Equidistribution of roots of unity and the Mahler measure
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Habegger, Philipp
Roots of unity of order dividing $n$ equidistribute around the unit circle as $n$ tends to infinity. With some extraeffort the same can be shown when restricting to roots of unity of exact order $n$. Equidistribution is measured by comparing the average of a continuous test function evaluated at these roots of unity with the integral over the complex unit circle. Baker, Ih, and Rumely extended this to test function with logarithmic singularities of the form $\log|P|$ where $P$ is a univariate polynomial in algebraic coefficients. I will discuss joint work with Vesselin Dimitrov where we allow $P$ to come from a class of a multivariate polynomials, extending a result of Lind, Schmidt, and Verbitskiy. Our method draws from earlier work of Duke.
 Okay, thank you, Robert, and thank you for introducing me and inviting me to this virtual conference in DEMINI. So the title of my talk is Equidistribution of Roots of Unity and the Mahler Measure, and I will talk about joint work with Vaseline Dimitrov. Okay, so let me just take some classical facts. Let's look at how Roots of Unity distributes in the complex plane. So what I'm looking at here, these points, they are Roots of Unity of order dividing 30. So they are complex numbers of the form e to the 2 pi i k over 30. And I let k run over all integers. Of course, it's enough just to go from 0 to 29. And I get these 30 dots here, and they lie on the unit circle, of course, and as you can see, they are very nicely equidistributed along the unit circle. And the theorem here is that if you let the order or the end here, capital N, go to infinity. So if you're looking at all Roots of Unity of order dividing N, then these points will become equidistributed around the unit circle. So let me recall what that actually means. How can we formulate equidistribution in a precise way? So the same picture with the 30 dots on top. And let me just... Okay, so what does equidistribution mean? It means that if I have a test function defined on the unit circle, this function f here, which has to be continuous, complex values, then if I take the average of my test points, of the test points, the Roots of Unity, take the average of this function on these points, and then take the limit as downstairs here, then I'll converge to the integral of the function along the unit circle. So the integral of f long, e to the... Well, we've got here e to the 2 pi i t. So we're running along the unit circle here in the usual counterclockwise direction. And this limit exists and equals exactly the integral. So this is a very classical result, and it's also quite easy to prove. Let me just give you an idea how the proof works, because we'll see arguments like this later on in a more sophisticated way. So let's see what happens if we take a very basic function, which is just f of z takes... f takes z to the lth power, where l is some integer. And well, now our average here simplifies to this geometric sum here, which is taking the sum k from 0 to n minus 1, and then upstairs here in the exponent, we have e to the 2 pi i k over n, and as one learns in the first year calculus class. Even earlier, how to compute sums like this. Well, if n happens to be divisor of l, of course, all terms will be 1, and so the average will also be 1. And if l is not a divisor, if n is not a divisor of l, then geometric sum formula tells us that the sum here vanishes. So in the limit, if we take n going to infinity here, in this case, then we will eventually be in this situation here. If l is fixed, and so the limit will be 0 for l not equal to 0, and that is exactly the integral of the function along the unit circle. So if more generally, our function is a finite linear combination of powers of z, so possibly negative powers of z, then as I just explained up there, the only term that survives in the limit is l equals 0, and that corresponds here to the term a naught. So that's what we get here. So for n goes to infinity, this average is exactly the constant term here in this Laurent polynomial, and that happens to be exactly this integral. So for these finite Laurent polynomials, liquid distribution follows just by considering a geometric sum. And in general, you can do some analysis to extend this kind of results to all continuous functions on the unit circle. Okay, so that's what happens with roots of unity and continuous test functions. Let me go on. So up to now, I've been looking at roots of unity of order dividing n. So there are always n different points in the complex plane of that type. Now I want to shift to roots of unity with order exactly n. And as you see here in the picture, there can be a lot less. Here we have only eight dots, and eight is the Euler function by evaluated at 30. So to get the number of roots of unity of order exactly n is just the order of this finite ring here, c modulo nz, sorry, this finite group, units in c modulo nz. So the number of elements in this group is given by the Euler function. So they're a lot sparser and it's not so clear by just looking at the picture that you'll get equidistribution because we've got a huge gap here, we've got a huge gap here, a slightly smaller gap here and a slightly smaller gap here, and also a gap over here. Nevertheless, this is also a classical result. So computing the corresponding sum here, where you take the average over the polynomial and the log polynomial z to the l of the roots of unity of order exactly n, this average can be computed and is also expressible up to sign in terms of the Euler function. As you see here, there's a to the minus one here in the exponent. So if n increases and l is nonzero, then the value here inside the Euler function will increase, will grow essentially like n, and we know also classically that the Euler function here, it grows almost linearly, it grows like n to the one minus epsilon. So this term here, if l is not zero, this term here will go to zero and we can also recover equidistribution just like in the first case. So what is the significance of taking order exactly n versus order dividing n? So the significance is of arithmetic nature. So here we're looking essentially at roots of unity of this type, e to the 2 pi i k over n, where k is coprime to n, these are exactly those of order n, and this happens to be exactly the Galois conjugates of any root of unity of order n. So this set is as arithmetic importance because it's exactly the set of Galois conjugates of roots of unity. So roots of unity of order exactly n, n goes to infinity, we also get equidistribution just like in the first case. So here I've got a picture with n equals 240, so now you begin to see that we get nice distribution, there are still some gaps here and here and here, but they don't matter in the limit. So what I'm going to talk mainly about today is what happens in this definition of equidistribution, what happens if you weaken the hypothesis that the test function is continuous. Remember you have to test your sequence against the continuous test function on the unit circle, what happens if we drop the continuity condition. And so there's a result I'd like to talk about by Matt Baker, Stephen E., and Rob Brumley from over 10 years ago, which is also motivation for the work presented today. So what they look at is they take an algebraic number alpha here and the test function they have is essentially of this nature. You take log of e to the 2 pi i t minus alpha, right, that's the test function. Now depending on the value of alpha, this may or may not be continuous, it may not be even defined on the whole unit circle, because if alpha itself lies on the unit circle, you'll have log 0 at 1.0, you have a logarithmic singularity. But nevertheless, they were able to show that if you take the average over roots of unity of given order, so of a Galois orbit like this on the left hand side, and you let the order go to infinity, then this average will converge to the integral. So the same thing you would expect with classical equidistribution. Of course, you have to be a bit careful. This may not be defined for us a fixed k and a fixed alpha, but as we're taking n to infinity, this difference will be nonzero for fixed alpha. For fixed alpha, eventually it's not going to be a root of unity of order exactly n, for n large enough, and so this will be well defined in the limit, and the limit exists and equals this integral. Now as I pointed out, it depends on the result or the proof of the result, essentially depends on the value of alpha. So the first case is if alpha is, for example, minus 1.5 as here, well in this case, the function, test function happens to be continuous again because there is no logarithm of singularity as t goes from 0 to 1. So I've plotted here the function in blue, below left, and you see there's a minimum here, it's negative, but it's okay. So in this case, if alpha is not on the unit circle, their theorem follows from the classical result I just mentioned before using those Raman-Uchand cells. Situation becomes a bit more interesting if our alpha is on the unit circle as here, alpha equals minus 1, and you see here the function, it has a logarithmic singularity at 1.5, e to the 2 pi i, 1.5 equals minus 1, and so that's where we have the problem. And in this case, it's not actually a problem as we'll see later on because e to the 2 pi i, a over n plus 1, which corresponds to alpha equals minus 1, this will be roughly, this cannot be too small if k over n, if the exponential there is not equal to minus 1. So this will be roughly the difference of k over n minus 1.5 if the left-hand side is non-zero. So whatever happens here in this logarithm is not going to be too small because this here is bounded from below by 1 over 2n, they're actually precisely, and so the n will appear here and log n is then negligible with respect to 5n because as we've seen before it grows almost linearly. So this case is slightly bad, but it's not really bad. The really bad case is the next one where alpha is on the unit circle like this one here, but it's not a root of unity. So e to the 3 plus 4i over 5 lies on the unit circle, it's an algebraic number, but it's not a root of unity. So the trick I just explained before does not work, and here you need linear forms in logarithms, the baker's linear forms in logs in their proof. So that kind of splits this case off from the other two cases, so this case is much deeper than these two cases here, this case is classical, the second case is fairly straightforward and this uses deep tools in transcendence theory. Okay, the three important cases we'll see later on