Do Koreans think that they are the superior Asian at least visually? Well, that's what this TikToker who lives in South Korea is saying. Oh man, this is going viral right now. There is a TikToker who is a Vietnamese American living in South Korea. Her name is Phong VV. Yeah, Andrew. She made some big accusations in this clip about South Korea. Let's drop it. We all know how Koreans feel about Vietnamese people. And it's because she is East Asian passing or Korean passing. Like when she first debuted a bunch of Koreans were like, Oh, I didn't even know that you're Vietnamese because you look Korean.  Long story short, Andrew. She's saying through K-pop, understanding the Southeast Asian K-pop members and how they're treated, you can see that South Korea is extremely racist and look at society. They look down on Southeast Asians and especially any Asians that do not look Korean and that all East Asians in general think that Southeast Asians are below them. So of course, Andrew, we got to get into the internet comments. Our own takeaways. Make sure you like, subscribe and turn on your notifications. Andrew, I will say this, the comments section went all around the world and back. It went way beyond the scope of what she was talking about, just like K-pop or whatever. I mean, why do you think that this just set the internet on fire? I think this is actually something that a lot of people have thought about. And I do think a lot of people agree with her to an extent. First of all, not all Korean people think like this. Not on an individual base. I can't speak for everybody. However, this is a stereotype about Korea and even a lot of Koreans would acknowledge, and if you've talked to enough Koreans, you can kind of see what she's trying to say, although her accusations were maybe a little too harsh and I don't fully know what she meant by, well, you know how South Korea thinks about Vietnamese people. I don't know. I don't know what they think about Vietnamese people. Yeah. I mean, I think she addressed some dynamics like you said that have been brought up on the internet for 10, 20 years. However, you know, there's just something about TikToks when you're just talking to the camera. It's so visceral. It's not a writer. It's not an academic. It's just like somebody saying it. So let's get into the internet comments. Somebody said, uh, it's true, though, a lot of Koreans do believe that. Yeah, I basically do believe that they are the superior Asian, at least visually. Not all, but obviously, and given what they've put out in the past few years, I could see why people think that. I mean, there's even other Asians that think Koreans are the best looking too. So guys, the work is there. I'm just saying, right? You're saying that what do you think about her comparison to saying that like the equivalent of being Korean passing in Asia is being white passing in America? Uh, yeah, I mean, I don't like comparing things to the dynamics in America exactly. But if we had to, it's because this is their country that Korea is more of a Korean country than even America is even a white country, in my opinion, way more. Yeah, I believe that Korea is 99% Korean, whereas America, obviously right now, I believe is only like 63% white. Yeah. So of course, yeah, Korea is even more pure blood than America is by far. I'm moving on to America. Somebody's like, yeah, see, it goes to show you everybody always complaining in America, saying how racist and bad a place America is. But racism exists everywhere. Hierarchy exists everywhere. You'd be a fool to think anyplace else is different. In fact, America does better with racism and lookism because there's more diversity here than countries back in Asia. So why is everybody always complaining about America? No, that's an interesting question. And I would ask people who think like her or would say it the way she does and say, Hey, well, what do you think about America's racism or discrimination history? Like what do you think about comparing it to do? Would you appreciate America more? I think it's very, very difficult to compare different histories with discrimination, prejudice and racism, though. It's like hyper layered and hyper complex. For example, Andrew, what if like Asian East Asian countries in particular are more judgmental, but they're less likely to act on any sort of violence in that judgmentalness and it more comes in stairs and comments or something like that. Which one will you take? Will you take the words and the side comments and the side looks or you want to take the potential where someone's going to like punch you in the face for looking a certain way? Also, it is difficult to compare new world countries that are like 250 to 300 years old with things that are like 3,000 years old with just one group of people. But, you know, everybody is free to make the comparisons they can. That was a point made.  It's just how we have been taught to act. It's not really like us trying to be that way against you guys, but that's how we are even are amongst ourselves. That's really interesting. I would like to hear the more historical or I guess cultural context for that, but that's an interesting point. Somebody said, come on, man. Every culture thinks they're the best, man. Literally, you go to Greece. They think they're the best. You go to Mexico. They think they're the best. Everybody thinks they're the best. Do you agree with this statement? So it sort of absolves everybody of thinking that they're the best in their own country or is there something where you go, nah, that's not true. Some countries definitely think more like that than others. No. First of all, some countries feel more proud than others. However, I think a lot of people who are not Korean that move to Korea have certain expectations about what their life is going to be like in Korea. Do you think they think they're going to get treated like a K-pop star? I'm not saying that they're going to get treated like a K-pop star, but I think that they view Korea a certain way due to the media and their expectations. And then they get there. And it is kind of like that, but not really. And I think that whole disappointment is what also draws out some of these emotions. Yeah. I mean, some people say that same thing when they go to Paris.  