 Yeah, I'm ready. Good morning, everybody. I see after the first lecture I scared some people away because the population is smaller. I also have an announcement to make. Yesterday was lecture one. Today's lecture two. And the organizers probably thought that lecture three will be quite complicated. So they asked me to give it twice. You can see it in the schedule. Instead, I'm going to ignore them. And I changed the title of lecture three, the second lecture three to be lecture four. So I assume that yesterday was kind of elementary. We did mostly ordinary quantum mechanics. Problem that almost all of it is regularly taught in the first course in quantum mechanics. This was the free particle. But then we managed to dig out of it some sophisticated stuff about anomalies. But in a very, very concrete setup where many of the usual complications of computations of anomalies do not arise, and everything can be seen quite clearly. If I get to it today, I'll present a high dimensional quantum field theory variant of this story, which will again be somewhat elementary. But we'll have more of the beef of anomalies that you will see. But before I do that, I wanted to give some lightning introduction to the theory of the Chern-Simons field theory. And this is a three-dimensional topological field theory where we write the action. And we limited ourself to u1. So it's a u1 Chern-Simons theory. And the action was written as k over 4 pi. Since we are in Euclidean space, we put an i. And then we wrote bdb. That's an integral over a 3-manifold. Or we can write it as an integral over a 4-manifold, whose boundary is m3. And we saw that the discussion would bifurcate to k even and k odd. So we saw that from the geometry of the 4-manifold and whether it does or does not need a spin structure. I just quoted the answer. But it can be shown in more completely. And today, we'll see that from a totally different perspective. So what do we compute? The first thing to do is to take a 3-manifold and just compute the partition function. So b is being integrated over. And we get a number which depends on the partition function, on the manifold, and perhaps a little bit more data that one needs to specify. So we get a number in the end, which is called the partition function. But we can also be interested in correlation functions. So what are the correlation functions? So in general, in quantum field theory, the correlation functions are gauging varying disturbances of the manifold. So they are either at points or along lines or along surfaces. So the natural objects to study here are closed Wilson lines. For example, e to the i n is a closed loop of b. So this is a Wilson line n. And it can be interpreted as the walled line. We'll be in Euclidean space here. So this is the walled line of a particle with charge n. And we see that it has charge n because that's what appears here in the Wilson line. Have you heard of Wilson lines before, or is this the first time? Who has not heard of Wilson lines? Who did hear about Wilson lines? So that's the vast majority. Well, if you didn't hear, that's what it is. When we compute in perturbation theory, so we can try and compute things in perturbation theory. And there will be lines. So this is the line. This is w. So this is the line l. And there could be diagrams where b, there is a b propagator like that. And these diagrams are typically the virgins. They need to be regularized, et cetera, et cetera. And the upshot of all that is that they give a spin to the line. So I'll just quote the answer here. This s is lower s. So this is big s. This is the spin of the line. And it is n squared over 2k mod 1. So this needs some explanation. So first of all, there's a quick way of doing it. If you ever studied two-dimensional field theories in two-dimensional field theories in the free field, the operator e to the ikx, or e to the ik5, has dimension proportional to n squared. And you put in the denominator the propagator. So the propagator here is 1 over k, because this is the coefficient of the kinetic term. And therefore, we have a k in the denominator. And there are n here and n here at the vertices. And therefore, we have n squared in the diagram. So all we need to compute is the 2 in the denominator. And that takes a little bit of work. The second thing is the mod 1. What this spin means is that we have a line moving in three dimensions. And we ask, what happens if we rotate space around it? So it transforms like it has some spin. But this is two-dimensional spin, not three-dimensional spin. As we move around it. Now, this Wilson line can always emit soft photon, soft B fields. So as it emits a soft B field, that can take some angular momentum away. That angular momentum will always be an integer. So we define the spin of the particle, or the spin of this line, only to be modulo 1. So these are some comments about this formula. And we'll come back to this formula soon. So imagine we have this line. And we've already renormalized the self-contractions by assigning it some spins. And now we can ask, what is the equation of motion when there is a line? So the equation of motion, we take the variation with respect to B, tells us that dB is 0, which means that the field strength is 0. This is valid at generic points. But it's not valid when there is a line going through our manifold. When there is a line going through our manifold, we have k over 2 pi from here. There is a k from here and a 2 pi because there are 2 Bs. We either take the variation with respect to this or with respect to that. So that gives us k over 2 pi. And then from the line, we get an n. So that gives us dB. And then we have plus. And I have a minus sign because it's e to the minus d action. So we always compute the functional integral with e to the minus d action. So that gives me n times