 with our afternoon session. So we'll have a next lecture by Chris Ackers about black holes and quantum error corrections. Thank you. All right, can y'all hear me? Well, this is working. Okay, good. So last time we set up this problem in the context of ADS CFT, where we had a Hilbert space of semi-classical gravity states in ADS and a Hilbert space of some CFT living at the boundary. And there was some, we didn't phrase it this way, but I would like to phrase it this way now. It's sort of the nice way of setting up the problem. There's some map V, some linear map that tells you for each ADS state, what's the dual CFT state? And in some sense, a question we're trying to answer is what are the, you know, what is this V and what are its properties and details? You know, it's something that we know in principle exists, but writing down a closed formula for it is tricky. We can sort of do this in certain examples, but even when we can write it down in certain examples, and especially when we can't, we would like to understand details about it. Such as the one given by this equation where, you know, we wanna know given some, say bulk state psi, so this is some psi in HADS, and some operator phi that's acting in ADS. If you act that phi on the bulk state and then map it to the boundary with V, for which, you know, when does, there exists some operator acting on the CFT and what does this operator look like when it exists? Such that this equation holds, where you could have first mapped psi to the CFT with V and then acted this operator O. So, you know, this is some compact way of asking the question we were asking last time about, you know, for which O's CFT is given some phi in the bulk, does this equation hold? You know, this, we would say that this CFT operator is reconstructing this bulk operator. And I'm gonna talk a lot more about equations like this as we go, but this is the basic question we're after. And we talked about reconstructing operators and what we know about it from the extrapolate dictionary and then what people figured out later, which is now called HKLL reconstruction, and these are nice, but they don't seem to be the full story. And we now think we understand more about the full story and the start of this is the so-called quantum extremal surface formula. So, let me explain that and then I can tell you how it's related to bulk reconstruction. So the idea is that this formula allows you to compute the von Neumann entropy of subregions of the CFT or the entire CFT in the case that the subregion is the whole thing. So, von Neumann entropy, V is a map from ADS to CFT. It does not commute with O, it's, it's, so, it doesn't, yeah, so the domain and the image are different Hilbert spaces, possibly different sizes. And so, it's an interesting question given some V, which O's satisfy this for a given phi. And this is sort of the question we're trying to understand. So, the von Neumann entropy is the following. So, let's say you're given some Hilbert space, HB. And on this Hilbert space, you have a density matrix. The von Neumann entropy is some quantity defined as follows. We might write it like the following. So, S of B with this little row subscript equals minus trace of row log row. This is some number and it has lots of nice properties. Is there a question? No, okay. Has lots of nice properties. You know, it's positive definite. And it's a nice measure of how much entanglement there is between B and whatever system purifies B. The purification, it's not very important for this, but it means the system for which B union and that other system is in a pure state. The QES formula is the following. So, the idea is let's say you, again, I'm gonna draw the cylinder diagram. Hopefully it's not too small. So, you're supposed to imagine, again, time is going up and we have some CFT living sort of on the boundary of this cylinder. So, this circle might be a time of slice of that CFT. And maybe I'll take this purple region to be some subregional called B. And I would like, you know, then the CFT is in some state that we'll call, we'll call capital Phi. So, this, the CFT on this circle is in the state capital Phi. And there's some, so the QES formula tells you how you can compute this quantity, where rho now is the reduced density matrix you get from starting with this state and tracing out B complement. So, B complement is the whole time slice is BB bar. And you're gonna compute it using the dual ADS. So, the formula looks like this. I'll write it and then I'll explain it. So, S of B in this state Phi equals, actually let me write this as Phi tilde for reasons that I'll explain momentarily. So, the CFT state is Phi tilde. And it equals the area of some surface that we'll call gamma divided by 4G Newton plus the entropy of a region defined by gamma. But I'm gonna call it B. So, this is, I might write this as, yeah, and this would be the entropy, the von Neumann entropy of some region little B. And a state we'll call Phi without the tilde. So, this will be a CFT state, this Phi without the tilde will be an ADS state. And what ADS state is that? It is, so let me draw a time slice. So, I'm gonna draw sort of, I'm just gonna pick a time slice in the bulk that ends on this circle that I drew here. So, B, remember was this region. The idea is, there exists some surface gamma that is homologous to B, so it ends on the endpoints of B. And the region, the homology region bounded by B and gamma is what I'm calling little B. And the total ADS state I'm just calling, I'll write it here, capital Phi ADS. So, capital Phi is the bulk state, capital Phi tilde is the dual CFT state. So, I'm gonna use tilde as in general to represent the CFT version of some ADS quantity. So, Phi bulk state by tilde, CFT state. And the idea is that this formula holds, gamma is the minimal quantum extremal surface. So, yeah, I'm just gonna use QES to represent quantum extremal surface. So, the idea is if you wanted to compute the von Neumann entropy of capital B, that could be difficult to plug it into this formula. So, you could instead, if you wanted to, look at the bulk dual state, find, you consider all surfaces gamma which extremize this joint quantity, this sum. So, when I say extremal, you're extremizing this A over 4G plus S bulk quantity. And if there's multiple gammas that extremize this quantity, you pick the one that minimizes this quantity among those. You're minimizing A over 4G plus S bulk, but only among the surfaces that