when we move to higher dimensional results. Okay, so let me first reformulate their theorem, so it's the same theorem as before in a slightly more general dressing. So s1 to set up some notation for my talk will be the unit circle, it has a hard measure. The infinity will be the roots of unity in the complex plane, and sigma will always denote an element in the Galois group of a psychotomic field, so which is classically isomorphic to this unit group here. So their theorem can be reformulated as saying if I take any polynomial with algebraic coefficients, p nonzero, then the average of the Galois orbit of the root of unity over log, so the test function here is log absolute value of p will converge to the integral as the order of the root of unity tends to infinity. So this follows directly from the statement I mentioned further up just because a polynomial always factors into linear terms. Now this integral here on the right has a nice interpretation, this is just the molar measure of a polynomial. So there's a Jensen's formula allows you to compute the molar measure without computing an integral, it's just given by the expression here, you take the log of the reading term and then you add the log of all roots of the polynomial that lie outside of the absolute values, logs of the absolute values of the roots that lie outside the unit circle. So this is the molar measure of a polynomial. So this is just a reformulation of the result that was on the previous slide using the molar measure, so this would be the molar measure of p. Let me continue. So what I want to talk about today, my talk is what happens in higher dimension. So higher dimension I just mean instead of looking at a single root of unity, I'm looking at a couple of roots of unity here, zeta 1 to zeta d and each of them has an order, we can find a common denominator n and so the whole tuple has an order and we're going to say that the order is n. I'm going to use a boldface zeta here to denote something in higher dimension and being of order n is the same thing as saying that the GCD of these exponents here together with one, with n is one. And computing the Galois orbits of such a point here, we can look at the Galois conjugates of this point, it's essentially the same thing. We look at powers of this tuple where the exponent is co-prime to n. Now the first thing that you see that is in higher dimension, just having the order go to infinity is not enough to have equity distribution because things can happen here that don't happen in dimension one. For example, I'm just I'm going to write this down here, take the sequence zeta zeta where zeta is a root of unity of order n. And this tuple here will also have order n, but it's clearly not equidistributed on S1 cross S1. So what is not so the Galois orbit is not equidistributed on torus as the order goes to infinity because both entries are the same. So that's something that can happen in higher dimension. You can also have something different like you have a square here. You have some power here and the same non-equidistribution holds. So you have to be a bit more careful in higher dimension, but the only thing you have to make after you worry about is if there are some relations among these coordinates that are pathologically small. So instead of the order, it's more reasonable to look at the following quantity, namely we're looking this in a sense generalizes the order in different direction, higher dimension. We look at for any given tuple, we look at the smallest non-trivial character that contains our tuple, whose kernel contains our tuple, right? We're just taking the least non-zero vector b in integral coefficients, so such that zeta to the b is equal to one and where this notation is quite useful. So instead of using a tuple to an integral tuple, it's just this character notation. So this turns out to be the more appropriate invariant than the order when looking at questions of equidistribution in higher dimension. So the classical fact or result, which is stated here, is that now if we have a test function on the d-dimensional torus, s1 to the d, that is continuous, then so there's a typo here. The gavel orbit, the average of f along the gavel orbit will converge to the integral of f on the unit on this torus and here we just have the product or measure. But the convergent tools as this invariant delta of zeta goes to infinity. So having the order go to infinity is not enough, but for delta going to infinity, that's okay. So let me also state that delta of n of zeta is always at most the order, right? Because if I take zeta to the nth power, I get one, so I can, for the nnnnn is a non-trivial character whose kernel contains zeta. So the delta invariant of zeta is always at most n. But of course, it can be much smaller than it. Okay, so what is the conjecture that we have for logarithmic singularities? So let's look at this test function here, given by coming from a polynomial with algebraic coefficients. The test function is then just for a tuple, we take log of absolute value of p evaluated at this tuple. And so the conjecture is, our conjecture is that if you average over this test function, you'll converge to the integral over log of the absolute value of p as this delta invariant here goes to infinity. Now here are a few words, here are a few comments are in order. First of all, this here is the higher dimensional molar measure of a polynomial in t variables instead of 1. So this integral, even though the function here is not everywhere well defined on the torus, perhaps this integral converges in as well defined. Here there's also a question of whether this sum here is actually well defined or not. It's a more serious question than in the one dimensional case, because the vanishing locus of p on the torus can be more than a single point or a finite set of points. You can be, some will see that later on, it could be something of dimension greater than zero. To show that this average is actually well defined, you have to show that for delta of zeta large enough, this Galois conjugate here will not be on the vanishing locus of p. And that follows from result of Michel Laurent, which is also a case of the money monford conjecture for the multiplicative for the algebraic torus. So actually even showing that this average here is well defined for delta of zeta large enough is a theorem. So that's the conjecture, the conjecture is open, and I want to report on some progress towards this conjecture. In the case v equals one, the conjecture is known by the result of Matt Baker, E, and Rommelie. So what kind of evidence do we have towards this conjecture in dimension greater than one? So we have a result by Gary Meyerson from the 1980s and then slightly extended by Bill Duke in 2007 for the following form. So here we're looking at roots of unity of prime order. So zeta will be a root of unity of prime order. And the tuple we're looking at here is the following form. We look at zeta, zeta to the A, zeta to the B. The whole packet has order p. And A and B are, well, they're only defined modulo p, which is good enough for us because zeta is a root of unity of order p. And A and B are the non-trivial elements in the unique subgroup of order 3 in the multiplicative group fp. And this, of course, requires us to suppose p is equivalent to 1 mod 3. Then there is a unique subgroup of order 3 in fp star. Take that group, we take the two non-trivial elements, and then we look at this tuple. What we average, we average the test function along this tuple here, along this sum. So the polynomial here is t plus t1 plus t2 plus t3. So it's, okay, it's in dimension 3, I guess, but you can reduce it to something in dimension 2 if you'd like. And by de-homogenizing, and the test function here is just logarithm of absolute value of p. So the limit here, so p minus 1 is just the order of the number of roots of unity of order p. And so the theorem was that this average converges to the molar measure, just as predicted by the conjecture, plus some error term that goes to 0. And delta here can be computed of this tuple. And this is actually even a more precise way of stating convergence here, that the difference goes to 0 as p goes to infinity in this way here. And let me just say, and I'm using fact about this molar measure, it was computed by Chris Smith to be equal to the value of the derivative of a certain L function at the point minus 1. So this value here can be approximated numerically, and let me see, it's a positive real number. Let me stop here, so this is some evidence towards the conjecture in dimension greater than 1. And so as next point, I want to state a result for a class of polynomials in d variables for which we know the conjecture is true. And for that, I need to introduce a notion of a toral polynomials. So a toral polynomials appear in the literature, I guess they appeared in work of Agar McCarthy-Stankus in 2006, and later in slightly different form in work of Lynch-Smith and Verbitsky. And we will introduce yet another variant of what it means for a polynomial to be a toral. And these three definitions are not completely equivalent, they are in but in similar spirit, I would say. We chose a definition that's more adapted to our method, or best suited for our method. So let's start out with any polynomial in t variables, and complex coefficients, and we're going to look at its zero set, not on the fine d dimensional space, but we're going to look at the zero set on the torus, the d dimensional torus. So these are just the points where the complex absolute value is 1. So this is a set, it's not complex algebraic because the unit circle is certainly not complex algebraic, but it's a real algebraic subset of c to the d. I'll show you some pictures later on how something like this can look like. And now the, of course, important feature that we have in the complex plane is that we can complex conjugate, and if we're on the unit circle, the complex conjugate of a point is just its inverse, right? So if we have a solution of this equation here, and if I take the complex conjugate of the whole packet, I'll be inverting here the coordinates because they're on the unit circle, and I'll get a complex conjugate here in the polynomial p. So we get actually, we get a second relation for any point here in this zero set, and this second relation may or may not be independent of the first relation. So of course, if, for example, p is symmetric with regard to real coefficients, and it's symmetric with regard to this inverting operation up to a monomial coefficient, then this will not give us a new relation. But sometimes you'll get a new relation, and generically you would expect this to happen. And