OK, OK, so first, oh, this is really, is that loud enough? Too loud? It's very difficult for me to gauge it. OK, first, I'd like to thank the organizers for inviting me to speak here. It's a great pleasure to be here. So what I'm going to talk about is also kind of isoprometric inequality, but more in the realm of re-money manifolds and metric spaces. So the problem actually is mostly considered in a field, two fields relatively far away from geometric measure theory, namely geometry and geometric group theory. But I'd like to show you how one can use techniques from geometric measure theory, mostly in metric spaces, to attack these problems. And so at the same time, it's basically showing how geometric measure theory can be used. On the other hand, it should also be a bit of an invitation to consider problems, these problems in geometry. As you will see, there's many, many open problems that we don't know how to solve. And I think geometric measure theory gives very powerful tools. And of course, in geometry and geometric group theory, people don't really are not so aware or not familiar with these techniques. Yes, so what's the problem? So it's a sample problem. I'll basically stick to the easiest case. And the first two hours, I will give basically an introduction to the problems and an overview. And then in the second lecture on Wednesday, I will provide the tools. We set up the tools in geometric measure theory. And then we're going to attack these problems using geometric measure theory. So what's the problem? The easiest problem of this kind is the following. You take a curve, a closed curve, say in R2 or R3 or wherever. And you try to solve Plato's problem. Plato's problem means you're trying to find a minimal surface with this curve as a boundary. So now you do this for every curve. And what I'm interested in is, for a curve of given length, what is basically the curve that is hardest to fill? If you take a long, thin curve, of course, you can fill it with very little area. So you have to make it round. So for a round curve, as we saw for the case of n equals the 2 of Alessio, there we have just a classical bias of metric inequality, which tells you that basically the area you need to fill it in is just 1 over 4 pi times the length squared. The same thing, of course, works in Rn. So there you have an upper bound by this amount. And you have equality if you have basically a flat circle. So now this is not really what I'm interested in. What I'm interested in is now, so consider all curves of a given length R. So now look at the curve that is hardest to fill. And you take this area that you need to fill it. For every R, that gives you a number. And this is so-called the filling area function in R of the space that you're in. For example, R2, R3, Rn, or a Riemann manifold, or maybe even a metric space. And the question is, what can you say about the behavior of this function, of this isoprometric function or filling area function? So now first, we will set this up. I'll give a definition of area and metric spaces like in a talk. And then we define this filling area functions. And then we will talk a bit about the growth spectrum. So what kind of growth behaviors? So we're always here, I'm always interested in when R goes to infinity. So large scale growth. We'll talk a bit about what can we get for functions. And then we will look at a special class. We will see that in general, it's very difficult, of course, for a given space to estimate. So one class that has been considered both in geometry group theory, and on the other hand, in geometric measure theory, are nilpotent league groups. And then at the end of today, I'll give a bit of a motivation to tell you why in geometry and geometric group theory do people study that. That's just a very small thing. So that's the part one. So let X be a metric space. If you don't feel comfortable with this generality, just always think of a remoney manifold already there. All the problems are interesting, and many problems are still open. So you can assume that. Oops, that was one too far. So as I said, we now, first, let's define area. So we take ellipses, a map, from the unit disk in R2 into your metric space. Forget for the moment this thing, this thing that came. So we define its area simply as this integral. So the integral of the number of points on the surface that are the pre-image points, the number of pre-image points, and we integrate with respect to the House of Two measure. So if you have a surface, a ellipse surface, that basically hits, folds up, then you count this with multiplicity. So clearly, when phi is injective, then this is just the House of Measure of the image. If you're in a remoney manifold, you know that ellipses' maps are differentiable almost everywhere by Radimacher's theorem, okay? You know that this area is actually, or this integral is, by the area formula, is just the integral of the Jacobian, okay? And the similar, so let's make this remark. And if we have a remoney manifold, then the area as defined before is just the integral of the Jacobian of the derivative, okay? And the similar formula, actually, you can make sense of in arbitrary, when phi goes to an arbitrary metric space. There you don't have differentiability, but you have so-called metric differentiability. We'll talk about that a bit later, okay? So now, let's take a curve in our metric space. And as I said, we want to try to fill it with a minimal surface, okay? For the moment, we'll just stick with so-called ellipses' disks, so just surfaces that are ellipses' maps from the unit disk of our two X, okay? And we want, of course, that at the boundary, this map coincides with C, okay? We call this the filling area. Later on, so the zero here, filling area zero, just says that we go from the unit disk. Later on, we will actually look at a more complex topological type. We will actually consider integral currents, so integral two cycles, but for the moment, I'd like to stick with this. And now, the isoprometric function, or the filling area function in X is defined as, so now we've taken minimal surfaces, basically, and now we're trying to look for the worst curves, curve to fill of length R. So here L of C is just a length of C. So as I said before, it's a two-step process. It's basically a min-max, right? Or a max-min process.  So that gives you a function. And clearly, in R two, the classic class prometric inequality tells you that every curve, close, simple curve, bounds an area of at most one over four pi times the length squared with equality if and only if it's the circle. We saw that today, like the most special case would be of what we saw today, okay? So in R two, and actually Rn, for n bigger equal to two, we get that this is exactly one over four pi times R squared. So what about if we have a real-money manifold of sectional curvature bounded by some number kappa, okay? So when we have non-positive sectional curvature, then we can fill it a bit better or we can fill it at least as good as in Euclidean space, okay? So the main observation or what you'll need there is that, so when you have sectional curvature, non-positive sectional curvature, then basically the main observation is that if you look at the triangle in your real-money manifold X and if you compare it with a triangle in R two that has the same lengths of sides, then this looks thinner than this one, okay? You can take this as a definition of non-positive sectional curvature. And now just intuitively you can actually make this precise. So intuitively, well if this in your space is thinner then this, well then the area here looks to be smaller than the area here. So in some sense you can expect that actually you should have something a little bit better, okay, and actually you can make this into a proof, okay, by, yeah, okay, we will see. So today I give only very few proofs and then we will go back to some of these results and then actually I'll give more rigorous proofs, okay? So if we have strictly negative sectional curvature, okay? If you have strictly negative sectional curvature then it is bounded not by something quadratic, okay, but by something linear, okay, the filling area function. So of course, you know, for small R it is still gonna be of that order because with R small then this is clearly better, right? But for large R this is much better than this, okay? On the one hand we have quadratic or most quadratic growth here and here we have actually linear growth at most, right? So that is, you already see that this filling area function that this detects something, okay? So here you can detect negative curvature, okay? So just note that the important thing is that we have quadratic versus linear growth. Okay, the questions that one can ask that have been asked a lot are, well, what are the possible growth types of this function as R goes to infinity, okay? In general, of course, this is a very difficult question. Well, one can first maybe ask it for general, but you know, for any X you can try to determine but that's extremely difficult, okay? So, well, we have two conditions that make life a lot easier. One is a curvature condition, okay? You already see here, you know, indication why non-positive curvature should help, okay? And another that is more similar to Euclidean space in some sense, to Euclidean geometry are nilpotent Lie groups, okay? So in a certain sense they're in group theory, you know, they're close to Euclidean groups, right? Okay, so you will see from geometry point of view this is actually gonna, is actually quite different, okay? So, for example, we can look at this class. Another question that we might ask is what does the growth of the filling function for a certain space X, what does it tell us about the space, okay? So the aim of the lectures here, oh, let me first maybe say what do I mean by growth of functions, okay? We will distinguish the following growth types, okay? So for any functions f and g, which are non-decreasing, you know, f, the filling area function is clearly non-decreasing, right? We say that f or g grows at least as fast as f if it is basically bigger or equal to f plus a linear term and plus a constant, okay? Maybe plus a shift in the parameter as well. And we say that f and g have the same growth if f grows at least as fast as g and vice versa, okay? So this, first of all, having the same growth type is an equivalence relation. Secondly, it can detect the polynomial growth, okay? So if, so r to the alpha grows like r to the beta if and only if alpha equals to beta. Of course, whenever alpha and beta are bigger or equal to one, you know, smaller than one because we add a linear term, it's gonna be the same, okay? We'll always be, you know, looking at growth at least linear, okay? On the other hand, it will not distinguish between various exponential growths, okay? So e to the two r in our definition because we have this shift, right, in the parameter gives, shows that it's exactly the same growth as e to the three r. So it's not so sensitive to things but it's sensitive to polynomial, to polynomial growths. So the goal of the lecture course, as I already mentioned, is to use geometric measure theory to study the growth of this filling area function. So the main tools, which I actually will introduce for those who have not seen them, are first currents now not in Euclidean space but in metric spaces. This theory, as I will explain, goes back to Ambrose and Kirchheim from about 14, 15 years ago. And differentiability of Lipsitz maps to metric spaces, a theory that was developed by Ben Kirchheim and we will also use a, we