Okay, so in the previous lecture, we discussed the conditions that we need in order for the horizon of a large black hole to be smooth. And we argued that we need to be able to identify a new set of operators, be tilde, with a property that they have to commute with the bees. They have to be thermally populated as the bees. And they have to be entangled with the bees in a particular way, which you can determine by doing effect theory around the horizon. Here, you see a two-point function between these two guys being a bit tilde, which is non-zero, which implies that these two operators are actually entangled. And then we presented this paradox, which was based on the commutation relations of the operators of Hamiltonian. And we found this negative trace, and then the question is, can we actually find operators between the CFT with the desired properties? Now, I will follow, I will present an approach that will give us these operators in a constructive way. And the idea is based on some work that is with Suvath Raju in the previous years. And the intuition is the following. If we look at the exterior of the black hole, we have this operator B. So this is the black hole in ADS. And here we have operators B which seem to be thermally populated. These guys are thermally populated. Now, the fact that they're thermally populated means that their two-point functions are thermal and higher-point functions approximately factorize, which means that there is no significant entanglement between the bees. So these bees seem to be thermally populated, but they're not entirely with each other, which means that there must be another part of the system with which these operators will be entangled. These other parts of the system we do not yet know how to represent the terms of operators, but what we're trying to do now is to use this entanglement to identify a new set of operators with these bee guys are entangled. So to repeat, we have a set of operators whose correlation functions suggest that these operators are entangled with something and we're going to use the entanglement in order to identify these other parts of the Hilbert space that purifies the operators B. So this second part of the Hilbert space that we'll try to construct will play the role of the bitildas and together the bees and the bitildas can be used to reconstruct the smooth horizon for the black hole. So now the idea is how do we start with a set of operators which seem to be entangled and how can we construct from them the remaining operators with which our operators are entangled? So there's actually a very natural mathematical construction which allows us to do this in a canonical way. Which goes under the name of Tomita's psychimodular theory. And let me describe what is the idea mathematically and then we'll try to apply it to the particular problem that we have. The idea is that we start with the Hilbert space and the particular state psi in the Hilbert space as well as an algebra, a phonomial algebra of operators act on the Hilbert space with a property that first of all, the state is cyclic with respect to the algebra which means that we can reconstruct the full Hilbert space by acting on the state psi with elements of the algebra. Now to be a little bit more precise, this definition means that we should be able to get a dense subspace of the Hilbert space. So we should be able to approximate any state we want to arbitrary accuracy. But I will just to write it more quickly, I will write it in this way that the Hilbert space is spun by the algebra act on the state. So this is the first condition. The second condition is that it's called that the state is separating with respect to the algebra which means that there are no annihilation operators in the algebra. You cannot annihilate the state psi with elements of the algebra. So just to give you some intuition, the second condition can be thought of as expressing algebraically the statement that the state appears to be entangled when you probe it with this algebra. Let me give you an example. If you have two spins, suppose we have two spins and suppose we take this state, take the state psi. Now there's no entanglement. This state is direct product. And suppose we define this small algebra to be the spin operators on the first spin. Now in this situation, when the spin operators of the first spin act on the state, it's clear that the state is not entangled. And in particular, we can see that we can't find annihilation operators. For example, if you act with a raising operator of the first spin of the state, you get zero. So if there is no entanglement, you can find annihilation operators in the smaller algebra. If on the other hand, you look at the same system for a state with entanglement, then you can convince yourself by trying all kinds of examples that it is impossible to find an operator in this algebra, in this algebra, which will annihilate the state. For example, if you act with this raising operator, it will annihilate this component, but it will not annihilate the other one. And more generally, you can check that there is no annihilation operator in the smaller algebra defined on the first spin. So when we have an algebra acting on the state, if we cannot find annihilation operators in that algebra, then it means that the state is entangled with another part of the system, and our goal is to identify that other part. So this theory allows you to identify the remaining part of the system with your algebra is entangled, and it can be done in the following way. So this theorem says that if you have this algebra acting on the Hilbert space, and if these two properties are satisfied, then the algebra has a non-trivial commutant, A prime, which means there are some new operators which commute with all elements of the algebra, and these operators act on the same Hilbert space that we started with. You don't need to enlarge the Hilbert space. Finally, this commutant, A prime, is isomorphic to the algebra A, and these two algebras are entangled in a very particular way, if these two conditions are satisfied. So to connect to what we're trying to do, for us, the operators in the exterior of the black hole will play the role of the algebra A, and the operators that we're trying to construct will play the role of A prime, the commutant. They're a commutant because they're space-like relative to A, so they have to commute. And as we will see, the entanglement pattern that you get from this theorem is precisely the one that you need in order for the horizon to be smooth. Yeah, I will explain what is the meaning of the state in the black hole context in a little bit. Yeah. So in this construction, you start by introducing an antilinear map, which is defined by considering all possible states that you get by acting with elements of the algebra on the reference state side. So this small A is an element of this algebra, and we consider all possible states that we get, we can get by acting with elements of the algebra on the state side. And then we define this antilinear map S by the following equation. We define it as the mapping which takes this state into the state A dagger times psi. Now, already at this point, in order to be able to define this operator S, we are actually using the statement that the state psi is separating, meaning that there's no annihilation operator for the state psi. Why is that? Well, for example, if you could imagine a situation where the state psi was annihilated by an operator A, but at the same time the state psi was not annihilated by a dagger. For instance, this happens, if you have a quantum field theory and you look at the ground state over free field and you consider an annihilation operator, then it annihilates the vacuum, but a dagger does not annihilate the vacuum. So if you try to define this operator S, in that case, you would run into a contradiction because you would be trying to define an antilinear operator taking a vector which is zero to something which is non-zero, which is inconsistent. So just to be able, for the starting point, to define this operator S, we need to have the condition that the state psi is separating. What I'm trying to do now cannot be done, for example, if you take A to be the algebra of Minkowski space operators and the state psi to be the ground state of the field theory. Then we take this operator S and we consider the polar decomposition of this operator, which means we divide the operator into, let's say, so normally, when you do the polar decomposition, you take an operator and you write it as the product of a unitary times something which is positive and Hermitian, but since this operator S is antilinear, in this particular case, this operator J is going to be anti-unitary and delta is going to be a positive Hermitian operator. Now, this delta, you can also write as S dagger S. In some sense, it is a magnitude of this operator S. And as we will discuss later, this object delta, or actually the logarithm of delta, the minus the logarithm of delta, is what is usually called the model Hamiltonian. So this is the setup, this is the starting point, and then the theorem makes the following statements.  