WEBVTT

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Yeah, thank you for hanging behind here. or he's correct.

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Tell us today about aspects of locality, unitarity and symmetries.

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We're going to be taking Ibrahima to dinner at Chef Tanz today. We'll be leaving early so he can make his train back.

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So anybody who wants to come, please come see me after the talk.

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Thank you for… nice introduction. It's always fun to come to Rutgers and talk to you guys. So this is a paper that came out end of last year, but it's part of.

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long time discussion that I've had with Shlomo Revimat, his student is a former postdoc of mine, but this is continuing in various directions. So this talk will be a little bit.

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unconventional from the usual way we at least think about symmetries in the modern world today. And the point is, I wanted to sort of.

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step back a little bit to try to understand the following.

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issue, which is that, as we all learn in quantum mechanics.

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very early on. We have Wigner theorem.

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which tells us that symmetry operators are either unitary.

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or anti-unitry. Now, if you have either an anti.

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Now, Unit 3 operators. 4.

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and the unitary symmetry operators. and that is it. Okay. So from this point of view, there's a rather interesting question in the modern day when we talk about QFT, we expect to get a full high effusion category to describe symmetries, which is a very exciting topic.

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that is near and dear to my heart. But from this, from the point of view of Wigner's theorem.

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Um, there is a bit of a clash, which is that when we think about higher categorical symmetries, we typically get symmetry operators that are non that are non-invertible.

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have an apparently seeming to be non-unitary and not compatible with this statement of Wigner.

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So to to say this more sharply. In quantum mechanics, of course, if I have a system and I have some symmetry, it commutes with a Hamiltonian, and we'll stick to just unitary symmetry operators. It can act, and the system could be in any representation of such a group.

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For example. and then I can imagine building a quantum field theory as taking different quantum systems, let's say, on a lattice, for example. And taking a tensor product and taking some continuum limit. Okay?

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So from this point of view, quantum field theory is strictly a quantum system in some limit.

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Okay. So from that point of view, Wigner's theorem should still apply that the symmetries of my system should be governed by unitary operators. But in the modern day, when we think about generalized symmetries in various contexts, what we see the phenomena of non-invertibility is ubiquitous.

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So what? What give? What? What is the missing of the 2?

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And I think. The heart of the issue, at least from my analysis, it will come come thin because of locality.

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Okay? So let me try to explain a little bit more. What do I mean by this point?

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So if I start with any. Let me take some system.

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let's say a ladder system, and I have. quantum systems that are defined in some cells of this time.

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Okay? Each one of these cells, I have a well defined quantum systems where I expect Wigner's theorem to apply in the usual sense.

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However, when I go to the full cell and try to before I take any continuum, I'm expected to take some tensor product of the symmetry.

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along the line, along the line. The key question you could ask is that if I want to think about the symmetries acting on the Hilbert space coming from the tensor product.

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Whether such asymmetry, such an operator that you would construct that commute with that Hamiltonian, whether they would necessarily decompose into a tensor product of something acting locally.

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Okay. So as soon as you extend. to some to a space or to a lattice, locality comes in and there's a natural question, in addition to Wigner's theorem, you have to ask the question of locality.

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Okay? And immediately you should run into the conclusion that a generic operator that commutes with the Hamiltonian, a priori have no reason to act locally. Okay?

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And I think this is at the heart of the issue of.

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Why, when we talk about non-inverted non-invertible symmetry versus unitary symmetry acting.

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Implicit to that, there is some choice being made, and the choice that is sort of implicitly being made is you prioritizing locality versus.

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unitarity, okay? And this talk is going to be about that. And how do you see it in some sharp way? And what are the things you can actually compute? But of course, we can say all of these words, um… But they become meaningful if there is something we can compute in some reasonable way. So this talk is about how to define some observable that can tell you.

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about the… that can measure the distinction of unitarity versus locality.

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Okay. Any questions?

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So to that point, first, st we should come back to this picture here.

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and we should ask, suppose I have some group G.

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with elements G alpha. acting in this way. And whenever we do this decomposition, there is an implicit assumption that the Hilbert space of my system decomposes into a tensor product.

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of some local Hilbert spaces, HI. Okay? And there are N cells. We'll use this n.

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So if you have some group element G, some symmetry that is acting. Is the size of this whole. Yes.

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The fight. I will label fight. Okay, so I will.

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For the time being, assume a very strict version of locality. There is various loosening we can do, and there are some questions you can ask about gauge theory, the theory of fermions. But let's pause for a moment.

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And that has to do with when we can expect this factorization. Okay? But let's assume this very strict version where I have where locality means that I can decompose my physical system into cells.

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And the five of the cells, we typically take it to be larger than the coherence length of the system.

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Okay. So we configure some group element G. And what we typically want to ask, we have rho.

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of G alpha, which is the representation over the whole fell. Okay? And.

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In general, this row of G alpha that's acting on the full Hilbert space may or may not respect this factorization.

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Okay. So there are 2 things that can happen.

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One, you can have u alpha. factoriz, meaning u alpha can be expressed also as a densor product.

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of you. alpha i where you alpha i is a representation of the of the group in each cell.

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Okay. Or 2. You can have that you alpha does not factor out.

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Okay, so these are the 2 possibilities. So you could. So then we'll focus on this one and see, does this buy me anything interesting for this system?

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Okay. So this is where what what we want to explore. So in this rather strict version. Let's suppose that each cell is in some representation RI.

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of this, then what I would like to compute is going to be.

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The trace. of the representation rho of G alpha.

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weighted with the phrase. of the identity operator.

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the identity of the system. So does this quantity have rather nice properties here.

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So I can split this identity to act on every single cell, and this quantity I can just simply rewrite as a product.

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over I of rho. I, which I should write as.

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trace of. you alpha i.

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over trade of the identity element. I will just write. Yeah.

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That's fine. Right? So this quantity, of course, you're assuming you're you're in the case. One. Yeah, I'm I'm in case one, which is the case that I'm going to talk about. And this, of course, is simply going to be.

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the product of. Of the characters.

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in the R representation of G Alpha over the dimension of the representation.

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This is your thing.

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The iceber space is the representation. Yes. Okay. What this? What I will say will actually not depend on the choice of R in a moment, as you will see.

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This is slightly more detail.

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Okay. So the question we want to think about here is that now, let me take this quantity.

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by I I of Ri. of the group element G over the dimension of the representation. And then what I want to do is I want to know what happens as I take n to infinity here. I can take n to infinity. So this is taking.

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some continuum limit where I have a large number of these states coming together.

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So if this group acts faithfully, for example, we have the following fact, which is that.

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IFRI. for all for the non-trivial element. G. It's strictly less than.

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dimension of the representation. So this thing then is going to go to 0.

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actually take n to infinity. Great. So there is something interesting that happens, which is that if the representation, if the symmetry operator is locally acting, meaning it factorizes in this way, you immediately observe that the character of the representation.

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for the full Hilbert space vanishes. Well, that depends on your value conditions for how G alpha depends on the U alpha i depend as you go. But then that would violate the locality constraint.

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Well, I mean, you cook with boundary conditions that they all do the warm.

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Yeah, but boundary conditions are. you know, we assume everything falls off at infinity. Let's see. What does it mean?

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The UI. Control David.

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So if you have to say something about what you're going to assume.

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about the U alpha i's as the sites i go out to infinity.

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So right. So the the when I say here, I I can still take this to be finite, right? Which is that I'm working. So this is this is not an issue. And when I'm taking n to infinity, I'm just partitioning my my space finer and finer.

