 First of all, I'm going to acknowledge the first Australians whose traditional lands we meet and whose cultures are among the most in human history. The Marlborough Lectures are a biennial activity of the Australian Mathematical Society, and are further jointly by the Society and by the Australian Mathematical Science Institute. They are furthered by a request from the late Furt Marlborough, one of the first mathematicians appointed at the ANU. In honor of Marlborough's research field, the Marlborough Lecture is usually chosen to be a Marlborough Lecture. This year we are especially pleased to have as our Marlborough Lecturer one of the world's most eminent mathematicians, Professor Peter Sarnack from Princeton University and from the Institute for Advanced Study in Princeton. Peter Sarnack was born in South Africa and moved to the U.S. Graduate School. After appointments at Stanford University and York University, he moved to Princeton University where he has been since 1991. He has had a major influence on the course of modern analytic math theory, both directly through his many research papers and through his numerous graduate students, some of whom are themselves major thinkers in math theory. Professor Sarnack's talk today will be on Marlborough Theory and Circle Patterns of Apollonies. Please join me in welcoming Professor Sarnack I won't use any microphone. I have a loud voice. I have a New Jersey accent. That's a joke. There are many South Africans in this country and I know you probably could recognize my accent. Just let me make sure this is on. Well, it's a great pleasure for me to be here. I feel like I've been making this trip from New Jersey around Australia up to the Mecca of where Marla came from. Marla was here and I've finally arrived at his stomping ground. He was a great number theorist, contributed to many different aspects of number theory. And one of the things that I will try to describe in this lecture is how we want to get rid of some of his work. Because it applies to problems that are too well known, maybe. So I'll try to explain this as we go along. Okay, so my topic is the circle packings of Apollonius. So that's Apollonius. I hadn't quite, until I prepared this lecture, I hadn't seen a picture of him. When I saw a picture of him, I saw that his nose is very much like mine. So maybe I always told people I have a Greek nose. A Greek god's nose. He was a god of number theory at the time. He's very well known for studying circles, conics. He coined the terms parabola, ellipse, hyperbola. Conic sections were certainly one of his favorite things. And parameter spaces of circles, which is what I'm going to describe in a second. So that's as much as we'll have of his picture. These are a bit distorted there. No, it's good. What you have here are three coins. This took me a long time to find. I was looking for three magical coins, and I looked in all denominations, even the Australian, mostly the South African, all over. I wanted them to be circular. And then, lo and behold, the ones that worked were the quarter nickel and dime. That's why those pictures are there. Now what was I looking for? I put these three coins together. They mutually tangent. If you think there's a size of a U.S. coin, I'm declaring the diameter to be 24 millimeter for the quarter. Is that the quarter? Well, anyway, that's the biggest one. So if you go to a website, you'll find that in some websites it's 24 millimeters, 0 something. I wonder if the people who actually think that there's a size to the coin that their coins are of rational diameter. I don't know. But this is extremely important for me that these be rational numbers. That's vulgar fractions, if you want to call them that, because I'm going to be doing some number theory, which is in the title. So one thing is for sure, these are accurate to the nearest millimeter. And since these worked after a long search, what do I mean by worked? So you take these three. I'm going to prove one theorem in this lecture. This is the first time in this entire series that I'm proving a theorem. So you'll have to watch for that. I'll prove that theorem, and that theorem will tell me there's a unique circle inside you which is tangent to all three. I put that circle inside there, and a miracle happens. That circle also has a rational diameter. I'm going to prove that for you. That was the miracle I was looking for, and I don't know any other coins that do that. Unless you take two coins to be equal, then it's much easier. All right. Now, because we're going to do something to do with whole numbers, that's what number theory is, I scaled that picture up here by 252. It's convenient, and it's also much more convenient not to talk about the radii or diameters of the coins, but rather their curvatures. Because I'm going to be filling in that space with more and more coins, and the coins are going to get, well, more and more circles anyway. There are no more coins left. I'm going to be filling in these circles, these loon regions, and they will be getting very small. So the radii will get small. I don't want to see small numbers. I'd rather see big numbers. So I'll look at their curvatures. So the curvatures just one over the radii. So what you have a picture here, which is just a restatement of the previous picture you saw, is that these three, that's what I started off with. In