So you generate a new set of operators which act on the same Hilbert space with a property that they commute with all elements of the algebra A. So what we want to do now in this context is to start with this operator S at the horizon. This will be the algebra A for us. We want to build up this structure and define this object J so that J, A, J will give us the algebra which is hiding behind the horizon. Moreover, from this definition, you can check that J squared is equal to one. So this means that this commutant is isomorphic to A. So in particular, the set of operators that you will construct on this side of the black hole are going to be, in some sense, isomorphic to the operators that you have on the exterior. So this is very natural from what we expect by the usual analytic continuation of the black hole to the Kruskal diagram. We get, let's say, a second copy of the exterior region. Then there are a few more statements. There's a statement that if you take this explanation of the model Hamiltonian and you think of it formally as a time evolution operator by some parameter S, then the theorem says that the algebra A is closed under this evolution and the same is true about the commutant. So I will explain later what is the physical meaning of the statement in the case of the black hole. And finally, there is some sort of formal KMS condition which is that if you take two elements of the algebra, A times B, and then you take one of them and you displace it by sort of evolving it using the model Hamiltonian, then this function F has some particular analytic properties and it obeys some sort of periodicity which is similar to the KMS condition which I wrote down a little bit earlier. So all these statements are mathematical statements about the situation where we have this algebra and the state which is set in separating and we try to apply the theorem to the case of the black hole in order to construct the copy of the operators which are hiding behind the horizon. Now, before I go on, are there any questions about this so far? Yes, the model Hamiltonian by definition is given by the logarithm of this object that we call delta. Okay, let me give you an example of how this works in Minkowski space. So we already talked about it in a previous lectures that you can take Minkowski space and you can divide it into regular wedges and then it is very, it's clear that we get two algebras, the algebra of operators which are localized in the right wedge and the algebra of operators in the left wedge. So I will call this A and this A prime where the reason we do it is because we sort of assume that, well, by locality these operators commute with A and we also assume that these are the only operators which commute with all operators on this side which is true for free theories, for example. And we will now start by working with this algebra and we try to see what we need to do in order to reconstruct the second copy. So this will be a toy model of what we want to do eventually with the black hole. So we try to apply this theorem in this case to see how it works and then we'll do a similar thing in the case of the black hole. Now, in order to apply the theorem we have to verify that the conditions that I wrote down before are satisfied. In particular, we want to make sure that, so the state that we will be working with will be the Minkowski vacuum. So the state psi for us is going to be the Minkowski vacuum. So we want to verify that the Minkowski vacuum is cyclic and separating with respect to the algebra of operators localized in one of the two wedges. So to remind you, these two conditions means cyclic condition means that we can generate the full Hilbert space by acting with operators on the right wedge and separating means that we cannot find any annihilation operators on the right wedge. Now, the fact that we can get the full Hilbert space of the quantum theory by acting operators on the right wedge is a well-known theorem in quantum theory. It's called the Ritz-Rieder theorem. And it is very easy to understand if you work with a free quantum theory and you write down the Minkowski vacuum in terms of ringlet modes. If you want, we can do it this afternoon explicitly. And then you can verify that by exploiting the entanglement between the two sides, any state you want to generate on the full Hilbert space you can generate is simply by acting on the vacuum with operators in the right wedge. So it's known that the Minkowski vacuum is a cyclic state with respect to the algebra of operators in this region. And you can also prove by using the same theorem that you cannot find any annihilation operators which are purely localized in one of the two wedges. The intuition is that, as you know, if you have an accelerated observer in one of the two wedges, this observer will detect a thermal gas of particles which means that he does not see the vacuum but rather a thermally populated state. So if you have a thermal state and if you are with a lowering operator, you do not annihilate the state. You just remove one particle from a heat bath. So an observer who is localized in one of the wedges cannot annihilate the Minkowski vacuum. So both of the assumptions of the Tomita-Takasaki theorem are satisfied. We have an algebra. We have a state with a cyclic and separating. So we can try to apply the theorem to see how we can recover the left side from the right side. Okay. So the starting point is to consider Lorentz-Boost on the TX plane. So for real parameter S, so the Lorentz-Boost generator is K. And for real parameter S, for real Lorentz-Boost, the coordinates are transformed in this particular way. Now, what we will do is we will consider this operator when we take the parameter S to be complex. So we'll consider a complex fight Lorentz-Boost. And then in particular, if you take the parameter of this Lorentz-Boost to be i times pi, if you plug it into these equations, you can verify immediately that T goes to minus T, X goes to minus X, while the transverse coordinates Y remain invariant. So a complex fight Lorentz-Boost maps points on the right wedge to points on the left wedge. So in particular, what we're trying to do now is to build up this operator S for the case of Minkowski space divided into the two regular wedges. And what we want to do in principle is to consider any element of the algebra of the right wedge. So a product of operators of the right wedge acting on the vacuum. And we want to find an operator S that will transform the state into a dagger times psi. Now, to make the presentation short, I will only show you how it works in case you act with only one operator on the vacuum. And you can look at the literature of how to generalize it to the situation where you have many operators. So we consider the situation where we act with phi T, X, Y on the Minkowski vacuum. And we want to find what is the operator S I have to put here in order to get phi dagger at the same spacetime point. So the first observation is that if I use the complex fight Lorentz-Boost that I introduced, it takes this state into a state where I have flipped the first two coordinates but the transverse coordinates remain the same. So this is not yet what we want. It's not the correct thing yet. But suppose that we take this operator and we combine it with a rotation by pi around the X axis, which rotates the transverse coordinates Y to minus Y. And in addition to that, we act with CPT operator which maps minus T, minus X, minus Y back to X, Y. Then we find that if we act on this state by the product of these operators, the complex fight Lorentz-Boost, the rotation around the X axis and the CPT transformation, then you can check that the spacetime coordinates are mapped back to themselves while this operator phi gets a dagger because of the CPT operator. So then it means that for the Minkowski vacuum and for the decomposition into Rindler wedges, this operator S can be taken to be CPT times the rotation times a complex fight Lorentz-Boost. Now, in the way I did the derivation, I was also very careful about the domain of electricity of this complex fight Lorentz-Boost. Also, I didn't do the case where we have many operators here, but this has been worked out in this theorem and you can check that all the steps can be made rigors. So all you know what we find is that in the Minkowski vacuum, if you take any operator on the right wedge and you multiply it by this object, you get a dagger acting on the Minkowski vacuum. So we were able to find what is S for this particular example, that we write here. So then we can proceed and calculate what is the Mohler Hamiltonian. I remind you the Mohler Hamiltonian is S dagger S and if you take that result and you calculate S dagger S, you find that this object theta squared is one, this object is unitary, so it drops out, so you're left with e to the minus two pi times k. So the Mohler Hamiltonian for this particular example is given by the exponent here where k is the Lorentz-Boost. So we derived what is usually done in the Ullur computation where you start with an accelerated observer and by expanding the field modes, after a while you prove that sort of the reduced estimates that the accelerated observer detects is a thermal density matrix where the Lorentz-Boost plays a role of the Hamiltonian and two pi is the inverse temperature