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Oh, I see. Okay, so that's one way you can think about it.

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So there is a… there are 2 limits you can take. There's a large end limit and the large volume limit. In fact, when we do this properly in a field theory, this is a question that we have to address in some suitable way.

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But for this example, this is the strictest version of this picture that I could say, and there's already something to learn here.

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Okay. So so what? What is the meaning of the fact that this is happening? The way you could? Did you assume representation was irreducible?

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Uh, yes. Yeah, but that point itself is not going to be super important. The key thing that's important is whether it's faithful or not. If it contains the identity representation.

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Then there's one term that doesn't go to zero.

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Sorry, sorry, sorry. So G alpha is non-trivial. So let me get to this point, and I think I will when I write the next thing, it will. It will make you happy.

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Okay. Very good. So to you can try to be more precise with this. So this quantity.

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This trace here, I'm going to call it C alpha, which is the trace of the representation. I can write it.

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in this way as a, I can try to decompose it into.

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1 over, let's D. Let me should I do this?

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I. For the sake of argument, I will take all representations at each cell to be the same.

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So that's not going to have any bearing of the argument. And non-trivial. The only difference is that.

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It's just going to change the details of the limit that I want to take. So I'll just take all representation to be the same, but I can try to compute.

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decompose directly this quantity, since it's a character of the representation into characters.

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of, uh… for each word A here.

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is labeling reps. Okay, so I can try to compute this.

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Because that's just the club for the decomposition of R to the end. More or less. Yes. So I can compute it explicitly. So I want to do this because it's instructive.

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This is going to be. I can compute this from the character products.

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the alpha. I mean, look at G Alpha.

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I r raised to the nth power times G Alpha and compute this exactly.

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So in computing this, I can pull out the identity element, which is what your question was about. But this, when I pull out the identity element, I can write this in this way, dr to the n.

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over G times the dimension of the representation that I'm looking at.

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buff a thumb. of G Alpha, not the identity.

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mindful character. of G alpha times.

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character representation raised to the R. of G alpha with the normal evation of dr.

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VR and everything raised to inspire. It looks like you're assuming GS finite. Well, you could probably just generalize to compact rooms. Yes, I have answers for compact groups, where you can compute it. So now, if you stare at this formula with this statement here.

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you immediately see that this terms. the cave fast. Okay? So what you obtain that this goes to simply dr. Of n.

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where the order of the group times the a.

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which then tells you this quantity. that I'm actually looking at at the very particular properties that in this large n limit, this quantity is simply 1 over G.

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times the sum of. DAIA.

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of G Alpha. Okay? So what this seems to say that when I take this.

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large end limit, which is a continuum limit, it seems that the Hilder space of my system is just many copies or infinite copies, how many.

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copy the factor of dr to the n of the regular representation.

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which is a rather interesting fact. But if I, even though I can start with individual quantum systems.

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And I think this larger sense with where each system could be in any representation that it wish to be.

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In a continuum limit, it seems that if the symmetry is acting locally, meaning it is factorized, it factorizes in this way.

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Right? Then it followed that the Hilbert space of the system is going to be some many copies of the regular representation. I don't see a word. Oh, I see. You're saying the Hilbert space factorizes as an overall Hilbert space times one copy of the regular representation.

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I just do the overall. That's the character in the regular representation. I don't see many copies of it. Oh, because.

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Because there is the… the normalization that I took.

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I started off with some normalization here. Right? So the more proper way to say it, the more proper way to say it is that this computation you can say is that in if I take the ratio of some representation, MA.

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Ma prime, and I take the large N. right? This goes to the ratio of the dimensions of the representation. I'm still confused. If I had n copies of the regular representation, I would have the thing that you call C of alpha. So the nth power.

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It's looking more to me like you have a neutral Hilbert space tensor one copy of the regular representation. No, no, no, no. This. So if so here I have normalized. So so let and and I picked a very particular normalization.

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So I can remove this normalization and drop it. I can still write this formula, but this formula is going to be formally infinite.

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It would be large, because there would be a factor of D of R that's coming in. So the the correct statement to say is you take a ratio of 2 representations of the multiplicity of 2 representations, and ask, what does that go to at large N?

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So in the large end, this goes to the ratio of their dimensions. It's not that you have one copy of the regular representation times something. It is that the whole spectrum that you have. If I ask.

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What is the ratio of one representation to the other is always going to be DA over DA prime.

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It's an infinite dimensional Hilbert space. But this quantity is meaningful.

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which is similar. It intensity of the regular representation.

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And the other is the female power of the regular representation.

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That's what you just presented looks to me more like the former principle.

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Maybe I'm misunderstanding here. Move on. Maybe I'll maybe I would be confused myself.

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because I just took the trace. I just computed the character of the representation of the Hilbert space. That's all I've computed right? And I'd say that that character is proportional to the character of the regular representation. Yes, exactly. So the ratio of any to the ratio of the multiplicity between any 2 representation.

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is that of the regular application. Totally agree. But that's also true. Take L 2 of R tends for the regular representation. Where L 2 of R is a singlet under G.

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Ah, ah, okay, so what you're asking, can I have just a bunch of singlets around? That's different from taking the n-tensor power of the regular representation. That is completely correct. But your statement there, what you're saying is that I have a factor of my Hilbert space where the symmetry doesn't act.

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right? But yeah, I agree with that. Yeah, but but but I started with the assumption that the symmetry acts non-trivially in the full thing. I don't have a sector.

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But what you're saying is completely sensible. If I have a sector of the Hilbert space where the symmetry doesn't act, then, of course, I cannot make any statement about the representation of that factor. But for the factor that it acts faithfully, this is a true statement. I don't want to derail you, but maybe just one question.

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So take the nth tensor power of the regular of this equation. Yes.

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What would you say is the character? I would say it's your C alpha to the nth power. That's correct. Oh.

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But then, if I… but the but the property of the regular representation, if I take the nth product and I decompose it, it also decomposes to a regular presentation.

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It cannot decompose to anything else. Right? So so that statement is already implicit.

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Okay. Right. Because the the point is that the regular representation is this sort of interesting central limit theorem for representation theory. If I take the nth power of the regular implementation, that also decomposes into.

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copied on the regular implementation. But so so that maybe that's a better answer to your question.

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Okay. Good. So the. Oh, we already are half time.

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So what does this do? So this tells you that there is an interesting interplay of locality and symmetry, which is that in particular, if a symmetry is acting locally, it has this rather interesting property that the Hilbert space cannot be in some random representation. It has to have this structure.

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And this is a property of locality. So this allows you then to try to think about a new object that you can define for your quantum system in quantum field theory, which is this function.

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In general, which I'll define as the limit, as a as a regulator, and I'll explain why I need to define it in this way.

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time U alpha e to the minus beta H.

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beta goes to 0. Over trace.

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of e to the minus beta H. So what am I doing? So in general, in the in the case where I started with some finite dimensional system and then take the limit, and so on. Of course, everything is well defined. Nothing is diverging. But if I go to a quantum field theory, I want to study the same sort of a property.

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In a quantum field theory, if I just try to take a trace.

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of some operator, even even when it is a symmetry operator, which are the nicest operators.

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formally, that is not a well-defined thing to do. Okay, so instead I can define a quantity which I regulate in some way. I regulate it using the Hamiltonian since you commutes with the Hamiltonian.

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So then you can ask in a given QFT.

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If you have a symmetry that's acting locally. Does this function satisfy this interesting regularity observation that we observe by building from some finances?

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Okay. And the answer here seems to be correct. Yes. If.

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G-Alpha. is locally acting.

