Now, we've just looked at the left and right distributive properties if we combine two operations. Don't get the left and right distributive mixed up with each other. Now, the way that we structure this course really remember is we're just building the stool set. We're just building definitions, understanding new concepts, and we put them inside of that to review them often so that we know about all of these. When we have our toolbox full, then we're going to start looking at the interesting things. At the moment, I do understand it's a bit difficult. It's all loose concepts. The next concept is actually quite a big one, isomorphism, but we're going to talk about isomorphism, but really we don't have any way to hang this yet. But once we've built our vocabulary, it'll be so much easier to look at the grammar when we learn this language of abstract algebra. Let's work through this next concept called isomorphism. Pack it away in your toolbox and we're going to pull it out later when things become really interesting. Now, I'm going to read to you, if I can, I'm going to put this on the screen as well. Listen very carefully. If we consider an algebraic system, S, that is our algebraic system, if we consider that to be the following, it's a set and all the relations and operations on those sets. So the set, all the relations and operations, that's an algebraic system. Now, if we consider two such algebraic systems, S and T, we consider them isomorphic, isomorphic algebraic systems. If there is a one-to-one correspondence between them, that's number one and number two, any relations and operations defined on the sets are preserved in this correspondence. So read that again. Stop the video here. It's going to continue on for you, but I'm going to write two such systems on the board and we are going to go through these things step by step. Now, let's tackle this as I said, piece by piece. I've got two sets on the board here. I've got my set A sub one and A sub two. A sub one has four elements. A sub two has four elements and they are unordered. In other words, it might have been two, four, one, three, four, three, two, one and the same with these. There's no particular ordering just because it's alphabetical and in order of the natural numbers don't read anything into that. They are unordered. Now, let's just consider these one-to-one correspondence. I'm going to do a mapping of a one-to-one correspondence. Remember, mapping, it's like what we used to say a function of, the f of x equals x squared. So for every x, I now map to an x squared and mapping real numbers to real numbers. That's a function, but now we're calling it mappings because mapping is much more interesting. Now, you can well imagine that there are 24 mappings with a one-to-one correspondence.  But the alpha mapping, it maps one to p in a one-to-one correspondence. So one to p, two to q, three to r and four to s. That'd be one of the 24 valid one-to-one, or mappings at least with one-to-one correspondence. Now, let's bring in a relation on a one and a relation on a two. My relation on a two, it is ordered by, and what I'm going to do with this relation is to order these two unordered sets. So for a two, I'm going to order it by the following relation. R to s, s to p and R to q as well. So it goes to q and s and s goes to p. I've just decided that is the relation and I could call this relation something, you know, that fancy R, two. That's my relation there. On a one, I'm going to do another ordering relation and that is going to be, it's divisible by, so divides. And let's start that ordering. So one, one definitely divides two because one goes into two or two over one. There's no remainder. Two definitely divides four and one also divides three. Three certainly doesn't divide two, two doesn't divide three, three doesn't divide four, four doesn't divide three. Divide means there's no remainder if I do that division. So I've got this ordering here and this is my ordering by this relation, a relation ordering and this is my other relation that gives me order. Now consider my following another one of my two mappings that give me one to one correspondence. So one to R, two to s, three to q and four to p. So you can well see that that is different from this one. Now does this mapping be preserved, the ordering that this relation has brought in? So let's do that. So instead of R, I can put one. Instead of s, I can put two. Instead of p, I can put four. And instead of q, I can put three. I could do it the other way around as well and what do I get? I get exactly that ordering. So this mapping has preserved my ordering. So there's one bit of isomorphism that you understand quite well now. Unordered set, 24 mappings with one to one correspondence. I bring in an ordering relation on them. I take one of my mappings with one to one correspondence and it preserves this ordering. I'm gonna clean the board because I need quite a bit of space. We're going to move on to the next bit of this isomorphism. Good, so here we are with the next part. So again, I'm starting with my two unordered sets, a1 and a sub one and a sub two, or unordered again, remember one, two, three, four, pqrs. And I have these two binary operations on them. These two binary operations and the way these two binary operations are set up is the stable and how do we read the stable? So we would have the following. So two with two, this binary operation with two and two is going to give me four. And on this side, say q and r, q and r, that gives me q and r gives me another q back. I've set up two binary operations, two binary operations, one for each of my set. Remember that I still have this ordering relation. I just want to bring this back. So I have this ordering relation one. This remember was divides one, divides two, two divides four and one divides three.  I've set up these two origins. Something, just let's put that in. So if I have u and v as elements of a sub one and I have x and y elements of a sub two with this one to one correspondence, u to x, v to y, which is what we've done with our one to one correspondence mapping, which was on the board previously, then if I have this relation between these two that would imply this relation between these two depending on this one to one correspondence. That's what we saw. So we've dealt with relations. What we are doing now is we're moving on to binary operations. So we set up these binary operations and imagine now I have the following of one of my 24 mappings, which have a one to one correspondence. So one to r, two to s, three to q, four to p. Now we can well say if we do this to this, if I put r in there, I have r, if I put two in there, I'll have s. If I put four in there, I'll have p and if I put three in there, I'll have q. So this holds this mapping that leads to one, this mapping with one to one correspondence, it is going to preserve my ordering relation. What does that do to my binary operations? So let's say for instance, we take on this side, we take two and three. So two and three, what does that going to give us? Well, two and three, that binary operation gives me a one. And let's look if we map this. Two maps to s and three maps to q and that better give us r. And that better give us r. So we've got s on this side, q on that side and that binary operation gives me r. So what have we got here? We've got an isomorphism between my sets, a sub one and a sub two. What has happened? I've set up an ordering relation. I've got this mapping with one to one correspondence. It preserves not only my ordering relations but it also preserves my binary operations. And if we have those two together, we have an isomorphism between sets a sub one and a sub two, not only the sets there, but what goes with the sets. The relations and their operations. So two algebraic systems. So algebraic system a sub one and a sub two are isomorphic. What I can do with the one implies what happens to the other one. I've preserved my relations and my binary operations. So there you have it, an isomorphism between two algebraic systems. As I say, you've got nowhere to go hanging this knowledge. We haven't looked at groups or anything yet, but put this inside of your toolbox, understand this concept of isomorphisms between two algebraic sets and we're going to see it again in the future.


Volunteers and local firefighters joined the battle against a long-lasting fire that has consumed trees and vegetation in northeastern Bolivia on November 15.According to a volunteer, the fire in the Rourena back area, in Beni, is a month old and has been difficult to extinguish.56% of the destruction of virgin forests occurs in Beni, a Bolivian department in the northeast, where eight of the 19 municipalities are on alert and people are affected as they have lost their homes. Water sources and crops, since the beginning of the year, 3 million hectares have burned in Bolivia.Bolivia is going through a period of extreme drought that is registering historic lows in large bodies of water. This fire that we have been under control for almost a month, we are with this kind of fire, and we have been able to control it for two weeks, and now it has come back to fire, we have seen it completely.


Former Isabaya actor Paul Anstladla speaks on his new role in the local mystery series D.A.M., thrilling new show. In showmax's South African series D.A.M., Paul Anstladla stars as Thumbot, a charismatic biker who comes to a small Eastern Cape town looking for his missing grandfather. The Isabaya actor co-stars with Lee Vivier, who plays Yola Fisher, a woman who also returns to the town to bury her father but finds herself tormented by spirits that may or may not be real. The peer meet at Yola's dad's funeral and quickly find they can trust only each other if they're to survive the threats they face. What Thumbot wants is to find a home, and what home symbolizes are the things that come with it, such as belonging, to love and be loved. He's always been in search of that. Starting from one place to another, he believes the grass is greener on the other side, Paul Anst says. Becoming Thumbot. Every time I build a character, I start by exploring what the character needs more than anything else, the 28-year-old says. I always have something to say and I say it through these characters. Need is something primal we all experience. It doesn't matter what corner of the world you come from, we all know the need to survive, to love, to be respected. All these things resonate with me. What Thumbot needs to do is to stop running. He needs to be still and to listen and understand his identity. Covet shooting. Filming during the pandemic was challenging, Paul Anst says. There was initially the uncertainty of not knowing whether we were going to shoot or not, but once we started filming on location in the eastern Cape there was a nice atmosphere on set in terms of being stuck in that world, because I wasn't traveling to and from the set. We were shooting for months, and I was there until we finished, so being trapped in that world created a feeling of camaraderie in the cast, a feeling of we're all in this together. A former motorcyclist himself, Paul Anst sold his motorbike years ago, but the chance to play a biker was one of the reasons I jumped at this series. My first bike experience was with my grandfather when I was very young. Paul Anst's grandfather died when he was in his early seventies, and the series brought back fond memories. Getting to play this character was like living my childhood again. It was fun. I kind of connected the feeling I experienced when I was a kid with my grandfather with the feelings I was trying to evoke on screen.


