 You can turn to page two now, Philip. Okay, so this is a talk mostly about inequalities between the regulator or relative regulator or regulator for a group of S units and the product of the V heights of the generators of the groups whose regulator is being considered. Parts of it are just described in the simplest language of Euclidean geometry. For example, the second page here that you're looking at. And then later there will be parts of it that will involve the basics of algebraic number theory, the construction of unit groups and so forth. So this first slide defines something in the middle of the page called Delta of bold face X where bold face letters are always column vectors in this talk. I've introduced a few other things from basic real analysis that I hope you're mostly familiar with X plus is self-explanatory X minus and so forth. The heart of the matter in this talk is a rather unusual norm that as far as I know was first appeared in a paper that Schinsel wrote in 1978. Schinsel's paper gives a upper bound for a determinant in terms of the what I call the Schinsel norm Delta of X defined here in the middle of the page. So the way it works is you start with a N by N matrix capital X and X1 bold face X1 bold face X2 and so forth are the columns of the matrix. And I'm interested in or Schinsel was interested in giving an upper bound for the absolute value of the determinant which is somewhat similar to Hadamard's inequality for a determinant. Anyway, I'm also going to be interested in something that's joint work with Shabnamaktari in which we replace the determinant of an N by N matrix with the L1 norm of a more elaborate object which occurs in the Grassman algebra or exterior algebra generated by the Euclidean space R to the N. So you might remember that the determinant of X could also formally be written as X1 wedge product, X2 wedge product and so forth until you get out to XN. That object in the Grassman algebra or exterior algebra only has one coordinate and that one coordinate is the determinant of the matrix X. But down at the bottom of this page, I describe if you're not too familiar with this what happens more generally if you had fewer than N vectors in your wedge product. So at the very last displayed equation I illustrate X1 wedge X2 wedge and so forth but just with a wedge product of L vectors. How exactly do you compute and what does it mean to speak about the L1 norm of that object in the exterior algebra? And the answer is, well, it's very simple. You just sum the absolute values of all the L by L sub determinants. The set of L by L sub determinants is often called the Grassman coordinates for this vector X1 through XL. And roughly what I would like to demonstrate here and indicate is that when you start with algebraic numbers and create the regulator, you could think of the regulator as a kind of height on groups. The regulator in its simplest form is the height of the unit group. If you expand to an S unit group, you get a height on the S unit group. This will take place here in a few more pages. Anyway, for the moment, I'm just working in Euclidean spaces. There are no algebraic numbers present. And the bottom line here is the important one. It's the recipe for how you compute the L1 norm of a wedge product of L vectors but where each vector has N coordinates and in general, N is larger than L. So the sum on the right down there has binomial coefficient N over L coordinates. Okay, let's go on to page three. Okay, so here is Schindsel's determinant inequality. This is kind of interesting. It's in the paper indicated there. It first appeared in 1978. I think I stumbled upon this in Oberwollfach at a meeting. Schindsel was present at the meeting. I was looking for something else in the same volume and my eye simply hit the first page which said a determinant inequality, something I was sort of interested in. And I looked at it and shrugged my shoulders and went on and gave it no more thought. Schindsel's inequalities, very simple. The determinant of the matrix X on the previous page is the product of the Schindsel norms applied to the columns, bold face X1, bold face X2 and so forth. Hadamard's inequality looks like that too except the delta becomes the Euclidean norm. It turns out that the Schindsel norm has a wonderful property which Shabnamaktari and I discovered, I think maybe 20 years or so after I first noticed this in about 2015. Here, before we go on to discuss the applications of Schindsel's inequality, here is a generalization that Akhtari and I proved. Suppose you start not with n vectors but with L vectors in Rn. Is it possible to give an upper bound on the L1 norm of the wedge product of the L vectors in terms of the Schindsel norm, the delta norm? And the answer is yes. And theorem one reports the news. I'm reporting the news here in two cases. One case is that n is in between L and 2L where you see a constant depending on n is a little different than it is when n is bigger than 2L. If you