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Jorge Pullin: Okay. So our speaker today is Eugenia Colafrancheski, who will speak about entropy and Hilbert spaces from gravitational path integrals.

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Eugenia Colafranceschi: Okay, so thank you very much for the invitation.

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Eugenia Colafranceschi: So today I'm presenting and work in collaboration with on gravitational entropy.

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Eugenia Colafranceschi: and so in particular, how to construct entropy from the gravitational path integral

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Eugenia Colafranceschi: and make sense of. The formula called the Rutakayana G formula, when we don't have all of the feed.

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Eugenia Colafranceschi: So let me start from the very beginning of this story, at least from historic point of view.

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Eugenia Colafranceschi: So about talking how to compute entropy through the path integral. So for a standard. The quantum system you can compute the

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Eugenia Colafranceschi: partition function, the trace of E to the minus beta h by doing a integral along close the s. One manifold

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Eugenia Colafranceschi: and the so from thermodynamic arguments you can derive the entropy for this standard the quantum systems.

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Eugenia Colafranceschi: according to this formula. So we can ask, What what should we do for a gravitational system?

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Eugenia Colafranceschi: And so you can make the following guess.

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Eugenia Colafranceschi: which is a also motivated by the fact that the gravity is a boundary term.

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Eugenia Colafranceschi: So the physical time translations are only on the boundary.

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Eugenia Colafranceschi: and so you can imagine to do a path integral over all possible geometries and topologies. Whatever receive you have before your path integral

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Eugenia Colafranceschi: by using as boundary a manifold which, as one manifold, so a direction in which your Euclidean time is periodic.

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Eugenia Colafranceschi: Now

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Eugenia Colafranceschi: this guess turns out to make sense in the case of a black hole. Thermodynamics. This is a work by working and

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Eugenia Colafranceschi: following that. Then the computation of the so we can consider, we can consider the path integral with boundary conditions, so

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Eugenia Colafranceschi: that are given by this manifold with a the topology of a circle, and whatever boundary you have for your space.

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Eugenia Colafranceschi: and interpret the the quantity that has been computed as the partition function of your system.

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Eugenia Colafranceschi: and then you can use the saddle point approximation, and so approximate this quantity with the dominant saddle. And again computing, what you would compute in the thermodynamic case. So

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Eugenia Colafranceschi: for the entropy. And this recipe turns out to give you the baggage time working entropy. So this is a a first hint of the fact that the

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Eugenia Colafranceschi: this, this guess makes sense. So this computing

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Eugenia Colafranceschi: Gravitational entropy through the Euclidean Path integral

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Eugenia Colafranceschi: can indeed be a meaningful technique.

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Eugenia Colafranceschi: Now, not that this is a special case. So this actually works when you have a killing vector field around your S one boundary

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Eugenia Colafranceschi: erez agmoni, and so indicate in the case in which you interpret the state prepared by Euclidean, but integral as an equilibrium state 150.

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Eugenia Colafranceschi: So the question would be, how can you generalize this story? So the computation of entropy through the freedom but integral when your S. One boundary manual does not have this time translation, symmetry.

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Eugenia Colafranceschi: So which boundary conditions should we use for the party integral? To compute the

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Eugenia Colafranceschi:  something that we can interpret as as a an entropy for our gravitational system. So let me formalize that the the gas I made before, so suppose that the boundary conditions which are relevant to computing some gravitational quantity

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Eugenia Colafranceschi: are given by whatever that integral would compute the same quantity in a non gravitational theory. So this is what we have done before. Let me go back just a second. So

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Eugenia Colafranceschi: we said that for a standard quantum system computing a path integral along an S. One manifold gives you a partition function.

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Eugenia Colafranceschi: When you use this as boundary for your gravitational part integral, you compute something that behaves like a partition function.

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Eugenia Colafranceschi: So this seems to be the the main guess. So let's say, if you want to compute some quantity in our gravitational theory, let's use as boundary conditions the path integral that would compute the same quantity in a non-gravitational theory.

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Eugenia Colafranceschi: Now, we are interested in gravitational entropy. Let me point out. So usually, it's very difficult to compute the phone because it's difficult to compute the trace of roll log.

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Eugenia Colafranceschi: So very often. What is use is the so-called replica trick in which you compute the phone. I'm an entropy as a limit of a rainy entropy. So Rani entropies are functions of

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Eugenia Colafranceschi: the trace. Of your density, matrix to some exponent. N, and this quantity is much easier to compute.

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Eugenia Colafranceschi: So let's use this replica tree. So compute the phone line and entropy as a limit of this Rani entropies. So which boundary conditions should we use to compute this trace overall to some power, and

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Eugenia Colafranceschi: according to the guests I stated before, we first need to figure out which quantity would compute something like trace of raw to some power and in a non gravitational theory.

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Eugenia Colafranceschi: So let's look at what happens in quantum field theory in quantum 50, or if you want to compute, for example, the second order, Rennie entropy.

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Eugenia Colafranceschi: you have a manifold like this. So this upper part is representing the that would compute the certain state side on the question surface, which is indicated here as the dash line

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Eugenia Colafranceschi: and imagine that on this question surface you have a partition into to register R and R. Bar. and you want to compute the entropy of the state for this region are

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Eugenia Colafranceschi: so. This upper part is preparing the state side the cat, so you can mirror it. To have something computing the Bra.

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Eugenia Colafranceschi: And then, if we want to compute the second order, any entropy, we need 2 copies of the system because we want to compute the trace of rho squared. So we take 2 copies.

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Eugenia Colafranceschi: We want to trace out our bar. So we sue together the the 2 our part regions. and then to compute the final trace. We we need to blue the adjacent regions of the our

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Eugenia Colafranceschi: sub-region. So

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Eugenia Colafranceschi: this is a very so this is the manifold that would compute this quantity

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Eugenia Colafranceschi: for a certain state, which is the state to prepare by this Euclidean fat interval.

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Eugenia Colafranceschi: Now, according to this, guess, what we want to to compute something that

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Eugenia Colafranceschi: to be interpreted as the trace of rho squared in the gravitational theory. We need to use these as boundary conditions for our gravitational.

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Eugenia Colafranceschi: So here it's just a cartoon not to to

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Eugenia Colafranceschi:  to convey the idea that this boundary manifold is going to be the input of our gravitational path integral.

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Eugenia Colafranceschi: Now this is what was actually done by Lacoix and Maldacina in 2,013.

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Eugenia Colafranceschi: So they show that the indeed, that the computation of gravitational entropy from the particular can be generalized to the case in which you don't have time translation emailed on your boundary.

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Eugenia Colafranceschi: And so they use the repeat and perform that calculation according to the gas. stated in the previous slide, and they were able to derive what is called the formula.

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Eugenia Colafranceschi: So given a certain region R

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Eugenia Colafranceschi: in your a certain region are in your space-time boundary, so the region R will be

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Eugenia Colafranceschi: codimension 2 in the full theory. So codimension one from the boundary perspective

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Eugenia Colafranceschi: performing this replica trick and using the gravitational integrals. So without any input from allography just I mean, at least

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Eugenia Colafranceschi: in the the technique to to perform the computation. Do not use holography

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Eugenia Colafranceschi: account with. So with this gravitational, you can compute the quantity that you want to interpret as the entropy of a certain region are.

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Eugenia Colafranceschi: and this entropy is given by the area of a certain surface in the bulk. and it turns out to be the surface that the minimizes the area functional.

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Eugenia Colafranceschi: Now, this was first confirm theory correspondence. But, as I said, the autography does not enter the derivation because you just use. Yeah.

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Eugenia Colafranceschi: yeah, you are still computing the entropy of some boundary region.

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Eugenia Colafranceschi: And indeed, the interpretation of what is computing is subtle and as I was going to say, you require at this stage holography to know what you are actually computing them.