will also use a differentiability into, we already said we're interested in nilpotent league groups, okay, so there we will use a different kind of differentiability, the so-called Ponzi differentiability of maps between nilpotent groups with a Carnot card to the metric, okay? So one thing that one should emphasize here is the following, so I already said all the problems that we're gonna consider, they will already be interesting in the setting of Riemann manifolds, okay? So then you would ask yourself, well, okay, why would we need if, say, and all results will already be new, okay? So you might ask yourself, well, why should we use, you know, not use, for example, the Federer Fleming theory of currents, you know, in Euclidean space, or then in Riemann manifolds, which you already know from last week, I think, why can't we use that, or then just draw a democracy, okay, they use a classical or a democracy. The thing is even, when we're only interested in the Riemann, in the smooth setting, we'll actually have to pass to metric spaces, okay? So that somehow also shows that, you know, this theory, even if you, these theories, even if you just want to study problems in a smooth setting, these theories can be extremely, extremely useful, okay? And maybe just one word, why that is the case, why do we need to think like metric space theory is the following, well, we're looking at the growth of the filling area function as R tends to infinity, that means we look at longer and longer and longer curves, okay? So in order to control what we want to do, we'll actually rescale our Riemann manifolder metric space, so we will rescale it, so as that a curve of a length thousand will just become a curve of length one again, okay? So then we'll keep doing that, and we rescale our Riemann manifold more and more and more, okay? And what we'll do, first of all, that's, you know, a metric construction, you can pass to a certain limit space, okay? We'll analyze this limit space, but this limit space in basically, you know, almost no case will actually be a Riemann manifold, even if you start with a Riemann manifold. Very often this will be a metric space, and we will analyze basically what you can say about fillings of curves in such a space, okay? So that's how these things will come. Please, if I go too fast, if you have questions, please stop me. Okay, so remember what we said already is that if you have strictly negative curvature, then you have the filling function grows linearly at most. So but the filling function only considers long curves, okay? So basically what happens, what you think, what happens is that, well, on a small scale, say curves of length one, that should not really matter, right? On a small scale, whether it is, say, zero curvature, if it looks euclidean on a small scale, that doesn't really matter, right? So Gromov, he developed a theory of large-scale negative curvature, okay? And so this is the following. So we start with the geodesic metric space. It's a space where each two points can be joined by a curve whose length is exactly the distance between the two points, okay? So now a geodesic triangle in a metric space is simply three points plus a choice of geodesics between these three points, okay? And so a metric space, geodesic metric space is called Gromov hyperbolic or delta hyperbolic if every geodesic triangle is delta thin. So delta thin, that means that any of the geodesic sites is contained in the delta neighborhood of the two others, okay? So you see, for example, this guy here, so the delta neighborhood of the black two is just what I drew in red there, okay? And so this guy here is exactly contained in these two. And he would have the same thing for this. So this black one is clearly up till about here. It's in the delta neighborhood of the blue one and then it becomes in the black one, okay? So from very far away, from very far away, geodesic triangles, they actually look almost like trees. You should think of this Gromov hyperbolicity as a coarse negative curvature, okay? So coarse, that means on a large scale, okay? So why is this? Well, let's give some examples. So first of all, any geodesic bounded metric space, so one with finite diameter is clearly delta hyperbolic, Gromov hyperbolic, where the delta is just exactly the diameter, right? So when X is a metric tree, metric tree by definition is just a geodesic metric space, such that every geodesic triangle is exactly a tripod, okay? It's isometric to a thing like that. That means three segments, okay? Plus, you know, the metric that if you want to go from here, you have to pass through that point, okay? So that is actually delta hyperbolic with delta equals to zero. So then, one can show that if X is simply connected by manifold with strictly negative sectional curvature, then it is also a Gromov hyperbolic. So but, as I already said, it should be a local concept. We saw already in the first example that it is in some sense a local compound.  So if you take a bounded strip in R2, okay, with Euclidean metric, then clearly, this is also a Gromov hyperbolic, since, you know, any triangle that you draw, okay, is always, well, the thickness is at most, you know, this width here, right? So Gromov proved that if X is a geodesic metric space, which is Gromov hyperbolic, and such that the filling function is never infinity, that means you can fill every curve, okay? Then it is gross at most, linear, okay? So this is a generalization basically, you know, of this result that we saw, that if you have a sectional curvature, a strictly negative sectional curvature, that filling function grows linearly at most, okay? Clearly, you know, this condition, you can't just get rid of it, right? Because you can always, you can always, for example, just here, you could punch holes in your strip, right?  And of course, then, you know, you can't fill at all, okay? So then this will definitely not grow linearly. Another thing, even if you can fill, what you could do is you could just add here, you know, very many tentacles, okay?   So you need, you need a condition like that. So actually I would just, to give you a little bit of an intuition of what this actually does, what this gramophibabolicity condition, this thin triangle condition does, I'd like to give you a quick proof of this, okay? Also because actually I've never found this proof anywhere like this. So I think it's probably the easiest proof that I know, okay?  So first of all, clearly it is enough to do the following, to show the following claim. So yeah, let's maybe first, so we will have to fill, we'll have to fill Lipschitz curves, right? So for a Lipschitz curve, what I will show is that you can take shortcuts, okay? So what I claim is the following, if you have a Lipschitz curve, there exists S and T in S1, such that the distance between the points is essentially shorter than the length of the curve between these two things, okay? A smaller equal to the length of C between S and T, minus delta, okay? Where delta is, so where delta is this gum of hyperbolicity constant that we saw, the thinness of triangle, okay? That hyperbolicity constant, okay? So clearly this is enough, right? So I start with a curve, okay? So I find such two points such that this distance is essentially smaller than this distance, oh, what I should say, there exists such that, okay? Length, sorry, is smaller than say 40 delta, okay? And, sorry, I want that this is also bounded, okay? So what do I get now?  So I have two points where this is much smaller than this, or just by delta smaller, okay? So this curve in here has at most length 50, 40 plus 40 minus, so 79 delta length, right? 79 delta, okay? So, and now, so because here I can just include, I can just take a geodesic, right? So I take this curve, I said that this has smaller equal to 40 delta, okay? This distance here is smaller equal to 39 delta, okay? So it's at most, yeah, 79 delta, okay? So now my assumption says that this I can fill in, right? By something, okay? So I can find elliptics map that fills this in, okay? That's no problem, and you know, yeah. Okay, so now I do the same procedure again with this curve here, with the rest curve. So now this rest curve has length, you know, at most the original length minus delta, okay? So I can do that again. And I just go in steps of delta, okay? And basically that gives me, every time gives me something, and so you get a linear amount, so one over delta times the length, basically, times this filling guy for 79 delta, okay? Is that clear? Okay, so that's the only thing we have to show, okay? So and this I'm gonna show now, so the proof of claim. So we fix a curve, I'm gonna do it like this. So let this be my curve, C, okay? And what I take, I take two points at distance exactly the diameter, X and Y. So X and Y on the curve, such that the distance of X, Y is actually the diameter of gamma, okay? And I call this D. So now I go on both sides of X, I go length 20 delta, okay? So I have here X1, oh, let me maybe do that with what? X1, and here I have X2, okay? So X1 and X2 are such that the length of, so now this will just denote the length from X to X1, okay? This is 20 delta, which is the length of X, okay? So this doesn't mean, this doesn't mean the distance between these points, but the length of the curve, right? Okay? So now I'm gonna look at two triangles, okay? So first of all, I would say, so without loss of generality, of course, okay? We have that these lengths here, okay? That they are at, sorry, without loss of generality. If I define DI as just the distance now between X and Xi, that this is bigger than 19 delta, okay? This is because if it was smaller than 19 delta, okay? Then we would have this already, okay? We would have that the length here is smaller than 40 delta, right? And the distance here would be smaller than 19 delta, here's the length, so we would have our shortcut already, okay? So that's already good, okay? So now I look at the geodesic triangle, so this we know has length D, and so now, okay, so we're not in Euclidean space, huh? Okay, so a geodesic triangle will look something like this. It has to be thin, right? So it will look something like this, okay? So now, you know, this side here is contained in the delta neighborhood of these two others and vice versa. So what I'm gonna do is I take the last point that is still in this delta neighborhood of this guy here, okay? So that means here I have delta, and from now on it's basically in the delta neighborhood of this, so I also have delta here, okay? So here I have both smaller equal to delta, and so this distance here is smaller equal to two delta at the most, right? So now, let me denote this by A1. We will do the same thing on this side as well, okay? A1, so then because this guy here is exactly D1, the distance, right? So that's a geodesic now, huh? So then we have here D1 minus A1. So here these are both geodesics. The endpoints are delta part, right? So that means you can bound this from above and from below by this minus a delta and plus a delta, okay? So and then this guy here, the whole thing is delta. So if you have an upper bound on this, you have a lower bound on this. If you have a lower bound on this, you will also have a lower bound on this, on this guy here, right? So let me maybe call this B1 here, okay? So now I need, I think probably it's not good to write down there, because otherwise you won't see it. So we know what we want to prove, right? So we just want to prove that there are shortcuts. Okay, let me do that. Okay, so what we get for B1, okay? This is bigger or equal to the length