for Inglis space. So let me just mention that this derivation that I gave here can be done for an interacting quantum field theory, so it does not rely on weak coupling, so it's more general. Now, we identify this object delta and we can also identify this anti-unitary operator J which maps A into a prime and in this particular case, this J has a very simple geometric form. It is a product of a rotation around X and the CPT transformation which clearly if you take an operator on the right wedge and you conjugate it by this object and you get an operator on the left wedge. So it allows you to map A into a prime. So this is how the theorem can be applied in the K-Tomir-Koski space and how it can, if you start with the right wedge, it can give you the left wedge. It can give you the operators on the left wedge. So before I move on to the black hole, are there any questions about this? So R1 is a rotation by pi around the X axis. So we're looking at the Rindler space, so we have XT and some transverse directions that they call Y. And the point is that the Lorentz boost that we started with was a Lorentz boost on the TX plane. So Y was invariant under this Lorentz boost. And by considering the complexified Lorentz boost, what we mapped was TXY into minus T minus XY. Now, then we took this new point and we transformed it by rotating by a factor of pi around the X axis, which leaves X invariant, but it rotates the vector Y into minus Y. And so all in all, we started with T comma XY, then we do the complexified Lorentz boost, we go to minus T minus XY, then we do a rotation around X, we go to this guy, and then we do a CPT to go back to the point that we started with. That's why we get this nice equation, which gives you, well, you start with a state where you act on the vacuum with an operator of the right waves. And by actually this combination, you get a dagger, I mean, find the dagger of this operator act on the vacuum, which is what we want to have. No, it can be generalized. If the background is symmetric enough, then, so the question is if you, first of all, formally you can generalize it, but there is no guarantee that this delta you will get is going to be something nice. Or this J that you will get will generally not be something that acts locally. So if the background has symmetry, for example, if you take ABS ringler, so if you take ABS phase and you divide it into ringler wedges, then you will probably get some generators here, which are nice and geometric. But if you do it for an arbitrary background, then you can define these objects, but they're not going to be nice. For example, it can be worked out in the ringler space, if you have a conformal theory, it can be worked out around diamonds, causal diamonds, ball shape regions and their causal development. But I also want to mention that this construction depends on the states, not only on the region, but also on the state.  So far, the state was the Minkowski vacuum. So all these nice, simpler results hold when the state is the vacuum and the algebra is the algebra of a nice region of space. If you take an excited state, formally, you can apply some of the steps, but again, the model of Hamiltonian we calculate is going to be something complicated. Probably, yes, yes. Also at maximum. That's right. Yeah. No, no, that's fine. That's a standard. That's okay. I mean, what we, so the question was, yeah, we want to define an anti-linear operator S, right? So if you start, even if you start with a field five, which is Hermitian, you can always multiply it by complex number. So we want to make sure that this map is anti-linear. That's why we have to go through this, we need to use this theta operator at some point. So this is how we apply this construction to this particular example, where we started with the quantum field theory and we divided space into two parts and we found that this part of the space, of the quantum field is entangled with the other part and then we use the entanglement in order to express these operators by, in terms of this delta nj and the operators like one a. Now, in the case of the black hole, there are some similarities, but there's a crucial difference, which is that we do not have a division of the algebra operators in physical space. So it's not that we are taking the CFT and the boundary and we divide the CFT into parts. Instead, if you think about it, we want to divide the CFT into the exterior of the black hole and the interior of the region behind the horizon, but this division is very symmetrical, the boundary, if the CFT is on the sphere. So it's not that we're taking a sub-region of the boundary, we're taking the entire spatial slice where the CFT leads, and somehow we want to define the sub-algebra operators acting on that region that will play the role of the operators acting on the exterior of the black hole. So the correct way to think about it is that the division is not done in physical space, but rather it's done in the space of operators. And basically, we define the space of operators into simple operators, which are defined as single trace operators of low conformal dimension, as well as their products, and all the remaining operators. So just to be a bit more precise, here we are considering only the operators which are dual to supergravity fields. So these are operators whose conformal dimension does not scale with n or lambda. And we are allowed to multiply them together with imposing the condition that when you multiply them together, the number of factors should not scale with capital N, where N is the rank of the gaze group. So in this way, we define a small set which will play the role of a small algebra for us. So and from the gravitational point of view, this small set is what is necessary in order to describe the effect of the theorem in the exterior of the black hole. So in this way, we define a small algebra, the product of single, small product of single trace operators. And we're going to apply the theorem using the microstate of the black hole psi and this object A, which will play the role of a small algebra. Now you may complain that if you impose the condition that you only consider small products of operators, this set is not properly in algebra. So you cannot properly apply the previous theorem. But actually, in the larger limit, this limitation that you have in the number of factors is not very important. And as I will explain a few slides later, the fact that this set is not an exact algebra, and as a result, we cannot apply the theorem in an exact sense, is actually an advantage of this construction rather than a problem. Because as we will see, this limitation is precisely what realized the idea of black hole complementarity that we introduced yesterday. So anyway, let me describe the construction and we'll come back to this point a little bit later. So we define the small algebra and we also define what we call a small Hilbert space, which is defined by taking the state, the microstate of the black hole and acting on it with elements of this small algebra.  This defines a subspace of the full Hilbert space of the theory that we call H of psi. And it is obvious now by construction that the algebra A, the state of psi is cyclic with respect to the algebra A if you restrict your attention to this small subspace. Because we define this subspace H of psi to be the span of this algebra acting on the state. So now we have this algebra A and the state of psi and we have the first condition that the algebra is cyclic with respect to the state psi. The second condition that we'll try to show is that the state is separating, which means we cannot find any annihilation operators inside this algebra for the state psi. Why is that? Well, that follows from the fact that the state psi appears to be thermal when you probe it by single trace operators. So we wrote down this approximation a few times earlier. If you have a correlation function of operators O, single trace operators on a very heavy state in the CFT, you can approximate it by the thermal correlation function up to one over N corrections. And in a thermal correlation function, you cannot find any annihilation operator because let me just show it. So let's take any operator in the small algebra and let's consider the norm of the state A psi. I want to check that this state is not zero, right? So what I need to do is to calculate a dagger psi. This object, which I can think of as a correlator of that form, and then I can approximate it by the thermal correlator, one over N corrections. And then now it is obvious that this quantity here is non-negative because this is a positive operator and this has positive eigenvalues, so this cannot be zero. So we cannot find any annihilation operator for the state psi inside the algebra. So what we have found is that this algebra A and the state psi have the properties that psi is both a cyclic and a separate vector for this algebra and then the theorem that I mentioned before implies that we can define a new set of operators acting on the same Hilbert space with a property that they commute with all previous operators. And in particular, as I already mentioned, this theorem implies that the entanglement pattern between the original operators A and the