 some examples, the most famous such example in condensed matter physics, the condo problem, how electrons interact with isolated spin degrees of freedom. In high energy physics, there's something called the Kallen Rubikoff effect, which is a beautiful problem to do with fermions scattering off magnetic monopoles that has some very interesting physics. Finally, and more recently, it's a problem that arises in trying to understand the physics of SPT phases, symmetry-protected topological phases in condensed matter physics. That's really going to be the focus of what I'll be doing in this talk, and I'll explain a little more about that as we go along. All right, so the basic question I want to ask in this talk is, what symmetries can you preserve in the presence of a boundary, and what the properties of those boundary conditions are? And in particular, I'm going to be interested in chiral symmetries. So for the most part, I'm going to be thinking about massless fermions. Massless fermions moving on a line. Massless fermions move at the speed of light, which means they have two options. They can either go that way at the speed of light or that way at the speed of light. And I want to think about boundary conditions that preserve symmetries where I'm going to get confused with the mirroring on Zoom. The right-moving fermions, for you guys, where the right-moving fermions carry different charges from the left-moving fermions. So to give you a complete example to have in your mind, suppose you have two Dirac fermions in one plus one dimensions, and give the two fermions going that way charges three and four under a U1 symmetry. And the two fermions going that way give them charges five and zero. And I want to put a boundary condition that preserves that charge under fermions. Now, you might be forgiven to think this surely just can't be possible. There's a very simple thought experiment. I put a boundary, what happens if I throw in a fermion of charge three? What can possibly bounce back? The fermions that bounce back have charged five and zero. It seems like there's no option. And it turns out that it's certainly true that it's not possible to write down simple conditions on the fermionic fields which preserve these kind of symmetries of the boundary. But it is nonetheless possible to have boundary conditions that preserve this symmetry. And there's various ways to do it. Perhaps the sort of intuitively most straightforward way, although it turns out not to be the way to actually solve these things, but the intuitive way is just to add some extra degrees of freedom on the boundary. So you could imagine putting on the boundary maybe some roto degrees of freedom. Just little quantum mechanical things that spin around. And when they spin around, they can soak up some charge. So if you have extra degrees of freedom on the boundary which can soak up the charge, then indeed something can come in and something with one charge and some different charge can come out. Now, there's lots of different ways to cook up degrees of freedom on the boundary. But if you look at very low energies, energy is much smaller than any characteristic energy scale to do with what's going on on the boundary. It turns out there's a universal description of these boundary conditions. It's a description in terms of conformal field theory or what's known as boundary conformal field theory. And that's going to be the kind of tools that I'll be using to describe these boundary conditions throughout this talk. All right, so that's the goal just to try and explore the kind of boundary conditions that one can put on. Before I jump in and start showing you some details, I just want to try and motivate why I care about this, because it's a fairly technical subject. So why would an earth would I be caring about this? There's really, well, there's sort of two motivations. In fact, there's a whole circle of ideas of motivations, but here's one motivation. It's the following very simple question. If you take mass as fermions and you give them a mass, what symmetries do you break when you give them a mass? I think this is a very interesting question in part, because for about 20 years I thought I knew the answer. And then in recent years, convinced matter colleagues have convinced me that really I don't understand this question at all. The reason I thought I knew the answer was that if you just look at the Lagrangian for free fermions and you add a quadratic term, the quadratic term breaks some symmetries. If you're even dimensions, it breaks chiral symmetries, axial symmetries. If you're in odd dimensions, it breaks things like parity or time reversal, or at least something related to that. And I just thought that was the end of the story. That is the end of the story for free fermions, or indeed for weekly interacting fermions. But if you have strong interactions, it's possible, and we know of certain examples, where you can give fermions a mass preserving symmetries that you might naively thought would be broken. And that's the kind of question that I'd really like to explore. It turned out to be a rather difficult question to explore. But the question about boundary conditions is sort of a nice halfway house. And they're related for the following reason. Suppose you have your massless fermions in a line for us, but more generally in any dimension of space, you can consider turning on a mass just in one half of that space. And then you send in the massless fermions from the gapless side. If you send them in with very low energy, they can't enter the gap region because they don't have enough energy. So they have to bounce off. And whatever symmetries are preserved when you gap one half of the space have to be preserved in the effective boundary conditions for these massless fermions. So there's this intricate connection between the kind of symmetries that can be preserved when you give fermions a mass and the kind of symmetries that can be preserved when you put fermions on a manifold with boundary. It turns out that putting fermions on a manifold with boundary is much easier to understand than giving them a mass. So that's one of the reasons that I'm looking at this. By the way, for what it's worth, the folklore in the literature, which is far from being proven, is that you can give a mass to any fermions, providing that they don't suffer an anomaly. That's totally striking. For example, the folklore, if you import it from condensed matter into high-energy physics, is that it should be possible to give all the fermions in the standard model a mass without breaking electro-weak gauge symmetry. That's a totally shocking statement because, let me remind you, the successful prediction of the Higgs boson came from the fact that we thought it was not possible to give fermions a mass without breaking electro-weak gauge symmetry. Nonetheless, it seems like there might be some things that we're missing, and I'd just like to understand what those things are. That's one of the reasons. There are several other reasons as well that may become important as we go through the talk, but that's the collection of ideas that I'm thinking about here. That's the end of my rambling introduction. I'm going to start showing you some slides, but I'm very happy to take any questions before we kick off with the formal part of the talk. David, I have a curiosity in terms of motivations. Is also the fact that adding fermions and rambling condition is useful, if you want to compare it with Laktis models, like Laktis gauge theories and so on and so forth? Actually, that's one of the main motivations that I have, but there's a problem in theoretical physics, which I think a wise man once told me when you're deciding what to work on in theoretical physics, you should do the following. You should take how important the problem is, and then you should divide by the number of other people that are already working on it. By this metric, I think the most important problem in theoretical physics is how you put chiral fermions on the lattice, and in particular, how you put the standard model on the lattice, which remains an unsolved problem after 40 years. These kind of questions go some way towards addressing that. If you put fermions on a lattice, you