First of all, for organizing this workshop and inviting me to speak here. It is always a pleasure to come back to ICTP. So in the last couple of years, I've been working on certain topics with my collaborators. All of them are here. OK, I don't see AD, but OK. So this is related to the work where we want to understand how the supersymmetrical localization, the technique which has been explored so much on the compact spaces, how does this apply in the case of quantum field theory on a non-compact spaces, in particular ADS spaces. So this is the title of my talk, Boundary Conditions and Localization on ADS. So these are the references for this talk. So most of my talk will be based on this second paper, which appeared a few months back. In first paper, we just looked at Churn-Saman's theory on ADS 2 times S1 background. This, the motivation here, was something different. Actually, it was related to some entanglement topic computation in Churn-Saman's theory. And it took us more than one and a half year to write the second paper. So in this paper, we developed some technique to compute the partition functions on ADS spaces, and in particular, incorporating the boundary conditions, especially normalizer boundary conditions. So most of my talk will be focused on this paper, specifically focusing on a simplest example, which is Karel multiplied in three-dimension. And near the end, I will mention some of the results, which we have obtained in this, which is still not a paper, but hopefully it will come soon. And also some results which we obtained in this paper. So in the last 10 years, there has been tremendous development in supersymmetric quantum field theory, in particular, gaze theories. And this is due to, thanks to the idea of supersymmetric localization, supersymmetric localization has provided us a powerful tool to evaluate or compute some of the observables exactly, and thus probe strongly coupled dynamics of a quantum field theory. This has led to various non-trivial checks of several conjecture dualities in several supersymmetric quantum field theory and defined in various dimensions. And also now we can compute various observables in quantum field theories using the technique of localizations. This has also provided highly non-trivial checks. Some of the work has been also applied. Some of these techniques have also been applied in the context of ADCFT. And it has also provided very non-trivial checks for this proposal. But most of this work has been focused mostly on supersymmetric quantum field theory, which is defined on the curved but compact spaces without boundary. Now rigid supersymmetric quantum field theory can also be defined on curved but non-compact spaces. And therefore, one can do the similar kind of computation, as people have done in the compact spaces and explore various kind of relations among supersymmetric quantum field theory, but defined on non-compact spaces. There are certain spaces which we are very much interested in. And these are spaces of this kind, ADS and SM. And these spaces are interesting because these appear in various physical examples, such as computing entropy of extramar black holes or entanglement entropy computation, conform field theory, or even as some surfaces describing Wilson loops in ADS. So therefore, these examples provide us motivation to study localization computation on such spaces. Now localization computation on non-compact space is a bit harder problem compared to compact spaces. And this is because of following two reasons. First, a simpler reason. And that is that now when we consider non-compact spaces, one is to worry about the boundary terms. In particular, the boundary terms which we get while considering the variation, supersymmetric variation of the action. And second, the most important one that now when we perform the path integral, we need to include boundary conditions or specifically, super suitable boundary conditions while doing the path integral computations. Now these boundary conditions, so these boundary conditions are integral part of any computation which we define on a non-compact spaces. In particular, non-compact spaces, boundary conditions defines the problem. If you choose different boundary condition while performing the same kind of computation, it computes different observable. And maybe in different quantum field theory. Now the boundary condition, when you impose these tell us in path integral which modes to integrate over. Typically, the natural boundary condition, one impose when you're considering a path integral or non-compact spaces is that a square integral mode, that is a normative mode with respect to standard L2 norm. And therefore, one would like to see whether one can do supersymmetric localization computation with such a square integral node. Now there are some motivation, the fact that we have been using this normalizable boundary condition with respect to this L2 norm. We have used this in the various other problems. And I mentioned some of them here, especially just a couple of them. And this will be for us as a motivation to use this normalizable boundary condition in the localization computation.  So there's a proposal which is called a quantum integral. So that may define some alternate quantization kind of thing. But here I'm focusing only on normalized boundary condition. Well, this depends on what you're interested in. Because if you impose some other boundary conditions, which makes action finite but which is not a square integral, this may define some different theory, et cetera. So that's what I say. So the boundary condition is important. You need to specify beforehand. So that will define my theory and observable. So that's what I here. So this is the, I define the normalized boundary condition. So this is the standard normalizer I'm using. That's what I'm saying. So you define your own, but I'm using the standard L2 norm, which we use in the, I'm giving some example now, where we use this standard L2 norm to compute some observable. And yeah, so that's what Atis was asking. So that will define some other theory and maybe you compute some different observable. Yeah, that's fine. So that's why I say the boundary condition on non-compact space defines your problem. So that's important to say beforehand. So we are going to use this and see how far we can use it. So this describes a non-compact open space. This will describe me the fall of behavior. And that's enough for me. So one of the problems which we are interested in. Yeah, one can impose a normal boundary condition. But possibly, yeah. Well, I don't know. You have to change the norm. So if I use this standard norm, so I'm not so sure whether that would be normalized or not. OK, so one of the problems which we are interested in, for example, a black hole entropy. So for external black hole, there is the proposal to compute the entropy, which is given a following path integral. So we have a couple of talks describing this function. So I'm not going to describe it here.  For example, in one of this work, there was a non-trivial computation, which is a logarithmic correction to black hole entropy. And this computation was performed with the normalized boundary conditions, the norm which I defined just in the last slide, using the normalized boundary conditions on all the fluctuations. And this led to this coefficient which matched with the microscopic answers whenever this available. There was another interesting work done for the black holes in ADS-4. All these people are here.  And for this, the entropy was computed from the twisted partition function, which are defined on ADS-2 times S1. This is a twisted partition function of ABGM theory. And they extract from this the entropy of this hyperbolic black hole. And in this work, they also impose a normalized boundary conditions. And they match the entropy of the black hole. Now, before going into localization with this L2 norm, normality condition, let me give you some brief, very briefly, what is localization. Since, again, this has been said several times in this workshop. So I'll not describe it here in detail. So just for the completion, so the basic idea of the supersymmetry localization is based on the following fact. That if you have a supersymmetry generator q, which is squared to some bosonic symmetry. And it is a symmetry of the theory in the sense that action is invariant. And the path integral measure is also invariant. Then the expectation will be of any q exact operator is 0. So Yanis, a few days back, talked about the subtleties in the path integral measure. I'm assuming here that the subtleties are not here, not there. If the subtleties present, then one needs to be careful. So using this fact, one immediately see that if you deform the action by any q exact deformation, such that square of q on such a deformation is 0, then the partition function is independent of the coupling constant of that q exact deformation. And therefore, one can evaluate this partition function for any value of t. In particular, for t goes to infinity. And if you have chosen v carefully, such that q v is positive definite, then this relates to the path integral over subspace of configuration. Where this configuration is defined by this equation. And this is the basic idea of localization. Okay, so let me, before going ahead, so we have applied this localization technique on two cases, two non-compact cases. One is ADS2 times S1, another ADS2 times S2. So for most of the analysis, I'll present the ADS2 times S1, but later on near the end, I would say something about ADS2 times S2. Now, so what do you want to do? So we want to have a background, ADS2 times S1, which is, I'll choose this metric here. L is some constant. And we want to put some n equal to supersymmetric theory, quantum field theory, with UNR symmetry on such a background. Now, so now this ADS2, so in order for a background, in order for a supersymmetric field theory, in order to have some super field theory on a background, it is necessary that the background must preserve one or more rigid supercharges. Now, typically we can achieve this by turning on some of the supergravity fields, present in my gravity multiplied. And such that they solve certain killing spinor equations. So this is, I'm using some formulation developed by these people. Now, so I'm not describing, these are some very super fields in the supergravity multiplied. So I'm not describing this very much here. But these are obtained, these are killing spinor equations, which we obtain by setting gravity norm and variation of gravity norm equal to zero. Now, ADS2 times S1 alone doesn't preserve supersymmetry. However, so what we want to do, we want to find out this, such some background value of these fields, such that ADS2 times S1 together with this background, satisfy this killing spinor equation. Also on the killing spinner, give us following killing vector. This will give me some equivalent algebra, which we can use for the localization computation. So these are the solutions. So here, these are very simple solutions. So these are the fields, value of the fields. So actually these are coupled to, this is later arithmetic case field, and this is some, I guess, relative to string currents and central charge current. Anyway, so these are the value of the fields and other components of the gauge field and HR set to zero.  