extremize that quantity. And let me be a little clearer by what I mean by extremizing this quantity. I mean, you literally consider, gamma is defined by some embedding function, some surface in ADS, and you can consider local perturbations of gamma. And under these perturbations local to the surface, you want this quantity to be extremized with respect to this. So, some surfaces have this property. It's a little more familiar if you forget about the S bulk term. You just try and extremize, say, the area of the surface. It's still formally defined, but a little harder to do in practice, although people have done it, to try and extremize this joint quantity. But that's the prescription. So, this is how your, this formula is supposed to hold. And it's a very big idea. It was first written down in this form, this formula in 2014 by Engelhardt and Wall, but it comes, so it was conjectured to be the correct formula that you could compute the left-hand side using this right-hand side. But it wasn't just conjectured out of nowhere. It evolved starting from some conjecture by Ryu and Takiyanagi in 2006, which was sort of a, it's a special case of this. So, Ryu Takiyanagi sort of didn't talk about this term, and it wasn't, and their formula said this equals this term for the minimal surface, which is true, but only in time-symmetric situations where you don't have to consider extremizing also in the time direction. So, starting from this paper, there was a series of generalizations, climaxing in this one, and since then, our trust in this formula has only grown because for two reasons. One is that people have come up with derivations of this formula using the ADSEFT dictionary. So, some people like to put derivations there in quotes because it's not like a proof of this formula because they haven't proven certain aspects of the dictionary that were used, but they're very plausible assumptions that went into driving this formula. The other reason people like this formula, perhaps the main reason, is that people have used it to compute answers to questions that we really like. So, it turned out to give us answers that we really like, so therefore, it's probably true. And this is sort of the logic behind these 2019 page curve calculations that you've heard of those, maybe I'll be able to mention them later, but they used this formula to compute the page curve. So, if you try to compute the entropy of the Hawking radiation using this formula, where B here is now the Hawking radiation, you compute it using this QES formula, you get the page curve, whereas if you had just computed it sort of the naive way, you wouldn't have gotten the page curve if you had gotten the so-called Hawking curve where the entropy goes up and up and up and you get an information problem. Anyway, so this formula gives you the page curve, that's really nice, so now I would say it's very trusted. The second term is supposed to be divergent, right? Yes, good. Let me comment on that, thank you. So, the second term, because the bulk is a quantum field theory, von Neumann Entropies and quantum field theories are divergent. So, the idea actually is that this quantity together is more well-defined than either separately, and that this one over genutin, the bear one over genutin is also infinite, but is the counter term, includes the counter term, with the divergence here, so that the sum is actually some well-defined finite thing is the idea. So, in string theory, the weird genutin is finite. Good, yeah, and then I guess the second one, it's not clear if it's finite, right? I would, so in string theory I would hope that probably you would end up getting finite answers from what you would, in that context, call the bulk entropy. Probably you would. I see, okay, so you think that it will be finite. Yeah, I think it would be finite, yeah. Very simple one. I'm a little uncomfortable with the definition of a quantum extremo surface because basically, it seems to me that you have this quantity which you identify with an entropy, and then you take the minimum of an entropy, and usually we take the maximum of an entropy. So, can you explain this oddity? Why is that? So, yeah, these contexts where you take maximums of an entropy, maybe what you're thinking of is some coarse-graining procedure, where you say, have some data about a state, and you pick the state that maximizes the entropy with respect to that coarse-grained data. That's different from what we're doing here, so where all of these are, in some sense, fine-grained quantities, where we're not throwing anything away, we're keeping all the data. Yeah, so it is an interesting question why the minimum shows up here. Why is it the minimal quantum extremo surface rather than the largest one? This is one that, this is a question that interested me for a long time, and in fact, one of the theorems we'll get to is trying to explain exactly this. What's the idea behind why it's a minimum? Yeah. Yeah, can you also ask a question? I think you said that the combination error plus the entropy of the fields in small b is finite, but I guess in general, an entropy in the CFT should be infinite, you always have a cutoff. Yes, so you're saying, good. Yeah, this is UV finite, but not IR finite, because this area goes all the way out to infinity. It actually diverges, and this diverges because there's infinite area as you go out, and it's this IR divergence on the right-hand side that matches the UV divergence you get on the left-hand side. I see it. Yeah, so it's a very interesting thing. People have understood in many contexts this sort of UV-IR relationship between the two sides, yeah. Sorry, very simple question. Given the region of small b, how to compute the bulk entropy? Yeah, so you have to just use your favorite scheme for computing the bulk entropy. So this is formally the answer. I think what you're getting at is, I said in some sense this could help you compute the left-hand side, but we also have an entropy on the right-hand side, so it doesn't seem very helpful. That's true. It was very helpful in the re-talking-augy version when we didn't have to worry about this. Now that we have this guy, we've just swapped computing that for this. That said, sometimes it is easier because when this is a strongly coupled theory, this can be a weakly coupled theory, and so there might be techniques available to weakly coupled theories. Another reason it might be useful is that if you consider, if you want an answer