get doublers. If there's some way to lift the doublers, leaving behind the fermions that you want, that would basically solve the problem. So there is a lattice motivation that actually is the big motivation underlying this. I'm not sure that was quite the direction you had in mind, but yes, yes, yes. Thank you. Other questions? All right, let me share my screen with you all. All right. Anna, can you see the screen? Is it good? And the arrow thing as well, the cursor? Yeah, yeah, it's good. All right. So everything I'm going to tell you is based on a series of three papers that have come out over the last year with my completely excellent student, Philip Oil Smith. I should stress from the beginning, everything is due to Philip. I know all the theorems that he proved, and it was just quite astonishing what he did here. To give you some sense, not only did Philip learn boundary conformal field theory to do this project, having learned it, he then taught me boundary conformal field theory. He's really a very, very impressive student. In return, I'm going to do you the favor of not teaching you boundary conformal field theory. It's a very technical subject, and everything that underlies the statements in this talk have proofs using boundary conformal field theory, but they're very messy at times. So I'm going to tell you the statements, which are very easy to explain, but I'm going to hide all of the dirty details from you, but you can find them all in these papers. All right. So this is where we're going. I'm going to start by telling you that all possible boundary states for fermions in two dimensions fall into one of two classes, a Z2 classification of boundary states in one plus one dimensions. It's related to a Z2 classification of SPT phases for those who know what they are in one plus one dimensions. It boils down to what's called a mod two anomaly, so I'm going to start by explaining that. Then I'll go on and tell you about the chiral boundary states that I care about, and then I'm going to go on and tell you about some properties of these chiral boundary states, and in particular about the ways they're connected to each other through what's called boundary anti-flow, all of which I'll explain. I should warn you, at the end I have a couple of cute little applications, one of which I think what I understand, and the other one of which I don't, one of which is to do with a two plus one SPT phases, and the other one is to do with the Kalan Rubikop effect, which I mentioned in the introduction. I gave this talk yesterday, and I ran out of time after section three, so I might get to section four and five, but it seems likely that I'll just run out of time. All right, so let me start with telling you about the mod two anomaly. Anomaly is a word that's just used way too much in theoretical physics, and the anomaly I'm going to talk about here is an inconsistency. You write down a classical theory and it looks very nice, but the quantum theory doesn't make any sense. Most familiar gauge anomalies, although this one turns out not to be. Here is what I think is the simplest example of an anomalous, meaning inconsistent, quantum theory. It's in quantum mechanics, zero plus one dimensions, and it's a single real fermion, so a single Majorana fermion lambda. I claim, gee, I, Witten claimed this, but following Witten, I claim this is an inconsistent quantum theory. All right, so why is it an inconsistent quantum theory? The simplest way to think about it is to take two of them, take two Majorana fermions, lambda one and lambda two. If you look at the commutation relations of two Majorana fermions, they form the Clifford algebra in two dimensions, but the Clifford algebra in two dimensions has an irreducible representation that acts on a two-dimensional Hilbert space. Another way to say it is you can pair two real fermions into a single complex fermion, which then you act as a creation operator. So the two states in the Hilbert space is no fermion or yes fermion. They're the two options, and you can't have two fermions because they're fermions and they evade penalty exclusion. All right, so two Majorana fermions act on a Hilbert space of dimension two, but if you then assume some simple tensor factorization, that means a single Majorana fermion must act on a Hilbert space of dimension square root of two, but that's nonsense. There's no such thing as a Hilbert space of square root of two. You can actually do the calculation of the dimension of the Hilbert space explicitly. You can take the path integral, you can compute the path integral with anti-periodic boundary conditions in Euclidean time. That path integral always computes the trace of one. Trace of one is the same thing as the dimension of the Hilbert space. If you do that, you find that the answer is indeed root two. So a single Majorana fermion in quantum mechanics does not make any sense. Two of them are needed to make sense, but by the way it was probably Kitayev who first really stressed this rather than Whitton. This underlies Kitayev's Majorana chain from, I think, 20 years ago. All right, so that's Majorana fermions in quantum mechanics. Majorana fermions in higher dimensions are perfectly fine. You can have Majorana fermions in two, three, and four spacetime dimensions. Nothing wrong with them. However, this inconsistency rears its head if you can consider them on manifolds with boundaries. So let me explain why. We'll take a Majorana fermion in one plus one dimensions. I'll call it chi. And for just the next two slides, I'm going to make the fermion massive. At some point I'm going to make the mass list, but it's useful just for pedagogical reasons to make them massive. So we have a massive Majorana fermion in one plus one dimensions, and I'll put a boundary. And there's two possible boundary conditions that you can put on. The left mover turns into the right mover, but it can turn into the right mover with a plus or minus sign of the boundary. So here are the two boundary conditions. Now just solve the Dirac equation in the presence of the boundary subject to those boundary conditions. You'll find that there's the following solution. It's the kind of solution that Jakeve and Rebbe found long ago when looking at fermion zero modes in the background of solitons. There's a characteristic exponential factor times a Majorana quantum mechanical mode, which is localized at the boundary. And this exponential factor does one of two things. It either decays as you go into the physical region of space, or it grows as you go into the physical region of space. If it grows, this is not a state in the Hilbert space because it's not normalizable, and you just throw it out and there's nothing wrong. If it decays, this is a normalizable solution to the Dirac equation, and it corresponds to a state in the Hilbert space. But it corresponds to a state in the Hilbert space, which is a single quantum mechanical Majorana fermion localized near the boundary. That's precisely the kind of system that we've just said is inconsistent. So if you have a massive Majorana fermion on the line, some boundary conditions have localized Majorana modes, and you have to worry about the fact that they may be inconsistent. There's various ways to deal with the inconsistency, but nonetheless, you have to worry that that single Majorana mode might be inconsistent. But by the way, we can see from here the choice of which condition is consistent and which is inconsistent depends on the sign of the mass, M. It also depends on whether it's the left or right boundary, so you're going X positive or X negative. All right, to just rehash this in a slightly different way, let me consider the story for a Dirac fermion in one plus one dimension. So a Dirac fermion is a complex fermion to two Majoranas. Again, there's two different choices of boundary conditions. One choice is I think the natural one. It's the left moving Dirac fermion turns into the right moving Dirac fermion. By the way, when I write this, I mean that these are conditions imposed at the boundary. As you can see, the boundary condition preserves the U1 symmetry, which rotates the phase of the Dirac fermion. That means that this can serve particle number or charging. In terms of the two