And everybody's saying, love, and they just find cigarettes. I would say this, any country that has all three things banging, their economics, their culture, which is like their foods and their shows, as well as their looks, anything that they got all three of these cylinders booming is going to feel a type of way about themselves. Because I remember America. Oh, bro, in the 90s, it felt like America was the only country that had the economics, the cultural power, and the looks just booming, which is like the soft power. Somebody said, I used to live in Japan. And let me tell you, they truly think they are the elite superior Asians. So I don't know what she's talking about when she says Koreans think they are. Yeah, I definitely still think Japanese do, but they have such a culture of being polite and not telling you that. But a lot of people would say who've actually lived in Japan, they would say Japan is even more low key racist than all the other countries. Yeah, it's just not high key. I think because of the Omo Tenashi. I would have this analogy, Andrew. Let's say, for example, Asia was the Asian Student Union at a university. Oh, and by the way, guys, everything we're talking about is Asia-Asia right now, not Asian-American. But let's say, for example, Andrew, you had the Asian Student Union. Like the Koreans, they might show up and be extra proud and say some comments that like make people go, oh man, they think they're the best. But the Japanese student group might not even show up because they're like, we would not go to a Pan-Asian event, we are different. By the way, guys, of course, these are just stereotypes. Somebody said, I lived in all three Asian countries and Japan and Korea actually have the same hierarchical ranking. But China, I'll tell you, that place is just real mishmash, man, you don't know what they like. Hey, China does not care as much about looks. Let's be honest. Oh, China's more about money. Yeah, now I think on a media level, a lot of the media stars will kind of have a similar look in China. But I think when you go to China, you actually see that everybody has seen all different types of looks. Like there's people who look Southeast Asian. There's people who look very, very East Asian. There's people who look like, oh, you get the whole thing. Oh, there's people in China who look almost like more Western, you know, more towards the Middle East. I would say this, Korea, if you look at it historically, is one tribe of people. Japan is three. And China is literally made up of 200 tribes coalesced over time. So obviously that is going to lead to a larger spectrum. But interestingly enough, she said, Fong Viviam, that she was more accepted in China. But also a lot of people would not want to move to China because it doesn't have all three, you know, the looks, the culture, and the economics, banging like South Korea does. Somebody says, from me observing Koreans in Asia, I have to say they're the most vain and appearance-focused people I've been around.  By the way, guys, I think this is a absolute stereotype. I think that maybe if you party in Hongdae or you party in Itaewon or Gangnam, you might get more exposed to like that segment of the population. Well, like why does everybody like to come to K-town? Why does everybody want to party at the Pochas? Everybody like, look at the scoreboard. Look at what we put out. Like you want to come hang out with us. So like, we're like the cool kids, you know? Yeah, I would say that Koreans, and even Korean-Americans, but Koreans in Korea, they sort of have this popular kid in high school persona right now. Andrew, what do you think? If Koreans have that stereotype right now, Japanese have sort of the robot, polite, pervert stereotype. Chinese have the annoyingly cheap and bad manners as international tourist stereotypes. Vietts have the one that is gangsta. Filipinos have the entertainer stereotype. What do you think about all of these stereotypes, essentially that exist in Asia for every group? Yeah, I mean, they come from somewhere. So obviously there's more stereotypes to go around. I think each country would have 10 or five, at least major stereotypes about them. But yeah, I mean. I mean, these are just the stories that emerge from that country. I'll say this, I think a lot of it has to do with the exposure that the foreigners have to your group. Like I said, when foreigners go to South Korea, they're going to go to the party zones. When you go to the party zones, you're more likely to meet a certain slice of the population. So that population mentally in people's minds gets overrepresented. Somebody said, man, there's so many different rankings. For example, Chinese are the dirty East Asians when it comes to cleanliness, but then they might also become the next world superpower. So they could either feel bad about that or they could feel good about that. Man, it's so complicated, man. You can feel bad and good for so many different reasons, to be honest. Somebody said, no one is more racist toward Asians than other Asians. I don't really think that that is true. I really, I always disagree with that. Always disagree with that. You know what it is? I think that they know the most about each other. I always say this, Asians are kind of like different brands of Android phones where like, you know how the Androids are always like, oh no, Samsung is better than LG is better than OPPO was better than OnePlus. But then to iPhone users, which are like Western people, all the Androids, they're just all the same. Yeah, I think giving like racism a grade or a metric is always very hard. Measuring it is very hard because do you go by the most extreme possibilities in that country? Which I think actually America is a country of extremes. So I think we have extreme racism, but also on the other end, people are very open and rewarding to each other, right? So I think it's just more, a little bit compressed in the middle for Asia. Yeah, America is very roller coaster. It's up and it's down. Somebody said, why are all East Asians just so hierarchical and just whether it's within themselves as individuals or as in larger groupings? Why are they so hierarchical? Is it Confucianism? Is it Tiger Parenting or Tiger Societies? What is it? Yeah, I think it's all those things, man. I think it's achievement oriented. I think it's the whole test taking culture. I think it's Confucianism. It's everything, bro, everything. I think it's the national exams too. Just that national exam dynastic system is really like actually sort of perpetuated over the years but of course in more modern formats. Somebody said, man, I'm beard, but I respect Koreans more than my own people, man, because they do the right things. They have the good manners.  So why wouldn't they be cocky, man? It's their place. They build it up. I think the truth is, listen, if you're like a non-Korean and you're into Korean pop culture, like why do you like all the K dramas? Why do you like physical 100? Why do you like all these pop stars? Because they are good looking because it's highly curated because they are highly trained because they go through a crazy skincare regimen. Japan and Korea are a very high-powered economic system. You like the product, but you don't like the process. Or you don't like how you're getting treated once you enter that product. Yeah, and so because you have all these extractations and yada, yada, I'm not saying that there's not a lot of fun things about it but it also comes with a cost. So just so you know that. So you're saying you gotta be more, I guess, aware or realistic about what you're buying into possibly, right? What I'm saying is like if we do the work and we put out the product and everybody liked the product and then you come to the factory and you're mad at the factory, like that doesn't make any sense. Somebody said, I'm Korean and I noticed that since we are doing so good lately are all across our own technical metrics. There is a lot of backlash hate against us. It is because we leapfrogged a lot of other countries in like our proximity and our ability to create like culture that is very similar but even more different and similar to America. Do you think that's