So this killing spinner is independent of time direction or the Euclidean time direction, the S1 direction. But this is anti-periodic Elingon theta. And using this super skilling spinner, one can realize supersymmetric transformation on the vector multiplet as well as current multiplet or charged current multiplet. And the supersymmetric algebra on this multiplet is something like this. So here, K is the same killing vector, which I mentioned here. And this lambda is some gauge transformation parameter, which is constructed out of the vector multiplet fields. And here we have some r-symmetric transformation with parameter L. Okay. Okay, so now we have a background. And we want to consider the partition function of chance-sample theory coupled to a matter in some representation. So for a moment, we'll focus only on fundamental matter, but one can generalize this to any arbitrary representation. So this theory is described by the following action. So here we have a chance-sample theory. And this is describing the matter part. Now in this paper, we looked at the only chance-sample theory. And one of the motivation was that this chance-sample theory is equivalent to, although it is supersymmetric, but it is equivalent to a bosonic chance-sample theory. And the reason is that the other fields like fermions and scalars are just auxiliary fields. Okay. So therefore one can compute the partition function of non-evalient chance-sample theory on ADS2 times S1, just by considering this one and considering and performing supersymmetric localization. But now we want to incorporate these matter fields. So here phi is some complex scalar field and there are different terms here. Like q here denotes the coupling of phi with the gauge field. And here delta is some r charge for the common multiplied. Here r is the Ritchie scalar and et cetera. Rest of the here dot-dot denotes the familiar terms which I'm not mentioning. Okay. So now as I mentioned earlier, so before going ahead to do localization computation, one important thing one has to do is to check that whether the action is supersymmetric or not. That means and where q of S equal to zero or not, okay? So it turns out that when you consider this chance-sample theory compared to matter, the action is supersymmetric up to a boundary terms, which is usually typically the case. In particular, the most important thing here is that the matter action itself is q exact up to boundary terms. So here I have just written down the boundary terms, which appear in the matter action. There is similar boundary term, similar kind of boundary terms for the vector multiplied. Okay. Now, so in order for localization computation, oh, I want to see matter action. Yeah. So this is the action, fermionic theory. What do you mean fermionic theory? Which is just this auxiliary field here. That is very fine. They're not dynamical, but they're very fine. Yeah, here f is auxiliary field in the current multiplier and g is auxiliary field in the vector multiplier. Yeah. Okay, so saying that the boundary terms, which we obtained, now we need to set them equal to zero by some condition. So one way to do it is that by adding especially a boundary term such that the supersymmetric variation of those boundary terms gives this term that is supersymmetric variation of those boundary terms gives exactly this term up to some negative sign. Or the other way to do it is that these boundary terms would go to zero with the boundary conditions. In particular, if you see these terms here, we find that all these terms go to zero with normalizable and smoothness condition on the fields. So with the normalizer boundary condition, I don't need to add a boundary term, but if you use some other boundary condition, then we may need to add a boundary terms. Okay, so let me make some more comment about the boundary conditions, especially the normalizable conditions and it's compatibility with the supersymmetry. So if you look the L2 norm, which I gave you earlier, so that fixes how the fields should fall off near the infinity as r goes to infinity. So r here is ads to coordinate and r goes to infinity as a boundary. So the normalizable condition requires that my fields like both scalar and the fermion should fall faster than exponential minus r over two or at most it should go like e to minus r over two. However, this condition is not consistent supersymmetry. And one way to say is by just looking at supersymmetry transformation. So what we find from supersymmetry transformation is that, okay, so, and this is due to the fact that the killing spinner which we had obtained earlier, which is given here, this killing spinner grows exponentially as r goes to infinity. In particular, it goes like e to the r over two. So therefore, this, since killing spinner grows like e to the r over two, we see that any normalizable fermionic fluctuation would give a bosonic fluctuation, which will not be a normalizable, okay? So thus we see that the supersymmetry, so we see also for the fermionic case also, in the sense bosonic, normalizable bosonic fluctuation will give a non-normalizable fermionic fluctuation. And this seems to be a generic fact for any ad spaces. In fact, if you look higher ad spaces, we have a similar feature of killing spinner, okay? So thus, naively, once it looks like the normalizable, if you want to do localization computation on ad spaces on a supersymmetric quantum filter, then we cannot use a normalizer boundary condition because it is not consistent supersymmetry. And the localization is just based on supersymmetry, okay? However, as I'll show you, the situation is not that bad. Okay, so as we usually do in the localization computation, we start with adding some q-exide deformation. So one such q-exide deformation is like this. So here psi me and psi are some fermionic bilinear constructed out of the fermions in the vector multiplied, okay? And these are the fermions of the carrier multiplied. This is chosen such that the qv, the bosonic part of qv is a manifestly positive definite. Now as a, so if you remember, I told you earlier that the matter action itself is a q-exide, okay? However, as localization argument suggests that the partition amongst the student depend on the choice of qv, okay? So whether we choose qv or whether we compute the partition function from the original as matter action, the partition function student depend on these choices, okay? Now, so the first step is to, in the localization computation is to find localization background and that is very simple. Just look at the, since it is positive definite, look at the minima of that. And this qv is minimized by the following configuration. So all the carrier multiplied fields are set to zero and the scalar and the auxiliary fields in the vector multiplied are given like this. So here alpha is some constant parameter. Let's say in a bilinear theory it is constant parameter but in non-ibilinear it will be some constant Lie algebra valued matrices, okay? So now once you have this localization background, so the next simple thing is to just evaluate the partition function on this background. So what we get here? So we are considering churn-simons coupled to matter. So first part here, exponential part comes from the churn-simons theory. So since carrier multiplied, all of the fields are set to zero, there's no contribution come from the action evaluated on the localization background. All the contribution come from just auxiliary field g and sigma. So that is this contribution. And the next thing we have to do is to evaluate this one of the determinants of the vector and the curl on this localization background. So yeah, okay, sorry. The action is what? This action, finite. Yeah, this is finite. This is finite. There's no word. On ad space here, this is finite. I don't know which other, but yeah, it's finite. Okay, so next we want to evaluate this one look determined for vector and the curl on this localization background. There are several ways to do it, but the way we, so the method which we are going to follow, which I am going to present here is something called Green's function method. And as you will see that in this method we're going to see some interesting features which probably we'll miss when we evaluate probably in some Eigen function method, et cetera. So in this method what we do, instead of evaluating the one look determined explicitly, we compute the variation of the logarithm of one look determinant, okay. So when you consider the variation of the logarithmic, we get the following thing. So here, the variation we do with some parameter. So here the parameter we choose was the localization, same as the localization parameter. So for the present discussion, we can think of this as a billion theory. So alpha is just one parameter, but if it is non-abillion, then one can choose any Cartian Eigen values. So this is a Green's function for the fermionic and bosonic part. And these are the kinetic operator for fermionic and bosonic fields, okay. So this involves, so this involves computation of Green's function for the fermionic fields and the bosonic fields. And then we need to, also we need to compute the variations of this differential operator with respect to this parameter alpha. Now, so just for the presentation, let me just give some basics of this, what is a Green's function and how do we use this to compute the partition function? So the Green's function very basically, so let's suppose we start with the differential operator D whose one loop determinant we are interested into compute, okay. Then the Green's function is actually nothing but the solution of this equation, which is if you solve this differential equation, we obtain the Green's function. So this is a differential operator which doesn't have a zero mode and we are computing the, I mean we are interested in computing this determinant of this D, okay. Now one way to compute this Green's function is as follows, very simply, there's a simple prescription to compute this Green's function. And that is that you start with the D and find the solution of this equation. This is nothing but solution of equation of motion, okay. Now since D doesn't have any zero mode, so there are no global solution. This global solution in the sense that the solution which is smooth as well as perfectly well behaved near the boundary, okay. So there will be no global solution to this equation that is just a statement that there are no zero modes. So let's suppose we have in the problem some second order differential operator, then we'll have a two linear independent solutions, okay. Let's suppose S1 of this, one of the linear combination, I'll call it S1, is smooth near x equal to zero. And the other one, let's say other, some other linear combination because since there are two linear independent solution, so there will be another linear combination. Let's suppose S2 and which is a valid solution that is in the sense satisfy the boundary condition near x goes to infinity, okay. Then the Green's function for this differential operator is simply constructed like this, okay. In our problem, in all these computations, whatever we are doing, all the normalizable and smoothness condition enters through this, okay. So by selecting the solutions which are admissible, which are not admissible, we are incorporating our boundary condition and smoothness condition in our problem. Now this Green's function which we constructed is very, these are some textbook material. So this very simply one can see that these are solutions of this equation, okay. And also it satisfy this user properties that first the Green's function is continuous at x equal to y. And also since we constructed four second or a differential operator, the first derivative of the Green's function is also discontinuous and this discontinuity will fix the constancy. But in the case when we are