second copy you generate is precisely the form that we will need in order to identify those operators with those living in the region, A prime, who's on the left side of this diagram. Now, just let me make one more comment. You can also check that in this particular example, you can try to see what this delta is. So this delta is this dagger times S and if you use the KMS condition, which I mentioned a few times at larger factorization, you can prove that this is a particular example. The modular Hamiltonian with a specular state psi and the small algebra A is nothing else but the CFT Hamiltonian, well, shifted by E naught, where E naught is the energy of the state, times beta, where beta is the temperature of the state. So we have a very simple expression for this modular Hamiltonian in the larger limit and then we can proceed and define these operators using the equation that I wrote down before and you find a set of linear equations which define how these operators, O tilde, which are defined by what I, so O tilde are defined as J O J, where O is any single-prime operator and O tilde is a second copy that we defined by this procedure and using this construction, what is guaranteed is that these O tilde's will commute with O's. So these O tilde commute with the operator O. So we found that it is possible actually to define a second copy of the operators of the conformal theory, acting on the same small filbert space, what we called this called subspace in the previous slide, with a property that they commute with the usual single-prime operators and if you look at this equation a little bit more carefully, you find that they also have the correct correlation functions, meaning if you look, for example, at the first equation, so if you take this equation and you multiply it by O, you find that the two-point function O tilde, O O tilde, where this is a usual two-point function of single-prime operator at finite temperature and if you go back a few slides where we wrote down the conditions for a smooth horizon and we had some two-point function between, sorry, between B and B tilde's, you find that if you compare this one to that one and using this relation, you find that they're actually the same. So we find that the operators defined by these equations have precisely the right properties in order to be identified with the operators living on the left part of the spacetime and then combining these two guys together, you can reconstruct the future ways as well as the pathways of the blackboard. Now, let me emphasize that these operators are defined only on this small subspace, H of psi, and not on the entire Hilbert space of the theory. So you can complain now, I mean, if they're not defined in the full Hilbert space, how can we work with these guys? The point is that all experiments that we're going to do in the bulk, we are trying to reconstruct the bulk at the level of effective theory. So we only want to come to reproduce, to reconstruct low-point correlation functions of the local field in the bulk and those low-point functions can be entirely characterized by correlation functions inside the small code subspace by definition. So we don't need to worry about how these operators are extended away from this code subspace. It's not relevant for computations that you do at the level of effective theory. Now, on the other hand, the fact that these operators are defined only on the subspace means that there is some sort of dependence of these operators on the state of psi that you started with. So remember, we started with a particular blackboard microstate psi and we defined the code subspace around it by acting with single trace operators on the blackboard microstate. If you start with a different microstate psi, you will generate a different subspace, h of psi prime, let's say, which is not going to be the same as this one. And then these operators of tilde for the new state will be defined to act on the new code subspace, h of psi prime. So in that sense, these operators depend on the reference states of the blackboard that we started with, and that's why we call them state dependent operators. So once you decide what is the reference state of the blackboard that you're going to work with, these guys are linear operators acting on this particular subspace on the full Hilbert space. So once you decide what's the blackboard microstate, they act as linear operators on the Hilbert space. However, there is some implicit dependence of these operators on the particular microstate that you want to study. Also, please pay attention that since we defined these guys only on this small subspace, even though I wrote down the equation before that the commutator is zero, we should keep in mind that this commutator is zero only inside this small subspace and not necessarily over the entire Hilbert space. So this is not an operator equation. It's an equation which is true only when you calculate correlation functions inside the code subspace. Now, another technical point that I wanted to mention when I wrote down this little argument and I told you you cannot find any relation operators, I was a little bit too quick because if you take modes A and A dagger which have very high frequency, the thermal expectation values of those modes are suppressed by Boltzmann factors. So in general, if this omega is this frequency is very large, then at some point this exponential separation will become very important and the state will be able to almost annihilate the state and then some steps in this construction break down. So the restriction that comes from this consideration is that when we fully transform these operators and we have this Fourier mode omega, we should impose an upper cutoff in the frequency that I call omega star, which can be very large, but it should not be independent. Yeah, I mean, so the question was when I identify this Autildas with the things which are on the left, do I assume HAC duality? So if I remember correctly, the statement of HAC duality is that in the case of Confield theory in Kowski space, that the statement is that the commutant of A is actually not larger than all the operations which are contained on the left. Yeah, we always know that all the operators here commute with A. It could be that there are some additional operators which commute with A which are not visible somehow here and I thought the statement of HAC duality is that these are precisely the operators that commute with A. Well, here unfortunately, I assume a lot more in the sense that this is a gravitational theory, right? The discussion you're talking about is a discussion in quantum theory. So here we are talking about gravity. We're trying to reconstruct the left region in the bulk, right? It's not the left region of the quantum theory. So there could be a lot of subtleties along those lines but for now I just want to have a, let's say a first approximation in defining these operators that can have the right correlation functions in order to describe the infalling observer. So I mean this question that you're asking is, I think it's very complicated to address in the context of gravity. So are there any other questions? Yes? The one-sided black hole. Correct. Correct? The one-sided black hole. Yes. Very good. Yes. Correct. Yes. So let me repeat the question. The question was, we started with the one-sided black hole. We have 150 in a pure state so there should be only one asymptotic ADS region. Yeah, but the way I'm doing it now, I'm producing a second copy of these operators. It looks like I will be able to reconstruct the full left sides, including the asymptotic region, which will be sort of equivalent to reconstructing a left CFD. We should not be there. Yes. Well, I don't know if it's, well, we shouldn't be able to do that. And it's precisely this restriction which doesn't allow you to extrapolate this construction or very far away towards the left in the asymptotically ADS region. So this, as I will explain in a couple of slides, this restriction and the frequency omega impose a restriction on how far towards the left you can go. So there's some limitation on how far you can reconstruct towards the left where this distance is determined by the cut of omega star. So how do we see that? Well, in ADS, there's a gravitational potential, right? So if you want to go very near the boundary, you need very high energies. If you want to write down a wave packet in ADS with very high, which is very close to boundary, you need to use Fourier modes with very high frequency because the wave packet has to climb up the potential well. So if you want to reconstruct the region very close to what would be the left boundary, you would need to use frequency omega with a very high. And those are not allowed in this construction because of this restriction. So we're not able to fully reconstruct the left region all the way to where the boundary would have been. There's one asymptotic boundary, but what we're trying to argue is that there's some space time here as well as there, right? Well, it sort of mimics the thermophil double, but not fully, right? I'm saying there's a cut-off. In the thermophil double, there's no cut-off. In the thermophil double, you can go all the way to the left. Yes. So