true? Because right now, Andrew, somebody said that there's a huge backlash because a lot of the K-pop sites are posting more negative news about Korea just to get clicks because that whole group of people whether you say positive or negative things is just like rabid and the click farms wanna get more engagement. Ah, that's really interesting. Ah, I guess, I mean it's kind of like the whole attitude of punching up, you know, anytime you get on or you get famous or everybody knows your stuff there's also gonna be, there's not only positive news about you. I mean, that's just what it is. And I think for a long time, if you talk to South Koreans like 10, 12 years ago they were like, oh, like everybody still thinks we're Chinese. I mean, this is even Korean-Americans. They were like, man, everybody still thinks we're Chinese. And then now everybody knows Koreans but now also everybody is gonna have a negative A as well as positive opinion about Korea. Right, because they just way more exposed getting way more pings. People thinking about it way more in positive and negative ways.  They run the world, right? But then America also is the coolest country. We have Hollywood. We have all the trends. We have a lot of innovation here. Somebody said, you gotta ask yourself, if your tribe was killing it as hard as Koreans are would you be any different? That kind of goes back to your like, if you did the work, would you be? I don't know. Yeah. Somebody said, why is she having a victim mentality? If she looked like a Vietnamese model she would not think like that. Because when people look like models all that like bigger stuff it doesn't even matter because they themselves is beautiful. Now I don't think that's essentially true. I think that even models are good looking people. It's not like they don't see the dynamics. They might feel differently about it. Maybe it doesn't affect them as much or maybe they even benefit from it in a different way a lot more. So it will skew, but it doesn't mean they don't see it. Yeah, that's a good point.  We're talking about Asian history. I'm talking about Japan. This was just a Vietnamese-American girl pointing out something that she noticed in her life. I'm not even saying it's true, but like true-ish or whatever. Like, and it just exploded. And it's still going viral right now. I think it's because it's partially true. It is partially true. And if you know any Koreans like to say that they don't talk about looks in a different way like I remember I've had conversations where I was like, whoa man these regular people are talking about looks in a whole different level than I was used to. You know? It's like everybody has a PhD in analyzing each other's faces. Now I think at the end of the day everybody can be very nice and respectful to each other. But yes, I mean that is a culture that is largely especially when you talk about things that come from the pop culture. They are focused on looks. And that's why their shows are so good partially because they're concerned with media. It's not that their stories aren't good or that the acting's not good, but they focus a lot on media and media is largely about looks. So that's what it is, you know? And I think that if you're gonna go there I mean you can say whatever you want about Korea but like I guess I'm not shocked at what she said. I don't, I think a lot of people disagree on the percentages and the extent of what it's true. Do you think they also disagree on why she pointed it out? Yeah I do, I think people disagree on her intent and the extent of how true it is but a lot of people agree that it is somewhat true. Anyway guys, let us know what you think in the comments section below. Like we said, always stereotypes, archetypes you know, percentages. These things are all real, right? For every, like the stereotypes are real. People believe in stereotypes. Let us know how true or untrue you think they are. We just went through the internet comments and this is going viral right now. So until next time, we're the Hop Hop Boys, we out. Peace.


elements and coming out this morning. So the next two lectures will be on special cases of the local Hamiltonian problem. So today I'm going to talk about the commuting case. And then tomorrow I will talk about stochastic Hamiltonians. And these classes sit sort of somewhere in between NP and QMA. So all right, so here's a plan for today.  And then most of the results that we know of rely on this structural lemma, which was introduced by Bravi and Viali in their initial paper on commuting Hamiltonians. So I'll state the structural lemma. And then we'll look at two different ways of applying it to show that certain special cases of commuting Hamiltonian lie in NP. And then depending on time, I'll sketch the proof of the structural lemma. So that's the game plan for today. All right, so let's define the problem. So the standard local Hamiltonian problem, we have Hamiltonian that's a sum of local terms. And the difference now is that every pair of terms commutes. So why does this make a difference? Well, if H is commuting, then all of the individual terms can be diagonalized in the same basis. So that means that every eigenstate of the whole Hamiltonian corresponds to a vector where the vector says for each term the eigenvalue of each term. So for every eigenstate of the whole Hamiltonian, it is also simultaneously an eigenstate of each individual term. So this kind of gives it a little bit more of a classical feel. The fact that I can sort of label every eigenstate in this sort of with this classical label. On the other hand, we have example, for example, the Torr code, which I think you've seen and in the later lectures, that show that commuting Hamiltonians can be the ground states can be highly entangled. So it sort of has a little bit more of a classical feel to it. But yet, the states themselves can still exhibit a lot of complex quantum properties. OK, so for the purposes of showing that things are NP, in NP, it actually suffices to consider the case where each individual term is a projector. So suppose I want to know, is there a state whose energy is less than some threshold T? Well, if I'm a classical prover, I can tell you the vector of eigenvalues corresponding to a state. And then I can give you that quantum state. And then as long as the sum of those eigenvalues is most T, and the state that I give you is, in fact, an eigenvector of each individual term with the corresponding eigenvalue, then yes. I've handed you a state whose energy is at most T. So the nice thing about this is that now I can actually give you state, but also give you the labels for each individual term. So if I tell you that information, well, each term is local. I mean, so it's a constant-sized thing. So I can manipulate it as I want. So it also means that there exists a vector of eigenvalues whose sum is at most the threshold T. And now I don't even have to consider the actual terms themselves. I can think of just projectors. So if I think of P a as the projector onto the lambda a eigenspace, I can just now take identity minus p. So what this pi sub a is, it's 0. Everywhere h a is 0. And everywhere h a is non-zero, pi a is 1. So it is a projector. And then if my state is in