from the CFT point of view and like a one over an expansion from the bulk point of view, some genutin expansion, this term being divided by genutin is very large in the semi-classical limit that genutin is very small, whereas this term in general, well, we will consider situations where it can be very large, like it's an entropy of a black hole, but often you can take this to be the entropy of some matter excitations, which is order one in this genutin expansion. So this is a subleading correction to this guy, which he's easy to compute. But that said, it's not in general a simplification, it's just the formula that was given to us by ADS-CFT, and it turns out that it's going to be useful for talking about reconstruction precisely because there's an entropy on both sides, actually. Do you need quite detailed information about the bulk, right? Because I mean, in the real technology formula, you just need like the metric, you, and it's not like geometric information, here you need to compute something. Yeah, this is much harder to use. Much harder to use. I think when it was originally written down, yeah, Netta and Erin thought that no one would ever be able to actually compute the left-hand side. I think that's why it was kind of a, no one would ever be able to compute it in a situation that was much more complicated than some expansion. I think that's why it was so surprising in 2019 when people actually used this, and this effect was very important, this term was very important, to compute something interesting. Is it clear that Gamma is unique? No, yeah, good, so it's not. So in general, there will be multiple Gammas that extremize the right-hand side, and sometimes there are multiple that have very, you can get in situations where it's degenerate, which one's the minimal. Maybe that's where you're getting at. Yeah, yeah. In those situations, this formula receives corrections that are in general order one, so sub-leading to this leading guy, but maybe at the same order as this guy, but generically, it can be very large, even whenever you need. So this formula actually should be, is not exact, it's always an approximate formula with non-perturbative corrections here, and these add up in a conspiratorial way whenever there's another Gamma that's very close to the one that you're considering, not close position-wise, close in this value, and those corrections can be very large in general, and so some of the lessons were, that's gonna be very important for some of the lessons we'll draw later. For the first lessons, you're actually segueing perfectly into what I wanted to say next. This formula and what it says about reconstruction were first understood in a context where we don't have to worry about this degeneracy, where we can just, which is pretty generic, most situations only have one Gamma that's the minimal QES, and others are very far away from it in this value, and those situations, there's a nice theorem that I'll write down that explains exactly what this formula tells you about reconstruction, so maybe I'll say that right now, unless there's more questions, these are great questions. The quantum extremal surface formulas lessons are easiest to understand, and were first understood in a restricted context, which will be the context for this lecture, and this restricted context is what I'll call a fixed QES subspace. I wanna be so clear about what this context is that I'm gonna write the whole definition, so it's a subspace, the bulk Hilbert space, so sometimes when I say subspace, people think I mean some region of ADS, like some region of this diagram here. I don't mean that, I mean literally a subspace of the Hilbert space of bulk states, so something like the vacuum state plus the first however many excited states. A subspace of HADS, position quantum extremal surface is approximately the same every state in that subspace, so I hate to erase this because maybe some of you have already copied this down. I'm just gonna say subspace, I don't know, let's call it S of HADS, such that the position of the QES is approximately the same in every state psi in approximately, I mean it doesn't have to literally be the same exact position, but the fluctuations in its position will be small relative to all scales of interest in the problem. So in particular, I'm gonna be okay, we'll consider situations where the deviations in the position are of order plank length. These subspaces are easy to find, it's not excessively restrictive, because for example, if you consider like vacuum ADS and all the states you can obtain by acting some order one number of light matter operators, it will be one of these subspaces. And this is what people sort of mint for a long time, whatever they said or it's like, let's consider a code subspace, if you've heard that phrase. So in this subspace S, I don't want there to be black hole states and vacuum ADS, things with macroscopically different geometries. I do want just like one, basically one background, like the vacuum ADS background, plus maybe some small perturbations on top of that. And you can construct these subspaces. And the reason why, let me just say in words, if you start with vacuum ADS and this is a time slice of vacuum ADS where I have some sub-subregion B, oh, I didn't say this here, but when I say fixed QES subspace, I really mean with respect to some sub-region B of the boundary. So it has some quantum extremal surface, maybe this quantum extremal surface gamma is the minimal QES in this say vacuum geometry. And however many light excitations I put, if I just put an order one number of them, they're not gonna change the position very much because they're just gonna back react and change the metric sort of higher order in genutin. And the surface is extremal, so it's gonna be an even higher order in genutin correction to its position. And ultimately if you just put like an order one number, so not scaling with genutin number of excitations, the geometry's not gonna change very much and the position won't change by more than order plank length or so. And we're only gonna consider reconstructing operators that are maybe here. I'm gonna consider some phi here and ask, can I reconstruct it in this sub-region B? And then this phi I'm gonna take to be further away from the QES than a plank length. So this distance will be much larger than this plank length. So these fluctuations in the position aren't going to concern us or change the answer. Later on we're gonna