that picture we had the one in the middle. In that picture, let me go on the outside now, there's a unique circle tangent to all three on the outside, and this circle has curvature 11. But because this is the only circle that's going to contain everybody else, there's an orientation in this problem, this guy's going to get a minus 11. Nobody else will have a negative number in front of it. So what these numbers are that you see in the circles are the curvatures of these circles. And now, this is where epilonius comes in. I'm going to prove this theorem of epilonius. This theorem is that if you're given three mutually tangent circles, there are exactly two other circles which you can place which are mutually tangent to all three. There's one of them, there's one over there. But anyway, in each of these loon regions we're going to put a unique circle according to epilonius and fill this picture in. So these four have, the miracle was I looked very hard and found that these four had whole numbers. The miracle is this will continue forever. So that is if you put these, this circle there, that circle there, that circle there, they also have whole numbers for their curvatures. This is highly unexpected. You should not have expected that. If you did, very good for you. I certainly would not have expected that. And now, once we have that we have all these new sets of holes. We can count how many holes there are and we continue to put these things on and we continue that forever and that is called an apollonian packing. This is an apollonian packing which clearly has whole numbers or diphentene aspects, diphentene means whole numbers especially when you refer to equations. It's another Greek diphentene that we, if you interested in theory of numbers you would ask what numbers do you see here? Is there a pattern? Is there some laws to which numbers are appearing here? The first thing I want to point out here is I wasn't aware of this apollonian packing until about six years ago. I learned it from a fellow called Jeff Lagerius. He told me have I ever looked at apollonian the integral structure, integral means whole numbers, structure of apollonian packings and I said, packings, there's only one apollonian packing, what are you talking about? Because I thought that, and it's true, over the real numbers they're the same. In a kind of transformation that I'll show you in a minute, certain inversions show that there's only one apollonian packing. However, if you're interested in these whole numbers it turns out, as he said, there were infinitely many and in fact he'd written many papers and it was a very famous set of ideas and I learned that from him by telling him that he doesn't know what he's talking about, running home and seeing that he knows so let me tell you about these integral apollonian packings. Firstly, there are infinitely many, we'll see that and the number theory of this is very challenging and I'm going to try to explain that and it takes us beyond Mahler and many other people. The kind of question that you ask here is which integers appear here? It is very typical in the theory of numbers that it's very easy to ask questions very hard to answer them. That's the beauty. That's the question. If you ask for Mahler's theorem it's a very simple question. Its solution is understood by a handful of people in the world. So the techniques to solve these kind of problems are often extremely sophisticated using all branches of mathematics but the beauty is the questions are always supposed to be explainable in a public lecture. So the question is if we had a second screen here, or we had used the second screen we might have put this up there and I would leave it up there for you to look for patterns that we're going to be studying along the way. Of course the numbers are so small here you can't see them but if you had a computer I guess you could make it bigger. Anyway, are there infinitely many primes? Are there infinitely many circles whose curvatures are prime numbers? We love prime numbers. Are there infinitely many pairs of circles whose curvatures are prime numbers? We'll call those twin primes. Is there actually a rule? Can you tell me which numbers are appearing in this packing? So it's a random question maybe but the techniques are really interesting as I hope to at least convince you of one or two things. The first person to discover this integral structure is actually a Nobel Prize winner in chemistry by the name of F. Soddy. I don't know why he was looking at it, he found this and he was very excited by it and even wrote a poem, a well-known poem about it, but I won't repeat that. I don't think the poems are great but it does rhyme. You can find that I'm sure easily by going on the web. Anyway, here's the first amazingly to discover this integral property and I'm going to, that's the one theorem I'll prove to you is that those things are whole numbers. There's a very beautiful paper, I mentioned Ligarius already, Bell Labs paper, this was done Bell Labs, they were all there working in teams as they do over there and they asked many of the interesting they set this up as a diaphragm problem and asked many of the interesting questions and what's happened in the last four or five years if we now have tools to answer pretty much all the questions and the tools come from all sorts of fields in mathematics. Modular