Jorge Pullin: Okay. So our speaker today is, I'm on Lagosite, who will speak about entanglement on holography and spin networks.
Simon L: Thank you for the opportunity to speak here.
Simon L: So today, I would like to give a sort of a big picture overview of some aspects of calculating entanglement entropy.
Simon L: particularly in gauge theories, and that is gauge views and spin networks, and in the end I would like to apply this to a holographic reconstruction
Simon L: in the context of reconstructing intertwiners from boundaries. Links in spin networks
Simon L: at
Simon L: this is partly based on work with Daniela Oditi and Diogenia called Francesca.
Simon L: So to begin this talk, I would like to give a
Simon L: sort of general overview of algebraic definitions and methods in understanding different contributions of
Simon L: entropy that can it go into entropy calculations in particular, choices of sub algebra, and how they affect what you're going to calculate.
Simon L: Then I would like to specify the discussion on spin networks, the types of algebra that you have there.
Simon L: I would like to into exemplify how you can
Simon L: see some parts of entanglement in spin networks as being inherent to the graph structure itself.
Simon L: and then also talk about the an entropy of a generic sub region in a generic state.
Simon L: and how to calculate this using techniques from random tensor networks.
Simon L: Then in the end I would like to apply the results from that section to see whether there's a sense of a notion of
Simon L: holographic reconstruction for the fixed spin spin network case, and then also hopefully discuss how the how that gets generalized to the
Simon L: a non-fixed spin case.
Simon L: which is more recent work.
Simon L: So to begin I would like to record the very basic notion of entanglement that we all know.
Simon L: which is that if you have a tensor product structure in a in a Hilbert space, then anything that is not a tensor product, state or separable state will be known as a tangled
Simon L: and attachment. Entropy is known as a measure that tells you how much a State deviates from such a separable state.
Simon L: This is all fine if you have a tensor product structure. But if your subsystems are and do not admit such a decomposition.
Jerzy Lewandowski: Should I see anything?
Simon L: Yes.
Jerzy Lewandowski: Except for the title, Boundary conditions, Algebra.
Simon L: Yes, this is still fine. This is still fine. Yes, don't worry.
Jerzy Lewandowski: Oh, okay. Yes.
Simon L: This is just a preamble
Simon L: to that section.
Simon L: So if this is, if you do no longer have a tensor product, search of Hilbert spaces. You rather need a notion of subsystems in terms of algebra.
Simon L: and the point of view that we will take here is that we describe a system in terms of sub in terms of algebras with units.
Simon L: So we will see those algebra as being the accessible operations on a given system
Simon L: in particular. If you have a subsystem, then it will be described in terms of a sub algebra with unit.
Simon L: So the operations that you can perform on a subsystem
Simon L: for entanglement. Then we will need a notion of complementary subsystems which, in this algebraic context, just means that you have an algebra Al and an algebra AR,
Simon L: with a letter of which is known as the as the commenting of the first.
Simon L: and of course vice versa.
Simon L: If you have that the algebra generated by these 2 algebra sub algebra's can be written as a tensor product. Then this setting gives you back the original naive, simple Hibbert Space definition.
Simon L: and that you represent it. And then you have a tensor product structure on Hilbert spaces, and you can see complementary subsystems as just tensor product factors.
Simon L: However, in general those 2 complimentary subsystems will have a non-trivial intersection
Simon L: of operators which commute with everything, and so which are known as a center.
Simon L: This in particular also means that the full algebra, that it cannot been written as a tensor product.
Simon L: What this means in terms of representations is that we can diagonalize all of the operators in the center.
Simon L: And so we can write the full representation as being of as consisting of block diagonal matrices.
Simon L: Each of the sectors in the decomposition, for example, on this layer level of filbert spaces
Simon L: will then have a tensor product decomposition.
Simon L: but will be a direct sum over all of the labels, all of the eigenvalues of the central operators.
Simon L: The the further furthermore, this is important because density of matrices in this subsystem, so the data that consists just of the 2 regions
Simon L: must commute with the center, and so they also feature, direct some decomposition, a block. Diagonal decomposition.
Simon L: which makes us makes it possible to interpret them as classical mixtures of states which live in each of the blocks separately.
Simon L: Then
Simon L: there's a certain type of subsystem which is very important in the case of lattice, gauge theories and gauge theories more general or field theories even.
Simon L: which is that of a finite region. For example, a spatial one, on which you impose some boundary conditions.
Simon L: Generically speaking, you can implement these boundary conditions by choosing a certain sub algebra of boundary operators. So bound operators which are supported on the boundary of the region.
Simon L: For example, in the case of a scalar field theory, you have on this corner
Simon L: operators given by the fields and their contributor.
Simon L: and if you remove from the algebra
Simon L: the momentum operators, you only have the commuting field operators left.
Simon L: and then, because they all commute. They form a center, and you can diagonalize them and give them fixed values. This is what you would call Dirichlet boundary conditions.
Simon L: So overall you can connect choices of boundary sub algebra with choices of boundary conditions on the system.
Simon L: In turn. Then.
Simon L: as we just see
Simon L: that the this implies that the Hilbert space has a direct Sunday composition.
Simon L: and that that similarly density matrices do as well.
Simon L: Now for lattice gauge degrees. There's so there's an additional complication in that there is a preferred gauge, invariant subalgebra of observables, of operations
Simon L: which is specified by having a center given by the Gauss operators.
Simon L: This gauge, and whether one calculates the entropy on the gauge, invariant algebra, or the full algebra will be a point of discussion
Simon L: later on.
Simon L: Now there are different choices, for what what can mean by a boundary in this case? In the lattice? One can, for example, mean that there is a given boundary, and there are some links piercing out of it.
Simon L: And in that case one will have about and a corner algebra generated by operators on those radio links
Simon L: on the other end one can mean a strict boundary.
Simon L: and then only tangential links will be in the corners of algebra
Simon L: hint.
Simon L: depending on what you choose for them for what you mean by a boundary, you will have different algebras, and therefore also different
Simon L: decompositions of the density matrix
Simon L: in particular. If you choose the first type. So radio links.
Simon L: and you choose the boundary sub algebra generated only by the electric fields.
Simon L: Then you have a center which is generated by the Casimirs
Simon L: of the of each of the links.
Simon L: This is what one would call electric boundary conditions in the latter sketch theory case.
Simon L: If one works with a non Abelian lettuce gauge theory.
Simon L: Then there is the further requirement on gauge invariant states
Simon L: that they are constant across the representation that is given by each, by the by, the boundary, Casimir.
Simon L: What this means in turn for the density matrices in each block
Simon L: is that they further decompose into a data invariant piece which I've denoted here with a bar
Simon L: and a piece which is a flat state. So in completely invariant state
Simon L: on the representation space itself, the finite dimensional one that's labeled by the boundary spin values. S.
Simon L: So in mind you in this case here the like, the sector label E, here is the set of Casimir values for the entire boundary of the graph.
Hal Haggard: Simon, can we briefly interrupt? There's a question.
Simon L: Yes, I see. Lauren has a question. Yes.
Laurent Freidel: Hi, Simon, just maybe a little bit of precision, this this about this magnetic boundary condition. So if I put it randomly, I would always kind of cut it in the middle of a link, whereas magnetic cut it relies being extremely precise where I cut.
Simon L: Yes.
Laurent Freidel: Tangentially in the app, so any little perturbation will destroy that
Laurent Freidel: that magnetic boundary condition. So in that sense, I would imagine that the first one is the one that survives the continuum limit, but the second one.
Simon L: That's a very good question, I think, in the typical. Let us gate theory literature. People would also consider the second type in that they just have sharp boundaries
Simon L: and also boundary holonomies.
Simon L: For example, in the
Simon L: in the paper
Simon L: up here
Simon L: you will find precisely that they discuss the led the second of the 2.
Laurent Freidel: Yeah, I know Cassini usually. But what I'm saying is this one is doesn't have any any.
Laurent Freidel: It doesn't survive the continuum, maybe because any little perturbation of your cut will destroy that boundary condition, whereas the other one is kind of stable, so.
Simon L: Move it.
Laurent Freidel: At.
Simon L: I I I understand what you mean. Yes, yes.
Laurent Freidel: Sort of ladies artifacts that.
Laurent Freidel: And that's that's why. Usually we use the electric one. Because.
Simon L: Yes, yes, I understand. Yes.
Laurent Freidel: No, you could. Yeah, you you could use the Teta term to rotate one into the other. But anyway, you you're good.
Laurent Freidel: Thanks.
Simon L: Now. With this decomposition of the density matrix, we can plug this into the for Neumann entropy directly
Simon L: and get 3 different terms in the in the
Simon L: Get 3 different terms.
Simon L: The first of these we will call the distillable entanglement, and is given as sort of an average in the classical distribute in a classical mixture distribution
Simon L: of the gauge, invariant data of the State.
Simon L: The second piece is a Shannon entropy of the mixture weights themselves.
Simon L: and the last piece is only present for non Abelian theories, and is associated to edge mode contributions directly.
Simon L: Of course, we can do this as well, not just with for Norman entropy, but also with rainy entropy
Simon L: and other types of entropies. If you wish.
Simon L: Now, with these 3 pieces we can combine them in several ways to have different interpretations.
Simon L: The first is the this syllable entanglement itself.
Simon L: This is known to not have contributions from edge modes.