a delta function along the lines. That's the equation of motion. Is this clear, or should I do it again? Should I do it again? How do we find the equation of motion? If you take a classical Lagrangian, you compute the variation with respect to the degree of freedom. In this case, B. This action is quadratic. So since it's quadratic, it's completely solvable. The integral is a Gaussian. And it's really dominated by the solution of the equations of motion. So what's the variation from here? So we have e to the minus s. The variation is there's a minus sign from here. There's an ik over 4 pi, ik over 4 pi. I copy. Then I take the variation with respect to B. So there's a factor of 2 because I can get either from this B or from that B. So there is a 2. And there is a dB. So that's the variation of the action. Now in the functional integral, we also have an insertion of that. So we also need to take the variation here. This gives us plus i in with the delta function of the line. And the equation of motion has to be set to 0. And I leave it as an exercise for you to go from this equation to that one. What does this mean? This means that the field strength, so it's 2 pi n over k delta of l. We'll not need all the. So this means that the field strength, or more mathematically, the curvature is 0 everywhere, except for a delta function on a line. That's what it means. So there's a delta function on a line. But it also means that if we compute the holonomy around the line, so take another line, l prime, such that this is the wall line of the particle, l. And this is l prime. So if we compute e to the i line integral of b around l prime, we're going to get a holonomy, which comes from here using Stokes theorem. So it's e to the 2 pi i by i n over k. This follows from this equation. Because we have an integral of b. We write it as a surface integral of db. And the surface integral of db just picks up the delta function when it touches the other line. So what do we learn from that? That we have a collection of lines. And because of the collection of lines, the field strength is 0 everywhere, except for delta function support on the lines. Each line has its own delta function support. And instead of discussing delta functions, we can just say that we remove these lines. But they are non-trivial holonomies around these lines. So as we go around these lines, we pick these Bohr-Maharon phases, which are this e to the 2 pi i n over k. This means, in particular, that the line wn is the same correlation function as the line n plus k. Because they have the same holonomy around them. Would I prove that, or is that clear? That's clear. So one might say that we don't need to allow all possible n's because some of them are related by this formula. So how many lines are there all together given this identification? OK, that's almost the right answer. This was a trick question. In order to understand that better, I want to introduce another notion, the notion of a monopole operator. So let's put this aside for a moment. And let me define a monopole operator. So we remove a point from spacetime, not a line, but the point. So now we have a sphere around it. So there's an s2 around the point. So a manifold is whatever, r3, or whatever. We just look at the neighborhood. So it looks locally like r3. And we remove a point. And now we have to specify boundary conditions for the field b around that point. So we can specify boundary condition that the integral db around that s2 is 2 pi some integer n. So simplicity, let it be 1. This means that there is one unit of flux coming out, one unit of magnetic flux coming out of that point. This is the definition of a monopole operator. In general, in quantum field theory, you can define operators either by writing them in terms of the classical fields. That's what you normally learn in school, write some function of the classical fields and the derivatives and so forth. But there are other ways of defining insertions, local insertion. Remove a point from the spacetime. And you specify something about the behavior of the fields or the degrees of freedom near that point. So this is a definition of a monopole operator. So we have a little s2 coming out of it. And it has one unit of direct, one unit of magnetic flux coming out of it. And we are allowed to put the flux wherever we want. And we would like the field strength to be generically zero, because that's what the equation of motion tells us. So we'll be able to satisfy the equations of motion almost everywhere. There is no way we can satisfy it everywhere because the integral of db has to be non-zero. So the most efficient way to do that is to say, this is our s2. And we take a Dirac string and taking it out of here. So there is a line emanating from this. So we are going to define this thing as a magnetic monopole for b. And we have to attach to it an open Wilson line, where we get the correct delta function. And if we look at this equation, we'd like the delta function to be exactly 2 pi. So n should better be k. So a good gauge invariant operator is this guy, e to that power. So this is the magnetic monopole. Another way of saying it is the magnetic monopole is 2 pi flux. If we just look at the Lagrangian, if we have db as delta function of 2 pi, we see that at that point b is non-trivial. So the magnetic monopoles carries electric charge. So it started its life as a magnetic monopole, but it carries k units of electric charge because of this equation. And if it carries k units of electric charge to make a gauge invariant, we have to attach this Wilson line. This is another way of seeing why the Wilson line w sub k, the Wilson line with charge k, has trivial correlation functions. Because remember this identification as far as the holonomy is concerned. n is the same as n plus k. They induce the