Ok, thank you, Fabrís, and thank you, the organizers, for the invitation to participate in the program and the workshop. So the talk will be mainly some review of non-results on the self-similar solutions of geometric flow. But this review is motivated by some recent numerical simulations we have done about the flow. And we have obtained some kind of what I think are nice pictures. And I will put the pictures at the end of the talk, ok? So the flow is the following. Here X is a vector in R3. And S as you will see in a minute is arc length. T is time, and this can be either in the continuous case or the periodic case. So for a fixed T what I have is the map from S to capital X of S, T gives me a curve in R3. So what I am giving is a law of motion of these curves. And I should better start saying what is the geometrical meaning. So in order to do that I have to define the tangent vector X, tangent vector T which is the derivative with respect to S of capital X. So then if I differentiate here, then when the derivative falls with respect to S, when the derivative falls here I get zero. And then the only term that is remaining is this one, which is called the Schrodinger map. In this case and to the case where I am interested is on to the sphere in R3. Because of course if I compute this which is the same as the derivative of the length square of the tangent vector. This is immediately zero. So if I start assuming that T has unit length then, and you have a good solution, regular whatever, then the arc length parameterization will be preserved along the flow. So then T can be visualized as a map from say whatever R or the torus into the unit sphere. And then of course because T is the tangent vector to the unit sphere and we have this cross product. Then this cross product has two implications. The first is that you are projecting the second derivative with respect to the tangent plane and the second one is that you are rotating. We are making a rotation of pi over 2 in this tangent plane. So of course you can rewrite this as where capital DS means the covariant to the derivative and J is just rotation in the tangent plane to the given point on the sphere. So this is the reason why this is called the Schrodinger map. So this is the first connection with say dispersive PDs at the geometric level. Then now from that equation is if I use, let me put it here, if I use the frenet frame which is T s equals the curvature times the normal. Sometimes people like to call the curvature kappa but I guess this is for plane curves. One of the papers we wrote we have to change everything because the referee decided that the curvature has to be called kappa. So this is the torsion and then B s equals minus the torsion times N. So the derivative of the tangent is the curvature times the normal.  And now the reason is this, that of course this is T cross product the derivative of x which is T s. So this is nothing but the curvature times the binormal. So the flow is telling me that if I have a point in the curve then this point is moving in the direction of the binormal with the velocity or the speed that it is given by the curvature. And as we will see in a minute there is no ambiguity assuming that the curvature is zero because that would mean that the velocity is zero in that case. So this is, as many people in the audience know this is a model, an old model that was obtained long time ago by the Rios as say an approximation of the evolution of vortex filaments where vortex filaments I guess there is no, you are in a equation quite likely you have a precise definition but if you want to see real fluids this definition is not clear but because you will have to assume always some kind of viscosity and then it will never be a filament. You can start with something which is a filament but any time afterwards it will never be a filament. Anyway, I did yesterday show the video where what is a vortex ring you remember that you have the cardboard and you have a hole and you use it as a smoke cannon and then you remember that you have these vortex rings of course what you saw in that experiment is that it is not really a filament, it is a tube but you can try to visualize say the vorticity in a better way by using bubbles instead of smoke and there are pictures where really when you look at the bubbles they look very much as a filament even for water. Anyway, so this is, so in a sense the story goes more or less as follows you have the velocity, you have the vorticity which is the curl of the velocity and you are saying that this vorticity is a vector but what it is supported is a singular measure which is supported in the filament. Then you want to construct, you want to compute the velocity through the Biot-Savart law, precisely at the filament and this is a singular integral and there is no original way of doing it and at some point you even have some log of epsilon where the log of epsilon was precisely the same that Didier had yesterday and then you kill it, you say a renormalized time and I obtain a relevant term, the first term of the Taylor expansion and this is the curvature times the binormal. So the reason I'm saying all this is because it's very very very unclear that this model at the end of the day is related to real fluids definitely will never be related at the quantitative level, right? A mother thinks this is a local model while Biot-Savart is a singular integral so it should never be a local model and there is the possibility that at the qualitative level maybe you can learn something about real fluids by studying this particular model in any case I think that from the mathematical point of view the system of PDEs that this geometric flow implies I think it is interesting and I hope I will support my statement with the pictures at the end of the talk, ok? So why do I think that maybe at the qualitative level maybe you are saying something reasonable is because there are some particular solutions of course which are straight lines of the binormal flow that are definitely guys that you recognize in real fluids that should be the tap, the vortex tap I mean the usual that you see in the bathroom the other one of course are the vortex rings and there is also the helical helices that are both solutions in the model and the oiler at least for anelics you can have elix configurations that probably they are not really helices but at least the shape is very much that one All right, so what was, as I said the motivation was this numerical experiment and the reason of doing something numerical and not just proving theorems is of course you, at least in my case I try to do numerics when I am not able to prove theorems and the reason why I cannot prove theorems here is because recently with the Laoth we look at the case where say the starting curve is a regular polygon, say like a equilateral triangle or a square and of course if you, when we submitted the paper we receive like two months later a very furious referee saying that we didn't know what we were talking about that this has nothing to do with anything that definitely nothing to do with fluids because the equation doesn't make any sense with this kind of singularity, right? Well, it is very unclear that you can or it is not clear at all what is the definition of the curvature times the binormal in a corner, right? So this is one of the difficulties but from the point of view this is the relevant thing I mean the sense that if I want to understand the singular dynamics this is the guy because he has the right scaling All right, so then it turns out you could ask can I make any experiment like the one that did I show you yesterday and of course you can be very naive and say well in order to create a smoke ring the only thing you have to do is to put a circle then you can say well, what happens if I put a equilateral triangle, right? or a square and of course this has done for many people you can go to Google and you will find many videos of that and if you do it I also did it with my daughter by the way is well you don't see anything at all I mean