 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 just look at the matrix A, which is m by n, and you look at all the m by m sub determinants, and then you add up the absolute value of all the m by m sub determinants, and that's the volume of the projection of the cube onto an m dimensional subspace. 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. Waste is one of those English words which, if spelled W-A-S-T means stuff you throw away, spelled W-A-I-S-T, it's the part of you where you wear a belt, your waste. 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. There are a couple of French people who work in functional analysis who have looked at slicing LP balls and LQ balls. 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.