Hello, everybody, and welcome to the Latin American webinars on physics. I'm Nicolas Bernal from the ICTB Safer in San Pablo. And I will be the host of this second webinar for this second season. Today, we'll have a great talk, which is in the line of actions and the real action. But before to go to the talk, don't forget to ask questions via the Google plus Q&A here, or via Twitter with the hashtag LWOQ. So the speaker today is Lorenzo Uvaldi from the University of Tel Aviv. Lorenzo plans PhD in the University of Santa Cruz, California. And after that, he relates a postdoc in the Bonn University. And now he's a postdoc in the Tel Aviv University. And the title of his talk is the Relaxation, Is Relaxation an Action? So Lorenzo, are you there? Hello? You can hear me, right? Very good. Hi, everybody. Well, let's see. So now I have to share my slides, I guess, right? Right, please. Let's see. Can you see it? Very good. Please go ahead. OK, good. Thank you, Nicolas, for the introduction. Very nice of you. I don't know if this talk is going to be a good talk. So you guys will tell me at the end, I guess.  OK, let's see how this goes. So this is some work that I've done in collaboration with these people at the Weissmann Institute. And they are Rick Gupta, who is another postdoc there. Zoarko Mardowski and Gilapérez were two professors there. And the paper is recent, so we put it on the archive at the beginning of this month. And I guess without further ado, I will just go into the slides and what I have to say. OK, so to begin with, oh, by the way, Niko, you can interrupt me, I guess, if you don't hear me anymore or something, right? Sure, I will do it. OK, great, thanks. And also, you told me that there are no questions during the talk, but if people are willing to ask questions during the slides, I'm totally happy to take the questions also during the talk, no problem. OK, so that's it. I wanted to start with a review of what this cosmological relaxation mechanism is about. So this is an idea that these people, Peter Graham, Kaplan, and Rajendra, have put forward a few months ago. They put out the paper in April, I guess. And I would say that it is genuinely a new idea, a new idea of how to address the hierarchy problem. And so first, I want to go somehow through the details of the mechanism. And then I would have some complaints, if I can say so, about some basic ingredients of their models. So here you can see in this slide the Lagrangian that they use in their paper. I use slightly different notation, actually, because in their paper the coupling G is dimensionful, whereas I defined it to be dimensionless. So their coupling G is just my G times capital lambda. This capital lambda that you see here is the cutoff of this theory. So you should think of this Lagrangian as an effective field theory. And now you see that pretty much the key point of this game is that you make the mu squared term, the quadratic term in the Higgs potential, you make it a dynamical quantity. So this mu squared now depends on the field phi, as you see on this slide. And the other crucial ingredient is that you need the potential for this field phi that is written here as an expansion in G phi, pretty much. So you have a linear term and you have a quadratic term. And this potential is pretty much responsible for phi to roll, okay? Phi will roll during inflation, I will get to that in a second. Another point is that as phi moves along field space, it scans the physical Higgs mass. So the idea is that you start from a point where the mu squared term is positive and as phi keeps rolling down its potential, it will reach the point where this mu squared will become negative. And when that happens, you know that you get into the broken phase, which means that the Higgs field develops a wave and you are in the broken phase. Now, another important term is the last one that I have on the first slide, which is the coupling of this field phi to GG tilde. And GG tilde is the, sorry, G mu nu is the field strength of SU3 color, okay? So what happens because of this last term on the first line, when you go below the QCD scale, so when you go below the confinement scale of QCD, this term generates a cosine potential for the field phi. And you can go through the whole song and dance of either computing like this potential through instanton effects, or you can compute that via like means of chiral agrangements. And what you find at the end of the day is that this cosine potential, which is generated, is proportional to the mass of U or the mass of the actually, in this case, times F phi cubed. So you see that it's linear in the quark mass U, which means that it generates a potential which is linear in the Higgs wave, okay? This V right here. Now, this potential is crucial because this is what is going to give us a back reaction to set the electric scale. So a cartoon of how the mechanism works is the following on this slide. So you have this potential, as I told you, it's linear up here, pretty much, okay? And the linear quark continues all the way down. But the thing is, once you hit this point where the field phi is equal to lambda over G, you start developing this oscillating potential, the cosine potential, right? Because so this is the point where the mu square term of the Higgs becomes negative. And below this point, you generate this cosine potential and so you start having these wiggles. And now the other important part is also that you see that these wiggles, this is just a cartoon, okay? So it's totally off scale, in a way, but it's just to give you an idea. But the point that you can see also from the cartoon is that the wiggles start small because the amplitude of these cosine oscillations is proportional to V. And V becomes larger and larger as you move to the left here of phi, right? Because as phi moves to the left, the mu square term becomes more and more negative and the Higgs wave becomes larger and larger. So as that happens, these wiggles grow. And at some point, you know, they grow enough that the rolling of the field phi is gonna stop, okay? So I put here the points one, two, and three, and as I told you, you start from very large field values, okay, values larger than lambda over G. Remember, lambda is the cutoff of our theory. Then the field rolls down the potential and it rolls down very slowly because this whole process happens during inflation. So there is some Hubble friction which is keeping the motion of the field very slow. And this is very important, actually, because if it wasn't for the slow rolling, the field wouldn't stop at this wiggle. It would just keep rolling down, okay? So it's important that the mechanism happens during inflation. And then as I told you, so when the rolling stops at this point three, we pretty much, at that point, we have, again, this is totally out of scale in this plot, so don't look at the field space in phi, it's totally out of proportion, in a sense. But the point is that you stop on a wiggle and at that point, you know, you're gonna have mu squared, which is much smaller than lambda squared. If you achieve that, that's pretty much a way to solve the hierarchy problem because it is a way to say why this mu squared parameter in the X potential is so much smaller than the cutoff. So in principle, if you wanted to solve the hierarchy problem, you would like to push the cutoff all the way to the plant scale. Okay, if you could do that with this mechanism, you would have totally solved the hierarchy problem. Now, we will see that actually in most models, the cutoff, you know, in the original papers, they claim a cutoff of 10 to the seven, 10 to the nine GV, depending on the model, so you don't quite get to the plant scale. Okay, so I want to list some of these conditions that you need during inflation.  Okay, I'm not gonna go through the derivation of these inequalities here, I'm just gonna state them. You can find kind of a derivation for them in the original paper, but not really, but anyway, they're not so hard to derive. I will tell you qualitatively what they mean. So the first one is easy, just it's a slow roll condition. So you're just asking that the speed of the field phi, or the kinetic energy is much smaller than the change in potential energy, pretty much, okay? And it gives rise to these two conditions. And they will be trivially satisfied, actually if you satisfy the second to last condition here, which is the classical Bitz quantum, I will get to that in a second. Another important condition is this number of E-folds. So this condition comes about because pretty much you want inflation to last long enough so that the field phi has enough time to scan the whole field range, the field range being this lambda over G, okay? So if you go through this condition, you find that the number of E-folds has to be larger than this quantity. I should mention because this is important that this capital H with the subscript i is the Hubble rate during inflation, okay? Another condition that you need to satisfy, you want the vacuum energy during inflation to be dominated by the inflaton field, which I haven't even written in the model, okay? It can be whatever, you will have an inflaton, that will drive inflation, and you want to make sure that the field phi, which is the field that rolls for us, doesn't dominate the vacuum energy. In other words, you want to make sure that phi is not the field that drives inflation, okay? Pretty much the reason for this is because at the end of the day, you want phi to be stable. So you don't want it to be the inflaton, the inflaton will have to decay. Okay, then there is this other condition that they call actually the condition that the barriers form. By the barriers, I mean the wiggles. So pretty much in order to generate this cosine potential, you have to go below the confinement scale. And maybe a better way to write this inequality instead of writing as I did, and as they did also, hi less than lambda QCD, maybe it would be better to write like the inverse, like saying that one over hi is larger than one over lambda QCD, because you can think of this in the following way. You can think that one over hi is pretty much the hub or radius during inflation. And one over lambda QCD is like, is the length, it's a confining length of QCD. So if the hub or radius was smaller than the confining length, you wouldn't have confinement. If you don't have confinement, you don't have instanton's effects and you wouldn't generate the cosine potential. So that's why we need this condition too. And then the last condition, which is