if you have your box here you should remember that the vortex ring was a very coherent structure that lasted for quite some time so if you do that with a shape like this you put the smoke inside and you just push then eventually what you will see is a smoke ring but if what we were saying could happen should have any possibility of happening it was clear that you were not be able to to see it with your eye right? so then you go to the video and instead of looking say far away you look close to the hole and then you see different things that are happening ok in particular something which is called axis switching phenomena is clearly seen so for example if I start with this structure then at some point I see something like this of course corners are not corners any longer because viscosity will kill everything and in the case of the square you will see something like this and this appearing and disappearing in the sense that for example if you go to the video and you just see one frame and one frame probably is in this position and the next one is in the opposite so then you go to Google and you discover that this is called axis switching and the keywords are non-circular jets and this is my definition of non-circular jets for example ok but then you will go to the references and this was studied with quite big detail in the late 80s and all through the 90s and it was mainly done for I guess experimentalists ok and there is plenty of data from experiments and also from real experiments and numerical experiments so the other phenomenon is that you have multiples of your symmetry of the starting symmetry appears so in any case the dynamic seems to be much more complex that one of the circle appears so that means that for example if I start with a triangle like this at some point I will see an hexagon and of course corners are no corners so you cannot go much farther because the next thing should be something with nine sides and then nine sides with corners which are no corners then it's very hard to see but for example so there are probably this is a reference 95-96 you will find a picture like this starting with a square and they really see like if you start with a four symmetry they claim they will see say coherent structures whatever this means with eight sides well if you use this model and you run the numerics and in fact if you do some analysis you explain these two things by the Talbot effect which is a phenomenon well known in optics so it's a wave phenomenon and in fact we claim this is a really non-linear Talbot effect meaning that it is not something that you can obtain by looking at the Talbot effect of linear optics and then from that one to the view something for this equation anyway so we kept doing some numerics and then we discovered that at time so if this is at time zero then we discovered at time zero plus if I look at the evolution of this corner of this one you don't see the influence of the other ones so the dynamics close to the corner and for infinitesimal times is dictated by the case where you only have one corner then what is the case where you only have one corner well this is something that lives so I would like to obtain something of this type to look at a configuration like this one where a plus and a minus are two unit vectors so you are doing something like this this is a minus and this is a plus and of course this is something that lives in the full space in the real line and it has a very precise scaling invariance so immediately what I am saying is that what the numerics suggest and you can quantify it in a very precise way there is a very clear geometric thing that is happening in the evolution of one corner some angle appears and you measure the angle and the angle is the same, that's what we are meaning so what I say is that at time zero this corner is like if I just look at this so you can really see the regular polygon as that you don't have just one filament that you have infinitely many one for each corner but of course where here you have to understand the close curve as a periodic one so you have infinitely many corners so then we decided to look more carefully about what we could say about so similar solutions and in particular I will review something which is a kind of hole that appears at the end of 2015 which is one is the continuation so this is one of the things I want to address continuation after the singularity has been developed and here the singularity means corner this is my first question the second question is about conservation laws and as Monika said this morning there was this program in Berkeley last fall where weak turbulence is in at least it's on fashion and it is very, at least it's not clear to me at all what is the meaning of weak turbulence but for example talking to different people more from the stochastic point of view you always try to look for some conservation law that it is not preserved you don't have good boundary conditions in this case the one I guess is sometimes is called the impulse or the linear momentum and this is one of the conservation laws that this model keeps coming from say Euler and Navier stocks in fact is the one very much related to the vortex rig is the momentum the vortex transports so that if you can use your smoke cannon to turn the candle off you see, that's the kind of thing and the third issue is can I say something about well let me put it in a function a transfer of energy this is of course a very strong words let me just probably to prove something like this lack of continuity of some no so these are the three issues I want to look at so let me start from the last one because it is not clear at all what is the good function space for the kind of singular situation I want to study so let's start from the beginning by saying that to use the frame is probably not a very good idea because you are using polar coordinates and then you are creating some singularity where it is not so it's much better to to use the so called parallel frame and then you have a system of normal vectors capital T of course which is the tangent and then E1 and E2 which are bases in the normal plane so that now the safe net equations become these other equations and now say the scalars which you need to that keeps the geometric information are given by a complex number which is xiveta e1 e2 e1 e2 e1 e1 e2 e1 e2 e2 e1 e2 e1 obtén o equáxio para o, obtén o 1D, cubic e ls e é concentrante porque o meu target é o sfeir e é o que eu interessa. Então, é unha half de psi squared minus a of t psi, onde a of t é real e é due para o variante gaucho, o problema. Eu sempre tenho o freio para usar as coordenades na equáxio para nada. Então, é unha sfeira interagora que tem infinitivamente má conservación. E a oftén é o l2, ao menos do positivo. Então, se eu fico con algo no l2, isto é sfeir, mas, afortunadamente, se eu quero considerar este tipo de situação, o curvo, nesta caso, é a funcione delta. Então, o que eu interessa, significa algo como seno x delta, que, claro, não é no l2. Então, o que eu empurro é a questão, ao menos a questão natural de, o below l2, e isto é algo que, então, você quer ver o positivo, dizer, para qualquer espaço below l2. E isto é algo que eu acho que o primeiro resultado en esta dirección era o que eu provei con Ana Vargas x Antainévao. E é que o espaço de funcione é muito melhor entendido, en termos do transfero de fúria da condición inicial. E você realmente quer ser uma variante de translación, porque quer o invariante de Galileia para ser preservada. Então, quer o espaço de fúria, algo que define o espaço de fúria, que é a variante de translación. Você também pode, mas, então, você significa que você não pode fazer algo, mas, então, você remaining quite far, eu acho que, se eu não estou errando, nós vamos para o quarta-quarta, no maior do que o quarta-quarta. Então, você pode continuar trabalhando nesta dirección. Então, assuming things like this belongs to some probably weight of the spaces and then Grunrock extended all the way up to say with some weights p smaller than plus infinity