same holonomy around it. So if we put 0 here, it's the same as k. 0 is just a trivial line. It's like the identity operator. And if this is the identity operator, it doesn't have interesting correlation function. And the Wilson line with charge k also doesn't have interesting correlation function. This gives us another perspective of it. We have some complicated correlation functions of various lines. And imagine one of them can end on a point. If one of them can end on a point, we'll just open it briefly here and here. We haven't changed anything. And then we can unwind this line in a continuous fashion. So every line that can end on a point, the line will have trivial correlation functions in this topological field theory. So we have just another perspective of this phenomenon that the line with n and the line with n plus k have the same holonomies. Because we can take one of them, just attach this monopole operator, change the charge by k units, and then unwrap it. But there is still an interesting subtlety. If we look at this formula, we can compute what's the spin of this line. So the spin of the line with charge k, so the spin is k squared over 2k, which is k over 2. And that's always mod 1. So if k is even, that's 0. That's good. So if k is even, the answer that was given here before is correct. This line with charge k is completely trivial. It has zero holonomy around it. It has zero spin. It's really like the identity operator. So this identification is correct. But for k odd, that's not correct. Because for k odd, the line has spin a half. Let me say it differently. For k even, the independent lines are 0 plus and minus 1, k over 2. So they are k lines. So that's the easy case. k is even. They are k distinct lines. They give us holonomy n over k. And the line with charge k has integer spin, and therefore we set it to 0. It's the same as the trivial line. But for k odd, that's not true. For k odd, the line with charge k has spin a half. It has trivial correlation functions as far as the holonomy is concerned. But that line has a spin a half. And therefore, it's not completely trivial. So such a line is being referred to as a transparent line. It's transparent because we can't see it. But we can detect that it's there because it carries spin. And this is an important distinction here in this series. So the discussion, we already said it yesterday, bifurcates to two separate discussions. For k even, we don't need a spin structure. This is an ordinary topological field theory. They are k lines. And the line with charge k is completely trivial. It can end on a monopole operator which has integer spin, which is therefore trivial. For k odd, the theory is not an ordinary topological field theory because the manifold on which we specify it needs a choice of spin structure. So we need to supply more information to define the theory. It has to be a spin manifold with a choice of spin structure. Correspondingly, there is this transparent line with charge k, a line that has no holonomy around it, but it carries spin. So that's an interesting thing to take into account. And correspondingly, the number of independent lines is k. It's 2k. Maybe I'm off by 1. I'd like to give a physical picture of this. One way of thinking about all these topological field theories is the description of what is known as enions. These are particles which have possibly fractional spin, fractional statistics, and possibly fractional charge. And the simplest example is the fractional hole effect where there are enions which charge a third and spin a third, et cetera. These guys are described by such a theory with k equals 3. Since k equals 3 is odd, this theory needs a choice of spin structure. And it has a transparent line. The transparent line is e to the i3b. And as such, this transparent line does not have any interesting correlation. It does not have a non-trivial braiding with the other particles in the problem. But it carries spin-a-half. So what is this particle in the fractional hole system that has spin-a-half, has relatively trivial correlation function, and the ordinary particles, the enions, have charged a third, this guy has charged 1. So what is this particle that does not braid with anything? Carries charge 1 and spin-a-half. And therefore, it's non-trivial. Sorry? The electron. This is the underlying electron. So this as physical as it gets, this transparent line is really physical. It's an electron going through the system. It does not braid with anything, but it's perfectly physical. Next, oh, my timing is terrible. Next, I would like to consider a system which has several fields. So that's going to be easy on the blackboard, but harder when you take notes. We'll just sprinkle some factors here. So I'll put here i and j. Put i here and j here, and that we will not even need. So since I'm running out of time, I'm going to use first trick in the book. When I'm running out of time, I'll give you more exercises. So the exercises would be, consider this theory. So it would be very similar to what we've been doing before. And the exercise is to repeat the same discussion with the indices. And what you need to compute are the lines. The lines are going to be ni line integral of bi. And my question for you will be, what is the spin of w ni or the w of ni? So this is a vector. And what is the spin of this line? Find the monopole operators, computer spin using this formula that you would get from here. And find when is it a spin theory? And when isn't it a spin theory? So what's the condition on the case such that it is a spin theory and when it's not? So I'll give you the answer. This should be the last thing. So I'll give you the answer so we'll know whether you got the right answer or not. What you need to have is all kiij to be integers. And then this is i and this is j. And