 Next speaker, Ashwin Vishnabhatta on topology correlation and superconductivity in war rate, super lattice protein. Which I won't have time to get to, of course, but yeah. Yeah, so thanks a lot to the organizers for inviting me. I picked a topic which I thought Piers might enjoy, interplay of topology and correlation effects and superconductivity. He's not here though, but I'll still talk about this. So this is work with my collaborators. So Senthil and Liujun, who's Senthil's student. Yejwong, who was a postdoc at Harvard now at San Diego. But actually most of the things I'll tell you was really done by my student, Adrian Po, who just graduated and he's going to be a popular fellow at MIT. So I'll be talking about twisted balear graphene. I think you already heard a couple of talks yesterday. So I'll try to make contact with those talks as well. You'll see there's a little bit of this problem is actually more complicated, even at the band structure level than we initially anticipated. So it'll be important to kind of make those things clear. So as you all know, there was this amazing discovery in twisted balear graphene. But to put it in context, let me compare it to some other correlated systems that you're familiar with, transition metal systems, like the ion-based superconductors, the cooperate superconductors. All those cases, you have atoms with some correlated orbital, de-orbital maybe. And the spacing between the orbitals is the atomic spacing, a 10th of a nanometer, roughly. And you get this kind of physics, spin density wave, superconductivity, and so on. Cooperates, you get a clear insulator at half filling. So here in this case, which I'll describe a little more detail, but I'm sure you all know roughly what it is, the effective atoms, the effective lattice is a factor of 100 bigger than the lattice in the atomic systems. This Moray superlattice is of over 10 nanometers. That's the separation between these islands that you can kind of see with your eye. And people were able to scan what happens to the system as a function of electronic density. And it's very easy to do that here, rather than chemical doping, you just tune the voltage on a gate. And what they see is at half filling, which one of the main confusions here is really the electron count. I'll try to clarify that. At half filling, defined in a particular way, you get a correlated insulator. Interactions are essential to getting this insulator. Let's call it a MOT insulator, but you have to be a bit careful with that word. You dope that small density of electrons or holes, and you see a superconductor. And the question is, how do we understand the system? How does it fit in with what we know about correlated electronic systems? So really the reason you get this physics, in fact, was theoretically predicted that this is an interesting regime to look at, this magic angle of around one degree, because the electronic bands are expected to be very flat. Kinetic energy is very small. Interaction energy is relatively large, so you expect to get correlation effects. So there are two kind of paradigms for strong correlation effects in electronic systems. One is the lowest lander level problem. You partially fill a lander level, you get all the exciting physics of the fractional quantum effect. The other thing is correlated materials. And it turns out that the recipe that you use to solve these two problems is actually very different. So for correlated electrons, what you do is you come up with a tight binding model that will describe the electronic band that's of interest, find out many of functions, and then proceed to writing down a Hubbard model. And then the entire question becomes, how do you solve the Hubbard model? But basically, this part is kind of assumed from the get go. But what I want to remind you is that that's completely different from the way in which you solve the lowest lander level problem. So in fact, you cannot implement the first step. You cannot write down a tight binding model with localized vanishing orbitals to describe a single lander level. So there's an obstruction because of the churn number of the lander band. And we know that churn number is essential to getting the fractional quantum whole effect. So instead, what you do is you work directly in the lowest lander level space of states. You're fortunate they have a relatively simple representation. And you project the interactions directly into this band. So the question is, this problem that we are confronted with, correlations in twisted bilayer graphene, which paradigm does it fall into? Is it something we have to directly work with maybe at the level of momentum space, introducing interactions in momentum space? Or can we write down something like a tight binding model? So what I want to convince you is that the answer is somewhere in between. The simplest version of a tight binding model cannot work. And I'll tell you exactly what I mean by the simplest version. We want to retain all symmetries. We want to come up with the minimal model. That turns out it's not possible. But