Simon L: but instead has an operationally an operational interpretation
Simon L: in terms of the maximum number of Epr pairs, you can extract from the State
Simon L: by transferring them using an algorithm doing, using a protot distillation protocol into an external cubic reservoir.
Simon L: There's another content combination which is known as the gauge invariant entropy which further includes the Shannon entropy.
Simon L: This
Simon L: you can also see as just all of the terms minus the non obedient piece.
Simon L: and is the entropy calculated when you consider only the gauge, invariant degrees of freedom, the gauge, invariant algebra.
Simon L: Lastly, if you use the full expression which is known as the full entropy.
Simon L: this will include all echoes. It will include the Kabat contact term.
Simon L: and is basically just what you would naively calculate.
Simon L: What should become clear from this is that any prescription, for when you calculate an entropy.
Simon L: must also come attached with the note of what it's actually calculating.
Simon L: The typical methods which cut, which are the replica trick via path integrals or the extended Hibs based procedure of Donelli Wall and Fridle, calculate the full entropy.
Simon L: whereas, if you use the replica trick and also fix gauge, fix some boundary conditions on the entangling surface.
Simon L: It will give you instead the gauge, invariant entropy.
Simon L: So in principle you have 3 types, and there are different, and there are ways to calculate either of them.
Simon L: and whether you prefer one over the other just depends on what question you want to ask.
Simon L: Now let us specify the discussion to spin networks. I don't think I need to go over the details too much in this, in this kind of audience
Simon L: what we will need is the sit with network vertex, Hilbert space of a single disconnected vertex.
Simon L: and the one of the of the whole graph
Simon L: in particular. What we will need is that the single vertex hip space decomposes into sectors labeled by the spins.
Simon L: and each of these sectors decomposes again into a tensor product of link spaces and at an intertwiner space
Simon L: for the graph boundary. We choose the standard choice here of just having a bunch of one valid vertices which are connected through boundary links to the rest of the graph
Simon L: for the corn algebra. Then we have holonomy, flux, algebras on all of the boundary links, and we choose electric boundary conditions, which means we only keep the fluxes inside the algebra.
Simon L: Then the corner center
Simon L: is generated again by the by the casmers on each of the links.
Simon L: and we will have a super a super selection sector decomposition of the Hilbert Space.
Simon L: Now there's already some entanglement which is included directly in spin network states on a given graph.
Simon L: This is a
Simon L: this is most easily exemplified on a single link which we can decompose into semi-links on the 2 vertices that it starts and ends on
Simon L: by taking the full, and by taking the space of a link and writing it as the space of the the tool semi links, but applying a projector on it.
Simon L: We can then realize these link states the glute link states as gauge invariant states on 2 semi-links. The gauge invariance in question is just write and write translations onto the 2 elements of the 2 arguments of the wave function
Simon L: on the level of the basis states. Then we can realize this sort of gluing operation of the 2 sides
Simon L: by taking the basis states on each of the vertices and vertices.
Simon L: and and applying to them, projecting them onto a maximally entangled link state such as this one here.
Simon L: What this means is that we can glue the at least for fixed spins.
Simon L: Vertex! Hilbert spaces together to form a glute
Simon L: spin network here. But space on a given graph.
Simon L: using a projection operator
Simon L: for fixed spins. This works all the way, and this goes all the way through.
Simon L: and we can write a spin. The spin network basis functions on a given graph
Simon L: as projected, entangled a pair of states.
Simon L: This is a type of tensor network state which is very common, and which simply consists of a tensor product state of some vertex states
Simon L: which are then projected onto maximillent pair states.
Simon L: this is, not so surprising. Many different types of bases in different contexts have this, and it's very useful in condensed metaphysics.
Simon L: Because they have a very clean entanglement structure.
Simon L: This can be seen, for example, that the products that, of course, has no entanglement in terms of the vertex factorization.
Simon L: but there is entanglement which is introduced on the links
Simon L: by the projection here.
Simon L: so we can extend this to a larger class of states, and specify a kind of corner of the Hilbert Space, in which we have a similar structure.
Simon L: So we would like to have a set of spin network states which still have this peps like structure in which you can talk of
Simon L: entanglement coming from gluing.
Simon L: And in order to get this, we take the fixed spin procedure.
Simon L: but we weigh all of the maximally entangled link states in the sectors
Simon L: by some arbitrary coefficients and superpose them.
Simon L: Then we can. We can do the same procedure. We can take arbitrary spin network vertic states with no restriction on the spin labels
Simon L: and project them onto these new, modified, maximally entangled pair states.
Simon L: this will give what I would give us what I would call a spin tensor network state.
Simon L: They are. They also feature very clean entanglement structure.
Simon L: and we can probably expect something like an area law in general from them as well.
Simon L: What's maybe important to mention is that the taking the full spin network, vertex civil spaces, and applying this sort of state to them
Simon L: is only a proper projection. If all of the coefficients are equal to one.
Simon L: The resulting States, if they are not.
Simon L: still have a similar interpretation in terms of spin network states. But
Simon L: it's just that the operation itself is no longer a projection.
Simon L: Now, before we get to calculating the
Simon L: reduce the entropy or entanglement entropy between 2 sub regions, I would like to give some definitions, some terminology
Simon L: of different types of entanglement we can have in spin networks by choosing different types of subsystems.
Simon L: The first type that would be relevant is that of link entanglement. For that we just choose a subsystem given by a single link, just as we had before.
Simon L: and then we can split the link into semi links in order to calculate the entanglement between the 2 sides
Simon L: for intertwine entanglement. It's similar. We can choose 2 vertices and a sub algebra subsystems, which was the operators on intertwining degrees of freedom.
Simon L: and then, by reducing the system to those into the into that subsystem.
Simon L: We can then check the entanglement between the 2 intertwiners
Simon L: intertwina spaces.
Simon L: Those 2 will be the most important for us, but just for completing, as I also mentioned, that there's a possibility to consider entanglement between boundaries and meetings, as well.
Laurent Freidel: So sorry. Could you go back and explain a little bit more? Your notation? What do you call E, and ge is the Casimir and E. What is it?
Simon L: G, in this case.
Laurent Freidel: So ge. Is the Casimir, the opposite.
Simon L: Yes.
Laurent Freidel: Yeah, I'm not sure. And what do you call E.
Simon L: He is the is just any link.
Simon L: any link in the graph.
Laurent Freidel: Oh, it's the edge. Okay.
Simon L: Yes, it's the edge. Yes.
Laurent Freidel: Okay.
Laurent Freidel: and what you call a peps is simply an arbitrary proposition of
Laurent Freidel: of edge.
Laurent Freidel: Of representation associated with the edge.
Simon L: Yes. So the so the so the way the link states here. Yes, those are just arbitrary super positions.
Laurent Freidel: Are you going to find a way to select the the coefficient.
Laurent Freidel: Or Gj.
Simon L: There are some conditions which come from requiring holographic reconstruction. But I'm not aware that you can get anything particularly explicit out of it. No.
Simon L: I think they need to be sort of
Simon L: peaked around some, but around some
Simon L: spin configuration, in order to support.
Simon L: to support holographic reconstruction.
Laurent Freidel: Okay. But what about? Okay, you're not going to discuss
Simon L: I'm not going to get into that today.
Laurent Freidel: Well. The modular Hamiltonian, for instance, which is the the central aspect of.
Simon L: I'm not going to get into that today. No.
Simon L: now.
Simon L: moving on we will consider
Simon L: and entanglement between 2 sub regions of a spin network.
Simon L: We will consider the sub regions to be split in in into by A
Simon L: by a set of links along a set of links inside the graph.
Simon L: and we will. This induces a new center that comes along the entangling surface, which again is given by Cassimir's on the edges of the links.
Simon L: This again induces a decomposition of the density matrix
Simon L: where now the sum is over, spin values on the. For example, if we reduce to the left-hand side.
Simon L: then it will be given by the boundary, but by the spins on the boundary edges and along the entangling edges. Here
Simon L: and again, importantly, we have this contribution here, which is flat and
Simon L: does, which comes from considering gauge invariant states
Simon L: that we can again
Simon L: without even knowing anything about the State, makes some statements about the entropy entropy overall
Simon L: the in particular, the non Abelian piece will localize on edges that come from the button that sit on the boundary.
Simon L: and if we
Simon L: for example, reduce to fixed the fixed spin set case
Simon L: where the classical mixture weight is a conical delta.
Simon L: then we have no Shannon piece.
Simon L: and we have a contribution. From the non Abelian piece.
Simon L: which is precisely just the logarithms of the dimension
Simon L: counted over all of the boundary links.
Simon L: This is actually the leading order, contribution,
Simon L: in what's and in what's known as the area law behavior.
Simon L: And it's also clear from comparison with the K with the single link case before
Simon L: that. This accounts for all link entanglement between the 2 regions.
Simon L: Yes, Carlos.
Carlo Rovelli: just. I'm confused. That what is a claim here? That
Carlo Rovelli: this particular kind of entropy as the structure, or that the
Carlo Rovelli: There can be entanglement with entropy between the 2 regions, which has nothing to do with this particular entanglement.
Simon L: The State. The statement is in, for example, in the fixed spin case.
Simon L: There the entropy consists of a contribution from the distillable part.
Simon L: and this non Abelian piece G,
Simon L: which has the form of something which sits on the boundary.
