The discussion of the problem in theories that we are doing on the reading last class. You remember we discovered that the operator product expansion was a platform that stands for hand and leg. OK. Now, I would like this class to complete our general discussion of the problem in theories with generalities. From next class onwards, we'll go on to study more detail specifically about it. OK. So the things I want to talk about in this class, center largely around the state operator map, which will be important for many purposes. But as a preliminary to do this, I want to tell you about, OK, so part of what we're going to be doing in this class is to understand the importance of this central charge testimony. So the first thing I want to understand is what it means for what this operator product expansion means, what it implies for how the stress states transforms under a conformal transformation. OK. So this is a very simple. We know that the generator of a conformal transformation parameterized by the function. So suppose we had z equals z prime plus epsilon z times z prime. OK. We have the conformal transformation of this state charge. We know that the charge that generates this conformal transformation, where we know that the charge that generates this conformal transformation is epsilon of z times z. OK. We also know that the change in the operator, OK, we also know that the change in the operator under such a conformal transformation is the residue of the form in epsilon of z, g of z, for the operator, I say no, in question. So residue. Now, let us see what this would imply. OK. The stress tensor with itself has an operator product expansion that has these things. Let me first start with the simpleture. Suppose we had an operator. Suppose we had an operator such that it's operator product expansion with the stress tensor. I lift the form, g of z with the form of 0 was equal to h of o divided by z squared, that's 0 divided by z. It's just the general form of any operator which we diagonalize this bit in the scale dimension charge of the operator. But what's special about this thing is that there are no hours in it. So it is to compute the residue of the pole in this expression. Well, what do we have? We have a pole here directly. And we have a double pole here, which we can get to make a pole if we get the first order term of the data expansion of this epsilon of z function. So we assume that epsilon of z doesn't have any pole to it. So with that assumption, we find that g of z of 0 is equal to epsilon prime of z of o, that's h, plus epsilon of z, that's h of o was the operator that we get in the path in table after making the change of variables. New operator, the operator o tilde. So the operator o tilde that we have to insert into. So if you had a path in table with an operator o inserted and the original z variables, then once you get the transformation to the z prime variables, you are supposed to insert the operator o tilde. My o tilde will be equal to 4 plus delta. My o tilde of z, but since this is z prime, that's called z prime, is o of z prime plus delta o of z prime. This is the insertion that we have to make in the z prime variables if you want to get the same result as the path in table with o inserted in the z variables. But let's expand this out. This is o tilde of z prime is equal to o of z prime plus h epsilon prime of z o plus epsilon of z delta. These two values can be combined into let o tilde of z prime is equal to o of z. Remember z was simply z prime plus epsilon of z prime. So this is just the first term in the table expansion of that. So this part of the transformation is very simple. It's just telling you that the insertion that was at the point z in one variable, if you make an insertion at a given point with a given value, let's say 1 in the first set of variables, the same insertion has moved to points with different values in the second set of variables. So the variables are different, not for the smallest set. It is very difficult. This part can also be promoted to a finite expression which has this term as its first term in the table expansion. It's simply del of z divided by del of z prime to the power h. z was z prime plus epsilon of z prime. So if you expand this out, this is just 1 plus epsilon of z prime, and then to the power h into the factor of h. There is a 1 plus epsilon prime of z. It's done a lot of stuff, so it's not h. I have to break the exercise. So this is a finite form of the transformation that will reduce to this infinitesimal transformation, 400 decimal epsilon. This is an exercise for you to convince yourself that you can get to this finite form by integrating infinitesimals and operating them such that the operator product expansion of the stress tensor with the operator had just these two severe instances. The 1 by z square piece, the 1 by z piece, then that edge of that operator transforms in a very simple way under the form of transformations. And we found that the form of transformation. We found that that simple transformation. Now, we could see that it's like this d z prime by d z to the power h over to the value of z prime. Simple transformation property under the stress tensor about primary operators. This operator product expansion would map to some nice property of representation theory of this. Well, this operator product expansion will imply that the state due into this operator, what that means, we can't do it in a minute. We can't do it any soon. But the state due to this operator is at the bottom of a tower of its representation under the video so go ahead and do that. But the real answer is we also can't do it in a minute. Sessions. I need to shift this. That is great. We see that this form of the operator product expansion implies that operators have a special role to play to enter that tension theory. That's basically the reason they call it primary. The primary or most major. OK, but at the moment, it's just me. We see that the stress tensor is not itself a primary operator. Because the operator product expansion for the stress tensor, while it has these two pieces, also in addition, but unless C is 0, the stress tensor also in addition has this effect, this is 0. Now, what does that imply for the transformation? What does that imply for the transformation of the stress tensor under the formula? So now, suppose you perform an infinitesimal formula transformation on the stress tensor in a second. Is the verification of the definitions? Yes. What is the difference between a primary and a quasi-primary? OK, a quasi-primary operator is an operator that has definite weights under state dimension and state. OK, I am state. And state. So any operator has a definite value of age in each part, OK? It's called quasi-primary. For an operator to be quasi-primary, all that is required is that its operator product with a stress tensor has this form, as well as 1 by z squared and 1 by z squared. We saw last time that that was the same statement as having definite transformation properties under under state dimension. A quasi-primary operator could in addition have other additional singularities in its operator property, OK? A primary operator has no other, no additional states. So the T would be a quasi-primary. It would be a quasi-primary. OK, so now the question we're asking is what is the transformation problem? How does T set transform under control transformations? OK, so that's 1,000, 1,000 into the test method. What we want is the recipe of the poem in this expression. Well, of course, these two parts survive. So T till and out z prime is equal to, as these two pieces in it, so it's 2 epsilon prime z T plus epsilon z then T. But that's an additional piece. The initial piece comes from taking this epsilon and differentiating three times, OK? So that would give us an epsilon triple prime of 0 divided by 6 from the data expansion. And we're also going to do the z cube. The z cube, we can't set against the z to the 4 here. It'll give us a 4, will it? And then we will have the c by 4 and then c by 12. And then we'll have the c by 2. So you get c by 12 and epsilon triple prime, OK? Now, you say that this piece is conceptually different from this piece. I insert the second t at 0. If I insert it at w, it would be epsilon triple prime and w. You see, because what am I doing? Instead of epsilon, let's look at this, OK? I'm taking t of z over 0, so the analog of that is t of z, t of 0. I want now the pole in z of this expression. So what I want to do is expand epsilon and I'll see this expansion in z. So I take this way epsilon and I want the pole of epsilon by z times w to the power 4 prime. Exactly. So there's no w. We'll put the other epsilon here. OK, so we multiply it by epsilon and we want the pole of that expression when we divide by z to the 4. So no, if you want to cancel, so we've got a singularity at the end of z to the 4, if you want to cancel that, you have to do data series expansion in epsilon of z. It should be epsilon prime over z to the 4 prime. Yes, yes, I'm sorry. Do it back up by just putting this thing in w. So let me tell you, OK, suppose I put this in w, I put everything in w. Everything will be out of shape now, OK? Now let's put everything in w. Then this, everything here is w prime. Should I try to clear out everything here? Maybe I should. OK, so everything here would be w prime and w. So this is 4 of w prime. And it's just a switched z goes to the u. Again, the same, the translation would be equal to the same. It depends on what name you put on it. Sorry, sorry. OK, I'll just find the answer since w is w by u and it's the same. I won't change that. And this is all it's going to be. It's all in there. Maybe we should put w equal to 0 because we had a problem with the extra piece which we had. The last thing, yes? If it hits over the left square, even if there's a rotation, you're asking what they will be? t of z, t of w is always h. But o, let's say o prime, o prime, w prime is z minus w prime. Always true. It holds for every value of w, in particular w equals 0. So there's no problem in putting w equal to 0. It's just sometimes obstacles things. But it's not wrong. This relation has been terrible. What I should really do is to clean it out and get rid of it in unitation. But you understand what I've done, right? Logic is clear as you can fill out the notation. Now, everyone, I want you to look at this transformation piece and notice that it's logically different. I mean, it's a big difference between this piece and this piece. Because this piece, this piece is all homogeneous in stress terms. All of them, degree 1 in stress terms. Well, this piece is degree 0 in stress terms. But what's happening here is that this additional piece, because it's not proportional to the stress terms itself, it's just picking up and telling you that you pick up. When you go into the corner transformation, in addition to what you would have got, in addition to the transformation that you would have got, had t been a quasi-prime, had t been a completely primary operator, what you get is an additional constant piece. So t shifts by something that's proportional to t plus a constant. Oh, so this is the ratio of t under the constant. Under the constant. Exactly. I think your physical interpretation of this is one really good case in a moment. It's basically, the physical interpretation will be the vacuum energy. You see, since it corresponds to a constant, it's only the constant. It contributes only to the 0. So this basically is telling you that the zero order of t is the vacuum energy, not the vacuum, it's the vacuum energy. So this basically is telling you that a formal transformation can change the vacuum energy. But this is a very familiar event, because vacuum energy has a Casamine part. So in a problem field, in any one of these areas. I mean, we have not put anywhere that you know, like, we are both in refining. Well, it could be. We'll see in a moment. And an example of this, such as a formal transformation, where we want the texture from the cylinder to the plane. So it takes you from working in a finite space to an infinite space. So the infinite space you have no gas. The finite space you do have gas. So