Good. We saw your signal, Walter. Good afternoon, everyone, to this basic notions seminar. It's a pleasure for me to introduce our speaker today, Professor Jeff Valer, who's an emeritus professor at the University of Texas at Austin. So, just beforehand, Jeff was my PhD advisor, this is a double privilege for me to introduce him today. He got his PhD in 1974 at the University of Illinois at Urbana-Champaign under the direction of Harold Diamond, and over these years he has done amazing work in the fields of number theory and analysis. So, his career has been very bright, with a good deal of it devoted to the mentoring of students. So, throughout the years he has had an impressive number of 35 PhD students, which I'm just one of them. He has published more than 60 papers, and he's definitely one of the brightest scientists that I've met in my life and a big inspiration for me, and I hope that you guys will enjoy for the next hour his talk today. So, it's a survey of volume inequalities, our basic notions title, and I pass the floor to you, Jeff. It's a real pleasure to have you here with us today. Thank you for your kind and exaggerated remarks. So, my intention here is, as Emmanuel described, to speak at a kind of general level. There'll be a few little technical pieces, but please do not hesitate to interrupt with a question if I have something up here that seems strange or mysterious and you just like to pursue it. I'm happy to deal with that in the middle of the talk at any time. So, in number theory there are quite a few situations where you want to count something, and oftentimes the something can be embedded into Euclidean space, and then the properties of the something tell you that these are all the maybe integer lattice points, points with integer coordinates in a usually convex and symmetrical set. Symmetrical here means that if x is a vector in the set, then minus x is also a vector in the set. And since it's convex it must contain the origin, and I'm going to put on some additional assumptions to simplify matters. So, I'll be talking about convex symmetric sets K, Minkowski called these convex bodies, and so I'll probably use that term quite a lot as well. Anyway, in number theory oftentimes you want to prove that a convex symmetric set contains a lattice point, a point with integer coordinates, but not zero. Zero is sort of automatically there, and if K is sort of big enough and round enough it ought to bump into some other lattice points in Euclidean space. Another common problem is as the space, as the set K expands, you want to be able to count the number of lattice points or at least estimate the number of lattice points in such a set. So these are both kind of fundamental questions and issues that come up in analytic number theory, and they represent, or that represents the sort of the background of why one would be interested in the volume of convex bodies. So, for example, if you have a convex symmetric set in R7 and it has volume 1,000, and you were to guess how many lattice points, how many points with integer coordinates it has in there, 1,000 would be a good guess. Roughly speaking, we expect the number of lattice points in a big, round, reasonable set to be approximately the volume of the set. Okay, let's see, I've just illustrated some very elementary possibilities here. B sub n, which will come up in a couple of situations later, well, that's just the unit ball because I'm using the Euclidean, or little l2, norm. In general, a convex symmetric set is the closed unit ball in a Euclidean space, R to the n, the closed unit ball associated with some norm on the space. The Euclidean norm being perhaps the most popular one to use. Sometimes it's useful to introduce a dual convex set, and from a functional analysis point of view, the dual convex set is the closed unit ball of the dual bonnock space, so to speak. But maybe more simply, you can define a second norm, the dual norm on vectors y in R to the n by this rather simple recipe, and a few lines can show you that double bar with a star attached to it is also a norm, and so we get a dual convex body k star. I guess one of the simplest examples, probably familiar to many of you, if you start out, for example, with a little lp norm on vectors with n coordinates, and then you choose q so that p inverse plus q inverse is equal to 1, well then the dual norm of the little lp norm is the little lq norm, and of course if p and q are both 2, then that tells you that the dual of a sphere is itself, roughly speaking. Well, I mean to maybe make one more example, if p is 1, then you get kind of a generalized octahedron as the unit ball associated with such a norm, so the convex symmetric set. Then the dual norm would be the infinity norm, which is also written as the maximum of the coordinates, and there the convex body is a cube, so both those have fairly simple volumes to compute. Let's see, I quoted some theorems here that are sort of relevant to the question of why one is interested in the volume of a convex body. The first one is an old one, a little more than 100 years old due to Minkowski. Minkowski's theorem asserts that if k is a convex symmetric set, I've also asked that k be closed, so it's a convex set, and to avoid things that are squashed into a subspace, I want k to have a non-empty interior. So some of those conditions are just put in to make the situation reasonable. Anyway, Minkowski's convex body theorem asserts that if the volume is big enough, bigger than 2 to the n, then the convex body k must always contain a lattice point, which is in my language here a point with integer coordinates, but not the zero lattice point. At first when you think about Minkowski's theorem, it's oftentimes the case that it seems like it's a theorem which is giving you information about lattice points that are sort of close to the origin. It's not easy for anybody to visualize r100 or z100, but I guess this tells you that if k is convex and therefore kind of round, as it expands to get a little bit bigger volume, it eventually bumps into a lattice point that's probably kind of close to zero in some sense. But actually that's kind of naive. It turns out, as I've indicated here on the bottom, that this is really a theorem about all lattice points in the sense that if you took a lattice point far, far from the origin, 100 light years up in this direction, and then you removed its negative way, way down below, far, far away. Well, as long as the coordinates of those lattice points are relatively prime, a modest request, then if you remove those from the set of lattice points, Minkowski's theorem would no longer be correct because you can actually find a convex symmetric set, which I call k sub eta here, where eta is the name of the exceptional lattice points that are being removed, and this convex symmetric set has the required large volume, 2 to the n, but the only non-zero lattice points in there are eta and minus eta, which were removed. Yeah, so it's a little bit more far-reaching than maybe it seems at first. Let's see, here's a second theorem much deeper than theorem 1. Theorem 1 nowadays would be viewed as a theorem that could be proved whenever you have, for example, a locally compact group with a discrete subgroup where the quotient is compact. Then the Haar measure plays the role of volume, so Minkowski's theorem can be generalized enormously, but the second Minkowski theorem, the so-called successive minima theorem, is much, much more specialized, and that's only been generalized in a few situations. Anyway, the way it works is you learn how to dilate the convex body k by a positive parameter alpha, and then you define, as Minkowski did, the successive minima associated to k to be some positive numbers increasing lambda 1 through lambda n, and the definition is pretty simple. Lambda 3, for example, is the smallest dilation that you can make to the convex set k so that alpha k contains at least three linearly independent lattice points. So these are well-defined, and lambda 1 is obviously positive because k is open, and lambda n is obviously finite because k is compact, and Minkowski also proved that the successive minima satisfied that inequality. But as I said, even in modern times there's really no simple proof of that fact. The first theorem can be proved in great generality using harmonic analysis, but the proof of theorem 2 still uses the basic idea that Minkowski employed. You can smooth it off in a few little places with modern language, but it's still a tricky theorem to prove. Let me now turn to some actual problems and inequalities about volumes. So a first one that I'll talk about is the so-called product formula or product inequality. A little experimentation will show you that if you choose the norm that creates the convex set k in such a way that the volume of k is rather small, then the volume of the dual convex body, k star, will be large. The dual of k star is back to k again, hence the word dual, and of course it goes the other way. If k has a big volume, then k star has a small volume. And so it's an interesting problem to try to determine approximately what the product of the two volumes is. A complete resolution of this is still unknown. As I've indicated here, the Blaschka-Santalo inequality, which was proved quite a while ago, I think in the late 1950s, gives an upper bound. So not too surprisingly, if that product of the volume of k and the volume of its dual is going to be large, well it's largest at the situation where the dual convex body is actually equal to the one playing the role of k, so in the case of a sphere, a ball produced by the Euclidean norm. So this one is, the upper bound is well known. So this inequality holds for all k, and obviously you get equality if the volume is the volume of a sphere. Kurt Mahler, who did much work in the geometry of numbers in the 60s and 70s mostly, conjectured this lower bound. That conjecture has never been completely proved. It's known to be true in many cases, I'll describe a slightly complicated case where it's known to be true, but it's not known just under the general hypothesis that k is a convex set. You get equality in Mahler's lower bound, if for example k is a cube, and then k star would be the generalization of an octahedron in higher dimensions. So the case of equality is known, but for all we know there might be some mysterious k whose k star has the property that that volume actually lies below for the anobran factorial. It's also known, as I've indicated on the bottom, that this inequality holds, this is a result of Reisner, whenever k or k star is a special type of convex body called a zonoid. I'm going to define a zonoid here in a minute. It's not a very simple definition, although it turns out that a lot of convex symmetric sets are zonoids. Now I'll give some examples here in a minute. Things will get a little more technical then. So what's a zonoid? Well, if you start with a norm k and k star, then you can make the two convex bodies, k and its dual convex body, and it turns out that k star is a zonoid if the norm that produces k star has a particularly, well, special looking definition. So if you were to put a measure row, a finite measure row, a Borel subsets of the surface of the unit ball in Rn, which you could do, and then you would compute the norm of x by integrating over that ball with respect to the measure, and then you could use that norm to produce a convex symmetric set. The convex