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Then. Phi Alpha of G alpha of phi alpha here.

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is equal to delta alpha identity. Okay, so this is a statement. It's… not exactly a theorem, because I haven't proved it. But this is something you can go and check as many examples as you would like, and see what it what it tells you, and I will discuss that. And in particular.

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This is a place where you can now try to explore something that is special about non-invertible symmetry than generalized symmetries, right? Symmetry that fit in some categorical language. It turns out that when the symmetry operator that's acting is categorical is coming from some fusion category.

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This statement seems to hold.

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This? Yeah. This is G Alpha. If G alpha is locally active.

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Any questions? So this, I'm not sure I understand the equation in the box.

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This equation. What do you mean by delta alpha? Meaning it is 0 for any for every every element of the group that is not the identity.

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They said what we got. Yes. Yes, the so this quantity is actually 0.

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and this quantity is a trace is the character of the regular representation.

00:27:57.000 --> 00:28:13.000
Right? So so this. So this was done to them. So so here this formula is actually 0. So this is one of the reasons why this is a computation that makes sense. What is the ratio of the multiplicity at large and the ratio of the multiplicity, the ratio.

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of the dimensions, therefore you get a regular representation.

00:28:19.000 --> 00:28:34.000
Well, plus 0 is delta alpha E. Right. Answering to the question I made the stipulation for the element does not the identity.

00:28:34.000 --> 00:28:44.000
Great. Yeah. And the fact that that that quantity is delta alpha E also just follows from the character product. Yeah.

00:28:44.000 --> 00:28:55.000
Okay, so this object actually has appeared in the literature in various contexts. It appeared in the context when people were studying asymptotic density matrices of, let's say, black holes.

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Right? Where you have some charge in your system, and then you want to know what is the asymptotic distribution of for the density matrix. So this appeared first in a paper of Uri and Harlow, where they explored such a function.

00:29:12.000 --> 00:29:29.000
We actually discovered this and then retroactively found out their papers were more or less the same. But in that context, they weren't trying to make a statement about the Hilbert space of system, which is of the QFT as we have, but they were trying to study is what is the asymptotic behavior of the density matrix.

00:29:29.000 --> 00:29:44.000
And the fact that the asymptotic behavior of the density matrix was regular in their work is exactly the statement that asymptotically, the ratio of the multiplicities are that of the regular representation.

00:29:44.000 --> 00:30:00.000
Okay, so there's a sequence of papers first by Carlo and Aguri, then Pal and Sun also studied this for 2D CFTs. And then this sort of statement or this asymptotic version.

00:30:00.000 --> 00:30:16.000
of taking the traits of a symmetry operator. You can also ask what happens if this is a non-invertible symmetry. This was explored by Lynn Okada and Yuji Tajikawa.

00:30:16.000 --> 00:30:31.000
Where the same sort of… Statements seem to be buried in their result that when you look at a quantum field theory and you have a symmetry that's acting locally, that somehow the Hilbert space has to organize in this way.

00:30:31.000 --> 00:30:35.000
Okay. So I can show you a couple of examples.

00:30:35.000 --> 00:30:40.000
But I'm not sure I have… quite a bit of time.

00:30:40.000 --> 00:30:52.000
And, you know, some of these examples are actually very good things to work out for students. You can, for example, take Heisenberg spin chain, where you have a spin one half.

00:30:52.000 --> 00:31:04.000
That's sitting at each place. So there's an SO3 that's acting. So this is an example with non-abeliant with continuous non-abelian symmetries. There's an SO3 that's acting on each site.

00:31:04.000 --> 00:31:18.000
And then you can write the standard Hamiltonian where nearest neighborhood interactions are present. And then you have the diagonal So 3 that's acting everywhere. Okay? So in that case, you can, for example, ask.

00:31:18.000 --> 00:31:26.000
In that case, for the Heisenberg spin chain.

00:31:26.000 --> 00:31:35.000
Uh, you have, what is the multiplicity?

00:31:35.000 --> 00:31:44.000
lucidity. of the eighth representation.

00:31:44.000 --> 00:31:57.000
I'll echo it free. So this you can compute exactly, for example, and if you compute this, you find that n.

00:31:57.000 --> 00:32:01.000
the multiplicity for the case representation where you have n spins.

00:32:01.000 --> 00:32:06.000
It's an exact formula. It's going to be given by.

00:32:06.000 --> 00:32:13.000
and factorial. 2k plus 1, which is important.

00:32:13.000 --> 00:32:18.000
over n over 2. of K plus one.

00:32:18.000 --> 00:32:26.000
And over to minus K. Authority. So when you fix N.

00:32:26.000 --> 00:32:35.000
Sorry, when you fix K. and K prime like 2 2 representation. And then you take nth.

00:32:35.000 --> 00:32:44.000
to infinity. You can explicitly check that. And K over.

00:32:44.000 --> 00:32:52.000
and K prime is going to go to. 2k plus one over 2k prime plus one.

00:32:52.000 --> 00:32:59.000
Okay, so you can explicitly check in this case and there that you do recover the ratio.

00:32:59.000 --> 00:33:12.000
that would be present in the regular representation. You get this regularity property again. But this is the case where you have an SO3 symmetry that is acting on the Heisenberg spin chain, but it has a local action at each site.

00:33:12.000 --> 00:33:23.000
Great. So it from that fact, you can. You can that allows you to compute this, and then you can see that the ratio tends to what you want it to do.

00:33:23.000 --> 00:33:35.000
Okay, you can also compute it for free scalar with a U1 action and a free fermion. Those are examples where you can do it explicitly by hand. And the statement seems to hold.

00:33:35.000 --> 00:34:02.000
to also highlight here, we've defined the trace using the Hamiltonian in simple symmetric theory. You can actually also do this right in super symmetric series. Let's say we take n equal to 2 degrees. So right here, I have to at some point say something about inequality theory.

00:34:02.000 --> 00:34:17.000
as being the limit. of Q equals one, where Q is a standard Q that appears in indices. And, in fact, what you will obtain, you can regulate this.

00:34:17.000 --> 00:34:24.000
In this way. minus 1 to the f times the element times Q.

00:34:24.000 --> 00:34:31.000
to the delta minus R, where R is the SU 2 charge.

00:34:31.000 --> 00:34:38.000
Then brace. minus one for the F.

00:34:38.000 --> 00:34:50.000
So this quantity here for if you know, this is the ratio of the Shure index right of the refined sure index with the sure index itself. So this.

00:34:50.000 --> 00:35:05.000
In this case, these quantities are actually computable for n equal to 2 theories explicitly, and in particular you can computer theory. You computer for class s theories, and so on. And you can take, for example, the SUR symmetry.

00:35:05.000 --> 00:35:10.000
Right? So in the SU2R symmetry, you can compute.

00:35:10.000 --> 00:35:18.000
You can take. So this quantity here is what we call the is a sure index, which has some function.

00:35:18.000 --> 00:35:27.000
depends on Q. and V, which you can compute. So this index.

00:35:27.000 --> 00:35:43.000
I, Q. be the refined index that you compute in this way you can decompose it into characters of the.

00:35:43.000 --> 00:35:47.000
You can put this into the characters of the Z.

00:35:47.000 --> 00:35:57.000
Yeah. Uh, you can dig more into the characters, where this is actually 2 characters.

00:35:57.000 --> 00:36:10.000
with representation and and n over 2. Okay, you can compute this explicitly. And you can show that these quantities, which is the multiplicity that appears, turns out, indeed.

00:36:10.000 --> 00:36:14.000
Cn of… uh, Q.