here, which is delta, okay? So it's basically, I said already, it's bigger or equal to this guy here minus two delta, okay? And so this guy here is bigger or equal to D minus this minus a little error, right? So we can do, we can just calculate. So this is bigger or equal to now D minus a D, D1 minus a1 plus two delta minus two delta, okay? So this just gives us now, this is D minus D1 plus a1 plus maybe minus, sorry, minus four delta, okay? So now we know also that this point here, because these guys here are at distance to the diameter. So this guy here is, the distance here is smaller or equal to D, okay? So what we get is that D is bigger or equal to DX1Y, okay? So but this is bigger or equal to B1 plus this length here or this distance here, right? Which is basically a1, okay? a1 minus a little error. So this is a1 minus delta plus b1, okay? So now you can just plug in b1 here, you have an upper bound, right? So and then I can calculate basically what a1 is, okay? So I just plug it in and so what you get is that a1 is smaller or equal to one half D1 plus three delta, okay? You see a1 appears twice, right? And so then we have this, okay? And since D1, we already said D1 is smaller or equal to the length which is 20 delta, so that we have smaller or equal to 13 delta, okay? So now we can do exactly the same on the other side, okay? So we can do exactly the same on the other side. We have a geodesic triangle here, which looks something like this. We have again the same situation. We look where we have this little triangle of size delta, right? And so we get here, if we take this a2, we get this, okay? So now in order to, I would like to show you that the distance between x1 and x2 is smaller than 40 delta minus delta, okay? 40 delta is the length, right? So now I just, so I'm gonna do it like this. The distance is bounded by this distance plus this, plus this piece here, plus this, plus this, right? So the only thing I still have to do, I have to go between these two points. Let me maybe call this eta. So I have to bound this, okay? But this is easy because eta, so we have here is basically d2 minus a2, here is d1 minus a1, and basically this distance is at most the difference plus some deltas, okay? So first, of course, without loss of generality, we can assume that a1 is bigger or equal to a2. So that just makes then, right, that this point down here is lower down than the other.  So the d1 minus d2, because they're both pinched between 19 delta and 20 delta, this is going to be smaller or equal to delta, okay? So this guy here is smaller or equal to a1 minus a2 plus 5 delta, and now we're practically done. We just have to add everything up, and that will give us what we need, okay? So let's maybe, so we get the distance between x1 and x2, as I already said, this is bounded by a1 plus 2 delta plus 2 delta plus eta plus 2 delta plus a2, okay? Now you plug in, you get rid of the a2 here, okay? So then this is smaller or equal, you get 2 a1 plus 2, 4, 9 delta, okay? a1 is smaller or equal to, what did we say? 13 delta, so you get 26 delta plus 9 delta, so this is equal to 35 delta. The length was 40 delta, right? From here to here, and you're done. So that's, you see, that is relatively easy. You just have to use in a very elementary way this triangle, triangle comparison things, okay? Okay, so that's not so surprising, right? To have negative curvature gives you linear growth. So however, there's a converse, which is much more surprising. So Gromov was able to show that if the filling area function in a geodesic metric space grows at most linearly, then the space, the triangles, they must look like this, okay? Actually Gromov proved something much stronger, okay? It's the following. So the theorem is due to Gromov, but then because it had, for this mostly geometric group theory, it had a big impact, there were lots of other proofs given by other people, by Bowditch, Drutu, Papasoklu, Short and others, yeah. So the theorem is the following. If x is a geodesic metric space, and if, so now it's not just, you know, the growth type is quadratic, but now we need an explicit constant, okay? If the filling function for sufficiently large r, oops, I'm sorry, is Bowditch by one over 4,000 times r squared, so just think very, you know, quadratic but very small growth, then actually x is already Gromov hyperbolic, okay? So, and as we already saw, Gromov hyperbolicity implies that you have linear growth, so that means if you have quadratic growth with small constant in frontier, then actually you get linear growth. So in particular, as a consequence, there exists no geodesic metric space where the filling function has growth between r to the alpha and r to the beta for alpha and beta between one and two, okay? So that is usually now, of course, referred to as the gap in the geodesic metric spectrum, because there's nothing between quadratic and linear. So now, of course, you might think, so well, what does, is this constant here, okay? What is this old constant here, one over 4,000? So Gromov actually proved when x is a universal cover of a Riemann manifold, okay? Then I can replace this one over 4,000 by one over 16 pi, okay? And you still get that, okay? So you see we have one over 16 pi for certain Riemann manifolds, for universal covers. So that's the best constant in that case. On the other hand, we can definitely not go beyond one over 4 pi, right? Because r, too, you know, the triangles there are clearly very thick, right?  So you can't go above that, and this is known. So where is the truth in here? And what we're going to do, we're going to give the following optimal strengthening of Gromov's result, okay? Using geometric measure theory, okay? And I think actually I'm not sure whether you can prove it without using these techniques from geometric measure theory. So it's the following. So if you have growth slightly below one over 4 pi times r squared, then already it has to have linear growth, okay? So that's basically the first result we'll be able to get from our techniques from currents in metric spaces and metric differentiability, okay? Clearly this theorem is, of course, optimal because we have for r2 or rn, we have this constant one over 4 pi. Then what we will also see is, at the same time, so see here, you know, for example, in the Riemann manifold, very short curves, right? I mean, near a point, basically any Riemann manifold just looks like Euclidean, almost like the tangent space, right? So that means there you have Euclidean and there, you know, you will never, for a fixed epsilon, you will never, for very short curves, round curves, they will never satisfy this, right?  However, what we will show, and for metric spaces, of course, that works, right? You can have this. So we will show that if this holds for every r, then your metric space will just look like a metric tree, or is a metric tree, okay? It can be very wild, and then, of course, the filling function in here, right, in a metric tree is just zero, okay? So you get from sub-codratic, you fall down to zero right away, of course, okay? That's something that the proof just will show. As I already pointed out, for certain Riemann-Emanivolt's co-compact group actions, for example, or universal covers, we have this constant one over 16 pi. So, of course, you might also wonder now, so what can we say if in the borderline case, okay? So we already saw when epsilon equals to zero. So we already saw that simply connected Riemann-Emanivolt of non-positive curvature. There you have exactly this inequality, okay? So now this question basically is, well, what about if we know that our space has such a thing? Can we say something about it, okay? For Riemann-Emanivolt, it's not difficult to show. If you have a Riemann-Emanivolt that satisfies this with epsilon equals to zero, then it has to have non-positive curvature, okay? For general metric spaces, it's not so clear. However, it's still true that it has non-positive curvature. At the moment, so this is joint work with Alexander Litschak, and at the moment we can only show it for proper geodestic metric spaces. Properties means that every ball of finite radius is compact, okay? And so the theorem says if you have the Euclidean one, then your metric space has to have non-positive curvature in a metric sense, which is equivalent to non-positive sectional curvature. So what does cat zero mean? My space X is cat zero if and only if, well, first of all, it has to be geodestic. I already showed the picture, basically. So whenever I take a geodestic triangle, so three points in my space and three geodesics, and now for any such three points by triangle inequality, you know that you have three points in R3 with the property that the distances are the same. So that means this distance here is this distance, this distance, here is this distance, and this distance is this distance. And so non-positive curvature just means the following or cat zero, if I take any two points on two sides and then since they have the same, I can find comparison points that means they have the same distance there. Then this distance here is smaller or equal to this distance here, okay? In Euclidean, in ReMoney-Manifold, for simply connected ReMoney-Manifold, this is exactly equivalent to having non-positive sectional curvature, okay? So as one of the examples is this. We will not, actually, I will not prove this because this relies on different methods from analysis. This is mostly harmonic maps into metric spaces and I'd rather concentrate on the currents in metric spaces theory to show you. So we will actually not prove, but I'd be happy to discuss this if you have things. Okay, this is actually from this year, so we're in the end of writing it up, so you won't find it on the archive. Okay, so let's go back to the isopropymetric spectrum. So as we have already seen, there's nothing between linear and quadratic growth, okay? So, of course, the question is, are there other gaps in the spectrum? Okay, so do you have between two and three? Can you construct no things? Okay, the answer is no, okay? I can ask more refined questions, but let me just mention this result by Grimaldi and Ponsi from 2003. If f is a smooth function with positive derivative and such that for any k in n, we have that f of k plus times r is bigger equal to k times f of r for all sufficiently big r, then if f grows at least quadratically, then there exists a surface of revolution with filling area function growing exactly like f of r. So what does that... So this is basically almost a necessary condition, or it's very natural in when you look at surfaces of revolution, two-dimensional, right? So basically what it says is that if, well, k times r, if you take a curve, a closed curve, if you go k times around this curve, right, then, well, you could, in principle, when you fill it, right, so you can just take k times the same filling, the same area inside, right? So this basically just forces you that you can't have that if you go k times around that this is actually, you know, I mean, yeah, this is at least the filling area, right, that you need. So that's a thing that you have to... Okay, so that means there's no other gaps in the spectrum. So now I think we're gonna... Sorry. So, I mean, what you have to... What you have to imagine is the following. I mean, it's basically something like this. So if you basically make it like a cone, almost like a cone, right, it will just have r squared, basically, and then if you make it small here, or not small, but, you know, if it doesn't diverge too fast, then, you know, you can make it a lot bigger.  