symmetric set you would produce in that way is a zonoid. There are other also kind of... Yeah, so k, I know this seems odd, k star is a zonoid if its dual norm has that description. Yeah, so x is a vector in Rn. Yeah, and we want to create a norm on Rn, so the way you do it is this way. Now, on the one hand, it might seem very, very special. Well, I suppose it is very, very special, but keep in mind you got a lot of measures. If row is a discrete measure, then this turns out to be what's sometimes called a polytope. It's basically the intersection of a finite number of regions in between parallel planes. Well, then you might say, well, that sounds like a discrete measure, and I could take a weak limit as that measure converges to some other measure, and probably the convex bodies will converge in some suitable sense that I'm not describing, and that can be more or less verified, but you have to be a little careful on what the sense of converge here is. Okay, well, so is there an example of something that's a zonoid? It turns out, yes, and I'll get to that. First, we're going to pass through some preliminary stuff. So here's a nice formula in theorem 3 for a volume. This is the volume of the projection of a cube. I'm going to make a little adjustment in my previous discussion. Instead of thinking of the cube as the little L infinity unit ball, I'm going to ask that the coordinates are all less than or equal to half. So think of the points in Euclidean space. All of those coordinates are less than or equal to a half in absolute value, and the reason I'm doing that is that then the cube Cn has volume 1, and that removes some extraneous numbers that would otherwise appear in the formulas. Okay, well, what can you do with a cube Cn in Rn? Well, here's something you could do. You could take a linear transformation, so just multiplication on the left by a matrix, and you could map that cube into a lower dimensional subspace, say a subspace of dimension m. So K sub m is contained in R sub m, and it's just the image of all the vectors in the cube under the linear transformation A. This actually came up in a problem that I was working on in the 1990s, and I was sort of interested in, what is the volume of K sub m? Well, obviously it must depend on the matrix A that you selected to do this mapping. I was amazed that this wasn't known until 1984. It seems like something that would have been asked or known much, much earlier than that. Anyway, Peter McMullen, a British mathematician, proved this very nice, tidy formula for the volume. Here's how it works.  You could ask this question, and I'll just mention this casually here. You can ask this question if R is replaced by the complex numbers, and A is then a complex matrix, and volume in C3 really becomes volume of dimension 6 with respect to the real and imaginary coordinates. I don't think this formula is known to hold in the complex case. It's one of these things, if somebody tells you it's true and you go home and try to prove it, you'll probably prove it eventually. Maybe the hard part is deciding that there is such a thing and that the sub determinants have anything to do with it, I suppose. Anyway, that's an example of the subject of this talk. That's a volume E quality in that case. So here's a much more elaborate one, which, believe it or not, the proof relies on McMullen's formula. Let's change the scene here quite a bit, just for a moment at least, and begin with an arbitrary measure space. So X is a set, M is a sigma algebra of subsets, and nu is a measure. I think nu has to be finite measure in order to really end up with what I want, there's probably an interpretation if nu is not finite as well. Anyway, I can make a rather elaborate convex set K by choosing points X in Rn such that the integral of this linear combination of my initial functions, when you integrate it with respect to this general measure, it's most one. Well, it can be shown that that rather elaborate and hard to visualize K and its dual satisfy Mahler's conjectured lower bound. That's because it turns out you can prove that K, actually K star, is a zonoid. And if either K or K star is a zonoid, then Reisner's result shows that Mahler's conjectured lower bound for all Ks, in fact, holds in the case of a zonoid. Now, this is K. There's a formula for the dual K star, but it's fairly complicated. And in fact, I don't know that it really improves on the general definition of the dual once you start with K. But it turns out that there's an actual formula for the volume of the dual body K star. And it's sort of inspired by McMullen's theorem. You start with determinant, looks sort of unusual, a determinant in which you take your n functions on a general measure space and you introduce n variables, each one running over the measure space. And that gives you a row index and a column index for a n by n matrix. Well, and then you could compute the determinant. And so you're going to be multiplying functions in this function space. It's at that point that you might want to make sure that the measure is finite. Yeah? With respect to the previous formula, the one of the equality. Can you please go just one slide? Yes. So maybe I fixed the matrix lambda once for all. Is it true that if I increase m, this volume refers to the theorem increase with m or not? It's not guaranteed. Could you describe the initial part of your question again? I have the matrix is m by n. Yes. Then you are fixing m and you are computing the volume. Yeah. I imagine that I do m equal to 3, 4, so I increase m. Yes. Is it obvious that the volume increases with m? No. Because it's possible to have a matrix with determinant 1, but it has a sub-matrix that's either very large or very small. So as the matrix increases, the determinants could behave in an unpredictable way. You can make them behave however you want, I think, yeah. It's not strange. Well. I don't know why. Yeah, I think in the generality of your question, there won't be any kind of inequality of the sort you propose. But if you put on further hypothesis, of course you could perhaps arrange for that to happen. Yeah. I mean it's just the phenomena that you could have a matrix with gigantic numbers whose determinant is epsilon in effect, yeah. So any other questions about anything I've said here? OK, so using McMullen's formula, it turns out that there's a formula for the volume of k star. Well, you might say, but you need to evaluate this matrix, or this integral. Well, yes, but the point is that you would use this in a situation where you knew what the f's were and so forth. This is just a general sort of formula. And then because k star is a zonoid, using Reisner's inequality, 2 to the n is less than or equal to the volume of k, which we don't know, times the volume of k star, which turns out to be given here. I think in the inequality, the 2 to the n over n factorial are combined with the 4 to the n over n factorial that appears on the left-hand side in Mahler's conjecture. And so that's why it looks maybe a little bit simpler. And lo and behold, you have a lower bound for the volume of k. And this is useful because lower bounds tell you that the volume is greater if the volume is sufficiently big than by Minkowski's theory. It contains the lattice point, or by his deeper successive minima theorem, it appears in an inequality about the successive minima. Anyway, this is one of the more complicated volume identities that I know. Now, let's set that aside and think about a very different kind of problem, or at least different in many ways. This concerns a subject that's come to be called cube slicing. The really simple version is the following. Start with a cube with volume 1 centered at the origin in Euclidean space. Imagine it has a large dimension, n. And suppose you were to slice the cube by intersecting it with a subspace of, say, dimension M. Well, the subspace of dimension M intersects the cube in a somewhat complicated convex symmetric set. Now, of course, it might be the case that you slice parallel to a face of the cube, at which time you would just get a littler cube. And so that little cube has volume M-dimensional volume greater than or equal to 1, in fact, equal to 1. It was conjectured that, in fact, that's how you get the smallest volume of a cube intersected with a subspace. So the conjecture was, if you intersect a cube of volume 1 through the center of the cube with a lower-dimensional subspace, then the lower-dimensional volume is greater than or equal to 1. Well, this is a problem I worked on very early in my mathematical life. And I've started with a slightly more general version than what I just described. You might begin with an N by M matrix. And then you could take a vector in a lower-dimensional Euclidean space and multiply A on the right by X. And this would, of course, be a vector in the higher-dimensional space. And we could restrict our attention to those Xs such that AX lands in the cube. Or to say it another way, the volume, of course, is also the value of that integral where chi is the characteristic function of the cube. Well, it turns out that you can prove these inequalities for the volume. So the integral here is also the volume in question. I just wrote it down once. The lower bound looks like this. And the upper bound is of the same kind, but with an extra factor of 2 to the M over 2 in front of it. I proved the lower bound a long time ago, and another mathematician, Douglas Hensley, proved the upper bound. I think the upper bound has proved a little bit before the lower bound. OK, so what about intersecting the cube with a subspace? Well, that's just the case where the matrix A has orthonormal columns. So suppose the matrix A has orthonormal columns. Well, then, of course, it spans a subspace, HM, and it's then kind of an exercise to show that the volume on the previous slide just becomes the M-dimensional volume of the subspace intersected with the cube. And determinant A transpose A, what's that going to be? Well, if you think about it, if A has orthonormal columns, then A transpose A is the identity matrix, and so it's determinant is 1. So we get the predicted 1 as a lower bound, and we get 2 to the M over 2 as the upper bound. Both inequalities here are sharp in the sense that there exist cases of equality. Here's a problem I formulated. It made I formulated it using exterior algebra, which maybe some of you aren't familiar with, but I can explain that easily enough. The wedge product here is something in the exterior algebra, and the exterior algebra is basically a larger vector space in which the terms, of course, have coordinates, and the coordinates of that vector are the sub-determinants, the M by M sub-determinants of the bigger matrix A. So what I'm asking for here is, does there exist a function little f so that when you want to evaluate that integral that came up in cube slicing, f does it by applying f to the sub-determinants of the matrix A. There are binomial coefficient N over M coordinates there, and I think there's a function f that goes in there. So f is independent of the matrix A, of course. f depends on the cube only. I don't know what that function is, though. I think you can do this if you just maybe have the way in low dimensions, let's say. Try it for two vectors in R3 or something. Now it turns out that while cube slicing is a theorem now, there's a conjecture formulated by Gromov, which, if it's correct, would be a tremendous generalization of cube slicing. This is related to a theorem of Gromov, usually called Gromov's waste inequality.  