Majoranas, this condition imposes the same kind of boundary condition on chi one and chi two. That means that with this condition, you either have zero Majorana modes localized at the boundary or you have two Majorana modes localized at the boundary. Either way, there's nothing to worry about. You've got an even number. However, there's a different kind of condition that you could impose, which is psi left equals psi right dagger. So it doesn't conserve a particle number or charge, but it is a boundary condition, which is important. It's responsible for what's called Andreev reflection. So if you have a wire attached to a superconductor, an electron goes in, the superconductor doesn't preserve charge because of the Cooper pair compensate. And Andreev reflection is the process where a hole bounces back. Andreev reflection is imposed mathematically precisely by these kind of boundary conditions. Again, if you look at what it means in terms of the two Majoranas, one of the Majoranas has one sign for the boundary condition and the other Majorana has the other sign. So with these kind of boundary conditions, you're guaranteed to get a single Majorana regardless of the sign of the mass of the Dirac fermion. So again, it's a boundary condition that you're going to have to worry about taken naively. It's anomalous. It's inconsistent, but you need something else to rectify that anomaly if you want to impose such boundary conditions. All right. Sorry, can I ask a question? Hey, Francesco, how are you? So about this inconsistency of the quantum mechanics of a single Majorana, so is this an inconsistency that can be cured or just, I mean, we have to give up the theory in the sense that, so as you see, you can compute the partition function, you get the square root of two, but you can compute it. So if you apply a part integral formulation, where do you see that there is something wrong? Oh, there are various ways. If you compute it with Riemann boundary conditions, it's zero. It's zero precisely because there's a zero mode. But the zero mode is telling you that what it wants to compute is an expectation value for a single fermion. That doesn't make sense. So it's not so easy to see in the path integral formalism, but it's certainly true in the canonical formalism that it's inconsistent. But by the way, it's very easy to cure. And I'll give you some examples on the following slide. And of course, if you start with a quantum field theory in higher dimensions, somehow, if that was consistent, it's guaranteed to cure it in some way. And we'll see various examples of that in this talk. Okay. Thank you. I've introduced the mass on the past two slides for the fermion. It was purely as a crutch. The nice thing about the mass is it shows the inconsistency of this localized zero mode. But as always, anomalies are always independent of the mass. And so moving forward, I'm going to set the mass to zero. But these anomalies are still going to survive. Oh, now you had a question. Yeah. So you computed the partition, basically computed the square of the partition function and took a square root, right? So what fixed this? Is there any reason to believe the sign to be plus? No, I don't think so. Yeah, it's a good question. I don't know what happens if you pick the other sign. I don't see an option for that sign choice in the consistent theories. Yeah, I don't know is the answer to your question. And is there any reason, is there any way you know that the square root of two can be computed just for one of them from a Hamiltonian perspective? No, no, no, not without taking the square root of, literally taking the square root of two. But by meaning taking two, Myranophermions and then taking the square root. No, I don't know of such. The thing you were saying to Francesco, if I understood it correctly, you said that for one of them for Ramon boundary condition, there is a anomaly of the minus one to the power F, right? So and I believe the statement in the Hamiltonian picture is that there is no non-projective representation of minus one to the power F on the smallest representation of the Hilbert space. So in that sense, I think the anomaly can be understood from the Hamiltonian perspective. But I have been asked, I don't know, is there a way to get the square root of two somehow from a Hamiltonian? Not that I know of, yeah. Okay, thanks. All right. Francesco said, how can we deal with this anomaly? There's various ways to deal with it. The real lesson I want you to take from the past, from the previous slide is the boundary conditions fall into one of two different classes. That was psi left equals psi right or psi left equals psi right dagger. One of those classes gives the Myranomode on the end. If you have now the system on an interval rather than a half line, if you put the boundary condition here, which has a localised Myranomode, you're obliged to put the boundary condition here that also has a localised Myranomode. Yeah, if you don't do anything else at least, if you don't do any of the other things on the list on this slide. Then you get two localised Myranomodes. It doesn't matter that they're separated in space, you still have an even number. So the system is perfectly consistent. The problem arises if you pick one boundary condition on one end and a boundary condition from the other class on the other end. That then gives you just one Myranomode on one end and then you left with an inconsistent theory. So you can patch it up in a very simple way on an interval with a boundary just by putting the same boundary conditions on both ends. There's other very simple ways in which you can cure this anomaly. If you have a single Myranomode here, you could just add by hand another Myranomode, which is localised on the boundary. It seems trivial, but we'll actually see examples where these kind of things are generated dynamically as we go on. Finally, there's a lovely way to cure this, which I'm going to skip over a little bit because it's not really relevant for this talk. But there's a topological field theory called the ARF theory. It's a topological field theory whose partition function on a general Riemann surface with a spin structure is minus one to a topological invariant called the ARF invariant. It turns out that that theory is ill-defined on a Riemann surface with a boundary and the anomaly it suffers is precisely the same anomaly that you get by putting the dodgy boundary conditions. Actually, this ARF topological theory enforces on you. By the way, for the string theorists in the audience, there's a beautiful story which I just want to advertise. It was in a paper last year by Yuji Tachikawa and friends, but they were really fleshing out an idea from the talk by Whitten from a couple of years ago. The GSO projection that you do on the string can actually be understood in terms of ARF invariant topological field theories that live on the world sheet of the string. Then the possible boundary conditions you can impose on fermions which really correspond to the kind of D-brains that one can have are related to this mod2 anomaly. When the dust settles, you can understand all sorts of beautiful things in terms of this anomaly, like for the fact that type 2a has d0, d2, d4, and even spatial world volume dimensions. Type 2b has the others, but the non-BPS brains which live in between have attention, which is bigger by a factor of the square root of 2. It's precisely that square root of 2 of an extra myrrh anomaly. A lot of the things that Ashok Sen and others understood about D-brains in the 1990s can be understood in terms of this topological theory. All right, I'm going to close the door as well. Let me now move on. That was for a single direct fermion. Let me now tell you about multiple direct fermions. Again, the real lesson is going to be that all boundary conditions for fermions fall into one of two classes due to this mod2 anomaly. This is the general setup I want to consider. We're going to consider two n-massless fermions. Now it's going to be massless for the rest of the talk, or equivalently n-direct fermions in one plus one dimensions. The question I want to ask is what symmetries can be preserved from the boundary? The answer which I think is folklore is the following. It