the situation is not as difficult as the problem of the lander level. There are ways in which you can sort of try to come up with something that looks like this, but you'll have to give in at some places. It makes it a harder problem, I would say, than the physics of typical correlated electronic material. And that's where the topology that I was talking about comes in. So this is monolayer graphene. This is two layers. And this is what we want to think about when you twist them by some small angle. This is not a small angle. This is a very big angle. This is 21 degrees. But it's somewhat harder to draw pictures of the small angle structures. But what will be very important is that the physics at small angles is actually quite different. And we'll have to be a bit careful about that. We'll see that there are certain emergent symmetries. And that's going to be very important for our story. So when you look at the structure, you see that the blue and the red are the two layers. There are places where the lattice points more or less align. This alignment would be much better in the small angle case. And there are places which look more like bilayer graphene. So in the usual Bernal stacked bilayer graphene, you have one side in the center of the hexagon. And then only the A sides, for example, are eclipsed from the top and the bottom layers. So there are regions. We'll call this A regions, where they're all on top of each other. We'll call this AB regions, because one sublattice is over another sublattice. But they're different sublattices. So that terminology will appear. So what is discovered several years back by these electronic structure calculations was that if you carefully tune this twist angle, you can get to certain angles where the dispersion of the bands is very small. And those were called magic angles. And the most relevant one, this is actually the largest magic angle, is about 1 degree. And in that case, if you make a simple comparison, you'll see that electronic interactions can be very important. So strictly in these models, the band width is literally 0. At the magic angles, if you do a slightly more realistic calculation example by these people, they looked into the effects of relaxation of the lattice. It has some bandwidth. So they obtain something like 8 milli-electron volts. There are others who might obtain about twice that number. But this energy scale is significantly smaller than a crude estimate of the Coulomb interaction. So an estimate of the Coulomb interactions, you imagine a pair of electrons separated by a lattice site. Take the dielectric constant of your medium, and you find it's about 30 milli-electron volts. So this is a situation where at least the two are comparable, if not the interactions are even greater. And experimentally, what is seen, which is somewhat hard for these numerics to precisely capture, is that there is a band gap. So this is really the beauty of this twisted by-layer graphene. There are bands that are separated out from everything else in your spectrum. On top of that, these bands are very flat. But the zero-thorough statement is that there are these bands that are completely separated out from the other bands. And they are active when your chemical potential is near neutrality. So experimentally, this gap is around 30 milli-electron volts. So one way of thinking about this problem, which I kind of like, is this angle is very small. So the right way to think about this angle is in radians. It's about 1 60th of a radian. So everything in our problem will be controlled by this small number. There's a small number in your problem, so you've got to keep track of that. So if you want, for example, to convert these energy scales into something you can compare with solids, my general rule of thumb is to multiply by 60. So that'll give you a rough estimate. For example, we said that the interactions were around 30 milli-electron volts. Multiply by 60, you get about two electron volts, which is roughly the u of the cooperates. Also, the transition temperature of superconductivity here is about 1 or 2 Kelvin. And you can do that multiplication, and you'll see it's not unreasonable. So we're really talking about something that's a high temperature superconductor once you make allowance for this rescaling of energy scales. So another analogy which may be useful is it's a bit like the C60 materials, which are also known to be superconducting under the right conditions. But instead of having 60 atoms in the unit cell, that's C60, this is more like having 60 times 60 atoms in the unit cell. And that's the kind of scales that we are talking about. So the length scale of this Moray superlattice is lattice constant divided by theta in radiance. So now let me remind you how people first looked at this problem. So because of this small parameter, you can make use of a continuum theory. So you know that graphene has this Dirac-like electronic dispersion near neutrality. And we are thinking of these Dirac electrons flying through this by layer Moray pattern. And their motion is affected by this