by new relation, I mean that we get a second polynomial that is coprime to the first one up to a monomial. Then we'll say that that's a situation we can deal with. So the precise definition of a torals as follows, we say that the polynomial p is a toral if there exists coprime polynomials, r and s. So r and s are allowed to have complex coefficients again. Or if we're working over some smaller field, then we'll ask the coefficients to be in the smaller field. Coprime, so they induce different relations, and we want the common set of zeros of the set of zeros of p on the detours to be contained in the common zeros of r and s. That's what we want. So if, for example, these two guys here, these two relations are honestly independent, then we can create polynomials out of these two guys. In general, they may not be, and there may be other polynomials, r and s, that are coprime such that a is contained in the common zeros of r and s. So this actually happens to be equivalent to saying that the dimension of a, so remember a is a real algebraic set, is at most d minus 2. So a is real algebraic, it has a real, there's a concept of dimension of real algebraic sets, and it's certainly something of dimension, at most d minus 1, because p is nonzero as a polynomial, so it's at most d minus 1. But if it's d minus 2, then that's the atoral case. In a sense, being of dimension d minus 2 is the generic situation, because if you look at it from the real algebraic point of view, these are actually two conditions, namely the real and the imaginary part of p vanish at the point z1 to zd, and so generically you're removing two dimensions from d. But in general, there can be polynomials that are not atoral, and I'll show you an example in a moment. So in dimension d equals 1, being atoral just means that the polynomial does not vanish on the unit circle. If there are two polynomials, co-prime, in one variable, then they will not have any common roots, and so this set here will be empty. D equals 1. For d equals 2, what is the common set of zeros of two co-prime polynomials in two-nit variables? That's a finite set, and so for d equals 2, being atoral just means that this set here, a, is finite. That's what it means for d equals 2. And for larger d, it's not so straightforward how to formulate that, but for example, this linear polynomial t1 plus t2, et cetera, plus td is atoral because you can cook up two co-prime polynomials from using this kind of relation here. So not all polynomials are atoral, even in d equals 2, or for d equals 1, there's obviously cases in d equals 2. You can look, for example, at Blaschke products, and I've prepared a picture or a small image. I hope you can see this. I've switched to a browser. I hope there's no problem with screen sharing at this time. So this is an example of a polynomial that you can construct using a Blaschke product. It's called as a form 2xy minus x minus y plus 2, and this blue torus here corresponds to the two-dimensional torus, s1 cross s1, and the intersection of the complex roots of this two-variable polynomial with s1 cross s1 is represented by this red line here. So this red line or this red curve would be something of real dimension 1, and so this is not an atoral polynomial. So this is something that's difficult. We'll see that this will be difficult to treat with our methods. So I'm returning to my notes. I hope there hasn't been a problem. Okay. Let me continue. So let me say something about what is known in this atoral setup. So this is using the definition. This is a result by Lin-Schmidt and Verbitsky using their definition of atoral, which is slightly different from ours, but which in this special case, I think, corresponds to ours, at least for irreducible polynomials. They show that if we average over this test function here, then in the limit we get the Mahler measure, right? We're not averaging over Galois orbits, but over finite subgroups of this torus here, this multiplicative group to the d. And they also have a notion of delta of g. This is just to mean that the points of this finite subgroup equidistribute in the classical sense as we run along this sequence. So they have convergence for finite subgroups under the atoral condition. So again, if the polynomial does not vanish at all on the torus, then convergence follows from the classical results because then our test function is continuous. And there was a previous result by the same three authors from 2010 under the condition that the vanishing locus of the polynomial on the d torus is finite. Now remember, atoral is weaker. That just means essentially that the vanishing locus is of real dimension at most d minus 2. So starting from dimension 3, this is a stronger condition than being atoral. And finally, Dimitrov and his PhD thesis, 2017, he dropped the hypothesis on p, so there's no atoral hypothesis, but he allowed only subgroups of this type. So the entorsion groups of the groups of unity. All right. So what is our result here? So we're back to the Galois orbit setting. This is joint work with Vessel and Dimitrov. And so we have an atoral polynomial. So with the same hypothesis as in work of Lynch and Verbitsky, but we're averaging over the Galois orbit as opposed to finite subgroups. And so we get the limit here equals the molar measure of p if, again, under the assumption that the Galois orbit actually becomes equidistributed in the limit. Just a few remarks. So we get a rate of convergence that is a polynomial with a small exponent in 1 over delta. And we have a slightly weaker condition on atoral. Actually, it's something that we call essentially atoral. And I can explain that to you in a short, in this short picture. Remember the one-dimensional case, there were these three situations. So this was somehow the trivial one where the function is. The function is actually continuous on S1. Then we have the case where the pole, there's a small fault, wrong picture here. So this is actually, this should be a singularity here. There is a case where there's a pole at a root of unity, which was somewhat easy to treat. And then there was a case where there's a pole at a non-root of unity. And this required Baker's linear forms and logarithms, or at least some quads, a version of it. So in the higher-dimensional case, we can have poles on the detourists. But the poles set is not allowed to be too bad. So it's at most of real-dimension D minus 1. So this is somehow the atoral case. We can also treat this case where the pole set is large of dimension D minus 1, but it comes from essentially an algebraic subgroup, like here minus 1 is an irreducible component of an algebraic subgroup. This is a case that is conjectural. The general case that in the non-dimensional setting requires Baker's linear forms and logarithms is at the moment open in higher dimensions. So this is what further ideas are needed. And also a small point, we don't need to really work with this Galois, or we can also replace Q by a number field. All right. Let me give you an example of what there's a nice application of these equidistribution results. Let's look at this polynomial, T1, T2 plus T3 plus T4, and this happens to be atoral. Let me just look at these two guys. You invert it, and then their coprime is Laurent polynomials, and then you get the two relations R and S. Also a nice feature that the molar measure here of P has been computed to this value. It's also, this is related to the zeta function. The actual value will not play a role for this work, but what's important is that it's positive. That will play a role. Okay, so let's say we have four roots of unity, zeta 1 to zeta 4, such that their sum is an algebraic unit. You'll find examples like this if you look. So you can take, for example, e to the 2 pi i over 6, e to the minus 2 pi i over 6, and then two general roots of unity from minus here, then the sum will be exactly 1. So there are examples where this is an algebraic unit. What happens in that case? Well we can apply the theorem. The theorem tells us that the average here will converge to the molar measure as delta goes to infinity. But if I'm a unit, then this average here is just the logarithm of the absolute value of the field norm, and so this will be zero. Right. And then of course we've got a problem. As I said, this is a positive value here, this smaller measure, and so something zero cannot converge to something positive when both things are constant. So that means that delta of zeta has to be bounded. So it's a certain finite result. So as a conclusion, if zeta 1 plus zeta 2 plus zeta 3 plus zeta 4 is an algebraic unit for some four roots of unity zeta 1 to zeta 4, then there has to be a multiplicative relation between these four roots of unity where the exponents are bounded by some constant b. And here the relation is clear. I mean there are many relations. So for example, you can take zeta 1 to the 6 equals 1 here in this particular instance. And there are other relations, zeta 1 times zeta 2 is more than so on. So this is a certain kind of finite result. You can get out of equidistribution statements like this connected to a conjecture of Su Yong Yi, which was also a motivation in the work of Baker, Ian Rumley. Okay, let me say a few things about the proof of our theorem in the last, I think, 10 minutes. So let's go back to the univariate variable case. So if we just have a polynomial in one variable, let's see what we can get there. So the proposition that we need to prove or that we prove in our paper is a following. So we're back to the one variable case. Later on we're going to try to reduce to the one variable case. We take an average over a gaol orbit of unity, plug it into our Q, and proposition says that in the limit, this is the molar measure of Q. And the point of this proposition is that we get an explicit error estimate here, which is explicit in all the data. Now there is an important hypothesis here. Namely, this is just the atoral hypothesis in dimension one. As you recall, being an atoral in dimension one means I'm not zero on the unit circle. So this is where the atoralness comes into the picture. And let me give you an idea how the proof works. So the proof is rather straightforward here. We can assume that Q is just linear by factoring the polynomial. And then what does the average look like? Well, there are contributions to the average that are harmless, namely those conjugates that are away from any pole here, any pole of this, that are not too close to alpha. So there are actually no poles here, but still alpha could be very close to the unit circle. So you want to sum over first those conjugates of zeta that are not too close to alpha. So being not too close is being