Carlo Rovelli: Okay, okay? So the following entropy can still have a bulk, bulk.
Carlo Rovelli: product, features and.
Simon L: Yeah, through the week.
Simon L: Yes, in fact, in fact, as as I was, as I would say, with the last line on the slide here.
Simon L: This kind of raises the suspicion that the bulk bike
Simon L: contribution is precisely only coming from intertwina entanglement.
Simon L: and that this corresponds directly to the distillable piece of the entropy.
Carlo Rovelli: Right? So the reason I'm asking this is because this is a sort of non holographic part of it
Carlo Rovelli: for Norman. Would you agree with that?
Simon L: It depends on what you mean by holographic, but definitely.
Carlo Rovelli: But that's that's what. Okay? Yeah.
Simon L: It's the only part that that quite that is like the leading order.
Simon L: area law term.
Carlo Rovelli: No, I'm I'm lost. Is that.
Simon L: The other way around. Yes, you spoke about.
Carlo Rovelli: That I intend to.
Simon L: Why not entertainment? Yes.
Simon L: that's the non-allographic piece. Yes.
Carlo Rovelli: Is a nonographic piece.
Simon L: Yes, but this here would be the area law piece.
Carlo Rovelli: Right. But somehow
Carlo Rovelli: the reason I'm
Carlo Rovelli: posing this question and I least get confused is that the the idea of holography?
Carlo Rovelli: Is often
Carlo Rovelli: associated to the idea to the sort of hypothesis
Carlo Rovelli: that the only possible entanglement between
Carlo Rovelli: the 2 sides is of this area kind, and that's not the case here. Right.
Simon L: Oh, I mean it. It is. It is. It is true that it's not in general the case, but we will find later on that only when the intertwine entanglement is very small, do you actually get to reconstruct stuff
Simon L: so holographic reconstruction really does correspond in a way to asking that the intertwine and intertwine entanglement is relatively small.
Carlo Rovelli: Right. But physically, why does it have to be small.
Simon L: We will get to it later. But essentially, it's just a quantum information reason.
Simon L: It's just that the boundary that the boundary space from which you reconstruct the bulk. Data
Simon L: is, relatively speaking.
Carlo Rovelli: And why should you
Carlo Rovelli: reconstructed bulk data from the boundary.
Simon L: Isn't that what holography is about.
Carlo Rovelli: That's exactly the point. So the question.
Simon L: I mean, we will get to this later on. I will give a concrete notion of what I mean by photography, and then I think it will
Simon L: be easier to discuss.
Carlo Rovelli: Okay, okay, so, but let's postpone this to the to the the end. Thank you.
Simon L: Yes.
FAU: Hello, I I want some clarification of Number 13, equation number 13. So you say, area law. But but this is not area, right. It's logarithmic of area.
Simon L: Oh, okay, it's area law in the sense of the graph area.
Simon L: So the number of boundary links.
FAU: Yeah, so it's a different sense of area.
Simon L: Well, it's it's a different sense. Yes, in that we are working with an abstract graph. And there is a priori not really
FAU: Not really, if you'll measure the area, because.
Simon L: It's not geometric area. No, definitely, not. But this is precisely the kind of sense that you would see. People claim area law behavior in tensor network stories.
FAU: Thank you.
Simon L: Okay.
Simon L: Yes, moving on.
Simon L: Now there is. Now let's look at the written. Let's look at a case of
Simon L: more typical spin network state. Let's first look at the fixed spin cases, but then also expand to the spin tensor network case.
Simon L: So what we would like to what we could do now is, we could calculate the the entanglement entropy between 2 fixed regions
Simon L: for an, for a typical, for a random state.
Simon L: So how we will do this is, we will look at spin tensor networks where we fix the graph pattern we fix the States on which we project, but we choose the vertex States randomly from the distribution.
Simon L: The way we do this is, we first fix a reference state, and then we rotate those States with some randomly chosen unitary on the vertex civil spaces.
Simon L: The Unitarians themselves will be chosen by the hard measure on the, on the unitary group.
Simon L: So just already giving you the caveat.
Simon L: This requires that the Hilbert space in question are finite, dimensional. You implement this by introducing cutoffs.
Simon L: Now.
Simon L: if you do all of this, and you choose your cutoffs, so that the representation labels, the spins are relatively high.
Simon L: Then, by using results from the literature in random tensor networks, we know that the distribution localizes on its average, so fluctuations can be neglected, and we can rewrite the, for example, second Revenue entropy, which is relatively easy to calculate
Simon L: in terms of a quotient of some numbers. Do you want it to be 0?
Simon L: Those numbers can be calculated using a mapping onto a set of random icing models. Okay, not random, but
Simon L: they will become random. So the idea is that we first use the replica trick to rewrite the product in the numerator of the Andrani entropy.
Simon L: and on 2 copies on 2 replica copies and an additional swap operator.
Simon L: Then by linearity, we can pull in the unitary average
Simon L: and calculate the average squared density matrix under the unitary average.
Simon L: and express that as some swap operators on vertices
Simon L: of the graph
Simon L: with what is.
Jerzy Lewandowski: What is replica? Trick.
Simon L: Of course.
Jerzy Lewandowski: I know that this is a famous term, but maybe I can learn now.
Simon L: okay, I wish I had written this down now. But the basic idea is that
Simon L: if you have a trace over a product of A and B.
Simon L: Then you can rewrite this on the say on 2 copies of the same Hilbert Space
Simon L: as a trace of a tensor B
Simon L: times a swap operator.
Simon L: So something which just swaps 2 copies of the Hilbert Space.
Simon L: I would try. I would try it out on paper. Just to
Simon L: just to get the idea across. But that's the idea. There you can
Simon L: products on multiple copies of hybrid spaces, using these swap operators.
Jerzy Lewandowski: Day. Thank you.
Simon L: Yeah. So the the tensor product here can then be expanded into a sum.
Simon L: And in the sum we will have a different number of swap operators
Simon L: on on some different sets of vertices that we can label the sets of where the swap operators live by some icing spins that point up or down.
Simon L: This way we can convert
Simon L: the the whole expansion of the object c. One and Z. 0
Simon L: into partition sums of an icing model
Simon L: in the icing model. The couplings are determined by
Simon L: by the spins on the given graph. So here we're still in the fixed spin setting
Simon L: and the temperature of the icing model is given by the mean spin across the graph
Simon L: in the high spin regime, then.
Simon L: which corresponds to the low temperature regime of the icing model.
Simon L: We can then, just take the ground state of the IC model as the main contribution. And this will be coming from something which is like something like a domain work close to the entangling surface.
Simon L: What we then get is that for fixed spins.
Simon L: the entropy of a entanglement entropy. Okay, there are any 2 entropy of the reduced state
Simon L: has 2 main contributions.
Simon L: the first coming from the edge piece, which is the area law piece.
Simon L: and another piece coming from the reduced Intertwina state. So this is the true bulk entropy.
Laurent Freidel: Could could you? Oh, you go!
Carlo Rovelli: This is for those particular States that you get averaging
Carlo Rovelli: sort of the sort of locally right. This is not a general statement about
Carlo Rovelli: average states of the full hipaa space. Is that correct? So the class of state? I want to say.
Simon L: Just said.
Carlo Rovelli: If you you're integrating over this use, but it's use a local sort of screen by spin, so to say.
Simon L: But there are several caveats in additions to that.
Carlo Rovelli: So the fool
Carlo Rovelli: unitary, sir, that would be completely destroyed, because you have all sort of.
Simon L: Yes and no. Yes, in that a prior you could expect so. But no, because when you actually do it, you find the same kind of result.
Carlo Rovelli: Well, I'm not sure.
Simon L: I mean the the calculation is out there in particular. You can do it without choosing this particular class of states, and you with doing a different kind of
Simon L: argument. Eugenio Bianchi has a paper on this in which he just looks at a generic subsystem state. And basically it's the same result.
Simon L: It's just.
Eugenio Bianchi: You get some volume low, right? You get some volume.
Carlo Rovelli: Really
Carlo Rovelli: you don't tell Gary. Hello! That's the point
Carlo Rovelli: in Eugenio's. Calculation is very clear. If you go that, the second term would be much larger in a sense.
Simon L: Yes, yes, so the the the calculation here does not make any statement about the sat relative size of the terms.
Carlo Rovelli: Okay, so
Carlo Rovelli: there's nothing
Carlo Rovelli: Ariello generic
Carlo Rovelli: so far. Right.
Simon L: In principle, no.
Carlo Rovelli: Okay.
Carlo Rovelli: It's a.
Simon L: For instance, it's.
Carlo Rovelli: I mean somehow to squeeze in a a subtle, graphic apology somewhere. You get to some holographic conclusion.
Carlo Rovelli: That's what I would insist if you don't sort of associately bring brings in somehow graphic hypothesis. You don't get an autographic.
Carlo Rovelli: a
Carlo Rovelli: definitely.
Simon L: I did not. No, okay, good
Simon L: Lahore.
Laurent Freidel: Yeah, it was just a question about what is your notation? Zen? One slash 0. What does that mean?
Simon L: This is just that it's an or it's sort of it refers to what and to what into z one or z 0.
Laurent Freidel: Where z is the swapping or no swapping.
Simon L: Z is the is these quantities that.
Laurent Freidel: Appear, when you, when you.
Simon L: The tracing and the unitary average.