worry a lot about lifting this assumption of working in these subspaces, but yeah. I have a question. So by this definition do you mean that all the states in S have the same geometry? Or you admit states with different geometry but such that the curve is the same? Yeah, let me do the more conservative option of having the same geometry at leading order in genutin. So I'm gonna consider metrics that can take some genutin expansion and the first piece of that, the G zero piece will be the same, but then the higher orders, the ones that come proportional to genutin or genutin squared, those can be different. Okay, so I have sort of a general question about the systematic expansion for this formula, meaning if you are going to ignore order one, I mean if all the order one fluctuations are going to amount to the same quantum surface, does it really make sense to include all the one correction in that? Good, good, good, yeah. So I will take into account how these order one excitations change the area and what they won't do in a way that will matter for this problem is change the position. So yeah, this is important. So because the area in this formula comes enhanced with this division by genutin, a change in the area at order genutin, as you're saying, will show up in this formula at order one, I will take into account those effects. And so it will be consistent with me to add order one contributions from this guy. But the position won't change at that order because, and the position I'm not considering any sort of genutin expansion where it's enhanced by genutin. Yeah, so if you make some metric perturbation, you could imagine the area changes sort of from two effects. One contribution is that, at leading order in this perturbation, the position is changing in the background metric and the other contribution is that the metric is changing at the current position and will contribute to this A over 4G at order one but the position, yeah, what do I wanna say? Yeah, what I wanna say is that if you take this background metric and you add perturbations to it that are proportional to genutin, those can change the position of the surface a little bit. So, and an amount proportional to genutin, but I'm gonna consider scales of the problem like trying to reconstruct operators that are sufficiently far from it, like some order one distance, that this change in position by order genutin won't matter for the question of is this guy still inside gamma or is he outside? Yeah, but the correction term will matter. The correction term will matter to this one. Yeah, I will add those in. So it's in this context that an amazing theorem holds and there's sort of a long list of papers that built up this theorem and I'll put them all in the lecture notes, but this theorem in its form that I'm gonna write it was written down in this paper by Daniel Harlow in 2016. So it's sort of, this paper takes the developments from some previous work over the couple years prior to this, adds in one piece and then writes it all as one nice theorem and so I'll write it. Let me say this theorem is worded for simplicity in terms of finite dimensional Hilbert spaces. This is the context that, this is how this theorem was written here and you might worry that this doesn't apply to ADS CFT because both have infinite dimensional Hilbert spaces like the CFT is some field theory and that's a valid concern. So there's two ways you might think about this. One is you could just place both theories sort of on a cutoff, you could put your CFT on a lattice and then this theorem would apply there. I don't think you would lose anything qualitative about what we wanna learn by doing that, but if you want you could also use the more appropriate language without placing any cutoff of working with these infinite von Neumann algebras and you could ask does a theorem like the one I'm about to write hold in this infinite von Neumann algebra setting and the answer is yes. So this theorem as I'm going to write does hold the infinite von Neumann algebras setting. There have been a number of nice papers that came out after 2016 that generalized it to the setting by Tom Faulkner and Monica Kang and David Kolchmeier and others. So the basic philosophy, nothing qualitatively seemed to change when they took this theorem and rewrote it in that context and generalized it to that context. So my philosophy is going to be to present to you the simpler to understand one that's about finite dimensional Hilbert spaces and just tell you if you want that this one generalizes to that context. Theorams I write later are not yet generalized to that context but I expect that they could be. But my philosophy for presentation will be just show you the finite one. It's easier to understand. So the idea is we're gonna let H be some Hilbert space that factorizes and this Hilbert space, you should regard as like the CFT Hilbert space. So I'm gonna draw a picture here sort of guide the eye for what this theorem means. So H is like the, so this is like some circle that's like the time slice of the CFT and B might be this subregion and then B bar is the complementary subregion. So there's this Hilbert space and there's also a second Hilbert space which we'll call H code. It's called H code for historical reasons that will become clear as we go. And it also is imagined to factorize into H little b tensor H little b bar. This you should think of as like the fixed QES subspace. This is a subspace of H ADS. And this is like morally little b is going to be that region contained between the quantum extremal surface in B and little b bar will be its complement. Both of these will take to be finite dimension Hilbert spaces. And also we're gonna let just like in the ADS CFT setting be a map from H code to H. And the claim is that the following two statements are equivalent. You'll have enough rooms right both but let me write one of them here. So the first is some statement about operator reconstruction and it's this statement. For all phi B and respectively to phi prime is just a different phi. It's not like some derivative of phi or anything. There exists their OB. So this is an operator capital B and respectively for all states psi in H code. The following equation holds. So let me write it here. Psi equals V phi B psi. And also the analogous thing when you replace and you have to put the primes on there, I guess, as I've written it. And then also with the Hermitian conjugates there. Yeah. Just a quick