actually not strictly necessary in a way, it just, it helps with the game, is that the classical rolling beats the quantum fluctuations. So for this, you can have again in mind, if you think of inflation for instance, I think let's think for a second about the inflaton. In the inflaton, you have this inflaton field, which is slowly rolling down its potential, okay? And at each point of the rolling, if it rolls down too slowly, you don't know exactly where the field is because there are quantum fluctuations around the point where classically the field is, okay? The same is true for our field phi. So as it rolls down, if it rolls too slowly, that would be quantum fluctuations that tells you, okay, maybe the field is not exactly there, but it could be slightly to the left or slightly to the right. And okay, in principle, this is fine, but in practice, it's nice to impose this condition, which just tells us that the rolling has to be slow, but not too slow so that it's pretty much dominated by the classical rolling. So that once we stop, we know exactly where we stop, okay? Whereas if there were quantum fluctuations, you could say, oh, maybe I stop in this week, or maybe I stop in the next one, okay? So it's just for convenience, let's say, but in principle, it could be relaxed, but let's keep it, it's okay. And then the crucial condition, which is the one that actually, at the end of the day, sets the electro scale, is the flat. So the condition for the rolling to stop is given by setting the derivative of the linear potential, which is in VG phi, equal to the maximum slope, the maximum derivative of this oscillating potential, okay? It means that at the point when the slope is flat, the field is gonna stop because it's rolling so slowly. So when it gets to that point, it will stop. So from this last condition, you get this equation here, which tells you the G times lambda cubed F is equal to lambda to the fourth QCD. And now if you put together all these inequalities that I have on this slide, together with this last condition for the stop of the rolling, you get the following cutoff, okay? You get this condition, the algebra is pretty easy to go through if you want to do it. And if you put it in the numbers, if you put it in lambda QCD, you put it in the plant mass, you get that the cutoff for this model would be 10 to the seven GV, okay? For decay constant F, oh, I should have mentioned also what F is, F is, in this case, it would be the axon decay constant. Pretty much what it is, is the scale at which the original symmetry is spontaneously broken. But the original symmetry mean like the axon, I will get to that also in a second, the axon will be the Boston boson of U1. So the scale of spontaneous symmetry of this U1 is this F, okay? I will, I'm putting here some more numbers on this slide. So if you have this cutoff, lambda 10 to the seven GV, which by the way would be not bad at all, you get like this coupling G that I had in the other slide to be 10 to the minus 33. So it's incredibly tiny, okay? So somebody might be even bothered by this number. I will tell you in a second that in principle you shouldn't, but then I will tell you in two seconds that actually you should. And another number, which is worrisome and that will not get into any of the worries related to inflation, but just to mention that, is that with this cutoff and with this game in this model, you need 10 to the 50 E falls of inflation. And remember that usually, I mean usually, usually people talk about 60 E falls of inflation. That's enough for inflation to do its job pretty much. So this number here is about 50 orders of money to greater than what inflation needs. It's okay, it just means that inflation lasts so much longer than what people usually have in mind, but maybe it's not a problem, except that I guess there are a lot of people thinking about also this aspect of the model. Anyway, I will leave it there because I'm worried about other things. So let me get to that. I want to get a closer look at the basic ingredients of the model, so I just rewrote the Lagrangian once again. And now what the authors, the original authors observed is that the term with the cosine respects this discrete shift symmetry, which means that you can shift the field phi by two pi F times an integer, which I call K. And then they say they call this symmetry global, and they say that the terms with G in the first line of the Lagrangian would break this symmetry explicitly. But then the point is now G can be tiny because as G goes to zero, you recover the discrete shift symmetry. So in that sense, you know, they say that the model is technically natural. So in the previous slide, I showed you that with the model, you had G to be 10 to the minus 33. If there is a technical naturalness argument, it's fine. G can be as small as you want it to be because as it goes to zero, you recover this shift symmetry, and then your model is natural. So it means that you have a technically natural way to set the electric scale if you play with them, okay? The model that I just described actually is already excluded because the author called the phi an axion, and they say this in the paper as well, by the way, if phi is an axion, when you stop at the wiggle, you actually introduce a strong CP problem rather than solving it as the original axion should. So the model is excluded in that sense. And then, you know, they have other alternatives with like, let's say an axion prime, so an axion that couples to G prime, G prime dual, so like some hidden SU3. So you can play different games, and in the end of the day, you get a similar cutoff. If not a bit higher, and you have the same, the store is pretty much the same, or at least the concept of how the mechanism works, okay? So I will not go into that. But again, so let's again say, what is this phi? Okay, in the original model, they say they use the axion to be phi, but what is an axion? An axion is a pseudo-Nambo-Golston boson, okay? And we know exactly what a pseudo-Nambo-Golston boson is, and we know how it arises. So for instance, for the axion, we know that you can start from a global a check-in symmetry, A1, and you can parameterize a complex field like this, like rho, which is the modulus times e to the i phi over f. So phi over f is the phase, and that will be our Golston boson in the end. It will be the axion. Okay, now the original U1, under which the capital phi rotates, translates into a global shift symmetry for the axion. And this global shift symmetry is respected also after the spontaneous breaking of the global U1. But now the thing is that a subgroup of this global shift symmetry is a discrete shift symmetry, which I wrote already beforehand right here, which is this phi going to phi plus 2 pi f times the integer k. And now you can see that by definition, by the way, the Golston boson is defined because a Golston boson is just a phase, just an angle. Whenever you go 2 pi around the circle, 4 pi, 6 pi, 2 pi times k, pretty much. Nothing changes, right? I mean, clearly if you use this discrete shift symmetry in the original phi, you see that the phi phi just remains itself, simply because it's 2 pi i is 1, okay? So in this sense, the discrete shift symmetry is just a redundancy. It means that, you know, every time you go around the circle, you have the same description of the physics. So in this sense, this is a gauge symmetry. And then, you know, there is no operator in a local quantum field theory that can break this gauge symmetry. The way to see this, you know, you can try to write whatever operator in terms of capital phi. Also, operators that explicitly break the global U1PQ, they will never break this discrete shift symmetry. An example, actually, we have already in the theory of the axon, and we have already used it. So we know that this U1PQ is anomalous, and it is explicitly broken by instantons. And these instantons generate this cosine term, but the cosine still respects the discrete shift symmetry. And you can try to write, you know, whatever terms you can try to in a local quantum field theory, you will never generate any term in the first line of this Lagrangian, okay? So that's pretty much the main point of our work, the main point of our paper, which is very simple. So these terms cannot arise, the terms with the g phi, which are absolutely needed in the relaxation mechanism. And if you insist that you want to put them in by hand, okay, maybe you can do it, but then you have to deal with the fact that by doing that, you're actually breaking a gauge symmetry, not a global symmetry. And at that point, you know, you cannot really call the smallness of g technically natural, okay? So again, this is the main message. I want to do just to make some brief remarks. So these conclusions are actually true for any pseudo-Nambu-Golson boson, as long as it arises from a compact symmetry group, okay? So it's not only true for the action, it's true for Golson boson in general. And I just want to reiterate on these three points here that I have here, that you can have, you can make these three statements and they're all equivalent. Saying that the global symmetry group is compact means that if you spontaneously break the symmetry, the corresponding Nambu-Golson boson is also a compact field. And it will remain compact, also if you break the symmetry explicitly, okay? As we saw before. So this is in other words, another equivalent statement is that, you know, whenever you have a compact group, the Nambu-Golson boson or the pseudo-Nambu-Golson boson will have the shift symmetry, which is a gauge symmetry of your theory. Okay, so these were the bad news. Now let's try to go back and do some positive news. And that's the question. Okay, is there a way that we can save this relaxation game? Can we do something about it? And I'm going to list two options in the paper we'll list four, but the two that I'm not listing here are actually very likely not going to work. Whereas the two options that I'm going to mention here definitely require some more thought, especially option one. So if you guys plan to work on relaxions, I think this would be a slide for you. I could give you some points to think about. So one option is that the relaxation is not a pseudo-Nambu-Golson boson. So it is a non-compact field, okay? Because in order to ride the couplings with the G in your Lagrangian, the field needs to be non-compact. Okay, so what can it be? Maybe it's a modulus, okay? Because you need something with the potential which is, you know, kind of flat. It cannot be a very steep potential, otherwise