wanted to, you could make one formula and simply use the minimum of the two expressions n over n minus L to the power n minus L and 2 to the L. It turns out that the constant in front of the product of deltas is the minimum of those two always. This inequality is very sharp. The bottom inequality always has cases of equality. And the middle inequality always has cases of equality whenever n minus L divides n. It is known that the best constant in the middle inequality, the one with n over n minus L to the n minus L power, it is known that the best constant is always an integer. And in fact, with more effort, you could show that the best constant is the integer part of the expression there. But I didn't want to clutter this up. So I put in expressions that were sort of the simplest ones to look at. Okay, so later in the talk, I will come to a point where instead of regulators, we're going to be talking about the analogous object associated with very proper subgroups of smaller rank of unit groups. Okay, let's go on to page four. Thank you. So now the Euclidean geometry is set aside. And these are things I hope are familiar to many of you from number theory. K is always a fixed algebraic number field of degree D at each place V. D sub V is the local degree. And that's enough information on the second displayed equation to define the V height of an algebraic number alpha in K as a sum over places. In fact, it defines, as you can see, the degree times the V height with slightly different renormalizations you would describe the V height. Okay, now I want to speak about some multiplicative groups. And let me start with the group of S units. So the S integers in K, here S is always in this talk, a finite set of places of K that contains all the infinite places. And so the ring of S integers, a generalization of the ring of integers is the set of points in K whose viatic absolute value is less than or equal to one for all V, not an S. And then just below it, you see the unit group. It's well known that the unit group O S cross of S integers, the group of S integers is a finitely generated group of rank equal to the cardinality of the set S minus one. So that will be always in this talk called little R or little R of K when that is necessary. Now I'm also going to use the places in S, the Vs in S to describe a copy of Euclidean space where most of the action will take place. So that's R, boldface R, the real numbers to the power little R plus one, and I'm especially interested in the diagonal subspace of all column vectors X which have a sum of coordinates equal to zero. The coordinates now are indexed by the places in S. This is the so-called diagonal subspace. So when you take a multiplicative group and use the logarithmic embedding, you always embed it into this diagonal subspace because of the product formula, okay? If anybody has any questions about notation or terminology, please interrupt, let's take a look at page five. Okay, so the logarithmic embedding is very simple. I take a point say alpha in the multiplicative group of S integers and I associate to that point a vector with R plus one coordinates, the coordinates indexed by the places in S and the coordinate indexed by the place V is the local degree D sub V times the log of the viatic absolute value of alpha. And that vector is in the diagonal subspace by the product formula. And so it could also be written as the sum using log plus minus the sum using log minus. That may seem sort of silly unnecessary, but in fact, it's exactly that formula that struck our eye in about 2015 as putting us in position to make use of the seemingly unrelated Schindsel inequality. By the way, Schindsel's old paper from 1978 has been referenced in the literature four times once by Shabnam Akhtari and myself. So it seems to be a kind of obscure and forgotten inequality, but as you will see, it's exactly what we need in this particular situation. Okay, so the kernel of the logarithmic embedding is the set of portion points in the S unit group. This doesn't occur in all treatments of this subject, but I find it sort of convenient to mod out by the torsion subgroup by roots of unity. And so my embedding into the diagonal subgroup is really gonna be an injective map because I'm gonna mod out by the torsion subgroup. And I'll be mainly talking about the S unit group down at the bottom, the German or Frachtur U sub S. So that's the S unit group mod torsion. That's now a free group of rank R. It's not a big deal, but the fact that it's free allows some very easy conclusions to be drawn later. Okay, let's go to page six. So here is the heart of the matter. Suppose alpha is in the S unit group or mod out by torsion, the group I call U sub S. Let's compute the Schindsel norm of the logarithmic embedding of alpha into Euclidean space. So that would be delta applied to boldface alpha. The coordinates are indexed by the places in S. And according to the recipe for the Schindsel norm, it's the max of exactly the