if all kiij, all the diagonal ones, are in 2z, it's non-spin, which means that there is no transparent line. But if one of them, this is if all of them, it's enough that one of them is odd, at least one is odd, it's a spin topological field theory. So I'll leave it as an exercise for you. It's actually a very educational exercise. Next, I'd like to discuss in more detail a very, very special case of this. So going back to a single field, so the action is i over 4 pi. There you go, bdb. Very special case. So this is u1 gauge theory with level 1, so k is 1. First question, is this a spin theory or a non-spin theory? This is not a trick question, unlike the other one. It's a spin theory. How do we know it's spin? Because k is odd. How many lines does it have? Two lines. So what are they? So first, we have the identity e to the i b. And what is the spin of this guy? A half. So we have two lines with spin 0 and with spin 1 half. All the correlation functions are rather boring, because there's no non-trivial braiding, no holonomies of that. All this is trivial, except that this guy is spin 1 half. So this theory is almost trivial. When you compute things, there isn't a lot to be computed. And we can write a formula for the partition function. So the partition function, which is a functional integral over all possible b e to the minus this action, i over 4 pi, bdb. It can write a closed-form formula for it. For any manifold, have minus signs. The signs are crucial. It's e to the 2i. It's given by an integral over a gravitational-churned Simon's term. And the gravitational-churned Simon's term on some m3 is defined again in terms of a 4-manifold, 192 pi. An integral over m4. So this is the boundary of m4. Place our wager. So you take your 3-manifolds. You attach a 4-manifold to it. You compute this integral that's a number. And that's the answer for the partition function. So this theory is almost completely trivial. Do you know what r is? Should I explain that? You do or do not know? No. So r is called the curvature. Should I explain the curvature? So we take these manifolds. We put this metric on it. And using the metric, we can compute the curvature. Have you taken any gr? Do you know what the curvature is? So you know what trace our wager is? I think of it as a form. So it's a metric in two indices, and it is a two-form. This is a standard expression. And all these things have names that I keep forgetting and confusing them. The important thing is not so much what it is, but it is completely determined by the metric. So you know the metric of the 4-manifold. But not only that, it's actually independent of most of the information in the metric. Because for a closed-form manifold, this integral is a topological invariant. So we have only the fact that it is the boundary. So it's a cousin of what we discussed yesterday. So we are going to use the following terminology. The Lagrangian with i over 4 pi bdb is dual to minus 2i, the gravitation of trans-Simon's Lagrangian. So what this double arrow means, left-right arrow, what it means is that they're not the same. These are different Lagrangians written in terms of different variables. But this is an example of duality. It's a very, very trivial example of duality. We have a quantum field theory on one side and a classical field theory on the other side. So computing correlation functions here, or putting this on an arbitrary manifold, it's the same as putting this theory, which is classical, on the same manifold. So this is not to say this left-right arrow does not mean that every configuration here matches to a configuration here. These are two different quantum field theories. It's only the integral on the left-hand side. It's the same as the integral on the right-hand side, except that in this particular case, there is no integral on the right-hand side, so this is a classical field theory. So this is an example where a quantum field theory, that's why this theory is almost completely trivial. It's a quantum field theory, but it's really a classical field theory in disguise. So it's a quantum field theory where the quantum variable is B, but the quantum field theory, the variable is B. But in the end of the day, we don't have to think about B because it depends only on the metric. And we'll find it convenient. So talking about this almost trivial theory, you will soon see why I spent so much time discussing this almost trivial theory is that we can modify it a little bit and put in the Lagrangian, not i over 4 pi BdB, but couple it to a background field, i over 2 pi, a background U1 field, BdA, ADB. So now we couple the source, but we have enough information to do the functional integral over B. Can anybody suggest how we do that? Complete the square. Thank you. So we complete the square. And then we have something that depends only on BdA, and the rest is the same as before. So this theory is dual in the same notation. So now I have to get the signs right because when I complete the square, I always get the signs wrong. So this is minus i over 4 pi, minus i over 4 pi, AdA, minus 2i, the gravitational transform. And I'm going to define that as minus i depends on the metric in A. So if we specify background metric and we specify big A, background U1 gauge field, the functional integral over B with this Lagrangian is given by this classical expression. So the last example I want, let me skip that example and move to another topic. I do want to get to some more interesting things rather than just give you background material. And on the other hand, I feel that the background material is essential because if you just jump to the state of the art modern things, if you don't have good foundations, then we'll eventually have to go back. I'm trying to find kind of a smooth path. So now I'm going to put some of these pieces