we only have the right safety. However, what we're saying here is that there is something like the left region but not the entire left region. There's part of space time there. Well, the left region is part of the right safety, right? The right safety describes both the exterior as well as the left region. No, no, you cannot study that region because we have imposed this condition that does not allow you to set up experiments where you will send particles very close to the region you want to study. Sorry, which second statement? So the quake. No, these are operators in the conformal field theory. So the question was, is it that this equation is valid only near the horizon, up the horizon? Well, no, these guys, O and O tilde, they're both of them are operators in the Hilbert space of the conformal field theory. So they're defined on the CFT. So this equation is true only if you evaluate this commutator inside a particular subspace of the full Hilbert space of the CFT. There's no left CFT. No, in the thermo tilde. Oh, good, good. So the question is, suppose that we actually have the thermo field double, you could try to reconstruct that region by just a conversion of left CFT or you could try to do something like this. How can this to be consistent? Actually, if you apply these equations, this commutator like a psyche theorem, all the right CFT, and if you're in a thermo field state, you automatically get that these mirror operators are the left operators. So the construction by itself knows how to deal with this complication that you talked about and I will present it a little bit later. No, for the old, there's no cutoff. For the old, there's no cutoff. Yeah, yeah, because, well, I mean, the CFT contains operators all with all frequencies, right? Then at some point you want to define a small algebra where you select some operators and you put them there and then you try to double them, right? In that selection, we do not include operators all with frequencies are too high. No, because, well, okay, not necessarily, right? Because this construction is needed if you want to define these old field operators. Once you define them, nobody prevents you from using all the O's then, right? Yes, yes, yes. Yeah, that's the point of complementarity, right? And this idea that the Hilbert space of the gravity doesn't factorize. So it looks like it's causally disconnected and at the level of the code subspace it seems to be causally disconnected. But fundamentally, as I will explain a little bit, actually you can rewrite these old fielders in terms of the O's in a very complicated way. No, I think you can, no, you will recover the factorization of correlators but not the factorization of Hilbert spaces. So complicated observables like, for example, entanglement entropy or things like that may not have the naive classical limits. Okay, I think I need to accelerate a little bit and then during the discussion session we can... There will be one hour, more than an hour discussion session this afternoon so even the time maybe we can finish what he wants to say and then... Okay, so now we define this operator that we'll call the mirror operators and now I'm coming back to this point that I mentioned earlier that this set A is not really an algebra and in particular, this theorem that we use cannot be applied in an exact mathematical sense. And this implies that this A prime is not an exact commutant. Now you may worry that this is a problem but actually as I already mentioned from the physical point of view this is a desirable feature because what it actually means is that this A prime, which you naively think is a commutant is an independent part of the Hilbert space as you said, in reality it's not really independent. It's somehow hidden inside the algebra that you started with. So, okay. So in that sense, as we discussed already there is some non-locality in reconstructing the interior in this left region which can be probed once you start considering operators which approach this cutoff of the algebra. So if you start multiplying many of these single trace operators or the end of those then you can start seeing this non-locality. And okay, this we already said. So the pixel that we're getting to is this one.  And if you combine them together you can describe for instance the interior, the black hole interior, which you can do more precisely by doing an SKL-like construction multiplying these operators by some wave functions and these other operators by some other wave functions. And then you can verify that correlation function of this guy reproduce what you would naively expect in the smooth black hole, in the hard-working state. Now, this cutoff on the left is determined by omega star. This cutoff on the frequencies. I already explained the physics of how it works. And as I already mentioned, this region should not be thought of as a fundamental independent part of the Hilbert space. It's already included in the Hilbert space of the right 50. Now, these operators are state dependent. Why? Well, they are defined in the equations which define these operators use a reference state psi. So they're manifestly state dependent. So what you could ask, could it be that we can sort of build up these operators for every single code subspace in the theory and then combine them together into a consistent globally defined operator? The answer is that it's not possible. The reason is that the number of microstates psi you can write down is exponentially large, right?  So there is a huge number of these psi's. So all these code subspaces have, if you combine many of them together, all of them together, they have significant overlaps. And then these equations cannot be applied simultaneously on all of those code subspaces. In fact, we already know that this cannot be done because we had these paradoxes before. Remember, like this negative trace that I mentioned earlier, which suggests that you can never find a single operator acting on the entire Hilbert space that will reproduce the desired correlators. So the way that we were able to avoid the previous paradoxes is by allowing these operators to depend on the microstates. So in particular, remember when I told you this, we had this trace e to the minus beta h bit tilde, bit tilde dagger. We found that this guy was negative, right? That's what we showed earlier. Now, what we're saying is that this guy, the way you have to select this operator, depends on the microstate. So then in that case, it doesn't make any sense to take the trace, which is the average over all possible states, because when you do that, you have to keep track of the dependence of this operator on the state. So this argument cannot be applied. So there's no, all the previous paradoxes can be avoided. Also, yesterday I mentioned that there's an intuitive paradox, which is that if we want to have a smooth horizon, we need to have very specific entanglement between the interior and the exterior. And I try to argue that it's very hard to understand from the point of view of statistical mechanics why a typical state would end up having the correct entanglement. Well, in this construction, we define the operator by the entanglement. So by construction, they always turn out to have the correct entanglement. So we talked about this complementarity a little bit. So we were able to define this all till does because we restricted our attention to a small algebra of operators O. Now, if you start enlarging this small algebra, there are some point this construction breaks down. How does it break down mathematically? It breaks down because this construction depends on the condition that for any element of the algebra, a is of psi must be nonzero, right? That was a separating property. Now, if you make the algebra too large, if you include, for example, all single trace operators without any restriction of how many of them you multiply together, then you can actually find some very complicated combinations which will annihilate the state. And then this entire construction breaks down. So this means that the notion of this O till does is meaningful only if you restrict your attention to a small algebra of operators on the CFT with precisely the spirit of complementarity, namely that the interior of the black hole makes sense only if we consider effectively theory of the exterior, but fundamentally the degrees of freedom of the interior are already contained in the degrees of freedom of the exterior. So that is nice because it is in some sense more mathematically precise realization of the idea of complementarity. And in the case of the black hole in flat space, if we extrapolate, it would imply that it is consistent to assume that the Hilbert space of the interior and the Hilbert space of the interior have overlap. Just that, in simple correlation functions, simple commutators between let's say one operator here and one operator there, the result is equal to zero or perhaps exponentially small. However, in principle you can find some very complicated operators in the exterior which will have significant commutators with operator on the interior. So this suggests that locality can break down dramatically once you start considering very complicated operators in the exterior. By complicated