fact an eigenvector of each individual term with the corresponding to lambda, then if I consider a new instance using these projectors instead of these individual terms, the state is in the ground state, and it's frustration-free. So for the purposes of NP, we can sort of consider the frustration-free case. Yes. Frustration-free means that, good question. I should have defined that. So frustration-free means that each individual term is satisfied. So what it means is that the global state that I give you is simultaneously in the ground state of each individual term. The corresponding thing with Boolean satisfiability would be an assignment where you've satisfied all the clauses. So if we think of each individual term as a clause, there's the global assignment that satisfies everybody. Now, each individual term is instead of a generic Hermitian, it's a projector. So it's either going to be 0 or 1. And there's a frustration-free solution if there's a global state that hits the zero eigenspace of each individual term. OK? Yes. So I'm giving you a hint, though. So if I'm trying to show that a particular class of communing Hamiltonians is NP, I have to tell you what those lambdas are. So there has to be some classical hint. So if I just give you a generic instance of communing Hamiltonian and you want to know, say, is there a BQP algorithm for it, you don't know the lambdas. So you don't know what to look for. But if we're talking about an NP protocol, I can give you the lambdas as part of the witness. And then you can do your transformation and just reduce to a frustration-free. But it requires this additional bit of classical knowledge to know those. So when we're actually talking about most of the results we'll talk about today is that certain special classes of commuting local Hamiltonians sit inside NP. And so then I really only have to worry about the projecting case, because I can give you the lambdas along with the hint. Yes? Yes, you need to know which eigenspace you're trying to hit, all right? Yes. But it could be that the ground state hits a non-zero eigenspace of H. So it may be not possible to hit a zero eigenspace of everything. In which case I have to, for some of those terms, I have to hit a non-zero lambda. And I need to know which one I'm going for in order to know which one to project onto. I can diagonalize, but I don't necessarily know which of those eigenspaces the solution lies in, the ground space. Am I? So maybe it's easier to talk about in the classical sense. So let's say I ask you, is there a solution that satisfies everything except one clause? You need to know which one clause it is in order to verify, right? And yeah, OK? Yeah, so I think that the point I was just trying to make is that this reduces to the case where they're all projectors. But if I just have a generic instance of Hamiltonian, I don't necessarily know how to do that reduction, because I don't necessarily know which eigenspace of each term my term should hit. So I don't know exactly which A to use, or which lambda A to use when I do this transformation. In NP? Certainly not, yeah. So generically not, yes, yeah, yeah, yes. I think that's what I'm saying, yes, yeah. So most of the time, just regardless of whether we need the NP witness or not, I will mostly be talking about the case where they're projectors. Now whether you need a witness or not to do this, I think, yeah, yes, yes, yeah. So the debate, so the question in people's mind is whether we need the classical hint or not. Is that sort of the question at hand? So the way Bravi and Viali actually introduced the problem was sort of a common eigenspace problem. So I give you an instance of a local Hamiltonian. I give you a vector of target eigenvalues for each term. And I'm asking you, is there a global state such that every term corresponds is an eigenvector of that term with the corresponding eigenvalue? So the frustration-free question would be if that vector were all 0s, for example. So you want to know, is the space corresponding to this vector of lambdas empty or not? So that's sort of how they originally phrased the question. We're kind of used to speaking in terms of minimizing energy, so I phrased it this way. Yes? Right, right. Psi could be a highly entangled state, though. So if you're not a quantum verifier, I don't know exactly how you would do that. So yes, there's a vector, but the ground state itself, as in the ground state of the toric code, could be a highly entangled state. So I don't necessarily have a classical description of that state that I can give to the verifier. Yeah, but how do I? And I guess that was the original question of Bravi and Viali. Just because I'm telling you this vector of lambdas, doesn't mean that there's a state that corresponds to, like if I think of each of these projectors that's projecting on to the lambda eigenspace, if I apply them all, that could be trace zero operator, and there could be nothing in that state in that space. So just because I give you a vector of lambdas doesn't mean that the space is non-empty. And that was actually the original question that Bravi and Viali posed. They posed it exactly that way. I give you a vector, and you have to figure out is that space empty or not? So when we call, sometimes we'll refer to a solution, and the solution is now a frustration-free state. So a lot of the times we'll sort of bounce back and forth between these two. So why would we be interested in the commuting local Hamiltonian problem? Well, as we've kind of talked about, it has a little bit more of a classical feel to it. So it sits in between the quantum and the classical world. It's not quite constraint satisfaction. It's not quite the full CMA. But it does have sort of a little bit more of a classical feel in that I can label every state with the corresponding eigenvalue. And as we talked about, the eigenstate can be sort of labeled, but it still can be a highly entangled state. Stabilizer codes are ground states of commuting local Hamiltonians, so they're sort of of special interest for that reason. It's also a nice test ground when you're working with proving different claims. So it's been an interesting test ground with sort of playing with different claims around the quantum PCP conjecture. And that is also a kind of nice, easier case to in the study of gapped Hamiltonians. So for example, there's been a lot of work on area laws in 2D asking whether gapped Hamiltonians have efficient descriptions as tensor networks. Well, and this is actually quite challenging to do. There's been a lot of interesting work and it's still unresolved formally, whether it's the case that every ground state of a gapped Hamiltonian has a tensor network representation, but it is true fairly handily for the commuting case. So it gives sort of a nice test bed to sort of examine other questions. So it's been useful for that regard. Okay, so I'm gonna state, when Bravi Vialli first introduced the problem, they developed this beautiful theory in this beautiful lemma called the structural lemma, which I'm gonna state. And then basically all of the work since that time has sort of made use of this in some way. So I'm gonna state the lemma, I wanna make sure we all understand it, show a couple