00:36:14.000 --> 00:36:21.000
over Cm. Of Q. You take the limit as Q.

00:36:21.000 --> 00:36:29.000
goes to one. This exactly goes to N plus 1 over M plus one.

00:36:29.000 --> 00:36:37.000
And you can check this in all integral to theories where you where you know this thing is computed, and you see that this works out. You can also equally do something.

00:36:37.000 --> 00:36:52.000
nice, which is that instead of decomposes into the S2 characters, let me decompose it into the carton U1 characters. In that case, the multiplicity that you would get, they go to… Everybody should know this answer.

00:36:52.000 --> 00:36:59.000
one right because the multiplicity of a representation. Yes, yeah.

00:36:59.000 --> 00:37:03.000
right? So you can just explicitly check that goes to one.

00:37:03.000 --> 00:37:09.000
Which… which is actually quite… I mean, so it's sort of surprising that this feature is just there.

00:37:09.000 --> 00:37:18.000
And, um… It hasn't been noted, at least in this way.

00:37:18.000 --> 00:37:36.000
Okay, so… So there you can also explicitly check this when you have. So since I don't have time, I will not go through it. It is a very instructive exercise to consider rational Cft, wherein rational Cfts you have symmetries that are acting.

00:37:36.000 --> 00:37:55.000
But the symmetries are given by Verlin. So those symmetries are characterized by some fusion category. And then you can check exclusively when you have, you can then take one of those Berlinde lines and taking the trace correspond to taking it on the spatial circle.

00:37:55.000 --> 00:38:11.000
Okay? And then compute that that trace in that way, and ask what happens at large N. Well, sorry, and ask what happens in those case. In those cases. It's already a field theory. It's a 2d field theory. Everything works.

00:38:11.000 --> 00:38:20.000
You can then argue from there that the representation of the Verlende lines for the Rcfd is going to be the regular module.

00:38:20.000 --> 00:38:39.000
taking the traits of the action of the frontal line with e to the minus theta H. And then what you use in that case. So there in that case, this highlights. Then you want to take beta to 0, but then to take beta to 0, you use modul invariance.

00:38:39.000 --> 00:38:50.000
Right? You get the quantum dimension. You get the Froggenius variant quantum dimension. Yes. You get the ratio of the Frobenius pair on quantum dimension.

00:38:50.000 --> 00:39:04.000
And that argument could be generalized in a more physical way. But so if you're curious, I'll be happy to walk through that computation for you. But the way you can generalize that argument, at least for 2DCFT, is you want to compute.

00:39:04.000 --> 00:39:13.000
the partition function. Uh, as a function of beta of, let's say, trace in the Hilbert space.

00:39:13.000 --> 00:39:28.000
of G E to the minus beta H. Notice that when I say this, G could be any operator, invertible or non-invertible in 2d. You can use modular invariance, where this thing.

00:39:28.000 --> 00:39:45.000
So this Z of beta. is going to map by using a cardi formula e to the minus one over beta times delta G. Okay? So what is delta G here? So delta G.

00:39:45.000 --> 00:39:51.000
So first, st when when I do modular transformation.

00:39:51.000 --> 00:39:56.000
here. So let me write this as this. And this is a modular.

00:39:56.000 --> 00:40:03.000
transformation. When I do a modular transformation, taking here beta.

00:40:03.000 --> 00:40:10.000
0 correspond to taking beta twiddle to infinity. Yeah.

00:40:10.000 --> 00:40:21.000
Okay? So now I'm computing the partition function where I'm taking the infinite temperature limit. Sorry, the 0 temperature limit. So that should pick out the ground state.

00:40:21.000 --> 00:40:38.000
Okay, so when I have G inserted right? Then doing this modular transformation takes the g, which was along the spatial circle. And then the modular transformation puts it along the temporal circle.

00:40:38.000 --> 00:40:47.000
So then that's a deformation of the Hamiltonian. that I explicitly have. And then what you're computing is that the lowest.

00:40:47.000 --> 00:40:57.000
state in the twisted sector of the Hamiltonian. Okay, so this delta G is exactly the energy that you get there.

00:40:57.000 --> 00:41:11.000
And then you can take now the ratio of VG of beta over z of the identity of beta, because that's the normalization that you would get. So this quantity is going to go ask.

00:41:11.000 --> 00:41:28.000
some number times e to the minus theta, uh… Delta G minus delta E, where this is the ground state in the untwisted sector.

00:41:28.000 --> 00:41:33.000
an unfiltered factor. I know.

00:41:33.000 --> 00:41:43.000
Mr. The zip quantity listed 0. Beta twiddle. Yes. So this goes to 0.

00:41:43.000 --> 00:41:50.000
Okay? So this goes to 0 when I take beta twiddle to 0.

00:41:50.000 --> 00:41:56.000
Sorry. You're right. Okay? And if.

00:41:56.000 --> 00:42:07.000
G is ironically, this is 1. I mean, he was giving the facilities unity because it could be Yamily where they can probably get the theory I'm assuming unit.

00:42:07.000 --> 00:42:18.000
Good. Thank you. Okay, so this is a a rather nice observation. This is the thing that tells you that this is a feature.

00:42:18.000 --> 00:42:28.000
actually of the the 2D theories will have this feature, and we expect this to extend to higher dimensions, where you can check it in many in many, many, many settings.

00:42:28.000 --> 00:42:36.000
So if you're if you want, I can show you many other examples that we do. But this also highlights something else that is interesting.

00:42:36.000 --> 00:42:46.000
Which is related to. Uh, what I mean, this argument seems rather special to the two-dimensional case.

00:42:46.000 --> 00:42:54.000
Yes. So let me say more about this. It's also pretty close to the original argument for calling the quantum dimension. Yes.

00:42:54.000 --> 00:42:59.000
But where you take the trace, you know, you define the dimension of the ith representation.

00:42:59.000 --> 00:43:16.000
It's the limit as Q goes to one of the character of the nth representation divided by the trivial representation, and then you get si0 over 00. Yes. So the lesson that I'm getting to here, the fact that that's true is inherent to what we… to when we talk about categorical symmetries.

00:43:16.000 --> 00:43:33.000
Okay, so there there. One thing here that that is that is true. So let's go back to the initial setup where I have some quantum system, and then I have some symmetry that I say acts on it. Okay, there are 2 things that could happen in the standard quantum field theory. When we have a categorical symmetry.

00:43:33.000 --> 00:43:39.000
We demand that, you know, whenever I have a symmetry operator acting.

00:43:39.000 --> 00:43:54.000
In any system where I decompose into a time, direction, and space, where I define a Hilbert space, that there exists a weak rotation where such a symmetry operator becomes a defect. So this is a.

00:43:54.000 --> 00:44:06.000
feature that is necessary to define categorical symmetries. Okay? And in fact, this is precisely the property that when I have a categorical symmetry, I talk about in a Euclidean.

00:44:06.000 --> 00:44:19.000
Field theory, where I'm allowed to insert this in any way I like, right? And the fact that you have this exact duality or this exact equivalence between the.

00:44:19.000 --> 00:44:34.000
symmetry operators that are acting at a Hilbert space at some fixed time, with the defect is because you can do this weak rotation, and this is guaranteed to you by categorical symmetries. And this is exactly the property that's used here.

00:44:34.000 --> 00:44:49.000
Okay, this is exactly the property of this use here. So this is the thing that it highlights. So now, suppose I did not know that I have this full Lorentz symmetry. I didn't have all of the beauty of fusion categories and so on.