we mean that we're multiplying order as black hole operators. Now the fact that, so this is a concrete realization of the idea that the Hilbert space does not factorize fundamentally. It appears to be a product when you look at low point functions in effect field theory, but fundamentally there's no factorization. And if there's no factorization, the previous argument of the monogamy of entanglement, the strokes of activity paradox we discussed yesterday cannot be applied because the strokes of activity paradox requires as an assumption that you have a factorization of the Hilbert space into A, B, and C, which is not the case. And as I mentioned, the one problem of complementarity is that it was not obvious that you could reconcile it with locality, but in this particular construction it is obvious that locality is a very good approximation for low point functions, even though it breaks down at high point functions. Now in the last three minutes, let me just mention the connection with the eternal black hole. So you probably know that if you have two identical copies of the CFT and we take them to be in a particular entangled state, the thermophil state, then this state is supposed to be dual to the eternal black hole. So now somebody asked the question, what happens if we try to define the mirror operators in this case? Well, it works out very nicely. So if you define your small algebra to be the algebra of the operators on this side, so if this A is defined as the algebra on the right, and you follow these steps, what you find is that A prime is precisely the simple operator of the left CFT. It comes out automatically, precisely because it's committed like a side construction, selects the other operators by their entanglement, and in this particular state, the simple operators here are entangled with simple operators there, so the whole thing is consistent. So if you define the mirror operators of the right, in this case, you will just get the left operators, which is what we expect. You can also test a more kind of sophisticated example where instead of taking the thermophil state, you consider another state of the two CFT Hilbert space, where you have the same amount of entanglement, but the details of the entanglement have been randomized. So instead of taking this nice diagonal state, we write down some generic entangled state between the two CFTs. We can take this to have the same mass as this one on both sides, and this represents a typical highly entangled state, but where the details of the entanglement are very different from the fine-tuned entanglement of the thermophil. And then you can try to define the mirror operators of this algebra, and you find that now they are not the left operators. There is a new set of operators, and you can define the mirror operators of this guy, and they are not those. So in this case, we find that the prediction is that you will have two disconnected interiors. So two black holes, each one of these has its own interior, but there is no wormhole connecting them, which what you would expect in any case by considering correlation functions on the states which I can explain later today. So what I want to say is that this construction works also in situations where we have two CFTs, not only in the original case, but also in more general situations. So amount of time, so there's two more slides. So more recently, Gaujafres and Wall found a nice way of probing the interior of the internal black hole by certain double trace perturbations which allow a particle to traverse the wormhole, and this is very useful because it allows us to probe the interior of the two sides of the black hole by correlation functions from the boundary without having to define local operators in the middle. It's a little bit like an S-metrics experiment. You send things from the exterior and you see what comes out. You don't need to define local operators inside the black hole. So it's very useful and provides evidence for the smoothness of the horizon of the internal black hole. So what we did is that we used these mirror operators to set up a similar experiment in the one side of the black hole where you start with a black hole in equilibrium and then you can excite the left sides. So normally this particle would not be able to escape, but by using an analog of the Gaujafres wall experiment, we can also generate these additional perturbations which allow the particle to escape. So we have done this, we have done some calculations in the SWK model as a toy model and this provide evidence for the physical that this left region is actually present and it has the interpretation that we just gave that you can have particles living there and you are able to also extract them provided that you perturb the boundary state with appropriate operators. So I think I have to close. So maybe we can continue with more details during the discussions. So thank you. Okay, so I propose no.


Okay, we're back at Moscone West in San Francisco, the Cube, one of the Cube's favorite locations actually. This is in 2010 we started in Boston, but then VMworld was our big event. We were set up in Moscone South. We were here in Moscone West last year at VMware Explorer. We had two sets. This is a great setup. We're really excited to be here on broadcast row all week RSA 2023. Wendy Whitmore is here. She's the Senior Vice President of Unit 42 at Palo Alto Networks.  I've been pouring through it, getting all the data, wrote about it for the last couple of weeks. Thank you so much for making some time for us. Oh, you're so welcome. I'm happy to be here and happy to see RSA back what seems like in full force and bigger than ever. Yeah, 2020 was, this was the last event before COVID. Everybody was kind of weird, not really sure whether or not we should mask up and uh-oh, I have a cough or a sneeze. What does that mean? Who knew though, that we were going to be shut down for the better part of two years. So, I mean, this is back, right? Wouldn't you say? It seems like it's as many people as ever, if not more. I mean, you can't even, when you walk from north to south, there's booths in between and- See people everywhere. Well, there's 4,000 plus companies in this business. I mean, it's, wow. It's so wild. Yeah. You're giving a couple talks here at RSA. I am. What are you talking about? So, I've got two panels tomorrow. I'm excited for both of them and they're both very different. So, the first is a crew of leaders from organizations like mine at Palo Alto Networks, but also Red Canary and also Dragos. And then, Lily Heynuman from Wired is going to be moderating the panel. And she's going to be asking us about really just what is life like as a responder? So, we've got the head of incident response from Dragos who specializes in industrial control attacks, head of threat intelligence from Red Canary who sees attacks from all over the world. And then myself who really sees a bit of an intersection of both of those, right? As well as our MDR partners, right? The data that we're seeing from our threat hunters throughout the world. So, we're going to talk about what it's really like. Maybe demystify some of the most exciting topics. Maybe what's some of the more mundane things that we see as well. And then the second panel is really focused on public and private partnerships from the government lens. I'll be working, I'll be co-presenting, excuse me, with Kevin Mandia who's at, you know, Mandiant slash Google. Also the cyber security ambassador, Nathaniel Fick. And then the ambassador of cyber security from the Netherlands. So, really talking about how organizations like ours work together, what we can do to continue information sharing, make it even more effective moving forward. Well, that sounds like, I mean, both, they both sound fantastic. I mean, we've had Kevin on before. You guys both have really good observation spaces, you know, if you will. What's changed since we last talked at Ignite in December? We had you on and you were sharing with us kind of the state of play. What's new, what's changed? Well, you know, it's interesting because I remember when we met in December in Las Vegas and my presentation at that event was focused around the cyber security threat landscape and the things we weren't hearing in the news. And in particular, with this overlay or this perfect storm, if you will, of stories from ransomware, cyber criminal attacks, from lapses to the Russia-Ukraine war, what we were saying was, hey, in the background, we're still seeing a lot of nation-state activity and in particular that coming from China. And if we look at the headlines today, you know, what are we seeing? A tremendous amount of potential diplomatic conflict at least brewing if not actually, you know, occurring between the US and China. And certainly them have taken a stance what's going on between Russia and Ukraine. And I think you're going to continue then to see that emanate in the cyber threat landscape and the headlines on the cyber criminal side. We continue to see a lot of ransomware attacks. Some of them are less successful than they have been in the past, so that's good. And we can talk about why that is if you'd like. But I think also what you're seeing is these threat actors put even more pressure and exert more pressure on their victims.  