clarification. So H code is a subspace of H ADS or the image of H, I mean, is H code the same as S? H code is the same as S. Okay. Not its image. It's really a subspace of H ADS. Yes. Yeah, I think I know why you're asking this question because sometimes people take it to me in the image. I'm taking it to me in this, but yeah. Yeah, sometimes people say, oh, well, they're isomorphic, so I might as well mean the image. And I think that's fine, but that's not what I mean. Is the space in the gravitational side? Yes, I have factorized both. And I think what you're pointing out is that this is a sort of illegal thing to do in gravity and also in quantum field theory. But I'm taking both. So at the level of this theorem, I'm just defining these to be factorizing Hilbert spaces. And you can take this to be a model for ADS CFT and you could then ask after I finished writing this theorem, well, does this hold when I make this and this more realistic Hilbert spaces? And the answer is, yes, people have generalized this to those settings, but nothing qualitatively changes, so it's easier to understand here. Two conformal feature in the left and in the right, but HB tensor to HB bar is for one conformal theory? Yes, right now I'm not thinking about having two separate boundaries and two separate CFTs. I just have this one CFT living on this circle. And yeah, I've factorized it into B and B complement. And so the statement here of condition one, I haven't finished the theorem. So the whole point of the theorem is that this is equivalent to a second statement. But this statement says that this region little b, any operator here can be reconstructed by an operator on capital B. And also any operator and little b bar can be reconstructed on capital B bar. So I need to write the second. So let me... Sorry, maybe this is the same question that you should ask, but I didn't understand. So is B injective in this theorem or B can be anything? Sorry, I did not say that. This V and this theorem is an isometry. So yeah, it's every guy. So there's a unique... Yeah, there's a certain image here. It's not one to one. This could be much bigger than this. Yeah, so this has some image. By isometry I mean all of the inner products are preserved. Okay. Yeah, so I could write, sorry. So it's an isometry on the image. It's... Yeah, let me... Sorry, if I can write it here, yeah. So it's like for all Psi and H code. Okay, so on the image. Yeah, Psi Vdagger V Psi equals Psi Psi. So it doesn't change the inner products. And this is only possible if H is at least as big as H code, but it could be much bigger. So yeah, let me try the second condition, erase this picture and just gesture maybe to this one. This is little b and this is a little b bar. So it's the same picture, so I just, I need to comment here the room. So the second condition in this theorem, the one that's equivalent to one, is that there exists some operator on H code such that for all density matrices on H code, the following equation holds. And it's supposed to remind you of the QES formula. So if you wanted to compute, so if you took row and you mapped it to state, you have the dual CFT state, you would write that as V row Vdagger. This V actually kept part of row to send it to the boundary, like the CFT Hilbert space and the Vdagger ax from the raw part of row to do that. So this is like the CFT version of row. And this equals, this is like the expectation value of a hat, you know, morally, you should think of as, this is like the a over 4g and then plus s of little b in the state row. And this is supposed to be like exactly the same thing. This is like the s bulk term. It looks so similar, I'm not even gonna bother saying what it's like. So this is supposed to be, and also likewise when you replace b and b bar, little b and little b bar. So this is the second condition, I've now fully written it. So the point is that if you took some state to find an h code, that defines some state on h by mapping it to h with v. And then now you can compute the von Neumann entropy of this factor, say, and it would equal this thing. So you could, this is a formula that you can use. It's like a bulk formula or like a formula using quantities from h code. And, you know, this a hat doesn't depend on the state. That's so important that I'll write it. So let me circle this a hat. Sorry, if it is getting jumbled, I just wanna say that this depends on v. If you change v, that changes what the a hat operator is. But a hat does not depend on row. It's independent of row. So this is, that's just like, the operator that measures the area of the QES is some operators independent of the state, right? It's expectation value can change in different states, but it's some operator. This is supposed to look like, I mean, it's supposed to be the QES formula. So what this theorem is supposed to tell you is that take a context like ADS-CFT now where we know something like this holds, then we can learn that one is true. So if we have some situation where we had our subregion b and then we could find its quantum extremal surface gamma, then we can learn what this theorem is telling us is that any operator that's in this homology region between b and gamma can be reconstructed using just capital B. That's the lesson. Which is a nice lesson because HKL formula told us something like using the Rindler reconstruction version, you can use, you can find the operator on b that reconstructs any operator that's in the causal wedge. This, the region defines by the quantum extremal surface is at least as large as the causal wedge, almost always bigger. So the causal wedge might be something like this, this dashed line, the quantum extremal surface will be outside of it generally. So some operator like this, phi that lies between them, this tells us that there is, there must be a way to reconstruct phi using just b whereas we and HKL didn't do that for us. So it wasn't giving us everything. This is telling us there's more you can do. And the amazing thing is that this tells you not just that there must be a way to reconstruct all the way up to gamma, there's no way to do better. It's not, no one's gonna later figure out that there's some way, if you're really clever, to, you know, there's some operator here. Sorry, like you said that you're structing two cases where this gamma only shifts by a plank length and operators which are much more than a plank length from gamma. So like, can you really say for sure that you