should be possible to preserve any symmetries that do not themselves suffer from a top anomaly. I say it's folklore because I know of examples of boundary conditions that I would love to construct where I do not know how. They're chiral boundary conditions but they don't suffer from a top anomaly. Nonetheless, I think this is probably the correct statement and we're certainly going to see in this talk that we'll be able to construct boundary conditions but only those that don't suffer from an anomaly. What's an example? The anomaly cancellation condition in one plus one dimensions is that the sum of the squares of the fermions for the right movers has to be equal to the sum of the squares of the fermions for the left movers. In the introduction, I gave you the example of two left movers with charges three and four and two right movers with charges five and zero. Those numbers weren't picked randomly. They're the smallest Pythagorean triple so that they satisfy this anomaly cancellation All right. As I mentioned in the introduction, there's a technology that underlies this talk which is the technology of boundary conformal field theory. I'm not really going to say as little as I can about it but I just want to give you some idea of how this works. It's an idea due to Cardi from the late 80s and it's the following. You start with this picture here which is your system on an interval with some boundary condition a at one end and some boundary condition b at another end. You want to compute the partition function and to compute the partition function you can pacify Euclidean time. That's this blue circle direction and off you go and compute. Cardi's idea was very simple. You do a modular transformation which graphically is just turning the thing on its side and you view what was the spatial direction as time and what was the temporal direction as a new space. Now in a quantum theory, the states in the Hilbert space live on a spatial slice so the boundary conditions a and b after this modular transformation or what string theory is called open-closed string duality, the boundary conditions are encoded in states in a Hilbert space that live at the bottom and the top of this cylinder. Cardi explained how if you want certain boundary conditions you can understand them in terms of states of the Hilbert space. The states have to have certain properties called Cardi conditions. Those Cardi conditions are not very easy to solve in general which is why it's not trivial to write down these boundary conditions but the whole technology is to write things in terms of states in a Hilbert space rather than actual boundary conditions. All right so this is the setup I'm going to be using. We're going to have n-derac fermions or two n-myaranas and I want to preserve a u1 to the n symmetry so it's a maximal abelian symmetry for for these direct fermions. The left movers are going to have charges q sorry right movers are going to have charges q bar and the left movers are going to have charges q under these symmetries q is a matrix i goes from one to n and labels the fermion and alpha goes from one to n and labels the particular u1 symmetry that we're thinking about. There shouldn't be any toft anomalies under these symmetries that means toft anomalies under each u1 but also mixed toft anomalies so the requirement that there are no toft anomalies is the statement that well these n squared statements here you sum over the n fermions but the alpha and beta labeling the different u1s are left floating in this equation so I want to pick a bunch of q's and a bunch of q bars where this statement holds okay for what it's worth if you have three four and five zero you could always augment them with four minus three and zero five going the other way so it's it's if you have a particularly u1 symmetry that you want and you don't care about the other n minus one you can always find an n minus one that obey this condition so you have a full u1 to the n. All right it's going to turn out that all the results i'm going to tell you about don't care about q and q bar individually but they only care about this combination so you take the inverse of q bar and multiply it by q one thing I should have said but didn't say q's are integer valued it's important the charges are going to be integers that means that this matrix r has rational elements because I divide by or take the inverse of an integer valued matrix and moreover this r is a is a rational orthogonal matrix the orthogonality of r just follows from that toft anomaly cancellation condition on the previous slide so to give you some uh simple examples if you do the the normal boundary conditions psi left goes to psi right this matrix r is just the unit matrix if you do andrea of reflection for every one of the n fermions this matrix r is is minus the unit matrix um however if you want to do um something more funky where the left moving fermions have very different charges from the right moving fermions um then r will in general be uh a more interesting uh rational orthogonal matrix all right the actual mathematical object that's going to appear everywhere is derived from r and it's the following charge lattice so you do the following uh procedure start with the lattice z to the n so in r m just draw a dot at every point uh with with integer value coordinates to big lattice um take every point in that lattice and rotate it by the matrix r if after the rotation by the matrix r a point ends up at another point on the lattice keep that one but if after a rotation by a matrix r it ends up just floating around somewhere in the middle throw it away so um given a matrix r and an original lattice z to the n you can construct a new lattice that i'm going to call lambda okay um get some intuition for what lambda is if the left moving fermions and the right moving fermions have the same charges r is the unit matrix and lambda is just z to the n but what happens is as the charges for the left movers and the right movers start to differ more and more r is typically a matrix that has um a big number in the denominator of uh of the entries that that means that it rotates you just a little bit the fact it rotates you just a little bit means that to find some lattice vectors which get rotated into other lattice vectors you have to go quite a long way out you're always guaranteed to find some just because r is an irrational matrix but you may have to go quite a long way out from the origin before you find one which when you rotate it gets rotated into uh another lattice vector um what it means is that the bigger the difference between the left and the right moving charges the sparser this lattice lambda but that's the the intuition to have in mind by the way everything i'm going to tell you for the rest of the talk is going to be in terms of this lattice lambda so it's important we understand this that are there any questions about um the interpretation of this yes uh David can you give us an example for the second type of boundary condition how this charge lattice will look so if we take psi uh real directly with the psi left equal to psi dagger right so psi left equals psi dagger right um this is also z to the n this is so it's always purely diagonal with the nothing else no sorry so in the particular case where it's andre of reflection for every single fermion r is minus and um but every lattice site in z to the n gets thinner uh you know it gets uh mapped to another lattice site so for normal boundary conditions and andre of this is z to the n but for boundary conditions like um uh three four zero five um this is a slightly sparser lattice um and so on and so on i'll tell you one of the properties of the lattice uh lattice is shorter i think thank you can i ask another question yeah please so here how important is it that you want to preserve the one to the n symmetry so suppose you want to preserve some other subgroup of son i mean so instead of just charges you have a kind of bunch of representations left and right measurements would it be a some analog of this art matrix in this case i don't know how to do it and in fact there are some very important problems i would love to solve that fall into that category yeah it's important that um you only that you have a