What's their social media account? How can I reach them? And I'm going to personally reach out to them. So we're seeing a 20% increase, or excuse me, 20X increase on what we saw in terms of extortion and threats year over year. So a lot of interesting behavior going on. You mean reaching out to family members to try to get them to put pressure? So just added pressure from- Honey, why haven't you paid the ransomware attacker yet? Because now they're impacting my Instagram account. I mean, this is literally going on and we're not talking just spouses, family members across the board, children.  Why have ransomware attacks been somewhat less successful? I think it's directly related to a couple things. One relates to what we're seeing in terms of Russia-Ukraine war and organizations sharing information more rapidly. So you've got a lot better threat intelligence. So how that translates into ransomware attack is we have decryptors faster. These attackers have to move more quickly to have the next iteration or variant of the software that they're using before law enforcement, for example, or public companies get access to it and share that pretty rapidly. You're also then seeing organizations who are simply more prepared. They've got their battle and response plans in place. They have backups, they know how to get them. They've exercised and tested their plans so that when these situations occur, they can react very quickly. They're also detecting the activity earlier in the attack, in the lifespan of the attack, which is always better for everyone. Are you seeing backup and recovery, so-called data protection, become a more fundamental part of a cybersecurity strategy? We're almost like the last line of defense. Absolutely, I think it's kind of more now seeing as like a first line of defense, right? In terms of preparation and organizations are allocating budget to do that. So before that was seen more as like a disaster recovery type of last case-needed type of situation. Now that's just a fundamental integrated part of many business practices, which is good news. Somebody told me the other day that the, I wonder if you can confirm this, that the sort of, the playbook has changed somewhat.  My understanding is now that's somewhat shifted to, it's not even about getting your data back, it's about them not releasing it to the public and maybe it's one of those both situations. Has there been a change there? Is that a misperception or? Absolutely, so a year ago at this time attackers were doing what we would call a quadruple extortion. So looking to, for ransomware attack, looking to essentially get payment for encrypting the data and returning it unencrypted for when they steal it, not extorting it for DDoS attacks and then for giving that same data to your most sensitive clients. What we've seen now is that they've, by and large, stopped doing that first part, which is the encryption piece. So there are certainly still our organizations who are victims of mass encryption, but attackers are seeing that that takes a lot of time and effort. I mentioned that we're able to get decryptors faster so that's starting to encroach in their business market and the effectiveness of that. So they have less ROI on the encryption part. It also takes a lot of time, a lot of energy for them to decrypt it, to make sure it works. So have a customer service elements of, did this, oh, did you have this partition on this hard drive field? Shoot, let's give you another key for it, right? There's a lot of activity there in interaction that if they simply steal the information and then extort you on the back end, much simpler business model. Okay, so it makes it harder for customers. So you can protect yourself with air gaps and immutability and that still might not obviously solve the problem. So you have to figure out that layered approach as well. So you've got to have it both covered. It would be a recommendation, I presume, right? I mean, 100%. Why don't more companies do those types of things like air gapping, like immutability? Is it just because that's a pain as well or it's just an added expense or they don't know how? You know, I think everything is, to be looked at from a business perspective and a risk calculus based on that, right? So there are many organizations who do that, but does that mean that, you know, you're not going to hear a story in the news of an organization who didn't do that and was impacted that way, right? So I think by and large organizations are really taking security serious and we're seeing a lot of them really close the gap, but unfortunately the cost analysis is still heavily weighted in the attacker's favor than it is in the defense favor. Now, I don't know if you know this, Wendy, but AI was invented in late 2022. We're in chapter two, right? It was 150 days ago, AI was invented, but so my question is how much AI activity generally but specifically foundation models like GPT, have you seen in signatures of attackers? Are you seeing obvious evidence of that yet? So we're not seeing it consistently across the board yet. That doesn't mean it's not happening though, right? And I imagine as we continue to move forward, you know, this few months past AI being invented, right? For that lens that we're going to then start to see more and more patterns that are very indicative of it. I think the impact that it's had, just this discussion, right? On like the fact that like we know what generative AI is and what are large language models, that's really then crept into the defender's mindset as well in the landscape of technologies we're building moving forward to accelerate some of these manual tasks that we're delivering today to really be able to make a positive impact on response and getting detection measures in place faster. I want to ask you about the public-private partnership. And I know there's sort of a narrative, there's certainly a lot of what I call finger wagging going on from the government, the company's got to do better. It's like, okay, you know, we know, we got to do better. But it seems like, you know, the government in some sectors wants to break up big tech, they block mergers, they're very sort of unfriendly to big tech, there's a real negative narrative, particularly around social media companies, but also other companies. What is the state of the public-private partnership? And you know, my sense is it's not horrible, but it's not great, it could be improved. Do you agree and what could we do to improve that? So I am an optimist by nature, but I would say absent that, I would still give you the same answer that I think that we are in a better spot than we've ever been as it relates to cybersecurity and the information sharing going on between the government and private and public organizations. A few concepts I think that are fundamental there. You know, one, cybersecurity as a field and as an industry in practice has now been around long enough that you have people in the government who have had hands-on experiences, practitioners in the government, but also largely who have been leaders in the private sector and have gone back and then work in the government as well. So they have a really strong understanding that there's not one organization anywhere in the world that has all of the data, right? It really is a combination of great organizations in the public sector, within the government, within the military, and within the private sector. And the ability to infuse all of those elements with data that's enriched is really critical for us being able to compete with attackers. And I think the war in Russia, Ukraine, has really accelerated that in terms of the information sharing component. It's never been better than it is before, not only between public and private entities, but also between organizations like ours who might have traditionally competed with other organizations in our space. We're sharing data much more rapidly than ever before. So I'm pretty optimistic about what we're looking at moving forward. How about sharing data with allies? I heard maybe it was on meet the press, I think. Netanyahu was on, he was doing the tour. And he basically made a similar point. He said, look, because of course, you know, journalists want to poke them. But he basically said the same thing, that the relationship between Israel and the United States has never been better. And he specifically mentioned sharing on cyber data.  