and the real good one, this is my first fúria space would be that I would like to consider data such that the fúria transform is in L infinity. This is for the say the curvature and the torsion and if you ask yourself in geometric terms what is the analogous of this would be to say that the derivative of the tangent, which means curvature times the normal to be so remember this is the same as curvature times the normal of you prefer you use the other notation will be this guy in an infinity where n is n1 e2. So this is for me today the natural space ok, I'm not going to enter if you think this is reasonable or not but this is it ok so what so this is about functional spaces so the next thing I have to to recall was out so similar solutions so they do some elemental ODE exercises and try to understand the similar solutions and of course I could try to do it here but it is not you can do it but it's much better to work directly in the binormal flow and then they write ansatz is something like this and you are looking for the capital G and T here is positive look that this this equation as Euler has more than one possible scaling but if you really want to preserve the arc length that in fact you want this is the only possible one and of course it's the one that also comes when you you start saying navier stocks and you pass to the limit in the viscosity so now I want to find a solution of that binormal flow problem under these settings so you do the usual things and you end up with this ODE and if you differentiate again you will obtain and I keep calling capital T the corresponding derivative then immediately you obtain minus S over 2T prime equals S over 2T prime equals T cross product T second prime and then you use in this case the net frame and you immediately see that then the curvature has to be a constant and the torsion has to be S over 2 ok so this is very easy and then you have a nice real analytic curve in in 3D and of course the starting question is in USPECT when the limit went to 0 plus of this guy is this guy this is the question that means that you really have to understand say for S good for S strictly positive you have to understand the asymptotics of G at infinity and for example if the asymptotics of this curve at infinity is given by a straight line this is very easy because if you just do this derivative this gives me G prime S minus minus G over S square but remember that this is the curvature times the binormal the curvature is constant so precisely this is minus 2 times the curvature times V divided by S square and if you are in the elliptic case meaning the sphere then the binormal has Euclidean distance 1 if you are in the hyperbolic plane you have to work a little bit more but not that much so that means that the derivative is integrable so that of course U of S has a limit when S goes to plus minus infinity although of course you don't know if these two vectors are the same or not so in order to have a corner you need to prove that these two vectors are not the same ok so in fact you can do much more this is something I did long, long time ago with Gutierrez Antigas is that in fact I always get confused about the angles but so this is a minus this is a plus and I think this is the angle I have to consider you can even obtain a very close formula you say that cosine of theta over 2 is e to the minus pi C squared over 2 so in fact this gives you uniqueness existence for all angles except when the two vectors are one the opposite of the other one and also gives you uniqueness in fact you can work a little bit more and with tricks similar to that one and you can say that T of S is precisely a plus minus plus 2 times C naught over S times the binormal plus O of 1 over S for say S bigger than 1 and that the binormal is e to the pi S squared over 4 plus some logarithmic correction that is of course is completely reminiscent that in here you are dealing with some kind of scattering this is scattering problem because I give you the information at zero and I ask something about what happens at infinity so this log correction is because we all know that the scattering for this equation is long range so for some b plus some O of 1 over S where this b plus minus is tangent to the a plus minus and of course this is some real part plus I imaginary part and they are orthogonal of unitary length ok so the picture is more or less as follows it's very easy to more or less see how is the picture of capital G because you know that close to zero the torsion is zero so it has to be close to say a horseshoe and then at infinity has to be two lines in between you have a cane you can kind of freeze and frozen the the S and then it will be an elix with a pitch that will be changing with S so essentially this is but you have to so this will be the a plus and this will be the minus and then of course the information at the origin is given by the curvature and this is the information at infinity alright so this is it so this is well known now let me try to understand now these three questions about this guy and the first one yes yes because of this formula and why this formula is true is well I guess it depends as you probably know to be able to give a close formula for these kind of problems is extremely unusual so you need to have plenty plenty of symmetry to have such a thing and of course this is this is not any similar solution you can have more other similar solutions for example the ones that comes from the principal value of one over S that are logaritmic spirals and those ones I cannot give you a formula like this the reason is that this one is Galilean invariant and it is the only one so many more symmetries anyway so concerning the first question this is a theorem there was several words with Barillabanica but at the end of the day we were able to prove ah ok so before I do this in order to conserve to answer the first question I have to really give you to try to extend the solution for say negative time and I will do it as everybody has done in the previous talks related to this I just plug the corresponding conservation law that goes for negative time so you have always this symmetry that all these are oriented curves so if I change the orientation I change time so in order to change the orientation the only thing have to do is to change this guy into this guy and vice versa with a proper sign so I could construct in say this kind of an artificial way the following guy which is my starting solution is a square root of T G S over the square root of T say for T positive and then a square root of absolute value of T some rotation of G of S over the square root of absolute value of T for T negative and the only thing I need is to change the orientation so I need a rotation that I think has to do this or something close to this and because the problem lives in varying rotations then I will have a solution for T positive a solution for T negative and that they have the same initial condition so I could wonder if this is a solution or not ok so then the theorem with Valeria is that yeah that takes still is a stable solution in a appropriate sense and that means that if I make a small perturbations I remain close to the to the curve in particular the corner is the same and here small perturbations are done at the level of the curvature and the torsion they are small in some space that is related to that kind of setting that means that so that means that I can if you want creation and relation of corners is stable and this is very important for the case of the regular polygon because corners appear and disappear in a precise way alright so there are two words about this and this is as I said the last paper appeared at the end of last year the first thing is that if I do it at the level of NLS then it is no way and this is because if I make say I started so let me see what is the I should have said this under this situation the corresponding psi of course is the one we expect ok so imagine that at time one I start with a small perturbation of C0 then we are able to construct a