maximal rank uh symmetry that that's preserved to construct the boundary state it's somewhat similar to um uh to the requirement or the statement that you can only solve conformal field theories for rational conformal field theories it sort of feels morally morally similar to that if i have time i'll give you an example of an open problem at the end that requires different boundary conditions that don't fall into this you want to the end class so i would love to be able to i don't know how any other questions all right um let me uh let me tell you some results ah before i tell you results let me very quickly um a slide only for those of you who know about boundaries and form or field theory if you don't don't worry um we want to construct the boundary state that corresponds to this um uh this is the boundary state uh here i should say that for the most part we're not the first people to to construct this it's been known for many many years you can find it in old d-brain papers by wreck and arglund schomeris um in the context of spt phases it was um stressed in a couple of papers uh by schoon say you and collaborators that the pcft was a useful way to think about spt phases um this weird barar type object with double uh double bracket is called an ishi bashi state um it's a state which has the property every one time and yes it's a state that that has the property that no current flows into the boundary that's really what this uh this state is telling you um it's labeled by um uh charge vectors for left and right movers under the various um various you want to the end symmetries and it has this coherent uh state type um type expression um the general boundary state is then an uh a linear superposition of these ishi bashi states and the question is what's the linear superposition that um there's uh uh i should just stress a couple of things just just so you know what all these symbols mean um the boundary state is not unique uh if you fix the charges uh there's a bunch of parameters um that you still have these parameters are called theta here um and a a better way to think about theta for the for the string theorists um if you're writing down a boundary state for a d-brain the analog of theta is is just the position of the d-brain or sometimes the wilson line around uh around circles um so the boundary state isn't unique in that case you can move it around this is the after going from bosons to fermions through the usual bosonization that this is what becomes just these phases uh uh theta there's a very annoying phase um that sits here um which uh actually this wasn't seen in the literature before as far as i'm aware we're the first people to um to understand this it's important for one of the things i'm going to tell you about but um it's horrible and it's absolutely horrible it's you can find it in the appendix of one of our papers if you really care it's one of these really annoying subtleties that you hope you never have to deal with and it just turned out we had to deal with it um for this paper the important thing however is is this coefficient in front so um whereas in quantum mechanics you don't care about the normalization of a state in the hillbett space for these boundary states the normalization is crucial um it's related to something called carney condition um and understanding this normalization um is one of the most important things in in this talk so uh let me tell you a little bit about this this normalization so the the importance of the normalization was first um stressed by aflec and ludwig um back in 1991 and this coefficient is called the boundary central charge um what aflec and ludwig realized was um was the following if you compute the partition function uh of this system on an interval then there's two contributions to the partition function or say the free energy there's one contribution from the bulk but then there's an extra contribution that comes from the boundary itself and the contribution from the boundary is proportional to um this number gr so it's called the boundary central charge the right way to think about it is that somehow it's capturing the degrees of freedom which which live on the boundary itself it in particular i i said in the introduction that one way to impose these boundary conditions would be to cook up little rotor degrees of freedom that lived on the boundary um this boundary central charges is somehow telling you the number of such degrees of freedom that you would need to write down at least at low energies after interactions have been taken into account to impose these these kind of boundary conditions so that that's the the meaning of the central charge um like other central charges um it's something which decreases under rg and that's going to be important later um so it's a measure of the degrees of freedom that live on the boundary um we can ask what's the central charge for for our states um well we we calculated this in the paper last year it has this very nice pretty expression um the central charge boundary central charge for um our chiral states is the square root of the volume of the primitive cell of that lattice lambda it's quite pretty it's also quite intuitive i i said that the wilder the more wildly different the left and right moving charges the sparse of the lattice lambda and the bigger the volume of of this primitive cell so that means if you want to impose really weird boundary conditions where left and right movers have wildly different charges you need to have a lot of degrees of freedom on the boundary in order to to do that this somehow is capturing um that number uh to to give you some sense if you impose normal boundary conditions or andrea boundary conditions this central charge is just one if you impose the kind of boundary conditions associated to a pythagorean triple um three and four and five and zero or in general a and b and five and c and zero um this central charge is the square root of the larger mochello you asked about the kind of lattices that you get in general that this is one characterization of of those lattices i think now i i see slowly and um we we thought we were the first people to write this down but actually costus backers got in touch with us and and and just pointed out there was a very similar formula that that he wrote in a paper with ilke brunner daniel rurgenkamp some years ago um they were computing the tensions of d-brains in certain models that this boundary central charge is also the tension of d-brains um their formula is the same as ours but it takes quite a lot of work to show to show the equivalence so um we were sort of scooped by by almost a decade by uh um uh by these people all right um i said at the very beginning because of this mod 2 anomaly uh there are two different classes of boundary states all boundary states fall into one of two different classes um given the charges that we preserve the next question is which which state which class do you fall into is it the kind of class that would give rise to a myerana zero mode or the kind of class that uh would not um that we also figured out in this paper from um from last year so there's the following um rather nice formula that there's going to be two formally in this talk that i'm i'm quite proud of but feel there should be a deeper understanding and this is one of them um this is the number of ground states in a system if you put boundary conditions are here and boundary conditions are prime here so a system on an interval with different boundary conditions on on the two ends and um it's the square root of the volumes of the primitive cells of the two lattices divided by the volume of the intersection lattice the the lattice which you only include points that are both in lambda r and lambda r prime and then you multiply by this square root of the determinant um factor where the the prime here means you get rid of any zero modes that happen to be in the determinant um yeah i feel like this formula is crying out for some simple explanation but but um we don't have one in fact in this paper last year most of the paper was taking up with trying to understand the properties of this this formula because it's not obvious it's an integer because of all these square root factors um and in fact