Anyway, Gromov's waste inequality starts with the Euclidean ball, B sub n, and Gromov proved the following theorem. If you start with the ball, B sub n, and you use any continuous function to map it into a lower dimensional space, R to the L, doesn't have to be linear, doesn't have to be a matrix, just any continuous function. Then, it turns out, there's a point Y in RL, such that the inverse image of Y under the map F is fairly big. What do I mean by fairly big? I wrote down volume, technically, you should use M-dimensional Hausdorff measure at this point. But to make it look kind of simple, I'm just going to call it volume here. Anyway, what Gromov proved was that there's always a point in the range whose inverse image has volume bigger than the M-dimensional volume of a ball. There's a formula for this using the gamma function, so that's not mysterious. If you'd like to read more about this, I recommend a paper of Larry Gooth, the Waste Inequality in Gromov's work, which contains a discussion of this and many, many other related things, and also a discussion of the next topic here, Gromov's Conjecture. So, let's start as I did in my discussion of cube slicing with a unit cube, but coordinates less than or equal to a half, so really it has unit volume, which simplifies some aspects of the subject. And let's suppose, as in Gromov's theorem, that you have a continuous function f that maps the cube down into a lower-dimensional Euclidean space, say r to the l. Gromov's Conjecture asserts, as his theorem did with the ball, it asserts that there exists a point y in rl such that the inverse image is possibly complicated, but not too bad, because f is continuous, subset whose m-dimensional volume, really m-dimensional hall measure, Hausdorff measure, is greater than or equal to the m-dimensional volume of a m-dimensional cube, which of course is one. Are there any cases known? There's only one, the cube slicing inequality. That's the special case where f is linear. In that case, we know exactly how to find y. It's the zero vector. Is this for the standard elliptical norm or is it the conjecture for every norm? Well, if you read Larry Gooth's paper, he discusses that point. No, I don't think it's likely to be true for every norm. These are definitely unsolved problems. As I say, in this particular case, the only general class of functions f for which the conjecture is known is the case where the map is just linear. Well, you could take e of x and then add any vector to it, and that just moves the body somewhere else and doesn't change anything about it. But the linear case is the only case that's known, and in that case, the inverse image of zero is just the cube slice, which we already know is greater than or equal to one. Yeah, so if this is true, it would represent a very striking generalization of the cube slicing inequality. It would say that cube slicing is just the easy case of a linear map and that in fact it's a topological geometric property of cubes that any continuous image into a lower dimensional space has such a remarkable point y that has a big inverse image, so to speak. Any other questions about this conjecture? Again, the paper of Gooth has much more about this in it. So far, I've been speaking only about convex bodies and convex sets. There are other situations in which you'd like to know the volume of a set that has some regularity but isn't necessarily convex. Well, of course, once you drop the convex, you've got all sorts of possibilities. Something that often comes up, especially with applications in number theory but also applications in physics, is to use what's called a distance function, usually a number theory or a gauge function, usually in areas connected to physics. So this is weaker than a norm. Norms are gauge functions but in general a gauge function is just asked to be continuous and symmetrically homogeneous. And again, to avoid degenerate situations, I'm going to assume that the set S, which is no longer necessarily convex, I'm going to assume that the set S has a non-empty interior. And as soon as there's a non-empty interior, then I know that the volume of that set is positive. Sets of this sort have a variety of names in the literature. In number theory, they're usually called star bodies. There's an example. This isn't a very good one because the star body has infinite volumes. It's not a very good one for my talk. It's the simplest example that's not a norm that might sort of naturally occur. Associated to star bodies is what I called here a sort of basic counting principle. It's still the case that one would expect if the star body is kind of reasonable and not too crazily defined, that the number of integer lattice points that are inside the dilated star body, so in other words, the gauge function of the lattice point is less than or equal to t, is still about the volume of the set of points where delta of X is less than or equal to t. But if you think about it, that's just S multiplied by the parameter t. If t is large, you expand S. If t is small, you crush it down. I have never seen a proof of that in this generality, but for many star bodies that actually occur that people have studied, there is usually an identity with an error term that depends, the big O here depends on the star body S, which is sort of part of work in counting points inside of star bodies. Okay, here's a star body that has interested me. So in number theory, there's a function that you can define oftentimes on polynomials called Mahler's measure, and it has a wide variety of uses, applications and so forth. I've defined Mahler's measure in the first displayed equation here, but I've done it in sort of a non-traditional way so that instead of thinking about Mahler's measure as something you do to a polynomial, I'm thinking about Mahler's measure as something you do to a vector X. And the polynomial arises because X is the vector of coefficients for some polynomial. And then the Mahler measure of that polynomial, well, it's the sometimes called the geometric mean of the function. Anyway, it's the exponential function applied to the integral of the logarithm of the function where here you're integrating around the unit circle, so e to the 2 pi i times a variable makes an appearance. That might look formidable difficult to deal with, but it turns out that there's at least one simplification that can be introduced. If the alphas are the roots of the polynomial whose coefficients are X0, X1 through Xn, then using Jensen's formula, there's this somewhat tidier formula for Mahler measure. Of course, this one, while tidier, depends on the roots. If n is large, the roots may be kind of mysterious. I mean, this one depends on the coefficients, which are sort of plainly there, but here you would have to determine the roots. Anyway, it turns out that the star body associated to the Mahler measure that's defined here in dimension n actually has a finite volume. That's not too difficult to prove. You basically have to prove that the star body is abounded set, and that pretty much does it. What was much more surprising to me when finding the volume was proposed to me by one of my PhD students is that the volume is actually a rational number. So I'm working sort of simultaneously in two situations, but really the real case is perhaps the more interesting. You can do everything I've talked about with a vector whose coefficients are complex numbers, and then you can ask for the analogous volume of the associated star body. Well, here's a formula for the volumes in question. In the complex case, it's not too bad. In the real case, it's fairly bad to actually work through the whole proof. It's a long, complicated argument. But if you look closely, you see that while this expression over on the right-hand side is a bit complicated, it's nevertheless producing a rational number. So it turns out that even though the exponential and logarithm function appear, plus an integration and so forth, that the volume in question is rational. In fact, this function is nicer than one would expect in yet another way. So one possible approach to volume questions, such as I have discussed here, is to introduce a certain analytic function of a complex variable. And if you can say a lot about that analytic function, you can extract the volume out of the behavior of that analytic function. I'm going to illustrate the ideas, just in the case of the molar measure function that has occurred here. So what you do is you... And remember, the molar measure function is defined on a polynomial of degree n, which has n plus 1 coefficients. So you set the first coefficient just equal to 1. Another way to say that is I'm just going to look at polynomials that are monic polynomials. Okay? Not so crazy. And then I'm going to raise it to the power minus s, where s is a complex variable. And then I'm going to integrate over Euclidean space. If I integrate it over r to the n plus 1 and left that as a variable, it wouldn't converge anywhere. Yeah, so if you're going to use a method like this, you have to get yourself in a situation where this really does define a function of a complex variable, s. Well, it turns out that, again, in spite of the relative complexity, the exponential function, the logarithm make an appearance, in spite of that, this really is an analytic function in the half plane where the real part of s is bigger than n, the integral converges, and it converges to this rational function of s with rational coefficients. And so by analytic continuation, this must be the value of that function throughout the complex plane. Here, cn is a certain complicated constant that comes up. Well, with a little more work, you can discover that the volume of the star body is also that analytic function at s plus 1 plus a little bit of extra factors that come in. Okay? So in the theorem that I proved with churn described on the previous slide, we were interested in the volume, but actually it turns out that you can learn much more about molar measure by using this more elaborate fact. So this contains more information than the volume. Where is the volume? It's just in that number. But of course, there are other complex numbers that are used to evaluate it. And that leads using the counting principle to a kind of interesting theorem. Suppose we wanted to count polynomials of degree n with coefficients a0, a1, and so forth, and we wanted to restrict ourselves to the situation where the coefficients are just integers. So they belong to integer lattice points. To count them means to estimate that sum. And by the basic counting principle, which happens to hold in this case, it's the leading volume term times the parameter t to a power. And that's the volume of the associated set here. And since you're counting and not measuring volume when you do that, there's an error term. The error term is big O over one lower power of t. And again, the volume I've just repeated here is this rational number. I think I have one more slide. I thought I should finish by mentioning an open problem in this field. This is maybe the simplest, really genuinely interesting open problem I know involving volumes. So here's a type of norm that so far we haven't talked about. I'm going to call this the infinity norm, but don't confuse that with the maximum coefficient type norm. This is different. You take a vector in Rn plus 1, and