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Eugenia Colafranceschi: But the techniques are purely from the bulk perspective. So you are not relying on a quantum feist theory, on on some duality with the content. 50 on the boundary

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Eugenia Colafranceschi: is only gravity, but to know which boundary conditions use to compute that grav, that gravitational quantity you

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Eugenia Colafranceschi: the hint that comes from what you would compute for a boundary theory. but I mean from the gravitational point of view. So I have my pass integral. I want to know which boundary conditions to put them, to get the number out of it.

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Eugenia Colafranceschi: and to know which boundary conditions to put to you. You use the the input from them. Non gravitational queueing. So allography does not enter the derivation, but it is required to have an interpretation of this quantity as a standard and through P.

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Eugenia Colafranceschi: Why? Well, because so which state are we computing from the gravitational perspective? You don't know which state you are assigning to the region are, I mean, if you have a dual boundary theory, then, is clear that you are computing

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Eugenia Colafranceschi: the entropy of the the state which the that's the non gravitational series assigning to our. But from the back perspective. First, the the space from, I mean does not factorize between the region R. And the complement that race is not well defined, and so a state for the region. R is not well defined.

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Eugenia Colafranceschi: Now this receiver, however, seems to have clear implication for the case of an evaporating black column.

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Eugenia Colafranceschi: Indeed! Has been shown that when we consider the coupling between gravity in a hunting de sitter space time with a non gravitational bath.

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Eugenia Colafranceschi: Then the phone 9 and entropy for the bus is given by what is called the Day Island formula, which is a special case of the formula with quantum corrections.

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Eugenia Colafranceschi: and this recipe so gives you an entropy that reproduces the page. Quote.

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Eugenia Colafranceschi: these are the the set of results from Pennington, Angela, Mara, Maxfield, Almiri, and from 2,019

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Eugenia Colafranceschi: so the first derivations that were in ads cft, and then that generalization to the case in which you don't need allography. You just use gravitational radiance.

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So

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Eugenia Colafranceschi: a possible solution in this case for the problem of the interpretation of what you're you're computing comes from a work by Marathon Max Field in 2,020 and so they show that what you are computing is the entropy of the working radiation in a given super selection sector of your space.

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Eugenia Colafranceschi: So

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Eugenia Colafranceschi: Now, this is just to to remark that this Utah kayana G formula, this derivation from gravitational replicas

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Eugenia Colafranceschi: as the problem of the interpretation from the bulk perspective. In this case of Eds gravity coupled to a non-gravitational bath, this story seems to

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Eugenia Colafranceschi: be clear, even without allography. When you consider when you add the this Max field perspective.

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Eugenia Colafranceschi: That's so. Again, the question is, can we generalize this story? So apart from now, we we talked about this special case of. But the question is, can we so generalize this techniques? of computing gravitational entropy from the past, integral and having something that is actually computing an entropy

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Eugenia Colafranceschi: in gravity. even if we do not assume holography.

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Eugenia Colafranceschi: so to be more concrete to consider a gravitational system.

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Eugenia Colafranceschi: We 2 asymptotic boundaries.

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Eugenia Colafranceschi: be let's say, left and right. So be L. And BR. Now the heatherspace. that you associate from the bark perspective! To a system with these 2 boundaries a priori does not factorize.

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Eugenia Colafranceschi: Of course, if you ever log in a fee. So if you have a an analogic dual theory associated to this bound, that is, then you have a factorization of the best base. And so there is a clear sense in which you can associate an entropy. To the left, the boundary to the right boundary.

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Eugenia Colafranceschi: But our goal is to not consider allography. And so our goal is to construct a here best space associated to, for example, the left boundary.

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Eugenia Colafranceschi: and to prove that the can actually be understood that as computing a standard entropy for the left system, so an entropy in terms of a standard trace on a Hilbert space basis.

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Eugenia Colafranceschi: And, as I said, this is not clear, because a priori, this in the space does not factorize.

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Eugenia Colafranceschi: Let me close the motivation part with the a different perspective which comes from the recent results in the phone line and algebra. So

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Eugenia Colafranceschi: context. So recent works. From Pennington and Week, and and many others. Ii mentioned only few of them.

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Eugenia Colafranceschi:  I've shown that in very special context. And again to be can be interpreted as an entropy of a type 2 phone line and algebra.

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Eugenia Colafranceschi: That's there.

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Eugenia Colafranceschi: Yeah, composes seem to type 2 factors.

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Eugenia Colafranceschi: Yeah.

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Eugenia Colafranceschi: so, and the

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Eugenia Colafranceschi: So, however, what we want for us, so what we have for a standard quantum system is an entropy in terms of a human space race. which is what provides the entropy with a State counting interpretation.

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Eugenia Colafranceschi: And the the heater space trace corresponds to what is a type? One trace. So type one phone line and algebra is the standard algebra of bounded operators on a Hebrew space. So what we want is actually an entropy in terms of the standard type, one trace.

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Eugenia Colafranceschi: So again, can we understand again, to be as computer through the gravitational integral as a state accounting end, to be in terms of a standard, the Hilbert Space.

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Eugenia Colafranceschi: And so what I'm going to show today is that if you have a a quantum gravity theory which is asymptotically, and your Euclidean path integral, satisfies a simple set of actions.

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Eugenia Colafranceschi: Then it is possible to us to say the standard, the phone line and entropy to your left or your right boundary.

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Eugenia Colafranceschi: And this entropy in the semi-classical limit is given by the formula, and there is no need to invoke holography. You just need a set of options for your

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Eugenia Colafranceschi: okay. So this is the line. I will start by stating the actions for the path integral. Then I will show how to define a heal. Best pace from your path, integral.

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Eugenia Colafranceschi: how to define an algebra operators. So from it.

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Eugenia Colafranceschi: then it will show that this algebra is actually composed of type one for 9 months. So and so in the end. that this structure for your operator algebra, send your heater. Space gives an interpretation to the entropy in terms of a standard Hilbert space. So as a state, the counting entropy

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Eugenia Colafranceschi: questions so far.

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Eugenia Colafranceschi: Good. Okay.

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Eugenia Colafranceschi: okay, so let's start with the actions for the Euclidean integral

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Eugenia Colafranceschi: now, before stating the actions. So let me just build some intuition for them.

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Eugenia Colafranceschi: So what is so an object that we might call in our theory. And Euclidean path integral is something that is a map. That associated to every closed boundary with some boundary conditions.

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Eugenia Colafranceschi: a complex number. So in this case

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Eugenia Colafranceschi: we can consider this is a so we don't want to. assign any receipt to our pass integral. So we just want to.

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Eugenia Colafranceschi: So it then we we just require the a set of actions for it. But just to to give you any day, or to motivate the actions. So now I'm considering, I think again, in terms of a sum over geometry and topologies, which is, what we usually

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Eugenia Colafranceschi: think of for for the integral. So in this case the path integral so is a performed of a set of bulk fields, and II

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Eugenia Colafranceschi: they noted them as 5, and this fine includes both the bathmatic and whatever matter Field, so you might have in your theory.

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Eugenia Colafranceschi: And so your potential is integrating over all possible bark field configurations

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Eugenia Colafranceschi: which satisfies a set of boundary conditions, so

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Eugenia Colafranceschi: which I denoted by M. So not that here M is not just the boundary manifold, but also the set of boundary conditions assigned to it. So

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Eugenia Colafranceschi: the the boundary conditions for your magic and for your matter, fields.

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Eugenia Colafranceschi: and in the following II will refer to these as source manifests to

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Eugenia Colafranceschi: Convey the idea that is the manifold in which you specify both the boundary and the sources for your matter fields.

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Eugenia Colafranceschi: And with this notation I mean that I'm integrating over all 3 configurations that satisfy these boundary conditions.

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Eugenia Colafranceschi: So now, what you expect from

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Eugenia Colafranceschi: like this with Samsung geometries and topologies? Well. you you want it to be finite for smooth boundary conditions.

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Eugenia Colafranceschi: You expect it to be continuous under small deformations of your boundary conditions.