different applications of it, and then if we have time sketch the proof, okay? So suppose I have two operators, they're commuting. A acts on particle x and particle y, B acts on particle y and particle z and they commute. Then I can take the Hilbert space corresponding to particle y and divide it up as a direct sum of different subspaces, okay? So that a and b are both invariant on that space. So when I operate on a, it doesn't move me if I were to start with something in one of these subspaces, apply a, I'd end up in that same subspace. And if I now isolate a and b to one of these subspaces, there's a nice tensor product structure. I can take that subspace and express it as a tensor product where a acts only on one part and b acts only on the other, okay? So let's just kinda, I'm gonna say it again a different way and then you can tell me if you have questions about this. Let's look at the first one. So I can take my Hilbert space for particle y and express it as a direct sum of different subspaces and if I look at a and b, it's gonna be block diagonal where every block corresponds to one of these subspaces. Okay, so another way of saying that is I can express a as the sum where I'm projecting onto a subspace and if I sum it up, so one, this one, does that work? Oh wow, that's really loud, okay. All right, so if I look at one of these terms on the sum, that corresponds to one of these blocks, okay? And this is true also of b. So what that means is if a solution exists to my problem, then it's gonna exist entirely within one of these subspaces, okay? So, you know, a flash forward, you can imagine that one of the hints that the adversary might, or the prover might provide is which subspace it lies in, for example, okay? And then if I look at, if I take one of these slices and I look at a and b, now restricted to that slice, I get this nice tensor product structure, meaning that a, restricted to this subspace is acting on part of the subspace and b, restricted to that subspace is acting on a different part and they're completely disentangled. All right, so questions about the structural lemma and what that means, yes. It's proven, I don't know if it is, I don't think so. It's sort of based on a theory of C star algebras and I'll sort of say a little bit more about that at the end if we have time, all right? Okay, I expressed it as a two term here but it actually works for multiple terms. So if I have multiple terms, then I can still divide up y into a direct sum where all three, I put three here to just to keep it simple, our invariant and within each slice, all of the terms simultaneously act on different parts of that slice, okay? So let's look at how we can use this. So right off the bat, one of the original results in the Bravaviali paper was that two local Hamiltonian is an NP, okay? And this is on an arbitrary dimensional particle but just as long as the terms are two local. And so the NP witness consists of which slice of each particle the solution lies in. And then once you know the slice, you have this complete tensor product structure. So I drew some pictures. So here's the set of particles and here's how they're divided up by the two local terms. The witness says for each particle which slice the solution lies in. And now if I look at all of the terms restricted to those slices, I get this nice tensor product structure. So the actual, the operators are operating on completely different parts of that space. And in particular, it tells me that the ground state is can also be expressed in this tensor product. So in this case, I don't have highly entangled ground states. So if the ground space of this term is gonna be within this part of the Hilbert space and similar. So the ground state itself can be expressed as a tensor product where the tensor product, each term in the tensor product lies inside these little subspaces that span a little piece of each particle. Okay, that's sort of proof by picture but any questions about how that works? Yes, let me. So once I tell you the slice, okay, then this thing here is just a constant size object. It's a constant size Hilbert space. So if I look at this term restricted to that space, I can then just compute exactly how that tensor product gets teased out. So what I can do if I take, if I know what slice to look at and I take this operator and I restrict it to that slice, I know exactly what portion of this Hilbert space it acts on. So once you know where the slices are, everything else is polynomial time computable because these Hilbert spaces are just constant size. Yes. So the slices, the slice is one, ooh, the slice is one of these guys. So I'm taking the Hilbert space and I'm expressing it as a direct sum and that's pictorially, that's this. That's this sort of sum of spaces. And then within a slice, I have this tensor product structure. So if I tell you that the solution lies here and here and here and here, now I can take all the operators and just restrict them to this subspace. I can compute exactly what they are. And in this case, the operators are completely in tensor product. They're operating on completely different parts of this subspace. Yes. Yes, because they're constant size things. You can do basically anything with a constant size particle, yes. The prover tells you where the solution is. If there's a solution, it has to be in some sequence of slices and then the prover tells you which ones. And that's also classically describable because every particle, again, is of constant size. Yes. Right, yes. So I can do this here and then I'm all simultaneously just gonna be doing this for particle W where anything that acts on W is also in this sort of tensor product structure. So simultaneously that works. Yeah. They can all act on it. The slice might be a higher dimensional slice. So you can imagine one of my slices is a projector on to say an eight dimensional subspace. They respect this structure. So they're all invariant on each slice. But they could all be sort of acting on it in a non-trivial way. Yeah. So for example, if one of the slices had dimension eight, one term could, I could think of that eight dimensional subspace as a product of three qubits. One term could act on one of the qubits, the other on the second and the other on the third. All right? Okay. So this is sort of a sketch of why two local. So then, yes. Yeah. So the slices are just the values, the classical standard basis states. Okay. And then I'm just telling you where the solution is. All right. So I'm either the identity on that or I'm, yeah. Okay. So now we'll move to a slightly different application and it's a more sophisticated application of the structural lemma. Yeah. Yeah. This one. Right. So if Y is a qubit, then there's, we'll look at that case, but there's really only two ways to slice it up, right? I can think of a one-dimensional space and a one-dimensional space or everything in a two-dimensional space. And that's all you got. It could be act as the identity or it could sort of be an X term. So, but certainly if this is a qubit and it happens to be that the slicing up of this qubit is a two-dimensional space, then only one of these guys is gonna act non-travely on that qubit. So if this particle is a qubit and it just so