solution all the way to 0 plus but when it goes to 0 plus what I obtain so if you want the asymptotics is like the C so it is 1 over the square root of T e to the Ix square over 4s sorry s square over T plus then you have the problem of the long range appearing epsilon plus plus liter of 1 so it doesn't matter what you do this problem is imposed and this is quite generic for almost all it doesn't matter how small it is I can I make a perturbation I construct the solution I go to the delta function in here but this one does not have a limit because of this guy of course this from the geometric point of view has no real implication because you always have the freedom of the gauge so you really want to know if this is a real problem or not and it turns out that this theorem is telling you that there is kind of a unique meaning of what of that has to be precise to be precisely written there is kind of a unique way of continuing the problem when you go back to the geometric setting alright and the second was that in order to do this, this information so what you are able to prove is that your solution goes to some guy T0 of this and at the limit you recover the important information which are the a plus minus b plus minus that comes from the similar solution and this is the one that allows you to construct the frame at time 0 and to continue and because you have a characterization of the similar solution then you can do it so this is what I wanted to remember and if this fits pretty well you will see in a minute with the kind of pictures that Manuel del Pino also in yesterday for the heat you will see in a minute why so this is about the first question the second question we use this blackboard is that in fact it's very easy to see that the linear impulse is not preserved because the linear impulse is just in this setting is nothing but this but this is G of the square root of T G of S over the square root of T cross T of S over the square root of T and you are iterating in S so this gives me T G S cross T but remember that G satisfies that equation so the only thing I have to do is and G prime is T so the only thing that remains is the the normal so this is nothing but the normal cross part with the tangent so this is T curvature times the normal which is the derivative of the of the tangent and then what you are saying is that the tangent plus minus infinity has this value so this is precisely T A plus minus A minus 40 positive and 40 negative will be the same but with absolute so the linear impulse so this is for all so the linear impulse is not preserved and it's of course because the energy or whatever is going to plus minus infinity and the third answering the third question it turns out that there is some small surprise at least for me which is as follows if I compute the L infinity norm of T S which is the same as the curvature times the normal this is 4 times 1 minus e to the minus pi C naught square this is for T equals 0 and for T different from 0 is 4 times pi C naught square and it turns out that this number is strictly bigger than this one as soon as C naught is not 0 so there is a lack of continuity in this particular norm which if I had to make the picture of what's going on it's exactly the same picture that you you made just you have a jump at 0 and then you recall of course you are able to do this because you have this nice formula and this is not so straight forward because for example if I go to these asymptotes I have to pay attention to the next term because I am computing the Fourier tanche so this was some elemental facts so now it's a very natural question what can I say when I start with a regular polynomial right? and these are the pictures so I need to so what do I do to turn this where is the linear momentum remember is here so what I am going to show you is what can I say for example with the linear momentum in the case of a regular polygon where you have many more symmetries and well if you start to think a little bit about what I have just written then you don't see this problem in the periodic case of course ok so the linear momentum has three components this would be the third one my my in this case I think it's a equilateral triangle I don't know what happens and this would be the third component so the third component will be this one and you see that essentially is conserved you see that all the numbers on the left are the same alright and then if you do the other two by symmetry they are going to be zero ok just by symmetry of the problem so what is kind of you want to see if there is at least when you look at the density of the of the conservation law if there is some transfer from one place to another one and then you do the instead of integrating in a closed curve you integrate in half of it and I show you the first and the second component this is what you see which is the classical picture of say for example the Riemann non-diferenciable function something say some intermittency or whatever and some fractality this is one of the components and the other one is this one ok so this is what you see and now let's see the other question is what can I say about the maximum of the Fourier coefficients of the tangent right and it is a symmetry in this case and it makes in fact if t3 where this so I am going to compute this in fact this and this other one and I am going to take the maximum and in fact by a very simple symmetry when one is not zero the other one is zero and so on and so forth anyway and this is the picture so this is for the t1 plus i2 so this is the maximum so I guess you see I was not expecting this to be this because these times are I'm sorry you are not missing the these are time this is time and this is the maximum of course these peaks are the rationales the smaller the denominator the bigger the peak of course well this I guess there is a good definition of intermitency I guess this is one and of course that means that you have transfer definitely I don't know how to prove this and this is for the other one this is for the t3 and you could wonder about what happens here right and if you do a plug of a log then it fits but poof we have to do plenty of more experiments to see there is really a log there that we don't have any idea so as a conclusion I do think that in the right function space the dynamics of this completely integrable system is much richer than what I would have expected at the beginning and that's it, thank you this is an effect of the of the periodic boundary conditions of course this is the Talbot effect so if you increase the the length of the do you still observe these peaks or do you have some dispersion coming well if you remain as long as you have periodic boundary conditions it's going to happen of course the peaks will become smaller and smaller but I bet it's a question of its case but the stability that result that you presented uses dispersion yeah but you do it I mean the the numerics of this is kind of delicate because you really have to capture the dynamics that happens at rational times and then you could wonder how are you going to be rational times in the computer here there the rational times of course but and then well we have for quite some time the laos of this is done with the laos we have plenty of numerical simulations but we were using the Schrodinger the stereographic projection thinking from the analytical point of view the stereographic projection is better because then you don't have the non-linearity of the biggest order but it turns out that for the numerics is extremely worse in fact we use in here you are using very strongly that you don't I mean you have not just this equation but also the other one so in the numerics you use first of all you can use a spectral method so from that point of view the numerics what you are saying I would say is very very stable it doesn't matter how long the period is and it works pretty well I don't understand why it works so well but you really you have seen the pictures you go to the rational times and you see your polygon of MQ sites it is there I would say so because of course the numerical scheme as I said is spectral which is the reasonable thing to do here of course yeah those are questions let's thank Luis again


