 Skupaj. Da. Skupaj. OK. Myslijo da podemos začati vsevno. Vstupaj smo začeli srečke lektu, na nekaj tukaj sebev, z njeljavnimi in različenimi vsevnoj 2-1-zimesnimi quantofil teori. Počin. Češo. Zateljno da smo na ICTP, začo je prvi visit. Vsah je prijevno, Zelo bom bila tako čutim v tem, da sem si zelo srednje. V moj boženiji manje drzili prej od diy v tju začutim na velime, in božem otvari. Puputim me iz temptedi, da po zdaj si neko razipravim, ne boš, da boš, da počutim, kaj si nekaj. Count det kako mora srečenik na kondensemözim fiziko. Srečenik na hrečenih energijov, všim njo. Z � CAM. Aha, nekaj je vrečenik na matematik. in in vsezivno, tako je hodnje energijne fizika? Daj, da se mi... In kaj so mnogo, nekaj je studenci? Vsezivno, nekaj je studenci. Zelo je zelo. Zelo, da je zaživaj, tudi je zelo, da sem vsezivaj, zelo, da sem vsezivaj, zelo, da sem vsezivaj, in zelo, da sem vsezivaj, ne zelo. If you want to read the references everything I'll be talking about was basically to my notes is copied and pasted from papers I wrote during the last 2 years and they have complete list of references. I start giving the references especially given that the presentation will not be in historical order, it will be very difficult. Because things that will look completely trivial from the point of view I'll present Pro blocks and them here, were completely non-travel when they were first discovered and it really made it very difficult. Second general comment is that the level of my presentation and the needed backgrounds would be total, totally uneven. There would be some things that look completely elementary like the first half of today's lecture, there would be some concepts there that I will use later s mešljenji, da sem bilo, da bo to zapotrati. InIf you think, this is to trivial, I know that, please, please, please don't decouple from the presentation, because various comments I'd be making in the trivial things would be connected to things later. The flip side of that is that some things will be quite sophisticated, maybe towards the end of this lecture načom vse. In ta, če energijame played, vse ni razdužili ga v tom, da so lahko sajno. If you find something too complicated or too fancy, don't think oh yeah, I lost, that's the end, I'm not going to pay attention any more because most of what will be afterwards, this will not be essential, although knowing the deeper stuff will really help. So, I am going to start with a very simple quantum mechanical system that I really love because you can learn an enormous amount of physics from it. And this is the problem of a particle in a circle. So, we have one quantum mechanical degree of freedom q, which is identified with q plus 2 pi. And we write a kinetic term, so there is a Lagrangian, ki je 1,5 q dot square. Zato sem si však odložite sistem. Zelo je zelo, da je vse simetri. Zelo je spektrum energijov. Zelo je momentum pi, ki je q dot. Zelo je energija, ki je 1,5 p square. Zelo je vse funkcije. Zelo je vse funkcije. Zelo je funkcije, ki je psi n, ki je q n, ki je 1 over 2 pi, ki je i n q. Zelo je energija, ki je 1,5 p square. Zelo sem vse zelo se zelo zelo. Zelo je vse komplikatno, da je zelo zelo elementri. Zelo je zelo, da je zelo v rovi, delta over 2 pi, q dot. Zelo je vse nekaj, da se vse nekaj, ki se zelo izvečilo? Vse svoje Mercuryo ne ga se vedimo. Zelo je vse, da je zelo, da je oklopo vse. Zelo je ovo, da je vse k glasbene, ki je x pi, ki je, da je zelo 2 pi. Zelo je vse, da je vse, kaj je zelo, da je zelo analiziračne z tem, da je tudi pravno, da ne se tako ne zelo. Zelo to je izgledo, da je vse. Tako je momentum. Tudi da se pričelim pa. To je ampak, da je vse vse. To je zelo energija. In energija je nekaj zelo, da je pi plus theta, nekaj ne boš vzela, ali da sem tukaj vzela, da je minus i dq minus 1 over 2 pi theta square on psi n equals en psi. The wave functions will remain the same, but the energy will be that theta over 2 pi 180 square. So what does the spectrum look like as a function of theta? So this is the energy. For theta equals 0, there are states, there is a state here. There are two states here. There are two states here, et cetera. And the states have energy as a function of theta, which live on parabolas. So if theta not equal to 0, the degeneracy here is lifted, and the degeneracy here is lifted. So we have a bunch of parabolas. And the physics is periodic in theta, but the square is here. Thank you. Thank you, at least one person pays attention. And so these are the energies. And the degeneracy that was here, you could ask, why was there a degeneracy here of the states? The degeneracy here in the states was that when theta equals 0, there is a symmetry that maps q to minus q. And for all theta, there is a u1 symmetry that maps q to q plus alpha. So these two states, this is the state with n equals 0. So this thing maps the state n to the state minus n. So for n equals 0, there is only one state, but the states with n equals plus or minus 1 are degenerate, and the states with n equals plus or minus 2 are degenerate, et cetera. So these two symmetries, which we call these p, or charge conjugation, or parity, these two symmetries together give us an O2 symmetry. So the global symmetry of the system at theta equals 0 is O2, and it turns 0 theta. P is broken, but the u1 remains a good symmetry. Furthermore, the Hamiltonian at theta is conjugate to the Hamiltonian at theta plus 2 pi. And I leave it as an exercise for you to find the transformation that conjugates h of