you make this polynomial in the complex variable z. And then the norm of that vector x is just the soup norm of the polynomial on the unit disk in the complex plane. Sort of common place where interesting analysis can occur. This is the best I can report. The volume I do not know, but I know that number is a little smaller, and I know this number is a little bit bigger. And it's pretty easy to figure out where that came from. For example, something larger than that would be the L2 norm around the boundary of the unit circle by the maximum modulus theorem. The maximum has to occur on the boundary somewhere. And so a slightly smaller number would just be to average the square of the function over that set. And that's where the lower bound comes from. A really easy upper bound would be, well, whatever that is, it's certainly less than or equal to the sum of the absolute values of the xn's, the little l1 norm, in other words. And that gives you a slightly bigger norm, which produces a slightly smaller convex set. So those simple ideas produce that inequality. But this is something that would be of interest in number theory, in applications. If you want to make a polynomial that has a particularly small sup norm, how do you choose the coefficients? Well, one way you could do it is to use Minkowski's theorem. Once you've established that the volume is big enough, so to speak. So if that isn't big enough, then replace the one by a parameter or just multiply the set by a positive number so as to dilate it. OK? I think that's maybe a good place to stop. Thank you, Jeff, for the very, very nice talk. Now let's open the floor for questions and comments. Does anyone have any questions or comments or remarks? Yeah. So in the formula, the factor n plus 1 or n1 factorial keeps occurring. Why does it keep occurring? Well, yeah, that's a good question. Factorials, of course, naturally occur in combinatorial problems. And for n large, oftentimes the set in question can be sort of chopped up into simpler sets to deal with in a combinatorial fashion. That's not a very accurate answer to your question. Maybe another one you might consider. If you work out for yourself the volume of the little l1 unit ball, I think that's 2 to the n over n factorial. And oftentimes an easy way to find that formula is by induction on n. And each time you raise the n, something happens that starts to cause the factorials to occur. Yeah. At a slightly different level, the gamma function, which sort of extrapolates the factorial, makes an appearance in many situations of this sort as well. Any more questions, comments? Yeah. Let me give you the microphone. I just want to make a small comment. I was very pleased to see that Santaló appeared in your earlier slides. Oh, yes. Santaló was a Catalan mathematician who immigrated to Argentina. And he actually was in the faculty when both Adriana and myself were students. Oh, I see. So you knew him. Yeah. I didn't know that. Yeah, I think there was a lot of mathematics in Argentina evolved because of the Civil War in Spain. So him and others immigrated. Yeah. Interesting. More questions, comments? I had a quick question, Jeff. Yeah. You mentioned the lower bounds of Mahler, of the four to the n over n factorial. But that is still conjecture, as I understood from your talk, right? Yeah. I think the best result is due to Borghain and Milman who have a slightly smaller value than four to the n over n factorial. But then it holds for all convex bodies. Well, the Milman-Borghain lower bound holds for all convex metric sets. I see. But it's smaller than four to the n over n factorial. That was one of my questions. Yeah. And the other one is that, in the cube slicing formula, you have two-sided inequality for the... And it seems to me that if you try to do, I mean, with other convex bodies, for example, a ball, if you just take a ball's license, it just becomes the same, right? My question is if there would be some sort of general principle that would apply not only to the ball or cube, but to a general, say, convex body with some mild assumptions, eccentricity conditions or something. You know, the problem is that you need some kind of symmetry to hope for something else, because you could just take a long thing like this. If you slice it one way, it's radically different than if you slice it another way. So you'd have to enforce some kind of symmetry condition. So for example, one way to do that is to make the norm invariant with respect to permutation of coordinates. So there are some inequalities.  I forget their names, but anyway, there are some efforts in that direction that have been made. Yeah, I was referring to some sort of bound that depends on the ratio of eccentricity from the largest ball and the smallest ball, something like that. And I noticed that when you mentioned Gromov's conjecture, there is no analogous upper bound, like you would have in the cube slicing. Is there something reasonable to expect that you will find also a point where it's not only bigger than something, but another point that is less than than something else? Yeah. I don't think Gromov formulated a proposed upper bound of that sort. I agree it's reasonable. You might look at Larry Gooth's paper. It's 25 pages, so he gets into the technical detail, shall we say. It's some sort of bizarre form of a pigeon hole. In a way. Okay, more questions or comments? Well, if not, let's thank Jeff again.


My name's Katie Altoft and I'm the coordinator of the Environmental Management Assessment Postgraduate Program in our School of Environment and Horticultural Studies. So our program is a one-year postgraduate. There are eight courses in each semester and it's intended to be really kind of finishing school. So we find that students come with a variety of different backgrounds, everything from deep environmental studies to things like communications, environmental governance, etc. And they know a lot of things but they feel like they don't know how to actually execute in the industry. So they've likely looked for work and haven't been particularly successful and even more concerning don't really see themselves in the job descriptions that they're looking for so they don't feel like they're qualified to do the jobs that they're applying for. So many of them choose to come back for an extra year in the hopes that they'll get some practical hands-on skills and be able to see themselves in those positions to be more competitive in the environmental industry and to be able to understand a little bit more about what they would like to do and wouldn't like to do in the environmental industry. So from that perspective we have a real obligation to make the program as practical as possible. So we're in the very fortunate position that we don't have to teach them the boring stuff because someone has already done us that paper. We get to teach some of the hands-on skills, the real finishing activities that would make them very attractive to an employer. So a good consultant, someone who would be effective in a policy position in the government, someone who could walk into a PR position or someone who could be the environmental coordinator of an organization. So to do that in the courses we always try to make them as practical as possible. Certainly from a student's perspective I want them to graduate here with a much more effective resume so they'll have a hands-on work placement but also a bunch of projects. So instead of I took this course I completed this project for this client. So they have a portfolio that speaks to what they can do. They don't have to say I can do this. They can say here I've done it. And so that requires a significant amount of work on our end because you need a constant supply of clients who are willing to let us do the work but if you think about it from a resource perspective it's an absolute tragedy to have 50 graduate students working on environmental issues that aren't real. So working on a project that's for the sake of school when you have a bunch of clients around the region who can't afford to get this kind of work done. So we've done an enormous number of projects across the region since the beginning of the program but I'm going to tell you today about one specific one that we're working on right now. The City of Welland was presented with a challenge a couple of years ago when the atlas landfill landed on their doorstep and so what this was is it was a hazardous waste landfill that atlas had been using for many years to dump slag as part of their production. One of the challenges is it was extremely close to the Welland River, a highly contaminated site and it became the city's obligation to clean it up. Sites of that magnitude were obviously very expensive to clean up and they didn't have the budget to do that so they looked for a creative solution and found one when Walker Industries came and said you know in a partnership scenario they'd be willing to clean up the site if they could use it then for their landfilling operations. So part one of the project was born and that happened and it's been ongoing since that time Walker's has been using it as a landfill. The landfill is very close to its useful life and as a result they're beginning to look at closure plans and that's where we stepped in. So this has been a site of challenge for the city and they were looking for solutions about what they should do with it. After the closure happened looking for a public acceptance of a potential conversion to parkland and looking at someone to help them walk through some of the legal steps that would be required for landfill closure, remediation of a contaminated site, conversion to parkland so that's where we came in. And so this project spans six of our courses currently. We are doing a phase one environmental site assessment of our contaminated sites project. We have done a PR project where we did a public open house to get public feedback on potential options for the project. We are doing an ecological restoration design for the project and we're doing the open houses now that the design selection has happened. So our intent is to be able to kind of walk this through everything from presentations to public outreach to technical sampling and analysis to a phase two environmental site assessment to a risk assessment to closure and at the end the students will not only have made a number of connections in many important partnerships around the region but also have a portfolio of complete projects that they can take to an interview and say here's just a sample of what I can do for you.


Motorcycle riders. With spring here and summer right around the corner, the time to schedule your motorcycle training course is now. The Navy's Motorcycle Safety Campaign aims to prevent motorcycle injuries and train sailors on the importance of proper riding techniques and safety gear. All riders are required to complete motorcycle safety training in accordance with the Navy traffic program. The basic rider course is a prerequisite for riding your bike on or off any Navy installation, and sailors must retake the course every three years. Even the most seasoned riders can miss minor defects in safety gear and routine maintenance. Remember to always wear your safety gear and practice proper riding habits to stay safe on the road. For more safety tips about riding, visit the Navy Safety Center website. To sign up for the basic rider course, visit the Navy Motorcycle Rider website.