sum log plus or the sum log minus. I think two slides back, we observed that the sum of the log plus minus the sum of the log minus is zero, which is another way of reporting the product formula. Therefore, in this max, the two expressions are the same. And what is that expression? Well, it's the degree times the V height of alpha. So this, I think is a remarkable norm, sort of produced by Schindsel in 1978 for a completely non-algebraic number theory purpose, sort of a combinatorial problem that somebody had asked him about. Or at least that's when I pointed these things out to him, he just shook his head and said, no, he had no application to heights or algebraic numbers in mind at all. But it turns out that if you use the logarithmic embedding and just apply this Schindsel norm to alpha, you get exactly the degree times the height. So that means that there's going to be essentially no loss in certain inequalities that come up. So here's an example. Let's start with a basis for the S unit group, froctor S, that's the S unit group modded out by roots of unity, which play no role in this. Okay, let's inject that up into the diagonal subspace D sub R. And so we trade in alpha one, alpha two for the corresponding vectors, boldface alpha one, boldface alpha two. And so forth. Okay, now when you compute the regulator, remember there's a step where you have a R plus one by R matrix and you remove one row and you are given the knowledge which can be worked out that all the sub determinants are the same so it doesn't matter which row you remove. That's how you compute the regulator. Okay, let's use that same idea to compute the L one norm of the wedge product. So that's just the set of all the R by R sub determinants. How many are there of those? R plus one of them. So the L one norm here in the middle of the wedge product of those vectors is R plus one times the S regulator. As an immediate application of theorem one, the upper bound is R plus one times the product of the degree height combination. I will cancel the R plus one from both sides and get the inequality down at the bottom of the page. You'll notice that there are no extraneous constants depending on anything that come into this. It's just a pure inequality because of the Schindsel norm inequality and the fact that the Schindsel norm of the logarithmic embedding just produces exactly the degree times the V height of the algebraic number alpha. Maybe I should ask if there are any questions at this point. Okay, if not, let's go on to the next page. So here is a slightly more general version that the proof differs in significantly but it's just a little bit more to work out. Suppose instead of vectors in the S unit group, I have a subgroup of full rank R and I have alpha one, alpha two and so forth multiplicity of the independent elements in this subgroup, Fracture A. So the inequality on the previous slide didn't have a index on the left side because I was assuming that the alpha one, alpha two and so forth were actually basis vectors. Now I've given the basis vector hypothesis up. They're just some multiplicatively independent elements and so they generate some subgroup of rank R and you get exactly the same inequality. Again, there are no extraneous constants. You get exactly the same inequality but instead of the regulator on the left-hand side you now get something a little bit bigger the regulator times the index of the group generated by the alphas inside the S unit group. Now theorem two raises the question. Is it possible to prove that there exist some alphas for which there is nearly equality in that inequality? That is a problem that can be attacked using the geometry of numbers. And roughly speaking what you always need to do in such a situation is to figure out how to apply Minkowski's theorem on successive minima. In order to do that, you need to know the volume of the unit ball with respect to the Schindsel norm. And there's a formula that I give for that volume there in the middle of the page. And once you have that volume then you can apply Minkowski's successive minima theorem and you get the result at the bottom theorem three in which the inequality is reversed. So theorem two is a universal upper bound on the regulator that applies for all multiplicative independent elements. And theorem three says there exists some special independent elements in the subgroup, Fractur A, such that you come close to an equality but you're off by a small constant that depends on R. And that constant comes as you can probably guess from the formula for the volume. The constant is a little smaller than R factorial. In this case. Okay, let me now describe this for the relative unit group. So let's go to page eight. So relative units are a much more recent development. I think they occurred really, well, they occurred with I think the most significant results attached to them in some papers of Antonio Costa and Eduardo Friedman that appeared in the mid 1990s. This will have to be