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Eugenia Colafranceschi: And the since the action should be a real function you expect

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Eugenia Colafranceschi: the path integral itself to to be a real function. So with this notation, I mean, the same boundary, manifold, but with complex, conjugated the sources.

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Eugenia Colafranceschi: And then, what else? So consider the path integral. and imagine to cut it into 2 regions, so to cut it along a certain surface, Sigma.

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Eugenia Colafranceschi: and then that. So

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Eugenia Colafranceschi: this means that you are cutting every configuration entering in your path, integral in 2.

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Eugenia Colafranceschi: And so the geometry and the topology of Sigma will depend on the particular configuration you're cutting. But the boundary of this surface. So the the one intersecting the space time boundary is actually independent on the particular configurations, because it depends on the initial boundary conditions. So

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Eugenia Colafranceschi: you put on your boundary manifold.

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Eugenia Colafranceschi: So this is why the Hilbert Space associated to this Sigma might be labeled by

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Eugenia Colafranceschi: this partial sigma, the boundary of this cushy surface.

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Eugenia Colafranceschi: and so, given a a path integral, you can interpret it as computing the inner product between 2 States. So the 2 States prepared

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Eugenia Colafranceschi: By the lower part and the upper part of your pass integral.

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Eugenia Colafranceschi: And so, in particular, when the the 2 sides coincide.

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Eugenia Colafranceschi: you are computing the norm squared of your state, and so you expect this

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Eugenia Colafranceschi: time reversal. Symmetric path, integral, to be positive, definite.

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Eugenia Colafranceschi: So, as I said that this was to give you some motivation for the actions for your path integral. So let me state now. Let me state them

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Eugenia Colafranceschi: so. The first one is the which is a Mac from boundary conditions defined by smooth, manifold to complex numbers, we expect it to be finite. So to give you a a complex number.

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Eugenia Colafranceschi: we expect it to be a real function of the possibly complex boundary conditions.

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Eugenia Colafranceschi: Now you expect the patent to get to be reflection positive

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Eugenia Colafranceschi: what what the I showed about to the norm computing the norm squared of the state.

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Eugenia Colafranceschi: then you require you might want to require continuity. This is actually a very weak condition. So

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Eugenia Colafranceschi: by continuity, women, that if the boundary manifold contains a a cylinder of a certain size, epsilon, the path integral is continuous under changes

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Eugenia Colafranceschi: of this epsilon.

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Eugenia Colafranceschi: so here the ceiling, that is the topology of the which is the the core dimension to surface the

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Eugenia Colafranceschi: still calling this case times some interval. And so if you deform if if you change the length of this interval, your path integral.

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Eugenia Colafranceschi: is continuous.

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Eugenia Colafranceschi: Now, the the last actions

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Eugenia Colafranceschi: which may be the last obvious one is factorization. So if you have 2 closed boundary manifolds.

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Eugenia Colafranceschi: then you ask that the department under this joint union of these 2 manifest factorizes. So it's given by the product of the first manifolds

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Eugenia Colafranceschi: times the particular integral over the second month.

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Eugenia Colafranceschi: Now, we expect in some situations we expect the integral to be equivalent to a collection of what is called the that'd be universal super selection sectors. So yes, you might expect your theory to decompose into such sectors. And in that case,

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Eugenia Colafranceschi: this factorization property holds sector by sector.

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Eugenia Colafranceschi: So this is a a property about what the the like. The contribution of the in your theory, because in each sector, this you are, you are basically

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Eugenia Colafranceschi: and

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Eugenia Colafranceschi: canceling any contribution from Walmart. So between these 2 space time regions.

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Eugenia Colafranceschi: and so as proved recently by Marathon Max Field. You can expect your theory to. They compose into these sectors. And so

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Eugenia Colafranceschi: your theory to satisfy these factorization property sector by sector. So our analysis will apply in this sense.

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Eugenia Colafranceschi: Okay, so now that we have the actions for the let's start with the definition of a Hilbert space from it.

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Eugenia Colafranceschi: So when we cut open the passive integral which is now I'm I'm I'm seeing that there's a way

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Eugenia Colafranceschi: informal procedure. So what we what we do is to cut the close boundary into 2 pieces. So which I'm calling here, and one and and 2,

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Eugenia Colafranceschi: and these 2 pieces are such that they they share the same

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Eugenia Colafranceschi: boundary. So this this red boundary here, which is a codimension to surface from the bath point of view.

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Eugenia Colafranceschi: And so these 2, as as I mentioned before, you can imagine your food but integral to compute the inner product between 2 States which are the States prepared by the path integral. With this n. One and and 2 boundary conditions separately.

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Eugenia Colafranceschi: So we might consider this set of these surfaces with these partial and the boundary. but when we glue them back together. We want the glueing to be uniquely determined that because if I glue back and one and and 2,

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Eugenia Colafranceschi: then I want to recover the the original manual from which they were cut, and so a way to ensure that the the gluing procedure is well defined is to leave that point. So

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Eugenia Colafranceschi: along these boundaries. but also we want to be. We want to ensure that the manifold, the close manifold that we obtain from the viewing is most.

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Eugenia Colafranceschi: and this is why instead of constructing the full quantum gravity per space, we construct only some sectors which are the one the ones in which this boundary camps with the rim.

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Eugenia Colafranceschi: So basically, we require that in a neighborhood of this partial and boundary

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Eugenia Colafranceschi: the geometry is, I'm sorry, is diffomorphic to the ceiling that I was showing before.

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Eugenia Colafranceschi: This is a way to regularize the the gluing procedure.

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Eugenia Colafranceschi: So

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Eugenia Colafranceschi:  Once you have your the set of your surfaces with this partial and boundary denoted here as this Y partial hand, so in this case, I'm assuming the bath to be B plus one dimensional. And so this.

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Eugenia Colafranceschi: no, no, nothing. I'm just so. Yeah.

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Eugenia Colafranceschi: so

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Eugenia Colafranceschi: so given this set of services. So we can associate to every sources and a state in what is actually appreciable space with the inner product defined by the path integral.

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Eugenia Colafranceschi: And then to get a proper heal best phase. So you need to consider the quotient to buy the new States and then complete the the result in the usual sense. And so you you get best pace

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Eugenia Colafranceschi: associated to this passage and boundary.

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Eugenia Colafranceschi: So we can construct. So not just the human space, but also operators.

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Eugenia Colafranceschi: Yeah. So before you have an inner product, you also need a linear structure. So how do you define? Yeah, yeah. Sorry I didn't. Yeah. Tha, that's true. II actually from by this set. I mean, not just the surfaces, but the only combination of them.

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Eugenia Colafranceschi: Picture the linear combination of manifolds.

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Eugenia Colafranceschi: I am just taking the sum over different surfaces with some coefficients that can be complex. So I define so these

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Eugenia Colafranceschi: conjugation operation, which is just the pro the reflection

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Eugenia Colafranceschi: of your surface and the complex conjugation of all sources on it.

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Eugenia Colafranceschi: So I can just take a linear combination in the sense of literally taking the some over this surfaces. So with arbitrary coefficients, and when I glue things together.

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Eugenia Colafranceschi: then I need to glue. Of course, all the I mean, if I am gluing 2

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Eugenia Colafranceschi: Samsung over surfaces than I glue all the elements of one sound to all the elements of the other sound

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Thomas Thiemann: and the elements of each sum are considered to be disjoint.

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Eugenia Colafranceschi: Yes.

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Thomas Thiemann: okay, thanks.

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Eugenia Colafranceschi: Yes. yes. So in this sector there are also the the linear combination of them. Thanks for the question.

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Eugenia Colafranceschi:  okay. So

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Eugenia Colafranceschi: so now,

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Eugenia Colafranceschi: we can specify. So so far, we consider surfaces with the generic boundary partial. And now we can construct the operators that have an actual action on our Hilbert space. Consider, for example, the case in which your boundary is actually composed of 2 pieces.