the game wouldn't work. So we know of fields like this in super symmetry, for instance, like the moduli. The Saxon, for instance, which is the scalar partner of the axion in super symmetry would be one of these moduli. Can you do the game? I don't know, we don't have an answer to that. The, one of the problems here would be that if it is a non-compact field, very likely you're not gonna get a cosine potential because this field has no periodicity. But maybe that's okay too. Maybe you can construct a model with a different back-reaction potential which is not a cosine, which is not oscillating and this still does the job. So some more, something to think about. And I would already say the third point there, I guess. The other option could be still under the category of option one of non-compact fields. Maybe the relaxation is a dilaton. The dilaton being a golden boson of conformal symmetry. And you know that conformal symmetry is a space-time symmetry, so it's a non-compact group. It's a Lorentz group. So if you can spontaneously break, sorry, it's not a Lorentz group, it's a conformal group. So if you can spontaneously break the symmetry and get this guy, maybe that can play the role of the relaxation. But then there are a few questions to address. So the first one is that the dilaton would be a scalar rather than a pseudo scalar. So it would couple to gg rather than gg tilde. So it's not obvious that then you're gonna generate the right potential. And then the other problem is that, at least I am not aware of many or non-suic examples of spontaneous conformal symmetry breaking. Well, there are some trivial examples, but it's not clear that they can be good for this game. Another option could be the following. Could be, okay, let's insist that we're not breaking this discrete shift symmetry. So maybe we can have two sectors, okay? One where you have a fundamental periodicity scale which I call FUV, and the back reaction sector, the one that produces the wiggles pretty much with another periodicity, periodicity F. And now you can take FUV to be N times F. So you can take the first periodicity in the UV to be much larger than the one in the back reaction potential. Now, just simply the fact that the action requires field excursion to be over the lambda over g, you need to ask that the larger scale of periodicity is larger than these field excursions. This just means that you don't want the action to roll over more than one period pretty much. For the reason that I said before, the action is a compact field, so rolling over more than one period doesn't really make sense for the action. So you have to impose this condition. Once you impose that, and you play this game with the two sectors, it leads to a cutoff lambda, which is proportional to this four pi V times N. So N is this guy that appears here, is by how much one periodicity is larger than the other. V is the electric scale. So you see that you cannot raise the cutoff much above the electric scale unless you take N to be huge, okay? And this is somewhat unappealing phenomenologically. But okay, this slide was a bit quick, but I was quick kind of on purpose because I have an explicit example in the next slides to show how you can play this game that I just outlined here in option two. So instead of the action, consider a different ghost on boson, consider a familon, okay? Phi, I still call that phi. So this is the ghost on boson of a spontaneously broken flavor symmetry, okay? And I'll consider this Lagrangian that I've written here. The Y1 and Y2 are Yukawa couplings. ML is a vector-like mass for the fields L and LC, and then MN is a Majorana mass, okay? So now this L and LC are doublets under the wheat group of the standard model, but N is a singlet. And these are new fermions, by the way. So they are charged under a Z2, just this new fermions, so that you can avoid couplings with the standard model fermions, okay? Otherwise they could mix with the lepton doublet of the standard model. Okay, so now you can notice, if you give it a little thought, that if you look at this Lagrangian, and if you set MN to zero, this Lagrangian respects a global U1 symmetry that we call U1 and L. And you can see this by assigning the following charges as I have in the table to the various fields, okay? So all the standard model particles are uncharged under this U1. The E to the I5 over FUV as charge one, and then notice here in the Lagrangian that I have like a two N here at the top, okay? And because of that, these fields N, L, and LC have charge N. This small N is the same N that I had in the previous slide, okay? We will see in a second why that is true, I mean, and how the whole thing works. Now take this MN to be non-zero. If that is true, I mean, if that is true, we're gonna impose that actually. So the symmetry, the U1 symmetry, which by the way, we have already assumed to be spontaneously broken at the scale FUV, okay? So at that scale is spontaneously broken, and now if we turn on MN, it will be also explicitly broken, okay? By the Lagrangian that I had in the previous slide. And this explicit breaking is going to generate this Kalman-Wember potential for the field phi, which has this form that I have here, okay? So again, we get a cosine. And the other important thing here is that we get a proportional to V squared rather than V, as in the action case. I could have more comments about that, but in the sake of time, I'm gonna leave them aside for the time being, just to keep with the logic. Now the F that you see inside this cosine at the denominator is this FUV divided by 2N, okay? So this calculation is actually straightforward. You can really compute the Kalman-Wember potential explicitly, and that's what you find. That's the leading term. I mean, there are other terms, but this is the important one. Now you see that the periodicity now is F, which is the FUV divided by 2N. And this is the back reaction potential. Now suppose that we have another sector, which I'm not gonna bother writing down, but you can have Lagrangian similar to the one that I have in this sector with other fields, okay? Instead of L and N, you could have other fields and concoct Lagrangian in such a way that again, you explicitly break this U1. But now you want to have these fields with charges of order one, rather than N, okay, under the U1, and you want these fields to be heavy. So their masses will be at a scale similar to the cutoff of our theory, okay? The cutoff lambda. So with that, we could generate terms like this, like the two that I've written here. So you can generate a terms proportional to M squared to sine of phi over FUV times H dagger H, okay? And now you see this is the oscillatory part for the mu squared, so you can go from positive to negative. And now from here, if you have this term, you can think of closing the Higgs loop with the HH dagger, and that will generate the term in the second line, okay? So we'll just generate a potential for phi, which is again oscillatory. But now the oscillation here, because they are divided by FUV, which can be much larger than F, will give you pretty much, that this V phi potential pretty much will give you like a very long wave in this cartoon of the potential for phi over phi, okay? And along this very long wave, you will have these small wiggles that are given by the VCW here on top, okay? Because the periodicity of VCW is much smaller. So those wiggles will be very tiny compared to the long wavelength of this V phi. And now pretty much you can play again the same game. So you will have the field phi rolling down its V phi potential. You have to satisfy all those rural conditions that I enumerated before. And then this field is gonna stop when again the derivative would be zero, okay? So it will stop on one of the small wiggles once you satisfy the condition. So if you work that out, you get this, together the cutoff then is gonna be given by this expression for this specific model. And now you see that F u over F is pretty much N or two N in this specific case. And now if you plug in the numbers, as we did as I did here below, you find that this cutoff can be of order, let's say five TV. If I take like this parameters ML and ML to be 900 JV, I take the Yucca was at the edge of the perturbativity limits. I just, I want to be optimistic and push this cutoff as high as I can. And I took this N, the charge of the fields in my back reaction sector to be 10, okay? Now you see that this behavior is N to the 1 fourth, which means that even if you take like really large charges, you know, you're not gonna improve so much on the cutoff. Okay? You can in principle, but it's not gonna buy you much. I'm pretty much done because I guess my 30 minutes are over, according to my clock. So this is the summary. So what I pretty much try to argue here, what we try to argue in our paper, is that the original relaxation, the original idea, it cannot be an action, simply because the action is fundamentally a compact field. So there is no way that in a quantum field theory, you can generate terms that will make this field non-compact, okay? It's compact by definition pretty much. Now if you want to insist that this relaxation be like a pseudo Nambo-Golston version, you can. One way we showed that you can is, you know, by introducing two sectors, this is what I showed in this very last example, that break explicitly this original symmetry with two different prejudices. But as I showed you, then the resulting cutoff is not gonna be so high, okay? The best that you can do is, you know, is that you can achieve a cutoff of order TV, okay, a few TV. So this is maybe okay to address the little hierarchy problem. It's like you are postponing, you know, the energy scale with respect to see new physics just to a few TV, okay? So maybe just barely enough that the LHC can barely discover something. But obviously, I mean, this is less ambitious than the original proposal, where, you know, the goal was to push the cutoff as high as you could. Another possible option that I put forward in this talk is that maybe the relaxation is indeed a non-compact field, okay? So it's not a pseudonymic Gaussian version. But as I mentioned, you know, model building in this direction requires more thought and it's definitely, it's not trivial, it's not easy. So it's not clear that this is even possible. But, you know, we thought about it not so much, so this is definitely something to think a bit more about. And I want to conclude that, you know, despite the fact that maybe I sounded negative with respect to this relaxing idea, I don't want