pattern. So you can sort of reduce the problem instead of writing down very detailed tight binding models. You can approximate all the physics by the dispersion of these Dirac fermions from the two layers that then move through this potential. So what's going to happen first and foremost is that you're going to get a valley symmetric. So the Dirac fermions and the opposite values will not talk to each other to a very good approximation because there's an enormous discrepancy in the length scales between the two. So for example, if this is the Brouvoir zone of the original graphene, we're going to end up with a mini-Brouvoir zone when you have this larger length scale super lattice applied. And there's a big mismatch between the momenta over here. This is the kind of momenta that you get scattering by when you have this periodic potential. There's a big discrepancy between this momentum and the momentum of the underlying electronic lattice. So questions like is this a commensurate or incommensurate structure, you can ask those questions. But they're probably not relevant for the problem. The system doesn't scatter sufficiently large number of times to recognize if it's commensurate or not. So that is the idea of this continuum theory. And we'll see that it actually is exactly the right thing to do when you are in the limit of small angles. This is really where the small angle approximation comes in. Try to apply this for a 20 degree Moray super lattice. It's not going to work that well. So how do we end up with this particular angle being special? So there's a simple way of thinking about it. So of course, there's this unperturbed electronic dispersion. It's a linear dispersion and it has an energy scale about 16 electron volts per when you put in the lattice constant. And now you have this procedure where you twist the electronic structure. So you inherit two of these drug cones which are separated by an angle theta. So if you look at the separation in momentum space, it's just theta times this Brouwer zone vector. So if you figure out what is the energy splitting over here between the drug cones, this has this value. And what we'd like to do is we'd like that the mixing between these drug cones, which is the interlayer tunneling. All of this is assuming that we have two independent layers. We just write down the free electron dispersion for two independent layers. When you allow for mixing between them, you'll open up gaps where these things cross. And you want those gaps to be large so that you push these levels together and make a flat band. And you can write down a simple criterion for that. The splitting should be of the same order as this number in order for that to happen. So if you put in the value for that coupling, about 0.3 electron volts, you'll end up with an angle that's very close to the magic angle. Of course, that's a bit fortuitous, but the calculation proceeds in a similar way. So that's how this particular angle gets singled out. So I already mentioned that there's a large separation of scales which leads to this internal symmetry, which we're going to call valley symmetry. Or if you like, there's a u1 symmetry corresponding to the independent charge conservation of electrons in this valley and this other valley, which is at the other end of the microscopic Brouwer zone. So this is going to be an internal quantum number like spin. And all our degeneracies are going to be doubled because of this internal quantum number. So the formalism that this continuum theory utilizes is you have two Dirac fermions in the top and the bottom layer, and they're coupled through the scattering matrix, which is essentially the interlayer tunneling. But this can have some wave vector dependence, and you just solve this problem. It's kind of like solving this empty lattice problem. You start with the free electron dispersion, and you put in a weak lattice. It's very much like that. The difference is that you're dealing with Dirac fermions to begin with. And that makes it very interesting, and gives you all kinds of topological properties. So this is just the statement that the scattering between the two valleys is suppressed in the angle 1 over theta. So if you solve this theory, so you've got to figure out what this coupling matrix is. I'll mention that in a second. But there's a natural coupling matrix that you can come up with. You allow for this coupling between the top and bottom layers, and you end up with a band structure that looks like this. I'm showing this for a slightly larger angle, two degrees. But qualitatively, the physics is the same. You have this set of bands near neutrality. And you see that if you go to the k point, you have a Dirac crossing. So in the mini-Brilva zone, you have a reflection of the original Dirac fermion, in that you have a Dirac crossing over here. So we have two colors over here. You see these are the two different bands corresponding to the opposite valleys. The dispersions are not exactly the same. The two are related by time reversal symmetry. So along some lines, the dispersions are