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Eugenia Colafranceschi: which I will call BL,

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Eugenia Colafranceschi: and so you have the surfaces with this to be boundaries, and as I'm going to define in the following here, just to to give you an idea what's going on. We will. The we will define operators up from the surfaces acting

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Eugenia Colafranceschi: on our Hilbert space. And these surfaces, which show, for example, be a boundary, will preserve a sector of the Hilbert space.

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Eugenia Colafranceschi: which is defined for 2 boundaries, in which one of them is vl.

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Eugenia Colafranceschi: so these elements will preserve the this sector of your Hebrew space.

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Eugenia Colafranceschi: So let me

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Eugenia Colafranceschi: so let me formalize this. So

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Eugenia Colafranceschi: first of all, we can define on this set of surfaces. So with 2 boundaries, a left product and the right product. So if you consider 2 elements, and you label the 2 boundaries as left and right, then you have a left product. When you glue the left boundary of B to the right boundary, or a right product for the other way around.

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Eugenia Colafranceschi: Now this act.

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Eugenia Colafranceschi: all the surfaces, but also linear combinations of them, was remark before, it keeps the way this left and right product defines what we call left and right. So face algebra.

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Eugenia Colafranceschi: Now we also introduce. There is a natural involution on this algebra, which is also an isomorphism between the left and the right algebra and the star operations is a reflection of the surface.

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Eugenia Colafranceschi: a complex conjugation of all sources on it, but also this working of the 2 left and right labels.

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Eugenia Colafranceschi: And so we will see that the there is a natural sense in which the left to surface algebra acts. On the left boundary of a hidden space defined with the left and the right and the right surface algebra. So the surface algebra with the right product acts on the right one.

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Eugenia Colafranceschi: So crucially, the path integral defines a trace operation on this algebra. So if you consider an element, so a surface a.

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Eugenia Colafranceschi: then a trace, for this element is just given by the path integral.

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Eugenia Colafranceschi: which has a as boundary conditions, so demanding for the obtained by gluing the let's say, left and right boundaries of A

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Eugenia Colafranceschi: and the from the dictionary that we saw before between the ring surfaces and states. Then what you have is that 2 states associated to 2 surfaces A and B.

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Eugenia Colafranceschi: The inner product between them

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Eugenia Colafranceschi: is the trace of 2 of the corresponding elements for the algebra, and is given

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Eugenia Colafranceschi: indeed, by the path integral performed by the close manifold obtained from the gluing of the A surface with the B surface. So, as you can see, this is style operation, and Tessa

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Eugenia Colafranceschi: in the map between your bra and the the corresponding algebra element.

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Wieland, Wolfgang Martin: And sorry. Excuse me, could I ask a quick question.

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Wieland, Wolfgang Martin: If you perhaps if you go back to the previous slide, it is easier to understand the question. So this trace should I think of it as a partial trace in the Hilbert Space, or as a trace really over the entire Hilbert Space. I'm asking this because I'm wondering if there are other like boundary conditions along the cylinder where you put like an X in your picture.

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Eugenia Colafranceschi: Yeah, thanks for the question. So first, so my surface comes with certain boundary conditions attached to it. So this is why I mean, you have different operators, so different elements depending on the boundary conditions you put on your surface.

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Eugenia Colafranceschi: The trace is a trace on the full. Hilbert space. So in this case if we are considering

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Eugenia Colafranceschi: this operator as acting on a hidden space associated to.

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Eugenia Colafranceschi: So let's name these 2 boundaries B.

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Eugenia Colafranceschi: So this is an operator that has a natural action on a human space associated to be.

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Eugenia Colafranceschi: And so this turns out to be the trace on the full.

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Wieland, Wolfgang Martin: okay, thanks. Thank you. That that clarifies your question. Thanks.

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Eugenia Colafranceschi:  okay? So in this case, of course, your state actually lives in a 2 boundary space, right? Because

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Eugenia Colafranceschi: your state so maybe scared from this point. So this is a state in the Hebrew space associated to left. And right.

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Eugenia Colafranceschi: So in this case. These States States, on a two-boundary Hilbert space.

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Eugenia Colafranceschi: and the trace in this case is not, is, is defined. For from the path integral

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Eugenia Colafranceschi: from from the gluing of these 2 surfaces together

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Eugenia Colafranceschi: now.

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Eugenia Colafranceschi: you can use the 2 boundary surfaces so to define elements of a a bigger set of a set of 4 boundary surfaces. So if you imagine.

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Eugenia Colafranceschi: Now, so consider the Hilbert space given by associated to 4 boundaries, that here I'm indicating as L. One so left, one right, one left, 2 and right? 2.

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Eugenia Colafranceschi: So this is a way of constructing 2 states starting from the same A B surfaces.

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Eugenia Colafranceschi: And so

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Eugenia Colafranceschi: here, what? What we can prove is that the trace defined by the path integral?

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Eugenia Colafranceschi: satisfies A trace inequality, thanks to the positivity of the inner product on this 4 boundary, Hilbert Space.

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Eugenia Colafranceschi: So consider the 2 States. I show here. and if you consider the inner product between them, you can see just from the picture that the inner product coincides because it's just given by the performed over these 2

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Eugenia Colafranceschi:  donuts. Let's see.

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Eugenia Colafranceschi: So from just the Kushi's vaccine equality, which is a consequence of adding a positive in a product on your 4 boundary, Hilbert Space.

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Eugenia Colafranceschi: You can see that the trace defined by the path integral satisfies this inequality, and this inequality is crucial because is crucial for the results we get.

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Eugenia Colafranceschi: as I will show later. But first let me talk about our representation of our algebra on the Hilbert space, because so far we only talk about an abstract operator algebras in the Hilbert space.

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Eugenia Colafranceschi: But, as I said, there is a natural sense in which this algebra acts on the Hilbert space, so in particular, we can associate to every surface a a corresponding operator.

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Eugenia Colafranceschi: and this operator acts on a certain state to be on our ill-bus space just through the gluing of the

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Eugenia Colafranceschi: of a boundary for your a element to the boundary of the B element corresponding to your State in the Hebrew space. So with this left product. The left operator algebra is another relaxa on the left boundary of your Hebrew space.

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Eugenia Colafranceschi: And as I said, is crucial the tracing equality, because, it allows you to prove that these operators are actually bounded.

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Eugenia Colafranceschi: and you can similarly define a representation of your right surface algebra on the right boundary of your Hilbert space, and again it is algebra will be composed of bounded operators.

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Eugenia Colafranceschi: Now we are going to focus on diagonal sectors of the best pace, meaning that we are going to consider, I think, best pace in which the left and the right boundary coincide.

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Eugenia Colafranceschi: and these are in particular, allows us to consider cylinders in our Hilbert space, which play the role of the identity operators.

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Eugenia Colafranceschi: And the these cylinder elements allows us to define a trace operation

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Eugenia Colafranceschi: on the integral, thanks to the continuity action, indeed, that we can define now a trace for our inverse space operators.

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Eugenia Colafranceschi: By this formula. So by considering the action on 2 cylinder states and taking the limit for the length of the cylinder that goes to 0.

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Eugenia Colafranceschi: So this is considering the diagonal sectors. Is is what allows us to have a natural extension of our trace operation on the representation algorithm.

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Eugenia Colafranceschi: Okay? So now. so what we get is a set of banded operators on our Hilbert space.

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Eugenia Colafranceschi: And this means that we can define algebras starting from our operator algebras by taking the closer of that algebra within the space of Banda operator on our human space in the week, operators topology.

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Eugenia Colafranceschi: And so we also show that the trace that we defined on the representation algebra can be extended to all positive elements of the phone women algebra.

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Eugenia Colafranceschi: Now. We can additionally prove that this trace is faithful, so the trace of an element a is 0 if and only if the A is the null element

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Eugenia Colafranceschi: is normal. and it's semi finite and having a semi finite trace basically means that not so ensures that not all elements have an infinite trace.