theta to h of theta plus 2 pi. It's completely straightforward. Give you a clue, it's like e to the iq. So if you conjugate the Hamiltonian by this operator, you find the same thing. And therefore, you'll learn that physics is periodic in theta, because the Hamiltonian is the same. We just conjugate the Hilbert space. We can also see it from here. Imagine time is Euclidean, and it's compactified. Then the integral of q dot is always a multiple of 2 pi, because the integral of q dot is the multiple of the periodicity of q. And that multiplies theta, and therefore physics is periodic in theta. Another way of saying it, this system has instantons, which in Euclidean time go around the circle in Euclidean time, n times. And configurations as we go around the circle n times have a weight e to the i n theta. But since n is always an integer, physics is periodic in theta goes to theta plus 2 pi. But that means that the theta equals pi. So we learned two facts. The physics of theta is similar to the physics of theta plus 2 pi. All we need to do if we look at theta, or we look here at 2 pi, the spectrum is the same. See, there's one state here, there's one state here. There's a degeneracy here, there's a degeneracy here. And it keeps going like that. And the theta equals pi, physics is different. The theta equals pi, physics is different. The states are degenerate. There's a degeneracy here, and a degeneracy here, et cetera. And this degeneracy also reflects a symmetry. So the fact that the theta equals 0 is the same as theta equals 2 pi, that follows from this periodicity. And the theta equals pi, there is also a symmetry p, call it p hat, as we use the same notation here. So there's also a symmetry p, which reflects, so p generates q goes to minus q. But if the theta equals 0, it maps n to minus n, and that's why it's responsible for the degeneracy The theta equals pi p maps q to minus q, but it maps n to minus n plus 1, which is consistent with the degeneracy here. This state has n equals 0, and it's mapped with a state, which is plus 1, similarly for all the other states. So the map of p, the map of p of the state, p is a symmetry. We say the theta equals pi is invariant under this transformation. You can call it time reversal or charge conjugation. So that's what the symmetry is. But what is surprising, so that's the standard statement that theta equals pi, there's a new symmetry. But I claim that there's something funny here. The element p here combines with u1 into o2. But it does not quite combine with u1. For o2, theta equals pi. The group is a little bit different. So let's analyze the group. So we still have the u1 generator by alpha. So we gave it a name here, o alpha. And that's the operator that performs p. So o alpha takes q to q plus alpha. And therefore, it maps n e to the i alpha n. This is a transformation on the states. So let's compute p with q alpha. So q alpha o, o alpha p, the other order, p o alpha equals i alpha o minus alpha. So if we didn't have the factor here, we would have said that's perfectly reasonable. This looks like o2. We do a rotation by alpha, we reverse orientation, and then we do a rotation by, and so we reverse orientation, we act with o alpha, and we reverse orientation. Again, this is the same as rotation by minus alpha. That's the standard o2 relation. So the parity and rotations do not commute. But here there is another pre-factor between them. So at theta equals 0, we got the standard o2 relations. But at theta equals pi, we don't. Notice that this is not some fancy map. This is almost the simplest quantum mechanical system you can study. There is a complete solution on the block board. All the wave functions, all the correlation function can be computed. There is really nothing sophisticated here. And yet, we see here something very subtle. The system at theta equals pi appears to be charge conjugation invariant, appears to be u1 invariant. And when we try to combine these transformations, they don't quite combine to o2. They combine into a central extension of o2. Let's call the central extension. In fact, this is something that was known already to Wigner. Wigner said that if a classical system has a symmetry group, in this case, it's o2. So he couldn't quite study this system, because theta does not appear in the classical physics. But given that, the system has an o2 symmetry. The operators are in representations of o2. But the Hilbert space might not be in representations of the symmetry group, so said Wigner. It might not be in the representation of the symmetry group. It might be in a representation of a central extension of the symmetry group. This is quantum mechanics 101. And we see it here. So there are a lot of things we can say about that. The first thing we could say is, let define the operators differently. So let us define v alpha to be this central element e to d, but with minus 1 half. So e to the minus i alpha over 2 times o alpha. If we do that, then I leave it as an exercise for you to check that we do have the correct relations. That's good. We basically pull this e to the i alpha and split it evenly between the two sides. The problem is that now v of 2 pi is equal to minus v of 0. So we have doubled the number of v's. Another way of saying it is that the symmetry that we have here is still isomorphic to o2, but it's a different o2. In other words, as we make in o2, when we make a 2 pi rotation, we come back to where we started. Now if we make a 2 pi rotation, we come back