constructing the Green's function for many case, then usually they are the first differential operator. So in that case the Green's function itself will be discontinuous, okay. So now as I showed here, so the basic strategy to computing this variation of one loop determinant is just to compute this Green's function, follow this prescription, select which solution you want, which solution you want to throw away and just plug in here to compute the variation of one loop determinant, logarithm of one loop determinant. And once you have this result then we integrate with respect to alpha to get the one loop determinant, okay. Okay, so if you have zero mode, then you have to subtract out that zero mode. You have to consider something called generalized Green's function. What, sorry? Here I have not used any kind of regularization. But actually you encounter such a zero mode, let's suppose when you can computing the one loop determinant of the vector multiplied because the scalar fields now have a localization diagram and they will serve as zero mode. So you have to subtract out their contributions. Yeah, so but one can construct this is the Green's function for that problem also. Okay, okay, so first let me tell you the solutions which we use to construct the Green's function. So now let me make some statement. And the statement is that if the boundary conditions are consistent with supersymmetry then the Green's function for Bosonic field and for many fields are related. So let me explain what does it mean. So this slide is a bit generic here but in next few slides I'll be more specific. So let me explain what does it mean. So let's suppose I have X naught, which is my Bosonic field. For example, for chiral it is just simple phi and it satisfies this equation, okay. Then it's for many partner Q of X naught will satisfy the similar kind of equation, okay. This is not a remarkable statement. In fact, this is just a statement of supersymmetry of the action, okay. This just follows from the fact that the action supersymmetry. And the other my for many field because typically we have two sets of for many field one which is super part of X naught and another just X one. So the other for many field X one will be related with this X naught by some operator M. I'll mention just in next slide what is this operator M in this chiral multiplicative case. But typically this M will be constructed out of this DIJ operator. This is in the language of piston but this is nothing but so you have V and you express your V in this form, okay. So my full set of Bosonic fields are like X naught and Q X one and these are full set of for many fields, okay. So this M will be constructed out of this. So what one find here is a very simple fact that is that a solution for X naught which is solving this equation is also actually solution for Q of X naught because it is solving the same differential equation. However, this solution will administer provided it is consistent with their respective boundary condition and smoothness condition, okay. In other words, if I pick up a solution it is all the solution of this equation but may not be admissible because it may not satisfy their boundary condition. However, if the boundary condition and smoothness condition is consistent with supersymmetry then a solution for X naught is also solution for for many system. So let me for example give us X some explicit form. So this D which I have mentioned here which satisfy by this X naught is just in the Kara multiplied case is this one. So here I have a D mu some covariant derivative gauge covariant derivative which is involved which involves all the connection field and it will archimetry gauge field, et cetera. So it will, yeah exactly. And these are my other terms for phi this give like some kind of mass terms. In particular if we interested on localization background so all this G and sigma we have to evaluate we have to put it on the localization background. So this will give some sort of mass term but this will be also function of R because G and sigma are functions of R. R is the radial coordinate of ADS2 and et cetera. And the other thing which I was mentioning here this M which is I said that considered out of this operator is nothing but for Kara multiplied is this one. In particular this operator which we see here is nothing but D10 operator. The D10 operator which typically appears in the index computation. So actually as we will see there is a very interesting link between D10 and the Green's function computation which we still don't understand but yeah there is some, I feel that there is some interesting link. But then so, okay good point. So, okay good. So no, so this I'm computing for the part whose one loop determinant I'm interested in. So let's suppose I'm computing QV the one loop determinant from QV. So the Green's function which I need to compute is for the differential operator which appears in QV. Okay. Yeah, yeah. So, okay. No, actually, okay. So what I have to say that for Kara matter the action itself is also QX act. Okay. So there are two QX act here. So one which we added or deformation you can say QV and also action itself which is also QX act. So we'll compute the partition function for both the cases for action also and QV also. And localization computing idea say that it should be same. Okay. So as I said, they don't have a zero mode here. No, they don't have a zero mode. So that's what I was pointing here that so this equation shouldn't have a zero mode because this is what we are completely different this operator we are completely different we are completely different determinant of this operator. So typically they don't have a zero mode. But however if let's suppose consider vector multiplied scalar which has a localization background they will have a zero mode. So you have to subtract out that zero mode computing. So I'm computing right now for current multiplied. So for the whole professor I will just show the current multiplied and the vector material part I'll just mention the result near the end actually. Okay. Yeah. If there will be normalizable solution there will be also non-normalizable solution and you have to select which one of them you want to keep. Actually I'll show. I'll come to this point actually here in a moment. Okay. So this is the explicit form. Okay. So now coming to your question which you're asking. So now what we are saying that we have to solve the equations. Okay. And find out what are the solutions and based on the smoothness and normality we have to keep some and throw some. Okay. So explicitly solving the equation motion we find following. So since we are dealing on ADS2 times S1 so it has two circle directions. So I'm representing tau and theta. Tau is S1 direction and theta here is a circle direction of ADS2. Okay. So my solution, so my fields in the Fourier mode will be labeled with two quantum number N and P. Okay. So since this N which is the measure which measures the mass of the KK mode coming from this. So this will dictates the follow behavior of the fields near the r goes to infinity because this N determines the mass of the KK mode. On the other hand P associated with the theta which shrinks to zero as r goes to zero. Therefore P will dictates the smoothness behavior of the field near r goes to zero. Okay. So we find the following. So we find that so for a scalar field we have two sets, two solution because we have a second differential operator.  We find that F plus is normalizable for N greater than delta minus 1 over 12 and F minus is normalizable for N less than delta minus 1 over 12. Now as I said earlier in this slide that given a solution one can construct one can construct solution for many field. Okay. So we find that we can construct a normalizable solution for for many field from the F plus only if N greater than delta over 12. On the other hand we can construct the normalizable solution for for many field from F minus that is so B plus and C plus are constructed from F plus and B minus C minus are constructed from F minus. So we can construct the normalizable solution from F minus only for N less than delta over 12. In others what it implies that normalizability is compatible with supersymmetry for N greater than delta over 12 and L less than delta minus 1 over 12. And in particular in this interval if there is any integer the supersymmetry is not compatible or normalizability boundary condition is not compatible with supersymmetry. And this is and here we see especially the breaking of supersymmetry. Okay. So for example, so here I've shown here what normalizability boundary condition imply for various fields here. On the other hand let's suppose just for the presentation if you had to use supersymmetry boundary condition then it will allow the like other fields to blow up as R goes to infinity then the situation would have been different. In this case we find that the one which is normalized for Bosnian field also give the normalizable solution for Fominic field. And the one which is normalized for in this range also give the normalizable solution in this range. Okay. Now there are two points which I'm missing here N equal to delta over 12 and N equal to delta minus 1 over 12. So this is normalized boundary condition. And this is supersymmetry boundary condition. So here, okay. So there can be different supersymmetry boundary condition. So here I'm using the one where the Bosnian field is normalizable and then using supersymmetry transformation to detect what is the boundary condition for fermions. So we get this is the boundary condition. So this is that the C field falls much faster than E to minus 1 over 2 but B field blows up as R goes to infinity. But one could choose other way around also that you require the fermions to be normalizable and then from supersymmetry find out what is the boundary condition of a scalar. That will allow the scalar to blow up as R goes to infinity. Okay, so how much time do we have? Since we have 10 minutes gap anyway, so, okay. So as I said earlier, so what you have to do, you need to compute this quantity. We have a Green's function available because we know solutions about the admissible solution. We know Green's function. We know Green's function and we know we need to compute this one. Just for the explicit purpose. So for color multiplied, we got this one. This is two cross two matrix because fermions are two component and this for a scalar which we get something like this. So now the most important result of our paper. The most important result is that it's falling. If when normalizable boundary condition are consistent with supersymmetry, that is in the range n greater than delta over 12 and less than delta minus 1 over 12, then the variation of one loop determinant is actually total derivative up to an equation of motion.  