there should be a slip in vacuum energy. That will be the physical interpretation of this effect when we see it. But let's have more to see in the example. No, but it's true it sounds funny and puzzling. And the key point is that it has to do with the fact to get the energy, energies are generally infinite. To get them right, you have to perform a subtraction. That subtraction has to be local. You set it to zero in one space, it becomes something finite. That's essentially what's going on. That's what I was thinking about. Gasoline energy. That's right. But it's correct, and it's visited. Now, once again, I'm going to leave as an exercise for you people. I think we should make actually a list of exercises. So let me promise that by next class, I'll give you a set of seven-eight problems to solve. Once again, I'll leave as an exercise for you people. And it's not an entirely true exercise as far as I know. I'm not telling you shows. I don't think I've ever wanted to do that with you guys. To integrate this to a finite space. Yeah, you know how much it changes the system. There's plenty that you have to work it out a finite expression. If the final answer is given in Kuczynski's book, it's given in terms of a short shit there. The final answer is supposed to be the following. I can't remember it. I'm sure it's not difficult. But you will explain to us. No, but OK, it's the following. It says that remember this was, so once again, plus your del z, z prime squared, z prime, equals the t of z, then minus c by 12 times this funny object, z prime z, where z prime z, f of z, f times z, is equal to 2 del qf del f, then square f, then square f divided by 2 del f. I did check, is that if you plug this in for n is equal to the z plus the decimal, and you work out the first order, the reason for it. The problem is integrated to a finite. If you take this expression, the claimers, that they're going to integrate this to a finite transformation this week. So this is how the stress tensor transforms under the form of transformations. It transforms like a primary operator plus a shift in its zero point in its constant bar. And the shift in its constant bar is given by this. That's all I wanted to say about transformation of properties of operators under the form of transformations. We come back to this in some middle part of this lecture. We come back to this in some middle part of this lecture. But before we continue now, and now I'm going to introduce to you a particularly looking form of transformation. That is a particular use of the study of the problem. So let us, this is, OK. When we were studying string theory, when we, the whole discussion of the problem in theory started because we were studying string theory. So when we were studying string theory, we were studying it on a cylinder. We had a variable sigma that went from 0 to 2 by. And we had a variable tau that defined a variable w. We're working in nuclear space now, unlike the discussion that we had in string theory, which was in Minkowski's space. Yeah, so metric of sigma and tau is the included metric. And let's define a variable w. So Kuczynski calls this guy sigma 2. Sigma 1 is sigma 2. Let's define a variable w, which is sigma 1 plus i times sigma. This is a complex variable that parameterizes where you are on the cylinder. This study of a particularly interesting conformal transformation of the cylinder. And the conformal transformation of interest to me is to a variable z, which is e to the power minus i times m, where which is e to the power minus i times m. Now, the first thing that we know is that the variable w are defined over the whole complex plane, because sigma 1 is equal to sigma 1 plus 2 by. So we get all the action that's happening in the w complex plane. It's happening between 0 and 2 by in this complex. w goes to w plus 2 by, z then comes back to itself. What a length. What sigma 1 is for the variable z is the fact that sigma 1 went from 0 to 2 by in the complex plane, where the z is not a restriction. That will restrict the region of the complex plane over which z. So we've already seen, we've already been able to understand what z is in terms of length. The phase of z is where you are on the circle. What about the radius restriction? The radius position on z is, well, let's look at it. So the minus i here, the minus i here and the plus i here. So the radius, we see that model z is equal to e to the power sigma 2. The radius goes to infinity, negative infinity. The radius goes to z. We've understood what's going on. We've got this w plane here, and we've got this z plane here. Line to constant w map the circles of constant modulus of z in the z plane. So higher up, these circles become much. When down to t equals minus infinity, the circle becomes extremely small. If you went up to t equals plus infinity, the circle becomes extremely small. The conformal mass cylinder is also an extremely useful thing. It's often an extremely useful thing to do. Firstly, you understand that there is a sense in which, if you know everything about the theory of the plane, you also know everything about the theory of the cylinder. Because these two theories can normally relate, because this theory is a conformal theory. And the map from the cylinder to the plane is a conformal. So you don't have to separately understand the theory of the cylinder and the plane. You can do whatever you want. Secondly, from the point of your complex analysis, the plane is a much simpler surface. Because you don't have to worry about any conditions. So it's often simpler if you're trying to use the techniques of complex analysis, which we will use a lot as we go on, to perform this map and work with it very soon, that we look up to very soon, that makes this map very good. We need that conformal transformation. This is a particular conformal transformation that will be of great interest to us. Immediately, we'll talk about the change of the stress tensor. What the change of the stress tensor is under this map. OK, first let's work out how primary operator of weight H would transform. OK, so primary operator of weight H, remember we had, which we didn't go up, we have that. Yeah, so there is z prime squared t of z, prime equals t of z, the one that's ready to see it. So let's choose z equals w, z prime equals z. And remember that z was equal to e to the power minus i. OK, so del z over del z prime, sorry, del, which is the subject here, is equal to minus i times z, which is the transient. This relationship becomes, remember z prime or z, minus i times z, all things squared, oh well, let's call this h, I think we're going to get it h. Minus i times z to the power h of o of z prime z is equal to o of w. That tells you what an insertion of an operator in the w brain would transform to as an insertion of an operator in the z plane. So suppose you had some function of, you were doing a path integral of the x, OK, and you inserted some operator with some function of x in the w plane. In order to get the same answer for the path integral, you have to insert a different operator and the same operator, the same function of the field times this one. So the operator would have to insert z, that way it would be of that. It would be the same operator as a function of the field. Divide that by minus i z to the power h. So look, the operator I said equal to 0 would, you'd only be able to take the limit because it would blow out, blow down. We will understand why that's the case. But the stress sensor, we have in addition a constant piece which you can work out by working out this expression on the exponential map. This is a question that if we have any operator, which is not the local operator, then we just replace the path integral in the cylinder with this particular touch. Is it always? This is true for any primary operator. You see, we derived this under the assumption that the transformation of that of a private operator. Any operator whose singularities with the stress sensor are simply those of the double pole in the cylinder. It will not be true if it's not a primary operator. But I think we'll see actually knowing everything for all primary operators in the theory is not to know everything about everything because representation theory allows you to go from a primary operator to anything. So primary operators are particularly important. We'll see this. But this statement's only true of a primary operator. Is this clear? Now, the next thing we want to do is to see what works out with the stress sensor. So the stress sensor, we're going to get the same reaction. So we're going to get T of, well, that's right. So T of w is equal to, in the case of stress sensor H2, so that's minus z squared T of z. Plus some constant, which will work out by working this thing out on the exponential map. And I'm not going to do it, I'm going to quote the result, which is k is C by n. It's the error operator will be the mode of this T. So this is relatively minus sine wave. So this is not going to work out okay, actually. But it will imply a shift with a minus sign when we do a more expansion. So the first thing we'll notice is a minus sign between these two, okay? Obsolve that minus sine wave, what we will get. And we will see in more and more detail as we go on, is simply the gas energy. I mean, the fact that the stress energy on the plane has no gas mean energy. And the stress energy then cell on the cylinder, that's what makes it work. Okay, we'll see this in increasing detail. Now, let me go. So this can form a map in the cylinder of the plane. We've understood how the stress transfer transforms and this can form. The second thing I want to understand is, suppose I was interested physically in some path integral on the cylinder, right? So I was interested in calculating some path integral which is from some state, sorry one here, some initial state, sorry one, to some final state, sorry two. Okay, with various operating answers. We've got initial state, final state, and operating answers. And I wanted to do the path integral. I want to understand what path integral this is and what it transforms to, what it transforms to on the safety. We have a question of what path integral calculates a transition amplitude with operating answers. What if you know very well from the general formulae. So suppose the fields that we were doing the path integral over are denoted by x, we have dx, and then we have exponential I times the action, suppose we were in the Minkowski states just to be homophobic. I times the action, then you put the operator as four, one, two, three, four, one. Four, one, two, three, four, one. That represents this part of the evolution. But now we're going to put in the fact, we want to put in the fact that we're making a transition amplitude from this state to that state. How do we do that? Even state is represented by a wave function. So let's say that this state is represented by the wave function psi of x. x, as in this case it is, that is the function of sigma, this would be psi of x of z. A Schrodinger wave function whose variables are x of z. And the final state, I got the phi one, so let's go to the psi one, the final state is sum psi two of x of z. And then this transition attitude. We want the time t two of the conditions that act time t one value of the function of x of z. So let me call, do you know that by time t one x of sigma is frozen to somebody here. We freeze it to some other value, let's call it y of z, not completely modified. With the boundary conditions not completely modified. So this is some function of, which depends on x of sigma, it arrives. And then in order to get the path into, to get the attraction amplitude in question, what do we do? We take this function of, and we integrate it against psi one of x of sigma one, of sigma, and psi two star of y. And then do the integral over all values of state to expression, of the transition attitude between state psi one, state psi two, where the operator is excited. Is this correct? I mean, if quantum mechanics has a familiar analog, the path integral where you demand that the initial particle is at x one, and finally the particle is at x two, is the transition amplitude for a particle