can now reconstruct operators that are like further than this causal boundary if you're only allowing that causal boundary to move by a plank length and you consider operators that are more than a plank length away from it? Yeah, yeah, good question. So this distance, so this boundary of the causal wedge and where the QES is, are generally extremely far apart, macroscopically far apart. So there's an enormous, like not at all related to the plank length region that this is telling you you can reconstruct operators in. And the scale of the wiggles of the QES are very small in this picture. So, yeah. And what I wanted to say was that this theorem is very nice because these are equivalent. It's not just like two implies one. One also implies two. What that means is if there was some operator here, five prime, if you could just cook up a reconstruction scheme that allowed you to represent five prime as an operator on B, that would contradict this theorem. Because this theorem tells you, it's saying whatever the region is that you can reconstruct, that's the region that shows up in this S of little b. So whatever, you know, so if you can reconstruct operators here at this point, then this point better have been included in the, you know, this bulk part of the QES formula. So it's an if and only if, and that's why it's so nice. It's giving us a guarantee about exactly what region of the bulk your subregion can reconstruct. It's this region bounded by the QES and no more. I'm sorry. Yeah. Since I'm still a little bit confused. So are you saying that since the theorem doesn't really characterize, I mean fixed capital B, the theorem doesn't really characterize what is small b. Are you saying that if small b is a little bit different than what you draw there? Then for example, statement one, you would never be able to find a V map that does what you want. Is this what you're saying? Ah, no, so good. So the setup includes a V. So V is given to us in the setup. So I could also choose a different small b than this one. Yeah, good. So let's say you chose a different, like, yeah. So let's say, yeah. You're looking at this theorem and you're saying, let's imagine, let me use a different color, some orange region, orange region here. It's smaller than gamma. Second? Also a different region. I mean, b is not really characterizing the theorem. So you just say that h code can be factorized. So I can also take a completely different small b. A different small b you're saying. Yeah, you're saying like maybe this region here. This could be small. Something like this, yes. Yeah, so what happens is if you take this to be your small b, neither one or two will be satisfied. Okay. And so that satisfies the theorem because they either both have to be satisfied or both not satisfied. Okay. Yeah. So it just so happens that they're both satisfied only in a situation where little b is exactly the region bounded by the QES. Which still is not correct. Like, yeah, it really means that if I take just a different small b, then like, even the first statement doesn't make any sense. Like I cannot find this b. Yeah. So if you take, yeah. So this gamma doesn't, yeah. I think what you're getting at is gamma doesn't show up here. I never wrote gamma. Exactly. So gamma here is characterized by, it's just in this theorem implicitly as defining the only little b in which one and two are both satisfied. Okay. Yeah. Okay. Yeah. Thanks. Yeah. So this segues very nicely into a point I wanted to say. So thanks. So what is, I've written this and I've said this is like the QES formula and it tells you exactly the region you can reconstruct. But you could complain because the QES formula that we wrote over there really has this condition that's important. It says gamma has to be something that extra-mises this quantity and among those it minimizes this quantity. And I didn't write anything about extra-mising or minimising here. And that's true. The reason it doesn't show up actually comes from the kind of restrictive assumption that we made where we were in this fixed QES subspace. So the QES gamma, the minimal QES gamma is just always the same for every state. So it's just sort of implicit in the theorem. There doesn't have to be any extra condition or wording about how you find gamma given a certain state. There's going to be a theorem too that I'm going to write down next lecture that lifts this assumption. So it's just the same theorem, lifting this assumption, which is very important because this doesn't allow us to talk about black holes very well. So we wanna talk about black holes. We lift this assumption. This theorem changes and you have to include now a condition that tells you it's gonna be the minimal surface. So there's gonna be a way to characterize gamma explicitly. Thanks. Okay. Can we understand this as just a theorem about finite dimension Hilbert spaces? Yes. Yeah, so if you could explain that, it seems to me that the operator A hat would be trivial in this case. Yeah, so yeah, good. In that, right, if you, the two entropies will just agree you don't need trace row A hat for finite dimensional. This term will be non-zero. I'll give you an example of this. That's okay, yeah. You could, yeah, so you're saying let's take a Hilbert space and a subspace of it. Yeah. And let's calculate. Yes. Yeah, so yeah, can you give an example? Yeah, yeah, so this is an example I was hoping to get to. But maybe it'll be very, it'll be formatted for me to do it right now. So yeah, a very simple example of this is the so-called three-q-trit code. And this is a code, this is a quantum error correcting code and code is showing up just like H code over there, where the idea is you have one three dimensional Hilbert space. This H is nine dimensional. So it's like we are embedding one q-trit into three q-trits. And I can tell you, so to finish telling you the setup, I need to tell you what the V is, the map between them. And it's defined by saying zero goes to zero zero zero plus one one one plus two two two. I'm using a basis for this, that's like three q-trits. So three three-dimensional systems. And I have to tell you, sorry, and this is also divided by one over square root three. And then one goes to, it's something, it's zero one two and then some permutation, I'm happy to, let me get the exact thing. I was gonna go through this anyway, so this is a good time. It's zero one two plus one two zero plus two zero one also divided by square root three. And then