And this is one of the most important, I mean, this is one of the important result of our paper. On the other hand, if the boundary condition is not consistent supersymmetry, then there's no cancellation, okay. In particular, we find that that now the variation include boundary term up to equation of motion plus a bulk term also. The bulk term, which we cannot write as some total derivative term, okay. And then the complete one loop partition function given by the two terms, we have to add these two term depending on the intervals. And this is a complete one loop, okay. Let me state the result. So we compute the partition function. So this is a computation from the classical action, which is Qx act, okay. So we find the following. The first two terms come from the supersymmetry reason. The reason where the variation becomes a total derivative. So we have already integrated out. We have already done the integration with respect to alpha. So we have just mentioning the result. And then these two terms come from the reason where supersymmetry is not compatible with normalizability, okay. So this parameter, which here, X hat and Y hat are given here, which is a combination of these quantum numbers and R charge, et cetera, okay. On the other hand, if we had done this similar computation from the Qx act deformation, instead of a classical action, but from Qx act deformation, what we find is following. We find that first three terms is exactly similar as this one. But this term, which is a bulk term, this is a bulk term. This term is now become something like this. This A twos, B twos are very complicated quantities. They are functions of this quantum number N and P and delta, et cetera. And so what we find that the part, and this quantity is not equal to, this bulk term is not equal to the bulk term presented here, okay. So what we find here, the result is that the one loop determinant computed from with the normalizer boundary condition is not independent of Qx act deformation. And therefore, if there is an integer N lying between this interval, then we cannot perform localization computation. However, if there is no integer lying in this interval, N greater than delta minus over two L and less than delta over two L, then there's no problem. One can perform localization computation with normalizer boundary condition, okay. So some comments. So as I said, I said that if the boundary condition is consistent with supersymmetry, then the variation is given as a boundary term. And this is one of the important result of our paper. And these boundary terms are independent of Qx act deformation. However, bulk terms are dependent of Qx act deformation. And we have done also some other consistency check, especially in the free column multiplied by evaluating the using some different method. And we found that our result is consistent with other methods, okay. So however, one can ask that, okay, fine. But one can use some other boundary condition in particular supersymmetry boundary condition, okay. So one can find that in the supersymmetry boundary condition. So again, not going to describe here, but in last few days there talks how to compute this. So basically in this case, the computation is reduced to computing some index of some D10 operator. And evaluating this with supersymmetry boundary condition, we find this one, okay. In fact, this result is consistent with our result as long as there is no integer lying in that interval, delta minus 1 over 2L and delta over 2L. So if there are no integers, so our result consistent with this. So this is for the column multiplied part. What about vector multiplied? So we have also done the vector multiplied calculation using the index best calculation, which is given by this. This result is same as S3 remarkably. So the partition function of vector multiple on ADS times S1 is same as S3. However, recently we have gone back again to do this computation, but now with normalizable boundary condition. And what we find that this index result agrees with normalizable boundary condition, only if L square is greater than three over four. So if L square is not greater than three over four, then we have again bulk term. And normalizable boundary condition not consistent supersymmetric. We have been doing this. Okay, no, no, no, okay, see, okay. So this index computation, I'm not using any fixed point formula here. So I'm explicitly computing the Hilbert space of kernel and co-kernel. And this result based on that computation. So this is of course a question that whether fixed point formula, which people have been using, whether applied to this, I'm not using that formula here. I'm explicitly finding the Hilbert space of kernel, co-kernel, and then calculating this. So this, no, here it comes out to a finite dimension. Computation come out to a finite dimension. Okay, so I just, okay, okay. So just this, I have two more slides to finish first. So we have also done this hypermultiple calculation on ADS2 times S2. This has not come out yet, but hopefully it come out in soon. So we have done this computation with equal radius.  So ADS term S2 itself preserves supersymmetry.  So we have a two different background, and we can put theory supersymmetric quantum filter on this background and compute the partition function. So in this case, we have done the computation for hypermultiple using normalizer boundary condition. Again, from supersymmetric transformation, one finds that this normalizer boundary condition is not consistent with supersymmetry. However, in this case, we find something different. In this case, we find that, that when the case gravi-photone background equal to zero, that means just pure ADS2 times S2, the partition of the hypermultiple is given by this. There are no bulk term and the boundary conditions, normalizer boundary conditions consistent with supersymmetry. In fact, this answer is same as S4. On the other hand, if you turn on gravi-photone background, which is for the black hole background, then we find this result. And there are no bulk term again, normalizer boundary condition consistent supersymmetry and this result is exactly the same as the result obtained in this paper. So ADS2 times S2 calculation seems to work well, even with normalizer boundary condition. Okay, so just to summarize, so what we have found that normalizer boundary condition is consistent with supersymmetry, as long as for carrier-multiple case, let's say as long as there's no integer in this interval. Of course, one can, and we can see for, and we can have some delta and L and where there will be no such integer and there's no problem. We have done also computation for vector-multiple and we find that this is consistent with index calculation as long as this, that condition on L is satisfied. And for hypermultiple, we have done this computation, where it's normalizer boundary conditions consistent with supersymmetry, okay? And so in last slide, so we have developed this Green's function method to compute the one loop determinant and incorporate the boundary condition ADS spaces. These boundary conditions, when these boundary conditions are consistent supersymmetry, then we find that the one loop determinant is total derivative and that's the one important result. However, we still don't have a general proof that in general theory, whether that is always true or not. In general, it is very hard, Green's function is very hard to compute for higher spin field and also this difficulty increases space and dimension. But one of the advantages of Green's function method compared to index computation where expressly finding kernel or co-kernel is that when supersymmetry is consistent in normalizability, one just need to evaluate boundary terms and the boundary terms just depend on the boundary behavior of the fields. Okay, so accept that, okay. Thank you.


Meat Boy is back! Today we are going to do my healthy take on classic barbecue short ribs. So this isn't necessarily going to be a purely carnivore recipe, but if you guys are keto or just want to follow a healthy nutrient-dense diet that is free from the negative aspects of our modern food supply, that is also nutrient-dense full of the vitamins, minerals, and fatty acids that our bodies need. This will definitely fit and this is certainly something you can do on a cheat day or even something that you can eat every day, especially something more approachable to give to the kids. So let's take a look at the ingredients. As with any meat-based recipe, the most important ingredient is the meat and the quality of that meat. So today I'm using our first light 100% grass-fed Wagyu short ribs that we have on Frankie's range meat and the reason I like short ribs so much as a cut in general is because it's one of the principally fatty cuts of an animal so even if you're getting grass-fed it will still have a lot of fat. You know normally if you're buying a ribeye from a grass-fed animal it might be very lean whereas certain cuts like the ribs, the brisket, the belly, you know the suet fat, the marrow, certain parts of the animal always have fat. Short ribs is one of them you know go to a local farm butcher. These will probably be pretty cheap, pretty affordable and they really are delicious. So let's take a look at these. Now wagyu doesn't just mean higher marbling. You know wagyu animals specifically have more tender meat and that's very important for cuts like short rib and belly. You know whereas normally a short rib is very tough, you know I don't normally cook these low and slow like I do today. You know sometimes I just throw these on the grill for a few minutes on each side and they come out really really delicious. So if you get these on Frankie's Surrange Meat this is how they will come. This is 12 pounds so each of these short ribs are about 6 pounds and you know I would literally eat this for a week. So you know it's a one-week supply of short ribs. You know you'll get plenty of fat, plenty of nutrition from these. And the marbling is the main difference here. You know you could see you know crazy crazy marbling. You're not going to see that on on regular short ribs. So you know not only do we have more tender meat there is ample intramuscular marbling on these short ribs. So definitely try these out if you guys want but you know any short rib from a local farm, super super delicious. This is just going to be a slight step up. So all I really want to do here is just pat the meat a little bit dry with the ever so valuable paper towels that we can't come by. You don't necessarily have to pat the meat dry. I'll just put a little more salt and pepper on there. What I would do is if you break open this package definitely get a marinade on the other rack of short ribs. You know hey just put some salt and pepper on it let it sit for a couple days. You know develop that flavor maybe some honey but once you guys take a look at the barbecue sauce we're making today you can even put some of that on the second pair of short ribs. So the total cook time for these ribs is going to be between four and five hours. The first few hours are just going to be salt and pepper and we're going to baste it with red wine vinegar. So if you don't want to use barbecue sauce if you want to keep this recipe you know 100% carnivore you can just do that. Salt, pepper, baste with red wine vinegar. Very simple. What we're going to do is get this in the oven with the salt and pepper and then I'll show you guys the barbecue sauce. So very simply you know two parts pepper to one part salt and you want to put the salt on first. So we'll get a thin layer of salt on the ribs and then plenty of black pepper and you know I'm sure some of you guys have heard about Franklin's barbecue which is a very famous Texas barbecue joint. This is his recipe and these wagyu ribs are so delicious I would go as far as to say don't even touch a barbecue sauce but I know how some kids are I know some people are very picky so you know the barbecue sauce is definitely optional. The same with the other side and it gets get a nice thin layer of salt on here. So we've got the salt on and we'll do the same thing pepper on this side. So this is all seasoned let's pop it in the oven. So the temperature is 275 and we're gonna pop this in. You know every 45 minutes to one hour we're going to baste it with red wine vinegar just to keep it moist and the vinegar the acidity is very important for cutting through all of that