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Eugenia Colafranceschi: And by using an extension of the 4 boundaries argument that I showed before

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Eugenia Colafranceschi: that you can prove that the trace inequality also holds on the on the formal man. Algebra

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Eugenia Colafranceschi: this trace inequality is a once again crucial because in particular can be so if you consider the trace inequality

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Eugenia Colafranceschi: apply to to A B elements which are the same projection. P. Then, you get a constraint on the trace of your projection. So you get that the trace must be bigger or equal to one.

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Eugenia Colafranceschi: So while this is important, well, let me recap some non results. About phone and algebra. So so first, we know that every phone line and algebra is a direct sum of factors. And these factors can be so. Factors are algebra with 3% and can be type 1, 2, or 3.

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Eugenia Colafranceschi: Now, there is no faithful, normal, and semi-fin on type 3.

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Eugenia Colafranceschi: And so our trace cannot be type 3

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Eugenia Colafranceschi: on type 2. For any faithful, normal, and semi final trace. There are non trivial projections with arbitrary small trace, and since from the trace inequality, we derive the a lower bound one for our web project for the trace of the projections. This means that we can only have type one factors.

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Eugenia Colafranceschi:  there is a question.

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Laurent Freidel: Yeah. Hello, Virginia.

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Laurent Freidel: again, you have a question about this this type one like, is it, you know? Is it an assumption or a derivation

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Laurent Freidel: in in the sense that you know here you have to assume that the projector is traceable.

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Laurent Freidel: Right?

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Eugenia Colafranceschi: But the project 3 traceable, because we extend that the trace operation to the, to the, to our algebra.

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Laurent Freidel: Yeah. But I mean, okay, because I have example in mind, like, you know, in tqft topological kind of theory, which is a way to satisfy most of the axioms.

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Laurent Freidel: If I take a TQFT. Which is rational, then the trace. You know, the projectors that is, the cylinder projector is traceable. But in in TQFT. That are associated, let's say, for instance, with Lauren's group

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Laurent Freidel: or or non compact groups which are not the rational, then we know that the trace operator. So, although your actions are satisfied. But then it's a little bit more different because the trace is not. The projector is not trace about there. So there's different class of topology, permanent fee theories which are example of this

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Laurent Freidel: axioms you're sticking. So right? So maybe for you, this should be an assumption. I mean, it seems it's an assumption that you're only going to look at the rational Topology board bottom p theoryy.

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Eugenia Colafranceschi: so these are condition on the project. Some come from the trace inequality which interns can from the positivity axiom.

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Eugenia Colafranceschi: the trace of P is finite. Right? I mean in, let me say, if I take is not necessarily finite.

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Eugenia Colafranceschi: the trace of P is bounded

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Eugenia Colafranceschi: as a lower bound. It's bounded by one.

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Laurent Freidel: Yeah, because on a on a phone line algebra, the trace is allowed to take to take infinite value. So we are not saying that to the trace indeed! Actually, there is no upper bound on the trace of P.

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Eugenia Colafranceschi: Integer values. But I'm bounded.

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Laurent Freidel: And in type 2, one, the trace of P would be 0, right

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Eugenia Colafranceschi: in type

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Eugenia Colafranceschi: can be 0. Yes. But the since we have a lower bound one.

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Eugenia Colafranceschi: You can have arbitrary. You can have trace 0, for example, the lower bound that comes from using applying the trace inequality to a projection fee.

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Laurent Freidel: Yeah, you you get the trace squared

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Laurent Freidel: minus or equal to trace. P.

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Eugenia Colafranceschi: Trace P.

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Laurent Freidel: And if trace P. 0, this equality is violated.

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Eugenia Colafranceschi: But trace peak cannot be 0, because we proved that the trace is faithful.

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Laurent Freidel: Your trace is faithful, for where? Yeah. But what is the

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Laurent Freidel: I mean a

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Laurent Freidel: where. The question is whether the projector piece part of your algebra right? Which in general it might not be. That's the I think it's it's related to this rationality. So in rational topology or kind of theory, the projector is part of the but in on rational is not part of the algebra. Right?

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Laurent Freidel: So I agree that Facebook means that elements of the algebra are represented with nonzero trace. But

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Laurent Freidel: it doesn't prove that projectors like cylinder project. I mean, it could be next class. You're putting that the cylinder projector is part of the algebra, and then you should continue. It was just other examples which are not like that. That's all.

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Jerzy Lewandowski: Okay.

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Eugenia Colafranceschi: because the the projection would be one of the surfaces we started from.

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Eugenia Colafranceschi: Well, okay, we we, we might continue

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Eugenia Colafranceschi: later. So

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Eugenia Colafranceschi: okay, so so we get that the our factors must be type one factors.

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Eugenia Colafranceschi: And then the commutation theory theorem for semi final traces. So then tells us that this left and right algebra are competence on the Hilbert space on our 2 boundary Hilbert space.

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Eugenia Colafranceschi:  then we can also show that the central operators of our algebra have a purely discrete spectrum.

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Eugenia Colafranceschi: And so we can simultaneously diagonalize all these central operators. And so we get the decomposes into this direct sum of sectors.

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Eugenia Colafranceschi: So as a result of this structure for our hit, best place, then the also the composer

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Eugenia Colafranceschi: into this direct sum of type, one. Factors, because we proved that there cannot be type 2 or type 3.

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Eugenia Colafranceschi: So in particular the fact that the the algebra for a given, so that a given factor in your algebra is a type, one factor, and as a a commutant.

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Eugenia Colafranceschi: the factor the corresponding factor for your on your right algebra means that the the human space actually has a decomposition like this. So you have a direct sum of a tensor product

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Eugenia Colafranceschi: of a Hilbert space. So for the left and a Hilbert space for the right.

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Eugenia Colafranceschi: So. we are now the last point. so so we

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Eugenia Colafranceschi: We got this so we derived the structure for the Hilbert space

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Eugenia Colafranceschi: from the structure of the phoneme and algebras acting on it now, the trace defined, a path integral is aware, different from a standard the Hilbert Space trace.

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Eugenia Colafranceschi: but what we know from the theory of algebra is that the faithful, normal and semi final 3 system on python algebra are unique up to another normalization constant.

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Eugenia Colafranceschi: So this means that the in each sector we have a constant depending on the new label for your sector

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Eugenia Colafranceschi: and the so we have this relation. So the trace defined by the path integral for a certain element, a multiplied by this constant must be equal to standard space

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Eugenia Colafranceschi: trace. For your element. A.

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Eugenia Colafranceschi: And now what you can do is that consider, is to consider as a one dimensional projection. And so you get to that to the trace

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Eugenia Colafranceschi: for this new

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Eugenia Colafranceschi: Aigan sector is actually one.

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Eugenia Colafranceschi: So these gives us a normalization for the Hilbert space trace. and in particular, we can use so a a generalization of the 4 boundary. Arguments I give you before that allowed us to bound

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Eugenia Colafranceschi: the trace of a projection by one, unless it is 0. But As a remark, our trace is faithful.

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Eugenia Colafranceschi: And the so from the positivity of the Internet product on a for boundary space. So we derived the this constraint. If we consider the positivity of the inner product

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Eugenia Colafranceschi: on a a larger space with for example, number N of pairs of B boundaries.

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Eugenia Colafranceschi: Then what we got is a bound like this for the trace of a projection. So by this recursive argument we get to that. The trace of our projection must be an integral.

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Eugenia Colafranceschi: so it can be 0 1, 2, 3, and so on.

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Eugenia Colafranceschi: So this means that the

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Eugenia Colafranceschi: for any nonzero finite dimensional projection

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Eugenia Colafranceschi: since the trace is a positive integer, we find we can constrain the value of the constant that sets the normalization of our trace.