actually different. But if you tell me the dispersion in one valley, I can always figure out what it is in the opposite valley. It's a simple transmission. So often, we will just look at one valley, the band structure of one valley. So one thing that's a little different over here is how many electrons it takes to completely fill this band. So what we're used to in usual band theory is that it takes two electrons to completely fill a band. The cooperates empty to completely fill these two electrons, half filling is one electron per unit cell. It's a little different over here for two reasons. One is we have this valley symmetry. So that gives you an additional factor of two. So you might think that there are now four electrons required, completely empty to completely full. But it turns out the number is actually eight. And the reason is that this band is kind of like the dispersion of graphene itself. So we know that in graphene, there are two sublattices and you need twice as many electrons to completely fill the band. So in fact, to go from completely empty to completely full, you need eight electrons. Two for spin, two for valley, and two because you have this kind of sublattice structure. So this is the number of electrons per unit cell of this Moray superlattice. So neutrality is right in between. This is if you didn't try to dope it in any way, then you need four electrons up and four below. And this Dirac physics is gonna be important. Okay, so now before I talk about more formal stuff, let me just show you some band structure calculations for this problem. So one thing that is important is, not the kind of universal physics we are talking about, whether there's a degeneracy of the k-point, but really if these bands are separated out from all the other bands. Okay, we said that that's important. That's seen in the experiments. You know, is that feature reproduced in these calculations? Okay, so it turns out that if you just do a naive calculation as done in the original papers by McDonald and others, it often happens that you get like this red line. Okay, where the band in the center is sometimes touching the bands above and below. Okay, but if you allow for a little bit of lattice relaxation, which is very natural, the two layers are gonna kind of relax mechanically in order to find the best structure. And what it likes to do is it likes to expand these AB regions, okay, where it looks like Bernal-Stadt graphene. That's mechanically the most stable structure. Okay, so if you allow for that, there's a nice paper by these people where they put in the relaxation and at least the quantitative aspects of the band structure changes. So you now have a nicely separated band that's close to neutrality. It was redone in this particular paper. There's this band that's close to neutrality. And you have some band gaps that separates it from the other bands. Okay, so effects like this may be important in order to get the quantitative features. And you can incorporate that into the continuum model. The continuum model has this coupling matrix we said that couples the two layers. It has two parameters. U couples the A and A sites and U prime couples the A and B sites. And usually you take them to be equal. They'd be different. But if you have an expansion of the A B regions you expect this U prime to be larger than U. And that's essentially what this relaxed structure band structure can be fitted with. You can fit it as a continuum model where there's more hopping between the A and B regions. Okay, so now let's see how this whole band structure that people came up with compares to experiment. Okay, we'd like to experimentally verify this band structure that we have. And to do that you want to go to somewhat larger angles. Okay, you don't want to be in the limit. You have a question? Yes, that's right. So it isn't quite DFT. It's a tight binding model where you have certain slater parameters, yeah. Ah, so the question was, is this at a commensurate angle and is it DFT? And rewind for the answer. Okay, so there's this very nice experiment several years back by Eva and Ray's group where they did STM on this system slightly larger angle. And they see that the electrons are mainly living on these so-called AA sites, form a triangular lattice. And then they can go and look at the tunneling as a function of energy. Again, this will give you access to this question of whether the bandwidth is very narrow near the magic angle. Okay, so they have a sort of a proxy for the bandwidth that's this vanhove splitting. It's not exactly the bandwidth, but it scales with the bandwidth. And what they see is that if they go to large angles, there is some substantial splitting, half an electron volt. But as they approach one degree, which is supposed to be this magic angle, this bandwidth drops precipitously and goes very close to zero, right? So this is sort of a nice demonstration that at least the bandwidth aspect of our story seems to be correct. Yeah, so that's the picture of the AA regions where the electrons seem to be mainly positioned. The AB