Simon L: Yes. So in this sort of setting.
Simon L: what you can extract is that you achieve maximal entropy for the reduced state.
Simon L: If you have, for example, small intertwana dimensions.
Simon L: So because the boundary in this case is the only
Simon L: so the the maximum entropy you. The maximum bound on the entropy you can have in this case
Simon L: is already
Simon L: saturated. Wait a second.
Simon L: Okay. I had an argument in my, but this is not the correct one for this case.
Simon L: okay, I will. I will keep this this remark here, for later on, where it's actually relevant. Actually.
Simon L: now. If we extend the calculation now to
Simon L: the spin tensor networks of before.
Simon L: So where we do not have a restriction on spins anymore
Simon L: the calculation is still doable, but it requires that the partition sums of the icons now have an additional sum over spin labels throughout the graph.
Simon L: one for each copy in the replica. So for
Simon L: the second venue entropy, there is now 2 sums.
Simon L: There's some additional factors in the sum which depend only on the dimensionalities of the representations
Simon L: which now at the point on of the classical mixture distribution.
Simon L: Then there's also, furthermore, some restrictions on the combinations of
Simon L: the spin sectors
Simon L: and the Isaac configurations.
Simon L: So some configuration, some combinations, are forbidden just by orthogonality of some spaces.
Simon L: and this leads to the inclusion of a of an additional Boolean factor in the definition of the of the icing models. Partition sums.
Simon L: What's important is that in this sitting there is no longer a universal notion of temperature like there was in the fixed spin case.
Simon L: The reason is simply that before we had a mean spin across all of the across the whole graph. But this is no longer the case. When we have multiple different assignments.
Simon L: we can still, however, cut and go ahead and calculate stuff by performing a cumulant expansion of this expression here.
Simon L: In the distribution given by the by 2 p. And 2 copies of the classical mixture. It's 2 copies here because we're calculating the second Renyi entropy. If it were K copies, then it would be
Simon L: care copies of the distribution.
Simon L: and this overall allows you to still find and to still calculate, approximately speaking, what the entropy is.
Simon L: Now with this technology in mind.
Simon L: I would like to turn now to the notion of holographic reconstruction that I'm interested in.
Simon L: and that this has been applied to in the past.
Simon L: So for the simpler case we will just work with fixed spins. So holographic reconstruction of intertwina data from the boundary links.
Simon L: The way that we that this is supposed to work is that given a state of the full graph. Again fixed spins.
Simon L: We induce a mapping
Simon L: from the Hilbert space of intertwiners
Simon L: to the Hilbert space of boundaries and meetings.
Simon L: and require that this is an isometry.
Simon L: The map itself is just given by taking a partial scalar product
Simon L: with the intertwina state. So this is a state of the full graph. This is the intertwinus, and if we contract all of the indices here, what's left is a state that sits only on the boundary links.
Carlo Rovelli: But you're restricting of all the States side those for which this is possible.
Simon L: What do you mean? Restricting?
Simon L: I mean at this point it's just a definition.
Carlo Rovelli: Isometry. Oh, sorry isometric doesn't. Isometric means only that the.
Simon L: Isometric as a map between Edward spaces.
Carlo Rovelli: Okay? So it's does not mean that they are.
Carlo Rovelli: This is something virtual.
Simon L: Invertible
Simon L: depends on dimensions.
Simon L: I mean
Simon L: in principle, you would like for it to be invertible. But that's not a requirement, a priority.
Simon L: Of course, what this means is at least that the dimension here is bigger than the one here.
Simon L: otherwise it cannot be isometric.
Carlo Rovelli: So this, this is crucial. So let me. This is your
Carlo Rovelli: You're going to the bulk to the boundary right.
Simon L: Yes.
Simon L: we're turning. We're mapping a bulk state to a boundary state.
Carlo Rovelli: So you you start from a much larger state to a much smaller state. That's right.
Simon L: In principle.
Simon L: If you, the scaling between the 2 is
Simon L: very different, and the bulk Intertwina stays will be very large. Yes.
Simon L: so what you have to do is either restrict to sets of spin labels where the intertwana spaces are small.
Simon L: or you have to restrict to what's called code subspaces
Simon L: in which you can actually do an isometric reconstruction.
Simon L: So I get what you mean in principle. The intertwined spaces are pretty large, and the boundary spaces are pretty small.
Carlo Rovelli: Yeah.
Carlo Rovelli: So why, this is like strong, ordinary
Carlo Rovelli: shrinking of the spaces be network that one usually consider.
Simon L: So
Simon L: so in principle, what the what this refers to is
Simon L: for certain.
Simon L: So this is a definition. It doesn't refer to how generic this is right.
Carlo Rovelli: That's what I'm saying. That was my question. It's not generic at all.
Simon L: No, absolutely not.
Carlo Rovelli: Okay.
Simon L: It's absolutely not generic.
Carlo Rovelli: I somehow want to avoid the impression that there is anything natural in olography.
Simon L: I mean, I'm not claiming anything of this vote right?
Simon L: And making a definition.
Carlo Rovelli: Alright. So this is a this is a definition that picks up a very
Carlo Rovelli: small set of cases in all the possibilities.
Carlo Rovelli: An old spinaker. But.
Simon L: Isn't that what you would call non generic.
Carlo Rovelli: Right? So it's highly non generic.
Simon L: Yes, so it captures what you want to. What you want to say.
Carlo Rovelli: No, because the the holography thesis is that there's something generic about being allographic.
Simon L: Where that's a hypothesis.
Carlo Rovelli: Okay. Good.
Simon L: Brett.
Simon L: good.
Simon L: Okay.
Hal Haggard: Simone had something to add also.
Simone SPEZIALE: Your gamma allows loops.
Simon L: in principle. Yes.
Simone SPEZIALE: Right. So there's really an infinite amount of information that is being thrown away with this projection right?
Simon L: In principle. Yeah, I I think so.
Simone SPEZIALE: In practice, not just in.
Simon L: In principle and in practice. Yes.
Simone SPEZIALE: Thank you.
Simon L: Yes, I mean, we would get late to to to to like an interpretation of what this kind of photography lets you recover, and
Simon L: I think on some level, it's quite restrictive.
Simone SPEZIALE: Okay. Thank you.
Simon L: So
Laurent Freidel: Maybe another.
Simon L: Yes.
Laurent Freidel: And I'm
Laurent Freidel: I mean here, maybe to to add to that discussion at some point. I mean, we we want. I think that's maybe where where you're going I don't know, but you know, at some point you want the State side to be a solution of the constraints. Right? So I mean in general, which means the admin constraints and the default constraints, whatever that means in.
Laurent Freidel: you know. And of course, so you know, here I'm assuming. So of course we don't have that that expression, but that that restricts right? So there's a restriction coming coming from the fact that site cannot be an arbitrary state, as you know. Is that what you.
Simon L: Yes, yes, that is true, but it's but it's due to the restriction of the definition.
Simon L: Yeah, more general definition. But yeah.
Laurent Freidel: This business. P. Network in the bark is very, very big, but the the subsets of States which satisfies the constraints like invariance of the support I mean translation, etcetera, is much smaller, right?
Laurent Freidel: I feel.
Simone SPEZIALE: We all agreed, since many years. That is not a subset, because Wilton and constraint is not generating compact orbits. So it's not.
Laurent Freidel: Yeah. But psi, psi is
Simone SPEZIALE: There will have to be some distribution.
Laurent Freidel: The dual sigh, is the dual state here.
Simone SPEZIALE: Right, right, so.
Laurent Freidel: That that projection map should be done with, you know, with something that satisfies the constraints, which, of course.
Simone SPEZIALE: Well, it's not an element, is is not a subset of the space. I mean. Of course, you could try to construct it by looking at the duo.
Laurent Freidel: Sure put put a dual in his age, and then you have the same
Simone SPEZIALE: Well, but then we are talking about much, much more general quantities than what he's been talking about, which were L. 2 States.
Laurent Freidel: I don't know. I mean.
Simon L: 'kay
Laurent Freidel: Comments, but.
Simone SPEZIALE: Okay.
Laurent Freidel: I mean so.
Simon L: Perhaps perhaps the the second definition I'm about to give a bit later in terms of algebra will give a bit more food for discussion later on.
Simon L: So
Simon L: in order to check how generic this is, and we would see that it's not so generic.
Simon L: we first want to transfer this state in this this condition of isometry
Simon L: into a statement that we can quantify using some numbers
Simon L: in particular, we can turn the isometry question into a maximum entropy question of the in the reduced intertwin state.
Laurent Freidel: Could you record the definition of being isometric? I don't think you write it. There.
Simon L: The one on the left here, basically.
Laurent Freidel: Yeah, that's what you mean. Okay, is proportion to the identity on on.
Laurent Freidel: Yeah, that's.
Simon L: The prefactor comes from it being being induced from a state. Basically.
Simon L: Now, we can again, verify this this condition on reduced into intertwana states by doing a similar mapping to an icing model calculation. So, using the same set of fixed spin, random tensor network
Simon L: states, we can calculate the entropy in that setting.
Simon L: and in which case the IC model acquires some additional bike.
Simon L: external magnetic fields
Simon L: which are which tell it where the grounds, and which modify the ground, state a little bit.
Simon L: What we find from this is, that if the intertwiners in the given sector, that we that we consider
Simon L: are smaller.