In principle, you could just take the curve at distance, you know, the certain distance here around, and, you know, yeah, is this actually the worst curve? Could be, I mean, then afterwards, you have to look at every curve, right, that... And you have to show that this curve here is actually, you know, easier to fill than a curve like this, where you have to go all the way up to the cone. OK, so that's our, yeah. OK, I think it's probably good to take a break, a 10-minute break now, and then we're going to go to a different class of metric spaces, too. Yeah. OK, that is nil-put-li groups, OK? We'll actually only deal with very special ones, for those who are not so familiar with. I'll just give, you know, as in the Gromov hyperbolicity argument, I'll give you one easy case to think about, and then we'll do more on Wednesday and Friday. So we start with a connected and simply connected legal group, G, with the algebra, G and bracket. OK, so G is nil-potent if there exists a K bigger or equal to one, such that the descending central sequence terminates after K plus one steps, where here G i plus one is the span of all the brackets where V is in one before and W is in G, OK? So in here, by this I mean that K should be the first one, the first one such that GK is non-zero and GK plus one is zero. This is called the step. And so now G, we will always endow G with a left invariant Riemannian metric, OK? So in analysis of metric spaces, very often you consider left invariant sub-Riemannian metrics, but for the moment we will just be concerned with the Riemannian case. So left invariant Riemannian metric means when you translate from the origin somewhere else, the metric looks like at the origin. So translations are isometrics. So Riemannian isometrics. So a difficult open problem is to determine the possible growth types of the filling functions of nilpotent league groups with a left invariant Riemannian metric. So actually you can also ask, of course, about other kinds of league groups, but there actually becomes even, for example, solvable league groups, and that becomes even much harder. So there you have a huge, vast open field. And yeah. OK, so let's maybe consider the first basic example. So, you know, the basic, the easiest example is, of course, Rn, right? Rn is a league group of step one, OK? So the easiest example of step two is the entheisenberg group, which is topologically just R2n plus one, OK? But where the multiplication, so a league group, remember, is just a manifold with a group structure that, you know, is smooth. So where you have basically the first two Rn's, so here we have x is in Rn, y is also in Rn, and z is in R. So the first two Rn's, we have just euclidean multiplication or addition, and here we have an extra term, the inner product of x with y prime. So this is non-euclidean group structure. So you can easily just clear that this is a league group, and the basis of left invariant vector fields are given as follows. So for i from one to n, you have xi is ddxi. So it's just the usual i direction in the first Rn's, and then in the last one, you just have a vertical, OK? And in the y's, you have in the y direction plus something in the z direction, in the normal direction, and times xi. So that would be at the point x, y, and z. So you can simply calculate the brackets, and you will realize right away that xi bracket yi is going to be z, and all the others, they vanish. So that means then that if you take your g1, it's g, OK? It's just the xi, yi, z's. But then if you go to g2, that's already there because it's just going to be r times z, OK? And the third one is going to be zr, OK? So you can directly check that hn is nil potent of step two, OK? We will actually see a bit later that it has some extra structure. It's a so-called Carnot group, but that will only come later and will become important later, OK? So now, topologically, this looks like R2n plus one. Yeah, it's diffeomorphic to R2n plus one, OK? But when you endow it with the left invariant remunimetric, OK? Then the filling function shows that this is actually quite far from Euclidean space, OK? So at least when n equals to one, OK? So the first Heisenberg group, h1, again endowed with a left invariant remunimetric, has filling, oh, OK, so that should be in h1, sorry, OK? Sorry, that should be h1, OK? Has cubic rather than quadratic filling area function, OK? So let me maybe just give you, again, for those who are not so familiar with this geometry, let me tell you, you know, at least the lower bound. Why we have at least cubic, right? OK, and then actually, we will give, much more generally, we'll give the argument, why it has to be smaller equal to cubic, actually, we will give that for any step k, we will do that, OK? So sketch of proof, at least cubic. So I should say, again, for small, right, for small r, this because h1 with a left invariant remunimetric is locally like Euclidean space. We have, of course, that it's locally, it's for small r's, it's r squared, right? So that is only, we're only interested in big scale, OK? So, OK, so I will give you a curve explicitly, you know, which I'll show that is hard to fill. So the curve is simply the following, and this already reveals a lot of the geometry, somehow, what's happening in this space, right? OK, so this, so we are in r, in h1, which is basically just r3, right? So we have x direction, we have the y direction, and we have the z direction. So now, I'll construct a curve that goes up for a given l, say bigger than 1, OK? I'll show you how to get here up to l squared, so 0, 0, l squared very quickly, OK? Much faster than, you know, Euclidean, it would take you time l squared, but I'll show you that because you go much faster, right? So how do you do that? Well, I first go l in the x direction, and that means here we're at 0, at l0, 0, 0. And now, I look at, so, OK, I should have said one thing first, right? It doesn't really depend, the filling area function doesn't really depend on what left invariant remunimetric I take, OK? Because at the origin, right, they're basically bi-lipschitz, OK, so then because, you know, so then, you know, any two are bi-lipschitz equivalent, OK? So I will, just without loss of generality, I will assume, so without loss of generality, I can assume that these vector fields here, x, y, and z, OK, they're at each point an orthonormal basis at each point, OK? So now, when I'm here at l0, 0, well, a unit vector, right, if I go in this direction y, a unit length, where do I end up, right? So the unit vector there is basically just 0, 1, and l, right? So that means if I go a unit into the direction y, right, in this direction there, I actually gain a lot of height. So let me, I'm very bad at drawing it that you already know probably, so I'll still try to make this picture a little bit, OK? So I go l in the y direction, OK? So where do I end up? I end up at the point l, l, and l squared, OK? And the length of this curve, OK, is because a unit vector here, right, is just 0, 1, and l, OK? The length of this curve here is, length is just, is l. So now I can go just back along the x direction. I end up here at 0, l, and l squared. And now I go back along the x direction, oops, maybe, I go back along the x direction. I get back to 0, 0, l squared, OK? And that's also, this direction is just unit, right? So when x is 0, then just y is just d dy, OK? So this guy here has length l, this guy here has length l, this l as well, and this l as well. So it takes me for l to get up to here, OK? So that means the distance in, you know, with respect to this lefty-varned re-money metric between the origin and this point up here is basically for, at most, for l, OK? So it's quite different from Euclidean space, OK? So now I would like to have a closed curve, right? That is hard to film. So I have to close this up somehow. I will not go just straight down here because, you know, this is going to take me l squared, right? So I would like to have something that's just linear in l, OK? But what I can do is I can just basically go now further in the minus y direction, then go in the minus x direction, and now I go down into the y direction, and, you know, basically then I can go down, right, at unit length. So let's try to draw that, but please, you need to use your imagination to understand my picture here, OK? So here, here. So I go further in this direction, OK? Then I go here, oops, I go down, and then I go in this direction. So that's not correct. Yes, that's correct. So you basically do exactly the same thing just in the other way around, OK? So now we get a curve, oops, we get a curve of length. So this gives us a curve. See, closed curve, length exactly one, two, three, four, eight l. So I will show that this has filling area, at least l cubed, OK? So this is basically calibration argument. So I just take the, so define a one form, alpha equals to d to y dz. So then if I take the exterior derivative, this is just dy dz, the two form, OK? And just by plugging in the x, y's, and so the big x, these are unit vectors, right? Just by plugging in pairs of that, you see that the point-wise norm is at most one, OK? So this is just, you know, you have dy dz, right? If you put x and z in there at zero, if you take x and y at zero, y and z, you get an extra d dz. But the d dz, you know, if you have a d dz already, that's just cancelled. So that's at most one, OK? So that's, that's, that's directly. So now I just use Stokes' theorem, OK? So if I go around, so, well, maybe I should start. So now let sigma be any surface, surface with boundary is just my, my curve C, OK? So then I have, so now you can, you can calculate, you can calculate that, I'll do that. So I'll do that in a, in a minute. So that l, that the integral of the one form over C is l cubed, OK? So then this by Stokes is just sigma d alpha, OK? But now since this is a form that is bounded by one, so this gives you that this is, has to be smaller or equal to the area. So, and this is simply a calculation. You know exactly what the curve is like, right? You know exactly the curve. So you have dc here, so definitely in this direction and this, and this, and this, you know, it's all zero, right? And you can just calculate directly that you get, that you get l cubed here, OK? So for this leg actually you get, for this leg you get one half l cubed, OK? And then for the other one going back here you get one half l cubed again. And then you're done, OK? Another simple argument is the following that you can actually do, which is even more elementary using, so maybe let's say that's the end of the proof, OK? A different argument you could do is the following. You just project this curve, OK? And the surface you just project onto the yz plane, OK? Then the curve here will basically just be this triangle. So this will be this triangle in the yz plane, OK? And this triangle clearly here has area l cubed, more or less, right? So this is just projection, projection along the x-axis, OK? Sorry? Ah, the other triangle, sorry. Yes, sure. Yes, the other triangle. Sorry? Yes. So this triangle has l cubed area, OK?  