How are you seeing the data sharing and the collaboration with United States and its allies? You know, I've had multiple meetings today with heads of foreign governments, right? Related to cyber information sharing. I doubt that five years ago even that volume of meetings would have occurred with an organization like Palo Alto Networks, right? So across the board, we're seeing our government really broaden their information sharing capabilities and continue to work with allies throughout the world. But we're then seeing those allies understand the importance of working with organizations that aren't also government or military and making sure that they're infusing those technology capabilities in their landscape. How's that come about? Why do they want to meet with you? What's the, you know, the outreach like, hey, we just want to sort of start a dialogue or as we want specific information or? You know, okay, so that's a great question because the dialogue has, the relationships and the dialogue has already existed for quite some time. And so these are really just natural, organic extensions of the conversations we're already having. We have teams of people, for example, throughout the world in allied countries, not only from Unit 42, but from larger Palo Alto Networks. And so we make sure that we've got those touchpoints existing consistently, that they're aware of what developments we have ongoing with technology, that they're aware for Unit 42, for example, that we're expanding. We had a press release this week that we've expanded into EMEA where we've been now for over a year, but what's new is the JPEC region. So, you know, we now will have capabilities throughout Asia Pacific. So we're really excited about having people on the ground there. Those relationships are really what kind of drives that, the ongoing discussion about, hey, what can we do to help and what can they do to help in reverse? So it was more, hey, we're going to be at RSA, let's get together kind of thing, and then- Certainly, but then to talk some specifics, yeah, about information sharing and what other type of work we can do together. The Unit 42 threat intelligence report came out and there's a great data in there. I have to say, I looked at it and said, it's like we're making the same mistakes over and over and over. That one chart that showed 80% of the alerts come from 5% of the rules. And obviously, organizations have trouble prioritizing, but it does seem, when I look at the data and I look at what practitioners are dealing with, it seems like a lot of the same things for years. And so, the more things change, the more things stay the same. It's obviously more complicated. Is that unfair observation? If not, why is that? And how do we move beyond that? Yeah, I don't think it's an unfair observation, right? There is a deluge of data that is occurring that responders need to look through and prioritize and figure out how do I manage my day so that I can effectively protect my organization. I think on the positive side, there are much better technologies in place that are incredibly more effective. So, organizations have the ability to leverage that. We have the ability to view single panes of glass now versus an analyst that is having to do a tremendous amount of manual work of this screen that I'm going to switch to this other. And then this, and I'm going to manually pull this information together. So, from the AI side, I think we're going to see a lot more infusion of that data, a lot more ability for analysts to move faster. So, in terms of being able to respond faster to detect these incidents sooner, those are some real positives. Yeah, one of the things in the threat report, I got, I have it right here, the bar chart that says the time it takes in days for organizations to resolve security alerts. And 60% of the organizations were four days or more. Now, the stat used to be it was 365 days to find, you know, an infant. So, that's getting better, but it's still, you know, not good enough. And so, basically you're envisioning a world where, you know, it's kind of real-time remediation, I guess my question is, will the machines, how far away are we where the machines will actually take action for the practitioners such that, you know, the practitioners actually trust the machines to do that? You know what I mean? Because automation sometimes, it's like unintended consequences. So, where are we in that spectrum? So, a great question. So, that report specific to the cloud, right? And so, what we see there are organizations who are challenged with having multi-cloud environments. They might have, you know, one vendor, one set of data for one part of their business in, you know, Microsoft and one in Amazon, for example. And those are not then coming back to the singular pane of glass that I mentioned. And then you have this hybrid environment where you've got on-premise devices and cloud. And so, all of that adds some complexity and that's relating to what we're seeing which are some slower response times. But to your point, those are fantastically better than what we have seen before, right? None of them though are at the speed of the attack, which, like you mentioned, needs to be a bit more automated in terms of the response pattern. So, that aggregation of the different data sets, being able to pull them in and infuse them together so that then you reserve human brain power for the pattern recognition, the very detailed analysis of is this actually bad or is it not, but you narrow that window down to such a small degree. So, to get to the automated remediation component, we have a lot of clients who are asking about it. Like you said, there are certain pieces of that spectrum which we will deliver, but there are some that you typically still want to make sure you've got human intervention on. I think we're probably still in the year's timeframe before we see just widespread machine decision-making at that scale. Is the state of the art today to take all that disparate, distributed data, I guess it's disparate, but distributed data, put it into a data lake and then be able to analyze it? Is that the, is that state of the art? Because it takes time, right? You got to bring it in, you have to clean it, you have to make sure it's of high quality, you got to ETL it, the whole data pipeline thing. But is that state of the art today? You know, I think it depends on the technology that an organization has. So certainly if they're one who has a huge backend, then that may be their process. What most, though, I think are doing is also analyzing it at the source of the data first. And so they're not having to pull in an entire huge mass of everything and then weed that out, but pulling in information that's most critical and then analyzing and working through that as quickly as possible. That's like you. Okay, so kind of real time, leave the data where it is and real time inferencing of the data in place, okay. So that you're reducing the amount of data that has to be analyzed and charged multiple times. Right, exactly. All right, well good luck with your talks tomorrow. Thank you. I'm so excited for you and thank you so much for spending some time with us, Wendy. It's great to see you again. You're welcome, always great to talk with you, Dave. Cheers. All right, and keep it right there. I'm Furrier, we're back with theCUBE, Dave Vellante. We'll be right back from RSA 2023 at Moscone.


Hi everyone, Joe for jaspyscasebreaks.com with 2019-2020 Panini Prism Basketball. It's a one-pack, sell-o-pack break, number 44, random team. Now at the end of this break, the whole point of this pack break is at the end I'll re-randomize everybody's names in a brand new dice roll and the top eight names are going to get into that basketball mixer. Pretty nice. Big thanks to all of these folks for getting into it and I think I'm going to do one more sell-o-pack and then sell out the mixer spots, remaining mixer spots straight up and then I think we'll be in breaks while we have time for that tonight if we want to squeeze that in. Let's roll it and randomize each list, two and a two, it's just for the break, four the hard way, one, two, three, four. After four times we've got Nick down to Cabot, hard four for the teams, one, two, three, and four. After four times we've got the Los Angeles Clippers down to the Charlotte Hornets. Alright there's all your teams right there, I'm not going to read off all these teams for this pack break but there you go.  So let's actually show all the teams all on one page, let's sort by column B, just a one pack break anyway, trade a window close, let's print, let's rip. We did get a Zion bass card though a couple packs ago so it could be some surprises in here, Zion red, white and blue card would be pretty cool in that pack, or John Morant or Hachimura so on and so forth, a lot of solid rookies in this class, Silver Rookie, no Thonmaker, Cam Reddish Rookie and Matisse Tybal Green, it's not bad, Sixers, Darren Mackenzie, Cam Reddish bass, for the Hawks that'll be for Pano, and we got Porzingus Red, White and Blue, Landry Shamet Red, White and Blue, and Justice Winslow Red, White and Blue. Alright, there you go, that's not that bad of a pack as far as just a one pack break goes. Now, once again, top eight spots, we'll get Mixer Spots, top eight after seven times, if you have multiple spots you can win multiple spots, if you bought multiple, yes you can win multiple sets, seven times, five and two, one, two, three, four, six and seven and final time, let's see who we got, thanks everyone for giving this a shot, and congrats to the following, Edwin, Rob, Diane, Pano, Chris, Jesse, Rob again, and Chris Erickson and congrats to the top eight, you're gonna be in that basketball Mixer, we'll put a little rooftop symbol next to your name, and we'll put CP44, Salopak 44 next to your name so you know what break this came from, where the winnings came from. There you have it, Joe for Jasper's Case Breaks.com, I'll see you next time, bye-bye.