a little bit quick here but you start with two fields K and L where K is contained in L, both number fields. And I described the same things that occurred previously. So OK cross is the unit group and K OL cross the unit group in L. And again, in this subject, it's easier, less messy to mod out by the torsion subgroups. So in the situation, FK is the unit group in K mod torsion, FL is the unit group in L mod torsion. And by just trading in the coset for the torsion subgroup of K for the torsion subgroup of L, you see that there's an injection of FK into FL. So you might as well think then of FK as a subgroup of FL and the ranks now are gonna be RK and RL. Okay, let's go to page nine. So the norm from L down to K maps the units in L to the units in K and the torsion subgroup in L to the torsion subgroup in K. And so that induces a homomorphism that I'm calling norm with a little N from the quotient group or free group, FL down to the free group FK. And in their paper, Costin Friedman defined the relative units in between L and K to be the set of units in OL whose norm was a torsion point in K. But since we've modded out by the torsion subgroup in our situation, we get the same group. We just do it a little differently. For us, the relative units is the group E sub L over K which is the kernel of the little N norm now. And that's a subgroup of FL. And the rank of that subgroup is R of L minus R of K, which I'm always gonna assume in this talk is positive. So we avoid trivial situations. And the only interesting case in which that's zero is the case where L is a CM field and K is its maximum totally real subfield. One of the thing we need is the image of the norm little N norm group. Okay, let's go to page 10 now and that will contain the definition of the relative regulator. So this is sort of mimics, but it's more complex. It mimics how you define the regulator of an S unit group. For an S unit group, you have to remove one row of the matrix to get a square matrix. To deal with the relative regulator as Caspta and Friedman defined it, it's a little more elaborate that I explain it here in some detail, this is perhaps not overly interesting to see the technical parts of it, but you start with independent elements in the group FL. And then you make a square matrix M sub L over K by carefully removing several places. So to make a normal regulator, you just remove one place to get a square matrix. Here you have to remove several places in the group FL and then you make a square matrix here you have to remove several places in order to get a square matrix. And when you're done with that activity, then the regulator or relative regulator of the relative unit group E sub L over K is simply the determinant of the matrix capital M defined up at the top of the page. And again, you can see that it wouldn't be so difficult to work with the wedge product in this situation because it turns out that while you have to remove more rows than you ordinarily would for the S unit group, it doesn't matter which rows you use because the resulting Graspen coordinates in the language of exterior algebra all have the same absolute determinant. Okay, there's another way of getting to the relative regulator E sub L over K and that's the equation down at the bottom of this page which shows the relationship between the regulator the relative regulator and the ordinary regulator of K and the ordinary regulator of L. And then there is this little positive integer on the side which is the index of the image of the K units inside of FK that also kind of comes along. Okay, let's go to page 11. So these are the two theorems that arise from essentially the same process as theorem two and theorem three, namely using the Schindsel norm when I apply the basic inequality that was stated in theorem one directly, we get theorem four which is analogous to the theorem I think I called it theorem two for S units but this is the analogous fact it looks exactly the same on the page of course. So it is the analogous fact for the relative unit group. It's more complicated to prove because of the way the relative regulator is defined and then the analog of theorem three for relative units again, looks pretty much the same what you would expect it's proved in the same way by using Minkowski's theorem on successive minima to show that there exist independent relative units in a subgroup, I think this is Frachter E contained inside the relative unit group and you come reasonably close to equality in the inequality theorem four but now the expressions are related in the opposite fashion with a constant that depends only on the rank of the group and there's no dependence on the field K and L and the constant only the rank. Okay, let's go to 12. So now for the rest of this talk I would like to speak about a new type of problem with the same basic ingredients. So far, we have always studied and talked about groups of units where we take the maximum number of independent objects in the group