two gets embedded as zero two one plus one zero two plus two one zero. And so you can check that this is an isometry. So all, you know, the left hand side guys are all orthogonal, of course. The right hand side guys also are. And so this is some three-dimensional subspace of the full 27, sorry, if I say nine, I meant 27, because it's three cubed. So this is some three-dimensional subspace of the full 27-dimensional Hilbert space. The point is that, well there's a lot I could say about this for now, just say the thing that directly answers your question, which is, so this V to be clear was a map that I might say is like a map from some Q-trit Q into map, into A, HA, tensor HB, tensor HC, and in particular maps, any state psi, which, you know, general is like, this is a general way for writing a state psi, maps it into some, you know, by V goes to some state psi tilde on ABC with the same coefficient CI, but now I'll write it as I tilde. So this guy here, let me define as zero tilde, one tilde and so on, and two tilde. So this, so V embeds state psi into ABC like this. And then now the question is, so okay, so I claim that if you take, say A and B, both one and two are satisfied in the following sense that, so for example, first let's see two. Yeah, so my decomposition is I wanna call this first guy A, this second Q-trit B, and then this third one C, sorry, yes. So H sub little b, this is like H sub little b, and then now I wanna take, I used B in both cases, sorry, that's gonna be confusing. I want AB to be what I was calling B there, and I want C to be what I was calling B bar there. Sorry, everyone. So I might make this, maybe a picture clarifies it or not, but if I were to draw this like ADS-CFT, so this is like a time slice, I might try and put HQ, so Q is like some Q-trit in the middle of ADS, and then is like this region, B is like this third, and then C is like this guy, and then I wanna take what I'm doing in that language is taking these two and calling them, calling them what I was calling B, B boundary. Sorry for the notational ugliness. And so my claim is that in this model, with this choice of capital B, both one and two are satisfied with Q as like little b. So that means in particular that if you were to compute the entropy of AB in any state, in any state, psi tilde, that it would equal trace of rho times some area operator plus the entropy of this little Q and psi. So this is my claim, and my claim is that this is not zero, it's actually log three for any state. What if it could be bar, I mean the... Yeah, little b bar in this example is trivial. It's trivial, nice. So this is my claim, let me try and just give you some intuition for why it's true, maybe without going through the full proof. So the intuition is that you have these i-tildes for each i as we go over there, and there's a very non-trivial fact about v's like this, which is also true of v's like the ADSCFT dictionary. The non-trivial fact is that there exists some unitary, it's a unitary operator that acts just on AB, so it's trivial on C, that puts this i tilde, when you act, I hope this is not too small, UAB acting on any i tilde equals i times some chi, where this i is on the A register, and then this chi is some fixed state. It actually here is the maximum entangled state between B and C. So this is an incredibly non-trivial claim that I, if I have time, I don't know if I will, when did I start, yeah, definitely won't have time. I would prove this, there exists a U that does this. And so the idea is that if you wanna compute the von Neumann entropy of A and B, acting a unitary on A and B doesn't change the answer. So you could take the reduced-density matrix row AB conjugated by any unitary you want, in particular this one, and then it would put it into this form, and this chi guy will factor out because he doesn't depend on i. And so then when you compute the entropy of AB, you're gonna get the entropy in A, which will just be whatever the entropy of Q was, because he just inherits the state of Q, plus the entropy of B, which in B is just maximally entangled with C, and that's what shows up here is this extra term. So this is all sort of trying to be a big answer to your question, where does this guy come from and why is it D zero? And the answer is that in general, when you have isometries like this, there's just, you're tacking on some extra part of the Hilbert space, an extra factor, and that extra factor can carry entangled, and that's the area of these. I just ask one thing. Yeah. So if B bar is trivial, then that A is B zero, is that right? Little B bar is trivial. Ah, good. It could be zero if the bulk was in a pure state, but I could also take some state in which Q was, say, the mixed state, and then. I see, so it mixed with something else, okay. Yes, yes, yes. Can I ask a question? I have not understood whether the theorem one assumes the existence of V before stating the theorem, or V is part of the statement of the theorem. It assumes the existence of this V. That's the structure that's input to the theorem. That's why I'm a bit confused about the content of the part one of the theorem. Yeah. If V is injective, it's an esomorphism with the image. So I could just define O B as V phi V inverse. And the. That would, in general, give you an operator on B and V bar. This is an operator on the image of V. But then this is a finite dimension on the space. It's an operator on the image of V, but this operator only has support on this factor. That's the non-trivial content of one. You can do it acting just on one factor and not the whole. I see, okay. So the non-trivial content is that the operator which are defined in this way is not supported on the full H, but only on the left factor. That's right, that's right. So the content here is that this operator phi can be reconstructed in just the sub-region. You don't need the whole. Can I ask, so basically V doesn't respect the decomposition necessarily, is this the point? V doesn't respect free composition, is that what you said? The decomposition into HB tensor HB bar. Like if you map HB, little b, with V, you don't know where it lands. That's right, yeah, because V in general, yeah, it acts on these two guys together and does something, who knows what, with it. So one thing I want to say is there's this theorem that we've been discussing, and it has a very important implication let me try and convey the idea of this in the next few minutes. So the idea, which I've