it turns out it's not an integer so it was quite a lot of work but after a lot of work we could show that um this number was either an integer or it was square root of two times an integer um remember this is the number of ground states in a compact system number of ground states should always be an integer um the fact that there's a square root of two is is this sort of recurring motif that square root of two is telling you that there's a myerana uh mode floating around somewhere so if you plug in um r and r prime into this formula and you find a square root of two it's telling you that r and r prime live in different uh different classes for these boundary conditions but by the way going back to something i said at the introduction um the uh the boundary conditions are closely related to gapped phases uh gapped phases in one plus one dimensions also have a z2 classification it's the classification the kit i have discovered uh 20 years ago and the question about which class you're in is a question of whether the boundary conditions are boundary conditions associated to a trivial spt phase or a non-trivial uh spt phase all right so here's a couple of simple uh properties of these um uh uh these boundary conditions um yeah on that yeah so uh what are the transformations that you can do on r and r prime which keeps this formula invariant is there some understanding of that i mean i mean that's basically changing somehow the both the boundary conditions am i correct in understanding that this changing both the boundary conditions without changing the ground state yeah i don't think there are simple um six simple transformations that you can do yeah and in fact given an r it's or an r prime it's not it's not obvious to us which class they live in you plug them into this formula you figure it out but actually for the case of n equals two we had a much simpler classification for two fermions um but to do with um the Euclide parameterization of Pythagorean trickles but we don't have any kind of simple classification for r and r for which phase they live in r and r prime for for more than two direct fermions other than plugging into this formula and figuring it out all right um the next thing i want to tell you about is how these various boundary states are connected um and i think um there's quite a pretty story here so um you have a system with massless fermions moving in on the line and then something going on on the boundary which is imposing these these boundary conditions and a very lovely story about boundary conformal field theory is that the zero plus one dimensional boundary acts in many ways uh just like any other conformal field theory um actually i said that incorrectly like any other quantum field theory in particular um there are operators that one can insert on the boundary um those operators can be classed by their dimension uh as irrelevant marginal or relevant uh the marginal is when the dimension is equal to one one because that's the spacetime dimension of the boundary um if you add an irrelevant operator to the boundary it doesn't change the the boundary conditions for low energy scattering if you add an exactly marginal operator to the boundary it changes the boundary condition in some smooth way so it moves you among a family of boundary conditions for example those parameters that i call theta in my boundary state there are marginal operators that are associated to changing theta um finally if you add an irrelevant operator to the boundary it disturbs your boundary state and it pushes it to a new one there's an rg flow that happens on the boundary between um the uv boundary state and the ir boundary state all the rg is going on on the boundary in in the bulk you always have mastis fermions and their mastisness is not endangered by this rg flow it's only to do with the question of the boundary condition um changing but by the way i should stress that um as with all rg flows it's usually hard to know where you're going to end up if you start from a particular uh uv fix point which ir fix point you end up with um in particular the condo problem which uh has been solved in simple cases only um is entirely a question of which boundary condition you end up with after enacting an rg flow in that case for a marginally relevant operator um but you know this general question of rg flows for boundary is very very similar to the general question for rg flows in general um in particular there's a beautiful theorem called the g theorem which tells you that under an rg flow the central charge must decrease uh in fact um this was the the first of these kind of theorems was proven by zanalogikov uh that's the c theorem um this was the second to be proven it was proven by fredan and connexially uh back in 2003 it was conjectured by african Ludwig um in their original paper um there's also a much more recent um very interesting proof using entanglement entropy by kassini lendea and uh toroba from a few years ago all right so the question is the following we have these boundary conditions that preserve certain symmetries i want to perturb them with a relevant operator and ask where you flow if you perturb them so the first question we need to ask what are the possible operators that i can use to um to perturb them this turns out to be very easy to to understand um you do it firstly by computing the partition function for the system on an interval uh reading from the partition function you can extract all the states their energies and their charges and then you use the state operator map which for boundary conformal field theory tells you about boundary operators not not bulk operators so um the upshot of that calculation is that uh there are operators um uh boundary operators associated to vectors which don't live in lambda but live in the dual lattice uh lambda star so all operators are characterized by a vector in the dual lattice um uh given a vector you immediately know what their dimension is and what their charge is but the dimension is half row squared and the charges is given by this okay again a little bit of intuition um boundary conditions that have very different left and right moving charges have very sparse lattices lambda but that means they have very fine grained lattices lambda star the dual lattice is fine grained if the real lattice is um uh is sparse um that means that um there'll be many vectors in lambda star close to the origin and in particular many operators therefore of low dimension so the more the more different the left and right moving charges the more relevant operators to perturb by so somehow though those wildly different boundary conditions are rather precarious in the sense that you can have more relevant operators so the question we'd like to ask is in general if you turn on um a relevant operator um you destabilize the boundary condition but you end up with a new boundary condition and what's the new boundary condition how do you relate um uh these things all right so we're going to perturb by a relevant operator labeled by some vector row um those operators carry a charge and therefore perturbing by them necessarily breaks the u1 to the n symmetry that I wanted to preserve uh and in general it breaks it to u1 to the to the n minus 1 um to figure out where we're going to end up um I need to make one assumption with this one assumption uh everything else follows I should confess immediately the assumption is not obvious um and um you you might think you know that there's no reason to believe this what I hope to convince you is things work out so nicely that that you'll you'll agree that this assumption seems seems very reasonable and the assumption is that you have a u1 to the n in the uv under the rg flow you have a u1 to the n minus 1 we're going to assume that when you hit the ir a whole u1 to the n is restored so there's somehow an accidental symmetry in the ir that you get a new u1 to the n the u1 to the n in the ir though is not going to be the same as the u1 to the n in the uv it's going to be a different u1 to the n um in some sense it's disappointing that this works you see if you ended up where you just had you a u1 to the n minus 1 symmetry this would be a new class of boundary states that we don't have any access to