happens that the way the, if these are all sort of operating on it as an X operator, then they're all respecting this one-one decomposition. Then the decomposition would be plus and minus, okay? But if they're acting, if they commute and they act differently on it and say that the decomposition is just a full two-dimensional space, only one of these can act non-travely on that qubit and the rest are identity on that qubit. So the qubit is a very special structure that there's only so many ways you can slice up a qubit. Yes. So that's an important part of the whole thing because otherwise that's the only thing that tells me that I can isolate to a particular subspace of that particle, okay? Great, any more questions? Okay, so now we're gonna look at a different application. Yes. I think you could actually compute what the decomposition is, I think given the terms and given the particle. The thing that you don't know is which slice you're in. Yeah, all right. Okay, so now let's look at a different example and we're looking at commuting local Hamiltonians on a 2D lattice, okay? And even with qubits, we get non-trivial structure here because we have this, the Torr code is a special case of this, okay? So for my purposes, I'm gonna think of the particle, so this is sort of the non-standard way to think of grids, but as sitting on the vertices of a grid and I have, and every face is a term, okay? Typically when you talk about the Torr code, you think about the qubit sitting on the edges and you have star terms and pluckhead terms. In the case of the 2D grid, it's actually equivalent because I can sort of rotate my grid and see it a different way, okay? But for my purposes, I'm gonna look at this particular way of discussing 2D Hamiltonians on a grid. Yes, for the Torr code, but we're gonna be looking at generic commuting local Hamiltonians, okay? So the ground state of these Hamiltonians already are gonna be much more entangled potentially than the case that the two local case. So we saw in the two local case that if there's a ground state, it has this very clean tensor product structure and there's not global entanglement in it. But in this case, there's no way to avoid entanglement because we have the Torr code as a special case. So for the Torr code, you can think of this as a checkerboard pattern. The red terms are the X terms and the blue terms are the Z terms, okay? And what we're gonna do is consider general commuting for local Hamiltonians on a 2D QT. So not just the Torr code, but any sort of commuting terms. So I will think about checkerboard patterns. I will consider the blue faces separately than the red faces. And I'm gonna sketch a proof by Norbert Schuch that this special case of commuting on the local Hamiltonian sits inside NP. And you know, okay, so here's my Hamiltonian. It's a sum of terms. PI is gonna be a projector onto the ground space of HI, yes. Not necessarily, it's not even translationally invariant. It's just arbitrary, it's just commuting. Yes, to kinda analyze it. If I look at the blue faces that has sort of an easier to analyze structure because the terms only overlap on a qubit and same with the red, but then we ultimately have to put it together but there's not necessarily translation variants in this picture. I'm asking, is there a frustration-free solution to this? Okay, so I will call that a solution, okay? Okay, so this is the projector onto the ground space and each term I can think of as a projector onto the non-ground space, onto the higher energy spaces and I can divide up the faces into blue faces and red faces. The, this PB is a product of the projectors onto the blue faces. This PR is a projector onto the ground spaces for the red faces. And I wanna know is if I apply both of these projectors is the space non-empty? So in other words, I wanna know is the trace of the product of these projectors greater than zero? Okay, so that's the name of the game and this is gonna be a non-constructive solution. So I'm just gonna provide evidence that the trace of this operator is greater than zero. You may not learn a lot about what the solution looks like in the process, okay? Okay, so the blue faces, the key point here is that they overlap only on a single qubit.  These PI and PJ are invariant on each of these subspaces. So if I, and I'm gonna think of P, Q is labeling the qubit and alpha is labeling the slice. If I specify which alpha, which slice I'm talking about, I can now, and I think of, this is, yeah, so if PI then is gonna be, this is just basically saying that PI is invariant on each slice, okay? And same for the PJs. Okay, so now I can imagine a vector of indices for each qubit and this again, we're just talking about the blue faces now and I'm gonna tell you for each qubit which slice the solution lies in, okay? Now for a given face, I can apply that projector to all four qubits and for a particular vector of slices, I can imagine this face restricted to each slice of the four qubits. Now I've expressed it as like a product of all of these Ks. So for all of the qubits, but really when I'm looking at face PI, I really only care about qubits one, two, three and four that are relevant to me. So what I'm doing is I'm taking PI and I'm restricting it for each of those qubits to a particular slice of each qubit. So for a vector of, I'm giving you now this vector is telling you for each qubit which solution, which slice the solution lies in, I can think of the blue faces as being completely restricted to those subspaces. And now the blue projector, the one I started out with is the sum of all vectors of the blue faces restricted to that vector. And I'm gonna try and think of a better way to say it. So what I'm doing is I'm summing up, let's say for example, the qubits get sliced up as the plus and minus as the subspaces. This is the vector of all possible ways to label each qubit with plus and minus. And for a particular vector, I can imagine restricting all of the blue projectors to those that set of slices. And then I'm adding up over all the vectors. And the same will hold for the red terms. Although it's gonna be potentially a different way of slicing up the qubit just like the core code. So with the Torah code, the blue faces are X, I'm slicing it up into X plus and minus. And if the reds slices are Zs, I'm slicing it up according to zero and one. So these different ways of slicing up the qubit are not necessarily compatible with each other. Okay, so the NP prover is gonna give two vectors now. One a vector for the blue faces and one a vector for the red faces. And we just wanna verify that when I restrict the blue faces to alpha and I restrict the red faces to beta, the trace is greater than or equal to zero. Okay, so it's actually instructive to look at what this means for the Torah code. So the witness doesn't say much about the ground state itself. So let's see what this would look like for the Torah code. The blue terms are XX, so I'm dividing things up according to plus and minus. And PI and PJ are certainly invariant over this way of dividing things up. And the alpha tells you for a qubit, particularly a qubit, whether you're in plus or minus. Similarly for the red terms, but now beta is telling you zero or one, okay? And all I have to do is give you one vector of alphas and one vector of betas, so that if I restrict the blue terms to the alphas and I restrict the red terms to the betas, the resulting trace is greater than zero, okay? So the vector for the alphas is gonna be a string of plus or minuses. The vector for the betas is gonna be a string of zeros or ones.  