and produce inequalities about those. Now what I want to do is go back to the S unit group of rank R and speak only about groups generated by say Q independent elements in the group where Q is strictly less than R. Q can be equal to R, that's the case already considered. So in sympathy with theorem one, I'm defining this constant C of Q R to be the minimum of the two expressions that appeared in the inequality in theorem one and using this generalization of Schindel's inequality we get the following new result. In this new result, I have alpha one through alpha Q multiplicatively independent S units. The bold face version, alpha J is just the image of alpha under the logarithmic embedding into the diagonal subspace D sub R. And then the inequality I get from theorem one is stated down at the bottom there. You will notice that there is no resultant in this inequality. And the reason for that is that Q is not R. So the word resultant as it has been used in number theory refers to certain matrices that occur when you have an independent collection of units where the number of independent units is the same as the group you're in. Here we're not doing that. So you should think of alpha one wedge, alpha two and so forth as a kind of height on subgroups. Maybe it should be called or could be called a regulator on subgroups as well. But I guess I've sort of started to think about it really as more analogous to a height on subgroups and any event, however you think about it, the L one norm of that wedge product is bounded from above by, and you can write it either as I have been doing as the product of the heights of generators or with slight variation as the height of the L one norms of the alpha vectors. Okay, at some point it seems the case that maybe putting everything in terms of the L one norms is a little bit more attractive. But any event, let's go on to 13 now. So what I wanna do is briefly just describe in the barest terms since I'm kind of running out of time here, how you produce the next theorem where you show that there's almost equality in the previous inequality. Again, you have to use Minkowski's theorem on successive minima to accomplish this somehow. And what we hope to do here when I'm describing on this page is the possibility that we start with the same vectors alpha one, alpha two and so forth that are coming from Q multiplicatively independent S units. I form the L one norm of the wedge product. That's a sum of many determinants. And now I would like to know, can I always show the existence of Q independent algebraic numbers in the group A of rank Q, such that there is almost equality in the inequality on the previous page. So either of the two bottom inequalities would suffice. And I've put in less than less than because I at this point don't know what constants should appear. Well, it's probably not a surprise. If you're going to apply Minkowski's theorem on successive minima in this case, you somehow have to interpret the L one norm of alpha one through alpha Q as a volume. If you can do that, then you're in business basically. And it turns out this is possible. It's quite a bit more complicated than in the previous cases, but let's go to page 14 now. So briefly, here is how it works. And here I've removed reference to algebraic number theory because this is just a problem of Euclidean geometry. And so the basics are the following. I go back to the matrix X that sort of appeared on the first page, but instead of having L columns, suppose it just has, instead of having N columns, I should say, suppose that has just L columns. So I use that to define the ball B sub X. B sub X is the unit ball of the L one norm of X, Y, where X is the matrix up at the top. It turns out that the dual unit ball is B sub X star. And we're now going to have a stroke of luck occur. The inequality in the middle of the page is a conjecture of Mahler, which is still open in general. But it turns out that Reisner, a mathematician, I think in Israel, proved Mahler's conjecture for certain special situations. And the good luck is that we are in exactly the special situation where Reisner proof applies. Roughly speaking, Reisner proved that the Mahler conjecture in the middle of the page is correct as long as either B sub X or its dual is what's called as zonoid as a special type of convex body in Euclidean space. And as luck would have it, there is a paper in 1984 by McMullen, which shows that the volume of the dual unit ball is related to the L one norm, exactly what we want of the product of the wedge product of the X vectors. So when you put McMullen's formula and Reisner's inequality together, you get a lower bound for the volume of B sub X, which is the bottom inequality there. And now the L one norm of the wedge product is going to make an appearance in the problem when we apply Minkowski's successive minima theorem because it appears in a volume formula. Okay, let's go on to 15. And this is the analog of theorem six where we start with