fat. Especially with a cut like short ribs you definitely want some acidity. So if any of you guys are familiar with a classic barbecue sauce recipe you know they're ketchup based they'll have a lot of brown sugar or regular sugar in there you have some mustard powder paprika some type of vinegar and I took some of those ingredients and I swapped them out for carnivore versions or higher quality versions you know so here I have you know organic ketchup instead of regular ketchup here I have fish sauce instead of Worcestershire sauce you know here we have an organic red wine vinegar you know this is a high quality smoked paprika this is a high quality French mustard instead of mustard powder you know here we have some raw honey instead of brown sugar so you know more expensive much higher quality you know it's gonna have more nutrition be less inflammatory and ultimately tastes better. So the first ingredient is one cup of raw honey and this doesn't really have to be raw because you know it is gonna be cooked in the oven at least some of it is so whatever you know barbecue sauce is gonna be off to the side that you might put on the ribs afterwards I guess if you want to keep the honey in that raw not too significant of a difference though one cup of raw honey we want one and a half cups of our organic ketchup and I've made ketchup before honestly just you know get a bottle of organic stuff and save yourself some time if you're gonna eat ketchup and you really want to make it you can give it a shot but you know this is only a few dollars and a lot less work the only problem with buying really high quality expensive ingredients is like if you leave this much stuff in here this is probably like a dollar worth of honey and ketchup so then we're gonna do half a cup of our red wine vinegar you could use apple cider vinegar you could really use any vinegar I think the classic recipe calls for white vinegar now you want about four tablespoons of mustard that's enough now you want one tablespoon of fish sauce this is actually a Coleture de Elysee I've been trying to get this on Frankie's range meat for you guys it's pretty much anchovy fish sauce the very salty lots of umami flavor really delicious stuff this is substituting the worst of sheer that normally goes in barbecue sauce I want two teaspoons of paprika this two teaspoons of salt two tablespoons of black pepper so that is our barbecue sauce give us a little taste that's pretty good stuff now when I taste this I think of that classic ketchup based barbecue sauce you know there's definitely a lot of room for experimenting but you know this is really close to what you would traditionally expect at any barbecue place just much higher quality barbecue sauce is done ribs are in the oven we're going to base them with red wine vinegar every 45 minutes for about three hours and at the three-hour mark we're gonna start putting the barbecue sauce on the short ribs so I'm actually gonna take half of this barbecue sauce and I'm gonna marinate the other short rib with it you know I think this is you know too much barbecue sauce for one short rib so you know this will preserve this and also give it the flavor you know for next time so it's been about an hour you know the top of this is drying out a little bit so what we want to do red wine vinegar on this to keep it moist just rub that on and normally you'll see them use like a spritz bottle for this but you know this isn't something I make that often so I don't have vinegar in a spritz bottle so we're gonna flip this over to the other side put some vinegar on that side as well just rub it in this is gonna go back in for another hour and then we'll base it again so it's been another hour we're just gonna put some more red wine vinegar on the ribs all right we got both sides so back in the oven another hour another 45 minutes one last base again the main goal is just to make sure that the ribs don't dry out because if you do this without spraying this with water or vinegar these will get very dry on the surface all right so back in the oven maybe like 25 30 more minutes and then we will get the barbecue sauce on it so it'll cook time so far has been about three hours so it's time to put the barbecue sauce on normally you would take the ribs and wrap them in foil with the barbecue sauce surrounding it but since aluminum foil is toxic you know you don't want aluminum in your food or anywhere near it and I don't really have a substitute right now we're just gonna pour the barbecue sauce on top of this and just let it go for another two hours it's gonna be about the same just a little bit on top but the rest on this side very important that the meat side of the ribs is down push that sauce to the to the bottom of the pan surrounding the meat on the bottom and then we'll just put the rest on top there was a little bit of fat and meat juice at the bottom here that I just poured off but it wasn't like a super significant amount so you could you could just leave that in there if you want you know so our ribs are coated nicely you know the bottom of the ribs are sitting in barbecue sauce there's plenty of barbecue sauce on top of here now we're gonna go back in the oven for an hour and a half two hours and we'll be done five hours later the masterpiece is complete so now these have to rest for about half an hour just so they're cool enough to eat don't stir my place I don't have insurance but down the knife Frank all right boys we have our meat all pulled off and I'm Gina Frank sister did you decide on a nickname yet I'm meat boy they weren't really a fan of sausage girl what's a bit better nickname idea I'm meat woman all right that works meat woman you want me to eat that actually off the bone so we have the inside of the rib is nice and juicy and red that actually looks really really really good I'm gonna cut a piece off I'm not gonna eat the whole thing you can have more later if you want this is pretty good I like this a lot it's so good the sauce is perfect unfortunately I only have good looks and not the cooking gene like my brother so you can imagine so if you were gonna serve this you would just like cut all the meat off the bone and it should really just pull off pretty easily and then you should be left with like a clean bone so you take a plate of the chopped up short rib I would just spoon some more of the barbecue sauce on top I would just eat the sauce straight up it's so good she's not really like she's not a big meat eater like she doesn't really like steak but I do like my ribs and I do like pork I like chicken and I'm gonna take a bite right in front of camera here so just to describe this from a culinary perspective for ribs it's super super juicy super super tender you know the temperature in the oven the long cooking time plus it's wagyu with a lot of marbling I think if you use regular short ribs that you wouldn't get nearly as good a result but you guys should definitely try this out regardless if you do have a smoker I would definitely use the smoker instead I just use the oven because I figured most of you guys don't have a smoker and it is raining outside today so good we think you know so if you have kids who don't like meat or you just want to try something different definitely give this recipe a shot as I said earlier you guys can get that wagyu short rib on Frankie's free range meat calm if you guys do want to support me further there's also a bunch of other stuff down in the comments of course guys please like the video subscribe hit that bell icon and share the video if you can guys let us know down in the comments if there are any other recipes you guys would like to see for next Saturday say bye everyone


Hello and welcome to this session in which we would look at an example that deals with treasury stock. Now what is treasury stock? Well treasury stock is when the company buys back its own shares. So simply put, company A buys stock of company A. So when you buy back your own stock, it's the opposite of issuing the shares. When you issue the shares, you sell your common stock to the public.  Now why would you do that? We'll look at an example. So let's take a look at this example to illustrate the concept of treasury stock. Adam Time's Inc. has 50,000 shares outstanding, $10 par value common stock, which were issued for $40 per share. Simply put at some time in the past, Adam Time's Inc. issued 50,000 shares and received $40 per share. Simply put, if you want to look at the cash, the cash for the Time's company was $2 million. We credit common stock for the number of shares, $50,000 times the par value, which is half a million and the remainder was paid in capital, common stock for $1.5 million. So this was the original entry. The company decided to buy back its own shares to send a positive signal to the market. The reason I called it Adam Time's is to illustrate the concept that at some point the New York Times, what they did, that's what they did in 2001, between 2001 and 2003. Between 2001 and 2003, the New York Times were so confident about their company. So what they did, they bought other newspapers and they bought back their own shares. So they had a lot of cash on hand. So what did they do? Well, they bought other newspaper, other physical newspaper across the country and they bought back their own shares. Why? Because they thought the future looks bright for the New York Times and at that time the New York Times did not really care about their online presence. So rather than taking this cash and invest this cash in their online presence, they wasted this cash on buying back their own shares and making lousy investments by buying physical newspaper which the business of physical newspaper is dying unless you are the Wall Street Journal. So to make the long story short, two years later, the New York Times had to file for a bankruptcy because their stock price fell below the price of an actual newspaper. At some point, you could buy the New York Times stock for less than the newspaper itself. The point I'm trying to make is not every time a company buys back its own stock, it's a good move. For example, the New York Times invested in their own company but they bellied up and they had to file for a bankruptcy and they came out of it but the point is it's not always a good choice. Now let's take a look at the journal entry. Before we proceed any further, I have a public announcement about my company, farhatlectures.com. Farhat Accounting Lectures is a supplemental educational tool that's going to help you with your CPA exam preparation as well as your accounting courses. My CPA material is aligned with your CPA review course such as Becker, Roger, Wiley, Gleam, Miles. My accounting courses are aligned with your accounting courses broken down by chapter and topics. My resources consist of lectures, multiple choice questions, true-false questions as well as exercises. Go ahead, start your free trial today, no obligation, no credit card required. We purchased 5,000 shares at $50 so Adam went to the market and bought back basically 10% of their shares. Well, we have to pay $250,000, we credit cash and we debit an account called Treasury Stock, Treasury Stock for $250,000. Now I highlighted 50 in yellow to illustrate this is the cost of Treasury Stock. This is how much we purchased them at $50. Now what type of account is Treasury Stock? Treasury Stock is a Contra Equity. So what happened in this entry, assets went down because we paid cash and equity went down because we added a Contra Equity. So both asset and equity went down. So we debit the Treasury Stock, credit the cash. Now we resold 1,000 of our