to minus that thing. So we have to go 4 pi around to come back to where we started. So this is a double cover of the o2. We can call that pin 2. That's standard notation, but that's not important for us. I'd like to get another perspective on the phenomenon we have seen here. So this business of the quantum mechanics of a particle on a circle with theta was fully understood in the 70s. Believe it or not, I was a graduate student like you and I learned it there. The only new twist that I'm going to add here today is that I'm going to present this whole story as an anomaly and going to interpret it in an interesting way that will be useful later. In order to do that, we're going to use a trick that is extremely powerful. Whenever we have a global symmetry in the system, we would like to couple the system to a classical background field. This is not a background, this is not a gauge field that we will integrate over in the functional integral. This is just a classical background field. For example, you have a system that has your one symmetry system of electrons. You can put it in a background electric field or in a background magnetic field. If it spins, you can couple to it and so forth. So we can couple it to a background gauge field. So our Lagrangian will be almost what it was before. It was a half q dot, but now we'll put an a0 and it would be a square and then there was a theta over 2 pi and there was a q dot. We'll put an a0 here. And so now the Lagrangian is invariant and the gauge transformations provided a0 transforms also. But a0 is a classical field. We are not integrating it. It's not integrated over in the functional integral. This a0 is used as a probe. We put the system in some non-zero background field. We don't sum over the background field and now this particle moves in a non-trivial background. The system is still solvable as we will soon see. When we turn that thing on, there's another term we can add, which we can call it, well, there are various ways of calling it. This is the term that would be most similar to the Chern-Simons term in higher dimensions. And one way to think about it is that everything here is gauge invariant. So before we had a transformation by alpha, we can now, since it's quantum mechanics, we can put a gauge transformation alpha of t and provided a0, the background transforms also. So these two terms are manifestly invariant. And this term is non-invariant, the Lagrangian is non-invariant, but it varies by a total derivative. And if we would like the total derivative to integrate to zero, then k must be an integer. So this is very much like what is common with Chern-Simons terms. Except that this is a quantum mechanical version, so it's so easy that people try to dismiss it, but a lot of the physics can be learned already here. We will refer to this term. So a person who was not interested in this a would not even have the parameter k. The parameter k enters the scene only when we turn on some a. And that's an ambiguity in how we couple the system to the background field. So we can couple the system. So I wrote the Lagrangian before and I asked how do I couple it to background field. So there's a canonical way of doing it, which is what these two terms do. And then there's freedom with an integer. And you and I might take the same Lagrangian, couple it to a0. This is the only gauge field in the problem. And you and I might pick a different value for k. That's right, that's perfectly right. What that would mean is that whenever we consider the response of the system to background fields, you and I are going to get different answers. But that's okay, they are labeled. This ambiguity is labeled by an integer. Now let's run the same story again with nonzero a. So now we'll see what happens here and why we got all these phenomena. The first thing I would like to say that the only gauge invariant information, a, okay, let me first do it with nonzero a. So the first thing I noticed when we were already here, we'll go through the same story. So what are the symmetries? There is a global symmetry, it became gauged. There was this symmetry we called p, or charge conjugation, and that flips the sign of q, so it should better flip the sign of a. So a0 goes to minus a0, this is this charge conjugation. The symmetry looks like o2, so far so good. Let's see whether this is right or not. If we shift theta by 2 pi, the Lagrangian is shifted by, to 2 pi cancels the 2 pi by this. This term integrates to an integer, so let's take it to Euclidean space, it integrates to an integer, and therefore it doesn't matter. But this one does not. So if we shift theta by 2 pi, this Lagrangian does not go back to itself. Even the action does not go back to itself. So under theta goes to theta plus 2 pi, the action, which is the integral of l with values of theta, and k goes to the action, which is the integral of l, but now 2 pi with k, which is the same as the action with theta, doubt the shift by 2 pi, but with k minus 1. Okay, sorry, plus 1. So when I said before that the system at theta, in the theta plus 2 pi is the same system, that was true when I did not turn on this background A. But when I turn on the background A, that's no longer true. The system with theta and the system with theta plus 2 pi are not exactly the same system. When I turn on A, they have different values of k. In fact, I said before that you and I might couple this system to a background field in a different way. We turned on this A, so the first