from point x one to point x two. But of course, if you've got some wave function, you have to convolve that amplitude with the wave function. Yes, you take that, take what's fixed entities, and convolve it against the wave function, the two wave functions, and that's how we do it. If you want the transition amplitude, you know what I mean, right? I mean, the path integral gave you x one, e to the power i h minus i h d x two. But if you are interested in psi one and psi two, what, let's call this one and two. Okay, what you will do, is to say that if I was interested in psi one, e to the power minus i h d psi two, and then instead of complete set of states, and get psi two x two, x two e to the power minus i h d x one, x one psi one d x one d x two. This quantity is the complex conjugate of the wave function, the final wave function, this complex quantity is the wave function, and these two entangles are the entangles of it, d x sigma d y. Right? So, if you were interested in computing some state-to-state transition amplitude with some operating sessions, this is the part of the amplitude, if you do. That's a scare, right, everyone? If you were interested in calculating a state-to-state transition amplitude, I chose my states to be eigenstates of, because we want everything to be nice and complex, let me use it to be eigenstates of both the Hamiltonian and the angular momentum. Initial state, for instance, I'm going to do the same with the final state, but first, for concrete conceptions, my initial state, okay, to be an eigenstate of the Hamiltonian and the angular momentum operator, and therefore, L0 and L0 bar, the way we do it. Okay, so this linear combination of left and right, 0.30. The next question I want to ask is, suppose I do this path entanglement, and I compare it with this path entanglement. Wherever the state of the final state, same operating insertions, in a state at an earlier time, I wanted to get the same answer for the path entanglement. What should the state at the earlier time be? The question is clear, because the path entanglement just generates that revolution. So if you start with that state, which if evolved, would have reached the state, and then do the path entanglement over this original region, you get the same answer for the path entanglement. We just have to say that what we will have to do is that we just have to say that if we started with a state psi, it was e to the power minus i h d of my heart. This state at time t is the same. So within the path entanglement, when you have done with convolution with that state at time t, and by that n time t, the final answer would be the same. That was true in the Minkowski space. When we are calculating the Euclidean space. Okay? You know that when you break the transition from Euclidean to Minkowski to Euclidean space, the time that we should operate have become e to the power minus h d. Okay? So in Euclidean space, if you wanted to get the same answer for the path entanglement, if you had a state that was my entanglement, and you wanted to get the same answer for the path entanglement, you made the insertion at different times, you should start with a state, Euclidean, that varies depending on your time of insertion, like e to the power minus h d, that's not the same. So the state was also a good eigenstate of angular momentum. It was also a good eigenstate of angular momentum at a particular eigenvalue and the eigen, and the angular momentum. We could say not just how the state varies with time, but also how the state varies with time. Because any eigenstate of angular momentum of eigenvalue l, which is a function of where you are on the circle, into the path by l sigma. Okay? So suppose we put in that condition as well, you know, we've got a star, which is e to the power minus h d, plus i is sigma. Sigma, sigma. And let's call this sigma two, taking your own notation. Yeah, this is just, okay. Well, this is on some fixed state. What I mean is that suppose I wanted to look at what the state became, if I changed my zero mode of what sigma was. Yeah, okay, this could be still, let's forget about it. It's not, it's not important. Okay, let's just deal with the time. Okay, now, now, is to understand what all the states that we've been making translate to the state. Translate to something that is simple. If I wanted to do a path integral with various operator insertions between one time and another, what happens to the z-plane integral with various operator insertions between one time and another? What operator do you get taking the operator that you start with and then performing the component? States, it's states, the functions of, you know, i is the function of where you are. It's just that the states become boundary condition of the path integral, not in time, but in radius. A lot of the second questions we have. That is, what path integral can I do where I change this state to be defined on at only a time to the cylinder? Cylinder picture, which means a smaller radius in the z-plane. What state do I have to define in order to achieve this? Well, the answer is this e to the power minus h sigma two on that state, which just translates to one over, over more z to the power, right? Because we had more z to the power z. More z became very large at sigma two large. More z was smaller. What's sigma two? Okay. We soon get an answer in your question, but what's the definition? The singularity of z, we'll get into that in a second. Okay, but you see that as long as we don't know all the way down to z, makes that any non-zero value ever it is. We have a well-defined path integral that's a path integral that was of interest to us. In the standard also we're considering states at different values of power, right? The states at different values of power. We are considering wave functions. We are considering wave functions? But how are we going to scale the wave function as we change this radius? Yeah. In order to get the same answer, it's totally well defined. If you take the path integral with one wave function, yeah, with more z to the power of h, that is the same wave function. Remember h is just a number, because we need a variety of states to have it. We didn't need to, but let's say this is a mistake. Okay, so it's just some, if the energy was something, we're more z to the power of that thing. But we hate to do that. Hate to do that. So if you repeat this, then I'm going to get off of scale introduction. It would be wherever this h was, under transportation. It would be the same way. It would not be the same attitude in this case. Yeah, okay. So it would pick up that extra fact. Yeah, yeah. The great thing about z plane is that there's a path integral. I could make z to be as small as I can. It's got a possibility that t equals minus infinity to z to the power of h. In that case, what does it make up? This becomes an arbitrarily small insertion. A wave function insertion arbitrarily smaller than an arbitrarily small insertion. That is the same thing as a local operator insertion in a path integral. A local operator insertion is simply a weight associated with a path integral that depends only on the value of the fields at that point and its name. Which is pretty different. Apparently, certainly the cylinder's very different because it depends on the value of x finite nature. But if you take this t to infinity, you say what we're doing is weighting this path integral by what x is doing by x of sigma, by what x is doing only in the name of a particular point in the meteorite, okay? And therefore in the limit that z goes to zero, this object, whatever it is, is some local operator. Yes, so this circle had a magnet radius. Then basically we are talking about the same path integral that you have given, but evaluated on that thing. Yes. With the initial and final shites. Yes, now. I'm currently thinking that as you are taking that, one of the weight functions out there that just becomes one of the operators in operation. Exactly, exactly, that's going to be it. You say the weight function, what is the weight function? It tells you to do the path integral in particular boundary conditions on the circle, give it different, and give the path integral different weights depending on what these boundary conditions are. Now, the place where the wave function is shrinking is zero. So what you're doing is doing the path integral with weights depending on what your field is like in the neighborhood of zero, then operate and search. Any function that weights the path integral, any weight, the path integral that depends only on the phase of the data, whatever, given points, whatever. If it is little local operation, insertion and weight, you can say that. We'll talk about that. We'll give you a say. We'll talk about that first. Sir, a particular point is mapped to a particular point there. I don't understand that. Just say that. Second? Okay. Okay. I can understand that the first one is equal to minus. Right. This is mapped to a point. Okay. But if it happened at that point. Wait, which point? This point? Yeah. Okay. Let's say I insert an operator at some point. Yes. Which is not zero, you're asking. Which is not zero, let's say. Yes.  That's correct. But what we're doing is establishing a map between operators and their functions. Okay. What I'm going to do is tell you what the processor given to you does is given an operator, given a way function, it defines a local operator. Once you've got the definition of local operator, you can insert it anyway. Okay, because what is a local operator? It's some function that feels locally at that point. Once you know what an operator is, you can insert it wherever you want. Wait. But the same function, not if it's at that point, but at some other point. Yeah, but what in mind about this? At 10 equal to zero, this mapping is not 1-1. There's the whole surface map to a point. That's why this good stuff is happening.  You see, because we're doing a bottleneck all over the place. So the generic way function is to put it at equal to minus 70. Ah, okay. I mean, when I map it, I will get something which depends on which... Which direction? Okay, but you see, we're also waiting by something singular. You know that the net effect is not singular. Because the answer to this process is the same thing as doing a pure, which is clearly well defined. You see, so if your body is that by this process where defining an operator is singular, that can't be. Because its correlation is a well defined. Its correlation function is a well defined. Are we using any property of asymptotic states that we make in states that are equal to minus 70? No, we're not using that. So, you know, it's conformal to the case. It's not going to happen. Ah, you know, in a subsequent lecture for the case of free bosons, work is a way of switching. Okay? Some of your body, I think, will act. Because you'll see how it works. Okay? Maybe I should do one. No, let's not. Let's leave that for when we get to the free bosons. Okay? For instance, we're sure that waiting by the operator del X to the power n is the same thing that lets you, is that if you act with the first oscillator excited eight times, you take that state and you do the procedure at the top. We'll show it explicitly that what this is is the same thing as waiting by the del X, del y. It's totally not singular. I think some of your body will act. All your questions would arrive in this context, you'll see how it works. But, even from the point of view of general logic, I think the algorithm is totally subtle. It's not subtleties. But things like normalizability of the states. You must start with well defined states if you want to. You want to go through this argument. When you're telling about the reverse, think in a moment. It's obvious.  