to be negative because I actually think this is a very clever idea, okay? In principle, it's very simple. So maybe it would be nice, you know, if more people thought about, you know, ways to refine it and to make it work. Okay, that's it. I'm gonna just leave you with a list of papers that are relevant to these various discussions about the relaxation. Thank you. Thank you very much, Lorenzo. I think now it's time for questions. Before that, let me remind you that you can, of course, ask questions via the Google Plus Q&A, so, here. Via Google, via Twitter, sorry, using the hashtag L-A-W-O-P. So I know a few guys have questions. And how do I see the questions, Nico? So there are some people here that, in principle, can ask questions live. Otherwise. If I may, I have a question. Sure, sure. I'll put it on the Q&A. So my question is if the cosine term is really invariant under the phase shift, because the amplitude of that term is dependent on the wave of the Higgs, which depends on the mass, which should depend again on the G coupling, right? So, the way you see it, if you make a shift of phi, you should also make, the wave might also change, right? So how does that get solved? Right, so there will get solved, for instance, like in, can I go back to my slides here? I guess so. Actually, you still see my slides, right? Yes. I see the list of relevant papers. Yeah, yeah, sorry. So let me go back to this, for instance. So now you see that the terms that determines the Higgs wave, right, in this VH, is also cosine. So that is also invariant under this shift symmetry. So now everything- Right, no, I was asking about the first expression. I mean, the original Lagrangian. Oh, in the original Lagrangian, right, no. So in the original Lagrangian, that would not be invariant. But then the question is, how can you possibly generate that term? So you see that the cost term is multiplied by the wave. Here, you mean, right? So the wave, yes, exactly, exactly. So the wave, in principle, will depend on the value phi. Yeah, that's true. So if you make a shift of phi, then that would also change the wave. That would also change the wave. And then you're saying then, not even that term would respect the discrete shift symmetry. That's my point, yeah. Yeah, that sounds even worse. Yes. I think you're right, actually, yes. The wave definitely will depend on the field phi via the mu square term. And then, yes, that would be non-invariant under the discrete shift symmetry. Right, so then it's not natural. The wave is given by that combination, right? If it's given by this lambda square, yes. Right, so then you have the dependence on G over there? Right. Yeah. Yes. Okay, so there's no solution to this so far. Okay. I thought there would have been some sort of way out. No, I think, well, people have not even addressed that. But I mean, the way we addressed it would solve that problem too, because now also the way the mu squared is generated, it's the way mu squared is defined, is also, it respects the shift symmetry. I see. So if everything respects the shift symmetry, then everything is fine, because then also the verb will respect the shift symmetry. But in this model, as it is on this slide, then it's not. I agree with you. Okay, thank you. Sure. Okay, great. I can see another question by Roberto Lineros. So his question is, can this idea work on other global symmetries like V minus L? Sorry? So Roberto, is it asking whether this idea could work with other global symmetries like V minus L? Yes, but with the same caveat, pretty much, right? I mean, so the symmetry that we use is a flavor symmetry. So sure, you can do whatever global symmetry if you want to, as long as we insist that you have to respect the discrete shift symmetry. Because indeed, if you write a model, nothing will break the shift symmetry. But yes, in principle, you can write our model with the family on field is just an example. I bet you can do this with the U1B minus L or whatever global symmetry you want to, as long as you cook up a model where you have both the spontaneous breaking, which I didn't even write that sector, okay, but that's easy to do. And you write also part with the explicit breaking. So then, yes, then it would be a very similar game, but I bet you can do it with that as well. Thank you. I have a question, actually, about the number of e-fold. You mentioned it's order of 10 to the 50 or something like this. Yes. Is that really a problem? Of course, we are used to something like 60 or something like this. Yes. But is that really, do we really have to worry about this or not to just... The first order answer is no, in the sense that, I mean, who cares whether inflation lasts for a short time or if it lasts like 10 giga years as we should, like with this huge number of e-folds. In principle, that's fine, although, I mean, it's not very aesthetic in a way, but whatever, that's another question. And I know that some people are bothered by this. So I think in principle, there is not much wrong with that, but then probably you can run into other trouble, but I'm not sure about it. At least also the original authors, they definitely mentioned this in the original paper and they say that in principle, it's okay. It's a weird scenario that we're not used to, but fine. It's just a super long inflation period, but it's fine. Or maybe some people will come up with a reason why it's not fine, but as of now, I cannot think of any. Okay, very good. Are there more questions? Eduardo. I cannot hear you, okay. Perfect, yes, I can hear you. So I just want to make sure I understand the basic complain on the original proposal which you have right there, right? Oh. Yes. Don't do that. Don't do that. Okay. What happened? Thanks. I see what happened. Okay, so let's see. So just to make sure, so you take this Lagrangian and you don't ask where the terms that break the ship cement we're coming from but just take G to be small by hand. Yes. Yes. So is there a statement that a large corrections to these terms will be generated radiatively or what is it exactly? So you're saying it's not technically natural but what does it exactly, what's the problem? Okay, no, that's a good question. So if you just, if you start from this model, as you say and you just think of this as an effective field theory, okay? And then you say, I don't know whether this, that ship symmetry is gauged or not, okay? So yeah, it's true that if you write those terms with the G's, they will only be multiplicatively renormalized, okay? So if you start with G small, it will remain small, okay? There are no radiative corrections that are gonna make it large. So in that sense, your theory is under control. But then, you know, it's like saying you could give two definitions of technical naturalness. So one is this, is just that, you know your parameters are renormalized in a multiplicative way, okay? And this model would respect that. And the other one is that, you know you have a global symmetry. So in the limit that your parameters goes to zero, you recover the global symmetry, your theory is actually technically natural. So in that sense, it would be fine. But then again, you know, we know actually that if phi is an axion, if it is a ghost on boson, we know that the symmetry is a gauge symmetry. It's not a global symmetry. So for sure, the second definition of technical naturalness wouldn't be respected because you're not recovering a global symmetry and you recover a gauge symmetry. I also asked about that. So let's say that now we imagine we can push to this, put in the axion, with some terms that break the global symmetry explicitly, right? Sorry, I didn't hear you well. Can you repeat that? Sorry. Yes, I'm saying, now let's imagine we consider the UV completion to this theory where you written in terms of the fulaxion, capital phi, right? Yes, yes. Which has only an approximate global symmetry. Apart from the instanton effects, you could imagine some other expertise breaking that you should think that people talk about by gravity, whatever, okay? So it's only an approximate global symmetry. Yes. And at those scales, let's say above the scale f, you care about not just the discrete part, the discrete subgroup, which you are labeling as gauge, but about the global part, the full continuous global part, right? So maybe at a point you can use the usual arguments of taking the naturalness, so you have an approximate symmetry that protects certain parameters. And then below the paycheck breaking scale, okay, you remain with this transformation, which is really not doing anything to the original scales, right? Which is what you mean by gauge. But at some point, this full global thing was buying something to you, it seems to me. And then if below that, you don't see any large rightive corrections, then I'm not quite sure I see that there is a big problem here. So I'm telling you about the reason. Fine, but if you want to do this, right? As you say, if you want to start from the UV theory, which I'm totally happy with, and you know, first you write things in terms of the capital phi, then you have to tell me what term you can write in the UV that in the low energy will give you this, you know, I agree, I agree on this important question, but also it seems to me, from what they remember from their paper, they only require some, so this V of G phi, right? They don't have to be this, so this linear plus quadratic and so on, because it's just some Taylor expansion. It was not important or super important that, so maybe your example's falling to that category. So you're generating a complete way, the explicit breaking terms, and I don't know if maybe they can be understood that some expansion where it's these linear terms that are playing some important role. No, actually, no, the point that I'm making is that in a quantum field theory, if you start thinking about the capital phi field, there is no way that you're gonna generate these terms for the small phi, the small phi for the axiom.  