together. We talked about anomalies. We talked about Sharon Simon's theory. And now I'm going to put them together and discuss another point. And this is the functional integral over fermions. And here, surprisingly, there is enormous confusion in the literature. So I'm changing topic. So the topic is integral over fermions. And my understanding is that the community has been using the wrong point of view for decades. And this was straightened out only in the last few years, mostly by Witton. So I'll present the modern interpretation here. So we have a Lagrangian or an action, which is i psi bar d slash a psi. With this Dirac operator, depends on the metric and on the background gauge field a. And we would like to compute the functional integral over psi. So the absolute value of the partition function, so we have a partition function, and it has the determinant of the Dirac operator. So that's what we should do. The absolute value is easy to define. But the interesting question is, is there a phase? And I'm simply going to state the answer here about the phase. And then I present various ways of thinking about the answer. Some of them are right, some of them are wrong, and I will explain that. So it's e to the minus i pi over 2, something called the eta invariant, which is a function of g and a. Where the eta invariant is a non-trivial function of g and a. And it satisfies that this eta is 1 over pi, the integral of what we defined before as i mod pi as an integer. So in particular, e to the i pi eta can multiply that by pi. So we get e to the i, the integral of i, the same i we defined before. So if we didn't have the i over 2 here, we would have said that the answer is an expert, the phase. So there's an absolute value, and there is a phase here. And we would have said that, aha, the phase is this term, Simon's level 1, with the gravitational term, Simon's term. That would have been easy. But we have this crucial 2 here, which tells us that that's not true. So some people would say, aha, this is term Simon's level 1 half. But term Simon's term level 1 half does not make sense, whereas this object does make sense. So I'll say more about that soon. This is the correct answer. And now we can ask, what happens under time reversal transformation? So under time reversal transformation, let's this be time reversal transformation, of this Lagrangian that we wrote here. We get the same Lagrangian, but we have to shift it by minus i times this object i. So this Lagrangian for the free fermion looks like its time reversal invariant. But in fact, it's not. When we perform a time reversal transformation, it is shifted by term Simon's level 1 for a, and a certain term Simon's term for the gravitational field. No, that's the whole point. I'll soon say what happens when you add a mass. You do need to regulate it. And the reason so you can say, why is there a phase anyway? Where did the phase come from? It came from the fact that the determinant is an infinite number of eigenvalues. And in the determinant, you have to multiply all of them. And there are positive eigenvalues and negative eigenvalues. So for the absolute values, no problem. But as you vary the gauge field, the whole spectrum of eigenvalues can slide. And the number of positive and negative eigenvalues could change. And when you multiply them, so what is the sign? The sign is the number of negative eigenvalues. That's infinite. So you cannot control that. But you can control what happens if some eigenvalue slide from positive to negative. And that's actually the key point. Because when a given eigenvalue slide from that, so you vary A and G and whatever you, and suddenly you hit a zero mode. What happens then is that an eigenvalue slide from being negative to positive, and the answer should flip sign. And that happens precisely when the determinant of absolute value vanishes. And this thing does exactly this. So this expression of Tia Potodic's finger does exactly that. It keeps track. It changes the sign when the eigenvalue goes through zero. I'll soon say more about that. So this comes from a regularization. Unlike the anomaly we discussed yesterday, where we didn't need to regularize anything, this can be thought of as coming from a regularization. But I don't want to go through this route. I would simply quote the answer here and state what's true and what's not true about it. So I'm going to say a lot more about it. So there is a spin connection in D slash. And there's a gauge field connection A. And when I perform time reversal, I really mean with the time reversal on the metric. The important thing is that this is a typical anomaly. The partition function, the theory looks like it's time reversal invariant. And therefore, it looks like the answer has to be real. The partition function has to be real. But it's not. It has a phase. However, it's not that bad. Because on the time reversal transformation, the Lagrangian is not invariant. But it is shifted by a term which is meaningful. I is meaningful. We can integrate it. This is a properly normalized John Simon term. So we can say that if we ignore the gravitational term, that when we perform time reversal transformation, the Lagrangian is shifted by John Simon's level one. Now one might say, ah, we have my one here. Why don't we split it evenly and put a half here and a half here? We said that yesterday. We are not allowed to do that because John Simon's level a half doesn't make sense. And I made a big deal out of it yesterday. Yesterday it was so easy that you thought probably, why is he making such a big deal out of such a trivial point? But now it's a little bit more subtle. Because now this John Simon term is that needs the quantization for more subtle reasons. But essentially, it's the same thing. So again, when we have an anomaly, what does it