polynomially bounded away here, one over n squared as a placeholder. And then there is a remaining term here of the bad guys, the guys we have to deal with later on. Now, because the proximity here is one over n squared, and these are roots of unity of order n, there can be at most one term here in this sum here. So there's a most one bad guy. And the rest, phi of n minus one guys we can handle. And using just the standard equidistribution techniques, this part of the sum here converges rather nicely to what you would expect, namely the molar measure of this univariate polynomial, which is this log of max one come of absolute value. So this sub part of the sum is harmless. What we need to deal with is this part here. And as I said, there's only one term here, but this one term can cause a lot of problems. Because even if alpha is not on the unit circle, we have to deal with the contingency that zeta or some conjugate of it is extremely close to alpha. And we'll spoil here this absolute value. And so this could be very negative in principle, at least, we have to deal with that. Okay, so what is the worst case scenario is when our average is log of max one comma. So this is the molar measure here, plus some pathologically negative term that comes from a conjugate that is extremely close to alpha. And then I'm not going to specify the error term here that comes from this one here. So recall that this sigma, this exceptional conjugate is just of the four meters of the two pi iq for some rational number q. And it's tempting to apply Baker's linear forms and logarithms at this point, just like in the work of Baker and Romney. Because we're dealing here with, if you look at the log of this, this would be two pi iq and the log of this would be the logarithm algebraic number. If these two guys are closed, and so the logs will also be, or choices of logs will also be, so it's an idea to apply Baker's linear forms and logarithms. But this was already observed by Duke in his work in 2007, that the current versions of even the state of the art of Baker's, even the two variable results, the dependency in the field of definition here of alpha is not good enough to get a result here. So the dependency of the field of definition is usually d cubed or something like that, and that will not, won't be able to dominate this, this denominator here. So that, that unfortunately doesn't work. So we have to do something else. And what we do is actually something, it's almost a sacrilege. So we have to bound, the thing to do is we have to bound this from below, right? And so what is the first thing you learn at university? Well, the first thing is probably the triangle inequality, and then there's a reverse triangle inequality if you just go the other way. But of course you pay a high price for that. And usually this price is much too high in any context, in any applications in numbers theory. But actually in this setting, it turns out to be good enough. So, so we bound the different, the distance between zeta sigma and alpha from below just by the distance of alpha to the unit circle. Yeah, this is may seem like a blunder, but it actually happens to work in this situation. And why does it work? Well, if alpha, the situation that we're interested in is if alpha is extremely close to the unit circle, and if alpha is extremely close to the unit circle, then the distance between alpha and the unit circle is roughly the distance between alpha and its complex conjugate inverse. So if alpha were precisely on the unit circle, then alpha to the minus one complex conjugate would be precisely equal to alpha. But they're, they're reasonably close to one another. So now remember alpha is not on the unit circle by hypothesis. No roots of qr are the unit circle. So this happens to be positive, which is already a good sign. We get a lower bound by something positive, but we need something a bit stronger. And for this, we return, we return to the tomorrow results on separation of roots of polynomials. So this is a result of model from 1964 five to distinct roots of a polynomial with here and future coefficients. Then I can, I can bound the distance between these two roots from below by something that is essentially linear in the degree. And with a dependency on the molar measure of F. And so if you plug in this result, you can, you can use this polynomial here with that has alpha and alpha. And it's the minus one is roots and you can bound this difference here from below. And the lower bound you get will be good enough for the application because it depends to here in D is essentially linear. And that will lead to an essentially linear dependency here in D. And the end, the five and we'll win. So for our result, we actually have to deal with different alphas because we have a product or polynomial is a product of linear terms as usual. And so we need a result like molar, but where we compare pairs of roots, like a separation result of pairs of roots. And so for this, there's a result of menyalt that strengthened molar result to pair to several pairs of roots, not just one. So we get a lower bound for several pairs of the same quality as more. So that kind of will give us something of this nature for the univari case. And then we have to go to the, we have to reduce from the, the multivariate case to the univariate case. So now let's assume that P is a toro. And we have a tuple of order n and of roots of unity. And so we can, we can reduce to the univariate case just by the basic observation that any tuple of roots of unity of order n is essentially generated. The coordinates are generated by a single root of unity of order n. So I can write this zeta, bold zeta as non-bold zeta to the, to some exponent a, which, which is a vector. And for Galois conjugates, I can do the same thing. And I also have some, some degree of freedom here because I can always add a vector, a multiple of n, because if n is the order of zeta, then that won't matter. So using this freedom, I can construct a polynomial starting from P using this exponent here. And the game is then to choose this exponent, choose tau and, and, and be in such a way that this exponent is as small as possible. Right. So this is going to be a polynomial of small as possible degree. And, but it will be univariate polynomial. So we can try to apply the proposition before to this univariate polynomial. The point, of course, the average doesn't change by doing this, this, this process. So we can reduce to the univariate case. The pay price we pay is that the degree of this queue here will be very large. That's why I was very concerned about getting good dependency in the degree in the last univariate case. So using Erdos-Turon-Coxma, we can get a bound for like an optimal choice here of this exponent, which will decay depending on this delta of zeta. Right. So we get, we get some improvement over the trivial bound, some power of delta. And then we plug in, we plug in the proposition, and then, well, at least if this delta here grows quickly enough, then the error term will be smaller one, which is what we want. Right. So if this delta here grows like n or some small power of n, then we get convergence here. So let me just, let me just briefly subsum what we have here. So if this delta of zeta grows quickly and often n is like some small power, then we get a convergence result. The molar measure here is again the molar measure of something univariate. So we have to, as a next step, we have to compare this with the molar measure of p, the original polynomial. And for that, we need a result of lots, or at least some, a variant of the result of lot that allows us to reduce the computation of multivariate molar measure to univariate molar measure. And there's a quantitative version of this that we show. We have to deal with this situation where delta does not grow quickly. And the first, the final thing is that we have to deal with the situation where q vanishes on the unit circle. Remember that was an important hypothesis in the proposition. Without this hypothesis, our method doesn't work. So we have to make sure that the polynomial q we constructed does not vanish on the unit circle. And just let me say a few words on that. So if q happens to vanish on the unit circle, that's where the atoral condition comes into the picture. If q vanishes on the unit circle, then we'll find a point on the detourus and on the vanishing locus of p of this nature here. And now we use that our polynomial is atoral. So this will be actually contained in the vanishing locus of two coprene polynomials, r and s. And then using a result of Bombieri-Massourenzanie, we can get a nice bound here on something orthogonal to this expression here. So this here uses a result of Bombieri-Massourenzanie and unlikely intersections. And let me just go to the final slide. So what happens if this nc is orthogonal to the exponent, well, then I just take, I can take c as an exponent in zeta and I get one. And that means that the delta of zeta is at most b. So that means if I have a root of, I have a point on the unit circle that's contained on the vanishing locus of my q, that means the delta of zeta is bounded. And that will mean that the proof is over because we can assume that delta of zeta is as large as one. Let me conclude by showing two or three pictures. Here this is what happens if this is a typical atoral polynomial intersected with detourus, this red point. It's a single point. The yellow line is the algebraic subgroup here that appears in Bombieri-Massourenzanie. We can make it a bit more complicated. Now it's x times y squared. And even more complicated, you see that some people always misses this point, this random point. Of course, it could also hit the random, the point in some cases, but the theorem tells us of Bombieri-Massourenzanie that this is somewhat rare and we can deal with this situation. Okay. So here, to close off, here's a picture of Lumini from Mont-Puget, who is marking the background three years ago. So I hope to return to Lumini soon. Thanks for your attention. Yeah, so thank you very much for the recent, very interesting lecture and for this nice picture from Puget. Maybe you know there is a fast track to go down just right off this place. You took the picture, which is a really very nice round tour of about three hours from the Institute. Yeah, it's a pity that we cannot do that. So we have still some time for very fast questions. Are there questions, comments? Yeah, there is a question. Can you read it, Philipp? Fabia? Fabia, I think you can hear the notion of atoral and the notion of tempered from Denninger. No, I've not. I'm not aware of tempered, but I'd like to look into that if there's a connection. Okay. Denninger is all interested in smaller measures. I'm back again.