And so now you just have to calculate, and that's, again, a very simple calculation that this projection, OK, this is not one-lipset. It distorts length a lot, but it actually doesn't increase area, OK? It doesn't increase area. And again, it's basically the same argument as why we have this. OK, so that would be another proof that you can give. OK? So this gives you why, you know, we have at least cubic. OK, we will actually give a bit later something much more general for the upper bound, OK? Any questions? So this is very elementary. Very elementary. It was first discovered by Thurston. OK, so when you take the higher Heisenberg groups, OK, and bigger or equal to two, then the behavior is completely different, OK? So then you have r squared, OK? I should say this should be, you know, the same growth as r squared, of course, because, you know, as soon as you go below, it's grown hyperbolic, but the Heisenberg groups are, of course, not hyperbolic, OK? So you have actually, yeah. So that was first proved by Gromov, and Alcock gave a very clear symplectic argument afterwards to prove that. OK, so we have these things. So now one can generalize Thurston's result for the first Heisenberg group, OK? So whenever you have a free nilpotent Lie group of step k, then the filling area function, of course, again, end out with a left-in-variant-dream-monometric, the filling function grows like r to the k plus one, OK? So one example is the first Heisenberg group, OK? So what does free nilpotent mean? That basically means that the Lie bracket doesn't have any additional relation than those that you actually need, OK? So for example, in the Heisenberg groups, right, there's many, so xi bracket yj is always zero. So this is a relation, this is a zero condition that is not necessary really to define the algebra. So that's what is meant by free, OK? I don't want to go into that too much, but it should just show you that for any k, not just, you know, for step two, but for any step you have groups that you... Lie groups that have this behavior, OK? On the other hand, that's also, in some sense, the worst that you can have, OK? You cannot grow faster than r k plus one, OK? So if g is nilpotent of step k and either contains a lattice or is a Karnag group, OK? I will explain what that is. Then the filling area function grows at most r k plus one, OK? The first Heisenberg group, all Heisenberg groups, they are Karnag groups, OK? And I will give, in the case for Karnag groups, to prove, OK, in the... So not now, but in the three lectures when I actually define... On Friday when I define... when we're going to prove results about nilpotent groups. So the question that was asked for about 20 years in geometric group theory is the following. If g is nilpotent, Lie group, again, endowed with a left invariant tree many metric, does the filling area function always grow exactly polynomially? Either, you know, polynomially, the sense r to the n, or if you want r to the alpha, where alpha is a real number, yes? Oh, you mean this one? The calibration argument. Oh, yes, oh, I see what you mean. All right. Yes, yeah, but then you only get... Yes, so the first Heisenberg group is embedded, of course, in the second Heisenberg group, but the filling, right, you have a lot more fillings that you can construct by going out of this, OK? Calibration won't work. No, the calibration won't work because calibration won't work because that's not small or equal to one, right? So you would take just dy1 and dz. No, sorry, now I'm confused. Can we discuss it afterwards? Yeah, that's better. Sorry, I don't know what it's on. Yeah, let's discuss this afterwards. OK, so the question that has been asked quite a long time and whether you always have exactly polynomial growth, OK? And basically why people believed this is mostly because, you know, all the examples that one has been able to compute, right? All the results they always give a exactly polynomial and always polynomial with, you know, actually an integer exponent. So, yeah, what we will do, we will show that actually the answer is no, OK? So there exists nilpotently groups of step two that have filling function small or equal to r squared log r and which grow, however, faster than r squared, strictly faster.  So the proof will again use tools from geometric measure theory will be again, we will need the fact that when you take a nilpotently group of step two or a Karnaugh group and, you know, you have longer and longer curves if you rescale this space then actually you get the space that you know what it is. It will be the same nilpotently group but instead of a left invariant remunimetric you'll get so-called left invariant subremunimetric, OK? And then we will analyze basically what happens in that space in order to show that if, you know, an example that we can construct explicitly if this had r squared then certain things would go wrong, OK? So again, in some sense similar technique to the one that I said before, we rescale, we pass to a limit and we analyze what we have in there. Of course you might ask now, so this theorem is just a first very modest step to disproving this conjecture here. So why? Well, you know, we will use a compactness argument and this basically just gives you in the end that this group will not have a quadratic filling area function, OK? And it will give you nothing more, OK? We'll just say you have more than quadratic but, you know, whether it is maybe even, you know, r squared log r, I don't know, OK? So there's still a lot of questions open. For example, you know, what is exact behavior of this? We'll see this is extremely easy group actually. The difficulty is just to actually, you know, calculate its filling function. OK, and as I already said, you know, filling functions for nilpotent league groups are not well understood at all. We, for example, don't know whether there's any gaps or whether there's any filling functions between r cubed on r to the 4 or anything like that. So besides this result, we don't know anything about non-polynomial behavior and I have the feeling that using all the geometric measure theory that one has in Karnov groups, OK, one should actually be able to say much more because their one actually, yeah. OK, so for the rest of the time now, I would like to just give you quickly some motivation why the filling function is studied in geometry and in group theory, OK. So the first problem comes from geometry and from geometric group theory and basically the question is, when do two spaces, to remind manifolds to groups, to metric spaces, when do they look alike from far away? How do I determine that? So what do we mean by look alike from far away? Well, the first trivial observation is that if two spaces are looked the same at any scale, also small scale, so that means they're bilipsid homomorphic to each other, then the filling functions, they have the same growth. Actually, of course, even more, right? I mean, we can say that one bands the other by a constant and vice versa, OK. So, OK, so that's just, I mean, just going from one space to another, that's absolutely trivial, OK. So the fact is that more interesting fact that this is true also if bilipsid homomorphic is weakened to, in some sense, bilipsids at a large scale, OK. However, this is not enough. You also need some local control, right? Because otherwise one could, somewhere could grow, could have a lot of area needed to fill and the other not. So what do we mean by bilipsids at large scale? That is the concept of so-called quasi-isometries, OK. Two metric spaces are quasi-isometric. If you have discrete subsets, subsets in each one of them, gamma x and gamma y, that are bilipsid homomorphic to each other and such that gamma x and gamma y are separated and dense in the following sense, OK. So any two points in, in, in, in, in gamma x, they're at distance at least a, OK. And b dense means that any point in x is in the b neighborhood of some point in gamma x, OK.  So clearly if two spaces are quasi-isometric, then that doesn't imply that their filling functions are, right, so for example R2 is, R2 is clearly quasi-isometric to Z2, right, OK. And so, but the one has zero and this one has quadratic filling function. So, but one still has something, OK. Gromov proved that Bryton gave, actually Gromov, well, he more hinted at it than Bryton gave a full proof that universal covers of close to many manifolds, close to many manifolds that means just that it's compact and no boundary, that if universal covers are quasi-isometric, then their filling functions are the same, OK. And the proof is actually a version of the theta-reflaming deformation theorem for currents where you basically take a cubical subdivision, OK, and of your space of Rn, right, and then you deform a curve or, you know, a higher dimensional thing into a, into the skeleton, the one skeleton, in this sense, OK. So you build curves that you can control, OK, and these you can transport via the quasi-isometric, OK. So there, that is again an idea from this, OK. Yes. No, that's actually not, that's not enough. So what you basically need is you need to, you need to be able to control, you have to have, in some sense, the same geometry at every place at the, at the given scale, OK. So just think, just think of the, of the following example. So you take R2 on the one hand, and on the other hand you take R2. So, but now basically I take balls, it's just a sequence of balls of radius A1, OK. And now I just, basically what I do, I wrinkle things up here to make, so for example you just add a lot of tentacles here, OK. So many that, you know, basically the area in here is large, OK. But such that the, the, the metric, you know, is still not so far from Euclidean, OK. So you don't want to go have long tentacles, OK. Because otherwise, suddenly this would not be quasi-isometric to this anymore, OK. So, but just, you know, just make things wilder and wilder and wilder.  Just to add more area, that's the only thing that you want, right. And now, of course, these two are still quasi-isometric, OK. This you should endow, of course, with the length metric, right. So, so, because basically you can just take the same, you know, integer points here. And if you place that right, it will never be in an integer point. And basically to go from here to here, it's not much further than, you know, in the Euclidean space, OK. So you need to have a control locally on a certain scale, OK. And that's basically what gives you this. You can actually generalize this, but it's, yeah. So if you have a remoney manifold with a co-compact group action, then this is already enough. Yes, I don't think so. Simply again, because basically it's, it also depends on the local geometry again, right. Because even, and this basically would be multiplicative one, but you can make them as different as you want, right. I mean, up to a certain, certain thing. You can just add these things at many places, and then you get worse things, OK. Now I don't think you can say anything. OK, so that's the first motivation. I mean, that's just a motivation. I would say people are interested in, you know, when, how to, to, to figure out whether two spaces look alike from far away. So in some sense, a large-scale classification of mostly that stunning group theory, but also, you know, in geometry with, you know, and, yeah. So what you can already say, for example, right, is now that the Heisenberg group, the first Heisenberg group in the R3, is not quasi-isometric, OK. Because they have, one has quadratic, and the other one has, has cubic filling area function. OK, the second motivation is, comes from, purely from, from group theory. So now, don't be scared. I know this is a course, should be a course in school geometric measure theory. So I will just quickly tell you about a related filling function, which appears in group theory. So we start with a finitely presented group. So this basically is just, you have a finite set of generators, and then a finite set of relations. So you take a finite set S, and you take the free group generated by S. Basically, what you