. So the thing that would join back of the sphere is not there. So this is not a planar graph. What? It can be made out of a... It's a sphere with a paper formula. That's the usual rule like that everybody next look at sphere with a paper formula. Now, if we looked, you see, so adding this extra line where it looks planar is not. Because it's not changing the tiling of the sphere. It's like a handle. It's like a handle. Now we have to understand how it's like a handle. Yeah, it's like a handle or a handle. Exactly. Exactly. Exactly. You can do on this. But first let's understand this. Now this thing, that this addition of this is not breaking up the tiling. It's of existing time. See, there's no guarantee that it would keep the genus unchanged. Have you understood that? There was no tiling at the back which would break up by putting these extra little things. Okay, now we want to regard this as a whole unit out of some surface. And give me one to say that today I'm using that. Try to... I'll give you a real visualizing text now. It will be as you said. Build the genus one surface and you take it home. Okay, that's how they could be handled. Not until you're able to see that. That's how it works. Yeah, isn't it? First, how vital is it, right? Since we showed the example of how we have the surface of the couple. There's nothing about the surface of the couple. It's just in this case. And then when you have a handle over it, basically you can improve the surface. Basically you can improve the pieces which weren't contracted earlier. Correct. So that's what the view on this is doing. Yeah, that's what the view on this is doing. Except that you want to see this not... You see, this is not handy because it's nothing. Yes, it's nothing. I understand that, but... So you want to see... Yeah, this is what must be going on. You want to see that clearly. Build up the pieces. The way you want to think of this is as was this guy? See, what you want to show? You want to show that it... Yeah, why is it difficult to visualize the spaces? Because there's nothing there above, even there. Like outside the wall line, that makes sense. There's nothing there. So that's why it's difficult to see the spaces. Yeah. Right. Now I'm saying the surface. See, the way the view is that... What would the surface have been? If there had also been... Yeah. That's the question about the Jesus. Okay. Okay. And if there had also been this stuff... Okay. This is sort of clearly one... Yeah, the problem there is that you would have another index which would then have the contract where you don't have that obvious. No, that's why it's a hole. So it's a hole eaten out of a genus one surface. So we want to understand the genus one surface and then we remove the green line. That gives us the hole. You see, both of these things, what hole is eaten out of a spring? This was a hole eaten out of a spring. And how do I put the green line here? It would have been out of a spring. This graph is a hole eaten out of a spring because if we added the green line, it would have been a spring. This graph is a hole eaten out of torus. Because if we added this green line, it would be a torus, except that I have to very clearly see so this green line would go like this. Let me do this next thing. Let me show it to you. I will work it out with you. We should do this. Can you see by direct counting? Instead of saying, because the lines in the double line animation are just indices, we can see the outline of the outline as the super green line and then make a hole inside that. That would also give you the same effect, right? But then you would have to join if we did that. If we did that, that's fine. We would have to join all the nuance in this. What I'm saying is that the internal nuance gives you a tiling. And then the complement of that tiling is the hole. So whether you go on the hole inside or outside, it depends on whether you put nuance here or there. It's the same. We are utilizing this. I'm very good at this. Any other questions or comments? Very good. So let us believe in this. This legacy we have next to this. So we don't have... I'm trying to take as many Wednesday classes as we can. We've done this Wednesday. So we'll do it on Friday. Okay.