two terms are canonical, but the last term, there is an ambiguity in the parameter k, and you and I might choose a different k. So there is a one parameter way of different, one parameter, different ways of doing it, labeled by this integer k, but this will come back if we try to shift theta by 2 pi. So naively shifting theta by 2 pi doesn't do anything, but if we couple it to background field we see that it does do something. It shifts this counter term by 1 unit. So the correct statement is that the system with theta and k is the same as the system with theta plus 2 pi and k minus 1. So as long as we don't turn on A, it makes no difference, but if we turn on A, it makes a difference. Now let's go back to the next step here. So that was a refinement of this statement. So when we turn on background field, this statement is not quite right. We have to add k. There is still periodicity, but the periodicity is more subtle. We shift theta, but we also need to shift k. Now let's go back to theta equals pi. If theta equals pi, why did we say that we have the symmetry P? Because we had the symmetry P because P maps theta to minus theta and it pi maps it from pi to minus pi, but then we use the periodicity to bring it back to pi and that's why we have the symmetry. But now we see that this is not quite right because when we shift theta, when we perform P, this charge conjugation symmetry on theta, it flips the sign of theta, that's okay, but now we want to bring it back to pi. We need to shift theta by 2 pi, but that's no longer a symmetry because it shifts the value of k. So the lesson of all that is that P, the way we define it, is a symmetry of the system if we do not turn on background A. But if we turn on background A, it's not quite a symmetry of the system because what P does, so if we act with P on the system, so it shifts theta, then we can undo the shift of theta using the periodicity, but then we have to shift the value of k. So when we perform P on the action, the action is shifted by one unit here. So the action is mapped to the action plus the interval of A naught with coefficient 1. Okay, so somebody could say, well, let's try and fix it. How can we fix it? Imagine we shifted here by putting a half here. If we put a half, we'll define something else. Times will be changed, et cetera, et cetera, but in the end the shift will go away. But the problem is that we are not allowed to add such a counter term. K has to be quantized. So if k has to be quantized, we cannot fix the symmetry by adding a half integer at this term, linear term with a half integer coefficient. This is the hallmark of an anomaly. An anomaly is a statement that we have a symmetry, in this case O2, but when we try to couple it to background fields, we somehow fail. I should pause here and clarify something about the terminology of anomaly. And this is a sin that was performed in the 1780s. There are three different things that go under the same name and cause a lot of confusion. One anomaly, which first appeared in three gauge currents, say we are in four-dimensional field field, three gauge currents, they tell us if such a thing exists, it means that the system is inconsistent. So we are not going to discuss that at all. The second thing, which is called anomaly, this is what pi zero decays. This is one global symmetry current couples to two gauge currents. Still the same triangle diagram, but on one side we have a global symmetry current. If this is non-zero, this anomaly means that the global symmetry is not there. The system is still consistent. There is nothing wrong with the system, but the global symmetry is not there. And then there is a milder thing, which is also called anomaly, not to be confused with the others, which is what we see here. And I think a proper term for that is a tuft anomaly, because he wasn't the first to study, to study, but he was the first to made a lot of views of it. And this is the statement that we have some global symmetry. So first of all, the system is consistent. Number two, no problem with the global symmetry. The global symmetry is there. But if we try to couple it to background fields, the effective action as a function of the background fields is not gauge invariant. And that's what we see here. So what we see here is that the theta equals pi, there is a global symmetry, o2, but we cannot couple it, but it has, but it suffers from a tuft anomaly. It's really nice that this elementary free quantum mechanical system exhibits such a subtle phenomenon, and I will see what we do about it. And then later in the series of lectures, I'll use the same understanding in more complicated system, which cannot be analyzed so easily. Here it's so easy that you might say, oh, who cares about these fancy words? I know what the system does. So anything to the right of this line can be taught in the first quantum mechanical, the first course in quantum mechanics. It's nothing here that goes beyond the first course in quantum mechanics. Yes? Okay, so event that I'm glad you asked that. So people often say that anomalies have something to do with quantum field theory, and they have something to do with divergences, and they have something to do with fermions. Yeah, okay, so here is a system in quantum mechanics, not quantum field theory. There is nothing divergent here. In fact, the system is completely solvable. We have the complete solution of the system on the blockboard. Hamiltonian is diagonalized. We can compute all expectation