mean? This is something I said very explicitly yesterday. And I'll say it again. And I might even say it a few more times because it's really important. If we turn off the metric and the gauge field, the theory is perfectly time reversal invariant in the sense that all the correlation functions are time reversal invariant. So time reversal is a good symmetry of this problem. What is not true? What is not true is that the contact term in correlation functions are not time reversal invariant. We need to define contact terms. And the contact terms are not invariant on the time reversal. Similarly, if we couple the system to background A, or background metric, and we start expanding in A, even for infinitesimal A, by the time we go to second order, we have A times the current. And these two currents can touch each other because we integrate over space. So we have an integral of J times A. And this integral of one J hits the other J. So as long as they are separated points, everything is time reversal invariant. But when they touch each other, there is a contact term. And the contact term is not time reversal invariant. So this is like in a tuft anomaly. The correlation functions at separated points are OK. The contact terms are not. And therefore, we say that this system is time reversal invariant, but with an anomaly. It's not that the time reversal symmetry is completely ruined. It's ruined only in a very subtle way because most observables are perfectly fine. They're perfectly time reversal invariant. And this is what we detect here. Not only that, even the observables that are not time reversal invariant, like the partition function is a function of A, transform on the time reversal in a simple way. So under time reversal transformation, the answer is shifted by something which is meaningful. The problem is that we cannot take something which is meaningful and take it apart and redefine what we mean by the Lagrangian to find something which is time reversal invariant. And every one of these sentences, I said yesterday. Except that yesterday, it might look too trivial. And now, it might look too subtle. But I really emphasize that this is conceptually identical to the example I had yesterday. There, the variation was a term linear in A. And so the variation was a term linear in A, but we couldn't split it between the two sides. Now, if you look at the literature, you will find a totally different description of this phenomenon. So let me repeat now what you find in the literature. And this is literature of the last 30 years or so. People say the partition function here is not gauge invariant. People refer to it as u1 level 1 half. The partition function is not gauge invariant. In order to make it gauge invariant, we add by hand a term Simon's terms level a half which is also not gauge invariant which cancels the anomaly in gauge invariance of this thing. Once we do that, we find the gauge invariant answer. That's imprecise. It's imprecise precisely because of this phase. This phase is almost the same plus the gravitational thing. Let me not bother with you. That's almost true. Now, this there is a half here, except that it's sometimes plus that and sometimes minus that. In fact, we have this formula here. This formula, let me flip the signs. Oh, it's actually correct. This formula is a correct formula. When we try to take what we need is to take the square root of this formula. The square root of this formula has a sign ambiguity. And therefore we cannot just say, ah, it's level a half. That's not true. It's true up to a sign. But all the physics here is about this sign, the sign that can jump. And when does it jump? It jumps precisely when an eigenvalue goes through zero. So what we could say is keep a generic. If a is generic and the metric is generic, there is no zero eigenvalue. And if there's no zero eigenvalue, we change a a little bit. There's still no zero eigenvalue. The variation of this thing in the exponent is the same as the variation of this thing in the exponent. So piecewise, this formula is true. And there might be an overall sign here that we pick once and for all. But that's only true piecewise. When we cross this point where an eigenvalue grows through zero, the sign here jumps in the formula. And therefore this formula is always true and this one is not. So it's a plus or minus sign. Also, we discussed yesterday that one could define the theory many different ways. That given the theory, we want to couple it to background fields. Who stops me from adding here i over k over four pi, a d a, it may be some gravitational terms. So when we discuss the free fermion with zero a, we don't see this and we don't see that. But we can add the background a here. And once we add background a, we have an ambiguity of which on Simon's term we add here. We have seen that before yesterday. So that when we had the particle on the circle and we want to couple it to a background field, there's a one parameter way of doing it labeled by an integer k. Same thing here. So you and I are interested in the same fermion and we want to couple it to a different a, but you and I are going to pick different k's here. There's no preferred value. The formulas are wrote so far with k equals zero. But since this is classical, that goes along for the right and we can just edit everywhere. In fact, if we decide to regularize the system in different ways, we might end up with a different value of k. This is a general rule in quantum field theory. Different regularizations are perfectly okay. The difference between them is in a counter term. So you and I regularize the theory differently. You have your own regulator and I have my own regulator and I refuse to listen to yours and you refuse to listen to mine. That's okay. In the end of the day, we'll compare our answers, but we are