Hi everyone, Sean with jazzy's casebreaks.com here doing 2021 hit parade Football platinum edition series 5 10 box case break random team number one where obviously all card ship and Looking for ridiculously cool stuff like this. Every box contains one autograph card featuring some of the best stars Includes rare numbered autograph cards autograph member Bailey cards dual and quad autograph cards only a hundred hand numbered boxes are produced So what they do is they take all of them they put them in the ten separate cases and then they They Individually number the cases and they randomly insert out of the hundred boxes ten of them in each case Here's some of the names that you can look for in here Hall of Famers Are all stars all that other stuff, too? so Pretty good chance. I think there's one two three four five six seven eight nine Salesheet hits on here. There's a pretty good chance that every case will typically have one of these sales sheet hits, which would be cool Here is our list of customers from Garrett Fontano down to Terry Cahill with last spot mojo and The teams from the Cardinals down to the Washington football team dice will randomize both lists and pair them up four to one five times Fifth and final time from Peter Lombardo down to Adam Levy and five times on the teams Of this we have a second case of this which is in the store now, so we'll get your spots guys We've already had a couple spots gone Detroit Lions down to the New York Giants Five times in the randomizer for the cut the teams five times on the cut randomized for the customers five times on the dice So Peter Lombardo with the Lions PJ McNeely with the Chargers Robert Citrulo with the Washington football team Tom Lee with the Seahawks Steve Wright with the Bears Douglas Renouffe with the Rams James Connell with the Chiefs Brandon Mueller with the Broncos Jared Salviak with the Colts Garrett Fontano with the Dolphins Michael Morelli with the Browns Victor Zamora with the Falcons Chad Coby's with the Cardinals PJ McNeely with the Cowboys Tom Lee with the Bills Peter Lombardo with the Patriots Brian Hayes with the Ravens Terry Cahill with the Panthers Robert Flores with the Saints and the Titans Robert Citrulo with the Texans Chad Dahl with the Vikings Brandon Mueller with the 49ers Steve Wright with the Steelers Howard Tong with the Packers John McGee with the Jets Paul Norden with the Eagles Michael Morelli with the Jaguars Robert Flores with the Buccaneers and the Raiders Mike Cody with the Bengals Adam Levy with the Giants So trade window is open Browns and Jags are up for trade Michael says Remember guys when you're trading on the product like this there is no checklist, so it is a blind trade So be careful while you guys are thinking about trades We're gonna flip over to this camera for a second Because we got to figure out which case we're gonna do so I've got two cases here The second case is already in the store under random team number two But we have cases one and cases six so one out of ten and six out of ten We're gonna roll a dice to see which case we're gonna do for random team one and which one will keep for random team two So one two three on the left case, which is run one out of ten and four five six on the dice for the right case Which is six out of ten, so we're gonna go with the case on the right number six case one will be four Random team number two. Let's pop back over here to the randomizer Give you guys a minute or so to trade So Michael if I tell him I'm done with this if there's nobody else in the chat for trades We're gonna go ahead and print the rip You have try to trade couldn't trade mojo Okay, so it looks like there's no other talk of trade, so let's close it up Michael Try to trade couldn't trade mojo for you print So we're doing a case six out of ten. So there are our ten boxes We're starting off with Mine my number 31 out of 100 Special box here for us Start with something good There is our final list as well CD lamb two-color patch and auto 27 out of 50 from plates and patches very nice patch there as well Dallas Cowboys PJ McNeely sharp-looking card Next box is box 83 Next up is CEH Clyde Edwards E layer That looks like part of the glove That's where you'd get that kind of stitching and that kind of material on there on a glove rookie patch auto For the Chiefs that is 49 out of 75 Kansas City Chiefs James Connell box number three Go Tigers Justin says Row in prison rookie auto Josh Jacobs For the Raiders and Robert Flores That is 96 out of 149 little red wave rookie auto for the Raiders and Josh Jacobs rookie very nice. I Like me some Josh Jacobs Can't wait till he gets to a real team Box 63 is next Jerry Judy rated rookie auto out of optic. That is an on-card auto 125 out of 150 for the Denver Broncos and Brandon Mueller very nice J. J Josh Jacobs Jerry Judy JJ 55 box next junior sale Could we have a third JJ in a row Justin Jefferson? It is not but it is a go Definitive collection two-color patch and auto Jerry Rice one out of ten on-card auto 49ers Brandon Mueller. That is from 2015 tops definitive collection. That's pretty cool Little bit of two colors on that That is pretty nice Box 96 Remember guys second case of this is in the store That Jerry Rice was the sales sheet hit Told you every case typically has a sale sheet hit not too bad and James Robinson Jersey and auto rookie for the Jacksonville Jaguars Michael Morelle try to trade couldn't trade mojo 153 out of 199 see Michael try to trade couldn't trade mojo always works Box 76 Now let's see if we can get your Browns to hit to Justin Jefferson Football patch and auto rookie from Phoenix 61 out of 99 from the Minnesota Vikings and Chad Dahl There you go, man So we've had three double J's and it's Justin Jefferson Josh Jacobs. Who was the other one Jerry Judy? Box 11. This one is Chase Claypool contenders optic rookie ticket auto On-card from Obviously contenders optic For the Pittsburgh Steelers and Steve Wright Box 20 say we have another prism And it's prism rookie auto current year JK Dobbins for the Baltimore Ravens and Brian Hayes Very nice and last box for the case box 44 Final hits coming from select and it's Jersey and auto chase young rookie for Washington and Robert Citrulo 36 out of 75 Defensive rookie of the year Very nice way to finish it off and there you go guys. That is the case You can see there's that sales sheet Jerry Rice. We just hit That was 2021 hip parade football platinum edition series 5 10 box case break random team number one from jaspy's case breaks calm I'm Sean jaspy's case breaks calm. Thanks for hanging out. I'll see you next time


The effectiveness of COVID-19 vaccine campaigns depends on more than just efficacy and safety. Vaccine acceptance among the public and health care workers plays a crucial role in controlling the pandemic. This review provides an up-to-date assessment of COVID-19 vaccination acceptance rates worldwide, based on 31 peer-reviewed studies from 33 countries. Among adults representing the general public, the highest acceptance rates were found in Ecuador, 97%, Malaysia, 94.3%, Indonesia, 93.3%, and China, 91.3%. The lowest rates were in Kuwait, 23.6%, Jordan, 28.4%, Italy, 53.7%, Russia, 54.9%, Poland, 56.3%, US, 56.9%, and France, 58.9%. Only eight surveys among health care workers were found, with acceptance rates ranging from 27.7% in the Democratic Republic of the Congo to 78.1% in Israel. In the majority of survey studies among the general public stratified per country, acceptance of COVID-19 vaccination showed a level of greater than or equal to 70%. Low rates of acceptance were reported in the Middle East, Russia, Africa, and several European countries. Addressing vaccine hesitancy in these regions is recommended as an initial step for building trust in COVID-19 vaccination efforts. This article was authored by Malik Salam. We are article.tv, links in the description below.