regions form a honeycomb lattice in contrast. That's gonna come up again. Okay, but what about the rest of the details? Do we have, for example, this Dirac crossing in this mini band? So there's another very nice experiment due to Pablo's group, but a couple of years back. And they looked at somewhat larger angles again, not these things very close to the magic angle, but something closer to two degrees. I think it was 1.8 degrees. And now what they see is, they see the insulating states, but only where you would expect them from band theory. So if you completely empty the band or completely fill the band, you see insulators, that's the red and the blue. They can count how many electrons there are over here and it's basically eight electrons, difference. And in the center, you see a dip in the conductance and this is, we think, due to the Dirac points. You still have some conduction. It's not an insulator like at these limits, but it's reduced compared to being in the middle of the band. There's better evidence for these Dirac points, these things over here, if you do quantum oscillations. So you put on a magnetic field and sweep the density, for example, and they end up with these Landau fans. You see the Landau fans near neutrality. They are multiples of eight, but they're offset, they start with plus minus four. That's characteristic of a Dirac fermion. So a Dirac fermion will give you Landau fan with plus or minus half per Dirac flavor. Here we have eight flavors. So we have K and minus K, that's two flavors. We have spin and we have value. So eight and all. You should get exactly this Landau fan diagram. So that's a very nice confirmation that this band structure seems to be essentially correct if you go away from the limit of very strong interactions. You go to larger angles where the interactions are weaker because the bandwidth is bigger. You also see the Landau fan diagram over here that's near the extreme of this band, almost empty and almost full. So now you get the normal degeneracy. You get a degeneracy of four and the four is just the valley and the spin. You have multiples of the valley and spin. It's a quadratic band, so you don't have this kind of offset that you see over here. Okay, so it all seems consistent with the band structure. Okay, so the new physics really appeared when they took it to this small angle, the magic angle or near the magic angle. There's some, you have some freedom there. You don't have to be exactly at 1.08 degrees or 1.05. You can move it on a little bit, but not too much. 1.2 is too far away. Okay, so now you see an insulator and the insulator appears over here. Okay, so this is the midway between the completely empty band and the drug point. Okay, so you said there are four electrons to come up here. There are only two electrons to come up to this, this small insulator. Okay, so this two electrons over here, if you take into account the spin and the valley degeneracy is a fractional filling. And you need electron interaction effects in order to explain this. And so that's why we think, given the success of the band structure, having an insulator at this filling tells you that electron-electron interactions are important. Okay, so I don't know if I'll have time to talk about the martinsulator itself, but rather we'll talk about how we'll try to model this physics. Theoretically, move towards a problem like we had for the cooperates, this Hubbard model, and ask what is the analog over here. Okay, so to do this, we first have to understand what are the symmetries involved in this twisted bilayer grapheme. Leonathan, how much time do I have? Okay, good, I got it. Okay, so let's look at this picture over here. So again, this is a fairly large angle so that you can sort of see this easily. So what we did over here is we took a pair of these sheets placed right on top of each other and we twisted it by this angle, keeping a point at the center of the hexagon fixed. Oops. Yeah, so we kept this point at the center of the hexagon fixed and we twisted it. Okay, so obviously this system has got a C6 symmetric. You can rotate by 60 degrees and it's obviously symmetric. This pair over here goes to this pair over here. It also has what we call a mirror symmetry. So some of the people call it a C2X. So imagine that you reflect along this dashed line. That's not quite a symmetry because you also need to interchange the two layers. Okay, so it's really, if you like a rotation along this line by 180 degrees. Okay, but if you think of this as a two-dimensional system, it behaves like a reflection. You also have that symmetry. I should say though that this symmetry is often broken in experiments. You put this thing on a substrate, this symmetry will be broken. You cannot do the layer exchange. Okay, but it's sometimes convenient to keep track of this symmetry. Okay, so all in all, you have D6, the D6 group. Okay, so of course, in addition, we'll see that because of the small angle, you also have a valley conservation symmetry. Okay, that's not apparent, of course, here. Yeah. So Oscar said it's encapsulated