What we've done is understand how you can get a state from an operator. You always have to worry about whether you can do something formal. If you've got some state that doesn't like you, it's not normal. Those things are always possible and you have to be careful when you respond. But what we've done is what we've done is a map from states to operators. States to local operators. Yes, reverse. Because I think the path in general is actually like this. I insert the operator at the origin and I do the path in general subject to boundary conditions that x of sigma is something. The answer to the path in general will depend on what the x of sigma is. They're going to be functional of x. That functional is a way of functional. And it's clear in the root. It is the way of functional of the state operator map. It does function psi. And then you act it by the way of function that is 1 over the mod z to power h.  What was the property of that operator? By construction, it's supposed to be no operator, no other operator. And I chose this to be mod z into 1. Especially in this case. By definition, it is that operator is that operator which when inserted and with no other operators, it's the state. Is this correct? This operator which actually corresponds to the state at minus, at t equals to minus infinity. Or it depends on the state psi at the you know, I mean at this because you're trying to go to that state. Let's look at the logic. What I wanted to show you that if you start with the state, you can get an operator. Now I was supposed to forget about this. Give you an operator which you know nothing else about. You don't know how it's done from the state. But to ask, what state had I got this operation to move? Would I even need this operator? The question is clear. All you have to do is to do the positive integral with the insertion of that operator at the order. And you can give you all your functions. How does it give you all your functions? But the answer to the positive integral depends on what the values of your fields are on the subject. But it is a set of operators of the theory. Now you know, for instance in the theory of free state of field these operators would be operators like del x del square del x all the local operators in theory they're not quite creation operators because they look not the creation operators of particles which are not but they won't be local in the sigma 1 and sigma 2. But they're local always you see it up. It's the local weighting of the positive integral. So no matter which values you're looking at local is local. So, you see what we've done is define a map from a state to an operator well defined operator which is very insertive so we can just do the your functions of the transmission that on the operator and we can use that operator Exactly, exactly. That's good. You've got the operator, how it is some function of x is you know, reverse from that I think you wait, if you wanted to work with it that's it. That's perfectly right. If your question is how are you going to use this let's wait for me to use it for every statement. And the statement is there is a one to one map between the local operators and the state. What are your problems with normalization? You know, the second map is not so clear because it could give you a normalization of the state. It sometimes happens not on the things we're going to deal with. This formula we just have to be looking at. What are your possibilities of basically the stabilization? There is a one to one map between operators and theory, local operators in the theory and states. Now, I first want you to see how surprising this is because in quantum mechanics a state is a problem and operators are matrix. There are many more matrices in the world. Number of matrices in the square is the number of words. How can there be a map between number of operators and between states and operators? See it's clear that you can get starting with any operator you can get a state. Act on that state. If you stick that's a nice denominator map and that's actually what we're talking about. It's because you see there are a lot of annoying people but enough to do it. It's not, you know for instance then that would establish the person one to one. Because if you add it to the operator any operator that I added to the map should give you the sense of it. So your point is basically chopping across. It's not a local it's not a map between all operators of the theory and states. It's only a map between local operators and states. A map between operators and states would just be a manifest operation. Can't be true. It's clearly a map between local operators and states. It's not true in every theory. It is true in conformity. In fact it's true in conformity. It's in every dimension. The plane goes through more or less uniform. We're interested in that at the moment. Two dimensions. You can ask him. Okay. Very good. Your question was your question was why is this thing essentially acting with the operator on the back view? But you see suppose you do an arthenthector. Suppose you do an arthenthector and it's a little more it's actually acting the lowest energy state that survives when you act. The operator. Let me say this thing. Suppose you do an arthenthector and let the cylinder think already clear. In any space depending on what boundary conditions you put here it could have any state in one. But in any space it could have any state in one property. But if you're doing this in euclidean space then no matter what boundary conditions you put down below as long as the boundary condition isn't it's not so funny that it projects out the back view. If you put the boundary condition far enough below the only state that survives at late an arthenthector is the back view and the boundary condition of it is e to the power minus hd. You can do an euclidean arthenthector and allow the arthenthector to go down to times t equals minus infinity. What you're doing is stopping with the back view in order to get our state is acting with so we've got the disc which is the angle on the cylinder that went all the way down to minus infinity. It's back view state because we can transform it translate it in some sense but it's hard to do because that's the singular point. Okay? The operation and definition we will always use is the pass and take. So the the corresponding operative of the back view is the irq in some theories in all unitary theories is the irq. Because it may be something else but just like the logic that you hear right now the operative acting on the back view I mean the back view is also an operative right? Right. If you want to do that operative at this point Right. So there has to be the back view at that point also. Which we are acting Right. Unless something funny to be used to spoil that logic what you're saying will be true and it will basically be generically true like what you're saying is totally right. There will be funny theories like what you just said which we will discuss when we get to this point. But basically if you are interested in unitary the operative everything is nice what you said is exactly right. We come to caveats when we get to that level. Basically right. The thing I said was not totally right. Let's get back. Essentially right the not always come back with me. But it's a good comment. It's good intuition also. Good. Any other questions or minds about this? I know this is a very subtle idea and it took me as a student a very long time to understand. I think the technology explaining it has improved basically because Pozinski wrote this book to me. Right. But you know the explanations pre Pozinski's book made absolutely no sense at all. But this is the language in which I think it is the best, easiest to understand. And to my taste. Many of the ways of saying it many people have said many strengths and weaknesses. Okay. Fine. Now. All that's very nice. Now we want to use this. Use this fact nor do you do two things. Okay we've got much slower in this class than I hope to. But it's okay. You can't get the idea straight. I want to use this fact in two ways. Let me first use it for calculation. And I'll next use it in a concept. Maybe I'll look at the concept.  And then the calculation. First the conceptual thing. I don't understand the operator product of this action. Important thing. Say you operate a map. But you're now making a clean statement of what the operator product is. And I have two operators. I say 4-1 and 3-2. 4 operators at the point 4. Even if the operator product is action cells. In this case. Make a clear statement. Draw a circle around a 1 whose radius is which passes through it. All operators inserted within the circle. There can be any one inserted inside the circle. All operators inserted within the circle. Okay. Then this product of operators can be replaced by by OI or JJ.  I'll say it's 2. And this this replacement gives you the same answer for the path in the table. No matter what other insertion you perform. The claim of the operator product expansion is the statement clear. That insertion of this operator and this operator expires by an infinite sum of insertions on the edge, the point Z2 of different local operators. The correlations of this expansion are independent of what insertion happens. Of course, for any human insertion you know it's not so surprising that you can do some sort of data expansion of the answer. Because the answer depends on Z1 depends on Z2. So you could take the answer for the difference in Z2 and Z1 The key point is that it's independent of what these other insertions are. Is this clear? Just in order to see how he understood the statement, there's a general quantum field theory reasoning. You can read the one verse explanation in his book on quantum field theory. I never quite understood it but maybe it's a complicated explanation. No, that's correct. But I'll just see how simple it is. What's the act? What's the act? Well, you see doing the insertion with this operator insertion and this operator insertion we can do the path integral to some radius just outside O2. The integral is a well-defined path integral which depends on the boundary condition of X of sigma on this radius. And therefore define some state. Why just make an operator map? I should have drawn better pictures. Let me draw the pictures then. I'll start it. So let me just do simplicity. It's not essentially the argument at all. It's a translation. But for simplicity, I'll put this first guy at the origin. When you call the origin, it also totally repeats up. So I'll write an idea. This is O1. Let me draw myself. And I can always circle the operator O1 with the circle that lies outside both O1 and O2 and it plays no off. So whatever I put this correlation function is the one path integral. The path integral is two paths. The path integral over fields here and the path integral over fields here. There are also path integrals over fields here. So three paths are just integrating over fields in the wave function. So doing the path integral in the other disk computes over the function. And then doing the path integral on the other side is the other path integral weighted by that wave. It's some wave function with no reach in completion of the path integral that I have of interest. But that wave function is by the state operator map. The same thing as some operator. I don't know which but some operator when we call that operator is going dark with nothing here on it. In the disk. And these operators play instructions. What this operator is depends only on the map between between that has replaced the insertion of these two operators the insertion of the single operator and which operator it is depends only on the O1 and O2 one and the one on the other are the insertions in the which I choose to expand my operators in the basis of operators that were dual or two I mean states of the Hamilton at any good state can be expanded in a convergent fashion in I mean states of the Hamiltonian you can even argue stronger properties of this operator you can argue to convert this problem that's not just actually convergent. This is true only of these properties of this operator product expansion basically by replacing it about statements about expansions of states in a basis in which it is and so using the state operator map that's what the operator product expansion map. So this is the first thing it's very simple argument not even very subtle all the subtleties land the operator that the statement of the operator product expansion is really clear. So if you understood this then you know what I mean I could make these statements like t or z