Znači je, da neo-libera je dobro, zelo je življenja, da je zombija. Kako smo pošli? Nisem nekaj, da pričočne investije bila pričočnje, da je nekaj držav, da je nekaj držav, da je tudi 30. seči. Nisem nekaj, da je dobro pričočnje, da je tradizionalne konečnjega era, da je genera pričočnje, pričočnje, et cetera, should be a solution, though I do not exclude that that can be the right-wing answer. I think that it is clear that the phase of indebted consumption may create the appearance of stability for a while, but it creates a capital which is unsustainable. I am Riccardo Bellofiore and I am affiliated to the University of Bergamo, a university in the northern Italy near Milan. I was lucky enough to know Minsky since the late 70s, because he came to Bergamo very often, he even bought a house. I have been intrigued by the two sides of Minsky. The most known is the financial instability hypothesis. Minsky died in 1996, but became a very well-known figure after the crisis of 2007-2008, because he was connecting instability to private debt. Of course, private bubble for household consumption was a key here. There is, however, the other side of the moon, which is not very well-known, that for me was very important, this other side of the moon is the fact that Minsky in the 80s developed a stage view of capitalism, speaking of five phases. There was the commercial, the industrial phase of capitalism, the financial stage, then the managerial stage, the Keynesian era and after the money manager capitalism. But the other very important thing, which is already there in the writings of the 70s for Minsky, is the heritage of the New Deal. It was very much for a radicalized view of socialization of investment, and also for the state as direct employer. So his view out of the crisis of the 70s, but I would say also the same can be said for going out of this crisis, was actually a view which tried to connect the Keynes heritage to the New Deal heritage. The New Deal was the world in which he studied and he made also his political education. So these ideas of the socialization of investment and the status that the employer was very much coherent with the secret theory of money, but he provided a much more stress on the financial side, on the financial stability, which was lacking in the theory of the monetary secret. Minsky had this stage view of capitalism. His stress was mainly if not only on finance. He loved to say that in capitalism money is not the most important thing, it is the only thing. My perspective is very sympathetic to this, but I have an origin also in the Marxian framework, so my stress is also on labor surplus value profitability. And as you know in Marxianism there is this view of the falling rate of profit, which is very often interpreted mechanically. I tried to have a non-mechanical interpretation of the falling rate of profit, so in my view actually the long depression of the late 19th century was a traditional falling rate of profit theory to low surplus value. But in the 30s the crisis was an effective demand crisis, which I would not define as under consumption but as under investment. Keynes plus Kaleski are very important here. Then there was the way out of this crisis, mostly because of the destruction of capital and the World War II. The following era was the era of the traditional Keynesian phase, and it ended again because there was a profitability crisis. Two low gross profits, but this time not for the traditional view of the falling rate of profit, but because there were social struggles not only on the wage, but also in the direct production process. This view has some depth with the regulation school, also authors by Andrew Glean, where the profit was a containment by workers of exploitation and productivity. After there was neoliberalism. I see neoliberalism, yes, as Miski says, as the money manager, capitalist phase, that I would rather rephrase in a Marxian way as the real subsumption of labor to finance and debt. Actually, the workers were traumatized in production, so we had this phenomenon that economists call the flat Phillips curve. You can reach full employment, but there is no rise in wages, productivity, exploitation is very high and so on, but also households were included through the money managers and so on into finance and went into debt against this fictional wealth, or because they needed to. Doing this, they became not only actors through the alienated figure of the money managers in the financial market, they became also in-depth consumers. So in-depth consumption was the new form of aggregate effective demand, the leading form of autonomous effective demand. In this case, the crisis came with the great recession and they very much agree with the interpretation of Adam Tooze in his book Crash, that this so-called financial crisis was not really due against falling rate of profit under consumption, this kind of thing, and not immediately from effective demand. It was due from the perversion of too short-term finance, a sort of crisis of funding in a world which is now transnational and there are interlocked metrics. The heart of the matter is North Atlantic finance, the system of banks, shadow banking, institutional investors, et cetera, related between US, the Eurozone and UK. So the crisis came from there, not from where everybody expected it, the global imbalances between US and China or the internal imbalances within the Euro area. This stage view is linked to the idea that each phase has something which propels it, but the forces leading up are the same forces that exhaust themselves, and so you have an upper turning point and go into a structural crisis. But in this view, where there is a kind of necessary upper turning point, there is no necessity in the lower turning point, there must be institutional innovation. This is very strong view in Minskij, but it is also in others very different as Schumpeter, et cetera. The German idea, so to speak, of making profits through net exports cannot be applied to the world, and it leads to trade wars again as we are. So what is the way out? I think the way out is a kind of different government spending, a targeted government spending in the Minskij meaning of the world, but here I also use a concept borrowed from the other secretists, the French secretists, Alain Parghese. He speaks of good planet deficits, because the point is that the monetaries or neoliberalist policies, which are different, but here it is not very, very essential to distinguish them, this kind of policies, they say they are against government deficits, but they create government deficits. They are self-defeating because they kill GDP and employment and so on. We want to have ex-ante planet deficit, which rise for a while, the debt over GDP, but these expenditures are good because they create tangible and intangible new resources, not only infrastructures, but also the green expenditures, but they think that we should put into investment things like health, education and so on. So after a while, the GDP should go up, and as Minskij wanted, Minskij was the case idea that in full employment you should have some kind of balanced budget excluding the investment. We reach a way out of the crisis.