allowed to change the value of this counter term. But this counter term has to be properly normalized. It is an integer. And the main point here is that this whole subtlety that we have been discussing cannot be absorbed in a redefinition of this k. It could have been absorbed if we allowed half integer k, but we do not allow that. And again, I emphasize this is identical to what I said yesterday. It's exactly the same thing, but in a slightly more subtle system. So what are we going to do about this anomaly? We faced the same question yesterday. So option one, we say, well, the theory is not invariant on the time reversal. That's a fact of life and we'll just have to live with it. This is option one. We had that option yesterday. There we had C and we could have done it also with T. Would have been the same thing. Option two, say that the gauge field was normalized incorrectly. So options. Option one, we don't have this time reversal symmetry. End of story. Option two, wherever we wrote A, write two B. And B is a properly normalized gauge field. Now there's no problem because we wanted to add one over half times four pi ADA. This is ill-defined, so I'll put it in quotation. But in terms of B, this is two over four pi BDB and that's okay. Since this is okay, I can add such a counter term. In other words, if we say that the fermion carries not charge one, but charge two, we can add the counter term and fix this problem with gauge invariance. Or fix this problem with time reversal. Again, very similar to option two yesterday. We had to go to some double cover. And we have option three, which we also had yesterday. And that is to say, we thought our system is two plus one dimensional, but in fact it is three plus one dimensional. And therefore we add fourth dimension. And if we recall, we defined here the term Simon's action of a three manifold by adding a four dimensional bulk and arguing that things are independent of the bulk. That's what quantized K for us. Now we're willing to relax this assumption. And we say that our three manifold is the boundary of a four manifold, but we are willing to contemplate dependence on what happens in the bulk. Now we have more options because now we can add this half integral to Simon's term. So I'm going to study now the Lagrangian, which is I psi bar D slash A psi. This is what we did before, but I'm adding minus I, and I'll put it in parenthesis, one over eight pi ADA plus this gravitational trans-science. And what I mean by this, this is half of what it is allowed to be, but what I really mean by this is that this should be viewed as coming from a bulk. So in the functional integral, we have e to the minus s, which is e to the minus an integral over I psi bar D slash A psi over the three manifold times or minus plus I one over eight pi integral of, sorry, I integral of DADA over eight pi plus D of the trans, the gravitation of trans-Simon's, and this is integrated over the four dimensional manner. And since this is a half relative to what we said before, the answers here depend on the choice of four dimensions. So when we extend the gauge fields to four dimensions, before we said, yeah, we should do that, but this is a trick. In the end of the day, we have to make sure that the answers are independent of the extension to four dimensions. Instead now, things depend on what happens in four dimensions. So that's the price we pay. We need to specify more data to define the theory. And the bonus we get in return is that the system becomes time reversal invariant. So again, the three options was keep the system three dimensional, but declare that it's not time reversal invariant. That's kind of the standard way of dealing with it. Second, change the normalization of the gauge field, say that the fermion has charged two. And three, keep the charge of the fermion as one. Restore time reversal symmetry at the cost, and the cost is that things depend on what happens in the bulk. So when we add all these terms, the partition function Z is still the absolute value of the determinant, the same way we had before. And we had this eta before. Had e to the minus i pi over two eta. But now we have plus i pi, whatever integral of m four, i integral of m four of this object. This is classical. And this whole thing using the athiapatode, this single index theorem is plus or minus the determinant of d slash a, depending on the configuration, it's plus or minus. And this plus or minus can also be thought of as a minus one to the index of the Dirac operator. And the key point is that it is real. So what have we done here? So yesterday somebody complained that my trivial quantum mechanical example was not quantum field theory. Where is he? So now you got it. You got exactly the same physics in a quantum field theory, but you see it's a little bit more subtle because it's associated with infinities and how we regularize the infinities, the phase got in through there. And we had these options of sacrificing one of the symmetries in the problem. In this case, it was time reversal or extending things to the bulk and characterizing the answer by what's going on in the bulk. The dependence on what's going on in the bulk is very mild. Most of the action there is topological. We take two of these objects and we put them together. It's only a plus or minus sign. So if we take this bulk action and we exponentiate it and we consider a closed form manifold, it's only a plus or minus sign. So small changes don't make any difference. It's only big changes and even then they only affect the sign. But when the dust settles, when we add this object, we get a real answer and therefore time reversal invariant. Another way of saying it is that for a closed manifold, the partition function with this action gives us plus or minus one. Calling it partition function is overkill