Unlike a red flag, a yellow flag means proceed, but with extreme caution. Let's talk about spirituality. I have a project called Christian by Default with the premise being that most of us as religious people are only religious based on how we were reared and socialized. So if you were born 50 miles to the north, you might be a different religion. If you were born 100 miles to the east, you might be a different religion. So I don't necessarily want to go into the religious distinctions. Some of the people are going to watch this are Christian, Muslim, the whole nine. It's irrelevant. What I will say is, be aware of people who self-identify as empaths. Like I said before, these people tend to think that they were tasked by God to take on the world's ills and rehabilitate the world's children. It needs correct. You don't know that. I'm the only one who knows that. Whether that's the shitty men that they date, whether that's the shitty friends that they take on. And now as her man, you are left to deal with the consequences of that. Because again, she is going to in some way shape or form going to be the sum of her friendships and her past relationships. So if she has an empathic disposition where she is not selective or discriminatory in who she allows in her circle and in her body, you're going to be left to pick up the tab. Also, we're seeing a growing community. I call them the Ericobadu disciples, the Sister Mother Goddess archetype.  When the reality is that the black man is God, black woman is earth. Because the sea doesn't come from earth. It comes from the divine. And then earth incubates the sea. But the people with the Sister Mother Goddess archetype tend to harbor this deep idea that they are the gift. They should be worshiped. Men exist to glorify them, which sometimes manifests as irresponsibility that's never justified and never apologized for or narcissism that's never justified and never apologized for. And same with the, you know, the chakra, incense, sage, crystal, astrology, women that we're seeing pop up. Even now on TikTok, there's a sector of TikTok called witch talk where you're seeing this soft core witchcraft. Be wary of these types of women. They are not rooted in reality nine times out of 10. Their feet are not planted firmly on the ground. Sometimes they believe that they originated from another planet. Sometimes they believe they can see people's auras and that despite their ability to articulate their decisions or their plans, they just believe things are going to work out. Which really when you investigate it is just immaturity and incompetence, but it's packaged with crystals and it's packaged with sage and astrology and vibes and confidence. But yeah, this is my room and these are my high energy crystals. Like they I'm serious. I don't care if you don't believe in it. It's true. They got power. What does that one do? Oh, well, this crystal is all about good energy, all about good vibes. So it keeps all the negative energy away from it. That's kind of cool. Yeah, it keeps all bad vibes. Like, Samantha, stop leaving your cocaine on the floor. Oh my God. I keep saying, Mom, get out. No, no, it's cocaine all over the place.  I'm not going to keep doing that. Crystals, get out. Samantha, the crystals don't work. Please. No, the crystals don't work. Well, sometimes even Christianity, there's now some crossover. You find that sometimes women who are super Christian now live the life that you might be disgusted by if you were able to see what they did. Women who are now celibate. Sometimes they used to celibate a pussy. It's also consistent with women who self-identify as free spirits, women who self-identify as hippies, or mermaids, or goddesses, or they're familiar with only one goddess of the Yoruba, Pantheon, Oshun.  These are things to be cognizant of. And actually, this entire spirituality category could fall under red flags. But for the sake of this, we'll leave it in the yellow and orange realm. Send this to your boys. Have conversations about it. Share stories. Father, send this to your sons. Have conversations about it. Share stories. And as we continue to elaborate on some of these conversations, my hope is that we can establish a strong and prosperous black community full of strong men who are intentional about the women that we choose and the behavior that we reward or that we stay far away from. Thank you guys for watching. Leave it in the comments. Run these likes up. Hit the subscribe button on your way out. Check out some of the other content. And I'll see you guys in the next one. Peace.


I'm Ryan. And I'm Allie. After falling in love, we decided to ride bikes across America, collecting love advice along the way. This is Love Cycles. No flatties, no wannies, no crannies. Hi, oh yeah, y'all. Good morning. Good morning babe. I'm a little nervous. I don't know why. What's gonna happen? We're gonna ride our bikes.  That's what's gonna happen. I've been humming the theme from Jurassic Park this morning. It's very triumphant. And I'm wearing my helmet because I slept in it. I'm really excited to ride today. Your final time putting on the bags. My rear tire is looking a little bit bald, which makes sense. It's been 3,500-ish miles, but it's gonna last all the way. And I don't want to jinx myself, but we have not had any flats this entire ride. And you want to know why? It's because we sing the song every day. No flatties, no flatties, no flatties. All right, love bug. 100 more miles. We are not riding alone today. Jason is coming with us and I think just about three-quarters of Allie's family is going to meet us at a rail trail in about five miles. Oh man, it's a beautiful day. Sorry if I'm waking up all your neighbors. We're just a little bit excited. And everybody keeps on asking, how are you gonna ride into New York City? I don't know. I'm just gonna ask Google Maps and it's gonna tell me. It seems a little overwhelming, but we're gonna figure it out. What's up, Nate? How you doing, bud? Nate's going without shoes today. He's hardcore. He's like Huck Finn. I have not worn this helmet. Look, since I was little. Good morning, Jen. Good morning. How you feeling? I'm feeling good. How are you feeling? Feeling great. How many of your family members are here? What are we looking at here? Oh, I can never count them all. This family does not hold still long enough to be counted. Good to see you. Good to be seen. High five, Leah. Boom. Are you excited to be up at 7 a.m. on a Sunday morning? Yeah, of course. How about you? Yeah. One, two, three, no wabbies, no crash, he's so flighty.  That's impressive. This kind of reminds me of that scene in Forest Gump when he's running across the country and people just start joining him and there's a huge crew. I'm pretty tired. I think I'll go home now. Feels so great. We're surrounded by love, which makes sense because love cycles, you get it? You be very careful. A lot of the alley crew is turning around now. It's like the shuttle. We had the booster rockets boosting us this far. Now they're peeling away, letting us go into our orbit. Shiloh, Jason and Wyatt are going to continue the ride with us for a little while longer. I like that we have one last hard day to finish this thing off. 100 miles, a little bit of hills, feels good. Feels really good because after today, we're not going to ride our bikes nearly as much. I'm excited to go running in fact. It's been three months since I've gone running. Join these country roads out here. It's hard to believe that we're going to be in the concrete jungle of New York City later this afternoon. New York is really impressing me with its amount of rail trails. We're up on mile like 550-ish now in New York and mostly trail. So thank you, New York. You're awesome. Worth County Rail Trail, and this is going to take us all the way to Van Cortland Park, which is just north of Manhattan. Thanks for riding with us, buddy. You're the man. Shiloh and her son, Daniel, rode with us for 65 miles. That means you have 65 miles to get home. Everybody in your family is a super athlete. That's awesome. Happy trails. That just leaves you and me, Ali. Can you believe it? We have 34 miles to New York City. What a feeling. 34 beautiful, awesome miles. Let's enjoy it. We are hidden in the jungle here and about 100 feet that way is a raging highway. I am so glad that these rail trails exist. In the Bronx, would you believe it? It certainly doesn't look like the Bronx on TV. It's like we're deep in the woods. We could be in Idaho right now. Well, we certainly aren't on a rail trail anymore. Welcome to New York City. On the west side highway along the Hudson River, bike path all the way down to the bottom of Manhattan. Look at that bridge. It's huge. We have 3,000 miles ahead of us. 3,000 miles of adventures and beauty and open landscapes and starry skies and wonderful strangers that will become friends, and it's going to be amazing. I'm very, very happy to be doing this with you. I'm so grateful for you for inviting me along. Yeah. Thank you. This is a dream come true. This is one of those trips in life, which is an absolute dream come true to be able to do something like this with somebody you love. Although this is not officially the ocean. It's the Hudson River, but eventually it connects in a roundabout way to the ocean. We started in Astoria, the Pacific Ocean, and this will go to the Atlantic. It is beautiful. We are so close to the end of our journey. I know. We're going to ride down this west side highway and go across the Brooklyn Bridge and be done in Brooklyn. Like this. Okay. And since we're riding our bikes across the country talking about love, we might as well be doing spinny, lovey things. This is what they do in the movies. Watch this. And these are the moments when I've done these bike tours and I've been alone. These are the moments where I'm like, God, I wish I could share this with somebody. I'm so excited to camp in Cuddle tonight. It is time to go across the Brooklyn Bridge and fight through a sea of tourists. This is more treacherous than riding some of the highways we took across America. So Ali, our goal was to find the hidden roads of America. What do you think? Oh, I think that so far we've succeeded. We are out here. Yeah, I am loving all the gravel and just long stretches of empty highway. It's really special out here. We have landed safely in Brooklyn Heights. We are going to be my good friend, childhood friend and his wife, Magina and his little daughter, Elodie. And we're going to eat ice cream. Because that's what we do when we ride 100 miles. We eat ice cream first before anything else. What's up everybody? Oh, I'm sorry to scare you. I'm sorry to scare you. Hi. Let's get some ice cream. Ice cream will solve all problems in the world. Oh, look at that. Yum. Yeah. Believe it. Look at that city behind us. That's the city we've been aiming at for three months. Oh man, it feels so good. We did it together. Do you remember that day leaving Astoria, Oregon? I do. I do. Remember the waves washing over our feet and filling my shoes. I remember having tears in my eyes. I was so excited. Then gosh, all the nights getting lost in Washington and Idaho, the motorcycle guys, hot springs. It feels like we've lived so many different adventures, like a year of adventures. I know. The past two months. Every single day seemed like two days. We got two days for the price of one every single day. It was amazing. Every single day we made it happen. Every single day we woke up not knowing where we were going to go or where we were going to eat or sleep and every single day we made it happen and we did it together. Then we'd fall asleep next to a state fair. That was awesome. I'll never forget that. Or in some random town park we were poaching. We did that a lot. Yeah. I'll never forget all the people. Even the people we didn't really document or capture, but just the short moments we had with the guy outside of a gas station that was so psyched about our cross-country ride and tells us that we inspired him. Our time with Dana and Xantha and Bobalicious and Michelle and gosh, your whole family and Ben Leroy. Love that guy. And here we are. Look. How many incredible humans. I keep spinning us around because the audio's better the other way. This is into the wind. There's the city. This is better for audio. This is better for audio. This is us. It's kind of surreal. I mean for so long New York seemed impossibly far away. And as we rode, as we got here today it was like are we really going to be in New York? Because we were like in the jungle for most of the day on these rail trails and it's like New York can't be around the corner and all of a sudden there it is. There's New York. Now we're at the tip of it and it seems like the perfect place to start a stop. It is a good place. It's the perfect end for a journey. Yeah and by the way this is your second century ride. 100 miles. Congratulations. That's true. Still doesn't come easy for me. Yeah. Now we saved a big day for the last day. I mean we're going to be processing this for months and months and even years. It was all so much so fast that we really didn't get to when we took it in as much as we could but I think as we go home when we chill out and we wake up at three in the morning we're like oh remember that one time? It's all gonna I'm so excited. We get to relive this forever. It's a really beautiful thing to kind of share with you. I know. It is. I feel so lucky. I feel really lucky to have found you. I feel so lucky I found you. I'm really grateful that we got to share the past two months together. Two and a half months. Two and a half months. We left June 11th. It is now August 20 something five? 25. My arm is hurting. I'm holding this camera up. I think our I think our final speech is going to come to an end before my arm breaks off. I guess I could we could say thank you to everybody out there. Thank you. Thank you for supporting us and sending us all of your encouraging comments and just all the love that you poured at us. It really helped us. We really appreciate it. So thank you all for following along.  Absolutely. Even though we don't know a lot of you in person you're very important and we did meet some of you. That was awesome. We did meet a few of you YouTubers that met us out in the middle of nowhere. We appreciate that. So man, here's to many more adventures with us and sharing with you and love and joy. And I think my final word is that I just have a renewed faith in humanity after doing this ride all the beautiful people we met. There's so much good in this world and I can't wait to keep on exploring and making new friends on this planet and just spreading the joy spreading the love. I would say that my patriotism has been renewed. My my belief in this country and the ability of Americans to cross divides and find that place where they can connect regardless of differences and love each other. Yep. All right.  Oh, there it is. Ladies and gentlemen, New York City Love Cycles has made it.