or 0 or 0 or 0 or 2 blah blah blah blah I could make these statements not quite simple I mean there's a no no of course when I make such a statement I need make it in the sense of the operator product expansion and the operator product expansion but in a way it denies to manage the field in the stronger sense no I have to come out the same it's conformal in that basically what's important is the state operator that exists for that that exists for that see basically what was important was that we take finance as a thing and shrink it to a point once you can do that you have your statement okay right okay after you use last class that we want to testify last class we talked about how it was always possible to choose a basis in the state of operators that were eigenstates that are definite eigenvalue unestated dimension and rotations using the state operator map for this state you remember last class we borrowed with these particular operators for quasi-primary quasi-primary operators are definite scaling to the scale dimension and definite rotational quantities but under the state operator map the scale dimension goes to energy end I should have set this more clearly you see what is the generator of time translations on the cylinder what does it map to on the z-prime see time is a radius precisely radius is e to the power time which is the same thing as x mu d by dx it's the generator of scaling transformations on the z-prime of the cylinder maps to the generator of scaling transformations of the z-prime so with given values of scale dimension and rotation about a statement of states it maps to the statement that there are any states in an interface in a possibly expanded in a basis of state of item scales of the item scale which is a manifestly true statement about any mechanical system in which the handle got inactive because we are choosing a basis in the space of operators just have good scaling dimension and good rotational quantities this thing is not a heuristic thing it's just true and we see the power of the state operator maps operators are such a large unnamed class of states are much simpler local operators are the same states so any statement of operators becomes simpler, easier to understand what does the statement of operators mean? to t ok we see it's you see it essentially it would depend on the theory in theory the free pose of the the bilinear in in creation operators because t is a very special operator yeah it's you're right you're right in representation theory it's very special you see the best way to say it is this in words that we understand that as we go on but all states of the theory can be arranged in representations of the the formal group representation of the the formal group are labeled by primary operators there is one very special operator namely the identity which is a primary the operator corresponding to the state corresponding to the operator t is a particular descendant of the identity so it lies in the module generated in a very particular way so from the point of view of representation theory it is of course very special state what it looks like in terms of some other discussion yes it will depend on the theory especially theory but from the point of view of representation theory it's very special all that much of this is coming from other questions but the state operator why we always do state things in some basis now let me quickly before we end this class let me quickly use we're going to study the representation theory in a little bit let's handle it the basic point is that you generate states by the action of t on the primary operator but how do you get the operator t you get by acting on t on the primary operator which is the identity because that's t that's the basic point we'll get rid of it in more detail now however what I want to do is to use these ideas to understand the Villas alright part of this lecture as compared to the last lecture was to try to understand things now until the states and Hilbert spaces and so on you know you should want to build theories of operators but they're not only they also have this nice Hilbert space structure isn't it and it's the interplanetary between these two structures that often gives you the most interesting things in the theory okay you remember that we had this nice Villas sort of argument I mean we had this nice pk operator problem t of z, t of w t by 2 over z minus w 4 plus t of w square root of z 2 by 2 which says that this is a function only of z and w because we derive it inside the particle table the variation of all correlation functions as we move the right hand side in these arguments let's say we make the following condition okay I'll define t of z is equal to sum over n equals minus infinity to infinity l n divided by z what this what this statement means is always understood inside the particle we can compute correlation functions of t at every value of z we know how to solve the topic so that correlation function is we define a new operator l defined by this by this equation computer correlation function if you want computer correlation function of t of l what you do is compute the correlation function of t and isolate the dependence on z that was like 1 over z by n plus 2 what is it from 2 over infinity to infinity well it is it's true it's true outside it's statement about a Heisenberg operator evolving in time it's completely true fine we will only use it if you know new correlation functions you can compute it's only true within correlation okay so what do I mean more precisely suppose I want to compute something involving the correlation function of t how do I get a correlation function of l well that's very simple suppose I want a correlation function involving an l acting at some particular time the time might be important because this analyticity is only true if there are no operator insertions you know this analytic behavior of t that t is an analytic function it's true provided you're not sitting on some other operator because suppose you've got some operator here and some operator here then the dependence of t in this region is given by the expression


Amazing. Good morning everyone. Hello. In this session, I will walk you through how our team designed a multi-tenant cluster to host different type of workloads with ensuring security across different tenants and applications. My name is Ahmed Bevorz. I'm a software engineer at The New York Times in delivery engineering and I'm so excited today because this is my first session as a speaker at KubeCon. As you can see from these little icons, I'm a proud dad. I write a lot of go for Kubernetes operators, applications, and other things and also enjoy building on top of AWS. I'm also a scuba diver, so if you have recommendations, please hit me up, but only warm places. I don't like cold water. So let's dive in. Before we start, let me tell you a little bit about The New York Times. Our mission is simple. We seek the truth to help people understand the world, and we are doing this by aiming to build the essential subscription bundle for every English-speaking, curious person who seeks to understand and engage with the world. And The New York Times news and journalism is the most recognizable product we have. We also have other products like games considered like crosswords, spilling bees, and wordle. Also have cooking for amazing recipes, and wire cutter, audio, and athletic as well. So to get into the same page here, here are a couple of distinctions to explain the upcoming sections. When I refer to a Blatrum team that I'm referring to, delivery engineering, that's my team where we build Blatrum tool and operate. When I refer to like a team, that's more of an engineering team that's our amazing engineering teams across the company who are building application and products that I just mentioned earlier in the previous slide. Here's our agenda for today. I'm going to cover a few topics. Why are we building internal developer Blatrum at The New York Times? How did we design our container runtime Blatrum? Why did we choose Selium as a C&I? They want abrasion, Selium setup, and how are we using it? They do challenges, and that's important. And what did we learn so far? So starting with developer journey. When we talk about developers, they do have a journey like customer journey. And here I try to identify the steps that most developer would go through when they start to collect their business requirements until they deliver their application. So as you can see here, there are a few steps that developer would go through. Starts by an idea followed by design and another action that each developer need to go through to deliver their applications. But you can notice here, I call it a few different steps because here where we can help and we can try to like get all the similarities between these unique steps and help the developer with seamless experience. So let me give you an example. Here are a few colors. Think of them as color ballots. And the ask is to mix them together. Imagine how your team to build this. So I'm going to pause here for about 10 seconds. It's going to be awkward while we go through this exercise together. And you can imagine like mixing all of these together in one canvas and how it's going to look from your perspective. That's awkward, I know. Okay, here we go. You can see here are the results. Every team builds their own perspective and their own unique image. So we give them some tools, but they build their own process that they go around. So the goal here is not to limit innovation because we know we have very smart people, but we want to help them to onboard to the process and make the process seamless and easier for them. So the goal here is to help engineering deliver their products to our subscribers. And we designed a platform to do that. And we start by like creating application template of the resources needed for your applications to get deployed. Then we give you like the space where you can innovate and code and like build all of your business logic that you need than our centralized CICD system where like you can build and deploy your application. That's our runtime, which is the focus of my talk today.  So we talked about why we built an internal developer platform. Now I'm going to talk about how we orchestrated our runtime setup. So we experimented with different setups for our cloud across counts. And we found out that multi account architecture is the one that we want to go with. That's for a few reasons. Basically we can group workloads based on like common business purpose in distinct account, avoid dependencies and conflicts. We also can apply different security rules between multiple accounts, development, production, number reduction, sandbox account. We can also limit the scope to impact, achieve resource and dependency and also manage a cost across and have like different methods of cost allocation. So when it comes to build Kubernetes clusters, there's often a dilemma between like two things and most of you may have went through that. First, are we going to use multi singleton cluster so each team get the dependency of like, hey, I can do my own stuff. I don't have to worry about others. There's no like noisy neighbor problem and all of the kind of resources. Or we go with multi tenant clusters which we can handle like more like less of less of operation overhead and like we give them the space to innovate but we can like optimize for cost. So we started to look into the main design consideration for building our Kubernetes clusters. Since we already have isolated accounts by default, we began considering network isolation. It's crucial to ensure that each workload has boundaries from other services. So we don't cross these boundaries unless it's defined. So by default, all communication between services is denied. Another factor we looked into was like role based access control and essential to have proper access. So when a tenant is important to a multi tenant cluster, they don't gain access to resources that they don't own. And last, we talk about resource management. That's important because like if we want to manage a multi tenant cluster, we want to optimize for