because this is a classical field theory. So for a closed manifold, this partition function is plus or minus one and it's time reversal invariant. The problem is that when we have a boundary on the manifold, the answer is not real. The answer depends on what's going on on the boundary and it's not real. There's a phase. Our fermion problem is the same thing. It also has a phase. So we say that both the fermion problem and that particular action, both of them have an anomaly on the time reversal. This one only when there is a boundary and when we glue them together, the anomaly in one cancels the anomaly in the other. So there were some questions. Well, there are many ways to, there are many physical ways to understand that but the main thing here is that the change is a function of this A. So when there is a boundary, so that's why it's minus one to the J. So the issue is when an eigenvalue jumps. Yeah, so the answer is yes, but this will take too long to explain. Any more questions? Okay, this term we add from the bulk should come from some quantum system, for example, which lives in the bulk. Do you know an example of a system which gives such a term? Yeah, so that's an excellent question. First of all, the term as it is is a classical term. So I'll reinterpret your question. Can we make the system more complicated by having quantum degrees of freedom in the bulk such that when we integrate them out, we find this? So let me give you the answer. So this is the bulk. Say you call that the z direction and z positive is the bulk and here's z negative, it's the vacuum. So imagine you take a fermion and give it a mass, a four-dimensional fermion. Here it is. So now we take a four-dimensional fermion. So forget everything I said so far and we're going to start the whole story again, but from another perspective. So we write a four-dimensional fermion and it has a mass which depends on this z and it's positive here, some m naught and it's minus m naught here. So you have a fermion and its mass depends on z. Other than that, everything is trivial. So this system, it's a four-dimensional fermion, not three-dimensional, three plus one-dimensional fermion. This system is manifestly time reversal invariant because everything is real. M naught is real and it jumps and if you don't like that it jumps rapidly, a discontinuously, you can smooth it out a little bit. Makes no difference. So this system is time reversal invariant and it has fermions in the bulk and it also has fermions in the vacuum outside and there's nothing special on the boundary except that the fermion changes its mass. But it suggests that there might be some massless fermion on the boundary at the point where m goes through zero. This is a little bit too sleek but gives you the right intuition. Now we can do something interesting here and say in four dimensions we can change the mass of the fermion by doing a chiral rotation. So we could take psi, so we have psi and can rotate it into the i alpha gamma five psi and that changes the mass term because the mass term is psi bar psi. So it changes the mass term appropriately and as it does that, it shifts the Lagrangian by what is called also an anomaly but there's an alpha f where f and I will not bother to get the coefficients here straight. So the whole thing is still coupled to classical electromagnetism. So electromagnetism is not dynamical here with just a classical background field but there is such an anomaly. So this system with a master that changes like this can be thought of equivalently as a mass being independent of z but there's a four dimensional theta angle which is zero outside and it's pi inside the material because that's what the chiral rotation does for us. Now you can solve the Dirac equation here and realize that when you decompose so you can decompose a system in the orthogonal direction x, y and time. So you solve the Dirac equation with such a profile for the mass and in the z direction you get two exponentials and you find that there's a zero mode. There's a three dimensional fermion localized near the boundary and it's massless. That follows from this description. From this description you see that theta is pi inside the material and theta is zero outside the material and you don't have to think about the fermion because the fermion has mass all the time but we've already derived that there must be a massless fermion here. So all these pieces now have a physical interpretation and the physical interpretation is that we have a four dimensional fermion with mass which does that, theta independent of z or equivalently mass being constant but theta jumps either way the system is time reversal invariant. And since I've said all that I might as well answer your question now. Imagine you take a magnetic monopole from outside. So I take a magnetic monopole from outside so somebody handed me a magnetic monopole it's all kind of theoretical, it's not real life but this is a thought experiment that we can do. So we take a magnetic monopole and it happily leaves out here, has no electric charge and we try to drag it into the material, into the bulk. In the bulk we have theta equals pi. We have theta equals pi, it means that this magnetic monopole carries now electric charge. Where did the electric charge come from? Electric charge is conserved. The only way the magnetic monopole can have electric charge in the bulk is that as it went through the boundary it deposited some electric charge on the boundary. So we take a neutral magnetic monopole, we bring it in, it is charged here and it leaves behind some electric charge on the boundary. That by itself tells us that there must be degrees of freedom that carry electric charge on the boundary and these master's fermions that I have been discussing do the job. I think I'm over time so I'll stop.