Right, the reason why you cannot generate them is because if you have the terms like G phi in the potential G phi square, that breaks that periodicity. So it's the same statement again, like phi is a... But there was something in the period that was... Some explicit breaking, right? But explicit breaking, again, like you can try to write whatever operators you want. You can write explicit breaking operators in terms of capital phi, right? And it's very easy, you can do that. And if you do that, all these operators will respect this shift symmetry. So everything in the end of the day will be given in terms of cosines or signs of phi where phi is the axiom. You can try, you can try to write down whatever operator you want. At some point you're saying that it seemed to me that in the proposal it was not really... They just did some potential and it had no g times phi. Yeah, but in the proposal, they don't want to think about the UV theory at all. They just say, we just don't know the new energy, right? And in that sense, again, the only thing that I can agree with is, as we were saying before, that if you write those terms, they will not be corrected radiatively or not much because it would be some multiplicative renormalization of those terms. So it's fine if you impose that g has to be small by hand, it will remain small. It's stable under quantum corrections. That's okay, that I will give you. But then, for this model, you really need this ingredient, for instance. You need some field that couples to gg dual. So the only examples that we can think of is like ghost and bosons, right? Do that kind of stuff. And if you, as soon as you think of a ghost and boson, you know where it comes from and you know that it is compact and you know that there is nothing in your quantum field theory that can generate those terms that make it non-compact. So let me just wrap up. So is it fair to say that your criticism is not so much that g is technically natural or not, but that they have not provided a way to get these kinds of terms that are necessary for the mechanism. So you need some UV completion and then you can, you will not get these things, right? Yes and no. So my complaint is actually that g is not technically natural in this model because in order to get, at least to get like the cosine term or so, I mean, we know how to get that from a ghost and boson, from a P and GB. Might as well be the property that is in the category of normalize. So that's what was important for the application, right? That they may have to take g to be 10 to the minus 33, but so what, once you do that once, it will not get that corrections. That was important aspect for their proposal, no? True, true, definitely. But again, but then, you know, these arguments of technical naturalness kind of breaks down in a way, in the sense that, you know, now with g you're breaking a gauge symmetry rather than a global symmetry. And then, you know, fine. Then you are explaining like the hierarchy problem with the crazy number, which is 10 to the minus 30 or 10 to the minus 33, which is not justified anymore. But it is not justified because you have because you have not seen how to get this term from the UV completion, not because, okay, it doesn't matter, maybe it's just definition. Yeah, yeah, yeah, I mean, yeah. I mean, for instance, like from the UV completion, what they also mentioned in the original paper at some point and people, you know, keep also talking about this, they can think, they can say, okay, we can think of some kind of action monodrome, for instance, which is this mechanism in string theory involving D-brains, which I know very little about, that indeed, you know, where you have this axion which becomes non-compact. But okay, so first of all, I don't understand the mechanism in string theory. It's not clear at all that it has like a straightforward or acceptable interpretation in just quantum field theory. And another way also of setting this problem is that if you have to claim like some string theory to provide something that will give you this potential in terms of G-fi, first of all, it's a bit of hand-waving, one. And two, it's like if you need to say that string theory will give you this, well, string theory might as well solve the hierarchy problem to begin with. So you're not looking anymore for a solution of the hierarchy problem just within the framework of quantum field theory. You're doing something else, you're going beyond. Yeah, but it might be that string theory is just giving you these mechanisms for something. We don't know about that. Yeah, no, no, no, no, no, no, well, whether string is totally fair, yeah, it's a, yeah, maybe there is a UV completion indeed that can give you something like this. And what we're saying is that UV completion will not be a compact pseudonym blue-rosin version. Pretty much, I think that's our statement, a way to put it. And maybe there is another UV completion where phi is a non-compact field that will give you, in the low energy, this kind of Lagrangian and then the mechanism goes through and that would be wonderful actually. All right, thanks. Sure. Okay, thank you. Cassia, very last question from Diego Restrepo. He's wondering about the two fields, the LNN that you introduced, you want them to avoid mixing with the solar model formula via this Z2 symmetry, right? Yes. So is the lights of these guys also stable because just because of this Z2 symmetry? It is actually, which is, in principle, it's a nice feature of the model because that would be a winged dark matter. So this is, indeed, like if you look at this model, this is exactly the singlet doublet model of dark matter. It's also well-studied in the literature. So that's exactly what it is. I mean, so in the singlet doublet dark matter model, they introduced the cell and the LC, the doublet and the SU2, and the singlet N. And then the lightest neutral state is stable because of the Z2 and the dark matter candidate. It's a good dark matter candidate because it's a WIMP, right? I mean, this is all at the electric scale, so it has weak impact. And it's the usual, you know, WIMP miracle, freeze-out. So this guy would lie in the GV scale or something like this, TV? Yes, exactly. Around the TV, slightly below, probably, let's say, 500 GB in that range, definitely, yes. Nice. It would be a WIMP dark matter candidate, yes. Very good. So I think there are no more questions, so thanks a lot, Lorenzo, for a very nice webinar. Thank you, guys. Thanks a lot. One more quick question? Yes. So maybe one or two slides after this one. No, let me see the slides. Yes. Got it. Two slides after this? Maybe. So when we put the final expression for lambda in this model that's like TV and so on, yeah, this one. So the reason there that you don't consider the ML and AM to be larger in this lambda is that it's because you want them to be below the cutoff scale, so some self-consistency condition or why. So I think this ML and AM, I think that's lambda, but only to the fourth root. Yes. So you took 900 GB because that makes it kind of self-consistently below the cutoff, is that TV? No. Yes and no. Actually, the issue is a bit more subtle, so the quick question might be a bit longer. So the point is something that I mentioned very quickly in passing. If you look at this slide here, you see that you have this V squared here, dependence of the Kolman-Weinberg potential in the back-reaction potential. So this is actually an issue of this model and of all the other models that also are in the other papers, in the sense. Because as I mentioned, the original action model is excluded anyway because of the strong CP problem. So what you have to do is you always have to couple this action to a hidden sector. As here, we couple the familon to, let's say, a hidden sector, these states that are charged under a Z2. So usually whenever you do that, you get the back-reaction potential instead of being linear in this quadratic in V, which means that I can substitute V squared. Actually, you can see this explicitly if you do the calculation by keeping the Higgs explicit rather than the VEV. You see that this V squared will be replaced by H dagger H. And that means that here, again, you can close the Higgs loop from this potential.  And now you want to make sure that that contribution. What I'm saying is that you have the Wiggles also before the phi, the field phi, goes through the point where the mu squared terms goes from positive to negative. So when mu squared is positive, you still have these small Wiggles because of this. And then you have pretty much to make sure that your theory is under control. So you want to cut the quadratic divergence of the Higgs loop with something. And there is a long song and dance. Actually, you write this in a paragraph in the paper that tells you that then all these states better be below the TV scale, ML and MN. MN is special. MN could be a bit larger. But MN, the Majorana mass, has to be definitely not larger than that than what I have on this slide. So it was kind of quick if you have all the reasons to be confused. So just say it in different words to make you understand that. So the reason you get this TV scale for these new states is closely connected to the hierarchy problems and quadratic divergences in some sense. Yes, definitely. Yes, it is. It is. For sure for MN, the way you can cut these quadratic divergences or the way you don't want to introduce these quadratic divergences, what you have them, you have to cut them, is to make, by making MN light, pretty much. And the way we do this in the paper is by introducing actually another state that we call NC.  So you have to work a bit harder. And then at the end of the day, there is this other mass parameter in the full Lagrangian where we add another extra state, which I haven't written in these slides. And that would be what cuts the Higgs loop pretty much, what cuts the quadratic divergence there. And as a consequence, this MN that is given by the CSO mechanism would have to be lighter, definitely lighter than a TV. Whereas MN could be a bit larger, too. But then the other thing is also remember that these guys are stable, at least the lightest guy of the combination of LC and N, the neutral guy. And that would be dark matter. So if you make it too heavy, then you run into problems with dark matter abundance, pretty much, as well. So that's yet another reason why you cannot take these guys to be too heavy. So I'll give you a long answer. The short answer, pretty much, is yeah, it's related both to the hierarchy problem. Because if you take them too large, you wouldn't solve it. For this issue of the Higgs loop, that would give you an extra contribution to the Wiggles. That's the first point. And the second point is also the dark matter. You would get overabundant dark matter if you make it too large. All right, thanks. Sure. OK, so thanks Lorenzo again. Sure, thank you, guys. You guys enjoyed this webinar. So let me remind you that in two weeks, we will have a webinar by Diego Ristisabal. He will talk about leptogenetics. And yeah, hope to see you soon. Thanks a lot. Bye, guys. Thank you. Ciao. See you. Ciao, Nico. Bye, bye.