And if I restrict this X term to this particular projector, I just get all pluses. It's just projecting to the all plus state. And if I restrict the red terms to projecting onto the zero subspace of each phase, I just get a vector of zero terms. And the only thing I need to verify is that the product of these two projectors, the trace is greater than zero, okay? And sure enough, the trace is to the minus end, which is strictly greater than zero. This doesn't give you a lot of insight as to what the structure of the ground state looks like. It's just a proof that the ground space is non-empty. Yeah. In this case, they are. Yes, well, yeah, not in general. So we have to get a little bit more sophisticated when we look at, yeah. And they are tensor products in this case because each of these terms slice up the qubit in a one, one way, one space, one space, okay? But that might not necessarily be the case. Okay, so like we've talked about, qubits are fairly simple. There's only two ways to slice them up. I could slice it up in a one, one way in which case the Hilbert space of that little qubit is just a sum of two different subspaces, one-dimensional subspaces. Or I could split it up in a two way in which, but in that case, only one of these terms acts non-trivial because remember within that subspace I'm gonna, one's gonna, they have to have a tensor product structure. The only way to have a tensor product structure within a two-dimensional space is if one is the identity, okay? So this gives me a lot of structure to analyze with even for generic commuting local Hamiltonians. So, and I claim, I don't know how much of this, is that if either one of, so P one prime and P two prime slice up that qubit one way. P one and P two slice up the qubit another way. And I claim if either one commutes, slices it up in a one, one way, then we can basically trace out that qubit because I'm gonna have, the adversary is gonna give me a one-dimensional projector for one of those terms and that's just basically gonna end up saying that the solution has to lie in that subspace and I can basically trace it out. I have a little bit more of an elaborate proof but I'm just gonna leave it at that, skip over so that I can say a few words about the structural lemma at the end. So, but it's not too hard to show that if either one of these pairs commutes in a one, one way, we can basically trace out that qubit. I've got a 1D projector in my trace and it's gonna get traced out. So let me skip over this. So after I've traced out all of those qubits that I'm only left with situations where both pairs commute in a two, two way. In a two, in a two way. All right, but that tells me something about it. So if P one and P two commute in a two way, one of them has to be the identity on that qubit. If the other pair projects, divides the qubit space up into a two way, then one of P one, primer P two also has to act non-trivially, okay? So I can now analyze things a little bit more. I've traced out all the one dimensional spaces and what I'm gonna do is for each face, I'm gonna put a dot in the corner if that term acts non-trivially on the qubit and I'm gonna put an X in the corner if it's the identity on that qubit, okay? We know that certain structures can exist and I'm gonna say that two terms overlap if they both act non-trivially on the same qubit, okay? And the key insight now is that overlapping terms form chains and I'm just gonna sort of give a little bit of a hint as to how you would prove that. So basically I can't have structures like this where I have this term overlapping with that term, overlapping with that term, overlapping with that term which gives me sort of a branching structure. So in the end when I look at terms that overlap I get these chains of terms and then I can use sort of a tensor product analysis to determine what the trace is. And the idea is that certain structures are disallowed in this case. So for example, I can't have diagonal dots here. If P one and P two commute in a two way they can't both act non-trivially on that qubit. If I have a dot here then it has to be an X up here, okay? Another forbidden structure that I have is the following. So if I look at these two terms they overlap on this qubit and this qubit, okay? If the term above acts non-trivially on the qubit on the left, okay? Then the only place that these two terms overlap in a meaningful way is on this single qubit. I can apply the structural lemma again and one of them has to be the identity. So basically I can't have this, I'm using up the batteries on everything right now. So I've used up the batteries on this one too. So basically I can't have this structure. Yeah, what I'm doing is I'm arguing that if, so I'm taking all of these terms and I'm saying which neighboring terms overlap on a single qubit, okay? So those ones I have to sort of think about and disentangle in some way. And I'm arguing that I can't have a really complicated structure of this overlapping pattern. So that if I look at terms that overlap you can think of it like a graph, like I have this grid and I'm putting an edge between two neighboring squares where they overlap. But it says I can only have chains and cycles, okay? And this is sort of a sketch of why that might be the case. And basically it says I can't have branching structures in this. So if I look at sort of the chain of different grid terms that overlap, they can't sort of branch out in two different directions. And it's because certain patterns of overlapping are disallowed and it's a little bit of a case analysis to see exactly why that's the case. But in the end what I end up with, now I have to go old school here and is this type of thing, okay? And that's actually easier to analyze because I just have a chain of things. I won't go into it in too detail, but I can imagine expressing each of these projectors as a tensor and taking the contraction of the tensor. All right, question. Yeah, so I can then sort of construct this sort of graph of overlapping terms and then I just need to look at every chain independently and that I can do using sort of tensor product techniques to sort of contract a chain. It basically has a matrix product type structure to it. Yeah, you mean the two squares on top of each other there. So you think about what's happening there on the qubit on the left, the top one is acting as the identity. So the only place where they overlap is the qubit on the right, okay? If they only overlap on the qubit on the right, they have to commute in a two-two way. I guess if there are one, one, then I would have the same thing. I could sort of trace it out. Yes, yeah, okay. Somehow I missed my slide. I mean I could, one thing I wanna do, I somehow missed the slide which has sort of the survey of kind of everything that we know about commuting local Hamiltonian and I wanna, unless I erased it somehow, there it is. This is what we