Treasury Stock at $55,000. So remember we purchased 5,000, now we're going to go out there and resell them. And we resell 20%, which is 1,000 shares of our own Treasury Stock. Well we sold them technically above what we paid. Well when you sell something above its cost, technically you have a gain. We cannot call this a gain, simply put we made $5 in, quote, profit. Now if we take this $5 multiplied by 1,000, technically we have $5,000 in profit. Let's see how we journalize the entry. Let's start with the easy part. We received cash of $55,000, 1,000 times $55,000. So you can always start with cash. The next thing you have to do is to remove the Treasury Stock. You would remove Treasury Stock at cost because we are using the cost method. There's the cost method and there's the power value method. We're not using the power value, we're using the cost method for accounting for Treasury Stock. So we accounted for the cash, we removed the Treasury Stock at cost, the remainder is $5,000. Again, technically it's a profit. It's an improvement in our economic situation. We are $5,000 better off because we sold them at $55,000. What do we have to do? We have to credit an account called Paid in Capital Treasury Stock, which is $5,000. Paid in Capital Treasury Stock is an equity account. And this is what we did. We increased our equity. We assumed that we had no prior balance in Treasury Stock. We assumed this balance was zero. Now we have a $5,000. Let's keep on going. We resold $1,000 at $42,000. Now again, you cannot book this as a gain. Be careful. This is not a gain. You cannot book a gain from your own stock unless you're a company called Enron, which no longer exists because they did stuff like that. So this $5,000 cannot go on your income statement. Now you resold $1,000 shares at $42,000. So again, you resold some of the shares. You remember you purchased $5,000. You resold $1,000. Now you resold another $1,000 at $42,000. Let's start with the entry. Start with cash. Again, cash is easy, $1,000 times $42,000. We received $42,000 in cash. We removed Treasury Stock at cost, $1,000 times $50,000 is $50,000. Now notice here, in this transaction, we are at a loss of $8,000 because we bought them at $50,000, sold them at $42,000. What do we have to do now? Well, here's what we have to do. When we have a loss, first we have to see, first we go in order. We have to see if we have any paid-in-capital Treasury stock, if there's any paid-in-capital Treasury stock can absorb some of the losses we park it there. So of the 8,000, we can see that paid-in-capital Treasury stock can absorb 5, therefore paid-in-capital Treasury stock go down to zero. You cannot have a debit balance and paid-in-capital. So we have the remaining of 3,000. So let's book first the 5,000. So we debit it paid-in-capital 5,000, paid-in-capital Treasury stock 5,000 is done. Now we have a remainder of $3,000 of losses. What do we do when we don't have paid-in-capital Treasury stock? Once again, we don't hit the income statement. We reduce retained earnings. So what we do is we reduce retained earnings directly on the balance sheet, skipping income statement. So yes, we are worse off, but we don't show it as a loss on the income statement. I believe if we show it a loss, it's even more conservative. But again, we don't want to mislead anything for the income statement. We just hit retained earnings and retained earnings will go down. And this is one of the few things that you need to be aware of retained earnings. Remember, net income increases retained earnings, net loss reduces retained earnings, dividend reduced retained earnings. And notice here, Treasury stock losses could reduce retained earnings as well. But the only thing that should increase retained earnings is net income or errors that would result in an increase in net income. So very few things, well, not few, only one thing should increase retained earnings and that's profit. And that's basically an example, an illustration of Treasury stock. What should you do now? Go to farhatlectures.com, work MCQs, true, false, additional exercises. Your CPA exam is worth it. Your accounting education is important. Invest in yourself. Invest in your career. Yes, you're going to be spending time studying. It will pay later. Good luck, study hard and stay safe.


I want to thank NordVPN for sponsoring this video. They felt like a really fitting sponsor given the nature of it. So if you want to show me some support go to nordvpn.com slash polyphonic and use the offer code polyphonic to start protecting your internet today. Rush have never been a band to go along with the norm. In many ways it's a miracle that they've had the mainstream success they have given their sound. They load their music with strange time signatures and complex musical phrasings and they're more than a little fond of 10 minute songs marinated in moral philosophy and science fiction. But this is exactly why people love them. Throughout their career Rush have been proudly anti conformist and anti-authoritarian and this philosophy is clearly reflected in many of their finest works. Let's take a closer look. In 1974 Neil Peart joined Rush taking over for John Rootsy on the drums. In addition to being an excellent drummer Peart also took over as the band's lyricist. Peart was a voracious reader and many of the texts he would devour made their way into his lyrics. Of all the authors he read though few impacted his lyrics like Ayn Rand. Rand was an influential philosopher and author. To dive into her would be an undertaking but a bit of context is necessary. Rand was the creator of the philosophy of objectivism. The idea that the most moral path is to pursue one's own happiness. It's easy to see this influence on Rush in a song like Anthem. That song was the first track of Peart's first album with Rush and it even shares a title with a Rand novella. In that song Rush succinctly sum up the crux of Rand's theory. While Rush have stated that they aren't objectivists the philosophy clearly influences them. In Anthem Peart takes the philosophy and applies it to the art world but the song also looks at the wider scope of things. Getty Lee sings of the impacts that people have had on the world and how they've bettered others lives by creating art that they wanted to create and seeking beauty by their own path. Rush lived by this philosophy within their own art. Their next album was a commercial failure due in large part to strange epic songs full of obscure lyrics. After its release Rush's label asked that they return to a more radio friendly sound for their next album. And their response? Opening their next album with a 20 minute high concept sci-fi epic. By pursuing the art that they felt would fulfill them Rush ended up creating one of the greatest albums ever made 2112. The epic song that drives that album continued to pull from Rand's Anthem. So much so that Peart credited her in the album's liner notes. Rand's novella tells the story of a dystopian future where the concept of individualism has been eradicated. In this world the main character discovers the suppressed secrets of electricity and helps spawn a revolution to bring back the concept of individualism and the self. Similarly 2112 is set in a dystopian future in the city of Megadon ruled by the temple of seerings. We open on an instrumental overture followed by a single biblical passage referring to those who are pure of heart and artistic. In the next section we are introduced to the priests that rule Megadon. They're an authoritarian theocracy built around the worship of technology claiming their interests are for the common good. By keeping technology away from the people however they've consolidated their power and rule the world a hypocritical inversion of their collectivist propaganda. The next section sees our hero discover a guitar in an abandoned cave. After learning to play the instrument he takes music to the priests and shows it to them. The priests realize the power that emotion and self expression can have on their subjects so they ban the hero from showing his music to the masses. Despondent the hero leaves and in the next section he's greeted with a prophetic dream of the elder race of men who left this world many years ago to create their own society built on hard work and beauty.  As our hero dies we get the ambiguous ending where a new voice comes in announcing that somebody has assumed control presumably the elder race coming back to claim their home. 2112 is a brilliant piece that shows why unfettered creativity is important and it paints a bleak picture of a world without it. Rush believed that art in its purest forms can truly save the world and this mentality didn't stop at 2112. 1980s permanent waves dives into the importance of individual freedoms with free will. The chorus of that song is a celebration of free will discussing the importance of being able to choose your own path and make your own beauty. Permanent waves also featured the spirit of radio which warned of what happens when corporations interfere with an artist's individuality. A year after permanent waves Rush released Moving Pictures, a milestone album in their career. That album featured more experimentation and was bookended by a pair of songs celebrating their individualism. The album's final track, Vital Signs talks about the importance of breaking free from conformity. The opener of Moving Pictures is perhaps Rush's best known song and it depicts another character who falls in line with Rush's philosophy on individualism, today's Tom Sawyer. Pierre wrote this song alongside Pie Dubois building on Dubois' lyrics that depicted a free-spirited modern rebel. The second verse of that song lays out the titular character's convictions. The chorus portrays today's Tom Sawyer as a mythic figure, an archetype through which you can see society. He is a reflection of the society he defies. By looking at all that he rebels against, you can see the ills of society. When you look at Rush's work, it's no surprise that they could never conform to the norm. They're just not made that way. And they were all the better for it. The reason Rush were able to achieve such greatness was because they knew what they wanted and they were able to pursue their own beauty fearlessly. And by creating the art that they wanted to create, Rush found true beauty that helps us appreciate our own lives and the world around us. I want to thank NordVPN for sponsoring this video. Go to nordvpn.com slash polyphonic and use the offer code polyphonic to save 77%. Individual rights are one of the most common themes in Rush's music and in today's day and age there are few rights as important as privacy. Internet privacy and security have become one of the defining issues of our day from data throttling to net neutrality to companies selling or leaking your data, it's becoming clear that we need to be vigilant about our internet security. And NordVPN will help you with that. Their virtual private networks allow you to connect to the internet remotely and hide your actions online. I've personally started using them and between their mobile app and chrome browser extension they've made me feel completely safe in my browsing. The experience is seamless but you know they're working in the background and they help prevent people like the temple of syrinx from finding out what I'm doing with my internet. So if you head to nordvpn.com slash polyphonic and use the offer code polyphonic you'll save 77% on your subscription. That means for a little more than a cup of coffee a month you'll be able to make sure that nobody is messing with your data.