Meat Boy is kind of back because today we are making a beef and barley stew. The reason I like this so much, it's a more nutritious and delicious culinary take on something like chicken and rice, a basic protein and carbohydrate. By using higher quality ingredients, you not only add plenty of flavor, you're packing it full of balanced nutrition. And this kind of replicates how a lot of recent indigenous people ate. I know a lot of you guys are familiar with Weston Price's work. To me, this embodies their local harvest depending on the protein and grain source and then they could have added whatever seasonings, spices, extra stuff they had on hand from that harvest. So let's take a look at the ingredients and go over their nutritional properties. There isn't really a concrete recipe for a beef and barley stew. So it's your own take, it's what you have access to and also what you would like to do, you know what type of seasonings and spices you enjoy eating. For instance, a mirepoix, which is celery, carrot and onion, is sometimes added, but I don't really like celery and carrot that much so I'm just using onion and from a nutritional perspective onions aren't really that great, probably the only thing here that doesn't have anything to offer besides the garlic, which is what we're using just to add some flavor. Again, every single thing here is organic. We try to remove as many negatives as possible, especially the plant foods for the meat. It doesn't have to be organic as long as you know it's grass fed decent quality. These are the chuck steaks we have on Frankie's Syri range meat. You could throw these on the grill, sear them, sear them in a pan today. We're going to kind of dice them up, brown them and cook them for an hour or two in this stew. The other plant components, I have some mushrooms and some seasonings. These are in a box because I got them from my restaurant supplier, Shiitake are actually really, really high in copper, magnesium, some B vitamins. But before I brag about the nutrition of mushrooms, I don't digest them. They kind of come out the same as they came in so although the paper value nutrition is great, I'm not really sure the body is utilizing it. The main component, the barley, we have organic pearl barley. This is actually reasonably nutritious and it digests and it's calorically dense. You have some B vitamins more importantly compared to other grains. This has magnesium and copper, small amounts of the other minerals as well, manganese which is really, really missing in things like white rice. And that's also reflected in the flavor, it's nutty, it's more complex. This is pretty cheap, I think it's like $1.50, $2 a pound depending on what supermarket you get it from. The other main component, the beef broth and I haven't been making my own broth lately, ideally that's what you would do but since this is a stew, a soup, we do need quite a bit of liquid. A lot of recipes do call for using water so don't feel the need to spend 10, 15, 20 bucks on high quality beef broth, you know, you could actually buy some meat to make it yourself with that much money but this is what we're using today. Substitutions could really be any kind of beef you'd like, any flavored seasoning vegetables you'd like, again you could use water but the quality of the ingredients, what you're putting in is going to determine the nutrition as well as how it tastes and although the base being beef and barley you're not going to alter the, you know, the mineral and vitamin component that much but you can add or subtract a little bit especially when choosing the broth because using water is obviously going to be way lower in the vitamins than beef broth. I'm using bay leaf thyme and just salt and pepper, some really simple seasonings, something I've always enjoyed the flavor of, you might see a lot of other stuff added ranging from like soy sauce to Worcestershire which, you know, from a quality and taste perspective I don't really like and that garlic we mentioned earlier, you know, this again was just some stuff from my restaurant supplier that was pretty cheap and I was doing some pickling so I have a little bit of extra garlic to throw in here. All we have to do is dice up this chuck steak, we need to dice up these shiitake mushrooms and this is obviously, you know, way too many mushrooms, way too many onions, way too much barley for one recipe. The actual ratios are going to be for every one pound of meat you put in here you want about three quarters of a cup of barley and for the flavorings I'm just kind of guesstimating the onions, the mushrooms, the seasonings, the garlic, that's not too important the amount you put in, you just want enough to add flavor and for the broth you want like two to three cups of broth at least for every cup of barley and then you can always add more liquid to give it that soup like consistency. So let's get everything prepped out and we'll show you guys. So we have everything all prepped and for that two pounds of grass fed chuck steak that we kind of chunked up I took a proportional volume of onions and then I did roughly double that of mushrooms so I'm kind of like looking at the plate of meat I eyeballed that amount of onions and then the reason I did double the mushrooms is because these shrink down a lot just some cloves of garlic and again we have our seasonings one and a half cups of barley two pounds of meat the two significant measurements so on the stovetop you can do this in one pot I do think it's a little bit nicer if you use two because then you can really caramelize and brown those vegetables really well so what I'm going to do is take you know our main pot put it on high heat coconut oil in both the pot and I also have stainless steel pan that we're going to use for the vegetables now if you do use pork lard beef tallow whatever fat that's animal based butter would be delicious as well you're going to add a lot more flavor to this the reason I'm using coconut is because it's antimicrobial it helps against things like candida you know what the gut issues I've been having since the iron overload coconut oil has kind of helped me out a little bit so after we get this initial 10 15 minutes of cooking done then it's basically just hang out for an hour or two until the barley is done cooking so we sear the beef in the pot get it nice and brown then we're going to add our beef broth and our barley to that pot in the other pan we're going to sear off the onions the garlic and the mushrooms until they're nice and caramelized and then add that to the large pot with the beef and the barley then we're going to add the spices and seasonings and everything's going to kind of mold together and build and develop flavor so let's put the beef in and the most important thing to do here is don't overcrowd the pan otherwise it's going to steam and you're not going to get brown color this is just about right just want to kind of spread it out evenly in the bottom of that pan so each piece has contact with the heating surface so that it gets brown now onions in the pan over here mushrooms as well so after about five minutes on high heat our beef is looking about done really brown really caramelized a lot of bits in the bottom of that pan I want to really deglaze that with some beef stock to get those brown bits up and to stop the beef from burning so we'll get our first carton of beef broth in which is four cups for you reasonable people out there that's about one liter our veg is starting to brown nicely probably needs another 10 minutes just because it's so much and this is the most important part you know if your beef wasn't really really deeply caramelized like that if you don't get a lot of deep brown color on your vegetables hey this is still going to taste pretty good but it's not going to be like oh my god this is the best thing I've ever had in my life is what we always aim for another carton of beef broth and this is just a brand I prefer you know from any ingredients and quality perspective so three cartons of broth in here what I want to do is I want to go with my spoon and kind of scrape up the brown bits on the bottom of the pan I'm doing this by feel which is pretty easy then we'll add our one and a half cups of barley and this is how you would kind of normally cook barley to it's just a lot less liquid you know you would still bring it to a boil simmer it for a couple hours and the barley would be done in this case since we want to soup the volume of liquid is just you know two times higher now as this vegetable starts browning you know we do have the brown bits on the bottom of the pan kind of want to scrape that up so they don't burn too much now I'm going to take a little more coconut oil put it in the bottom of the pan here which I kind of emptied out and that's where I'm going to put our garlic you know the garlic the garlic's a lot easier to burn than the mushrooms and the onions so the garlic's nicely browned the rest of the vegetables actually getting a little too dark so I'm gonna take some more beef broth and all that brown caramelized vegetable flavor is coming up off the bottom of the pan we're gonna really scrape that up get our nice flavorful mixture into our soup now if you really want to you could put these spices in here with some oil just to like activate them a little bit but it's really not necessary so we have two bay leaves two teaspoons of thyme maybe a teaspoon of black pepper and a teaspoon of salt and we can always add more seasoning afterwards this is just the initial batch so this has been on high for you know about 10 minutes it's almost at a boil but this is basically where you know we bring this to a boil and then let it simmer for maybe an hour an hour and 15 minutes and then the barley should be cooked now we're gonna go for a moderate simmer for maybe an hour and then we'll test the barley and we'll see where it's at you know everything's really done here we're just tenderizing the chuck a little more we're cooking the barley and we're letting all the flavors meld and incorporate so our soup has been simmering for and maybe an hour and a half I haven't really been keeping track but it looks like it's done so we're going to assume so and take a bowl we have that barley it's kind of all nice and puffed up now there's not actually a crazy amount of liquid in here you know you could definitely add some more we got those chunks of garlic mushrooms some beef this looks good although I just had a cheeseburger so I don't know how good it'll taste so depending on you know how liquidy you want this to be if you want it to be warm like a soup definitely and more you know this is a lot of grain a lot of meat a lot of veg compared to a lot of soups that people usually used to I will say when I've made barley in the past with just water and salt I don't really like the flavor like if I make some white rice even brown rice I can kind of scarf it down but the barley not as palatable