Simon L: So there's a lot of inhomogeneity in the spins, and the dimension of of the intertwined spaces are very small, like one dimensions, 2 dimensions, 3 dimensions. Perhaps.
Simon L: then, it's relatively generic to find holographic and reconstruction behavior.
Simon L: But of course, what this means is. If the intertwina spaces are small. Then there's not really a lot of stuff to reconstruct. In the first place.
Simon L: Now I will.
Simon L: continue this to the generous spin tensor network case where we have spin super positions.
Simon L: in which case we will, we can no longer identify the center.
Simon L: the super selection sectors.
Simon L: and here we no longer have a Hilbert space decomposition
Simon L: between a bike system and a boundary system. So what we can do instead is replace it by a mapping between algebras.
Simon L: The algebra here in question. Would be chosen to be this
Simon L: to be the operators acting on.
Simon L: intertwine us.
Simon L: and the fixed label here is just the boundary spins.
Simon L: so we will still sum over bulk, spins
Simon L: off the intertwine into the spaces
Simon L: and the intertwinos here are really the ones of the full graph.
Simon L: Similarly, on the boundary end of the story.
Simon L: we take all of the but all of the boundaries and meeting spaces.
Simon L: and we
Simon L: and we sum over all of we take a direct sum over all of the operator spaces for all of the boundary spin values.
Simon L: Then, with this choice of
Simon L: bikes and boundary subsystems.
Simon L: We can verify that they are, in fact, complimentary in the sense of the beginning of the talk.
Simon L: and the center is given by the boundary spin, and by the boundary spins.
Simon L: Further, there is there is a notion of extension and partial trace maps.
Simon L: So here is an extension map
Simon L: share the eyes for injection.
Simon L: So this takes an intertwine operator and extends it to the into the boundary links by simply doing it sector-wise.
Simon L: Similarly, on the partial trace site, we take a general operator on both boundary and bulk.
Simon L: and reproduce an operator on the boundary. The output system
Simon L: from it, by taking the partial trace in each sector where we remove the bulk.
Simon L: we can use these partial trays and extension maps
Simon L: to define what's known as a Choi mapping.
Simon L: But in this context. I would like to call it the transport super operator transport, because it transports an operator from the bulk to an operator on the boundary
Simon L: from input to output
Simon L: and super operator because it maps operators to operators.
Simon L: So in particular, what this does is extends the operator to the full system
Simon L: takes a product with something written with a partial partial transpose of the full graph state.
Simon L: and then trace this out so that we get an operator on the boundary system.
Simon L: This is a complicated definition, but for a trivial center in the algebra that we chose.
Simon L: and for pure States this actually reduces to the Hilbert Space formulation. Before
Simon L: now. The more generally kind of definition of what I mean by a holographic state, then.
Simon L: for given fixed choice of sub algebra, in which you can choose the sets of
Simon L: spins you include.
Simon L: If you want, you can choose that and manipulate that by hand
Simon L: in that setting. In in that setting I call a State holographic. If the induced transport super operator is asymmetric in the Hilbert Schmidt sense.
Simon L: So once again, Hibbert Schmidt, we we choose because we are on. We suppose that the hip that the spaces and operator algebra here are finite, dimensional.
Simon L: So we once again need to impose some cutoffs.
Simon L: With this definition in mind, we can once again transfer all of this into an entropy, calculation.
Simon L: and the expansion calculations are a bit difficult because of this whole cumulant expansion thing that I've mentioned before.
Simon L: but we can look for some necessary criteria for the for the State to be holographic.
Simon L: The first is that the total dimension of all of the boundaries and meetings
Simon L: which is, in a sense which is equivalent to asking. The total area of the state
Simon L: of the system needs to be constant across all of the sectors you include in the super in the
Simon L: in the full algebra.
Simon L: So you're sort of looking at. An algebra which has a lax boundary condition where the total area is fixed.
Simon L: This introduces, then, a notion of scale to the system which you can use to rewrite the icing model partition functions, and then you can take again a low temperature limit.
Simon L: The low temperature limit in this case means you look at fixed large total area.
Simon L: So if you do this.
Simon L: if you reduce, if you restrict to this kind of setting, then the isometry condition just reduces to one that goes per boundary sector.
Simon L: So in each of the boundary sectors.
Simon L: So each of the
Simon L: sort of local. So what we fix is the total area right.
Simon L: But the local area densities
Simon L: are still not not fixed.
Simon L: And so each of these local area density still labels a super selection sector, and in each of those sectors we require, again, isometry to be there.
Simon L: So with that in mind, those are the necessary conditions we know about. We know about this by boundary reconstruction
Simon L: and the conditions for it we can reduce to the fixed spin case again.
Simon L: Now for some interpretation of this
Simon L: what we are doing here is that we are reconstructing intertwiner degrees of freedom of the public
Simon L: from pure gauged mode, edge modes, degrees of freedom on the boundary.
Simon L: and the constraint of fixing the full total area
Simon L: very easily understood, as that this is not actually part of the edge modes. This is not part, and the and this is not part of the
Simon L: sort of soft degrees of freedom, but rather this is part of the global gauge group which fits together with whatever spin defects you have in the bulk.
Simon L: So you can take many different conclusions from this.
Simon L: 2 that I'd like to highlight. Here is the first that we have a preferred corner of states here which do seem to be close to holography. We have identified something here which is which is specific enough so that we can do bulk reconstruction. That's one possible conclusion. The other conclusion is that somehow, because we need to restrict the intertwana degrees of freedom, so much.
Simon L: those are difficult to reconstruct from those edge modes alone.
Simon L: So, in a sense, what we can reconstruct from edge modes is anything but the intertwining degrees of freedom. And that's those are very, very simple. So with regards to the earlier question about what if I have loops in the bike? What if there's some curvature?
Simon L: Well,
Simon L: I'd imagine that it's difficult to encode this in the edge modes alone.
Simone SPEZIALE: You knew it was coming right. So this preferred the corner of the Baltimore space. It's safer to assume for the time being that it excludes loops.
Simon L: I don't know if there's
Simon L: Eugenio's already there.
Simon L: Perfect the answer. So I would say I'm not sure.
Simon L: but more prudently, probably not.
Simone SPEZIALE: Let's see if your journeys will.
Simon L: Daniel.
Simone SPEZIALE: Prudent, or bring.
Eugenio Bianchi: I just wanted to ask, have you checked the volume? We are seeing. The dimensions of Delbert. Spaces are either one or drop.
Eugenio Bianchi: maybe 2.
Simon L: But I checked.
Eugenio Bianchi: The volume of the interior.
Simon L: Oh!
Eugenio Bianchi: Volume of the interior hits the area. Is it 0.
Simon L: It's it's not a sharp criterion, right? So it's sort of the dimensions need to be small relative to the total scale that you fix for the system.
Simon L: and
Simon L: the the larger the Hibots, the larger the dimensions, the worse it gets. It's sort of an approximate isometry. In the first place, because the entropy is never quite maximal. It's only close to being maximum.
Simon L: So it's such a.
Eugenio Bianchi: Really, can you get something closer to 3 dimensional exuding space
Eugenio Bianchi: where that the area goes like length, square and the volume like the
Eugenio Bianchi: let's you.
Eugenio Bianchi: or is it that it's just a crumpled surface? It has an area that 0 volume is.
Simon L: I feel like it's close. It's probably close to a crumpled area of sorts.
Simon L: or perhaps or perhaps one interpretation that I like to at least have in my head with this
Simon L: is that
Simon L: if we have something which is entirely reconstructed, but from the a small degrees, gauge group, degrees of freedom alone, then it's probably because there's somehow flatteness inside.
Simon L: And with that in mind
Simon L: it may not be that the volume is necessarily so constrained.
Simon L: but it will still have some.
Simon L: It will still be very rigid.
Eugenio Bianchi: Okay. Thank you.
Simone SPEZIALE: I related to a genius question, but what fixes the valency of the interior nodes.
Simon L: In this case I just fixed them arbitrarily to the.
Simone SPEZIALE: Right genius question cannot even be asked. I guess right, because the whether you're if they're for Valent, or 5, 5 or 6 volunt, the properties of the volume will depend. So you have to first make the sharp, and then you will be able to. It can be asked. In order to answer it. I guess one needs also to have control over these right.
Simon L: Yes. So in the case that I've been presenting here, you have fixed a graph, and I just studied a fixed valency. But there's nothing that changes in the formalism. If you just let the balance see vary, so you can study both the 3D gravity case where everything is trivalent, but you could also do something where everything is for balance. You could mix everything you could have arbitrarily high balances.
Simone SPEZIALE: Just to make sure. I understand. You mean that your autographic reconstruction would be a fixed bulk graph.
Simon L: Yes.
Simon L: so all of this is very limited in that you need to have a fixed graph structure. Otherwise you cannot apply the random tensor network methods. As far as we are aware.
Simone SPEZIALE: So even in this small corner we are really far from holography. I would say right.
Simon L: In principle, but.
Simone SPEZIALE: Because you're already assigning a huge amount of information about the bulk by specifying the graph.
Simon L: Yes, but in the same way it's sort of
Simon L: It's sort of washed out once you perform the average.
Simon L: so I would not be surprised if you can choose different sorts of graph structures
Simon L: and most of the results.
Simone SPEZIALE: But yeah.
Simon L: Sort of.