values, and there are no fermions. And yet there is an anomaly. Okay? You're not disputing what I said. You just said that somebody else said, there is a whole confusion in the literature of whether anomalies are ultraviolet phenomena or infrared phenomena, or they come from this or come from that. I don't even know what's ultraviolet and infrared here. It's all trivial, and it exhibits the anomaly. And you can see it in many different ways. One way is just computing the correlation functions, the relations of the symmetry generators here. They don't satisfy the correct algebra. You can redefine then such that they satisfy the correct algebra, but then you need to go to the double cover of the group. And you can couple it to background fields, and you can see that the response is not right. Well, that's kind of overkill because in the classical system, we don't have theta at all, and the whole thing comes from theta. So in the classical system, we don't have theta, and therefore there's nothing to discuss. So yes, it comes from quantum mechanics, and it comes from the subtle object theories theta, which is... That's correct. But it's more than our run of bomb because it combines with this Z2 of charge conjugation. Well, we'll compute the path in a go soon. But yeah, there's nothing wrong. No, it's... Well, it's so simple that... So in the first course in quantum mechanics, you learn everything here, right? Menom and the wave functions, and you act on the wave function. So up to here, it's really elementary, here it's slightly... So as we move to the right, it becomes less and less elementary, but I think it's still quite elementary. So let me demonstrate it differently. Let's take the system to Euclidean space and turn on nonzero a. So if we turn on nonzero a, the only thing which is gauge invariant is the integral of a. So imagine time is compact. So we put the system in Euclidean space and we compactify it on a circle. And the only thing which is gauge invariant, so when we say a coupling to a, that's more complicated than necessary because only one number is meaningful here because we can always do gate transformations. And we can call this mu. And this is identified with mu plus 2 pi because the gate transformation is taking us around the circle, maps it like that. Okay, so that means... So what do we compute when we put the system on a circle? So on the Euclidean circle, what does this thing compute for us? Yeah, I heard the right answer. So it's the partition function or the trace over the Hilbert space e to the minus beta h, h is the circumference of the circle and we can put some chemical potential for this mu, mu mu, something like that, depending how you define things. Okay, so we compute such an object. So let's see what it is for our system. So since the system is completely solvable, I like to say, there's really no excuse to compute it. So let's define that z. And for simplicity, let's just be interested in the ground state because we can write the sum of all of them. This is not going to change what I'm going to say. So we're long looking at the ground state and I'm going to suppress various factors. So this and I'm leaving myself only through the ground state. Ground state. So there are two states there, the n equals zero state and the n equals one, which gives me one plus i mu. So these are these two states, the state with n equals one and the state with n equals zero because n is the charge. The charge is n. So this is the state n equals zero and this is the state n equals one. So going back here, we can see what really happened. The u1 symmetry multiplies the state. I wrote it somewhere here. The u1 multiplies the state by, oops, I forgot an n here. Sorry about that typo. The u1 symmetry multiplies the state by e to the i n alpha. So the charge is n. The state has charge n. So this is the state with n equals zero. Becomes the generate with the state with n equals one. So we see that when we act with charge conjugation which tries to map one to minus one, that's not going to bring us to the other state. The symmetry that really does the job here is a different symmetry. We needed to define it over there. There's a symmetry that maps n to minus n plus one and as we keep moving, we need to define this symmetry a little bit differently. And we see it here. We had two states, one with charge zero and one with charge one. They happen to be degenerate, but these two states are clearly not related by this charge conjugation symmetry. They don't have the same charge. So what are we going to do about that? Option one, say the charge conjugation is not a symmetry. That's option one. That would be too drastic. Option two, so option one, say the charge conjugation is not a symmetry. Option two, we take this partition function and define a partition function prime, which is e to the minus i mu over two times the old z. And that would tell us that we have minus i mu over two plus e to the plus i mu over two. Now, this charge conjugation symmetry is this. Yeah, this is all the theta equals pi. So I have, what do I, this is also, so option one is to say, the state, there's a state with n equals zero and the state with n equals one. They happen to be degenerate, but they're not related by any symmetry, so don't confuse me with charge conjugation. That would be too drastic and we're going to miss a lot of physics if we do that. Option two, that you will often find in the literature, say, well, let's multiply the partition function by this