I kind of don't like the natural flavor of barley but let's get a spoonful with a little bit of everything so let's get some meat let's get some mushrooms a little bit of onion very hot the garlic I usually add for flavor because what I've noticed is if I physically eat the garlic it does give me brain fog and that seems to happen with a lot of people there's certain compounds and garlic that do that I feel like this is something my girlfriend Victoria would like a lot although ever since she went vegan you know she's making like mushroom barley soup now she's omitting the beef but you know we get her some beef in some other ways that's really good this is exactly what I expected you know it's good it's not like the most amazing thing you're going to ever have but the flavors are balanced it's complex it's delicious you couldn't eat it with every meal there's not a crazy amount of fat so it's almost kind of like palatable and light what I really like about this is you can't overcook that barley if you've ever had barley before it has almost like a chewy texture naturally so even if you cook it for like five ten fifteen minutes extra it still has that texture and that chew to it and despite just eating two cheeseburgers liking this a lot it's really good mushroom onion beef seasoning spices really simple but if you do it right really amazing that Chuck got so tender it's so tender you know I didn't think the the choc roll would be that tender after just an hour of cooking but you cut it small enough that it really does so if you guys haven't had beef and barley soup before definitely try a recipe like this even if you don't want to use barley just try to mimic those ingredients with really any type of grain I mean even pasta I think it's going to be completely delicious so thank you guys for joining me today I think my sister is really going to enjoy that for her lunch tomorrow I've done recipe videos every single week for like as long as I've been on YouTube so you guys can go back check you'll be really surprised about what you see and I think I'll do a playlist update in the very near future so you guys can navigate my thousand videos a little bit easier you could please leave a comment down below drop a like on the video make sure you click that notification bell so Frank he doesn't get sad when he looks at his view count after posting his video every day and Frank dash the follow calm for anything else you guys would like to see so thanks again for joining me guys I will see you for tomorrow's video and we're going to do a live stream later on Frank to Fano check it out


Wireless charging is getting more and more popular with every new mobile phone released, and it is pretty clear why. It is comfortable, easy to use, it may even make your device more reliable. However, wireless charging doesn't have to be limited to mobile phones only. Other devices would benefit from it as well. Let me show you some of the examples and advantages of wireless charging. Nowadays, wireless charging is often used in smartwatches and earbuds. Other devices, previously reliant on wireless charging like health tracking monitors, spot sensors and smart glasses, are also being augmented by wireless charging. We may help you save up on PCB space as well, since our circuits are highly integrated, requiring very few external components. A notable advantage of wireless charging is that we can remove charging ports entirely. This helps avoid corrosion due to harsh environments, but this is not the only advantage. A rotating device, such as this camera, would normally be powered by a slip ring. Here, our STWBC86 transmitter and STWLC38 receiver power the camera wirelessly. This decreases mechanical degradation and maintenance costs, while also improving durability of the solution. We are also using ST60, a high-speed wireless data transfer chipset, providing both wireless data and power. Recently, she got a major update, introducing MVP, a new wireless charging standard utilizing magnets to achieve precise co-to-coal alignment and even better charging performance. With the update, also came new features and requirements, and our STWLC89 supports them all. This wireless power receiver can deliver a 15W inch MPP mode or up to 30W with our proprietary ST Supercharge protocol. Not only that, ST also offers a 7-year longevity for all wireless power ICs. To ensure fast time to market, our support includes co-design insimulation, software for tuning and firmware customization for key customers. For more details on the idea of these products, visit our website ST.com and have a nice day!


My name is John Anthony Luna, and I am the developer of the game Colonialis Jamestown. It takes place in the early 17th century North America where you place the English Running the colony of Jamestown, and your only objective is pretty much to survive My former science school teacher informed me of this Competition and encouraged me to join way back in May. I used scratch and it's my first time using it I pretty much taught myself just put the blocks together and see what happens. I think that's the best way of learning scratch Well, aside from games and game making, I also tried to make my own Drawing I like reading books on history watching documentaries and some anime when I have time


Let us now see some implications of this rank nullity in the context of one-to-one and on-to maps right. So I think it's by now okay let me leave it there for the moment and now let us say that you have suppose phi is one-to-one what is the equivalent condition of one-to-one that we saw just a while back the kernel of phi must be the 0 subspace what is the dimension of the kernel in that case 0 right. So that means by the rank nullity theorem what must we say what should we be able to infer that dimension of v from where you are taking objects and passing it through phi must then necessarily be equal to dimension of the image of phi is it not right just straightforward application of the rank nullity theorem to this condition okay. Suppose phi is on-to surjective what do we know? What do we know about the image of phi in that case it's w right. So then dimension of v is equal to dimension of kernel of phi plus dimension of w but now what sort of values can this number take non-negative integers is it not. So what can you say for on-to-maps about the relations between the dimensions of v and those of w's see dimension of v minus dimension of w is equal to dimension of kernel phi which is what greater than or equal to 0. So therefore dimension of v in such case can you not say this the dimension of v what is it yeah greater than or equal to we have to you cannot discount the possibility that is also equal what about this can we say anything about it of course we can where is this image of phi residing inside see image of phi is residing inside w. So it is a subspace of w can the dimension of a subspace be greater than the space inside which it resides of course not. So this is less than or equal to dimension of w right straight away in view of this now I urge you to think in terms of matrices back again so we will keep going back and forth with phi and matrices at special cases of these kind of phi's or linear transformations now think about it what do you think where do you think do fat matrices and tall matrices fit into the picture. So can tall matrices be both one to one and on to if at best they can be one of them then which of these can they be what does a tall matrix do suppose you have a matrix A which is m cross n can I not think of this as a mapping from mapping from where to where yeah. So what kind of matrices do you think can be one to one sorry tall tall means these numbers if you will think of them m is greater than n. So that means fits in here right because m is where you are mapping to n is where you are picking out objects to map from right similarly if it is on to where it must be a wide or a fat matrix right. Now the only way that you can fit both of these pictures at the same time not just for matrices but for mappings in general if I want a bijection what am I basically talking about. Now you have to impose both of these restrictions and therefore for a bijection what do I require that the dimension of V must equal the dimension of W. In other words in the language of matrices I am talking about square matrices yeah straight away right. If you have to invoke both of these then equality is the only possibility. So if equality holds is it guaranteed I mean is it an if and only if condition not really right this is a necessary condition for bijection but just because you have this I can cook up several linear maps which are mapping from vector spaces of equal dimension does not mean that the map is always going to turn out to be or the linear map is always going to turn out to be a bijection of course not. In fact if you do manage to find a bijection between two vector spaces independent of your knowledge about their dimensions if you found that they are indeed bijections that is an independent way of proving that the dimensions of the two vector spaces must be equal you follow what the point is right. Just because I have told you that the dimensions of the vector spaces are equal not every mapping that takes fellows from V to fellows in W is a bijection just because the dimension is equal it does not guarantee that but if you have found a bijection from a vector space V to a vector space W then you can definitely be sure that V and W have the same dimensions. In fact this is a very important class of mappings and we say that these are isomorphisms. So morph means morphology means structure iso means similar right. If you find a linear bijection then that is the same as finding an isomorphism you have already seen one isomorphism though I did not probably give it that name at that point in time can you think of what I have described in this class which is an isomorphism the way we assigned coordinates yeah the way we assigned coordinates is a clear cut case of an isomorphism why well here is another interesting cute little result that will tell us or show us in a very easy manner why it must be true why what we have shown is an isomorphism. So the claim is this okay suppose dimension of V is equal to dimension of W. So I have already started with this that dimension of V is equal to dimension of W does not mean it is a bijection but under these given circumstances surjection implies injection and the other way round. In other words when you know a priory that the dimensions of the two vector spaces V and W are the same if you manage to show that a particular mapping is an injection or a one to one mapping you do not need to separately prove that it is onto and the same way if you know somehow if it is easy to show that it is an onto mapping you do not need to separately prove that it is an injection it automatically follows right. So just verifying one property brings the other property