Simone SPEZIALE: Th. There's you mean washed out when you module by defamorphisms.
Simon L: No, when you perform the group averaging. But I guess that's not really.
Simon L: It's just about like I mean.
Simone SPEZIALE: Existed, you should be able to, from your boundary it alone to reconstruct the actual graph inside that.
Simon L: So I find that to be difficult, at least from given from the data that we use here.
Simone SPEZIALE: Agree, though so, if.
Simon L: If you don't augment the boundary with some extra degrees of freedom which allow you to reconstruct more stuff, I would find it difficult to reconstruct anything more than.
Simone SPEZIALE: Right.
Simon L: They were funny.
Simone SPEZIALE: Difficult to even.
Simon L: Instruct a single bulk, spin.
Simone SPEZIALE: Right? Yeah, I agree.
Simon L: So.
Carlo Rovelli: Simon. This. This obviously answers the sort of question that I was raising when you were giving definitions. There is.
Simon L: Yes, sir.
Carlo Rovelli: Then there are different.
Carlo Rovelli: different a priori assumptions people have in their hand, in their head.
Carlo Rovelli: But I think you your your what you present is very clear.
Simon L: Yes, that's that's what I meant. The the whole, the whole step up from giving the Hibbert Space definition to the algebra one.
Simon L: I think, having that in mind makes it more clear what I mean.
Laurent Freidel: But you have. Can you commit now on the role of the constraints which here it's? You know what your your discussion is when you don't talk about any
Laurent Freidel: any constraints
Laurent Freidel: right in the bulk.
Laurent Freidel: so we haven't studied that.
Simon L: And all the.
Laurent Freidel: And let's say, the translation, constraint.
Simon L: Haven't studied this at all so far.
Simon L: So.
Laurent Freidel: Hey? We'd happen.
Carlo Rovelli: If I may. Lauren Lauren. What you're saying is that there's a there's an interesting hypothesis here which is that preferred corner of the open space. Photography might be precisely
Carlo Rovelli: where the constraints project, the
Carlo Rovelli: the the kinematical illustration 2. That's the that's another formulation of the holographic hypothesis. I I I would say, or for which you might have intuitions, or or or or
Carlo Rovelli: or indirect arguments, and and.
Laurent Freidel: Yeah, I was. I was maybe making a statement in between. It's clear that here he's working on some kinematic a little bit space. So a lot of a lot of these States that you count as difference are completely redundant.
Laurent Freidel: So once you impose the constraints, the question, there's a drastic as reduction of what our admissible bug states. Whether thatuction it's drastic, right? It's like a, you know, because there's there's whether that drastic reduction means that you project onto the edge mode. That's that's another extra purchases. But we have to appreciate that. What we're talking about here is is vastly over counting in the box, right? Because we we are not positioning by the action of the gauge group.
Laurent Freidel: The by, the yeah. So
Laurent Freidel: there's there's a lot of space in between. Yeah.
Laurent Freidel: you know, counting speed network states as valid states.
Simon L: Smoke.
Laurent Freidel: And looking at the physical ones.
Simon L: I have 2 more slides.
Hal Haggard: I was. Gonna say exactly that, Simon, wedding
Hal Haggard: up, and then that'll.
Simon L: Yeah, then we can discuss more.
Hal Haggard: Can continue discussion.
Simon L: Yes, so to add on the on the fire of the holography. And what is the right notion story?
Simon L: I would like to add a further complication in that. It's also possible to just talk about entanglement between boundary regions. And this is an entirely separate question.
Simon L: One way you can do this is, you can take one fixed graph state and you fix some intertwining data. For example, if it's fixed spin, then you just fix one state.
Simon L: But if it's multiple sectors, then you fix one for each separate state and you sort of condition the boundary state on the spike data.
Simon L: So you now have an induced boundary state
Simon L: which only sits on the semi links.
Simon L: And this is now in a factorizing boundary head space. So this is very easy from the point of view of trying to figure out what entanglement means.
Simon L: There is no necessary necessity for an area law or something like that because there's
Simon L: no center, there's no center which would give you the kinds of contributions we saw earlier.
Simon L: So
Simon L: this is more similar, perhaps, to what people do in Ads. Cft, where they say that the entanglement between different regions of the Cft, which sits entirely on the boundary
Simon L: is what gives rise to bike connectivity and other properties of the bike space time.
Simon L: So maybe, in a sense, the actual geometry from entanglement that we should be looking for
Simon L: is actually of this kind, and is something that also would need to be studied in order to better understand what kind of stuff we can reconstruct.
Simon L: But it's more difficult to compute this sort of thing in general, so
Simon L: so far we don't know that much about it, especially not in the case of spin, tensor networks
Simon L: and
Simon L: observables would also need to be covered, calculated in this kind of context, in order to have to have an idea of what the criteria really mean in terms of, say, geometry.
Simon L: So
Simon L: let me just wrap up with some take-home messages.
Simon L: The first is that
Simon L: whatever entropy you calculate, given a density. Matrix depends on what algebra the density matrix sits in
Simon L: what algebra you choose to work with the gauge invariant one, the kinematical one which kinds of gates invariance, for example, just gals, or all the way to diffu invariance.
Simon L: Those different types mean different things in particular. If you ask a quantum information purist. They might tell you that only the distalable entanglement is really one that they are interested in
Simon L: depending on who you ask.
Simon L: Different types of algebra can also be. That's the second point assigned to regions.
Simon L: So you have different choices in that, and if have different choice of boundary conditions.
Simon L: it needs to be kept in mind.
Simon L: Furthermore, spin network states have a super selection
Simon L: by the boundary values of the spins.
Simon L: That's very, and that's very important for getting the edge mode contribution, this a boundary term in the entanglement, entropy.
Simon L: and in particular they are universal, in that, if your state is gauge invariant, it will have this contribution.
Simon L: Furthermore, the last point is that if you have, if you have a spin network state, and you impose some restrictions on what kind of State. It is
Simon L: then, on average, the typical State will support this kind of holographic reconstruction for match modes. So, of course, in the last point I was, and being a bit sparse with the details. But no, I think the details should be clear as well.
Simon L: Okay, there was an empty slide there. Okay.
Simon L: thank you.
Hal Haggard: Thank you, Simon. Thank you for the nice overview. Carlo.
Carlo Rovelli: Only me says nobody else would definitely be.
Hal Haggard: Yeah, so.
Carlo Rovelli: Anyway. Okay, so first of all, that was very clear. I think a follow. I'm not sure I got the last
Carlo Rovelli: of your comments, but that's maybe just number 5. But let me let me make 2.
Simon L: Just the stuff from before, but more not more short.
Carlo Rovelli: Let me make to so I I think I understood you. I think it's very nice, very clean. I think I got it. So it's very good, very useful, I would say.
Carlo Rovelli: I want to make 2
Carlo Rovelli: to side comments.
Carlo Rovelli: one may be related to your last, the the very last end of your very vague end of your your talk.
Carlo Rovelli: It might be useful
Carlo Rovelli: to think about the spin network state.
Carlo Rovelli: the super space you're talking about.
Carlo Rovelli: Not does the
Carlo Rovelli: the same of a sort of a cushy surface?
Carlo Rovelli: But rather as a boundary of 4 dimensional region.
Carlo Rovelli: namely, not the analog
Carlo Rovelli: of the
Carlo Rovelli: sort of help, the spaces of no relativistic quantum mechanics. So.
Carlo Rovelli: but the analog of the sort of tensor product of the in state and the outstate.
Carlo Rovelli: The what I call the boundary state in my book.
Simon L: Like a boundary state for spin foot models.
Carlo Rovelli: Yes, exactly, which is not the sense in your sense. It's not the sort of in physical terms. It's not the 2 dimensional bandwidth, the 3 dimensional band, the band of
Carlo Rovelli: Because if it is so
Carlo Rovelli: if you think in these terms, which is just definitions perfectly possible. And I think that's clean way of viewing, of viewing the
Carlo Rovelli: then the dynamics is given by just a brown, the state just a linear function of this state. Right?
Carlo Rovelli: That's it.
Simon L: I couldn't. I couldn't acoustically hear you. Sorry.
Carlo Rovelli: Oh, sorry!
Simon L: By by given by a brown State.
Carlo Rovelli: If you do so, then the dynamics is just Abra on this state.
Simon L: A brow on the state. Okay? Yes, yes, yes.
Carlo Rovelli: Linear function, this state. So you don't have to protect anything down to to a subspace. You just have to write right.
Carlo Rovelli: which, in the no relativistic case would be
Carlo Rovelli: just transform is transforming sort of psi final U evolution psi initial
Carlo Rovelli: into a bra on the
Carlo Rovelli: acting on some initial tense or sci vinyl dagger may start.
Carlo Rovelli: You think of these terms, then? This what you're sort of. Yeah, what you're sort of hinting in these slides
Carlo Rovelli: is that it is in this sense
Carlo Rovelli: that the
Carlo Rovelli: boundary through this place, in in my sense.
Carlo Rovelli: We should give different names because Bandar, hipaa space no has.
Simon L: Foundry versus Corner.
Carlo Rovelli: Yeah. Boundaries. This is called alright. Thank you. The boundary hit. The space is
Carlo Rovelli: include something about the book.
Carlo Rovelli: because in the classical case it always. It's clearly does right, because if you give that down the boundary you have.