 Okay, thank you, Robert, and thank you for introducing me and inviting me to this virtual  conference in DEMINI.  So the title of my talk is Equidistribution of Roots of Unity and the Mahler Measure,  and I will talk about joint work with Vaseline Dimitrov.  Okay, so let me just take some classical facts.  Let's look at how Roots of Unity distributes in the complex plane.  So what I'm looking at here, these points, they are Roots of Unity of order dividing  30.  So they are complex numbers of the form e to the 2 pi i k over 30.  And I let k run over all integers.  Of course, it's enough just to go from 0 to 29.  And I get these 30 dots here, and they lie on the unit circle, of course, and as you  can see, they are very nicely equidistributed along the unit circle.  And the theorem here is that if you let the order or the end here, capital N, go to infinity.  So if you're looking at all Roots of Unity of order dividing N, then these points will  become equidistributed around the unit circle.  So let me recall what that actually means.  How can we formulate equidistribution in a precise way?  So the same picture with the 30 dots on top.  And let me just...  Okay, so what does equidistribution mean?  It means that if I have a test function defined on the unit circle, this function f here,  which has to be continuous, complex values, then if I take the average of my test points,  of the test points, the Roots of Unity, take the average of this function on these points,  and then take the limit as downstairs here, then I'll converge to the integral of the  function along the unit circle.  So the integral of f long, e to the...  Well, we've got here e to the 2 pi i t.  So we're running along the unit circle here in the usual counterclockwise direction.  And this limit exists and equals exactly the integral.  So this is a very classical result, and it's also quite easy to prove.  Let me just give you an idea how the proof works, because we'll see arguments like this  later on in a more sophisticated way.  So let's see what happens if we take a very basic function, which is just f of z takes...  f takes z to the lth power, where l is some integer.  And well, now our average here simplifies to this geometric sum here, which is taking  the sum k from 0 to n minus 1, and then upstairs here in the exponent, we have e to the 2 pi  i k over n, and as one learns in the first year calculus class.  Even earlier, how to compute sums like this.  Well, if n happens to be divisor of l, of course, all terms will be 1, and so the average will  also be 1.  And if l is not a divisor, if n is not a divisor of l, then geometric sum formula tells us  that the sum here vanishes.  So in the limit, if we take n going to infinity here, in this case, then we will eventually  be in this situation here.  If l is fixed, and so the limit will be 0 for l not equal to 0, and that is exactly  the integral of the function along the unit circle.  So if more generally, our function is a finite linear combination of powers of z, so possibly  negative powers of z, then as I just explained up there, the only term that survives in the  limit is l equals 0, and that corresponds here to the term a naught.  So that's what we get here.  So for n goes to infinity, this average is exactly the constant term here in this Laurent  polynomial, and that happens to be exactly this integral.  So for these finite Laurent polynomials, liquid distribution follows just by considering  a geometric sum.  And in general, you can do some analysis to extend this kind of results to all continuous  functions on the unit circle.  Okay, so that's what happens with roots of unity and continuous test functions.  Let me go on.  So up to now, I've been looking at roots of unity of order dividing n.  So there are always n different points in the complex plane of that type.  Now I want to shift to roots of unity with order exactly n.  And as you see here in the picture, there can be a lot less.  Here we have only eight dots, and eight is the Euler function by evaluated at 30.  So to get the number of roots of unity of order exactly n is just the order of this finite  ring here, c modulo nz, sorry, this finite group, units in c modulo nz.  So the number of elements in this group is given by the Euler function.  So they're a lot sparser and it's not so clear by just looking at the picture that you'll  get equidistribution because we've got a huge gap here, we've got a huge gap here, a slightly  smaller gap here and a slightly smaller gap here, and also a gap over here.  Nevertheless, this is also a classical result.  So computing the corresponding sum here, where you take the average over the polynomial  and the log polynomial z to the l of the roots of unity of order exactly n, this average can  be computed and is also expressible up to sign in terms of the Euler function.  As you see here, there's a to the minus one here in the exponent.  So if n increases and l is nonzero, then the value here inside the Euler function will  increase, will grow essentially like n, and we know also classically that the Euler function  here, it grows almost linearly, it grows like n to the one minus epsilon.  So this term here, if l is not zero, this term here will go to zero and we can also  recover equidistribution just like in the first case.  So what is the significance of taking order exactly n versus order dividing n?  So the significance is of arithmetic nature.  So here we're looking essentially at roots of unity of this type, e to the 2 pi i k over  n, where k is coprime to n, these are exactly those of order n, and this happens to be exactly  the Galois conjugates of any root of unity of order n.  So this set is as arithmetic importance because it's exactly the set of Galois conjugates  of roots of unity.  So roots of unity of order exactly n, n goes to infinity, we also get equidistribution just  like in the first case.  So here I've got a picture with n equals 240, so now you begin to see that we get nice distribution,  there are still some gaps here and here and here, but they don't matter in the limit.  So what I'm going to talk mainly about today is what happens in this definition of equidistribution,  what happens if you weaken the hypothesis that the test function is continuous.  Remember you have to test your sequence against the continuous test function on the unit circle,  what happens if we drop the continuity condition.  And so there's a result I'd like to talk about by Matt Baker, Stephen E., and Rob Brumley  from over 10 years ago, which is also motivation for the work presented today.  So what they look at is they take an algebraic number alpha here and the test function they  have is essentially of this nature.  You take log of e to the 2 pi i t minus alpha, right, that's the test function.  Now depending on the value of alpha, this may or may not be continuous, it may not be  even defined on the whole unit circle, because if alpha itself lies on the unit circle, you'll  have log 0 at 1.0, you have a logarithmic singularity.  But nevertheless, they were able to show that if you take the average over roots of unity  of given order, so of a Galois orbit like this on the left hand side, and you let the  order go to infinity, then this average will converge to the integral.  So the same thing you would expect with classical equidistribution.  Of course, you have to be a bit careful.  This may not be defined for us a fixed k and a fixed alpha, but as we're taking n to infinity,  this difference will be nonzero for fixed alpha.  For fixed alpha, eventually it's not going to be a root of unity of order exactly n,  for n large enough, and so this will be well defined in the limit, and the limit exists  and equals this integral.  Now as I pointed out, it depends on the result or the proof of the result, essentially depends  on the value of alpha.  So the first case is if alpha is, for example, minus 1.5 as here, well in this case, the  function, test function happens to be continuous again because there is no logarithm of singularity  as t goes from 0 to 1.  So I've plotted here the function in blue, below left, and you see there's a minimum  here, it's negative, but it's okay.  So in this case, if alpha is not on the unit circle, their theorem follows from the classical  result I just mentioned before using those Raman-Uchand cells.  Situation becomes a bit more interesting if our alpha is on the unit circle as here, alpha  equals minus 1, and you see here the function, it has a logarithmic singularity at 1.5, e  to the 2 pi i, 1.5 equals minus 1, and so that's where we have the problem.  And in this case, it's not actually a problem as we'll see later on because e to the 2 pi  i, a over n plus 1, which corresponds to alpha equals minus 1, this will be roughly, this  cannot be too small if k over n, if the exponential there is not equal to minus 1.  So this will be roughly the difference of k over n minus 1.5 if the left-hand side is  non-zero.  So whatever happens here in this logarithm is not going to be too small because this  here is bounded from below by 1 over 2n, they're actually precisely, and so the n will appear  here and log n is then negligible with respect to 5n because as we've seen before it grows  almost linearly.  So this case is slightly bad, but it's not really bad.  The really bad case is the next one where alpha is on the unit circle like this one  here, but it's not a root of unity.  So e to the 3 plus 4i over 5 lies on the unit circle, it's an algebraic number, but it's  not a root of unity.  So the trick I just explained before does not work, and here you need linear forms in  logarithms, the baker's linear forms in logs in their proof.  So that kind of splits this case off from the other two cases, so this case is much  deeper than these two cases here, this case is classical, the second case is fairly straightforward  and this uses deep tools in transcendence theory.  Okay, the three important cases we'll see later on when we move to higher dimensional  results.  Okay, so let me first reformulate their theorem, so it's the same theorem as before in a slightly  more general dressing.  So s1 to set up some notation for my talk will be the unit circle, it has a hard measure.  The infinity will be the roots of unity in the complex plane, and sigma will always denote  an element in the Galois group of a psychotomic field, so which is classically isomorphic to  this unit group here.  So their theorem can be reformulated as saying if I take any polynomial with algebraic coefficients,  p nonzero, then the average of the Galois orbit of the root of unity over log, so the  test function here is log absolute value of p will converge to the integral as the order  of the root of unity tends to infinity.  