have to think about is an element, an element in the free group, is just going to be a finite word of letters in S, OK. And basically, what you eliminate is, you don't want to have S and S minus one, OK. So that should be, that should be, that should be the identity, OK. So, so you have just words in that. And so you have a finite set of, of, of relations. This R is a relation. And then G is isomorphic to the free group, quotient out the smallest normal subgroup generated by F. So as a quick example, Z2 has two generators, A and B, and one relation, OK. So it looks as follows. So now you should just take the integer points, but it's easier to visualize, OK. So you have, say, here, you have your identity, then you have one generator A, one generator B, OK. And the relation that you have in the group is just A, B, A minus one, B minus one should be the identity, OK. And every, yeah. So, so now the, the, the, the length of a word. So we will basically, you would like to, you know, define something like a filling area function. There we need the length and area, right. So a word, or a, just an element of the free group, as I said, can be written as a word like this, uniquely, up to this not allowing S minus, S minus one, OK. And its length, we just, you know, the number of letters that we have in there. So now for a word, a word in, or an element in the free group, which I will call the word, represents the identity in, in G, if and only if it can be written in this way as a product of, of conjugates of relators, OK. So, for example, yeah, OK. And so if it represents the identity, we just define its area as the minimal number of, minimal number of, of, of, of, of relators that we need to write it, OK. Let's maybe make an example, OK. So let's, let's look at the word, at the following word. I'll just do it geometrically already. Then you see easier what's happening, OK. I'm not sure if you can see this. Sorry. Ah, yeah. So, this word, what is this? This is just a, b, a squared, b squared, a minus one, b minus one, and then a minus a squared, and b minus b squared, OK. So now I would like to write it, we already said, remember, c2 is just two generators plus one relator, a, b, a minus one, b minus one, OK. And so now I would like to write it as a, as a product of those, OK. So now I should use another color. So how do I do that now? Well, we see already here what we could do. We can write the word. We can just insert stuff, right? So what we'll do is the following, a, b, a minus one, b minus one, OK. So we added two things that we should not have done. So we go back and then continue, OK. So that means we write the following. This is just a, b, a minus one, b minus one. That's already one of our relators. And now we've added two things too much, b and a, OK. And then comes the rest. So basically we split off this, and now we have a squared, b squared, and so on, OK. So basically what we've done here, we have this, OK. Now the second word, what do we do? Well, we could, well, we can do various things, OK. We could, for example, go further here, further here, until here. And we can add a thing in here, OK. Sorry, that's maybe not the best thing. No, let's not do that, sorry. I like it better when we do. Let's go till here. All right, then we add two things in here, OK. So basically we added this guy here and this guy here, OK. So then we added, again, a thing, right? So, oops. So we have one relator, and then we go, we have this, let me call this r, this relator. Then we have b, a squared, OK. And then we have a, b, and so on. We have, again, the relator, right? So we have a, b, a minus one, b minus one, OK. And now we have to go, we have to go back again. We have just b and a, all right, OK.  And you see what happened already in the first thing, right? We basically, we used, for this square, we used one, we reduced our word, OK, to get a word that includes one less, one less square, OK. The second time we do that, we lose one square again, OK. Then we lose this square, we lose this square, we lose this square. So the area of this word, if you just continue this process, will just be one, two, three, four, five, OK. And as you see, basically, a word represents just the, represents a curve in the skeleton here, and the area is basically just the area included in here, OK. So this shows then that when you have any word that represents the identity, so for any word, that is the identity. Then this is smaller equal to, I think it's something like, smaller equal to, well, a constant times the length. OK, so you have basically, and you can actually, yeah. So now we have this word, OK, and we define the Dean function now of what's called the Dean function in exactly the same way we did that with the filling area function, OK. Instead of curves, Lipschitz curves, we take words that represent the identity, OK, of length, you know, smaller than n, and, you know, we take its filling area, OK. And so, yeah. So that gives us an analogous construction to the filling area, OK. And the main result of Gromov and Brighton says that if G is the fundamental group of a closed-free money manifold, then the Dean function of the group grows the same way, has the same growth as the filling function of the universal curve, OK.  So you can actually reduce it to something geometric where you can use geometry, geometric group theory, whatever analysis, whatever you want, which seems to be much easier because you have much more tools, OK. You go from a discrete setting to a geometric setting, and, yeah. Yes, so I think, what's the time? OK, I think I have one more slide. If I correct, correct, OK. So, yeah, here, I mean, you basically see already, right, if you take a square curve, right, basically you have that the area is exactly the length squared times a constant, right, OK. So one last motivation is also from geometric group theory. So if you have a finitely generated group which generates a G1 through GK, and you look at words, right, so again just an element of the free group, there is the so-called word problem which asks, does there exist an algorithm which determines whether a given word represents the identity in G. So you would like to, you know, have an algorithm so that you can feed your group or, you know, group your presentation into the thing and ask then whether, you know, a given element is the identity or not. And the result is the following. If G is finitely presented as before and is the fundamental group of a closed-room manifold, then the word problem can be solved, that means there exists an algorithm, if and only if the dean function or the filling function is computable. That means, yeah, OK. I think I'll stop here. And so on Wednesday we're going to set up the machinery from, I'll say some words about current in metric spaces and differentiability of maps into metric spaces, and then we'll start proving these theories. OK, thank you.


And to appreciate hard work and not just strategic development, the Chief of Neville staff, Vice-Admiral Emmanuel O'Galla, Admiralty Medal, has encouraged the Nigerian Navy to reward excellence in service. For this end, Admiral Superintendent and MDCEO Neville Dokiat Limited has given long-time achievement awards to personnel who have shown diligence and co-consistency for over 30 years and above. This award was given on the sidelines of their annual end-of-year celebration at Neville Dokiat Limited, attended by flat officers commanding of various commands, general officer commanding 81 division, maritime big weeks, as well as, rear-admiral Hamza Koje, group managing director, Neville Dokiat Limited and special guest of honour at the occasion. There is a general encouragement package for the entire year, which is something we do yearly. We have a lot of incentives throughout this week. We've been involved in our end-of-year activities in line with the plan of the naval headquarters. We're just keen to that. And we've done a lot to encourage them. We've had the volleyball tournament. We've had our own in-house phone games. We've had so many other activities. And this is like the crescendo of everything we are doing for the week. So we try to encourage everybody, but of course, some people have to be singled out for recognition. In fact, we wanted to include those who have done 20 years and above. We found out that there were so many. There were like 25. So we said, OK, let's limit it to 30. So to encourage those who have been working so hard that they are not forgotten and they will always be remembered. In every organisation, you always have outstanding people. You also have areas of deficiencies. That is why they said that a chord is as strong as the weakest link. So what we try to do in the Navadokiat is to manage that weakest link to the point that that chord is not broken. You keep the chord running. And what that simply means is that for a huge structure and establishment like this, the systems and the structures must be kept functioning. They must be kept running continuously. Thank you.


Hey friends today, we are hanging out at the Magic Kingdom Disney invited me out so I can introduce you to some Remarkable cast members and then also the Crystal Palace brought back their character dining So we're gonna ride some rides eat some food and have a beautiful Magic Kingdom kind of day Anyhoo, let's go do this It is absolutely gorgeous out today The temperature is in the high 60s and it's like the perfect day to come to Magic Kingdom I woke up this morning. I saw the weather and I was like, oh, yeah, it is a Magic Kingdom kind of day Like I said in the beginning of the video Disney invited me out so I can introduce you guys to some Remarkable cast members and we're gonna get a chance to talk to them and hear some of their stories and learn a little bit about them Also, I'm very excited to go to the Crystal Palace and see Winnie the Pooh and Eeyore and Tigger So it's gonna be an amazing day Our reservation for the Crystal Palace isn't for another hour or so so we're gonna have to come back But I am so excited that the characters are back and also they decorated for Halloween and fall they have pumpkins actually look at you got little pumpkins in the planters pumpkins in the planters We're gonna make our way over to Adventureland so I can introduce you to our first cast member we're gonna talk to you today and She is a jungle cruise skipper one of the best roles I feel in Walt Disney World and she's gonna share with us some of her stories of being a cast member being a Jungle cruise skipper and then maybe we'll go ride the jungle cruise afterwards. So I'm very excited Here is Kenzie from Sycamore and she is a jungle cruise skipper, but you have many roles, right? What is your favorite role?  I love that. That's awesome. What other roles do you have? so I'm also over at the Haunted Mansion and I do the guest flow team in Liberty Square sometimes. Oh That's very cool. Well, what is your favorite joke or line to say as a jungle cruise skipper? That is a hard one, but I think my favorite would have to be when you enter the elephant pool There is some beautiful Visual gags done by Mark Davis, of course, so we have the classic two elephants that are right in front of A bigger elephant turned around so we get to say it's two babies bathing under a full moon Silly goose. I like it. Very cool Have you do you have a special memory where you got to make magic or you had magic made for you? Absolutely, actually my favorite magical moment. I ever got to experience I had a magical moment task Which is something we need to do sometimes where they kind of send us out in the land to try to find a family that We can make a magical moment for. So I was given a Pirate coin and I went over into Caribbean area trying to find someone who is like very into pirates I can make a magical moment for take them right on a pirate to maybe take them right on the jungle I'm gonna end up finding this little kid who was decked out in all Pirates gear I even got Captain Jack Sparrow to be in on it with me.  