cost and we want to make sure like everything is running efficiently. So after careful consideration, our design requirement, we came to the conclusion that multi tenant clusters fits our use case. And we recognize this approach will help us achieve our goals and also to support creating our runtime environment. In order to do that, we started to build like Kubernetes cluster across multiple regions for disaster recovery and making sure that we have efficient workloads that can fail over between regions. And it's important to understand that there's no one size fits all. This design worked for us. It may not work for you. You may have to figure out a different design. So we're talking about network isolation, but we're still running in multi tenancy. So we need to ensure that we have the right networking tools in place to ensure that there's no boundaries being crossed here. And that's where we started to look into different CNI options. And Cilium was the one we decided on. First, let's talk about performance. As you can see here, like this is from the CNI benchmark blog on Cilium. And like you can see a few components over there are being removed in favor of Cilium. And that's of course mean like there's performance overhead. Improvement that we can see and using BB filter for routing means we shift traffic filtering to the kernel and that's improved the process. And Cilium promised this originally. And there are different CNI benchmarks that chose this. And you can go and use our original benchmarks on Cilium blogs for more detail comparison between different options. On top of that, one of the other considerations between network isolation, we want to make sure that we have the right things in place. So Cilium network policy is a great extension of the Kubernetes network policy API. It brings such different policies from like L3, L4, and also L7 and support for DNS and also have identity for services. But also we have policy enforcement modes. The default mode is good to for most cases where like everything is allowed by default unless you specify something and then like other things are restricted. And there are more like always mode. I believe that's where like everything is restricted by default. And you can enable this for like more secure environments. But you can also see here like there's cluster wide policy which is streamed up process of you and apply something across all of your applications despite where your application is hosted and it doesn't matter which name is base you are in. Then observability is important. So like you heard about Hubble. So here's like a small diagram from Hubble UI. That's where like you can see the service meshes and network, how it works, traffic routed between different services. But the most important from the Hubble UI is that flow itself.  And that's important. So we can build like any understanding of like how traffic is flowing between services. But also like it comes with a lot of metrics out of the box. So you can see like we have here in points states we have like traffic drop traffic allowed all of the kind of metrics that allow us to observe more for our clusters. So how are we using Cilium? So as any organization we are using Terraform for cluster provisioning. But there are a couple of things. When we start to use Terraform like EKS where we build on top of that comes with CNI itself and comes with Kube proxy. So we need to remove that and then we need to install Cilium as a CNI and then have all the processes. So there are two ways we can do that. We can even like just provision our clusters, do the process manually or we can just script it in some way. So we don't prefer to do it manually. So we created just like a small script that does all of that for us when our cluster is being provisioned. So after the cluster is being provisioned there are a couple of things that we need to go through. So that's a resource we'll like we'll provide some base home values and like a script where like we'll get all of that sorted and we'll install Helium for us and mark all the nodes as ready. To go through the script what we need to do. So first we need to remove that AWS CNI and after we remove that we need to remove the Kube proxy and then install our CNI configuration which like you can see here we have some ENI config and also like some other subnet tags and then there's like some masquerading configurations that we need to install. So you all heard about IB exhaustion and that's where we use the ENI mode for Cilium on top of EKS. So there are a couple things here. First on that side you'll see like we have like 10 IB space where like this is where it gets attached to the Cilium agent and the nodes himself and then like because we are running a multi-account architecture if we kept using the 10 space we're going to run out of IBs which is like going to happen quickly. So we started to provision like other subnets which like 100 space that allow us to communicate and have more buds and because like we are in a shared environment that we need to ensure that like we can scale. So couple things here to unbox this that's another subnet but like you can see couple things here you can see like couple ENIs and you can see multiple IB addresses and brief fixes. So one of the things that there's a limitation when you run EC2 instances that you have limits of the number of ENIs that you can attach to the instance and depending on which type of instance that you can run into that may be problematic for like shared clusters. So one of the things that Cilium supports is IB brief fixes basically like you can get the same number of like ENIs by initiative having like a single IB or a few IBs very ENI you can have multiple brief fixes which allow you to expand I believe 16x the amount of buds that you can have on a node and we run different size of nodes as well. So how the process starts like you get an account the process is completely automated after you get that there's like a CRD will get created we have an in-house operator that will take the CRD and convert it into like Kubernetes resources you start by default namespace roles and all of the things that tie it back to your cloud account that gives you access and then like more important we talk about Cilium network policy so we start by a network policy that only allow traffic back to your accounts so like because we are running in shared clusters we want to ensure isolation so all of our accounts are connected together because we are shared but we still want to ensure that your services only can talk back to your account when you need to extend capabilities but there are other possibilities of using Cilium network write policy where like things like if you are in AWS you are familiar with this IB address which is the instance metadata service so we want to make sure that we block this we also want to make sure that we block other IB range that you are not allowed to talk to unless like we give you access to do that so so far so good everything is great and then you have the column before the storm that's where like our day two operation gets in so everything is running okay and like we are onboarding services and everything is great before like telling you about what happened and how we survived that let me tell you about our ingress model first it's a critical service that we have all of our traffic going through envoy and then goes to upstreams so we just onboarded our critical service into our cluster and then like traffic is coming there but then this happened and then you can see like traffic is still coming but like there's something happening and like we can't understand what's really happening and like we looked through it and like seems like one of the budgets not working for envoy we can't blame envoy for sure but uh we aren't sure so we removed the bot everything works and then like we're going forward but we weren't sure like what exactly is the problem so we started to investigate first is it in void so in void possibly like have misconfiguration and despite is not behaving as expected because once we remove it everything works as expected but then we discovered something that also might be DNS because we started to look into that specific bot and like we figured that it's not getting all of the DNS requested it needs but like it was a weird situation because not every time it's not going to get DNS sometimes yes sometimes no so it seems to be like more of a networking and that's where we decided to look further into psyllium to understand like what's really happening so the bug hunt will start and we'll try to use all of the tools at our disposal to understand where's the problem and psyllium gives us like a says dump tool which allow us to dump everything that happens from logs and configuration into like log file it's huge so don't like try to look into it but it's big uh so we went through like reading and understanding like how things are working there and then we found like a couple things like psyllium node CRDs which is like the ways that psyllium manage resources in in uh kubernetes cluster that's where like it has like uh as you can see here like in i addresses it has like the pool of ibs that it's using and all of that kind of stuff and what we found is there's a missing ib so here what happened like we have an invoid proxy that has an ib in the 100 space but then like we look at the node and this ib is not there so there's something wrong and we're trying to understand like what's really happening there and looks like this was happening so the use case there was the ib gets attached to the bud and then it gets marked as it should be released so in psyllium there's a feature that allow you to release xs ibs so what happened is like a bud get created and ibs being attached and then like psyllium think like this ib should be removed so start to remove the ib and that's where like the bud was already starting successfully passing all the props but then gets in a bad condition because it can't communicate so it goes bad and then like to fix that you just like simply go to the iads ib and everything works but like that's not going to work for us because not every time at 2am we can go and like add this ib back so also like when you look at here you simply can disable the xs ib release but we are talking about ib exhaustion so we want to make sure that we release all of the ibs we don't want to keep them we are running in a shared um ib bowl even in the cluster so we want to make sure that all ibs are released back and like you can i don't know if you can see the logs but like you can see like we're talking about bud creation time and then like ib is getting released after so that was the intermittent fix which like okay let's turn it off for a second let's make sure everything is stable and then move forward but then we went in to look into the code to understand like what's really happening and read through and try to identify where it's about and that's where community helps so looking into the code trying to understand what's really happening uh we found like data dog already contributed to fix to that i think they have a talk today i'm not sure if it's about the same thing but basically there was an edge case here that happens when like bud getting assigned and then getting released so like what's being introduced here and this is from the uh get hub request to fix this issue is like some sort of handshake or like your sign your signs ib when the bud get terminated there's like a handshake between the psyllium asian and the operator to update the cache and make sure that like the ibs gonna get released and it's not gonna be available in the cache to the asian within via signs ib again and that was our hunt so what do we learn so far we learned that ebbf is powerful so we started just by using it as a psyllium as a c and i am networking and have like all of this network policy and looking forward for more other opportunities where ebbf can be helpful observability is very important to us to understand like all of the traffic that we need to get uh from our clusters and platform and at the end like we appreciate the community like we found something wasn't working for