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Eugenia Colafranceschi: And this is particular. This is crucial, because, once we, we have this positive integer, we can actually define an extended space factor. So which is given by the tensor product of your

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Eugenia Colafranceschi: factor associated to the left of the right part tensor with the what we call our Eden sector, which is a huge space. So we dimension given by this integer here

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Eugenia Colafranceschi: in such a way that on these so we can define. We have a natural trace on this extended in the space, and then the trace the defined. By how our path integral

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Eugenia Colafranceschi: coincides

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Eugenia Colafranceschi: with the trace defined on this extended the Hilbert space. Of course we need to consider. I mean this map. You have that for any A in our phone line and algebra. Then you consider the extended element with just the tensor product of the original A with identity on the hidden sector.

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Eugenia Colafranceschi: So the the main point is that

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Eugenia Colafranceschi: adding these Eden sectors allows to interpret the path integral trace as a Hilbert space trace. Because, let me remark once again, the the the path integral trace.

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Eugenia Colafranceschi: Is not the sum standard. Some were on to normal basis.

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Eugenia Colafranceschi: So to to have something like this. we needed to find the right normalization for our trace. We found that the

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Eugenia Colafranceschi: this normalization actually corresponds to positive integers. And so this allowed us to to include the these hidden sectors. And consider an extended inverse space on which

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Eugenia Colafranceschi: the trace coincides with the trace defined by the Pathinder.

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Eugenia Colafranceschi: So now we can use the so in the the past, integral case can be used to define an option of entropy on the States on our

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Eugenia Colafranceschi: so you can start with the an element on our Hebrew space. Construct the corresponding density matrix

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Eugenia Colafranceschi: and embed it in the extended Hilbert space. And so you can now compute the standard phoneman entropy by using this path integral trace, which is standard. The Hilbert Space trace on the extended

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Eugenia Colafranceschi: hid their space. And

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Eugenia Colafranceschi: and so, thanks to this relation between these 2 places, now, the entropy computed through the path integral as a Hilbert Space interpretation.

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Eugenia Colafranceschi: And you will have a contribution,

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Eugenia Colafranceschi: given by standard phone line and entropy on your sector, plus a mixing term which depends on the probability of being in a given sector.

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Eugenia Colafranceschi: At this point we can compute the entropy by using the replica trick, so we can

353
01:01:14.820 --> 01:01:32.919
Eugenia Colafranceschi:  procedure we discuss at the beginning and compute. Also consider many copies of our States.

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Eugenia Colafranceschi: And compute the the Ranian to be in particular the trace of the density, magic to some power and

355
01:01:41.150 --> 01:01:48.830
Eugenia Colafranceschi: by using the the manifold given by the gluing of the many copies. As input for our past integral.

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Eugenia Colafranceschi: And so, if the theory emits a semi classical limit which is described by Einstein, but gravity or Jt gravity.

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Eugenia Colafranceschi: Then we can as I said, we can follow the procedure, and we can argue that in this limit the anthrop is given by the

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Eugenia Colafranceschi: okay. So let me summarize. So

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Eugenia Colafranceschi: We have shown that any gravitational path, integral which satisfies

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01:02:24.020 --> 01:02:44.309
Eugenia Colafranceschi: a set of actions so defines for my man algebras of observable so which are associated to asymptotic codimension to boundaries. So, and we call them the left and right algebras. We show that this algebras contain only type, one factors.

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Eugenia Colafranceschi: As a result of this structure the Hilbert space on which the Algebras act decomposes

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Eugenia Colafranceschi: as a a direct sum of tensor products associated to the live left and the right part.

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Eugenia Colafranceschi: The past integral trace is equivalent to a standard trace on an extended Hilbert space.

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Eugenia Colafranceschi: and this provides a huge space interpretation for the entropy, the entropy defined to the path integral trace. Even when the gravitational theory is not known to have an holographic dual theory. And we expect this

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Eugenia Colafranceschi: entropy in the semi-classical limit to be given by the Rutakayana formula.

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Eugenia Colafranceschi: okay, thanks for

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Eugenia Colafranceschi: okay.

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Jerzy Lewandowski:  thank you. And now we start the discussion.

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Jerzy Lewandowski: I'm Daniel. So in Daniel, go ahead.

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Eugenio Bianchi: Hi, Hello, Jania, yeah, this was very nice. II yeah, I have a very simple question can you describe, in your calculation?

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Eugenio Bianchi: over what degrees of freedom did you trace over? You're looking at the entropy? What degrees of freedom and what are the degrees of freedom that you are tracing over, looking at the entanglement with?

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Eugenia Colafranceschi: Hmm, thanks for the question. So so you can imagine adding, so consider a pushy surface

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Eugenia Colafranceschi: which has 2 asymptotic regions. And so what we are tracing over is the full. for example, if we call them left and right, we are tracing over

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Eugenia Colafranceschi: the right part.

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Eugenia Colafranceschi: for example. So

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Eugenia Colafranceschi: so is like, looking only so. So you have these 2 subsystems given by these 2 asymptotic regions of your cushy surface, and you look only at one a synthetic regions.

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Eugenia Colafranceschi: So you taste over

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Eugenia Colafranceschi: the full

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Eugenio Bianchi: So if I understand correctly, you still always need that. Your space time as to a synthetic regions is 2 sided like the cruiser extension. All right. We were doing a calculation of a gas in a box. So we would need that 2 boxes so that maximally entangled.

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Eugenia Colafranceschi: Yeah, exactly.

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Eugenia Colafranceschi: exactly need them. I mean, you can use as an example

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Eugenia Colafranceschi: JT. Gravity. And so your States are, yeah, Eigen states with the same energy. So the Thermophil double states but the case. So we so in particular, yes, in the case, we consider we only have 2, as in 33 Johnson.

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01:05:53.010 --> 01:05:55.640
Eugenia Colafranceschi: we close and compact the boundaries.

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Eugenio Bianchi: Do you see this type? One algebras can appear also, if you have one single boundary instead of 2,

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Eugenia Colafranceschi: and I mean, if you have one single boundary, and you look at the sub region of that boundary. So if you are partition in this case, our the partition was between 2 synthetic boundaries, right? So R, and the complement to where the 2 asymptotic boundaries, if you have only one, and your sub region is your region are is a sub region of their synthetic boundary. No, you don't expect type one, respect to type 2 or type 3.

386
01:06:32.610 --> 01:06:36.790
Eugenia Colafranceschi: So this type one is

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Eugenia Colafranceschi: as far as understand the

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Eugenia Colafranceschi: crucially related to the fact that you are taking a one full asymptotic boundary.

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Jerzy Lewandowski: Thank you.

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01:06:46.850 --> 01:06:47.760
Eugenia Colafranceschi: So

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01:06:52.270 --> 01:06:53.000
Jerzy Lewandowski: no.

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01:06:59.270 --> 01:07:00.680
Eugenia Colafranceschi: And whoops.

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Eugenia Colafranceschi: yeah, Danielle. Hi.

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01:07:05.880 --> 01:07:14.550
Daniele Oriti: so thanks for the talk. So the question is related to the early discussion with Lauran.

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01:07:14.600 --> 01:07:20.270
Daniele Oriti: can can you say more about the comparison between the axioms to a

396
01:07:20.370 --> 01:07:26.080
Daniele Oriti: posing on the I think, and the ones of topological feature is.

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01:07:26.110 --> 01:07:34.099
Daniele Oriti: and that generalizations suggest Barrett and Crane for quantum gravity long time ago.

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01:07:34.850 --> 01:07:38.400
Eugenia Colafranceschi:  

399
01:07:39.470 --> 01:07:42.100
Eugenia Colafranceschi: so what can I say is that

400
01:07:45.200 --> 01:07:46.470
Eugenia Colafranceschi: outside.

401
01:07:48.480 --> 01:07:51.020
Eugenia Colafranceschi: yeah, so

402
01:07:51.800 --> 01:07:54.750
of course I mean this i mean from

403
01:07:55.800 --> 01:07:57.919
auto fist theory perspective.