know so far. So I think it's useful to sort of get the lay of the land. I don't think I walked through this slide yet. So two local, so basically for commuting local Hamiltonian we know that a bunch of special cases lie in NP. But the rest is kind of wide open, okay? So the cases that we know that lie in NP is two local, three local on qubits and Q-trits for certain nearly Euclidean. In 2D for qubits, Schuch, we just sketched his proof. This second paper actually gives a constructive solution. So meaning that the adversary gives you a way of constructing the ground state using a circuit. Recently we extended Norbert's result to be Q-trits. So, and then Bravi Viali actually showed some special cases for factorized commuting local Hamiltonians. And factorized means that each term is the product of operators on individual particles, okay? The general commuting local Hamiltonian is completely up for grabs. We don't know if it's an NP, it's in QCMA or QMA heart. So that still remains an open question. Yes. No, that's actually any locality. As long as it's factorized and they're qubits, yes. So that's actually like non-local Hamiltonian, yeah. Yes, so the follow on work that we did was factorized in 2D, which limits the locality but it's for arbitrary dimensional particles, yeah. Yeah, question? You still need quantum because the ground state could be entangled. So yeah, so QMA one, yeah. Yeah. No, but okay, now I'm confused. So if I give you the eigenvalues, I mean you can verify with probably one that it's frustration free, right? That's a good question, yeah. What are you saying? Commuting local Hamiltonian, the frustration free version. I can tell you what I can space everybody lies in and then I can give you this QCMA. I still have to give me the state. The state could be highly entangled and in order to verify it, I need a quantum state in order to be able to do that, yeah. Okay, okay. Yeah, and we don't know for these ground states whether they can be their quantum circuits or whether they can construct them or not. Yeah. Oh, QMA one was the question, yeah. Yeah, yeah, yeah, yeah, yeah. I don't know how much time I have left. Should I do a little sketch of the structural lemma? Let's see if I can just give a hint at, this was a lot of slides, okay, here. Okay, so it rests on a theory of C star algebras. For our purposes, you can express these very abstractly. For our purposes, a C star algebra is a subset of linear maps on a Hilbert space closed under addition, multiplication, complex conjugate, and contains the identity in scalar multiplication, okay. And the center of a C star algebra is the set of all things in the set that commute with everything inside it, okay. And this was sort of a known fact about C star algebras before. If I have a C star algebra and it has a trivial center, meaning that the only thing that commutes with everything in my set is the identity, then A can be expressed as the set of all linear operators on a portion of the space tensored with the identity on everything else, okay. And that's sort of where this structure is gonna come from, okay. And so they show the following lemma that if I have some operator in the center of my C star algebra and it's non-trivial, meaning that I can express it as the sum of projectors onto different portions of the space with different eigenvalues, so this is sort of, I have some non-trivial expression of projecting onto different portions of the space with different eigenvalues, then everything in the algebra is invariant under those projectors, okay. And the way to see that, I need to show that the projectors themselves are inside the center. And so the way to see that is if M is inside the center, then any polynomial of N is inside the center because it's closed under addition and multiplication. So what I'm gonna do is pick a polynomial where lambda I is equal to one and all the other lambdas are equal to zero. So if I apply that polynomial to M, I get exactly that projector. So if the projectors inside the center, everybody commutes with it, they're all closed on that part of the subspace. So the name of the game then is basically to keep dividing up the space. So I can basically take something inside the center that's non-trivial, that's non-identity, use it to divide up the space and then look at each subspace independently. Does it have a trivial center? If not, repeat. So this is giving me a way of dividing up the space so that everything in my C star algebra is invariant on this partition and I eventually end up with something where if I restrict my attention to each subspace, it has a trivial center. And then we can apply that lemma with that nice tensor product structure. So we end up with some partition of the space. Every operator in the algebra is invariant on each subspace and each subspace has a trivial center if I look at the restriction to that subspace. And so therefore it can be sort of expressed as all linear operators on part of the space and the identity. Now if I look back to this picture of commuting terms, I can express A as sort of a linear combination of independent operators on Y. So I just can basically, I'm taking A and I'm expressing it in this block structure, okay? So I'm expressing it as a sum of terms. I can do the same for B. And because AX is the identity on Z and BX is the identity on X, if I look at the corresponding C star algebras, they have to commute. That's the only way all of those terms can commute with each other, okay? And the idea is that X is not gonna help them commute and Z is not gonna help them commute. They have to commute because their actual terms are commuting. And now if A, if these two algebras commute, then I can use A to divide up the space as we just talked about. So I can use A to divide up the space into this product, this direct sum picture. A is invariant on each subspace. And if I look at A restricted to the subspace, it's basically the set of all linear operators tensored with the identity. And then if B commutes with A, then B also has to respect that division of the subspace. That's the only way that B can commute with all of A. So B is invariant on each Y. And if it commutes with A, then it has to operate on the part of the Hilbert space that A doesn't operate on. A is all linear operators on part of the space. Therefore, B has to operate on the other part of the space. And that's where you basically get that structure from. So this was a whirlwind sketch of that proof, but I don't know if I can answer any questions about it. Yes. I think you need the C star algebra to give you this structure that if you have a trivial center, then you're essentially the set of all linear operators on part of the space. I think you need that additional sort of closure in order to get that you get all of the linear operators on a portion of the space. So you're saying, why do we need the closure under, say, like complex conjugate or, I think, yeah, so I think in order to get that sort of, the original thing that I quoted is that if you have a trivial center, if you don't commute with anything, then you're essentially the set of all linear operators. In order to get that, you need to be closed under all of those different operations. All right. Thank you.