00:44:49.000 --> 00:44:58.000
And I just say that, OK, I have some t equal to 0, and at t equal to 0, I have some Hilbert space.

00:44:58.000 --> 00:45:05.000
and on this Hilbert page I have a set of operator view that can meet with the Hamiltonian.

00:45:05.000 --> 00:45:10.000
and all of these operators form a symmetry. Okay?

00:45:10.000 --> 00:45:27.000
So in general, from this point of view, of course, whenever I have something that commutes a Hamiltonian, I can sorry the u of group representation, or just more general. So let me come to this point, right? So if I have any. So what we're instructed to do in standard quantum mechanics.

00:45:27.000 --> 00:45:35.000
If I have any operator that commutes with a Hamiltonian, I can make it Hermitian if I want.

00:45:35.000 --> 00:45:44.000
And then I can exponentiate it. Okay, so that's what we're supposed to do from a Wigner point of view. And then when you exponentiate, you get a unitary operator, which makes sense.

00:45:44.000 --> 00:45:54.000
to talk about. But the difference now is that when such a unitary operator acts locally, then this regularity condition is satisfied.

00:45:54.000 --> 00:46:01.000
And such unitary operator does appear as a topological operator.

00:46:01.000 --> 00:46:09.000
in the standard sense. But you can also have a unitary operator that is exponentiated that becomes highly non-local.

00:46:09.000 --> 00:46:31.000
Those guys you cannot rotate. Okay? So this feature of being able to rotate operators from the from from fixed time to fixed space is in a sense, the magic that where locality is somehow entering in a way that I haven't made exactly sharp.

00:46:31.000 --> 00:46:39.000
But we see it here play out explicitly in this calculation. So do you have an example of a non-locally acting synergy? Yes.

00:46:39.000 --> 00:46:52.000
Yes. So the simplest I can. So I have many examples in many different directions that you might that that you might be interested.

00:46:52.000 --> 00:46:56.000
So let me show a simple example of this.

00:46:56.000 --> 00:47:10.000
Okay. And then I will try to tell you that this quantity is actually quite interesting, and that it encodes a lot more information.

00:47:10.000 --> 00:47:23.000
Then it seems at face value.

00:47:23.000 --> 00:47:40.000
So the simplest example that I can try to tell you is, let's consider the Ising fusion category. Right? The Ising fusion category is composed of 3 elements. You have the identity. You have a Z. 2 invertible line, and then you have D is not invertible.

00:47:40.000 --> 00:47:50.000
Okay? And then we have the. Fusion rule that E squared is equal to E plus eta. So this is the Eisen fusion category.

00:47:50.000 --> 00:47:54.000
You can, of course, by the discussion that we just have.

00:47:54.000 --> 00:48:07.000
You can take explicit model. Take the term of transfer field Ising model at criticality. Okay, you. So you have an explicit model there, and you can explicitly check that you can compute trace of eta.

00:48:07.000 --> 00:48:25.000
which is trace of D. Which is 0. Okay, so these are locally acting symmetries in the sense which we want our symmetries. So here, what I mean by locally acting for fusion, what I mean here, it has to mean that locally acting means that they satisfy.

00:48:25.000 --> 00:48:41.000
The local property that we expect topological lines, topological field theory to do. This ultra locality that nothing interesting happens when they touch, and this locality property that is defined by the Etia Hitchen.

00:48:41.000 --> 00:48:52.000
definition of a topological field theory. Thank you. All right. What? What is this fusion of category acting on?

00:48:52.000 --> 00:48:59.000
What is this, Chase? What do you mean by the trace? So I can do it. So one example I can take the transverse field Eisen model.

00:48:59.000 --> 00:49:06.000
at criticality, right? So I can have an explicit realization of this, and I can take a continuum if you like.

00:49:06.000 --> 00:49:13.000
Okay, there's also two DCFTs where the IV infusion acts. So you can check in those cases.

00:49:13.000 --> 00:49:19.000
So here is an example of a symmetry that is unitary, but it's going to be non-local.

00:49:19.000 --> 00:49:34.000
Okay, so from this thing, I will do something that is illegal from the point of view of a fusion category, but it is completely legal from the point of view of decomposing my physics into one plus one spacetime. So if I treat these as operators that act.

00:49:34.000 --> 00:49:45.000
On my Hilbert space, then I can define. new symmetry operators eta plus minus, which is 1 over root 2 times D.

00:49:45.000 --> 00:49:55.000
plus minus E minus Ada over 2. So here, look, I'm taking. I'm defining a new operator which involves.

00:49:55.000 --> 00:50:09.000
Plus and minuses and irrational numbers. This sort of a thing is actually illegal from the point of view of fusion category, because there is no sensible way by which I should add and subtract or multiply by irrational numbers of.

00:50:09.000 --> 00:50:15.000
uh, when I'm talking about defects, because defects define Hilbert spaces. Okay?

00:50:15.000 --> 00:50:24.000
But from the point of view of thinking about operators that are acting on a Hilbert space at a fixed time, this is a completely allowed operation.

00:50:24.000 --> 00:50:39.000
And indeed, if I do this, then I have, instead of talking about this non-invertible nonsense. Some people would say non-invertible nonsense. I like non-invertible stuff. Then I would say my system instead admits.

00:50:39.000 --> 00:50:48.000
V. 5 2 3.

00:50:48.000 --> 00:50:58.000
Okay, so the first one is AIA, and the second one I can take you to the Eta plus eta minus, you can check if this is diagonal of T.

00:50:58.000 --> 00:51:05.000
But the fact that D was not invertible gives you a constraint. It tells you that.

00:51:05.000 --> 00:51:14.000
If I look at any Hilberg case where this is acting following projection has to be satisfied.

00:51:14.000 --> 00:51:19.000
Okay? That if I multiply this projected times that projector, it must be 0.

00:51:19.000 --> 00:51:32.000
acting on any acting on the Hilbert's place. And either plus or be unitary? Yes, by construction. Sorry. Right. So I should say this by construction, eta plus minus.

00:51:32.000 --> 00:51:42.000
8 or plus minus dagger. Okay? So you do get a Z 2 times Z 2 symmetry with a constraint.

00:51:42.000 --> 00:51:55.000
So what does this constraint tell you right? So Ada that's acting here. I can think about the Hilbert space, and you can check this in several examples. You can check this explicitly. But the Hilbert space.

00:51:55.000 --> 00:52:01.000
For example, for Ada. All right, let me write this way.

00:52:01.000 --> 00:52:06.000
If I have the state Ada, I will get plus and minus states.

00:52:06.000 --> 00:52:13.000
Okay? So and this would be 50%.

00:52:13.000 --> 00:52:31.000
You're here. And another 50% of the states would be here. So this would be the fact that Ada has this regular representation and it's a Z2 is an abelian, so the dimensions are all 1. So half of them are in the plus representation.

00:52:31.000 --> 00:52:34.000
The other half are in the minus. Okay, the minus is the… so the plus is actually that entity.

00:52:34.000 --> 00:52:50.000
But if you then look at this constraint, you can see that if I am in the plus representation, then I solve this constraint. Then the plus representation is going to split under Ada plus into 8 plus.

00:52:50.000 --> 00:53:02.000
And I might half half. Okay? But then, if I am in the minus representation, then I have a constraint here which tells me that everything is in the plus.

00:53:02.000 --> 00:53:14.000
Okay, so you can see exclusively. Here you have a 3 to 1 ratio in terms of the representations that are appearing right? So you can explicitly see that, um.