404
01:07:58.070 --> 01:08:04.169
Eugenia Colafranceschi:  I mean these. We certainly expect

405
01:08:05.250 --> 01:08:10.439
Eugenia Colafranceschi: these the actions to be satisfied. And the

406
01:08:11.760 --> 01:08:15.490
Eugenia Colafranceschi: so in in particular, I mean the the

407
01:08:17.399 --> 01:08:30.059
Eugenia Colafranceschi: the factorization, which is I guess very, very powerful from the back perspective. Because, as I said, this is I mean, it's a way to control the effects of

408
01:08:30.529 --> 01:08:39.520
Eugenia Colafranceschi: of Walmart. So so this in particular, is a straightforward, if you consider to

409
01:08:39.720 --> 01:08:46.200
Eugenia Colafranceschi: for example. format 50 years associated to to disconnect the boundaries.

410
01:08:47.040 --> 01:08:53.929
Eugenia Colafranceschi: So I mean, these actions are are are not what from a fifth theory, perspective.

411
01:08:54.960 --> 01:08:58.549
Eugenia Colafranceschi: The point is that the I mean

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01:08:59.569 --> 01:09:16.759
Eugenia Colafranceschi: we are not saying that these are necessary for having a good bulk, gravitation, or theory. The fact is that I mean, what is interesting is how much you can derive just from assuming them. Just

413
01:09:17.939 --> 01:09:19.120
Eugenia Colafranceschi: So

414
01:09:20.600 --> 01:09:27.639
Eugenia Colafranceschi: are are you suggesting that from what we know about the topological quantum field theory, we

415
01:09:28.630 --> 01:09:32.519
Eugenia Colafranceschi: we should know something more about the barker.

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01:09:35.330 --> 01:09:52.529
Daniele Oriti: Well, I was just recalling that the the idea was, floating around for a while that the full gravitational, possibly including also the some of the topology, would be fine. They generalize the topological.

417
01:09:52.899 --> 01:10:00.209
Daniele Oriti: And so we're just wondering if in your set of assumptions, we have also that possibility in mind. And what was the

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01:10:00.310 --> 01:10:11.519
Daniele Oriti: comparison between the options you on would have in that case. But I don't know suggesting anything specific about what should be mobified or what could not work, or

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01:10:11.660 --> 01:10:18.200
Eugenia Colafranceschi: I see. So the the idea here is what's, let's say, take the minimum assumption. So

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Eugenia Colafranceschi: to do what we wanted to do, so like to to request from the past integral. What do we think? And well, except, of course, for the last action about factorization. The other ones is, that's what you might want for for any

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01:10:41.020 --> 01:10:44.150
Eugenia Colafranceschi: for a good bulk theory.

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01:10:48.200 --> 01:11:10.449
Laurent Freidel: maybe that is time time for us. Yeah. I mean, maybe I can expand a little bit on that, maybe, I think. And then I have a question about you, too. But what Daniel is talking about is that in the eighties, you know at? Yeah, I mean, people study topological theory as a function of between cobaltism, geometry, and and little bit spaces. And

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01:11:10.520 --> 01:11:35.590
Laurent Freidel: you know, people put together a list of axioms for Tq. Which are kind of the ones you have there, right so on top of it, condition that the trace of the projector is finite dimension, or which is what I was asking about. And then people generalize these notion of Tqt to non national with the trace of the projector might not be.

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Laurent Freidel: you know, might be more complicated. So

425
01:11:38.430 --> 01:11:44.659
and then and then people look at so what he was saying is that then people looked at this axioms in conjunction with

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01:11:44.690 --> 01:11:54.010
Laurent Freidel: with a quantum. Iv. So what I'm saying is that these actions are the actions of Tq. As far as I can say right, tha. That's what the question of Danielle so.

427
01:11:54.120 --> 01:12:02.019
Laurent Freidel: And and there's a very rich literature about this connects to to. I think that's that's kind of a fair.

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Laurent Freidel: Yeah. But indeed, we are not claiming that these are, let's say, the pool. This is the full list of action, so that

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01:12:10.410 --> 01:12:22.140
Laurent Freidel: so our approach is more start with season. So let's see what we can say about the gravitational entropy.

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Laurent Freidel: The question I had was more at some point in your in your lecture, you you you talked about this hidden sector, and I wanted to understand, what is this hidden sector? Do you have an understanding geometrically? Is it about topology change, I mean, cause you you should. You have good space, and then you have to add that

431
01:12:43.260 --> 01:12:48.599
Laurent Freidel: so what is this standard space? And is it geometrical? That's the question.

432
01:12:49.370 --> 01:12:52.340
Eugenia Colafranceschi: So the the the answer is, I mean

433
01:12:52.950 --> 01:13:03.679
Eugenia Colafranceschi: this, this Eden sectors comes from the fact that the the Hilbert space we constructed might not be the full Hilda space.

434
01:13:04.010 --> 01:13:14.079
Eugenia Colafranceschi: So we can, of course, with the algebra as we construct, we can probe the full you best pay the fine from the

435
01:13:14.590 --> 01:13:29.159
Eugenia Colafranceschi: but you might have a I mean th. There can be degrees of freedom that you did not take into account during the construction, so I mean can be dramatical, but can also be

436
01:13:29.570 --> 01:13:42.250
Eugenia Colafranceschi: but not necessarily so, is more like a say, Ed. they represent a a lack of knowledge about the the, the.

437
01:13:42.530 --> 01:13:45.960
Eugenia Colafranceschi: the full quantum gravity. Hilbert space.

438
01:13:46.130 --> 01:13:55.829
Eugenia Colafranceschi: So they are saying. I mean, there are cases like in Jt gravity, which you can show that this are 3 yellow

439
01:13:56.170 --> 01:14:00.299
Eugenia Colafranceschi: but then there are other examples.

440
01:14:00.530 --> 01:14:14.450
Eugenia Colafranceschi: I think one is the topological model by Marilyn Maxfield. in which, the sectors are not trivial, related to one of the world brains. So

441
01:14:14.550 --> 01:14:25.600
Eugenia Colafranceschi: it's yeah. I mean this even sector. There's the degrees of freedom that you don't capture with this construction. But

442
01:14:26.110 --> 01:14:27.230
Eugenia Colafranceschi: And

443
01:14:27.240 --> 01:14:39.079
Laurent Freidel: but you're saying you don't necessarily have to have this sector? No, you need to add them to have a Hilbert space interpretation for that. So to have a trace.

444
01:14:39.260 --> 01:14:42.789
Eugenia Colafranceschi: So the trace constructed from the path integral

445
01:14:42.830 --> 01:14:50.519
Eugenia Colafranceschi: by itself, is not the standard, the Hilbert space. So you cannot express that, not as a somewhere to normal basis.

446
01:14:50.920 --> 01:14:55.499
Eugenia Colafranceschi: You can do that if you consider an extended, the hidden space.

447
01:14:56.210 --> 01:15:01.169
so is a is a huge space that you're algebras. So

448
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Eugenia Colafranceschi: is the part of the Hebrew space that your algebra cannot probe in a way. So

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Eugenia Colafranceschi: you yeah.

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Laurent Freidel: And in the end you get some folks base. So there's a restriction on on. I'm trying to understand. What is it? What is the freedom you have? Is it completely free? Is there? Is there a unique extension

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01:15:26.450 --> 01:15:35.640
Laurent Freidel: to have this trace property. Is that enough to determine what this spaces, is it?

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01:15:36.280 --> 01:15:41.920
Eugenia Colafranceschi: Yeah, yeah. You need to determine the full quantum gravity, address space to know which

453
01:15:42.640 --> 01:15:47.870
Eugenia Colafranceschi: degrees of freedom you are missing. The only thing we we can constraint here is the dimension

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Eugenia Colafranceschi: of these hidden sectors.

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Laurent Freidel: Thank you.

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Jerzy Lewandowski: Okay, Florida, Atlantic.