us but other people who already contributed to the upstream and solved the problem as we upgraded to the next version of it and i believe that's about all and i'm joined here by my colleague so we're gonna have like other talks from the new york times talking about argo and how we deployed this uh in the new york times and also we're gonna talk about oba later on thursday i believe uh and uh using this for testing and get ops and that's all thank you testing one two yeah hey um marco hemkin here um so uh just quick question about your pod network so you're not using an overlay network for the pods you're using the vpc network for the pods that's correct okay yeah we're using a secondary sider as in the vpc where like a sider get assigned to the node that's the original sider across all of our network cross accounts but we have a special secondary sider with the hundred space with like just assigned to the pods so when traffic gets back outside of the account gets with the 10 space but like inside the clusters it always has a hundred uh i d addresses hello uh for multi-tenant purposes do you plan to use different uh c idr private uh addresses for the pods part because you you seem to use only the 164 uh subnet for the pods and do you plan to split it to get really uh isolated c idr uh ips for your tenants like i'm not sure is that i heard the question can if you don't mind repeating it please yeah for for the private pods uh you you've got a 164 c idr uh address plan the do you split it and isolate it for every tenant you have or you just use the 164 in the for all the pods so if i understand the question correctly do we split the ciders between tenants across in our clusters no we don't so like if you have your own account you get your own cider that's in your vbc but in our runtime environment we have like a big ciders that we use and we just like give you the space in any address that we need to and then like we use like that your name space your identification of service of the candidate to understand like where are you going to and like understand like how we can egress and ingress traffic to your close to your services yeah so does this mean that cilium does natting on individual nodes like normally when you use as your cni or as your network plugins it's like a one-to-one mapping so it is already allocating the ip addresses while the building of the cluster right but when we have uh i think we've met or some other cni i don't remember which does a natting inside so when you see there's a multiple nodes you can spin up it's not dependent on the fixed network like you know it's like a one-to-one mapping kind of stuff so the cilium does the same kind of strategy like natting of the ip addresses so when you want to troubleshoot you will see some different ip address it's not the one the actual ip address right so if i understand the question like you are asking if cilium is natting uh our traffic from buds to other accounts is that is that the question yeah okay so when i mentioned i referred to that earlier so all of our nodes get 10 address space so that's like if traffic is crossing boundaries to a different account that's what the ip address will be showing out so like you get like 10 space ip on the other account but if traffic is aggressing between different buds we will only see the 100 ip address there so cilium is not like natting cross accounts but it's natting around that nodes themselves so we can see traffic flowing between buds on a specific range which is 100 space and then like the 10 space it just like cross account and if for sure if we crossing public internet there will like be like public ip addresses when how it's going to get egressed out okay thank you another question so as you said that you have removed the kube proxy demon set and things while deployment of e-cares so after uh implementing the cluster and it's live and there's some issues on the pot-to-pot connection on any worker nodes right or any problem so that means that aws will not take responsibility to support that it's us or it's like the developer has to find the root cause with this and has to provide the fix or go to the community check with people but that's a time consuming right so does cilium also have a licensing kind of stuff that you get support normally so if i hear correct uh you are asking if once we remove kube proxy and components from like eks like what's rule for aws and cilium in this place so like i'm not sure about like aws but like since we removed things that's already provisioned earlier they wouldn't support that out of out of this box and for cilium we basically support that so we understand the process we understand what's happening and like we need to work out with cilium i know that cilium has other offering as well so i think i've been out of time so thank you all


The shock trauma platoon at Ford Operating Base, Edinburgh is a roll-to-medical facility. They receive the patients from the point of injury, stabilize them, and then get them to a higher level of care. We sort of sit on the outskirts of our area of operation. There are many guys that are injured that they didn't have a place like Bob Eddie for roll-to. He might not have made it back enough. Made up of 30 sailors from Alpha Surgical Company, Second Maintenance Battalion, Second Marine Logistics Group forward, the team has cared for more than 150 patients with injuries ranging from lacerations to gunshot wounds to improvised explosive device blasts. As a roll-to facility, they can do basic medical care which includes IVs and blood work, but they can also perform surgeries for more critically injured patients. You see a lot of patients, but it's the one out of a hundred that comes in that is literally a death store that we do an intervention on that allows them to make it to a higher level of care and stand the art treatment and the best chance possible to hang out. Before patients can leave Bob Edinburgh, the doctors, nurses, and corpsmen must work together to ensure they are comfortable and stable enough to make the flight to a roll-three hospital. The credit to comforting our patients truly goes to the nurses and the corpsmen. Because when we're busy looking at extremities or looking at neck injuries or looking at various essential things, the nurses and the corpsmen are always right there. They're the ones that are providing that constant comfort and care. And my hat is off. Reporting from Forward Operating Base Edinburgh, Afghanistan, I'm Marine Sergeant Rachel Moore.


Hello and welcome to the session. My name is Mansi and I am going to help you with the following question. The question says integrate the following function x square log x. So let us start with the solution to this question. We have to find integral x square log x dx. Now according to the I-late rule we see that log x becomes the first function because this is a logarithmic function and x square becomes the second function because this is an algebraic function and logarithmic function is given preference over algebraic function. We call it I. So I will be equal to integration of second function that is x square dx into log x minus integration of dy dx of log x into integral x square dx. Now this is equal to log of mod x into integral x square dx is x cube by 3 minus integral dy dx of log x is 1 by x integral of x square dx is x cube by 3 and this into dx. Now this can be written as log of mod x into x cube by 3 minus. Now this x gets cancelled with one of the x and the numerator we have x square now. So we have minus 1 by 3 integral x square dx. Now this is equal to log of mod x into x cube by 3 minus 1 by 3 into integral of x square dx is x cube by 3 plus a constant c. This is equal to x cube by 3 into log of mod x minus x cube by 9 plus the constant c. So our answer to this question is x cube by 3 into log of mod x minus x cube by 9 plus c. I hope that you understood the question and enjoyed the session. Have a good day.


A new type of black phosphorus, BP, based metasurface has been developed which can be used to study its reflection responses and sensing capabilities. This metasurface consists of multiple bulges which allow for the coupling of dipole and polypole resonances. By adjusting the size and shape of these bulges, it is possible to tune the reflection spectrum of the metasurface. Additionally, the reflection spectrum of this metasurface can be tuned depending on the polarization of the incoming light. Furthermore, the metasurface can be used to detect changes in the environment due to its high sensitivity. This article was authored by Ning Xiu and Yang, Yixuan Wang, Ling Xiaoli, and others.


Welcome back to the AI for Good Global Summit here in Geneva, but of course at the ITU headquarters. And I'm joined now by Kriti Sharma, who's the Vice President of Artificial Intelligence at the SAGE Foundation. So Kriti, you're one of the younger members who've come and sat here. Why is a young person like yourself here? It's really important for people from all backgrounds, especially young people, to be part of not just consuming AI technology, but building it and creating what our future looks like. And this is why I would love to see even next year even more young people and even potentially a youth track where young people come in and show what they've built. I've had a bit of experience working in AI since I was a teenager really. I used to build my own machines, my own robots, my own computers, and I think it really helped me develop my technical skills, take out all the fear, and now I'm solving big global challenges using AI. What are the challenges that the UN has a lot of sustainable development goals? What's the link with what you're doing? A couple of areas I would like to focus on. One is what AI means towards women and equality, gender-based violence, abuse, harassment, these words that we're just recently starting to hear more about with the Me Too movement. But taking it a step further using AI to solve it, which is really interesting, using data-driven decision-ink to define policymaking and education, which is an area where I focus a lot of my time on. How do we create, make education of today ready for the future of tomorrow? A lot of young people growing up today will be doing jobs that don't exist yet. Indeed, a big area and foundation at yours is trust in AI. Are young people, as I say, used to use social media and big data being used publicly? Is that a concern for young people or not? We talk to them, a lot of them, we do a big research survey. What we learned is one in four young people are interested in AI in the future of AI and the rest of them have quite skeptical views, which we are able to turn around. The top three reasons that they give us about their skepticism towards AI, number one, they think it's not very inclusive because they feel like you've got to be a rocket scientist to be a participant in the AI movement, which I believe is kind of sad. I need to do something about it. Number two, they think it's not creative. It's probably just somebody sitting on a computer writing a quote all day, which again, as you've seen at this event, it's absolutely not the case. And number three, that the schools in the education system is not giving them the skills they need. So to our viewers, you built a robot. What are you going to build next? I'm going to tell you what my first robot did and that would give you some context. It solved a very important problem for the world, which was to fetch Snickers chocolate from the snack bar at 3 p.m. every day. But more importantly, it taught me how machines learn on their own and make autonomous decisions. And now I use a built robot to automate mundane tasks and make humans life more productive and convenient and also solving problems that humans can't. For instance, the project we are doing, building a bot in South Africa for women facing abuse, it's really interesting when they say they find it more convenient to talk to a machine about these very personal matters than to other humans because machines don't judge you, humans do. Kristi Sharma from the Lease Sage Foundation, thanks very much. Thank you. Looking forward to your next project. Thank you.