00:53:14.000 --> 00:53:30.000
Now, so this operator makes complete sense on the lattice. You can compute it. You can construct it in the transverse realizing model. You can play with it. You can do all sorts of things that you want. But then you clearly see that when you write it, that the representation of the Hilbert space.

00:53:30.000 --> 00:53:45.000
It's certainly not balanced in this way. Go ahead. Within the statement that the delta operator is not exactly defined for transverse realizing because it makes it translation. Good. So here I'm implicitly taking the continuum.

00:53:45.000 --> 00:53:51.000
But you can define the appropriate operator. So. I can show you.

00:53:51.000 --> 00:54:00.000
Yes. Okay. Question. Could this all be done in the CFT instead of the transverse realizing? Yes.

00:54:00.000 --> 00:54:07.000
I haven't done the explicit computation in the CFT, but it's doable.

00:54:07.000 --> 00:54:17.000
Great. So this is an explicit example where you can see this regularity playing out right? And and and so these.

00:54:17.000 --> 00:54:23.000
Hey, the plus and minus don't act on the local operator.

00:54:23.000 --> 00:54:44.000
Uh, good. They act on a local operator algebra, but they map it to a non-local operator. That's what I meant. Yeah, they mix local operators to non-local ones. So in fact, if you… in our paper, we completely computed the eigen operator of eta plus minus, and you can see that they're non-local in general.

00:54:44.000 --> 00:54:50.000
They have to have some Ada lines attached to them.

00:54:50.000 --> 00:54:55.000
Okay. Okay, good. So what is this going to be good for?

00:54:55.000 --> 00:55:01.000
Let's try to see if this is good for something.

00:55:01.000 --> 00:55:13.000
And I will be very fast in schematics.

00:55:13.000 --> 00:55:21.000
Okay, so now you can define the following quantity. So you can take. You can define the following quantity.

00:55:21.000 --> 00:55:26.000
B of G. which is equal to the norm.

00:55:26.000 --> 00:55:35.000
of this trace. Okay? For any operator G. So why do I want to do this? I want to do this for the following reasons. Suppose.

00:55:35.000 --> 00:55:40.000
Suppose you have some quantum field theory. some quantum system.

00:55:40.000 --> 00:55:50.000
and I give you the Hamiltonian of the system, and then you hand it to the student, and then you say, go and find all the operators that commute with the Hamiltonian that are well defined.

00:55:50.000 --> 00:55:56.000
Okay, that's a that's a that's a reasonable thing to do.

00:55:56.000 --> 00:56:15.000
So in general, if you compute all session operators from this Hamiltonian point of view, you're supposed to construct unitary operators. How would you know that you have a fusion category? Is there a way to get… to go from computing all to the operators to figuring out that you have a fusion category or knowing which one acts locally, which one doesn't act locally?

00:56:15.000 --> 00:56:34.000
But but the set of all operators is the set of all projections on the so in the talk, there is the part where I where I wanted to discuss about projectors, but I cut it out because of time. Good. So projectors inject. I'm glad you asked this question. So projectors, if I just take a random projector.

00:56:34.000 --> 00:56:45.000
It's not generally well defined. If I… let's say I have some lattice system and I diagonalize the Hamiltonian, and I look at the projector on the nth state.

00:56:45.000 --> 00:56:56.000
So that object actually is not. well defined. Certainly, when I tried to go to Continuum. So let's say then I go from N lattice to n plus 1.

00:56:56.000 --> 00:57:16.000
Right? The change of that projector. Sure. It's by no means smooth energy interval is well defined. If yes, good. So if you smear, then you might make it well defined. But such a projector are highly non-local thing. Absolutely. They're highly non-local things. So there's still a question of what are the right adjectives from this this setup.

00:57:16.000 --> 00:57:26.000
That I just give you. But I want to study this, and then I'll tell you some nice properties it has. Okay, before we get there.

00:57:26.000 --> 00:57:32.000
So suppose I tell you that I have a bunch of operators, they're all well-defined in some way.

00:57:32.000 --> 00:57:47.000
What you can do is you can take all of those operators. They're unitary, and let's consider the most general unitary operator that you can get. I'll call it G of some angles.

00:57:47.000 --> 00:57:53.000
Beta I. Right? You can construct such a thing.

00:57:53.000 --> 00:58:04.000
where this theta i is going to be a manifold, which basically maps out the directions corresponding to these different operators that you construct.

00:58:04.000 --> 00:58:14.000
It turns out this quantity B of G. have very interesting properties. The first, this quantity is bounded between 1 and 0, okay?

00:58:14.000 --> 00:58:21.000
at 0 is going to be all. uh, what we would call simple.

00:58:21.000 --> 00:58:30.000
operators. that are unitary.

00:58:30.000 --> 00:58:42.000
So if you have an operator where B of G is 0, then that's an operator. Basically, that is unitary, and that's also locally acting.

00:58:42.000 --> 00:58:52.000
from the discussion that we just had. If it is 1, then this would be the identity.

00:58:52.000 --> 00:58:58.000
Great. But then, in general, this this bead can take values along this whole thing.

00:58:58.000 --> 00:59:03.000
So observe that this is a maximum, and this is a minimum.

00:59:03.000 --> 00:59:10.000
Okay? And then you can ask, are there anything interesting in the intermediate regions?

00:59:10.000 --> 00:59:18.000
Great. So. So you could. You can try to ask, are there anything interesting in intermediate reasons?

00:59:18.000 --> 00:59:30.000
What you observe is that. Let's say I I started off with this Ising fusion category. So she uses the fusion as an object of the fusion together.

00:59:30.000 --> 00:59:42.000
Yes, I can take. So so what I'm going to say now is going to be taking explicit examples of fusion categories, and then I can say my statement now.

00:59:42.000 --> 00:59:50.000
So suppose I consider some fusion category C, right? You can think of, for example, the Isling.

00:59:50.000 --> 01:00:09.000
Ada and B. Okay, it is clear that. So if I so I can construct this B right? It's going to be some unitary operators constructed from these 3 lines. So this would be. So B here would be something like a 1 times identity.

01:00:09.000 --> 01:00:16.000
Plus, let's say a 2 times Ada plus a 3 times D.

01:00:16.000 --> 01:00:25.000
Okay? And then I pick A1, A2, and A3, such that this is unitary. And in fact, the simplest way you can construct this — sorry, G.

01:00:25.000 --> 01:00:40.000
In that you can show that G, you can write it as E to the i eta times e to the times theta times some phi times D. You can construct this operator where this defines for you.

01:00:40.000 --> 01:00:52.000
a manifold of two variables. Okay. Do you expect to be able to do this just for Tambara Yamagami, or just I can do this for any any symmetry category. Yeah.

01:00:52.000 --> 01:01:06.000
So here there is no I. There is a good reason for that, but you can take my word for now that this is going to be… so this is a function of two variables, and this defines a unitary operator. So I can compute.

01:01:06.000 --> 01:01:13.000
Bog. of Theta 5.

01:01:13.000 --> 01:01:27.000
Of course, when you compute B of theta phi, I mean, I have an explicit form of the answer, but it's not so interesting to write it. You can show that it has an extremum, which is 1, and at the extremum, when you evaluate it, exactly gives you the identity operator.

01:01:27.000 --> 01:01:39.000
It had the minimum which is at 0, and at the minimum at 0 it gives you the Ada operator. Then you can ask, can it actually know about the non-invertible symmetry?

01:01:39.000 --> 01:01:46.000
Then you find that B. of G have.

01:01:46.000 --> 01:01:53.000
a saddle point.