This paper examined the potential of Sentinel-1 synthetic aperture radar, SAR, to monitor crop conditions. It found that the cross-ratio, CR, which is calculated from the ratio of VH to VV backscatter, was the most effective indicator of vegetation dynamics. The CR was able to explain up to 87% of the variation in vegetation water content, VWC, for corn and 63% for winter cereals. Additionally, it was shown that the CR could be used to predict VWC with an accuracy of 0.34 for oilseed rape. Furthermore, random forests were used to analyze the relationship between CR and other variables such as soil moisture and microwave indices. These results suggest that the CR is a reliable indicator of VWC and can be used to accurately predict crop growth. This article was authored by Mariette Rugdenhill, Wolfgang Wagner, Bernhard Bauer-Marschlinger, and others.


On this 75th anniversary of the Universal Declaration of Human Rights, we honour the wisdom of its drafters. People like Bertha Lutz of Brazil, PC Chang of China, Minerva Bernardino of the Dominican Republic, H.J. Mehta of India, Eleanor Roosevelt of the United States and Jesse Street of Australia. All were seized of this simple fact. Human rights apply equally to all people, no matter who you happen to be or where you happen to be born. This was a global charter for the whole human family across east to west and north to south. They enshrined in this declaration that every individual should be able to live equally and that no one should be held in slavery or servitude. Article 23 guarantees that freedom. It affirms that everyone has the right to work, to free choice of employment, to just and favourable conditions of work and to protection against unemployment. Modern slavery is widespread and insidious. It is woven into our supply chains and our economy, exacerbated by disparities of power and opportunity in our global economy. And there is no way it can be eradicated unless wealthy developed countries take the lead by ceasing to consume goods made by forced labour. Australia recognises our responsibility to act at home and we are pledging to strengthen Australia's domestic regulation. The Modern Slavery Act, we are also establishing an anti-slavery commissioner. These pledges are matched by support for work internationally led by Australia's ambassador to counter modern slavery, people smuggling and human trafficking. This pledges one aspect of our ongoing commitment to protect and promote universal human rights. We do all we can in the ways that we can, motivated by principle and acting with purpose. We pledge to work with our region and with the whole UN membership to improve human rights for all. By listening to each other and by acting together all of us can choose to shape the world for the better.


This message is to all staff from the SCP Foundation occult study's department. You are advised not to look at the sky under any circumstances. This involves looking at clouds, the sun, stars, or the moon. If you observe the sky in any way, you may experience the following effects. Headaches, stomachache, confusion, crying, visitors, a itchy back. Visions, extreme anger, extreme sadness, vomiting, murderous intent. If you look at the sky and experience these effects, isolated even at E. You are likely under the influence of a dangerous chloropent entity. The SCP Foundation is working on a solution to this situation. Until such a solution has been reached, be strong, be safe, and stay inside. The Foundation thanks you for not looking at the sky and cooperating.


Welcome to NPTEL NOC, an introductory course on points at topology part 2, continuing with the study of filters, today module 33, ultra closed filter, filters and ultra filters on a given set, have nothing to do with a particular topology on x, though they control the behavior of all the topologies on x to a large extent, namely via the notion of convergence, right. Now we shall introduce a subclass of filters which depend on the given topology tau on x, so that is the difference between these filters and ultra filters and ultra closed filter, they have the name says that closedness comes because we are referring to a particular topology tau on x, a filter f on x is called a closed filter, if it has a base consisting of closed subsets of x, so I started the topological space, so closed subsets make sense, it is called an ultra closed filter, if it is maximal in the correction of closed filters, so that is the definition of closed filter and ultra closed filter. Now if an ultra filter is closed filter, then it is ultra closed filter because it is already maximal in the collection of all filters, on the other hand you start with an ultra closed filter, this may not be an ultra filter because it is maximal only in a subclass, there may be larger ultra filter which are not closed, so this difference you have to keep in mind that is all. Now let me have some examples, singleton capital X is always a closed filter because singleton X is always closed in whatever topology you take with respect to any topology, on the other hand every filter is a closed filter on a discrete space because every subset is closed also, so you can take the filter itself as a base if you like, no problem. So these are some easy examples of closed filters, the co-finite filter on an infinite set is not a closed filter, for if it were then it would contain a non-empty finite subset because the only non-empty finite, non-empty closed subset are the whole space or the finite subset, so if it contains X as a base then it has to be, if X is a base then X is the only element, if it is a co-finite filter X cannot be a base, so it must have finite closed subset as a base but the moment it is non-empty finite subset that is contradiction because it is complement will be also there, as a member all the open subsets are here, non-empty open subsets are here that is not possible, so conclusion is that this co-finite filter which consists of all open subsets other than the empty set in the co-finite topology is not a closed filter, so I have given you both examples, easy examples of closed filters as well as not closed filters, okay, let X be a P1 space, then I can give you more examples, for each X binomial to X consider the ultra filter FX, singleton X being a base for it and singleton X is closed, this FX is a ultra, I know it is a closed filter, already it is ultra filter therefore it must be ultra closed filter, also note that FX converges to X and X alone, why because FX contains an X, alright, so these are the important ones and easily available ultra closed filters, of course you may expect that there are and there may be many other ultra closed filters on an infinite set, okay, every closed filter is contained in an ultra closed filter, so this is similar to every filter is contained in ultra filter, of course again the proof is John's lemma you have to verify something, the verification is very easy here also, the proof is as usual appealing to John's lemma, take A to B the collection of all closed filter containing a given filter automatically it is non-empty, if lambda is a chain inside this inside this family A, each element of this chain has a closed base B I, B I is a closed base for F I, F I is a chain, okay, then take the union of all the members of all the elements in B I, so union of all B I that family will be a closed base for union of F I, this is what you have to verify, okay, because it is a chain this is impossible, okay, again union of F I is a filter is easy just like in the earlier case, so it is a closed filter, therefore it is a member of lambda, if member of this A and it is an upper bound for all members of this one, so that means condition for John's lemma is satisfied, therefore conclusion of John's lemma says that there are maximal elements, so every filter is contained in an ultra closed filter, okay, similar to the theorem 5.7.5 to 0, which is nothing but characterization of filter, ultra filters, the 3 characterization we had given, okay, we have the following characterization, it is not so strong a little weaker because we have only 2 of them here, okay, A and B, exactly similar, F is an ultra closed filter, the second one is not arbitrary subset but only for open subset, every open subset U in X either U or U complement must be inside F, if this condition is satisfied it must be ultra closed filter, start with a closed filter of course, you do not prove that it is closed filter, only ultraness is proved, okay, the proof is more or less same but you have to see why the openness is coming here, this condition is obviously weaker condition, it is not for every open subset, right, after all it is maximal is also inside the family of closed filter, so that is the trick, so let us see, let us go through the proof correctly, A implies B is not very easy, we are not very difficult, start with a closed filter, ultra closed filter, let B be a base for that, B consists of only closed subset and it is a base for F, that is the meaning, given an open subset U of X, consider the family B union complement of U, that is a closed subset, so it is a closed family, family of closed subset, right, if F union U complement has finite intersection property, in fact you do not need the whole of F, if B union F, this is finite property is enough, then B union U c also has finite intersection property, so these are actually, these are different only, and hence it would generate a closed filter F prime containing F, okay, the moment you assume this as a finite intersection property, there will be a filter F prime, that will contain this F, because it contains B, okay, now F is ultra closed, therefore F must be F prime, which just means that U complement is enough, okay, if it is not the case, means what this thing A has, it does not have finite intersection property, then what happens, that is otherwise, there exist an A inside F, such that A intersection U complement is empty, okay, that is the finite intersection property is violated, this just means that A is contained inside U, therefore A is inside F, so U is also inside F, so either U is there or U complement is there is what we have proved, okay, so B is proved now, now let us prove B implies A, okay, suppose there is a closed filter F prime, such that F is contained F prime, we want to show that F is equal to F prime, let B prime be a family of closed sets, forming a base for F prime, okay, there is one, because F prime is a closed filter, for any B inside B prime, suppose this B is not in F, okay, I am just supposing this is not in F, then the complement of B must be in F by condition B, but F is contained F prime, so B complement is in F prime, B and B complement both are in F prime, that is a contradiction, because intersection will be in F prime, therefore this B must be inside F, since this is true for all B, this B prime, the curly B prime must be inside F, okay, but B prime generates F prime, so F prime is inside F, so therefore equality holds, okay, so the proof is more or less the same, but you know the flavor is different because you have to use the openness and closeness and so on here, that is all, but condition C is missing here, okay, see now you understand why last time we put