I have found some awesome finds on Amazon. Very quirky, different things that I wanted to share with you while it's still gift-giving season and while there's still enough time to be able to order these to get to your friends or family. Hello friends! I'm excited to bring you this video because I've found some of the coolest things to be able to give people that I think they will really love. Not just beauty lover things because I do have a couple of those but things that I think that guys will like if you have like grandparents or parents in your life that you don't know what to get them. I think they'll like some of these things. So let's get into what I think and feel are really great oddball things that I would like to have myself. Actually most of these things I ordered so they're mine now or I'm gonna be gifting them as well. So let's get into that video. Let's start off with your beauty lover because I have this lighted mirror that I could not wait to tell you about. Now this is a U-gift if you really like it because this is a great mirror. Now I know a lot of us need to have the the magnification but I could never find a magnified mirror with lights on it and it seems like if I have the lighted effect on my face with a makeup mirror I need that magnification less. So this is a travel mirror. It sits very thin and as you can see right here it has the ports with rechargeable bonus and it has this flap right here that you can stand it up so it can sit on a table if you need it to or you can just use it as a handheld mirror which is what I've done. Of course mine already has some of the lovely fingerprints on it but let me show you that it does light up and then it has three different settings on it. So you have a daylight, a warmer and then just a natural light one and I love this. I think it's great and I am so excited to have it because I do my makeup usually on my bed because if I'm having a bad day but that's usually where I end up doing it anyway so this is perfect for me and I am just really loving that. Another one for maybe not so much a beauty lover but this would be something that I think that like I thought of my mother the minute that I saw this because it's just so unique. So it comes in this round like little hat box and then inside are these beautiful flowers. They're just so pretty and they come in different colors of course. I got the lavender one. Everybody knows how much I love purple but these are soaps so you can take these out individually and you can run them under water and they turn into a soap. Now I don't know it's so pretty. I don't know that I could ever do that. I feel almost like a paper. They don't feel like soapy or anything like that. They don't feel like silk flowers. They just feel like paper type and they have a lovely fragrance on them too but it's not overpowering. So I just thought this was a really cool gift for somebody. My mother loves flowers. Anything flower she loves it. I can't wait to show this to her and she's going to really love that. Okay let's say that you have a quirky friend. Let's say that you have a guy friend. Somebody that you just don't know what to get for them. I feel like this is a great gift. This is a bonsai growing kit and it's a starter kit for the little bonsai trees. So this is something that they're going to have to nurture and they're going you have to think of the person. Are they going to be able to you know sustain this and nurture this? So it comes with the seeds.  So you take the little starter part you put it in there you have the seeds. You do you do all of that and it has the little stakes with which tree this is on it. It comes with instructions and I just thought that this was a really great gift for somebody that you absolutely did not know what to do with. I am going to be doing this myself because I love little bonsai trees but I've never grown any. And then the next thing remember I talked about grandparents. This could be grandparents this could be a couple that you don't know that what to really get them. This is from a company called Nut Haven and they have a ton of sets like this. Different ones with different themes in them. You can get all nuts. You can get some that are like these. Now I haven't opened this because I wanted to wait until today to open it. So this let me just pull the little sheet out of here without pulling out everything. So this is dried fruit and nuts and it's very well packaged. The presentation is beautiful and I think that this is something that just about anybody that likes fruit and nuts would like to get. It's a gift that's really unique. I had never seen dried kiwi before and I just think that this is something that is presented really nicely and the gift itself is something that's really nice. So I like that as well and thought that this would make a great gift for somebody maybe that's older in your life or somebody that you really don't know what to get for them. Now I did put this on my Instagram and I believe that I also put this on my community tab here on YouTube. This is a survival kit. This is Vitord. I don't even know how to say that. Anyway it's a survival kit and I have opened this and I have really been impressed by what's inside of this thing. So there is so many different things that you wouldn't even think about that you needed in a car but I'm going to be putting this in my car and this one I'm going to be buying for my son's cars, my daughter-in-law's car, everybody's car I can think of because I think this is something that we all need to have that just in case of emergency. This is filled to the brim with things that you might need if you were in a situation where you were stranded or you know if there was a time when we had an emergency you know natural disaster or anything like that. You have all kinds of stuff in here. You have a little tiny flashlight. You have the emergency blanket that folds up into nothing. You have a fishing hook and line. This right here is a saw in case you needed to saw through something. There is also a pin in here that acts as a, let me find it, so this is a pin that you can use but on this side is the one that you can use to bus through a window in case of an accident that will bus through a car window if you need it to. This is a compass right here that fits on your wrist like this one right here. This is a knife right here on this side with a very sharp edge but that is a spoon fork and I was super impressed with everything that is in here. There's a regular knife in here like I said there's all kinds of things in here that I would never have thought about that you needed in an emergency. Go over to Amazon check out what's in here and take a peek at it and I'll make sure that I have everything listed and linked below but this is something that's going in every person that I love's car because I think that's great and then this right here this my husband loves and this is he likes that survival kit too but I was surprised that he was like oh my goodness I want that and it is a manicure pedicure nail set so you have everything in here that you could possibly need my husband gets really bad hang nails this little tool right here is really great for just nipping that hang nail without you know pulling and making it bloody there is a big pair of clippers a smaller pair of nail clippers scissors nail file tweezers just about anything you could imagine in here for what you need to do your nails or what you need if you're doing your toenails on your own which is what I want my husband to do so this is a great kit and something that is very easy to travel with too if you want to take it with you on the go the nail kit was the last thing that I had to show you I always bring you beauty things and I thought this was a great way to also bring you the things that I love from Amazon that are just kind of quirky fun things different things instead of just beauty all the time so I had fun finding these things and bringing them together for you in this video I hope that you enjoyed it do any of you have something on Amazon that you have recently discovered I would love to hear about that I'm always looking for things on Amazon I swear I'm such an Amazon addict and I have a package coming from Amazon daily all the time anyway I hope that you did enjoy it please give it a thumbs up and hopefully everybody is doing well I love you very much and I'll catch you all in my next video love you friends goodbye


The task of angels is that they protect by directing They protect by directing Protection through direction One of the ways that the angel of the Lord protected Hagar is by giving her the right directions Not by necessarily giving her a miracle We see exactly same thing how Jesus was protected is when the angel came to Joseph and gave Joseph directions He didn't took Jesus supernaturally from the place where Jesus was and relocated him to Egypt The angel came and gave him direction and Joseph got up during the night He didn't wait for three days to get confirmations. He didn't consult everybody He didn't just simply say maybe I ate too much pizza Joseph got up quickly and he got protected Because he followed directions Wise men were about to send sensitive information to the wrong people and the angel of God came at night and Protected them from getting into the shady business with Herod by giving them direction We see that everywhere in the Bible the way God protects With his angels is by giving you direction Not by just shielding you you can't go to a club expecting God's protection if you're struggling with drinking You can't go to pornographic sites expect not to fall into pornography expecting God's protection We can't put yourselves in a compromising situation and expect God's blessing when we simply disobey his direction It's kind of like Lot being in Sodom and Gomorrah and the angels come and say lot This city is going to burn You know and what we would do in lot shoes we say Lord You want me to get out of this city, but I want you to save this city I want you to bless this city God I want you to bless my mess and God says abandon it Because it's so broken. Sin is so twisted. It can't be blessed What can be blessed is if you abandon it and you follow my way then I can bless it can somebody say man God can protect you if You will be open to his direction The moment you reject his direction when it comes to your single life When God says to value purity He's not trying to take your fun. He's trying to protect you Protect your heart from being heartbroken protect your body your finances protect your future When God is asking you to put him first in your finances.  It's not Geico. It's not all state It's direction That's why when Paul met God his first question to God was what would you have me to do?   You're gonna join them and Paul says yes, sir And that's how he went many people today. They want God's protection rejecting God's direction rejecting God's leading Rejecting because God's direction is painful because God's direction might not make sense must understand every time God leads you Every time God's leads you he's always protecting you You might not get an extra blessing, but you don't know how much mess you will avoid you may say But I got into complicated situation. Listen, you don't know what complicated situation is When you follow God, he protects you. Amen