factor. We multiply by this factor, we say we have two states with charges plus and minus a half. Now they can be combined into an O2 because they have the opposite charges. So that looks good, but we can't have charges which is plus or minus a half. The global symmetry is U1 and our normalization was such that we only have integer charges. So what we need to do with this option two is go to the double cover. So option one, we sacrifice P. Option two, we say we thought the symmetry is U1, but in fact we have to be in the double cover of U1. That's also a little bit unsatisfactory. We have a system that had a global U1 symmetry. We try to couple it to U1 gauge field A and we get an answer which is like that. It's not seen variant. Option two, we get this answer, but this answer is not invariant under U1 gauge transformation. Under U1 gauge transformation, mu is shifted by 2 pi, so this thing will get a minus sign. So the partition function gets a minus sign under U1 gauge transformations. So we might think that's not a good solution. Both are valid. We can declare that the system doesn't have P or we can say that we have to go to the double cover. Both are valid solutions and various systems really realize, physical system realize one option or the other. We'll need to know, go very carefully, check which of the two options is the right one. Yes? On? You can compute it. It depends through, we'll have to go through the canonical transformation, but it definitely depends on this one because this is just an add-on. The other thing is that that would be the new P, etc. But we can work it out if you want. There's a third option and the third option, that's typical of anomalies, there are various things we would like to have. We wanted to have this O2, so we failed, so either we went to the double cover of the O2 or we throw away the Z2. These are the two options we discussed. And there is a third option, which is also interesting. We saw that the only problem was that we can trace back the problem is that this coefficient K had to be an integer. We said that we could fix it up if the coefficient could have been half integer. Then we would have said that there is no anomaly. An anomaly is something that you could have tried to fix by adding a counter term, but the appropriate counter term is illegal for some reason and therefore you cannot fix it up. But there is a third option. Say that our system, and it's always the third option, is that you relax the only assumption that was not yet stated. This is a general rule in life. When you are in a problem, there's always a hidden assumption that you forgot. Do you know what the third... Can anybody guess what the third option is? One assumption I didn't bother to state explicitly. Okay, see that everybody knows. So a system we thought is quantum mechanical and therefore it has only one dimension, zero plus one dimension. But instead our system is two-dimensional. If our system is two-dimensional, then we can solve the problem. We can preserve all the symmetries. So this is our Euclidean time that was compactified. Imagine we attach a surface to it. Arbitrally complicated surface. So we knew that if we only could add a half integer here, we would be in business. But we couldn't add a half integer here because that would not be gauge invariant. But we can add a half integer if we write it as what happens in the bulk. So add, so call that m2. So we define an m2 and add to the action. So the action was originally an integral d time of the Lagrangian. And now we add to it right the same s plus instead of that, I'm writing a half f on m2. So that's nice and gauge invariant. But it depends, I need an i because I'm in Euclidean space in this discussion. So that's nice and gauge invariant. But it depends on f in the bulk. So we had a system that didn't have a to begin with. Then I made it slightly more subtle by coupling it to an a. That allowed me to detect the problem more clearly. And it looks like all it depends on was what a does on the line. This formulation of the problem did not have any manifold attached to it. Now I'm attaching a manifold m and I'm adding this term. Now the problem is solved because I effectively added this half integer term term with k equals a half. The price I pay is that the answer depends on what a does in the bulk. So you often hear the statement and you will see it later in this series of talks that you add another dimension and then you work very hard to make sure that the answer does not depend on the added dimension. Here it does. When I wrote this term so that's an aside before that we could have written the integral of k with a. Things were one dimensional. We could have written it as k in integral of f on m2. But if k is quantized the things are independent of what a does in the bulk. Because how do we do that? This is a standard trick in this game. We add a bulk. We extend a to the bulk in some way. Now you choose a totally different bulk. So in this case it would be not a Riemann surface with one handle but it would be a Riemann surface with two handles. And you will extend your a the same a in a different way to the bulk. So we are going to form a new manifold which is the combination of the two. So my manifold was this one and your manifold will be this one two handles. And the difference between our two answers will be e to the i in integral of f on the combined system. So the difference between us will be e to the i k in integral of f over this whole