So, I think let us once again revise the diagram rules. So, we were specifically talking about the Huygen-Holtz diagrams. So, the rule number 1, I will let me rewrite the rules little bit more clearly. So, we have each dot contributes to a matrix element, either 1 or 2 particle. So, basically 2 particle means 2 electron integral and if it is 2 particle then it is anti-symmetrized. So, for each of these either F or V, there is a dot. V essentially is of course again 2 particle minus 1 particle. So, 1 particle does not contribute. So, basically 2 particle. Then the second is that all these dots have to be joined. So, the question is that how many dots are there? So, the perturbation order, I can also write here that with 1 or 2 pairs of, so each pair consists of one incoming and one outgoing line. So, I have either 1 pair for 1 particle or 2 pair for 2 particle dots. And then energy diagrams are represented by closed diagrams involving these dots. So, I just join all the dots in a manner that they are completely closed. Then an imaginary line between the dots, let me also specify right here that this matrix element is a numerator. So, that these are constructed with sum over e of holes minus sum over e particles as the denominator. But the sum is the number of hole lines that this intersects, the imaginary line that intersects, the number of particles that this intersects. Is it clear? Then number 5 is if you have in the entire closed diagram, if you have a number of holes is h, number of loops is l, loops can very often be seen from the diagram or from the algebra. Then you attach a factor minus 1 to the power h l, h plus l. So, multiply by this factor. Then number 6 is 2 to the power minus k, where k is the number of equivalent pairs. So, what is an equivalent pair? It is either a pair of holes or a pair of particle starting from the same vertex and ending at the same vertex. So, starting from a vertex and ending at the another vertex, which is basically have to be all same. And of course, at the end it is important to write that sum over all holes and particles. When you write the algebra, we have to ensure that all internal holes and particles are summed up. So, we did some practice problem. So, one was the MP2 itself. So, we said this is MP2, sorry, this is MP2 in the Hugen holes, which is 1 by 4. This is A, this is B, B and Q, 1 by 4, PQ AB. So, the matrix element that this contributes is basically the pair of incoming on one side and outgoing in the other side. Or since they are all real, it does not matter. You can have outgoing here and incoming here. So, this is this. Then we did the third order diagrams. So, we saw there are three third order diagrams. So, these are three third order diagrams.   This is a hole particle ladder. So, when you draw, of course, you can write the expressions. If I ask you to write the algebraic expression for this, I hope all of you can write it. So, just I am asking you a question. How many, what will be the factor? 2 to the power minus, what is the factor here? 1 by 8. So, here also 1 by 8. Here there is no factor. There is no equivalent line. So, just plus, just 1 for the hole particle ladder. And then you have minus 1 to the power H plus O. What will be the sign here? Number of loops. The number of loops is the only thing that it will count properly. And when you write the numerator, actually from there also you can calculate the loops. As soon as you write the numerator, you can calculate the loops without looking at diagrams, whichever you are comfortable. So, how many loops are there? Depends on how you are calculating. Depends on how you write the numerator. Let's try to write one of them. That's easier. So, let's write this A. Yesterday I did not write the expression. A, B, P, Q, RS. The first one I am writing. So, these four are particle lines. We will put this loop rule little bit later. First let me write down the numerator. So, this is sum over A, B, P, Q, RS. 1 by 8. Then you have RS. Now it is a question of how you write it. RS anti-symmetry A, B or RS anti-symmetry B, A. You have two choices. But write it in some way. And accordingly loops will change. So, RS anti-symmetry A, B. So, you have to now remember A is going to R from the diagram. And then you have RS anti-symmetry P, Q. And P, Q, just one bit. P, Q anti-symmetry A, B. You can write any way you want. Now you divide these two imaginary lines. Remember, when I wrote that rule, please modify an imaginary line between two vertices, every pair of vertices in succession. Not every pair means not between 1 and 3, every dots in succession, every adjacent vertices. So, then you have epsilon A plus epsilon B minus epsilon R minus epsilon S divided by epsilon A plus epsilon B minus epsilon P minus epsilon Q. Up to this point, now we will count the loops. So, let us start from somewhere. Let us say A. I have to come back to A, A to R, R to P, P to A. So, one loop. Then I go from B, B to S, S to Q, Q to B. So, I have two loops. See, A to R is R to A I can do. That is unimportant. That does not matter. So, you can say A to P if you are saying A to P, P to R, R to A. So, let me change it to a BRS. Because it does not matter. It is symmetric. There is no convention as long as they are real orbitals anyway. It does not matter. They are identical. So, we are all talking of real orbitals here. You have to keep track. As long as the convention is same. So, I think that will now cause A to R, R to P, P to A. So, it does not matter. You should not get confused by this. So, the point is here you have two loops. So, the loops are very clear. That is one is this loop. So, if you look at diagram actually, one is this loop, goes here, comes back. So, that is one loop. The second is the other loop, goes from here, goes here, comes back. So, there are two distinct loops. So, this is second one. I can label with the blue line. So, exactly the same way I can do the whole-hole. So, this is the first loop, this, this one comes back here, back here. And the second loop is follow this blue line, come here and back here. So, again there are two loops. And of course, there are two whole lines. So, this is the plus sign. And I can draw exactly the same way the whole-hole, except the symbols will now change. I will have A, B, C, D and two particular lines. Let us try to have another diagram. So, you can draw this. Let us try to write the third one. The third one, I can make three loops. For the third one, for example, this can be a loop by itself. This can be a loop by itself. This can be a loop by itself. So, there are three loops. Each of these vertices, you can have a pair of lines can be a loop. But again, it depends on how you want to write it. If you interchange, then loops can be two. I can interchange one line. But if I follow this and that is the more ready convention, so A, B, C, P, Q, R. So, let me label this again. So, let us say this is R, this is A, no, sorry, this is P, this is A, this is Q, this is B, this is R, this is C. That is easy. So, I have three particular lines, three whole lines. So, look at the first vertex. Outgoing is C, A, anti-symmetrize, then R, P, then you have B, R, anti-symmetrize, Q, A, no, sorry, this is Q. The second one is B, R, then A, A, C, sorry. So, B is going out, R is going out, and then A is going in C, A, C. Then the last one is A is going out, P is going out, not A, Q, Q, Q, Q. Q is going out, P is going out, and then B comes in A comes in A. So, you can see one loop is A to P, P to A, and R to C, C to R, Q to B, B to Q, in the same way. This is Q, sorry. So, that second one should be also Q. So, I am drawing this diagram again here so that everybody can see. So, the labeling that I have done is PA, outer line, this is P, this is A, then you have Q and B. So, this is B, this is Q, this is R, this is C. So, that is the label we have done, and this algebra corresponds to this. So, now you have to see the sign. How many internal holes are there? How many loops are there? So, how many loops are there? Three. The way I have written, and of course, the internal hole lines are also three. So, it remains plus, and no 1 by 2 factor. There is no equivalent pair of lines. So, that becomes the algebra. So, if I give you a diagram, perturbation diagram, you should be able to write the algebra using anti-symmetry matrix. Any problem? Oh, numerator. I am always forgetting the numerator. Numerator is easy. Of course, you just write it down. So, you have epsilon A plus epsilon C, A and C, P and R. See, numerator, you do not have to bother. There is no order or anything. Just see the lines. It is very easy.  