Carlo Rovelli: if you think in terms of Hamit, Hamilton, Hamilton, Jacobi logic, Hamilton. Function, logic, that sort of
Carlo Rovelli: define something in between. So that was my, it's very big. It's not something.
Simon L: When you say bark. In this case you mean the 4.
Carlo Rovelli: To Mr.
Simon L: Facetime. Bug, okay.
Simon L: okay, so.
Carlo Rovelli: Maybe we should determine, we should think, in terms of 4 dimensional geometry, not 3 dimensional geometry. When we think of the bulk, because that's.
Simon L: But do you.
Carlo Rovelli: After all, that's yeah.
Carlo Rovelli: That's one comment.
Carlo Rovelli: the second comment. I'm just repeating something which I know it's controversial. But I am.
Carlo Rovelli: I I think it's correct. And so it's a it's a
Carlo Rovelli: I think it's important to remember the there is a way of interpreting the now 3 dimensional here to space for spin networks as something which doesn't know at all about the the dynamics and Newtonian constraint.
Carlo Rovelli: but knows already fully about the
Carlo Rovelli: The diphomorphic constraint, the three-dimensional dermorphic constraint.
Carlo Rovelli: because at even given a graph, this is a truncation.
Carlo Rovelli: is not the full theory.
Carlo Rovelli: and as such it can be seen
Carlo Rovelli: a sequentization of a truncation
Carlo Rovelli: off the
Simon L: I mean.
Carlo Rovelli: Of of of of 3 dimensional geometries, not metrics. So
Carlo Rovelli: this is the way that Eugenio has been thinking for for a long time. So you can take this piece of.
Simon L: And see it as a top of.
Carlo Rovelli: Up to exactly up to different offices that they just think to.
Simon L: Eventually up to December.
Carlo Rovelli: And in this space you just look for a for a subset. That's a truncation.
Carlo Rovelli: You quantize that. Then you get this thing here. So that's a logic of Regent. That's regular logic of the.
Simon L: But easy. Yes.
Carlo Rovelli: I'm not saying that that's the only way of doing things or thinking things so one can think differently and do something differently.
Simon L: Yes, I think it's okay.
Carlo Rovelli: So what is missing from this mathematics to the physics from this perspective?
Carlo Rovelli: It's just a dynamics
Carlo Rovelli: I just wanted to throw there. This 2 comments are not directly related to your talk, but your talk.
Simon L: Was this.
Carlo Rovelli: It gives a. It gives sort of the mathematics of these things and the relation between book and boundary in a way. And I think that this 2 different perspective may help using all this going toward the physics.
Simon L: Okay? So just so, I understand your first comment correctly.
Simon L: you're saying that looking at the results of the failure to reconstruct a generic spin network evidence base from the edge modes alone.
Simon L: We should rather see the kind of holography that might be more generic, as the kind that reconstructs for the data
Simon L: from a boundary through which is code dimension. One.
Simon L: Okay.
Carlo Rovelli: Yes, and in in doing this case.
Simon L: Would that? But would that not be almost too generic in the sense that if, for example, we were in classical physics then, and if I don't have a sort of timeline boundary and just have a 7 of sorts.
Simon L: And I have initial data and final data, then the 4 of the data is sort of specified just by having a unique solution to the Cauchy problem.
Carlo Rovelli: It is but the the
Carlo Rovelli: for me. I I I've always hoped not in terms of Tamil boundary, but the 2 space like surface that meet at corner.
Simon L: So the.
Carlo Rovelli: Of the corner. I've always it might think I've always neglected the role of the corner and the corner. The role of corner cannot be neglected.
Carlo Rovelli: Obviously so, the corner is crucial here.
Carlo Rovelli: And
Carlo Rovelli: somehow, the the the generatistic evolution is multi. Finger devolution is local, is all global.
Simon L: Yes.
Carlo Rovelli: So given a cushy surface, you can evolve just a bump.
Simon L: Yes.
Carlo Rovelli: That you forget all the rest. You just look at the bump. You have a 4 dimensional thing, and you have a sort of initial and final piece of the cushy surfaces.
Simon L: Yes, but there's no time involved. Yeah.
Carlo Rovelli: Right, but of course there's a corner whether to meet, and I've been neglecting all my overall. My book is sort of neglected corner, and there all this.
Carlo Rovelli: looking at the corner, is crucial, and that's why I find it's it's.
Simon L: Thank you. Okay.
Carlo Rovelli: Meredith.
Simon L: Okay, I see what you mean.
Carlo Rovelli: Honest as as in in this.
Carlo Rovelli: What I'm suggesting is that now remember that what we have to do we have to do the dynamics. But there's a way of simple way of doing the dynamics.
Carlo Rovelli: Which is
Carlo Rovelli: the way just it.
Simon L: Okay. Thank you.
Simon L: The home. Yeah.
Laurent Freidel: Well, maybe Jersey first, since he hasn't asked.
Simon L: Okay.
Simon L: Jesse.
Jerzy Lewandowski: Yes, so thank thank you. So I have a very general and stupid question, so why do we actually need
Jerzy Lewandowski: the edge modes. It seems that all the Hilbert space is defined by the Hilbert spaces of intertwiners that are
Jerzy Lewandowski: additionally
Jerzy Lewandowski: reduced by so so, so, so, so I would say that edge
Jerzy Lewandowski: edges don't introduce extra degrees of freedom. They introduce extra, and they reduce the degrees of freedom. So so why do we need to count edges, as also.
Simon L: Let's say.
Jerzy Lewandowski: You think, to the Hebrew Space degree.
Simon L: It's
Simon L: So as far as the hipaa space of speed networks is concerned.
Simon L: the hologies live on there so I wouldn't argue for removing them. But, as far as the entropy is concerned, to raise a valid question
Simon L: in principle, the data that is there, once you have fixed the graph is just the data of this Hibbert space here, and the data on the internal edges is fixed. It's fixed to be the maximally entangled state.
Simon L: for example in the States that I considered here.
Simon L: And then you can actually just say, Okay, I don't want to count them. They don't. They are just constrained. They are just fixed already, so they are not. There are no degrees of freedom.
Simon L: You could say that.
Simon L: And then you count a certain type of entropy.
Simon L: Then you count
Simon L: basically just the distillable entanglement.
Simon L: But if you count also the contribution coming from the edges, then you're calculating a different type of entropy.
Simon L: So in the end. It's sort of you can do both.
Simon L: and whether you want one or the other depends on what you want to do with it, I would say.
Simon L: because the I don't know, having having just the distillable entanglement has an operational interpretation.
Simon L: But in many ways there are reasons to say that the edge MoD contribution is what you get. Naturally, if you just, if you just calculate it, via the replicatric.
Jerzy Lewandowski: Okay, thinking.
Simon L: So if you dislike edge modes, you can remove them.
Jerzy Lewandowski: Yeah, this is what I what I understood. Thanks.
Simon L: Okay. Simon.
Laurent Freidel: Sorry I had a question before.
Simon L: Yes, yes, Laurent go first.
Laurent Freidel: Yeah, I wanna go back to this. Maybe this comment about the Defamorphism cause. I think it's very
Laurent Freidel: appreciate that
Laurent Freidel: in the traditional picture of Lqg, we think that you know, deformism doesn't create new entanglement across the spin. But
Laurent Freidel: but in fact, that's
Laurent Freidel: you know, I would say that that is not correct. and so there's this work that we did with with Daniel. And and I think it's an important work for for what you're discussing. And then we, of course, we're going to continue. Is that this is what we did with that nearly there are. And then
Laurent Freidel: where we kind of looked at the at the, you know, covariance, phase, space, action of the defamism on the face, base, variable. And then you what you can prove there is that this defamism can be written as some kind of translation constraint. So there's there's a part of the deformism which act as a notion of local translation on the spin, and of course, a local translation, in fact, you know, changes the spin.
Laurent Freidel: and therefore it means that you know there is a part of the of the constraints. And hopefully, we'll discuss more with Sima about that. But it's very important to appreciate that the defam
Laurent Freidel: action as a quantum operator, not as a kinematical things on the embedded graph, but as a as a quantum operator acting on the facepace. Viable changes. You know the relationship between the flux and autonomy, etc, and that relationship can be quantized and will and does, in fact, lead to a a very interesting extra measure of entanglement.
Laurent Freidel: because what it means is that now you you you know, anytime you cut, or if you look at the edges, then this edges needs to carry now a representation of some kind of local point of group.
Laurent Freidel: which is what it should be like even in a local space. You know, you don't have it just local, Lawrence. You also have some kind of form of local point of it. So implementing that local point cafe entanglement is going to be absolutely crucial in understanding what is the spin network entanglement? And and I think it's very important for the field that we move
Laurent Freidel: towards understanding this this translation, and then, before even we, we understand
Laurent Freidel: the the, the constraints and thing of it.
Simon L: Thank you.
Simon L: Simon.
Simone SPEZIALE: Yeah, I just had a quick taking a question. I was wondering if these
Simone SPEZIALE: projector to the peps, as you call them. I can obtain it by doing
Simone SPEZIALE: you know, an integral implementing the area matching constraint
Simone SPEZIALE: in the spinorial representation.
Simon L: I think so. Yes, in principle. Yes, because all it's doing really.
Simon L: wait, this is the wrong one.