manifold. But the integral of f is always 2 pi times an integer. So that's always plus one for quantized k. So before I went to option 3 I could have said that this term can really be thought of as extending a to the bulk but it doesn't matter how we extend it to the bulk provided k is properly quantized. Now I say, well I need something head to give because we had this anomaly. So we extend it to the bulk and now things do depend because it's not properly quantized. So you and I are going to extend it to the bulk differently and then we compute this thing and now since k is a half we're going to get plus or minus one. So with this way of defining the system so we're still in option 3 now it depends on the extension so z the partition function will have plus or minus sign but that's okay because this plus or minus sign depends on how we extended it and we knew that something had to give so that's what it is. In other words we have a quantum mechanical system that has an O2 symmetry this is the system of theta equals pi somebody asked me before only a theta equals zero none of these problem arise we have an O2 symmetry, no problem a theta equals pi there is an O2 symmetry but when we try to couple it to background fields we couldn't quite do it but we had various ways of dealing with it one say okay we can't do it we can't do it, that's the end of it or we sacrifice one symmetry or we sacrifice another by going to the double cover or we extend it to the bulk and the answer depends on the bulk. What I've demonstrated here is a very, very common thing except that here it's really was boiled down to this is almost a simpler system there is a simpler version of this with fermions in one dimension but just as you always hear that anomalies have something to do with divergencers and quantum field theory also here they just have to do with fermions so I thought I would present a system with bosons which looks like nothing can get easier than that to conclude just to extend it to a bulk we extended it to the bulk in such a way that for a closed bulk we have a closed bulk the partition function that we compute with these added terms it's not a partition function just the value of this added term is completely O2 invariant when we have a boundary it's O2 invariant but depends on what the fields do in the bulk when we have a boundary it depends on the bulk it's not gauge invariant it's not invariant under this Z2 but the mistake in the O2 the lack of invariance is exactly canceled by the fact that the boundary theory is not invariant under the Z2 so we have an anomaly in the bulk we have an anomaly on the boundary and the two anomalies cancel each other so that we have a nice invariant system in other words the system of this quantum mechanical system at theta equals pi just one dimension attached to a bulk where in the bulk we have only classical fields classical fields with half unit here a background field with half unit this is what also called theta equals pi but that's a different theta so I do not want to use this terminology this system together is invariant under the O2 symmetry but each separately is not the anomaly in the quantum mechanical system is canceled by the anomaly inflow from the bulk are there any questions? depends on the topology of the bulk the topology and the bundle that you pick on the bulk but not on anything else so if you make small deformations of A in the bulk change the metric of the bulk a little bit it drops out so it's a very mild addition so when I modify the system by adding the bulk I've changed it in a very, very mild way but it is important that I had to change it no, it depends also on the first-term class, you raise your hand with math yeah, so it depends also on the first-term class does this answer your question? you said that different physical systems would employ option 1 or option 2 does option 3 capture all of them or will it not be able to capture some physical? so in a physical system it depends on what the physical system is so, for example the physical system might actually have a bulk yeah and it might actually have the full auto symmetry and it might have the bulk and then it will use option 3 the system doesn't have the bulk something we'll have to give so either you say that there are two states at theta equals pi which charges plus or minus one so charge conjugation is a symmetry but with u1 we had to go to the double cover that's extremely common and this is how you usually get fractional charges of various particles or you say, no, charges have to be quantized symmetry is u1 without going to the double cover but then we don't have charge conjugation this can also happen but it's important to know the menu of options and when you use one which one you use and not do half the system one way and half the system the other way because that would not be sensible yeah, it's actually quite common it could not matter people they don't present it this way but they do exactly this in systems that are actually in a way simpler and in a way more complicated than this so if you vary theta smoothly to pi is there anything special in the gauge group in this case it's not a gauge group it's a global symmetry we are only talking about global symmetries the system has a global symmetry of the global symmetry it's convenient for us it's a tool to couple it to a background field and once we couple it to a background field once we couple it to a background field we would like that global symmetry and that's what