this condition at the last, okay, and then proved A implies C, C implies B and B implies A, this condition is nothing but given A union B inside F, A is there or B is there, in the case of ultra filters, now similar condition but slightly different can be expected, but no, this only one way, one way implication is there, okay, the corresponding condition C is not equivalent to being ultra close filter, so let me state it separately, do not get confused with the earlier theorem, this proposition says suppose F is an ultra close filter, then given U and V, two open subsets of F, this condition holds, what is this condition, what is this statement, this statement is if and only statement, U Union V is inside F, if and only if U is inside F or V is inside F, actually one way is obvious because F is a filter, okay, if U is there or V is there, U Union V being a superset is there, so what we have to show is that this larger set is there, why the smaller, one of the smaller one is there, okay, under the hypothesis that F is a close filter, ultra close filter, okay, so one way is obvious, this implication is F is an ultra close filter, U and V are open subset such that the Union is inside F, we have to show one of them is inside F, this is not true means what, let us examine that, suppose this is not true, then it follows that the complements U C and V C must be inside F, this is not true means what neither U is there nor V is there, but then the complements C must be there because U is an ultra close filter and we can use the previous theorem, okay, therefore the Union, U Union V complement which is nothing but U complement, intersection V complement that must be there, but now you have a problem, we already assumed U Union V is inside F, but you are assuming that intersection, U intersection V intersection that is also inside F, this is empty, okay, it is complement and this one both of them cannot be there, that is all, by a simple induction we get the following important result, if X is written as a Union of finitely many open subsets V i and F is an ultra close filter on X, then one of the V i must be in F, okay, so this is very easy because what we have proved is that if U and V are open subsets and U Union V is inside F, then one of the U and V must be inside F, right, to begin with every filter contains the entire set X, therefore you can just write X as I raise to 1 to n minus 1 V i Union V n, apply this criteria, either V n is there, then we are done or I raise to 1 to n minus 1 V i must be there inside F, now you apply induction, okay, so one of the V i must be inside F, alright, so from this one let us reduce another important result here which is characterization of compact spaces in terms of ultra close filters, remember in earlier theorem we had characterized the compact spaces in terms of ultra filters, every ultra filter is convergent, that is the condition which ensures that X is compact and conversely, so exactly same result is known, space X is compact if and only if every ultra close filter in it is convergent, okay, so what is the idea, let X be a compact space and F be an ultra close filter in it, suppose F is not convergent to any point, then you will get a contradiction, very easily, namely it does not convert to any point means none of this nx, the neighborhood is contained inside F, that is the meaning of that X does not convert to F, this entire neighborhood contains at F is the same thing as X converges to F, F converges to F, what does this mean, for each point X inside X we have a neighborhood Vx of X such that this Vx is not in F, one neighborhood for each point, but when you vary this point you get a covering for X, but now X is compact, so you get a finite covering, okay, as soon as you have this one, this corollary says that one of the VI must be inside F, but that is a contradiction just now none of the VI Vx is inside F, okay, so compactness implies that every ultra close filter converges to some point, okay, it may convert to more than one point also, nobody ensures the uniqueness, now let us put the converse, assume that every ultra close filter on X is convergent, let C be a family of closed subsets of X with finite intersection property, okay, we have to show that the entire intersection of members of C is non-empty, so that is enough to show that X is compact, right, so let F be the closed filter generated by C, because any family which has finite intersection property generates a filter, but that will close filter because this will be members of all this here they are all closed subsets, okay, so we get a closed filter contained in a ultra filter, ultra close filter, so let F prime be an ultra close filter contained by John Slema, now X be a limit point of this F prime, because we are assuming that every ultra close filter is convergent, so let X be a limit point of limit of F prime, this means that an X is contained inside F prime, if X is not in C for some C inside C, that would have meant that complement of C is a neighborhood, that complement is a neighborhood, the neighborhood is contained in F prime, that means complement is in F prime, okay, but all the members of C are inside F and hence inside F prime, therefore X must be inside C, for every C inside C, that just means that the intersection has the common point X, so that is non-empty, okay, so characterization of a compact space in terms of ultra close filter comes out, see we have to prove this F fresh both ways because if every ultra close filter is convergent, this does not mean immediately that every close filter, every filter, every ultra filter converges, right, because there are many more of them, on the other hand if compact space shows that every ultra filter is convergent, but none of the ultra close filters may be or may be a ultra filter or some of them may be some of them are not, right, ultra close does not mean that it is ultra filter, so either way it is not true, but so we have to prove it fresh, but on the other hand if you look at the proof, it is not all that different, of course this is little harder, we have to use the comp, we have to use the closeness and generate close substrates and so on, okay, a non-convergent ultra filter is supposed to indicate the presence of a hole in the space which is making the space non-compact, okay, because if it were compact then every ultra close filter would have been convergent, okay, so this ultra close filter there is no space for it to be convergent because that point is missing, that is the idea of that there is a missing point in the space, so and that must be making it non-compact, so this idea is similar to the presence of non-convergent Cauchy sequences in an incomplete metric space, by the very fact we know that in incomplete means a metric space incomplete means there is a Cauchy sequence which is not convergent, so not convergent Cauchy sequence indicate there are holes inside a metric space, following a similar track as in the case of you know completion of a metric space, we may try to fill up these gaps by including all ultra filters on X, okay, so that could be one way of looking at it, but then you are warned already by the above theorem that you do not need all ultra filters, if all ultra close filters are convergent then the space will be compact, therefore it should be possible to get a compactification by just taking care of all the ultra close filters, so they should be enough to fill up all the holes, so these are all just surmising, I mean we are just you know loud thinking may say trying to do oh this may be true then finally you have to prove all this, let us concentrate our attention on the set WX of all ultra close filters on X, okay let us not give up this idea that is all we have to work of course, so WX is nothing but you know set of all ultra close filters on X, okay, next going back to the analogy again of completion of a metric space recall that eventually recall that eventually constant sequences correspond to the points in the space itself, right, if you have some a18 an and then all an plus 1 etc are equal to X, X, X, X, X that sequence automatically converges to X and then what we do in the completion the element X was identified with this Pochise sequence or the other way around whichever we will do, so that was the map from the space X to the collection of all sequences and then of course we introduce there some equivalence relation and so on, right, so that was the idea, so we can try to do that namely instead of sequences first of all you have to go to next eventually constant nets is the key word now here, on the other hand we do not want your nets but we will use the bridge from nets to the filter which you have defined earlier, right, right, so what do you get if S is an eventually constant net the corresponding FS what is that that will be the singleton atomic filter FX, okay, so you see we are coming to this atomic filter FX it is supposed to represent you know a prototype of eventually constant sequence that is the whole idea, oh that is how it is convergent also all the time, okay, in our definition it is also an ultra close filter provided X is T1 till then we do not need that, okay, all FX are closed filter if it only if X is a T1 space that brings us to make a blanket assumption that our space X is a T1 space because we just do not want to give up this WX, right, ultra close filter, so we better assume X is a T1 space before we go further, okay, in which case all happens if this subset of all atomic filter will just correspond to the points of X, okay, so this way WX can be thought of as a set which contains X by this embedding by this injective mapping by this identification X going to FX, so we have enlarged our space what remains we must put some topology on this one so that it becomes compact that is the first thing then we must examine that the map X going to FX is injective that is okay continuous and an embedding so that the image is dense this all we have to, okay, right, it is also clear that each FX converges exactly to one single point that is very important for us, right, that is an extra bonus though we are we were not bothered about that, okay, but that is because of some strange reason there is no T1-ness or T2-ness involved we do not want to bring the T2-ness here usually a filter will convert to a unique point if you have a host of space, okay, that is not the reason these are very special filters FX they will convert to only one single point we cannot expect such uniqueness behavior by other ultra filters unless we are ready to make further restriction that X is host of, right, so therefore it seems that we may need to introduce some kind of equivalence relation on the set WS, okay, so that the equivalence classes may be better qualified to become a compactification just like in the case of completion of metric space, okay, luckily it turns out that we do not have to worry on this point this process of compactification is much simpler than the construction of a completion of a metric space that is going to be the topic of discussion for us now which we shall carry out in the next module, okay, so this is the motivation for considering the so-called Wallmann Compatification as we will do next time, thank you.