Simon L: Yeah, this one here, all this is doing like the one on the top. Here is just taking the
Simon L: that say in Su, 2 data, the Su 2 representation on the left, and the 2 representations on the right and sort of matching them together to like have the same action.
Simon L: So in. So that includes the area matching constraint. I think it's basically equivalent to that.
Simone SPEZIALE: Okay, good. And another similar question very quick. If you remove the party team map you have in 7,
Simone SPEZIALE: you lose the maximal entanglement, or you still have it.
Simon L: No, those are related by Unitarian transformation.
Simone SPEZIALE: So you don't lose it. You said.
Simon L: You don't lose maximum entanglement.
Simone SPEZIALE: It's.
Simon L: Really that you can use any. It just leads to a change in parity.
Simone SPEZIALE: Right, yeah, yeah, so.
Simon L: Entanglement of the State itself doesn't care about that.
Simone SPEZIALE: Okay, right? Exactly. That's what I was expecting would have sounded weird. They just wanted to confirm. Thank you very much.
Simon L: Welcome!
Hal Haggard: Any further questions or comments.
Carlo Rovelli: maybe there's nothing else a technical question can can you say, can can you flash again your definition of the
Carlo Rovelli: of the States? Demographic states.
Simon L: For the here!
Simon L: Yep.
Carlo Rovelli: The state, the one, the the one with the map.
Simon L: Yeah, I would. I would use the general one, if you don't mind. But we can also take the one here.
Carlo Rovelli: This one, this one.
Simon L: Yes.
Carlo Rovelli: Because that's where you sort of use starting from
Simon L: Yes.
Carlo Rovelli: So metric the
Carlo Rovelli: the reason you
Carlo Rovelli: right? Why?
Carlo Rovelli: just trying to formulate question. Right?
Carlo Rovelli: what? What is exactly? What is the
Carlo Rovelli: what is the action on on the elastic quality on the left hand side?
Carlo Rovelli: Yes.
Carlo Rovelli: What is? What is the mathematics here? Exactly, I understand, said what is
Carlo Rovelli: seal site?
Simon L: So this doesn't just doesn't have the index. Psi, but it's the same thing.
Simon L: So it's just a linear operator from intertwina to boundary spin. And then the adjoint going from boundary spin to intertwine.
Carlo Rovelli: Right you don't use
Carlo Rovelli: you you!
Carlo Rovelli: Why don't you write it in the same notation as the
Carlo Rovelli: what are just trying to figure out.
Simon L: And the same notation as spot.
Simon L: I'll be. Ca, I just forgot to write the index lowercase. Sorry, that's really it.
Carlo Rovelli: No, no, this understand? But
Carlo Rovelli: phi! With a with index. I it's Bra. Is that what you're thinking?
Carlo Rovelli: It's a little.
Simon L: It's a it's, it's well, it's
Simon L: I mean, it's a. It's a linear map from from the intertwina space to the.
Carlo Rovelli: Oh, I see what you're saying. Okay.
Carlo Rovelli: okay, okay, okay, fine. And the
Carlo Rovelli: right is that
Carlo Rovelli: is that to?
Carlo Rovelli: Is that the only way of capturing the idea of this
Carlo Rovelli: allography, both to bulk or or this can be done differently. I guess that's what I want to do.
Simon L: Apologies.
Carlo Rovelli: End of the detail of this definition, the
Carlo Rovelli: the world conclusion.
Carlo Rovelli: Is there something else that one could require? Instead of that? Besides.
Simon L: Instead of isometry, you mean.
Simon L: I mean, what kind of photographic reconstruction, or
Simon L: what? What do you have in mind?
Carlo Rovelli: No, I'm just
Carlo Rovelli: wondering whether
Carlo Rovelli: so.
Simon L: So this is this is basically
Simon L: so maybe a bit of historical context for this the same the same exact definition is what was used in the context of tensor network ads. Cft.
Simon L: In that they have some bike Hilbert spaces which they reconstruct through a tensor network state from some boundary links.
Simon L: And the isometry question is, really it's it's it's used in that.
Simon L: If you have the boundaries. If you have a boundary state, and you know it's coming from the bike, and you have some operators acting on the bike. Then you can reconstruct it from all operators on the boundary.
Carlo Rovelli: Okay, okay? So that maybe that's the answer, the answer. What I wanted to know whether it's just
Carlo Rovelli: your way of capturing as a general idea, or it was deriving from
Carlo Rovelli: from a
Carlo Rovelli: specifically analogous formulation.
Carlo Rovelli: Oh, it's
Simon L: Latter, but I do think it's sort of a very concrete implementation of a general idea.
Carlo Rovelli: Okay.
Simon L: If you if you want, you could make this more generally by replacing the naive boundary, Hibbert Space by something bigger.
Simon L: Yeah, by placing some additional degrees of freedom on on the boundary vertices or something.
Simon L: Then I could imagine you can just be more. You can just reconstruct more stuff.
Carlo Rovelli: Okay.
Carlo Rovelli: Okay. Thank you.
Laurent Freidel: I mean, this is just the definition of Alo Yoshida, etc. You know.
Simon L: Exactly.
Laurent Freidel: That's how it is.
Laurent Freidel: They kind of formalize.
Carlo Rovelli: Okay, okay, so there's a history behind this definition. Okay.
Laurent Freidel: Yeah, yeah, that's that's kind of the quantum information definition of holographic intensive network.
Laurent Freidel: I do have a question, though, about 20
Laurent Freidel: because on the right hand side. You write the the identity on I but
Laurent Freidel: you mean I is a code subspace, right? It it cannot be the entire sets of, so I would. I would imagine that 5 5 dagger in general is a projection right.
Simon L: Yes, yes, I mean.
Laurent Freidel: Projection. So maybe for people we're not familiar. That's maybe the confusion here is that you know in in this, in this photographic map, what what you're imagining is that of course, you're not looking at the space of all possible
Laurent Freidel: bulk intertwiner. But you're looking.
Simon L: Yes, that's true. Yes.
Laurent Freidel: Call a good subspace, so maybe it should be called IC, or right, which is a subspace of intertwiner for which you have isometry right.
Simon L: Yes, I've been cavalier about including that part here, but you're completely right in principle. You can only reconstruct a small subspace of the full intertwining space, and that's the only requirement you have really.
Laurent Freidel: And and the idea about the constraints when I keep saying about the constraints differ, and a Miltonian is that that space
Laurent Freidel: is maybe supposed to be the the image of the projector onto the physical little bit space. Right? So that's.
Simon L: Hmm.
Laurent Freidel: So it's clear that there's a code subspacing gravity. And okay, that's right. That's the one thing which is clear.
Laurent Freidel: Yeah, whether that could subspace is reconstructable from the edge. That's the central question.
Simon L: Yes.
Carlo Rovelli: Yeah, agree? That's a that's a super interesting question. And it's a it's a
Carlo Rovelli: and now it's in this language is, there's a nice sort of technical way of formulate. Anyone could still turn it around in one way or the other is not
Carlo Rovelli: unique. But that's and I think that puts clear where the hypothesis is.
Carlo Rovelli: This is this is the this is in this language.
Carlo Rovelli: One way of formulating the holographic hypothesis. A lot. A lot follows from that right
Carlo Rovelli: black. What is black pollen to be, for instance.
Simon L: Hey!
Hal Haggard: Good. Shall we thank our speaker one more time? Are there any last comments.
Simon L: From the people at Western.
Hal Haggard: I think they they just had the video on.
Simon L: Yes.
Western: Sorry. Just a small one, just.
Simon L: Yes.
Western: Partner. Discussion.
Western: if I make correctly, for instance, indograph is, I think, most of the box reconstruction only concern the tongue. Language right.
Western: So not the bulk. Do you have an intuition of whether something similar is happening here, and whether your restriction on the Internet space
Western: somehow sending us that.
Simon L: So
Simon L: there are hints towards that. For example, from the Ad. Cft discussion. They also find that it's
Simon L: possible to
Simon L: to reconstruct bike operators only in some sort of causal wedge
Simon L: which is included in the entanglement, which
Simon L: but it's difficult to say that really, because even on that level.
Simon L: it's really only a given single time. Slice.
Simon L: and those co entanglement and causes wedges are usually constructed from having a full spacetime picture.
Simon L: so the interpretation of that stuff is a little bit. I would take it with a grain of salt.
Simon L: and
Simon L: on the side I don't, I think, in the formulation I give on this slide here. What you could do to test this sort of thing is
Simon L: you sort of reduce the algebra of inputs
Simon L: and the algebra of outputs
Simon L: to be sub regions of the of the bulk.
Simon L: So, for example, for the output regions, you just say, you see. But take a boundary segment.
Simon L: and then you take some other bike region of intertwiners.
Simon L: And then you ask, how do I have to modify these regions so that I get reconstruction?
Simon L: So that's something I could imagine you can do, but that is hampered a little bit by the question of What do you do when the state in question here, when the row is a mixed state.
Simon L: and it seems like there. The and the analysis with this sort of
Simon L: with this sort of stuff is quite difficult.
Simon L: So far I've not been able to
Simon L: make it work so well. It seems like the the criteria with like maximum entropy, only seems to work so well when you have a pure state.
Western: I see. Thanks.
Hal Haggard: Alright with that, let's thank Simon again. Thank you so much, Simon.
Simon L: You're welcome.
Simon L: Thank you for inviting me.