So this follows directly from the statement I mentioned further up just because a polynomial  always factors into linear terms.  Now this integral here on the right has a nice interpretation, this is just the molar  measure of a polynomial.  So there's a Jensen's formula allows you to compute the molar measure without computing  an integral, it's just given by the expression here, you take the log of the reading term  and then you add the log of all roots of the polynomial that lie outside of the absolute  values, logs of the absolute values of the roots that lie outside the unit circle.  So this is the molar measure of a polynomial.  So this is just a reformulation of the result that was on the previous slide using the molar  measure, so this would be the molar measure of p.  Let me continue.  So what I want to talk about today, my talk is what happens in higher dimension.  So higher dimension I just mean instead of looking at a single root of unity, I'm looking  at a couple of roots of unity here, zeta 1 to zeta d and each of them has an order, we  can find a common denominator n and so the whole tuple has an order and we're going  to say that the order is n.  I'm going to use a boldface zeta here to denote something in higher dimension and being of  order n is the same thing as saying that the GCD of these exponents here together with  one, with n is one.  And computing the Galois orbits of such a point here, we can look at the Galois conjugates  of this point, it's essentially the same thing.  We look at powers of this tuple where the exponent is co-prime to n.  Now the first thing that you see that is in higher dimension, just having the order go  to infinity is not enough to have equity distribution because things can happen here that don't happen  in dimension one.  For example, I'm just I'm going to write this down here, take the sequence zeta zeta  where zeta is a root of unity of order n.  And this tuple here will also have order n, but it's clearly not equidistributed on S1  cross S1.  So what is not so the Galois orbit is not equidistributed on torus as the order goes  to infinity because both entries are the same.  So that's something that can happen in higher dimension.  You can also have something different like you have a square here.  You have some power here and the same non-equidistribution holds.  So you have to be a bit more careful in higher dimension, but the only thing you have to  make after you worry about is if there are some relations among these coordinates that  are pathologically small.  So instead of the order, it's more reasonable to look at the following quantity, namely  we're looking this in a sense generalizes the order in different direction, higher dimension.  We look at for any given tuple, we look at the smallest non-trivial character that contains  our tuple, whose kernel contains our tuple, right?  We're just taking the least non-zero vector b in integral coefficients, so such that zeta  to the b is equal to one and where this notation is quite useful.  So instead of using a tuple to an integral tuple, it's just this character notation.  So this turns out to be the more appropriate invariant than the order when looking at questions  of equidistribution in higher dimension.  So the classical fact or result, which is stated here, is that now if we have a test  function on the d-dimensional torus, s1 to the d, that is continuous, then so there's  a typo here.  The gavel orbit, the average of f along the gavel orbit will converge to the integral  of f on the unit on this torus and here we just have the product or measure.  But the convergent tools as this invariant delta of zeta goes to infinity.  So having the order go to infinity is not enough, but for delta going to infinity, that's  okay.  So let me also state that delta of n of zeta is always at most the order, right?  Because if I take zeta to the nth power, I get one, so I can, for the nnnnn is a non-trivial  character whose kernel contains zeta.  So the delta invariant of zeta is always at most n.  But of course, it can be much smaller than it.  Okay, so what is the conjecture that we have for logarithmic singularities?  So let's look at this test function here, given by coming from a polynomial with algebraic  coefficients.  The test function is then just for a tuple, we take log of absolute value of p evaluated  at this tuple.  And so the conjecture is, our conjecture is that if you average over this test function,  you'll converge to the integral over log of the absolute value of p as this delta invariant  here goes to infinity.  Now here are a few words, here are a few comments are in order.  First of all, this here is the higher dimensional molar measure of a polynomial in t variables  instead of 1.  So this integral, even though the function here is not everywhere well defined on the  torus, perhaps this integral converges in as well defined.  Here there's also a question of whether this sum here is actually well defined or not.  It's a more serious question than in the one dimensional case, because the vanishing locus  of p on the torus can be more than a single point or a finite set of points.  You can be, some will see that later on, it could be something of dimension greater than  zero.  To show that this average is actually well defined, you have to show that for delta of  zeta large enough, this Galois conjugate here will not be on the vanishing locus of p.  And that follows from result of Michel Laurent, which is also a case of the money monford  conjecture for the multiplicative for the algebraic torus.  So actually even showing that this average here is well defined for delta of zeta large  enough is a theorem.  So that's the conjecture, the conjecture is open, and I want to report on some progress  towards this conjecture.  In the case v equals one, the conjecture is known by the result of Matt Baker, E, and  Rommelie.  So what kind of evidence do we have towards this conjecture in dimension greater than  one?  So we have a result by Gary Meyerson from the 1980s and then slightly extended by Bill  Duke in 2007 for the following form.  So here we're looking at roots of unity of prime order.  So zeta will be a root of unity of prime order.  And the tuple we're looking at here is the following form.  We look at zeta, zeta to the A, zeta to the B. The whole packet has order p.  And A and B are, well, they're only defined modulo p, which is good enough for us because  zeta is a root of unity of order p.  And A and B are the non-trivial elements in the unique subgroup of order 3 in the multiplicative  group fp.  And this, of course, requires us to suppose p is equivalent to 1 mod 3.  Then there is a unique subgroup of order 3 in fp star.  Take that group, we take the two non-trivial elements, and then we look at this tuple.  What we average, we average the test function along this tuple here, along this sum.  So the polynomial here is t plus t1 plus t2 plus t3.  So it's, okay, it's in dimension 3, I guess, but you can reduce it to something in dimension  2 if you'd like.  And by de-homogenizing, and the test function here is just logarithm of absolute value of  p.  So the limit here, so p minus 1 is just the order of the number of roots of unity of order  p.  And so the theorem was that this average converges to the molar measure, just as predicted by  the conjecture, plus some error term that goes to 0.  And delta here can be computed of this tuple.  And this is actually even a more precise way of stating convergence here, that the difference  goes to 0 as p goes to infinity in this way here.  And let me just say, and I'm using fact about this molar measure, it was computed by Chris  Smith to be equal to the value of the derivative of a certain L function at the point minus  1.  So this value here can be approximated numerically, and let me see, it's a positive real number.  Let me stop here, so this is some evidence towards the conjecture in dimension greater  than 1.  And so as next point, I want to state a result for a class of polynomials in d variables  for which we know the conjecture is true.  And for that, I need to introduce a notion of a toral polynomials.  So a toral polynomials appear in the literature, I guess they appeared in work of Agar McCarthy-Stankus  in 2006, and later in slightly different form in work of Lynch-Smith and Verbitsky.  And we will introduce yet another variant of what it means for a polynomial to be a  toral.  And these three definitions are not completely equivalent, they are in but in similar spirit,  I would say.  We chose a definition that's more adapted to our method, or best suited for our method.  So let's start out with any polynomial in t variables, and complex coefficients, and  we're going to look at its zero set, not on the fine d dimensional space, but we're  going to look at the zero set on the torus, the d dimensional torus.  So these are just the points where the complex absolute value is 1.  So this is a set, it's not complex algebraic because the unit circle is certainly not complex  algebraic, but it's a real algebraic subset of c to the d.  I'll show you some pictures later on how something like this can look like.  And now the, of course, important feature that we have in the complex plane is that  we can complex conjugate, and if we're on the unit circle, the complex conjugate of  a point is just its inverse, right?  So if we have a solution of this equation here, and if I take the complex conjugate  of the whole packet, I'll be inverting here the coordinates because they're on the unit  circle, and I'll get a complex conjugate here in the polynomial p.  So we get actually, we get a second relation for any point here in this zero set, and this  second relation may or may not be independent of the first relation.  So of course, if, for example, p is symmetric with regard to real coefficients, and it's  symmetric with regard to this inverting operation up to a monomial coefficient, then this will  not give us a new relation.  But sometimes you'll get a new relation, and generically you would expect this to happen.  And by new relation, I mean that we get a second polynomial that is coprime to the first  one up to a monomial.  Then we'll say that that's a situation we can deal with.  So the precise definition of a torals as follows, we say that the polynomial p is a toral if  there exists coprime polynomials, r and s.  So r and s are allowed to have complex coefficients again.  Or if we're working over some smaller field, then we'll ask the coefficients to be in the  smaller field.  