don't understand using boundary conformal field theory as I said in response to a question we need to assume the full u1 to the n even just to write down these these boundary states um if you however assume that you you restore a u1 to the n we're back in the class of states that we understand and then we can start to explore the properties of uh of this assumption see where indeed you end up um so uh what happens well uh the u1 to the n has to be non-anomalous um has to satisfy that tough denominator cancellation condition it turns out if you have n minus 1 non-anomalous u1 symmetries and you want to add an extra one there's only two ways to do it one is what you had in the uv and the other one is is where you must end up in the ir so if you start with a bunch of symmetries in the uv you perturb by a particular operator the ir charge matrix has to be um by anomaly conditions alone has to be the uv charge matrix reflected in the plane uh perpendicular to this vector load it turns out it's the only way of solving the the tough denominator cancellation condition um if this were or true uh well if this is where you end up um you might naively think that the infrared boundary central charge is given by the following formula um and then there's a very simple check of whether our assumption is correct the simple check is uh is this smaller than the uv boundary charge because the g theorem says g has to decrease as you uh as you flow um it turns out that this is not the right answer there's a bunch of subtle tears and caveats that we have to take into account which means that this naive central charge is a lower bound but the actual central charge can can be significantly higher in the infrared um as we'll see the end result is going to be that the u the infrared central charge is always smaller than the uv central charge but just by the skin of our teeth um so somehow we pass this g theorem constraint but but barely uh you know we a few times we thought we would fail it turns out uh it turns out we didn't let me very quickly tell you what these um uh some of these subtle tears are all right um i've told you that all boundary conditions fall into one of two classes this spt classification and when we were first looking at this we assumed that under rg flow you would remain within the same class it turns out that doesn't happen that uh you start with a boundary condition you undergo rg roughly half the time you flip spt class from one to the other now um this confused us how can it possibly be um consider the following experiment you have an system on an interval where the two ends have to sit in the same class the two boundary conditions that's perfectly consistent now you perturb the boundary conditions on this end and you undergo rg where you end up now is a boundary condition with different boundary conditions from different classes on the two ends but that's an inconsistent situation how can it possibly be the only way that this can be is if the system somehow dynamically generates an extra myeranomode um to cancel the anomaly i said this was a you know a boring trivial way to cancel the anomaly is just add one by hand here the rg must be generating an extra myeranomode for you so we claim that um if you flow from a boundary condition in one class to a boundary condition in another class the system must generate an extra myeranomode that extra myeranomode increases the boundary central charge by this famous square root of two factors okay caveat number one um caveat number two i'm going to gloss over actually it's a slight slight digression um caveat number three is the following sometimes you perturb by an operator and you don't break u one to the end to u one to the n minus one but there's an extra discrete symmetry which is is left and if you look at the candidate boundary state uh in the infrared that candidate boundary state is not invariant under that discrete symmetry now again that that's problematic you had a discrete symmetry everywhere down the rg flow there has to be a discrete symmetry that survives in the in the infrared um the only thing to do is to sum over the images um of that uh that boundary state again this this raises the central charge by the way for string theorists in the audience there's a very simple analog of this if you start with an unstable debris it's quite possible that an unstable debris a single debris decays to multiple lower dimensional debris um that's a kind of situation that that in the boundary conformal field theory language has has exactly this sort of uh sort of picture David sorry I have a maybe a bit naive question I mean is there a way that when one solves these boundaries rg flows one is somehow making an assumption that there is like bulk boundary complete decoupling so that what happens at the boundary never changes really what's happening in the bulk the the assumption is um that you're looking just at low energy physics at the end of the day that that's the assumption of boundary conformal field theory so there is a decoupling but the decoupling is the usual kind of rg decoupling that if you if you really wanted you know cogs and wheels on the boundary to impose these boundary conditions they come with a scale and you're looking at energy scales of scattering way below the the dynamics of what's happening on the boundary so that's the only sense in which there's uh there's a decoupling so since the bulk is gap but then this discovery is kind of controlled no bulk is gapless bulk is gapless yeah bulk is gapless massless fermions in the bulk okay I have to think about it all right so there's a bunch of subtleties and we you know we we stumbled upon them one by one we had to sort of figure them out um when the dust settles we end up with the following I think rather remarkable formula um the infrared central charge is equal to the uv central charge times the square root of the dimension of the operator by which you perturb um I've never seen a formula like like this before again it's crying out for some explanation um and I don't know what that simple explanation is that the proof we have is you you do the calculations you figure out every subtlety that can possibly happen and at the end of the day you you you get this um it's very rare it's very rare that if you if you just know the uv central charge and you know the dimension of the operator you're perturbing by you have enough information to figure out after the rg for where where you end up um it's certainly not my usual intuition for the way rg works um if anyone has any ideas or anyone's seen anything like this if I would I would love to know um notice it immediately satisfies the g-theorem it satisfies the g-theorem because by definition a relevant operator has dimension less than one um on the boundary because one is the spacetime dimension of the boundary um all right this is the end of of the rg part um what's left is a couple of other um sort of comments and applications one on a z8 classification of spt phases the other on the callan rubricoff effect but given that my my time is up I think I'm just gonna stop here and uh take questions and if there's questions about the applications I'd be happy to answer them um but thank you very much for your attention thank you okay what are the applications all right um let let me uh there's so there's rather lovely classifications of interacting spt phases one of which occurs for two plus one dimensional field theorists so there are two plus one dimensional field theorists where if you're on a manifold with boundary there are direct fermions that live on the edge and the z8 classification has a very nice interpretation in terms of these direct fermions um and why ah okay good um this is by the way this is work of fidkowski and kiktai a very famous work from a little over 10 years ago and what I'm going to tell you is a slight rebranding by you and yang and chi um the question is the is the following take uh fermions in one plus one dimensions massless fermions give them a mass and ask is it possible to give them a mass preserving left and right fermion parity now um obviously you give them a mass you always preserve fermion number but if you just write down a quadratic term in the action very clearly you break individually