457
01:16:00.790 --> 01:16:12.119
FAU(Florida): So I have a question, about the entropy formula. And can you go to your entropy formula, where? Where you have 2 terms, Shannon part and the mon Norman part.

458
01:16:13.020 --> 01:16:20.340
FAU(Florida): Yeah. And and what's what? Exactly the relation between this formula and and the Rt formula divided by Manasina.

459
01:16:20.840 --> 01:16:32.759
FAU(Florida):  in the semicol? I mean, you argue that these formulas relates to Rt. Formula in the semicolon. But what exactly these 2 terms relates to rt formula.

460
01:16:33.930 --> 01:16:47.980
Eugenia Colafranceschi: the so the the what the the Shannon entropy is, that's the taking into account the the the I mean. It's just the contribution from the probability of being in one sector, the other

461
01:16:48.740 --> 01:17:01.599
Eugenia Colafranceschi: understand that the so the the full quantity would the would the give you the formula. So the area of this surface some which is invariant under replicas symmetry.

462
01:17:01.690 --> 01:17:15.720
Eugenia Colafranceschi: so is the full quantity that would be given by in the semi-classical limit. But from the point of I mean and in this finite regime.

463
01:17:16.000 --> 01:17:19.740
Eugenia Colafranceschi: You have these. 2 contributions.

464
01:17:20.900 --> 01:17:31.480
FAU(Florida): What I'm wondering is is whether in certain limit or semicol limit. These one of the two-term standards were contributing dominantly

465
01:17:31.590 --> 01:17:33.660
FAU(Florida): or

466
01:17:34.190 --> 01:17:47.919
Eugenia Colafranceschi: both of them contribute. So the fact that okay, so the fact is that when you do the semiclassical limit. You are actually you lose the the

467
01:17:48.220 --> 01:17:58.799
Eugenia Colafranceschi: type, one structure of your algebra. Usually, I mean, you expect that these actions to fail. Some point in the semi-classical Limited.

468
01:17:59.310 --> 01:18:13.080
Eugenia Colafranceschi: So this is why the State counting interpretation, I mean, you have it defined at the regime. Huh? But the face of this 2 contributions in the semi classical limit. It's

469
01:18:13.240 --> 01:18:20.829
Eugenia Colafranceschi: I mean, it's unclear just by the fact that your full type, one structure, and so keep the space. Interpretation

470
01:18:20.940 --> 01:18:24.560
Eugenia Colafranceschi: is not clear anymore when you take the semi-classical limit.

471
01:18:26.310 --> 01:18:32.930
FAU(Florida): So but then, what is the use of of this formula? Respect to the Rt formula.

472
01:18:32.960 --> 01:18:36.870
FAU(Florida): What are you? What is the use of these 2 terms with them?

473
01:18:36.920 --> 01:18:48.899
FAU(Florida): the formula is telling you that what you are computing with the with the I mean is an entropy in the standard sense.

474
01:18:49.670 --> 01:18:53.740
Eugenia Colafranceschi: which is something you don't have. Unless you're holography.

475
01:18:54.780 --> 01:19:06.970
Eugenia Colafranceschi: So this formula is telling you that what you get in the semiclassical limit, as you take a 90 formula in the finite tabling regime is actually a standard

476
01:19:07.120 --> 01:19:11.399
Eugenia Colafranceschi: standard entropy given by this Channel term. And

477
01:19:13.800 --> 01:19:15.110
Eugenia Colafranceschi: and language that.

478
01:19:15.920 --> 01:19:17.350
Eugenia Colafranceschi: Thank you.

479
01:19:18.680 --> 01:19:20.660
Jerzy Lewandowski: Question from Dipan

480
01:19:20.940 --> 01:19:22.590
Jerzy Lewandowski: detail piece.

481
01:19:23.420 --> 01:19:25.280
3453 Deepan Betal: am I? Am I audible?

482
01:19:28.180 --> 01:19:31.060
3453 Deepan Betal: Yes, go ahead

483
01:19:31.470 --> 01:19:39.939
3453 Deepan Betal: in He had described some matters which is a part of the which is which is a part of the

484
01:19:39.990 --> 01:19:45.140
3453 Deepan Betal: or the manifold. Somehow, in the, in the definition in the, in, the, in the initial pages.

485
01:19:45.170 --> 01:19:51.779
3453 Deepan Betal: What kind of what kind of matter can be considered here? Can they be only scalar, or what kind of matter.

486
01:19:52.060 --> 01:20:13.199
Eugenia Colafranceschi: Oh, we so that was one was just to to give an intuition together. We are not assuming, I mean, can be any kind of matter. So scholars? But I mean the the object that we call includes integral is I mean fully data. I mean, just by the actions we listed them.

487
01:20:13.200 --> 01:20:27.230
Eugenia Colafranceschi: We are not assuming any, I mean, even sending over geometry sent to apologies. I mean is what you usually have in mind. But we are not

488
01:20:27.410 --> 01:20:43.500
Eugenia Colafranceschi: assigning any receipt to compute the the past integral. So that was just part of the let's say, 2 2. Yeah, to build up some intuition of what you might want from up. I think there are some motivations for your actions. But

489
01:20:44.120 --> 01:20:49.580
Eugenia Colafranceschi: we are not. We. We don't. Constraints, are the

490
01:20:50.690 --> 01:21:02.020
Eugenia Colafranceschi: we. We don't assume any any particular receipt for your. We just assume that whatever you are defining as your satisfies the actions.

491
01:21:04.540 --> 01:21:05.640
3453 Deepan Betal: Okay, thank you.

492
01:21:07.680 --> 01:21:17.409
Jerzy Lewandowski: It seems that there are no more questions, so let us sync the speaker again. Oh, sorry. There is one more by Daniela. Daniela, go ahead.

493
01:21:19.080 --> 01:21:20.580
Daniele Oriti: Sorry. No problem

494
01:21:20.650 --> 01:21:36.470
Daniele Oriti: I was just wondering about the relation between these 2 results. Specific assumptions about the gravitational in particular, about the asymptotic boundary conditions and allography.

495
01:21:36.750 --> 01:21:49.989
Daniele Oriti: I was wondering if, you know, one could try to read it together around that, you know. In a gravitational button with those boundary conditions so satisfying the actions you listed.

496
01:21:50.460 --> 01:21:55.379
Eugenia Colafranceschi: This result shows, that. Somehow, the theory is holographic

497
01:21:55.760 --> 01:22:05.050
Daniele Oriti: by reproducing the same assuming it.

498
01:22:05.390 --> 01:22:10.740
Daniele Oriti: And this sort of resonates also with the results. who do you buy

499
01:22:10.920 --> 01:22:23.249
Daniele Oriti: the on and others that you know, once you have that type of boundary conditions. you can argue for an autographic behavior. Ju Julie. Name from quantum gravity considerations.

500
01:22:24.230 --> 01:22:31.179
Eugenia Colafranceschi: You you're you're completely right. Yes, it seems to the exact same seems to say that if your theory is allographic.

501
01:22:31.430 --> 01:22:38.909
Eugenia Colafranceschi: it is as much as the idea Cft, as you expect. I mean that if it's holographic, is

502
01:22:39.550 --> 01:22:43.549
Eugenia Colafranceschi: indeed, with the structure of 50 way on the boundary. Yeah.

503
01:22:44.190 --> 01:22:47.510
Eugenia Colafranceschi: yeah, yeah. You can see in the other way around.

504
01:22:50.760 --> 01:22:51.620
Daniele Oriti: Thanks.

505
01:22:52.520 --> 01:22:53.510
Eugenia Colafranceschi: Thank you

506
01:22:55.050 --> 01:22:57.959
Jerzy Lewandowski: any any more questions or comments?

507
01:23:00.720 --> 01:23:10.399
Eugenia Colafranceschi: No, okay, so, Virginia, thank you. Thank you. Once again. Bye, bye, everybody.