sort of already alluded to by talking about this example that is a quantum error correcting code, will be that quantum error correction is important for talking about properties of this V, this holographic map, that if we want to understand how the holographic map has the properties it has, a nice way of doing that is using the language of quantum error correction. The idea is the following. So what we've learned here is that if you have some, this is a time slice, if you have some subregion capital B of the boundary, and let's say we're starting in the vacuum state or something, this is its QES, if you have an operator inside, you can find a reconstruction on capital B. If you have an operator outside, say this phi here, you can't find a reconstruction on capital B, but you could find a reconstruction on this region, B bar. Now let's divide it into thirds, just for fun. But maybe I'll call this B, this A, and this C, and now it's gonna look a lot like this, this example as sort of a model for what we're about to learn. So the idea here is that just like we said before, when we just had B, B can't reconstruct phi alone, A can't reconstruct phi alone, and C can't reconstruct phi alone. I've drawn all of their quantum extremal surfaces. But if you took, say, A and C, their quantum extremal surface, so the quantum extremal surface of the union of A and C is this guy, it's the same as B. So A and C together can reconstruct phi but not either individually. The same is true for A and B, they can reconstruct phi together but not individually, also B and C. So no one can reconstruct phi, but any two can, this is actually a big deal. Sorry, maybe I can just have two extra minutes. So if we let their reconstructions be OAB, OBC, and OAC, you could ask, are these the same operator in the CFT or are they not? Assume for the sake of contradiction that they are the same. All are the same operator O in the CFT. Then you can derive a contradiction by saying, well, OAB clearly commutes with any local CFT operator already this capital phi because they're just on different factors, they're space-like from each other. And likewise, OBC clearly commutes with any operator here, so this might be phi C, this would be phi A, OAC commutes with any operator on B. If they were all the same, then this would be an argument, if you're careful about it, it would be an argument that tells you that this operator commutes with every local operator in the CFT. And that's a problem because by very rigorous arguments, using Scherzlemma, it would tell you that this operator must be proportional to the identity, which is bad because we don't think that all of the bulk operators are trivial and proportional to the identity. We think that their bulk operators don't commute with each other, for example. So that would be a contradiction. So this tells us that we shouldn't think of all bulk operators, sorry, we shouldn't think of these three as the same, they're all different reconstructions of the same bulk operator. This is, so why is this, why did, how did we end up learning that one bulk operator has different representations in the CFT? The lesson is supposed to be, let me just say it, the resolution is that we should only expect these operators, they're different CFT operators, they're the same acting on this subspace that we started with. So what we're calling H code or this S, which is this fixed QES subspace, you should regard these O's as all being reconstructions of phi that do the right thing when acting within S. We're not guaranteed by this theorem or anything that these operators are the same or do the same thing when acting on states outside of S. So they're the same when you've restricted your attention to just this subspace, S. So the point here is supposed to be that this theorem had this assumption of having like this small fixed QES subspace. And then it promised us that capital B could reconstruct operators all the way up to the QES. And you might say or wonder, do these operators I get actually just reconstruct phi perfectly? Are they just, we had this assumption of fixed QES subspace, but maybe they work outside of that subspace. Maybe they just always work. And this contradicts that. It tells you, no, you gotta be careful. These operators that you're guaranteed reconstruct phi can't work in every state in the CFD Hilbert space. They only work in some subspace. So you should really only take them to be the operator reconstruction in this H code. If you want to start talking about reconstructing operators more generally that work in subspaces larger than this fixed QES subspace then you're gonna have to be a little smarter. Maybe have a theorem that doesn't have that restriction placed in it. And that's what we're gonna do next time. So we're gonna realize that there's more to life than fixed QES subspaces. When we have black holes, that's gonna be very important because general black hole states, if you have a black hole, different states of the black hole will very often have different QES's. And so if we wanna have reconstruction in the context of black holes, we don't wanna limit ourselves to fixed QES subspaces. So we're gonna have to prove a whole, we're gonna have a whole different theorem without this assumption to tell us what we can and can't reconstruct. That's next time. Thank you. Okay, maybe we can have one quick question and postpone the rest to the discussion session. So is the HKLA reconstruction different from all of these? Yes, so the HKLA reconstruction will in general not allow you to reconstruct all the way up to close to the quantum extremal surface. HKLA uses the whole boundary, right? You can do this granular reconstruction to use a subregion, but then the subregion that you are able to use has to be quite large relative to what this tells you you can use. If you use the whole, yeah, if you wanna use the whole thing, then HKLA will often work just fine. But it won't work for us whenever we, yeah, I didn't mention this, but some context that we'll care about are like context in which we have the black hole information paradox where we have say a bunch of radiation over here. And we wanna ask, when can we say use this radiation to reconstruct the interior or like when can we use this CFT to reconstruct the interior? And HKLA won't help us do that, but this will tell us that it's possible and give us a different way of doing it. Okay, let's increase again. We have a break until 3.30, and if you can...