We got them to like be in line for when Captain Bones came down He like walked along and I was like making sure that they knew who was boss and we took him out in the boat And he was a hoot like he just like got so Neeming. Wow. Captain Bones is my favorite kid ever. That is like probably one of the coolest magic moments I've ever heard of awesome. That is so cool. Thank you. Kenzie. Thank you for being a cast member. Thank you for making magic I hope you have a great day Bye Kenzie is a rock star and it was so awesome for her to share her stories with us and to be able to say that you Are a jungle cruise skipper and work at the Haunted Mansion that is incredible. I mean I want to be at that point one day I think that is absolutely amazing and Captain Bones If you got to see this that is so awesome that you got a shot out there And now I think we can actually make our way over the jungle cruise and go for a little cruise So thank you Kenzie. That was really amazing and you're you're really wonderful Here comes our boat looks like we're gonna be on the Nile Nelly the Nile Nelly I wish we can ride with Kenzie, but I think she left for the day So we're gonna have to keep our eye out for the elephants because she said that was her favorite Let me know in the comment section if you were a cast member What would be your dream role like where would you want to make magic at would you want to work at a certain attraction in a Restaurant or at a show let me know what yours is. Let me know in the comment section I think that's a fun thing to ask and now it's time for a little jungle cruise I can tell he's an African elephant The baby's underneath the full moon Okay, guys ready jungle cruise was so much fun and it was awesome when the little girl was yelling out elephant So the skipper I think her I think it was skippy at a skipper alley She kept on asking everyone to yell out their favorite animal noise and the one little girl just kept on yelling out elephant elephant literally hysterical As soon as I got off the jungle cruise I got a notification that it was time to check in and here we are a crystal palace of buffet with character We're gonna actually see Winnie the Pooh in there. We're gonna see Tigger and we're gonna see Eeyore We're gonna have some food one of my favorite things is the fact that they have tons of Walt's favorite food items like Walt's favorite dishes in here and it's 59 99 a person Mainly because it's a character dining buffet and I've been here before and I love it And I'm just excited to see the characters After dinner, we're gonna go meet some more cast members And I'm excited because I think we're gonna be talking to a caretaker at the haunted mansion And then we're gonna ride some rides and still enjoy our day at Magic Kingdom That's me And just like that we are at our table. I'm gonna give you the grand tour show you the buffet all the amazing food I'm not gonna dive into like too much detail because I've done videos in the past here It's all about the characters today. Like I'm so happy that they're back. Winnie the Pooh is my mom's favorite character So I'm gonna say hi to him for her and also I love Eeyore But I love Eeyore so it's gonna be fun, but we're gonna head up to the buffet and get ourselves a plate The crystal palace is just so beautiful like the inside I love it in here and and I've been here for when it first I was actually here the last night the parks were open and then I came the first day it reopened Then I came again the first day it went back to the buffet And now I'm like a week and a half late till when the characters come back But it's in full function in the full way. It was meant to be Looks like we got some country fried chicken with black pepper gravy honey Tabasco sauce and this is one of Walt's favorite Number of Walt's mashed potatoes and Walt's green beans with bacon and onions all of those are everything I like So we're gonna grab some right now Look at this piece right here.  I think it's shrimp creole, but look at them shrimps in there I'm all about the shrimp in business and grab a couple of them add them to the plate Give me some shrimps and then also I love these rainbow carrots. Look at these Laced rainbow carrots everything looks phenomenal and also some grits We also have oven roasted chicken and Italian sausage with peppers and onions I'm all about the peppers and onions and sausage here Oh boy This is much too good. This is everything. I love right here They used to have chicken wings on the buffet, but I think they took those away and one of my favorite things though Is the prime rib they actually have a carving station with prime rib and it is so phenomenal We'll definitely be back up to get some of that time rib and they also have some pork loin too I got a full plate, so I can't really get anything else. I also want to try Walt's chili Walt's chili is right there. We're gonna be a lot of food. We're gonna eat. We're gonna get a lot. I Don't want to go too hard because I don't want to get full because we're gonna hang out Magic Kingdom afterwards But there are a lot of things on this buffet that I love and the price changed because when I came last There was no characters. I think it was like $39 a person now It's like I said $59.99 that character price tag is heavy and that's everywhere I mean, it's a universal price either at any of the character dining in fact I think Some of them are more expensive like I'm pretty sure storybook dining is more expensive than $59.99. I'll have to check that But let's dive in Just look at this plate We've got the chicken with the country gravy and the hot honey Tabasco We've got some rainbow carrots some shrimp Creole some sausage and peppers and some of Walt's mashed potatoes Look at that Me personally, I feel like this is my favorite buffet. I mean, I know a lot of people don't like it They want to see like special things on the buffet, but you give me prime rib Fried chicken sausage and peppers Walt's mashed potatoes Walt's chili all on a buffet I am all in and like I said they used to have wings on the buffet And also they have the shrimp Creole a lot of things I love and I cannot get enough of these rainbow carrots Looks like our first character is here. It's piglet. Look at that. I piglet. How are you? Are you having a good day? Yeah It's a friendship day Well, I'm so happy to see you here again. It's been a while and oh, yeah, I like I like your ears Yeah, and it's very good and the food's great Thank you piglet my mom says hello She told me to tell everyone so I have to make sure to tell all of the friends that my mom says hello Can we have a picture? Yeah. All right I'm sure some of you guys think it's kind of corny that I say hi to the characters for my mom Let me tell you something my mom is probably sitting at home in her recliner Watching this video on her iPad right now and smiling. I guarantee it I guarantee it.  Look it. This is awesome You look for Dora you got the Roosevelt on you got the hat I actually have that hat too, and I love it and I have that shirt the shirt is phenomenal You got the glasses.  No way. I thought there were lenses. I love it. Oh, wow. Well, you made my day Thank you so much man. I appreciate it. Yeah, nice to meet you. See ya How amazing was that he's dressed up like me and I thought he was just wearing glasses There's not even lenses in there He literally put glasses on to match the look and I love that so cool. What a great day Everything I've eaten so far has been exceptional the sausage the rainbows. I ate all the shrimp I haven't dived into the chicken yet I should be getting there soon because I don't want it to get soggy I wanted to have like a little crisp and it's looking like it is still. Yeah, so we're good But I think ticker is coming.  So we need this quick There he is there is ticker. I'm so excited to see you. I've been very excited. I saw you from the outside Yeah, I saw you outside like you stand out. You definitely It's the stripes. Yep, but I'm so happy to see you again.  Gotta gotta let you know. She's a big fan and Yeah, I'm excited to see you. Oh You're the best Seriously, thank you. Can we do a photo? Yeah. All right back to the chicken business. It is amazing It's not hot because it does say like a hot honey sriracha, but it's not that hot at all And I love it very very good Walt would be proud Hi, you're Are you having a gloomy day today? No That's a good day. It's a friendship day. I love it. I like to go to your gloomy place sometimes It's a good place to go and be gloomy, but I love it. Thank you. We are you're my favorite I'm so happy that I got to see you today Seriously making my day my mom says hello Looks like we're clear to head back up to the buffet before food comes over I want to get some prime rib and show you the salad bar and then also the kids section because I like to get food too And then don't forget we still got dessert and there is a lot of dessert and one of my all-time favorite desserts here Here is the kids buffet and I like how they have it down low like it's a natural kids buffet And we got some chicken nougnougs mac and cheese Cheesy pizza bread and then miniature beef quesadilla and then over here We've got the salad section in the bread section and they got the good pretzel rolls Which I'm going to grab one of those to eat with my prime rib And then they got a nice selection of salads and all the different dressings and then over here You got peel and eat shrimp all you can eat peel and eat shrimp. Isn't that cool?  I saw you later earlier. Oh, yeah, I was waiting to come back And they didn't have room for the prime rep, but oh That's the way Savory peach chutney on both sides of me my friend. He's chutney. I'm all about that On game, thank you Will do I almost miss Winnie the Pooh. I got to make it back to my table. Hi Winnie seeing a little bit I had to rush back to the seat cuz I couldn't miss Winnie the Pooh and I seen him like I was watching him I was watching who from the buffet like waiting to see if he'd come by the table And then I walked past them and I wish word. I'll see you soon Winnie and he looked at me. He looked at me So he's coming Oh Winnie literally just got to my table and then left. I think he's going to do his dance. I missed him by a second I rushed back over he literally made him right here and then he has to leave, but he'll be back around He's coming soon I think they're actually going to do their friendship dance or friendship march The hip hip hooray The hip hip hooray was really fun, I love that now because Winnie the Pooh They did their hip hip hooray. They're gonna go away for a little bit and then they'll come back out So we have time that we can try Summer waltz chili right here. Look at this It tastes like Walt Disney The chili is phenomenal now, we're gonna take a little prime rib here and we're gonna let our prime rib sit in our Aju this is the way There we go. It's gonna sit in here. Let it just sit for a little bit Here we go Soaked up all that Aju and it's ready to go. I love prime rib. It is so good And this one is to die for There is Pooh bear So happy to see you my mom is a big big fan gotta say I am too But my mom loves Winnie loves you loves you And I'm so happy to be back here. I hope you're having a great day I seen your hip hip hooray parade and that was fantastic. It was very very nice and The food is amazing friendship day is the best. I love it. Well, thank you, Winnie. Oh You're awesome Now let's head back up and show you guys that the desserts and one of the desserts is the Butterscotch popcorn and it's one of Walt's favorite desserts and I love it It's one of my favorite desserts too, so I'm definitely gonna be getting some of that Here are the desserts and of course we've got the Butterscotch popcorn I mean the Butterscotch pudding and it comes with popcorn definitely get one of those we got a blueberry dome Strawberry shortcake a s'mores tart which is over here and then we've got French silk cupcake and I think that's it Yeah, and also we have some hot stuff over here, too. I think they have like a hot apple cobbler and some cookies Literally all those desserts and all I got was the Butterscotch pudding. It is so good. I love it they even serve this over at the Firewood dessert party, but you can actually get this during the Firewood dessert party and I love it We are just about finished here at the crystal palace But before we go I wanted to show you how cute the little restroom waiting area is Look at this little sign right here, and then they have the restrooms here and look at this. I Love it in here.  