Coprime, so they induce different relations, and we want the common set of zeros of the  set of zeros of p on the detours to be contained in the common zeros of r and s.  That's what we want.  So if, for example, these two guys here, these two relations are honestly independent,  then we can create polynomials out of these two guys.  In general, they may not be, and there may be other polynomials, r and s, that are coprime  such that a is contained in the common zeros of r and s.  So this actually happens to be equivalent to saying that the dimension of a, so remember  a is a real algebraic set, is at most d minus 2.  So a is real algebraic, it has a real, there's a concept of dimension of real algebraic sets,  and it's certainly something of dimension, at most d minus 1, because p is nonzero as  a polynomial, so it's at most d minus 1.  But if it's d minus 2, then that's the atoral case.  In a sense, being of dimension d minus 2 is the generic situation, because if you look  at it from the real algebraic point of view, these are actually two conditions, namely  the real and the imaginary part of p vanish at the point z1 to zd, and so generically  you're removing two dimensions from d.  But in general, there can be polynomials that are not atoral, and I'll show you an example  in a moment.  So in dimension d equals 1, being atoral just means that the polynomial does not vanish  on the unit circle.  If there are two polynomials, co-prime, in one variable, then they will not have any  common roots, and so this set here will be empty.  D equals 1.  For d equals 2, what is the common set of zeros of two co-prime polynomials in two-nit  variables?  That's a finite set, and so for d equals 2, being atoral just means that this set here,  a, is finite.  That's what it means for d equals 2.  And for larger d, it's not so straightforward how to formulate that, but for example, this  linear polynomial t1 plus t2, et cetera, plus td is atoral because you can cook up  two co-prime polynomials from using this kind of relation here.  So not all polynomials are atoral, even in d equals 2, or for d equals 1, there's obviously  cases in d equals 2.  You can look, for example, at Blaschke products, and I've prepared a picture or a small image.  I hope you can see this.  I've switched to a browser.  I hope there's no problem with screen sharing at this time.  So this is an example of a polynomial that you can construct using a Blaschke product.  It's called as a form 2xy minus x minus y plus 2, and this blue torus here corresponds  to the two-dimensional torus, s1 cross s1, and the intersection of the complex roots  of this two-variable polynomial with s1 cross s1 is represented by this red line here.  So this red line or this red curve would be something of real dimension 1, and so this  is not an atoral polynomial.  So this is something that's difficult.  We'll see that this will be difficult to treat with our methods.  So I'm returning to my notes.  I hope there hasn't been a problem.  Okay.  Let me continue.  So let me say something about what is known in this atoral setup.  So this is using the definition.  This is a result by Lin-Schmidt and Verbitsky using their definition of atoral, which is  slightly different from ours, but which in this special case, I think, corresponds to  ours, at least for irreducible polynomials.  They show that if we average over this test function here, then in the limit we get the  Mahler measure, right?  We're not averaging over Galois orbits, but over finite subgroups of this torus here,  this multiplicative group to the d.  And they also have a notion of delta of g.  This is just to mean that the points of this finite subgroup equidistribute in the classical  sense as we run along this sequence.  So they have convergence for finite subgroups under the atoral condition.  So again, if the polynomial does not vanish at all on the torus, then convergence follows  from the classical results because then our test function is continuous.  And there was a previous result by the same three authors from 2010 under the condition  that the vanishing locus of the polynomial on the d torus is finite.  Now remember, atoral is weaker.  That just means essentially that the vanishing locus is of real dimension at most d minus  2.  So starting from dimension 3, this is a stronger condition than being atoral.  And finally, Dimitrov and his PhD thesis, 2017, he dropped the hypothesis on p, so there's  no atoral hypothesis, but he allowed only subgroups of this type.  So the entorsion groups of the groups of unity.  All right.  So what is our result here?  So we're back to the Galois orbit setting.  This is joint work with Vessel and Dimitrov.  And so we have an atoral polynomial.  So with the same hypothesis as in work of Lynch and Verbitsky, but we're averaging over  the Galois orbit as opposed to finite subgroups.  And so we get the limit here equals the molar measure of p if, again, under the assumption  that the Galois orbit actually becomes equidistributed in the limit.  Just a few remarks.  So we get a rate of convergence that is a polynomial with a small exponent in 1 over  delta.  And we have a slightly weaker condition on atoral.  Actually, it's something that we call essentially atoral.  And I can explain that to you in a short, in this short picture.  Remember the one-dimensional case, there were these three situations.  So this was somehow the trivial one where the function is.  The function is actually continuous on S1.  Then we have the case where the pole, there's a small fault, wrong picture here.  So this is actually, this should be a singularity here.  There is a case where there's a pole at a root of unity, which was somewhat easy to  treat.  And then there was a case where there's a pole at a non-root of unity.  And this required Baker's linear forms and logarithms, or at least some quads, a version  of it.  So in the higher-dimensional case, we can have poles on the detourists.  But the poles set is not allowed to be too bad.  So it's at most of real-dimension D minus 1.  So this is somehow the atoral case.  We can also treat this case where the pole set is large of dimension D minus 1, but  it comes from essentially an algebraic subgroup, like here minus 1 is an irreducible component  of an algebraic subgroup.  This is a case that is conjectural.  The general case that in the non-dimensional setting requires Baker's linear forms and  logarithms is at the moment open in higher dimensions.  So this is what further ideas are needed.  And also a small point, we don't need to really work with this Galois, or we can also replace  Q by a number field.  All right.  Let me give you an example of what there's a nice application of these equidistribution  results.  Let's look at this polynomial, T1, T2 plus T3 plus T4, and this happens to be atoral.  Let me just look at these two guys.  You invert it, and then their coprime is Laurent polynomials, and then you get the two relations  R and S. Also a nice feature that the molar measure here of P has been computed to this  value.  It's also, this is related to the zeta function.  The actual value will not play a role for this work, but what's important is that it's  positive.  That will play a role.  Okay, so let's say we have four roots of unity, zeta 1 to zeta 4, such that their sum  is an algebraic unit.  You'll find examples like this if you look.  So you can take, for example, e to the 2 pi i over 6, e to the minus 2 pi i over 6, and  then two general roots of unity from minus here, then the sum will be exactly 1.  So there are examples where this is an algebraic unit.  What happens in that case?  Well we can apply the theorem.  The theorem tells us that the average here will converge to the molar measure as delta  goes to infinity.  But if I'm a unit, then this average here is just the logarithm of the absolute value  of the field norm, and so this will be zero.  Right.  And then of course we've got a problem.  As I said, this is a positive value here, this smaller measure, and so something zero  cannot converge to something positive when both things are constant.  So that means that delta of zeta has to be bounded.  So it's a certain finite result.  So as a conclusion, if zeta 1 plus zeta 2 plus zeta 3 plus zeta 4 is an algebraic unit  for some four roots of unity zeta 1 to zeta 4, then there has to be a multiplicative relation  between these four roots of unity where the exponents are bounded by some constant b.  And here the relation is clear.  I mean there are many relations.  So for example, you can take zeta 1 to the 6 equals 1 here in this particular instance.  And there are other relations, zeta 1 times zeta 2 is more than so on.  So this is a certain kind of finite result.  You can get out of equidistribution statements like this connected to a conjecture of Su Yong  Yi, which was also a motivation in the work of Baker, Ian Rumley.  Okay, let me say a few things about the proof of our theorem in the last, I think, 10 minutes.  So let's go back to the univariate variable case.  So if we just have a polynomial in one variable, let's see what we can get there.  So the proposition that we need to prove or that we prove in our paper is a following.  So we're back to the one variable case.  Later on we're going to try to reduce to the one variable case.  We take an average over a gaol orbit of unity, plug it into our Q, and proposition says that  in the limit, this is the molar measure of Q.  And the point of this proposition is that we get an explicit error estimate here, which  is explicit in all the data.  Now there is an important hypothesis here.  Namely, this is just the atoral hypothesis in dimension one.  As you recall, being an atoral in dimension one means I'm not zero on the unit circle.  So this is where the atoralness comes into the picture.  And let me give you an idea how the proof works.  So the proof is rather straightforward here.  We can assume that Q is just linear by factoring the polynomial.  And then what does the average look like?  Well, there are contributions to the average that are harmless, namely those conjugates  that are away from any pole here, any pole of this, that are not too close to alpha.  So there are actually no poles here, but still alpha could be very close to the unit circle.  So you want to sum over first those conjugates of zeta that are not too close to alpha.  So being not too close is being polynomially bounded away here, one over n squared as a  placeholder.  And then there is a remaining term here of the bad guys, the guys we have to deal with  later on.  Now, because the proximity here is one over n squared, and these are roots of unity of  order n, there can be at most one term here in this sum here.  