left and right fermion parity because it's psi left psi right is the mass term um what um uh fidkowski and kiktai as I said this rebranding by others showed is that if you have eight myerana fermions it's possible to give them a mass while preserving this left and right uh parity it's really rather a striking statement in terms of boundary conditions it becomes even more striking um it's a statement that if you have a boundary condition um left moving fermions turn into right moving fermions but it's possible to have a boundary condition where um the not the the fermion parity of left movers and right movers is conserved in other words the mod two number of left movers and the mod two number of right movers is conserved it's very weird because our usual boundary conditions left mover turns into right mover that clearly changes this uh this left and right moving fermion parity nonetheless it's possible to write down boundary conditions that uh that do this um in fact they were first written down in 95 by maldicina and ludwig this was maldicina before maldicina was was maldicina um they didn't realize anything about spt phases uh and the connection was made in a paper by a choice yuzaki ryu and ludwig a few years ago um but uh the general question you can ask is of all the boundary conditions we've written down which of them have this remarkable property that they preserve both left and right moving fermion parity um it turns out that uh this holds if um the lattice is even it even means that that every point in the lattice the uh the length squared to every point is an even number not not an odd number and so then there's a question you can ask when can you have even lattices of the type lambda r um remember lambda r is defined by one lattice being rotated into an into itself by this reflection matrix um that's a problem which we we ask many mathematicians they all agreed it was really interesting and they all said they had no idea how to how to solve it which is sort of mathematicians speak for they can't be bothered i suspect um but but they did give us enough encouragement so that um uh philip spent spent a couple of months and came up with a beautiful proof uh of when this can happen a proof involving some short exact sequences um and it can only it can happen only when this lattice has a dimension which is a multiple of four um now remember the n was the number of dirac fermions that means that it can only happen when you have eight myerana fermions or a multiple of eight um which is precisely the expectation from from the classification of spt phases so that's one of the um uh uh the applications this is a result from our paper that came out on monday actually um the the other one is an open problem about the kalan rubikov effect um and in modern language it's the following uh take a chiral gauge theory in three plus one dimensions and put in a tough line a tough line is like a heavy magnetic monocle um the claim i want to make is no one knows how to do that it's uh it's not at all clear how one can insert a tough line in a chiral gauge theory the problem is that a tough line is defined by imposing boundary conditions on fields the boundary conditions that you have to impose on the chiral fermions in three plus one dimensions turn into the kind of chiral boundary conditions that i've been describing here but not within this class that a u one to the n is is is satisfied and in fact they're boundary conditions that that uh just aren't known in in the literature so here's a here's a particularly simple example um this u a single u one gauge theory in three plus one dimensions with vial fermions with charge one five minus seven minus eight and nine this probably doesn't look familiar but it's the simplest non anomalous chiral gauge theory in three plus one dimensions the gravitational mixed gauge gravitational anomaly also cancels that that's uh that that's important um the chiral boundary conditions you want to preserve both s u two rotation around the top line and the u one gauge symmetry it turns out that the left and right movers have to fall into um representations of this which sit in the following representations and you want charges but but because it's an s u two times u one and not back some carton u one uh this doesn't fall in our bunch of uh class of boundary conditions and no one knows how to write down these boundary conditions so i i think there's a very important open problem here in in three plus one theories uh um that falls very firmly into the the kind of things that i'm discussing but that remains unsolved um all right now i've finished my talk so uh thanks for the question other questions yeah i'm just okay okay uh yeah i have uh some some some confusion so maybe my question is just is a stupid one uh so at some some point so you have this um you are saying there is a z two sp t classification of uh this of the quantum mechanics um so when i have sp t uh classification i usually think that this is classifying top anomalies so i have some global symmetry and this sp t phase in the bulk is classifying for me top anomalies of of the boundary but here you're saying that so first of all i will ask to ask you in this case what i mean is the z two asymmetry and if it is why you're saying that it is a um inconsistency rather than a top anomaly it's it's all tied up so let me tell you the full story for for one plus one dimensions um take a single myeranothermium okay i give it a mass um you can give it a mass with a positive sign or a negative sign okay but implicitly there's a regulator of howly villas for me on in the background but um with a positive sign it's in the trivial phase and with a negative sign it's in a topological phase and the topological field theory is minus one to the arson variant of the ream and surface okay okay that's that's the sp t phase in in uh one plus one dimensions now um put that on a manifold with a boundary okay it's the top anomaly is supposed to emerge in the field theory of of the boundary um if you put uh the myeranothermium with a positive mass in the manifold with boundary there's there's no zero mode if you put the myeranothermium with negative mass in the manifold with boundary you get the myerana mode on the boundary but the myerana mode is inconsistent but so too is the topological field theory minus one to the arse defined on a manifold with boundary but the combination of minus one to the arse on a manifold with boundary and a myerana mode stuck on the boundary together is is consistent so that's that's the uh relationship with the top anomaly it's not quite a top anomaly because it it's somehow you know I don't think a single myerana fermion you could think of as it as having a top thing oh maybe it is maybe it's a top anomaly in minus one to the f maximum in that case so could I somehow uh I don't know give up one minus one to the f and try to formulate a quantum mechanics that doesn't have minus one to the f and then have a consistent quantum mechanics or or or I don't think so yeah I think minus one to the f is special and that's why this is an inconsistency rather than the usual case of a top anomaly if you just it's just a top anomaly yeah I think it's I think it's minus one to the f which is yeah more crucial to keep than other scenarios okay so for this one dimension myerana fermion which action is lambda lambda dot uh the assumption is that it's a spinner in the sense that it's odd under minus one to the power f if we don't impose that I mean usually in higher dimension this has been statistics but in in this zero plus one why why you have to impose that it is odd under minus one together with a statement that lambda lambda is is vanishing because these are lambda lambda dot the action is yeah yes but but because it's a fermion you can't write down lambda lambda yes I agree with that yes that that that's really the the essence of it but it's the fact it's a fermion not a boson so it's a base power you are saying that it's a grass manian variable yeah but the grass manian variable and the relation with minus one to the power f is the spin statistics right yeah although quite whether that's uh that means in quantum mechanics I'm I'm less sure about so really it's the statement is question