as a guarantee right. So how do we show this suppose of course this is a base that we have to build it on so this is going to be true anyway assume injection it is just what we have done a while back just for the sake of this proof I will have to rewrite this assume injection means what the equivalence condition is that the kernel of phi is 0 right. So dimension V is equal to dimension kernel phi plus dimension image phi but this of course is 0 yeah what do I know about dimension V dimension V is equal to dimension W. So this means this means what dimension W from this given condition I am invoking here and from this given condition that it is an injection I am invoking here is equal to dimension of image of phi but look image of phi is contained in W if its dimension is equal to W we've been over this step in the previous lecture to we've seen that in a subspace has the same dimension as the original vector space then they must be one in the same you don't have to show containment both ways likewise to show that they are one in the same. So this means that W is equal to image of phi which is surjection next if you assume surjection first again it is the rank nullity to the rescue all the way. So if you assume surjection first then what happens of course just write the base step again that rank nullity which is why I haven't erased it dimension V is equal to dimension kernel phi plus dimension what is this image phi W right W but again by the given condition this and this must vanish. So therefore dimension kernel phi must be equal to 0 but we know of only one vector space which has dimension 0 which is the 0 subspace right. So kernel of phi must be that is phi is an injection. So when you are dealing with vector spaces whose dimensions are the same verifying one property is tantamount to verifying both of them at the same time you don't have to separately verify them. So why is this useful what were we discussing just before this yeah isomorphisms right. So now think about isomorphisms and think about the case where we assign this coordinates and we claim I just claimed a while back that it is an isomorphism. What is the one claim that I had made then think about that matrix that dictionary that I called from one language to the other what is in the kernel of that matrix we have already claimed that matrix is non-singular it is invertible why because we saw that the 0 gets mapped to the 0 the 0 of vector space V the n dimensional vector space gets mapped to the 0 of the n tuple f to the n and nothing else. So then it means what dimension of V is n dimension of f to the n is n. So they are definitely vector spaces of same dimension that is what the coordinate assignment does. So this checks out that V is equal to w and I have already shown injection because I have shown you that the kernel contains nothing other than 0. So that means it is a surjection if it is both injection and surjection it is a bijection if it is a bijection then see nothing is like carved in stone it is all basically part of the definitions and just pushing your way through. So it means it is an isomorphism they are the same structure which means that you can basically deal with them like you deal with Euclidean spaces that is why the idea of matrices is so powerful because once you have a finite dimensional vector space once you can figure out what its dimension is you can always say there is an isomorphism through coordinate assignment through choice of an ordered basis there is this isomorphism and then you can deal with it like you deal with n tuples of numbers right. So that is the power of this isomorphism right. So it is very important that you find out in any way whatsoever some one to one onto linear map so the three things that you need in a map it has got to be linear it has got to be one to one it has got to be on two. If you can guarantee those three things you do not have to check the dimensions of those spaces separately you can rest assured that the dimensions of the two spaces will be equal yeah. There is another beautiful thing that this allows us to do which is when there is an isomorphism you can always cook up inverses yeah you can always get inverse maps and the fact is that even this inverse map is going to be linear right. So not only does it guarantee the existence of the inverse but it also says that the inverse map is also going to be linear that is the question that we shall now try and investigate for matrices you already know this to be true you have this formula we are not going into those formula based things like determinants and all but you know that A is a matrix A inverse is also a matrix. So of course any matrix is a linear transformation so if you are operating between vector spaces or n tuples of numbers where n is equal on either side then it is a square matrix and any square matrix you know the explicit formula for those invertible square matrices right. But what we shall see is that in general for any linear transformation this is true right. So that is the main value of this kind of thing and then we shall see something very interesting even when you do not have square matrices that is to say even when you do not have the dimensions of these two sides to be equal you can still have some limited kind of inverses yeah in a limited sense you cannot have inverses like you do for square matrices but you can still push your luck and get certain kinds of certain special kinds of inverses right. So that is going to be the object of our investigations next. Any questions or doubts so far? Before we move over to these studies of these inverses there is a very important property that will set the tone for our study which is that what is it that is so good and nice about linear transformations? Why do we prefer them or why do we feel that they are tractable? See when you are dealing with finite dimensional vector spaces if you want to characterize what a linear transformation does to each and every object in that vector space and that can mean in finite number of objects in the vector space you only need to know what it does to a finite number of objects in the vector space and that is it you do not care about anything any other piece of information right. So if phi which is a linear mapping from vector space V to vector space W these are finite dimensional vector spaces right okay. So phi can be completely characterized by its action on any basis or V right which is to say that suppose BV is equal to V1 V2 Vn till Vn is so suppose this is a basis for V then knowledge of phi V1 till phi Vn suffices in evaluating phi V for any V in V. This is going to be the theme for any linear system okay that you know what it does to a few of them and you can basically extend that and figure out what it does to every one of them. So this is why it is so important to understand what a linear transformation does to a basis set. So let us try and understand why this is true any V in V can be written as V is equal to summation alpha i Vi i going from 1 through n consider phi V which is nothing but phi acting on the same object summation alpha i Vi but now because of the linearity I can pull out these terms one by one separate them. So this is nothing but summation alpha i phi Vi I have done two operations at one go okay. I first said this is phi of alpha 1 V1 plus alpha 2 V2 plus dot dot dot till alpha n Vn which is equal to phi of alpha V1 plus phi of alpha V2 plus phi of alpha V3 and then I have subsequently pulled out the alphas outside the scalars and said it is alpha 1 phi V1 plus alpha 2 phi V2 plus alpha 3 phi V3 so on till alpha n phi Vn right. So I have done both of those at one go but then notice what are these objects these are exactly the action of phi on the basis. So irrespective of whatever V you pick out I did not choose this V to be anything special it is an arbitrary fellow just an ordinary fellow sitting inside V which is representable in terms of this basis. So all that you need to understand in order to understand the action of characterized action of this phi is to understand what it does to a basis set. Once you know what it does to a basis set you basically defined it for good right yeah. So it is then important in this slide to investigate how we can say that when there is a one-to-one onto or a bijection mapping between two vector spaces V and W the inverse will also turn out to be linear and then how do we define the inverse. Of course defining the inverse that recipe is going to be very straightforward once we have figured that the inverse is also linear because just like the original map the inverse map is also going to lend itself to easy description once we know what it does to particular fellows therein right ok. So this idea is clear suppose phi which is a mapping from V to W is an isomorphism first we have to understand what we mean by this idea of an inverse. So I will try and explain that by means of a diagram. So this is V this is W and this is V again. So suppose I go this way using phi right then I go this way using psi. If all that it does is basically maps V to itself in what we call the identity mapping this is the id means identity, identity mapping but identity mapping in vector space V right what do you think is the cumulative effect of these two yeah no but if I am to compose these operations together isn't this the same as psi composed with phi being the same as the identity in the vector space V. On the other hand if I had the following W and W right so shock so this is my phi right. So this phi takes me from V to W and this is basically the equivalent of okay have I used the same color no I should use the blue color probably right yeah. So there is a fellow W here which gets mapped to W note that this is the identity in the the identity mapping in W it takes objects in W and maps it to the same object back yeah but then I am seeking maybe I should have used a different color for the object that I am seeking which is in this case psi and in this case this psi. So this is the much sought after entity but their actions are different you see because in this case in this case what is the object doing just notice what is happening first psi acts on it followed by the action of phi so isn't this the same as phi composed with psi giving me the identity operation in W is that clear see it is very important to understand if I don't put this subscript here you might be led to believing oh it's the identity no it's not so that's the significance of the left operation and the right operation in either case the object I am seeking is psi all right so that's the idea of inverse one sided inverses may exist but both sided inverse will exist for isomorphisms yeah that is that is the claim so with this picture we would like to bring this module to a close and we will discuss the idea of inverses and how they can be defined given a map how we can actually go about constructively defining the inverse of that linear map okay