by boron nitride on both sides. In principle, if you did it very carefully, you could have the symmetry if you made sure there were no electric fields. So that's not done in this particular experiment. There's another experiment which verified this. I don't know how well known it is by Corey Dean and Andrea Young, where that symmetry is clearly broken. So they apply an electric field and they see very different response for positive and negative electric fields. So most likely that symmetry is not there unless you really work hard to keep it there. Okay, but good to keep in mind, we'll of course, simplest case, we'll keep that symmetry, but we'll always think about what happens if you break it. Okay, on the other hand, the C6 is simply a symmetry of this structure. Okay, so there's a different structure which you can look at, which I think Oscar must have talked about, which only has D3 symmetry. Okay, so now again, you have these two layers that are superimposed. Imagine twisting them, but now keep the common site fixed. You see, this does not have 60 degree rotation symmetry because this site didn't have 60 degree rotation symmetry. You only can rotate by 120 degrees, okay? So there's that rotation symmetry, C3. That's a symmetry and there is again this mirror symmetry sometimes called C2X or C2Y. We'll call it the MX mirror symmetry. Okay, so now we seem to have these two different implementations of symmetry and which one should we use? Okay, so of course, if you have a bilayer graphene when you put them on top of each other, there's no reason why you should get one or the other structure. In fact, in general, you might think that these things are somewhere in between and you don't have any rotation symmetry. Okay, so the nice thing is that if you're in the small angle limit, it turns out it doesn't matter. Okay, so in the continuum theory that I talked about, this displacement symmetry where you line up the sites is completely irrelevant. It doesn't affect the electronic structure. Okay, so one way to see this, sorry, I don't think I need this right now. Okay, so really this, even if you started with the C3 symmetry, if you're at small angles, you will have an emergent C6 symmetry. What we believe is that it's important to keep the maximal symmetry that's present, then you do not have accidental degeneracies. You have a degeneracy, it's always explained by some symmetry or the other. Okay, so to kind of, in those pictures, I showed you those structures look kind of different. Okay, so you may worry, is this really a symmetry or how good a symmetry it is? Okay, so Adrian came up with this eye test. So some of these structures have C6, some of them don't. And actually this is at a fairly large angle. It's not one degree, it's at six degrees. But you can see that it's actually not that easy to tell which one is six-fold symmetric and which one is just three-fold symmetric. And if you're just curious to know the answer, anyone wants to guess, someone has really good eyes. I kind of get hypnotized when I look at this, so I don't want to look at it too long. But so these two don't have six-fold symmetry. Okay, it's very hard to tell from just looking at it and if you cannot tell, probably the electrons can't tell either, right? So the electronic structure of these different things are look very similar and you can see that in abolition calculations. Okay, so now let's talk about the details of this band structure and the very important part of this are these draft points. Okay, so what we're gonna do is we're gonna look at a single value, so we're gonna fix on either the k or the minus k value. Okay, so when you do that, you essentially break time reversal symmetry because you picked a value. And you can ask what the symmetries are. So you have this combination of time reversal and rotation. You can think of this as a combination of C2 and time reversal. Rotate by 180 degrees followed by time reversal. That keeps you at the same value. Okay, so that is a symmetry of the single value problem and let's just look at one value dispersion. Okay, the opposite value you can get by simply some symmetry transformation. Okay, and you look at this, you see these draft points and you can ask what is it that's protecting these draft points in the continuum model and it turns out that it's essentially these symmetries. Okay, so if you give me these two symmetries, what you can show is that the bands at this point, they have a eigenvalue under C3 rotation which is omega, the cube root of unity. Then the C2T symmetry, if I just take the cube of this symmetry, the C2T symmetry forces it to be degenerate to another representation which is omega star. So you get a two-fold degeneracy and the degeneracy between the two different values is ensured by some other symmetry, okay? Combination of time reversal and another symmetry. Okay, so it's really the combination of these symmetries that gives you this particular degeneracy. And if you lose any one of them, you're allowed to write on a mass term that gaps out these draft fermions. Okay, of course, you