01:01:53.000 --> 01:02:08.000
Right? There's some value where B of G has a saddle point. And and and you can ask, what is happening at that saddle point at that saddle point. You find that the operator that it evaluates to it B. So G here.

01:02:08.000 --> 01:02:19.000
is going to be given by some. Uh, let me call this eta c. And what eta c corresponds to.

01:02:19.000 --> 01:02:31.000
Eta is. one minus some projector, which I will be more specific, where this projector is going to be equal to E.

01:02:31.000 --> 01:02:38.000
plus. Ada plus.

01:02:38.000 --> 01:02:45.000
Wait, sorry. Root 2 times D. Over the thumb.

01:02:45.000 --> 01:02:57.000
of the dimensions court. as a function of the theta and phi. Yes.

01:02:57.000 --> 01:03:08.000
They're the saddle point which exists, and at the faddle point B evaluates to this G. And that evaluates to this eta, where the.

01:03:08.000 --> 01:03:11.000
So you need to essentially critical. Yeah, the critical point.

01:03:11.000 --> 01:03:27.000
It had the critical point in it where the storms. Oh yeah, the B is right. And the symmetry there. There is from Z 2, and the Z 2 is 1 minus times 2, this project in this projector is given here. This projector knows about D.

01:03:27.000 --> 01:03:34.000
And in particular, when you compute. B of G explicitly.

01:03:34.000 --> 01:03:39.000
You get that this is 1 minus 2 over the sum.

01:03:39.000 --> 01:03:47.000
over the dimensions of all the lines squared.

01:03:47.000 --> 01:03:54.000
Sorry. Yeah, this. So you can show explicitly that B at the saddle point is this.

01:03:54.000 --> 01:03:59.000
To B knows about the dimensions of the noninvertible.

01:03:59.000 --> 01:04:13.000
symmetries. So you're sending over the simple objects of the fusion category, the quantum dimensions of the simple one. What is that? Yeah, the quantum dimension of simple objects. So you can. You can show that at the saddle point.

01:04:13.000 --> 01:04:22.000
or at this extreme momentum. And a fee B is 1 minus 2 over this. And you can prove this exactly.

01:04:22.000 --> 01:04:30.000
And in general, what you find. Uh, is that… If I take a random fusion category.

01:04:30.000 --> 01:04:50.000
And I thought of just promote all the all the lines of being some operators that are acting. What you can show is that whenever you have an invertible symmetry operator, right? Sorry, you can take all of the symmetry. You can construct the most general.

01:04:50.000 --> 01:04:58.000
Unitary object you can do. So this gives you some non-trivial manifold. And then this manifold has extremas.

01:04:58.000 --> 01:05:08.000
So at the extremas where it's 0, right? So every extrema where it's 0 corresponds to a unitary.

01:05:08.000 --> 01:05:16.000
locally acting surgery. So this would be the invertible symmetry operators, and you can see this explicitly.

01:05:16.000 --> 01:05:21.000
And then what you could. But then also B will have some saddle points.

01:05:21.000 --> 01:05:30.000
at other places. So at at the location of the saddle points you can show that for every saddle point.

01:05:30.000 --> 01:05:36.000
You have some projector. which I'll call C prime.

01:05:36.000 --> 01:05:47.000
which is going to be a gauging projector. And this is this we can make sure.

01:05:47.000 --> 01:05:53.000
D of GA. A over.

01:05:53.000 --> 01:06:00.000
Ga prime. D. Prime.

01:06:00.000 --> 01:06:13.000
squared, where c prime is a subcategory in the general infusion category. And then whatever you can define this projector super a sub.

01:06:13.000 --> 01:06:21.000
Sorry. sub into the focal fission where you have one plus 2.

01:06:21.000 --> 01:06:39.000
Eoc minus. And then what you can show explicitly is that B of eta c prime is a saddle point.

01:06:39.000 --> 01:06:47.000
with the solution explicitly being one minus 2 over some of D.

01:06:47.000 --> 01:06:53.000
of GA. squared, where the sum is taken.

01:06:53.000 --> 01:06:57.000
here. Right.

01:06:57.000 --> 01:07:11.000
So the statement is then the saddle points correspond to symmetries.

01:07:11.000 --> 01:07:17.000
different projectors, gauging projectors that you can construct. This is the gauging projector.

01:07:17.000 --> 01:07:34.000
Great. And what? And and what we have been observed so far is that the number of saddle points plus the number of of of invertible things that you can make all sum up to actually be the number of operators that exist.

01:07:34.000 --> 01:07:45.000
in diffusion category. So you're saying this PC prime is always a projector. It always first? Yep, you can prove this.

01:07:45.000 --> 01:08:01.000
I have to know something about the future. That's right. So you can use the so this actually uses the facts of the fusion algebra that DA times DA satisfies.

01:08:01.000 --> 01:08:12.000
Yeah, yeah. So then you can prove that. So this is a projector, and it's a gauging projector. This is P just correspond to different gauging projectors that you can construct.

01:08:12.000 --> 01:08:20.000
Ah, because this project you into. the the states that are going to be.

01:08:20.000 --> 01:08:32.000
uh uh. Sorry, either projects you into it projects you to the subspace, which is going to have projector. I get that it's a projector.

01:08:32.000 --> 01:08:39.000
So then what? So because it's a projector, the A to C prime squares to one. Yes.

01:08:39.000 --> 01:08:44.000
Okay. and it's unitary by construction.

01:08:44.000 --> 01:08:51.000
But it's non-local. And you can show B of eta C is always going to look like this.

01:08:51.000 --> 01:08:59.000
And moreover, the value where eta c is sitting is a saddle point. It's a special point.

01:08:59.000 --> 01:09:14.000
If not any random place on this manifold. What do you assume? Good. So the thing that the only thing that I assume for the fusion category is the fusion algebra.

01:09:14.000 --> 01:09:30.000
Was this special to 2 dimensions? So this was general so far the reliable way we can talk about all of these things, because we talk about 0 form symmetries, which where I can talk about them as operators, and I can make sense of it in this way.

01:09:30.000 --> 01:09:46.000
or for two dimensions. But the question is, how does this generalize to higher D, and how to show it? And we expect that it should. But we don't have a paper yet worked out.

01:09:46.000 --> 01:10:01.000
But anyways, I'm well over time, but the point that I want to highlight is that this quantity B that we proposed seemed to have a lot more information about the system that it had the right to do.

01:10:01.000 --> 01:10:18.000
Okay, so first we started out as being a thing that can tell you whether symmetry operator is local or non-local. But then, when you when you evaluate it for actual fusion categories that you see that there is some version of a maximization that's emerging.

01:10:18.000 --> 01:10:28.000
right? Which is the thing that we would like to try to make sharp in some way. The critical points of your B function for the set of unitary output.

01:10:28.000 --> 01:10:37.000
unitary teeth, line defects. Yes. as these critical as critical points that.

01:10:37.000 --> 01:10:45.000
What? Generate a subfusion category. They have critical points that get associated to its z2.

01:10:45.000 --> 01:10:54.000
non-locally acting symmetries, which have their origin. from some from the projectors that you could construct.

01:10:54.000 --> 01:11:03.000
And the number of those, it seems in examples that we checked, the number of those is always enough that you actually recover all the different lines.

01:11:03.000 --> 01:11:11.000
all different. All the different simple operators, simple simple lines, invertible or not invertible.

01:11:11.000 --> 01:11:18.000
Because once you… So you're saying the C prime like I'm confused. You're saying that you are.

01:11:18.000 --> 01:11:26.000
But what I'm saying is that there's a. a small C prime of exhibit.