You just have to keep track of your symbols. Yes, total energy is some. So, MP3 is this plus this plus this. So, that is why I have written plus, plus, plus. So, up to third order, the diagrams, if you look at the third order diagram, they are also very similar to this diagram. And that is why they are called ladder diagrams. And there is no other possibility of writing any more lines. In fact, it can be seen that such diagrams actually come because of W-excited determinants that MP3 also contains only W-excited determinants. I said that even in the first quantized, if you look at the third order energy, it is Psi 0 0 V Psi 0 K, then Psi 0 K, sorry Psi 0 K, sorry Psi K 0, Psi K 0 V Psi L 0 Psi L 0 V Psi 0 0. So, this is basically one of the major, and then of course, you have a denominator. So, here you can see that you have only Hartree-Fock to doubles. And of course, this K must also be doubles because I have to come back to Hartree-Fock. So, you have doubles to doubles and back to Hartree-Fock. So, you have no other terms. Now, you should be actually able to see the first quantized MP3 diagrams. They are not the algebraic expression. You will realize that the actual expression is not so simple. The actual expression is a combination of two expressions out of which a part of first expression cancels the second expression. And we have only these particular terms which in diagrammatic notations are basically just the unlinked diagrams. And that was the basic idea of what we said is that the link cluster expansion. All parts of the diagram must be closed and linked. So, if we refine that H 1, all diagrams, one of the rules that we wrote, all diagrams are not only closed, all energy diagrams, but also they must be linked. So, this is something that I thought I would not bother so much, but I can at least show algebraically where the problem comes up. So, you can go back to the algebraic expression of perturbation theory and we can write E 0 3. So, let us start from E 0 3. I hope you remember how to write E 0 3. Psi 0 0 V Psi 0 2. So, if you remember E 0 3 expression, it was Psi 0 0 V Psi 0 2, second order in the intermediate normalization, second order. So, we have to now get what is Psi 0 2. So, how do I get Psi 0 2? I go back to my second order equation H 0 minus E 0 0 Psi 0 2 plus V minus E 0 1 Psi 0 1 minus E 0 2 Psi 0 0 equal to 0. What we need to do is to find out what are the expansion of Psi 0 2 in terms of Psi 0 k's. Because remember, exactly like first order, we will write Psi 0 2 as a linear combination of all k not equal to 0, which I now call C k 2 Psi 0 k. The same expression that I use for the first order except that the coefficients will now be called C k 2. So, I have to find out the C k 2. I have to find out the C k 2. So, a specific C n 2 is nothing but Psi 0 n Psi 0 2. So, this is the matrix element that I have to find out. If I know this matrix element, then I know the entire expansion of Psi 0 2. I will put it here and I will start to get E 0 3. So, this is also Psi n 0. So, let us now find out this what is the value. So, what do I do? I start with this expansion and project with a Psi n 0. So, if I project with Psi n 0, I will get E n 0 minus E 0 0. So, I will get first E n 0 minus E 0 0 into Psi n 0 Psi 0 2. This is what I want to find out. So, that is what I get from the first expression. I am projecting with Psi n 0 from the left. Then I have Psi n 0 v. I want to write this as two terms Psi n 0 v Psi 0 1 minus E 0 1 Psi n 0 Psi 0 1. This is not 0 because this is the first order. So, this is the expansion that I get. So, I write this as this is bracket. So, I have Psi n 0 Psi 0 2 equal to, I will have two terms. One is Psi n 0 v Psi 0 1 minus E 0 1 Psi n 0 Psi 0 1 divided by the denominator changed. Instead of changing sign here, I have just changed the denominator sign. Is it alright? So, now you can see that when I put this back here, I actually have two terms. One coming from here, one coming from here. So, if I explicitly write the E 0 3, I will first get Psi 0 0 v. And now I can expand this by putting an n here or this is basically the R, the solvent kind of thing. So, I can expand this by writing this as sum over n Psi n 0 and then C n 2, which is basically this term that I wrote. So, this would be now Psi n 0 v Psi 0 1 minus E 0 1 Psi n 0 Psi 0 1 divided by, it looks big but it is actually not big. What I have done here only, because I have to write Psi 0 2, I have written Psi 0 2 as sum over n not equal to 0, Psi n 0 and this coefficient. This coefficient is just here. So, I have just put this whole thing here including the denominator. So, it is really not big. So, just the denominator I have written here. And then you can see that I have two expressions. One is this, another is this. Each of the expressions have Psi 0 1, which I now know how to expand and I will put that Psi 0 1 expansion in each case and then I will complete the algebra. So, let us write the Psi 0 1. So, what is Psi 0 1? This I already know. So, Psi 0 1 is sum over some m. I am using a different dummy index from n, sum over m against C m 1 but now I know what is C m 1. So, can somebody tell me again Psi m 0 v Psi 0 0 Psi m 0. I can write Psi m 0 here divided by the denominator E 0 0 minus E 0 0. So, all I have to do is in this expression of Psi 0 1, I have to substitute that. So, you can see now there will be two double summation and because Psi 0 1 also has a denominator, there will be two denominators. So, if you are just little bit careful with the algebra, you should be able to write the entire expression now. So, then your E 0 3 becomes, let us do it carefully, sum over m, sum over m both not equal to 0. Sometimes you just write prime here to indicate that 0 is excluded. You have Psi 0 0 v Psi m 0. That is the first term. Now I am multiplying this Psi m 0 v and now I have to bring Psi 0 1. So, I do an expansion of m and then I write Psi m 0, close this. It has first a ket vector so with which this will be closed and then another factor will come Psi m 0 v Psi 0 0 divided by E 0 0 minus E n 0 which is already there and one more which is coming from here E 0 0 minus E n 0. So, you have a double summation, but the story is not complete because I have a third term, I have a second term which is E 0 1 which just comes out, which is E 0 1 which can come out and then product of this with this. So, E 0 1 times again there is a summation, remember E 0 1 is a number so I can simply take outside. E 0 1 is just a number so I have a sum over m and then I have Psi 0 0 v Psi m 0 which is the first term multiplied by this into Psi m 0 and now I write Psi 0 1 which is Psi m 0 Psi m 0 v Psi 0 0 divided by again these two denominators. So, it is very similar expression except that v is not here, it is directly Psi m 0 Psi 0 1, yes, we will come to that orthogonality, I am coming to that now, so these are the two expressions. You can see here that the expression becomes simplified for the second term because this is now delta n m. So, n must be equal to m, so although there is a double summation it actually becomes single summation, remember here it did not happen because there is a v, otherwise the expression exactly identical is very easy to write, take out the v from here, instead of that an E 0 1 has come. So, if you look at the dimension it is same, you have 3 v here, 1, 2, 3, third order, here you have 1 v and 2 v but the third is this E 0 1. So, if you look at dimension in terms of perturbation order it is still 3, instead of v coming here it has gone out. So, this is of course this can be taken out so n is now equal to m, so I can put this as Psi n 0 and the summation over m can be knocked out and I can call this also E 0 square, right, okay, you write, now it is very easy just write it, just put n equal to m, Psi 0 0 v Psi n 0, Psi n 0 v Psi 0, it is basically square of this just like second order energy but denominator is not second order, like second order denominator is a square, otherwise what is multiplied with E 0 1 looks like exactly E 0 2 except for this denominator, otherwise E 0 2 was exactly same, Psi 0 0 v Psi n 0 or Psi k 0 back to Psi 0 0, if you see this Psi 0 0 v Psi n 0 back to Psi 0 0 divide by the denominator but now you have a square denominator that is only difference, otherwise I would have written as E 0 1 into E 0 2, it would have been even simplified which is also dimensionally third order but actually it is not, this remains.