Welcome to All Hands Update, here's what's happening around your Navy. A U.S. Navy destroyer operating in international waters experienced several close interactions by Russian aircraft April 11th and 12th. While conducting landing drills with an allied helicopter, two Russian Su-24 jets made numerous close range and low altitude passes. The following day, a Russian helicopter conducted circles at low altitude around the U.S. warship, and two Russian jets conducted low altitude close range passes. That caused Donald Cook to secure flight operations. U.S. officials are using existing diplomatic channels to address the interactions, while the incidents are also being reviewed through U.S. Navy channels. USS Port Royal and USS Coronado successfully completed an independent deployer certification exercise along the coast of Southern California. The exercise is designed to provide realistic integrated training to assess independently deployed units while exercising advanced level mission skills. You can read more about what's happening around your Navy at Navy.mil.


Patriot Prime Reviews is a channel for adult collectors and may not be suitable for children under 13 years of age. Viewer discretion is advised. Hey, what's going on guys? Patriot Prime here once again with another ToyHacks.com sponsored video. But before I get started, I need you guys to do me a favor. If you're watching this video right here and are not a subscriber of Patriot Prime Reviews, please consider hitting that subscribe button right now. It won't cost you a thing but will help me and my channel out tremendously. Now, on to the decals. And the set we're looking at in this video is for War for Cybertron Kingdom Pipes. And this decal set includes three decal sheets on a combination of foil and vinyl backing. So now, without further ado, let's take a look at pipes both before and after ToyHacks decals. And welcome to Patriot Prime Reviews. In robot mode, straight out of the packaging, Pipes is just a little plain. I know the original Generation 1 figure didn't have a lot of paint applications, but I expected a little bit more with this more modern version. Now, taking a closer look at Pipes, he does have a red Autobot logo right there above his head, just like the original Generation 1 toy.  He's got silver paint there for his pipes. I believe that's paint. Yes, that is paint. And that's really all the paint applications on the figure. Lots of great molded details, but like I said, he is just really, really plain looking. So there you go guys, there's one final look at Pipes in robot mode before. And after. And this is what I'm talking about. Pipes looks stunning now. Taking a closer look at the figure. Starting here at the top, he gets a brand new Autobot insignia right there above his head. New chrome decals there on the chest. And yes, each one of those are applied separately. There's also a new gray decal right there. And yellow and red decals on either side. Underneath the arms, we get new decals right there that show some type of circuitry as he raises his arm up and down. There's also new blue highlights right there, there and there if you can see it. Moving on down the waist area gets these new gray, yellow and red decals. The knees get new chrome decals as well, along with some diamond plating right there that's mainly for vehicle mode. Here on the ankles, there's new decals right there inside the ankle tilt. New blue decals there on the feet. And of course the main decals are the chrome ones that go around his pipes that make those look awesome. So shiny, so chrome. I love how that catches the light. There's also new silver decals on top of the fists. Now a couple Easter eggs, at least I think they're Easter eggs, is right here this red decal. I think that's an homage to the original G1 Pipes eye visor. I mean that may just be me. And there's also hidden decals here inside his backpack. If you spin this around and flip it up, there is a missile pod. And I'm wondering if that's a homage to Transformers Cybertron armor hide, who turned into a truck just like Pipes here and had the hidden missile pod. So that is really cool. I love the looks of this guy. He looks so much better than what he did before. So there you go guys, Pipes in robot mode after. In vehicle mode, straight out of the packaging, Pipes has a little bit more detail than what he had in robot mode. Taking a look here at the front, he's got a silver bumper, silver grille, white headlights. I love the fact that they actually painted the windshield wipers. You don't see that often, that looks really good. Not a big fan of the translucent windows because you can look in and see all the robot junk there on the inside. He's got painted white lights here up top with black right there. The wheels, he's got black tires with silver rims. See, there's the silver pipes on the back. Love the texture here with that steel plating look or diamond plating look. And the back, we've got red brake lights. And that's pretty much it. So there you go guys, there's Pipes in vehicle mode before. And after. Just like robot mode, Toy Hacks adds a ton of new details to Pipes vehicle mode. Taking a closer look at the truck here. The first decals you're probably noticing is the new window decals that look really good. I love how it blends the gray into the black. It just looks awesome. Also, the front of the vehicle gets new chrome decals right there in the grille. New headlights, new chrome decals right there along the bumper, which really looks awesome along with a vanity license plate. On the side of the vehicle, you get the M that was very prominent on the Generational One Toy, nice throwback right there. Also, more chrome here to extend the bumper and a gray decal for the fender. Moving on here to the smoke stack. He does get decals right there that kind of show vents. And there is gray decals right there on top. Along the side here, there's also new gray details right there. And that's on both sides. Moving on to the back of the vehicle, we see the chromed pipes again. But let's take these off so we can appreciate these decals. New gray decal right there for the hitch. Blue decals with some red lights right there. Hazard striping. You got extra details right there. Excuse me, I got tongue tied. Some diamond plating right there. And brake lights with another pipes vanity plate. There's also blue decals right there on the inside to kind of extend the blue look of truck mode. So I love that. That looks so much better than what it was before. So there you go guys, pipes in vehicle mode after. Toyhacks did a phenomenal job on Kingdom Pipes here. Granted, they pretty much had a blank canvas to work with, but they absolutely killed it. I love the looks of this figure now. He is going to so pop on my display shelf. I did forget to point out these decals right there on top of his shoulders that kind of give him the illusion of having the smokestack arms. And that leads into one complaint, well not complaint, maybe a suggestion for Toyhacks in the future. I wish they made his entire arm chrome just like the G1 toy. But granted, this figure is an homage to the original Generation 1 animated look. So I guess I can give that a pass. After all, I did get the chrome pipes. So yes, if you have a Kingdom Pipes, I highly recommend this set. It vastly improves his look. Now guys, if you enjoyed the video, don't forget to like, don't forget to subscribe, and don't forget to click that bell icon to get notified when I upload new videos. Also, if you're in any position to help out the channel, I do offer channel memberships here on YouTube, and I gotta give a huge shout out to all my current channel members because it's support like yours that helps keep this channel growing. Once again, guys, this is Patriot Prime, signing out. Hula!