we run into trouble but the trouble that we run into in coupling to background fields is identical to the problem I presented here without ever mentioning background fields so it's a matter of taste whether you would like to set A to zero and study the operators more carefully and see whether the relations in the group are satisfied or not or to say we need that or just add background fields in keep track of all these subtleties using what happens with the background fields I don't know if I answered your question but the answer is A question enhancement of the group not the gauge group but the group there's no enhancement but from U1 to O2 or from SO2 to O2 yes that's a property of the Lagrangian is that different if you vary theta towards zero or theta towards pi yes, completely different because in both cases the group is O2 but in this presentation theta equals zero we had O2 on the nose theta equals pi we found the central extension of the O2 I don't know if you want to call it enhancement or not there's one more element either we have to double the range of alpha the parameter or we say that where did I write this comment here or we have to add this central element so it's up to you how you want to describe it but you have more element it's not a bigger symmetry it's a central extension normal questions am I done? I have four more minutes I've just finished the first 15 minutes in the first lecture so I'm going to change topic and I think this quantum mechanical example was worth presenting even though I will probably because of lack of time present much less than I planned because I can assure you that later on in life you will see this adding the extra manifold say that our space time is really the boundary of something else and to discuss the anomalies what happens in one dimension more or using this in various context for example, example 3 is what is known as a topological insulator although the connection might not be obvious at this point but I think this was useful so the next topic I want to talk about is moving to quantum field theory but it will be topological quantum field theory and I would like to make some introductions to transimons theory because we'll use it later who here has ever studied anything about transimons theory about half and I know that you haven't ok, so for simplicity so I'm changing gears and I will have a u1 gauge theory in three dimensions and I'll denote the u1 gauge field by b so I have a u1 gauge field and when we say it's u1 it really means that on every close two cycles that's in 2 pi z so all the fluxes are quantized and we can write an action so my convention is always that the action is like this and the actions since I've been in euclidean space will be i k over 4 pi integral bdb so this is the transimons action but to properly define it first similar to what we had before except in one dimension it's linear in the gauge field and here it is quadratic in the gauge field and one way to write it so this is a three manifold so call it m3 is to write i k over 4 pi integral db db on m4 where the boundary of m4 is a three manifold m3 so we take our three dimensional space time we extend it to a four manifold we extend the gauge field b to the bulk and we compute this thing in the bulk so this is nice engaging variant but we have to check that it's independent of what happens in the bulk so again the same story we had before this is our m3 and we can extend it to the bulk one way or some other way so this is m4 this is m4 prime but with the opposite orientation so you and I picked a different m4 different four manifold and we extended the gauge field differently how do we check whether it's independent of the extension we construct a four manifold which is the union of these so now it's perfectly smooth there is no boundary perfectly nice manifold so we compute this thing on the big thing on the whole thing so that gives us an answer and this is a good definition for the three dimensional theory if the partition function or this thing is independent on the closed four manifold is independent of the extension it's not even a partition function this has to be exact configuration by configuration in the functional integral so we're doing the functional integral over little b we extend it to the bulk sure it's independent of the bulk and the condition for that is that k over 4pi integral db db for this whole thing I'll call it m tilde 4 which is this whole thing four dimensional manifold this should be in 2pi times an integer and if that's the case it's independent of the extension it turns out that the answer depends on some properties of the four manifold correspondingly the three manifold if the four manifold is a spin manifold is a spin manifold and we choose a spin structure for it then the allowed configurations are restricted and that would tell us if it's spin that k is an integer if it's not spin if it's not spin then k has to be even so we'll see in the discussion that depending on there are two, the discussion will bifur k, depending on where the k is even or odd if k is even we don't need to choose any spin structure and so forth and if k is odd we do have to choose such a spin structure that's a tool so if it's non spin there are more options and if there are more options there are more restrictions on k and k has to be even so whatever I can do I'll do in parallel for the two cases and when the discussion bifur k I'll split to the two separate cases how much time do I have thank you