alpha one through alpha Q. We form the alpha J bold face vectors in the diagonal embedding. And now I'm hoping to prove the existence of a subgroup of Fractur A, the span of the alphas such that the product of the L one norms is relatively small. And I obtain a result down at the bottom by applying the inequality in theorem one. Well, I guess I'm mixing it up a little bit here. So theorem six applies the inequality in theorem one. And now I'm showing that that inequality is pretty sharp by appealing to Minkowski's theorem and the inequality of Reisner and the volume formula of McMullen. Let's see, in the last few minutes, I'd like to say a little bit about the conjecture of Fernando Rodriguez-Velegas. So let's go to 16. I think there are only a couple more slides here. So Fernando Rodriguez-Velegas was a colleague of mine at the University of Texas for about 15 years. And I think it was around 2007 or eight that he proposed a remarkable conjecture which interpolates in between the Lamer problem and an inequality that basically bounds the regulator from below the first version of which was proved by Zimmert. I think it goes back a ways, maybe the 1930s. Anyway, the Rodriguez-Velegas conjecture is the following, that there exist constants C0 and C1, C0 greater than zero, C1 greater than one with the following property. If I take any number field K and its unit group, here S sub infinity, of course, is just the Archimedean absolute values. If R is the rank of the unit group in the field K and alpha one, alpha two, alpha Q are independent points in the unit group, then the Rodriguez-Velegas conjecture asserts that the L1 norm of the inner product of those vectors should grow exponentially with Q. Now, that may not be surprising. What is deep and makes this a deep and difficult problem is that C0 and C1 are absolute constants here. They don't depend on a number field K. It may be that C0 is 0.05 and C1 is 1.2. So there's no dependence assumed or proposed for C0 and C1. They're just absolute constants. So for example, if Q is one, if Q is one, then this just says that the L1 norm, which might as well be the height, is not too close to zero, which is essentially the Lamer problem. When Q is R, this inequality is known and it's a theorem of Zimmert. So the Rodriguez-Velegas conjecture interpolates between the Lamer problem, the case Q equal one, and the Zimmert inequality, which is a known theorem. That's the case Q equal R. Now, if this conjecture is correct and we apply theorem six, we would be able to replace the wedge product of the alphas with just the product of their L1 norms, which is also a height. I have to minimize confusion, tried to just use the L1 norm now instead of the heights, but you could use the heights as well. Okay, so that suggests a second conjecture. Let's go to page 17 now. I think that's the last page. Yeah, so the second conjecture is the following. Let us just, I'm not formally making the conjecture, but just to have something to talk about, let's propose the following problem. Does there exist a positive constant, say E0, and a constant E1 greater than one with the following property? And now I'll just repeat the Rodriguez-Velegas conjecture. K is a number field, Q is between one and R, R is the rank of the unit group, and alpha one through alpha K are independent points in the unit group. And then I will propose that the constants E0 and E1 always satisfy this inequality. And again, if you wanted, you could replace the L1 norm with a product of heights. Well, we have seen that the Rodriguez-Velegas conjecture, conjecture one implies conjecture two. That was the last inequality on the previous slide. And now what I would like to just briefly point out is that we almost have conjecture two implying conjecture one, not quite. It may, but maybe theorem seven wasn't sufficiently efficient to do it. But at the bottom of this page, I observe that if alpha one, alpha Q generate a subgroup of the unit group of rank Q, then by theorem seven and conjecture two, we discover the existence of beta one, beta two so that we get this inequality down at the bottom. That's almost but not quite conjecture one. It's not quite because we are forced to accept a Q factorial there rather than a power of Q. So we're off by that amount. It may be the case of course, that in fact with more efficient use of the geometry of numbers, the factorial can be reduced to a power of Q. The Q factorial is very natural because in Minkowski's successive minima theorem, the successive minima generate a lattice whose index is really factorial in the original lattice that you start with. But it's not at all clear that that sort of situation can occur when you have special lattices which come from the diagonal embedding. So it's possible that that Q factorial can be reduced to a power of Q. And if that is so, and the details work out then the conjecture one and conjecture two would be equivalent. Okay, I think that's a convenient place for me to stop.