forms you should have expected that's sort of the main bulldozer in modern number theory. Egotic theory that seems to be something probability theory, it plays a role. Hyperbolic geometry, spectral theory, additive combinatorics, all these are entering in understanding this problem. So of course I'm not going to go into proofs of what I'll say but I will give you some demonstrations of some truths except for this theorem of Apollonius since he's in the title I want to give you the proof and the proof from the book. So his theorem says given three circles, mutually tangent they are exactly two circles C and C prime which are tangent to all three. To do that I'm going to use some geometry which I hope is taught in Australian schools but it wasn't in South Africa I can tell you that when I was there anyway and hopefully it's a bit better than now. Anyway this is a kind of transformation which preserves circles and preserves angles. So I'm going to use group theory well really I'm not going to actually use it yet but I want you to focus on the following. If I have a circle, we have one circle radius R and if I invert every point by which I mean I'll take p to q where the distance from p to the center of the circle distance from q to the circle the product of those two is the square of the radius. So this p q the original go to infinity to find point and you can write down a formula for that transformation it's very easy and if you use that formula you will see that circles go to circles and if two curves cross at a certain angle the image will also have that same angle. So tangencies are preserved, angles are preserved and circles are preserved. There's only one lie there a circle can be a straight line. A straight line through infinity is a degenerate circle so some circles will go to straight lines. This is a Mobius transformation in modern mathematics. Okay so you use this transformation that's all I will use. I'm not going to prove for you Apollonius's theorem without any calculations just this thing but I didn't show you that those were preserved that you have to make a calculation. So here are my three circles C1, C2 and C3 the dark circles and I want to find the unique circle there that one and C prime C prime and C double prime which are tangent to all three. Okay let me take this point C which is mutually tangent to C1 and C2 and let me invert so I'm going to take C to infinity so I'm going to invert in C remember I'm just going to take any circle who sent this C any circle in the world and invert in that circle and C will go to infinity, circles will go to circles tangencies will be preserved alright so C2 is now going to be going a circle through infinity I'll call it C2 prime it may not be parallel this way it might be parallel that way that's a minor point so that's my one image C1 prime also goes through infinity and it touches at C at infinity that's the tangency point so that's where these two circles go. The third circle C3 goes to C3 prime it's got to be tangent to both of these and the only circle is tangent to both of these just by looking at it with your eyes is a circle like this so those are my three circles and now it's plain and obvious what the two circles we're looking for are there's one on the right and there's one on the left this guy and that guy you can just stare at it and see there aren't any other circles around that can be tangent to these given three circles so in this configuration we found the answer and now because that operation was invertible I just invert back and I find the three the two circles that we were talking about so this is Apollonius's theorem and that was the heart of the construction of the Apollonian parking was to put circles which were mutually tangent to and as you can see there's a symmetry would you choose this one or that one the minute you don't know which one to choose you know that there's something important about it a group symmetry a goal asymmetry all the same thing here and that's going to be the group that dictates the answers to everything okay Apollonius is not enough for us we also need Descartes, Descartes the guy who loved the coordinates and he wrote everything down in terms of Cartesian coordinates and equations and this is a famous theorem of his which he wrote the I don't have a proof from the book nobody has a proof from the book the previous proof I will declare from the book meaning I don't want to see any other proof this is the best proof you didn't have to calculate all proofs of this theorem that I know do require some unpleasant calculation I have it at the end of the lecture but I'll purposefully run out of time so that I won't show it to you so this is the remarkable statement of Descartes he wrote it according to Coxett I didn't have a proof Coxett appointed out a slight flaw in the argument it's actually in a letter of his to some princess I guess he thought this theorem would be by something anyway it's a beautiful theorem it's like Pythagoras' theorem it's a little more sophisticated so the theorem is the following suppose I have four mutually four now not three four mutually tangent circles like the starting four that I had there coins in the one on the outside and with that sign convention then those four curvatures the A's are always the curvatures A1, A2, A3, A4 satisfy a quadratic equation and the quadratic equation is