So what are you doing here? The crystal palace was phenomenal so good and all the characters were great It's so nice to see them back. It's so nice to see everything coming back You know what I mean? I think they announced the Norway character dining is coming back and then also they Talked about Kona coming back and another one and I kind of just noticed there was a squirrel almost crawling up my back there So I lost my train of thought but anyways now we're gonna go meet another cast member I'm trying not to move, but I am surrounded by squirrels. I got squirrels to the front squirrels to the back This one's got an acorn in its mouth. You ever see a squirrel eat before? Here is Fletcher from Michigan, and he is a haunted mansion caretaker How did you end up going from living in Michigan to being a caretaker here at the haunted mansion? Well It's funny. I was actually supposed to stick around and study animation for a little longer But my brother just graduated his college with musical theater, and he wanted to come to Disney He said he didn't want to do it alone. So he kind of dragged me along Yeah, I ended up here, and it's a great time. That's awesome What's one of your favorite little details about the haunted mansion about the haunted mansion? Oh, it was the first one ever built here in Disney World, and I think that's pretty cool. Yeah first ride ever built I don't know that's a that's a good question I think my favorite thing about the mansion is when you're in the stretch room The very first like the pre-show and you disappear and you reappear and there's this second where full-grown adults Forget that you just walked they just believe in magic for like two seconds pretty much the coolest part that is really cool That is awesome Have you ever had a magical moment either happened to you or did you have an opportunity to make a magical moment for someone else?  They're like we park our mom right here.  I was like if you want to park your mom here I guess you can me and this little lady talked for an hour and a half about all sorts of art She's from Indiana. She was a fine artist. I'm an animator, so we just like clicked and it was fun I just talked we exchanged Instagrams really that is that is literally a magic moment right now That yeah, she's gonna remember that forever, and I'm sure you already do too. That is so Mrs. Look at that. That is so awesome. What what would you say is your favorite thing about working for Disney about working for Disney? Well, it doesn't hurt getting into the parks for free. Yeah, go to the parks whenever I want is a really good time Honestly for me. I'm a really big into storytelling. That's why I went into animation I think Disney does a really cool job with that. There's like fitting the theme. I like being on stage versus backstage and Really having the overall story that people are being told as they go through the parks like passes through time and like Overall stuff like that that makes any sense. No, I I get it totally well. Thank you Fletcher Thanks for being a cast member. Thanks for making magic, and I hope you have a great day. Yeah. Yeah. See you So Fletcher is awesome I mean that is so cool some of the stories he was talking about and also he had a whole bunch of different like Keys and a pocket watch all for his costume And it's really cool how hot and mansion caretakers get to like kind of add to the experience and dress up as long as It's a part of the theme and one of the things he had was a bunch of keys And he said he likes walking around the stretching portrait room And you can hear his keys like jingle as he like walks around and that that is so amazing and Yeah, very cool. Very awesome Haunted mansion is one of those roles where I would absolutely love to do so both seeing a Skipper and a caretaker and actually She was actually a skipper and a caretaker which is mind-blowing and now I think we're gonna go ride the Haunted Mansion hop aboard a doom buggy and Have fun I am so excited to ride the Haunted Mansion Especially after I was sitting there talking to Fletcher about some of the cool details and facts and I just like looking at it Look at it. Isn't she a beaut Clark? Time to hop aboard one of the doom buggies. I wish Gracie would hop in her doom buggy And eat this one Awesome was that we actually get stopped in the graveyard So I was able to stop and focus in on some of the animal Animatronics and that was really cool. I love I love getting stopped in certain areas of Haunted Mansion I just like staring and looking like you know what I mean I'm obsessed with animatronics So like being able to just stare at them without having to move fastly as you go by that was very cool And now it's time to make our way back down Main Street. We got one more cast member to talk to and Then I think we're gonna call it a night Can we just take a second and appreciate how the Sun is hitting the castle right now? Isn't that gorgeous and then we've got the pond here. You can see the reflection of the castle Look at that. What a beautiful day today is honestly. I'm loving the weather. I'm loving the Sun It's perfect a perfect fall day at Magic Kingdom The next cast member we're gonna go meet is at the front of the park because she is with guest relations And that is something really awesome I mean, we've got to you know talk to a skipper from the Jungle Cruise and a caretaker at the Haunted Mansion But guest relations is really Deep like that's a big big role So I'm very intrigued to hear some stories that she's gonna share with us and we're gonna head up to the front of the park and meet her now Alright and here is Katie from Kentucky and you work guest relations, right? Well, that's really awesome. So how did you end up going from Kentucky to working for Disney? My family but I actually got to come for my new trip in high school and whenever we were there We got to go on a backstage tour of the safari and animal kingdom and the cast was talking to us about how Disney hires from Every background every job every opportunity. There's so many different things that you can do here And I was like, okay, that's where I'm going And so I went to college at the University of Kentucky and then I did a college program down here I love that. What's your favorite Disney ride? If I have to pick one, I would say probably Guardians of the Galaxy Really? Going new school here? Controversial I know I mean no guardians is a great ride though. Absolutely love it And what is your favorite thing about living in Florida moving down working for Disney? And that's what I think is exciting Coming into work every day, and I have the opportunity Do whatever I want as far as like make magic for the guests and help them have you know a magical time a magical vacation That's what we're all about. So it's really making magic. Yeah, I love that. What what is your favorite making magic moment? I love making magic with the guest experience team because I'm actually out in the park And I can see what issues are kind of happening in the park So it's nice to you know kids crying because they really want to get on a ride I think they'll put them on that ride or if they're crying because I spoke about form I can get them in a pop form, you know, I'm free to do as much as I can for the guys I have like very small limitations. I love that. That's the best. Well. Thank you, Katie I appreciate it so much. I hope you have a great day Katie from Kentucky was really cool great to hear some awesome stories She got to make some magic and Guardians of the Galaxy is her favorite ride and that's really cool I mean, it's a really awesome ride I do love Guardians of the Galaxy and overall my whole entire day today has been fantastic Getting to meet all of these cast members and hear a little bit of what it's like to work for Disney You know, eventually that's my goal in life I want to I want to retire and I want to work for Disney I said this from the beginning back in the days even when I was just a restaurant manager a cook I wanted to Move to Florida work for Disney be a monorail pilot or boat captain and that's what I want to do I mean end game is that that's my end game. So a lot of people, you know You know, a lot of people always ask me what am I gonna do?  I think it's gonna be great I I'd be happy like I said monorail is probably my dream my dream job my dream my dream role is definitely probably the monorail and It's really awesome to talk to some cast members today So thank you Disney for inviting me out and I really hope you guys enjoyed the video because I Enjoyed making it and we'll see you next time Bye


Matt here for Layton Sports Cards ripping two boxes for Edward E. We have one box of 2023 Tops Chrome Baseball Hobby, one box of 2022 Tops MLS Renaissance Soccer. All right, here we go, Edward. We have this and then 20 gold packs for Gabriel and then we'll be done for the day. No personals tonight over on the Layton 2 channel. Kyle will be here on the main channel doing group breaks. Grab your spots up for those. Also, grab your spots for tomorrow's daytime group breaks. On the main channel, I go live at 10 a.m. Eastern with those. Darvish Refractor, Cal Mitchell, Austin Riley, 88, Titans, Yordan, Baldachuk, Verlander, Brett Beatty, Rookie, Jonathan India, Aqua Refractor to 199, Ryan McMahon Prism, Tristan Kossas, 88, Freeman J. Rod, Future Stars, Bob Thompson, Drew Rasmussen, Refractor, Jones, Pasquantino, Albert Pujols, Prism Refractor, Small, Ezekiel Tobar, Rookie Refractor, Volpe Rookie and Carol Rookie, Altuve Corbin Burns, Christopher Morel, 88, Soto Titans, Jeremy Pena, Refractor, Alves and Chris Sale, Blue Refractor to 150, Trevor Story, Tyler Stevenson, Rookie Refractor, or not Rookie. That wasn't Rookie, that was just a Stevenson Refractor. CJ Abrams, Prism, Aaron Judd, J.D.A. Devers and Rookie Refractor Otto to 499, Tyler Freeman, Incarnacion and Kodaisenga, Nolan Gorman, Rookie Refractor, Rookie Prism Refractor, Ezekiel Tobar, O'Neill Cruz, Refractor. Here is your renaissance. This is Patch Otto, 2 of 10, Danny Leyva, Seattle Sounders, that is Otto, 16 of 20, Kevin Paredes. All right, there you go, Edward, that'll do it for your boxes. Appreciate it. We'll get these right out to you.