01:11:26.000 --> 01:11:31.000
Yeah. You write AC program, PC fund. That's all C program.

01:11:31.000 --> 01:11:45.000
He wrote C prime is a subset of C subcategory of C. Yes, you probably made a capital C prime. Yes, that should be a capital C. Okay, so now you're saying that the critical points of your functions.

01:11:45.000 --> 01:11:52.000
select correspond to objects and generate some subcategories. Surely the thing.

01:11:52.000 --> 01:11:57.000
The capital C frame is generated by this swallowsing frame.

01:11:57.000 --> 01:12:02.000
Yes. Okay. Yes. So the the the statement is that.

01:12:02.000 --> 01:12:10.000
If I take a fusion category and I take the most general unitary operator that I can construct, I get this function B. Okay?

01:12:10.000 --> 01:12:24.000
then I can… then the statement is, how many saddles do you have for this? Right? You can count all the different saddles. So how many minima did it have to give you the number of simple operators.

01:12:24.000 --> 01:12:28.000
Simple operator that are unitary, and they form a subgroup.

01:12:28.000 --> 01:12:41.000
And then you ask, what are the different critical points that you have inside? If you have n critical points, the claim is that for each one of those values, the critical point, there is a projector associated with this.

01:12:41.000 --> 01:12:55.000
and for and and then for every non-invertible symmetry, you can construct the projector.

01:12:55.000 --> 01:13:09.000
The YFC primer subcategory. I think maybe I just abused language a bit. I just this is a selection of operators such that they define for you a projection. They generate the whole category.

01:13:09.000 --> 01:13:20.000
If not, it's not a single C prime. Not a single. But if I know all of them, yeah, that's exactly why I'm complaining about the small C prime and the big C prime.

01:13:20.000 --> 01:13:34.000
The small C prime is laid the critical points of this non-negative function. Yes. Okay. Now, then you can use select subset of objects. But asking, does the set of objects generate the entire fusion category? No.

01:13:34.000 --> 01:13:51.000
It doesn't have to generate the entire fusion category. Yeah. Does it generate a future? Yeah, to generate a subfusion category within the system. Yeah, a subfusion category. Yes, which is which is how how close is. Yeah, that's right. That's right.

01:13:51.000 --> 01:14:00.000
I mean, it could be the whole thing. But then there are things in the fusion category C, which corresponds to your.

01:14:00.000 --> 01:14:05.000
your symmetries, your symmetry defect lines. that are not critical.

01:14:05.000 --> 01:14:23.000
for the beef option? No, good. So the statement is that the critical points have all the information that you need. Why don't they generate the entire category? There are different critical points. No, but if I put them together. Yes, yes, that's right. That's a statement.

01:14:23.000 --> 01:14:38.000
If I go and find three critical points for each critical point, I can associate it to it the projector. Yes. And then NYT linear combination. I take the category generated by those.

01:14:38.000 --> 01:14:43.000
Yes. Does that is that the full category? Yes. That's a claim.

01:14:43.000 --> 01:15:00.000
Do the projectors canoe. Do the projected commute doesn't that's a good question. In general, they should not. So it's not a decomposition. No, it is not a decomposition. It's a set of subcategories which together generate. Yes.

01:15:00.000 --> 01:15:21.000
the full thing. Yes. So far, it's all an observation that, like, I haven't, like, what you can do is that you can ask and say, okay, B has a critical point. What's happening at the critical points? First, it was surprising that the critical points, like, when you evaluate the example of the critical points could only be written in this way.

01:15:21.000 --> 01:15:39.000
the value. And then you can then ask, what is the symmetry? The symmetry looks like that associated to some projector. Right? So we've discovered this other way around. We haven't proved it the fully right? So in order for this to be a sort of like litmus test.

01:15:39.000 --> 01:15:47.000
for obtaining fusion categories of a system that I don't know, you have to prove the other way.

01:15:47.000 --> 01:15:55.000
So you've just done this by examples. Yes. And then you can see that whenever you have a fusion category that acting, this thing is true.

01:15:55.000 --> 01:16:07.000
So I don't know yet what this function is and what is it doing? But it's very curious. But it sort of comes out. It's it's sort of surprising that trying to understand.

01:16:07.000 --> 01:16:16.000
this sort of weird tension of locality of unitary implies that something like this exists.

01:16:16.000 --> 01:16:22.000
This is why I said the talk was weird, and it's incomplete, but it's work in progress.

01:16:22.000 --> 01:16:29.000
I think I'll stop here.

01:16:29.000 --> 01:16:31.000
You said it's work in progress. Do you have something on the archive?

01:16:31.000 --> 01:16:47.000
Yeah, there's one paper on the archive. Yeah, there's a paper on the archive. Currently, one of the things we're trying to do is see if we can actually use this as a machine to figure out fusion categories, where you can basically… you have a bunch of lattice models.

01:16:47.000 --> 01:17:05.000
That you can study where you don't know what the fission category is, but people have to MPO constructions of the operator that commute the Hamiltonian, so you can actually go there and try to study whether, you know, so there are examples where a priori, you don't know what diffusion categories, can you figure it out?

01:17:05.000 --> 01:17:16.000
Can you figure it out. And then we work. And that's something we're trying to do with some students. Yeah, that's in two dimensions.

01:17:16.000 --> 01:17:28.000
So you've used this exponential formula to make unitary operators. In principle, there are other ways to put things together to make unitaries.

01:17:28.000 --> 01:17:37.000
Would you get the same results if you did something like that? I mean, if just a coordinate transformation.

01:17:37.000 --> 01:17:51.000
That's what I'm really asking. That's right, that's right. It is a coordinate transformation. Yeah. Oh, another way to think about it, you have to use the operator, they have some algebra among them, and what I'm really doing is computing a Lie algebra.

01:17:51.000 --> 01:18:06.000
And then I have this function that's defined on that sorry, computing a Lie group. And I have this function that's defined on that Lie group, and I'm just parameterized. It gives coordinate from that point of view. So however you construct this B.

01:18:06.000 --> 01:18:14.000
Then off you go. But but but but what some of the sort of subtleties that we are still trying to.

01:18:14.000 --> 01:18:23.000
understand it, what are actually the… so if I'm starting from scratch, I don't know anything. What are the rules? Do projectors count in this game?

01:18:23.000 --> 01:18:40.000
The answer should be no right? Because projectors themselves are not going to be things that are well defined when I take continuum. So so another question that we don't have complete answers of is that if I give you a random quantum system and ask you to go find all the symmetry that commutes.

01:18:40.000 --> 01:18:56.000
With the Hamiltonian, what are the rules of what applies? If I take any function of the Hamiltonian and commit to the Hamiltonian and I have infinite version of them. So basically, there's… Questions like that we're trying to sort out in in some detail. Yeah.

01:18:56.000 --> 01:19:01.000
What about integrative Hamiltonians? Correct.

01:19:01.000 --> 01:19:16.000
You can have fusion categories acting on that. But the fact that they're integrable just means that you can solve it, but it's actually nice, right? Yeah.

01:19:16.000 --> 01:19:26.000
Yeah, so good. So that's the case where you have continuous symmetries, completely continuous symmetries. Even fusion categories for continuous symmetry is not a settled point.

01:19:26.000 --> 01:19:32.000
even before you get there. But you have to be careful.

01:19:32.000 --> 01:19:36.000
Well, let's thank you very much. Well, thank you so much.

01:19:36.000 --> 01:19:44.000
It was a kooky talk, and I hope you enjoy. It was really, really interesting. Thank you.