could argue that mass term is gonna be smaller. This is a good approximate symmetry but it's best to start with a point where you have the symmetry present. Okay, so if you just stare at the representations in this figure, you know, based on some earlier work, we figured out that this is only compatible with sites on the honeycomb lattice, okay? Yeah, so that's one piece of information that you have and you can ask, can I reproduce just those two bands by some padbinding model on the honeycomb lattice? Okay, just a pair of bands living in a single value. And it turns out that there's a problem doing that. The problem is that these two direct points are kind of strange. Okay, on the face of it, it looks like some dispersion, like graphene dispersion. But if you go and examine these direct points in more detail, you find out that they have the same chirality. Okay, so if you go around them, you get a better phase which could be plus pi or minus pi. That sign, it turns out, makes sense in this particular symmetry setting and you can compare them and they have the same sign. Okay, this is assuming some smooth gauge and so on, you have to be a bit careful. But there are many different ways of seeing that these two have the same chirality. Okay, so if you started with the tidebinding model, even if it's on the honeycomb lattice for these two bands, you would always get opposite chirality. There's some essential part of the physics that you're gonna miss. And this may be important or not important for the actual physics and the material, but you'd like to be accurate when you capture these things. Okay, so this is really the obstruction. If you wanted to write down a valley filtered, spin filtered band structure just for the two bands, while keeping all the symmetry that preserves the drag points, the rotation symmetries and so on, you run into the obstruction. Okay, so let me kind of explain the obstruction in a physical way. So this goes on to the notion of a flipped halden model. So let's say you had some band structure. We'll argue by contradiction. Okay, you have a band structure that gives you a pair of drag points with the same chirality. So now imagine that I put on a staggered potential. I change the potential on the A and B sides in opposite directions, just an on-site term. What you can show is that this is going to gap the system. It breaks one of the protecting symmetries, it gaps the system. You get two bands that are separated by a gap and you can calculate the churn number of the bands. And you'll find that the bands have churn number. Okay, remember we already broke time reversal by selecting a value. Okay, so this staggered potential actually gives you a churn number. Okay, this is the same effect of the halden term. If you actually had a honeycomb lattice and you had this halden second neighbor hopping, you'd also get churn number. But here you can accomplish that and the only way to accomplish that is by an on-site term, staggered potential. Okay, you get a mass term that's independent of K. Opposite chiralities, they end up giving you the same, they give you a non-zero churn number. Okay, but this is actually a contradiction because you can imagine increasing the strength of this potential so that the electron just sit on one of these two sides. But that model cannot have churn number. It's just electron sitting on a side, right? So this combination of having this staggered potential giving you churn number and having just a two-band model. There's no additional stuff over here that can avoid this conclusion. In the limit of large potential, you're really on-site. That's what generates this kind of no-go argument. Okay, it tells you that simple tight-binding model like this cannot capture the physics of this really flat band. Okay, we've got to do something a little more involved. So there's kind of a menu of options you have. Simplest thing fails, what can you do? The first thing you can try to do is to lose some symmetries. You just drop some symmetries, either try to implement it later or just don't worry about the symmetry. All these symmetries are ones that are gonna gap the drag points. They're gonna do something bad to your band structure, but that's kind of the trade-off. Okay, and you can do that and that's essentially what these people did, which that's how I understand Oscar's construction, which is you take one of these D3 structures, you lose the C2 symmetry, and then you can proceed. There's no obstruction and you can complete this program. We had a different way of doing it. Initially, we lost the value you want. Okay, so forget about the value you want. You can come up with some model. You can try to implement it in the end. We have some procedure to try to implement that, but you get some four-band model which does not have the symmetry. And these are different implementations. You lose different symmetries, so you cannot directly compare these models. So that's one option. Let me just give you a little more detail before I tell you the other option which I actually prefer. Okay, so we're