So there's a most one bad guy.  And the rest, phi of n minus one guys we can handle.  And using just the standard equidistribution techniques, this part of the sum here converges  rather nicely to what you would expect, namely the molar measure of this univariate polynomial,  which is this log of max one come of absolute value.  So this sub part of the sum is harmless.  What we need to deal with is this part here.  And as I said, there's only one term here, but this one term can cause a lot of problems.  Because even if alpha is not on the unit circle, we have to deal with the contingency that zeta  or some conjugate of it is extremely close to alpha.  And we'll spoil here this absolute value.  And so this could be very negative in principle, at least, we have to deal with that.  Okay, so what is the worst case scenario is when our average is log of max one comma.  So this is the molar measure here, plus some pathologically negative term that comes from  a conjugate that is extremely close to alpha.  And then I'm not going to specify the error term here that comes from this one here.  So recall that this sigma, this exceptional conjugate is just of the four meters of the  two pi iq for some rational number q.  And it's tempting to apply Baker's linear forms and logarithms at this point, just like  in the work of Baker and Romney.  Because we're dealing here with, if you look at the log of this, this would be two pi iq  and the log of this would be the logarithm algebraic number.  If these two guys are closed, and so the logs will also be, or choices of logs will also  be, so it's an idea to apply Baker's linear forms and logarithms.  But this was already observed by Duke in his work in 2007, that the current versions of  even the state of the art of Baker's, even the two variable results, the dependency in  the field of definition here of alpha is not good enough to get a result here.  So the dependency of the field of definition is usually d cubed or something like that,  and that will not, won't be able to dominate this, this denominator here.  So that, that unfortunately doesn't work.  So we have to do something else.  And what we do is actually something, it's almost a sacrilege.  So we have to bound, the thing to do is we have to bound this from below, right?  And so what is the first thing you learn at university?  Well, the first thing is probably the triangle inequality, and then there's a reverse triangle  inequality if you just go the other way.  But of course you pay a high price for that.  And usually this price is much too high in any context, in any applications in numbers  theory.  But actually in this setting, it turns out to be good enough.  So, so we bound the different, the distance between zeta sigma and alpha from below just  by the distance of alpha to the unit circle.  Yeah, this is may seem like a blunder, but it actually happens to work in this situation.  And why does it work?  Well, if alpha, the situation that we're interested in is if alpha is extremely close to the unit  circle, and if alpha is extremely close to the unit circle, then the distance between  alpha and the unit circle is roughly the distance between alpha and its complex conjugate inverse.  So if alpha were precisely on the unit circle, then alpha to the minus one complex conjugate  would be precisely equal to alpha.  But they're, they're reasonably close to one another.  So now remember alpha is not on the unit circle by hypothesis.  No roots of qr are the unit circle.  So this happens to be positive, which is already a good sign.  We get a lower bound by something positive, but we need something a bit stronger.  And for this, we return, we return to the tomorrow results on separation of roots of  polynomials.  So this is a result of model from 1964 five to distinct roots of a polynomial with here  and future coefficients.  Then I can, I can bound the distance between these two roots from below by something that  is essentially linear in the degree.  And with a dependency on the molar measure of F.  And so if you plug in this result, you can, you can use this polynomial here with that  has alpha and alpha.  And it's the minus one is roots and you can bound this difference here from below.  And the lower bound you get will be good enough for the application because it depends to  here in D is essentially linear.  And that will lead to an essentially linear dependency here in D.  And the end, the five and we'll win.  So for our result, we actually have to deal with different alphas because we have a product  or polynomial is a product of linear terms as usual.  And so we need a result like molar, but where we compare pairs of roots, like a separation  result of pairs of roots.  And so for this, there's a result of menyalt that strengthened molar result to pair to  several pairs of roots, not just one.  So we get a lower bound for several pairs of the same quality as more.  So that kind of will give us something of this nature for the univari case.  And then we have to go to the, we have to reduce from the, the multivariate case to  the univariate case.  So now let's assume that P is a toro.  And we have a tuple of order n and of roots of unity.  And so we can, we can reduce to the univariate case just by the basic observation that any  tuple of roots of unity of order n is essentially generated.  The coordinates are generated by a single root of unity of order n.  So I can write this zeta, bold zeta as non-bold zeta to the, to some exponent a, which, which  is a vector.  And for Galois conjugates, I can do the same thing.  And I also have some, some degree of freedom here because I can always add a vector, a  multiple of n, because if n is the order of zeta, then that won't matter.  So using this freedom, I can construct a polynomial starting from P using this exponent here.  And the game is then to choose this exponent, choose tau and, and, and be in such a way  that this exponent is as small as possible.  Right.  So this is going to be a polynomial of small as possible degree.  And, but it will be univariate polynomial.  So we can try to apply the proposition before to this univariate polynomial.  The point, of course, the average doesn't change by doing this, this, this process.  So we can reduce to the univariate case.  The pay price we pay is that the degree of this queue here will be very large.  That's why I was very concerned about getting good dependency in the degree in the last  univariate case.  So using Erdos-Turon-Coxma, we can get a bound for like an optimal choice here of this exponent,  which will decay depending on this delta of zeta.  Right.  So we get, we get some improvement over the trivial bound, some power of delta.  And then we plug in, we plug in the proposition, and then, well, at least if this delta here  grows quickly enough, then the error term will be smaller one, which is what we want.  Right.  So if this delta here grows like n or some small power of n, then we get convergence here.  So let me just, let me just briefly subsum what we have here.  So if this delta of zeta grows quickly and often n is like some small power, then we  get a convergence result.  The molar measure here is again the molar measure of something univariate.  So we have to, as a next step, we have to compare this with the molar measure of p,  the original polynomial.  And for that, we need a result of lots, or at least some, a variant of the result of  lot that allows us to reduce the computation of multivariate molar measure to univariate  molar measure.  And there's a quantitative version of this that we show.  We have to deal with this situation where delta does not grow quickly.  And the first, the final thing is that we have to deal with the situation where q vanishes  on the unit circle.  Remember that was an important hypothesis in the proposition.  Without this hypothesis, our method doesn't work.  So we have to make sure that the polynomial q we constructed does not vanish on the unit  circle.  And just let me say a few words on that.  So if q happens to vanish on the unit circle, that's where the atoral condition comes into  the picture.  If q vanishes on the unit circle, then we'll find a point on the detourus and on the vanishing  locus of p of this nature here.  And now we use that our polynomial is atoral.  So this will be actually contained in the vanishing locus of two coprene polynomials,  r and s.  And then using a result of Bombieri-Massourenzanie, we can get a nice bound here on something  orthogonal to this expression here.  So this here uses a result of Bombieri-Massourenzanie and unlikely intersections.  And let me just go to the final slide.  So what happens if this nc is orthogonal to the exponent, well, then I just take, I can  take c as an exponent in zeta and I get one.  And that means that the delta of zeta is at most b.  So that means if I have a root of, I have a point on the unit circle that's contained  on the vanishing locus of my q, that means the delta of zeta is bounded.  And that will mean that the proof is over because we can assume that delta of zeta is  as large as one.  Let me conclude by showing two or three pictures.  Here this is what happens if this is a typical atoral polynomial intersected with detourus,  this red point.  It's a single point.  The yellow line is the algebraic subgroup here that appears in Bombieri-Massourenzanie.  We can make it a bit more complicated.  Now it's x times y squared.  And even more complicated, you see that some people always misses this point, this random  point.  Of course, it could also hit the random, the point in some cases, but the theorem tells  us of Bombieri-Massourenzanie that this is somewhat rare and we can deal with this situation.  Okay.  So here, to close off, here's a picture of Lumini from Mont-Puget, who is marking the  background three years ago.  So I hope to return to Lumini soon.  Thanks for your attention.  Yeah, so thank you very much for the recent, very interesting lecture and for this nice  picture from Puget.  Maybe you know there is a fast track to go down just right off this place.  You took the picture, which is a really very nice round tour of about three hours from the  Institute.  Yeah, it's a pity that we cannot do that.  So we have still some time for very fast questions.  Are there questions, comments?  Yeah, there is a question.  Can you read it, Philipp?  Fabia?  Fabia, I think you can hear the notion of atoral and the notion of tempered from Denninger.  No, I've not.  I'm not aware of tempered, but I'd like to look into that if there's a connection.  Okay.  Denninger is all interested in smaller measures.  I'm back again.
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