Hi, this is Dr. Don, and I'm back with part three of this series on Excel pivot tables. In the last video, we cleaned up our data and created this blank pivot table. Recall in the first video in part one, I said you could use a pivot table to select records for analysis, such as you need to do when you run hypothesis tests. But here we're going to build a report. So we begin by identifying the variables or fields we want in the report. This report is about the average salary in each of the offices for four specific departments. And we need to break down the salaries by gender, but we also need to be able to select only full-time employees to include in the report. Pivot tables make this relatively easy because you can just drag and drop and see what the report looks like as you go. And you not have to worry about messing up your original data. Let's start with the offices. I'm going to go up here in the pivot table fields area. I'm going to find offices, which is cities actually, highlight that with a left click, hold down the left mouse button, and drag that field, that variable down into the rows area. And now I look over here in my report area and you can see that's headed in the right direction. I have each of the offices listed there in a column. So now I want to build out the rest of my report, and I want to get salaries in here. Well, how do I get salaries in the body of my table? I'm going to go over here and I'm going to find the salary variable again, click it to select it, hold down the left mouse button. I'm going to drag that into the columns and let go. And whoops, that's not going to work. It's showing each of the salaries for these 9,000 employees across the top. So let's try something else. We're going to take that variable and we're going to drag it down into the summation values area and let it drop there. And now we see that we have the sum of the current annual salaries. When you're building a pivot table, Excel will look at your variable and determine whether it's a categorical, a label, a name, or like gender, or a quantitative variable. If it's quantitative, it will try to perform an operation and it defaults to the sum. Now we change that by clicking on the value in the table to select it and then right click, open up this dialogue and look for value field settings. And you can see that there's the default, which is the sum. We want average. Now there are a number of things you can use this value field settings for. I'll show you another one in just a second. Right now I'm going to click average and click okay. And now we've got the average salaries in these offices. But that doesn't look like dollars. That looks like a lot of decimals there. So I want to format that. Now when you're working with pivot tables, you don't want to use the formatting up here in the number area because every time you update your pivot table, you'll lose that formatting. What you need to do is again, click to select a cell, right click, bring up the value field settings. And now I'm going to go down to number format. And I want to click currency to get this into American dollars and click okay and then click okay again. And now my averages are in dollars, which is what I wanted. Next I want to show the genders, male and female. So I'm going to go over here into gender and I've got a choice there. I can put it in as a row and have it shown that way. But I really like having it shown the other way so I can put my departments in the column. So I'm going to go over here, I'm sorry, I already rose into the column and put gender there. And now I've got male and female gender here in this particular table. And very briefly I want to point out to you something that you should be aware of. When you're building a report, you can use filters to filter your data. For example, if I wanted to filter the offices, I could drag these cities up here into the filters. But that takes away the cities down here in my table. You can't have a variable both as a filter and as a row, as a filter, as in a column. So you have another option fortunately. If you look here in the table, you can see column labels, if I click that little funnel, it will show me that I've actually got three types of the gender variable. I've got selected for the male and female. There's another there. I'm going to click that blank and open that. And it shows us that we've got some records that don't have a gender indicated. And that's normal. When you have a lot of records, in this case again over 9,000, it's not unusual for whoever key is a daddy in to forget the key a cell. So we're going to leave the ones that don't have an indicated gender out of our analysis. But you should be aware that I'm going to go back up here to my column labels. And I'm going to uncheck blank, just as I just showed the male and female salaries. Now I need to include the counts. Now how do I get the counts in this table? The way I like to do it, I'm going to use something that is unique. Now the names you would think would be unique, but there may be two employees with the exact same name, but there won't be employees with the same employee ID. So I'm going to select that and drag it down into my values field area. Now you can see that it's given me a column called the sum of employee ID. Now the reason it's labeled that as sum is because the ID is a number and it's trying to count those. I'm sorry, to add those up. We're going to change that. I'm going to click there, bring up the value till settings, and I'm going to change it to count and click OK. Now I've got the headcounts for each office for males and females. You can see we've got the 9,200 plus records minus the 30 or so that are blank of gender. We need departments. So I'm going to go back up here and click on department and drag it down into my table under the city. And now we've got the cities and the departments organized that way. And you can play it around with that. I'm going to drag city below department. Now you can see we've got an arrange by departments by city. I prefer the first way, so I'm going to reverse that again to put city above that. We need to filter out those offices that we don't want, but how do we do that? Remember, we can't have the variable both in a row or column and a filter area. So what we need to do is recognize, again, that we've got these row labels here. And I'm going to click in this dropdown, giving the default of the offices. I'm going to click cancel that. But if I click down here in a department and then go to the row labels, now it's got all of the departments there and I can filter out the ones I want. I'm going to deselect the select all. And I'm going to go down here. I need administration. I need controller. I need finance. And I need finance, HR. I've got the four and click OK. Now I've just got those four departments that I need. And now I've got pretty much everything I need for my table to present to my boss. Except remember, she wanted to only have the full-time employees. Now when I get that, I'm going to use the filter this time. That is the assignment category. I'm going to drag it down here as a filter. And you can see it's got all. If I click the dropdown, there's regular and part time. She wanted regular. I'm going to select that so it will filter out the records. And you can see that we've got 1,917 people in these offices in the departments that we're interested in. We've got the salaries across here and the headcounts as well. And that's pretty much the report we need. All we've got to do now is to format it to make it look the way we want. And I'll show you that in the next video.


total success. Hello, welcome to Do Try This at Home, brought to you by the Institute of Physics. We're making these videos because we want parents and carers all the way around the UK and Ireland to be able to get their kids really excited and curious about the world around them, even if you can't leave the house. My name is Imogen and you might be thinking, Imogen, what are you doing in your bathroom? The reason is this is not my bathroom. I'm staying with my in-laws at the moment and this trick has got the potential to go a bit wrong. So I'm presenting from somewhere nice and waterproof. This trick is called the waterproof hanky and I love it. The first thing you're going to need is a glass. The next thing you're going to need is a hanky. Now I don't actually have a hanky so I am using one of these cleaning cloths. Don't use a paper hanky or a kitchen roll, it will not work. I speak from experience. The next thing you're going to need is a plate and finally you're going to need a source of water. The first step is to prove to your family that your hanky is not waterproof at all. So what you want to do is just drape your hanky nice and loosely over the top of your glass and then pour water through it.  The next step is to show to your family that the hanky can be made waterproof. So what you need to do is now stretch it over the mouth of your glass. If you're doing this trick with your family you might want to get slightly smaller glasses so they can get a really good kind of grip on the material around the outside. Then put your plate on top and you're going to aim to flip your glass hanky plate combination in one really quick motion so that it's completely upside down. You should be able to take it away and the water does not leak out. Once you've practiced this a few times bring it to the help of a volunteer. Sam can I borrow you for a second? What are we doing? Take your glass and your plate put it over the top and remembering to keep that cloth pulled nice and tight flip the whole thing over your totally willing volunteer's head and then take away the plate. What's going on here and how can you explain it to your family? If you look at your hanky really closely there are loads of teeny tiny holes in the cloth. This lets the water travel through it and air go through it as well. Now get your hanky nice and wet. Can you go through it? It's actually much much harder and that is because we have water blocking up all of those tiny tiny holes so for the air to go through it's got to push the water out of the way and that's actually pretty difficult. How can water block the holes? I hear you crying. Have a look at this picture. The surface of the water looks like it's covered in a stretchy skin almost like the surface of a balloon. This effect is called surface tension and that surface tension is strong enough to be able to hold up my pond skater so it doesn't sink. It's also a really cool thing that you can try out with your family with this paperclip. With a glass of water that's completely full the surface tension over the top is strong enough to be able to hold up something that should sink like a paperclip and as soon as we break that surface tension my paperclip sinks to the bottom. Surface tension is a bit like a balloon stretched over a hole.  Now with a glass of water without a hanky when I flip it over the surface tension is nowhere near strong enough. It breaks easily across the large mouth of the glass and the water pours out. When I add the hanky water sucks into the hanky and blocks the holes in my cloth.  If you're not totally sure about the science don't worry. Investigate with your family maybe look it up with your family if you need to and have fun along the way. Have a lovely day and see you next time.


This study compared the performance of four different machine learning algorithms when applied to sensor data collected during walking. The results showed that two algorithms, random forest and convolutional neural networks, provided better predictions than the other two algorithms. These algorithms also had a lower computational cost and could be used to accurately predict the target variables. This article was authored by Shema Mohamedi Mogadam, Ted Young, and Julie Quarson.