gonna take this approach. Forget about value you want. We have these two different bands from opposite values. Let me combine them into a single problem. I'm not gonna distinguish them. I'm gonna try to write down Vanier functions for the set of four bands. Yeah, and it turns out that you can do, there are no obstructions then. And from the representation physics we talked about, we know that these Vanier centers have to fall on the honeycomb lattice sites. Okay, we're gonna get two Vanier orbitals for every honeycomb site, a total of four. And that much is fine, but we saw in the experiments, for example, that most of the electronic density lives on the triangular lattice sites. Okay, so that seems to be a problem. How do you reconcile the two? But when Adrian went and calculated these Vanier functions, it has exactly the right form. It has this form. It's centered on a particular point, which is this, let's say this point over here, honeycomb lattice site, the AB site, that's the center. But most of the weight of this wave function is actually on these three lobes. Okay, so I thought of calling this a maple leaf, something like that, but Adrian says that most people know about fidget spinners more than they know about maple leaf. So anyway, we call this the fidget spinner Vanier states, and I think Oscar and this paper by Kushino, Liang, and Yuan also have a similar kind of picture with the different symmetry representations. Okay, so that's one possible resolution. You forget about the valley and then you're gonna do something at the end in order to try to keep track of that symmetry. Either you fine tune your model or you have to try to implement the symmetry in some non-local way. Okay, so we don't quite know how to do that when you have interactions. Okay, so maybe that's something to try to work out. It's interesting to work out, but there's a different option. Okay, the second option over here, which is to extend the model to include more bands. Okay, this is kind of familiar to people who do cooperate physics. You have these PD models of cooperates. Okay, you add oxygen. Is that 10 or five? It's over, all right. Okay, yeah, actually I just have five slides. Let me quickly go through. Anyway, so this is like this PD thing and it has all the symmetries. So I won't really talk too much about it, but we can get good agreement with the continuum model and our tight binding model with these extended bands. So you've got to include some more bands. The minimum is four additional bands, but you can get all the physics in one valley with all the symmetries and it has some interesting connection to the so-called, this idea of fragile topology. Okay, so usually you have a topological band. You should have another one which has the opposite topology to cancel it. You have a churn number band. You have to have anti-churn number to cancel it. Here it turns out that the kind of topology over here, this, you know, having the same chirality, you don't need that kind of cancellation. That's why we call it fragile. You can add bands, which is what we do, which are simple atomic insulators. Things that are close to being just localized on sites and they are able to resolve the obstruction and allow you to write down a tight binding model. Yes, anyway, this is to appear soon. So let me just end with just two other things which I didn't have time to talk about. The first question is, what is the mod phase? Okay, and we have some suggestion in terms of inter-valley, it's not internally current, it's inter-valley current. So it breaks the value once spontaneously. And we also had a different paper looking at the superconductor in the completely opposite limit, which is the limit of very weak interactions. If people do this in the cooperates all the time, it's been quite successful for the ion nictides as well. Surveys people including Andrea have worked on this. Usually you need some kind of nesting to drive this kind of superconductivity. So if you told me you had a triangular lattice at half filling, there's no nesting really to drive that. But this model is different. If you look at the Fermi surfaces, they have two valleys. Each valley is not very well nested, but the inter-valley Fermi surfaces are pretty well nested. Okay, so you can use that to drive some interesting physics at weak coupling. And you start, you do an RPA calculation and then you use that to figure out what the pairing is. Can we predict this kind of topological superconductors? It's very natural on a triangular lattice geometry to get a P plus IP or a D plus ID superconductor. Can we get that? Both of these are actually spin singlet. And it may seem strange to say a P plus IP is spin singlet, but you've got to remember you also have a valley quantum number that can absorb the minus sign. And the real question I think is which of the two is it? Is it P plus IP or D minus ID? These are the same symmetry, but they're topologically different, different number of edge states. But that's really doing the physics in the opposite weak coupling limit. Okay, so let me stop here.