that F of A F is this quadratic form twice the sum of the squares minus the sum of all the numbers squared the quadratic expression in the A1, A2, A3, A4 that F of A is always zero and if and only if if you have four numbers which satisfy this equation you can find four circles with that property that's Descartes' theorem and I don't have a really proof from a book so I skip that now I want to prove to you Soddy's observation and explain to you that what Soddy had realized or what he's doing has really to do with that lack of choice of which circle you want to do that's the heart of the whole problem all the 13 properties also come from there is where that's where they come from so suppose I have four mutually tangent circles with curvatures A1, A2, A3, A4 they must satisfy this equation I'll come back to this quadratic equation later but for now suppose I have such a story and now I use Apollonius's theorem one of these, so I fix A1, A2, A3 or C1, C2, C3 the fourth circle has curvature A4 and there's another fourth circle which is also tangent to the first three and it's got curvature A4 prime well according to this theorem both A4 and A4 prime these are fixed satisfy this quadratic equation that is to say that A4 and A4 prime are solutions to the same fixed quadratic equation in one variable because now the A1, A2, A3 are fixed so this is a high school for the formula for the solution to a quadratic equation it's minus B plus minus the square root of B squared minus 4AC over 2A not that jet lag and you get a relation that the sum of the roots is this expression you just calculate it and in fact the two roots are and you compute the minus B and you get A1, A2 A1 plus A2 plus A3 so this quantity is really the most important in terms of the questions of rationality of the circles because remember I start with three circles as I did with my three coins which are declared to be 21, 24 and whatever it was if this number delta so if I have now the fourth guy I'm looking at A4 and A4 prime I want to see if the new guy circles are two of them I can't distinguish them if the first three are whole numbers it would be nice for the fourth one to be whole numbers and that would depend on whether that square root is perfect whether that's a perfect square delta and that's the miracle of yeah not sure who made it okay so thank you anyway in my example which was 21, 24 28 you compute what delta is it's 42 squared and that was the miracle and that's why the circle in the middle was a rational number so that's what I went through these coins looking for three coins such as this delta that expression corrected is a perfect square and that's very unlikely but if you look long enough hopefully you could find it and of course if you can cook up the numbers yourself you'll find it alright so that's the miracle and once that happens then notice the following once I have that four of them which are whole numbers then if the four starting ones have whole numbers and I make one step in this Apollonian move which is to put one new circle in the new guy A4 prime from this expression here is minus A4 plus 2A1 plus 2A2 plus 2A3 it's a great property of the integers that if you add integers you get an integer and if you multiply integers you get integers and that's really the only property that we have at ease so this is very convenient and that's the proof of Saudi's theorem from now on each time you put in a new circle the expression in terms of the four circles the three circles and the two ones you're playing off against each other the new guy if the first four are whole numbers then each time you put a new guy it's a whole number and each newborn guy is a whole number so the curvatures from now on are going to be whole numbers and that's his observation and he was very happy with it and it is rather remarkable okay whatever wrong computer now the minute thank you this young man typed this for me you know I said if I run into trouble he's just going to take over okay I'm now going to describe these moves in terms of matrices so I hope you all at least have some vague idea what a matrix is this is an array of numbers like this that you can multiply matrices you can add them in this case we're going to multiply them so the statement that A4 prime was minus A4 plus 2A3 plus 2A2 plus 2A1 was a statement that can be written in one line with a matrix notation as follows that A4 prime this should be A prime the new vector of curvatures so the first three are unchanged but the fourth guy changed by this combination now I notice I'm multiplying on the right because it's much nicer to multiply rows than matrices to get back a row a vector should be a row not a column okay that's the only reason I'm multiplying on the right which is rather unusual I guess anyway so this is A prime and that's A4 prime so this is a little misprint there it's not serious so you get the new guy from the old guy by multiplying by S4 and that's if you take the first three circles to produce a fourth one now there are four ways of choosing three circles out of four circles and you can take any three to produce the new element and so they are generated by these four matrices so I'm now going to define simply these matrices they are the matrices which the epillonian packing follows and in a way that I'll describe here and there will be a little more notation here is I take S4 it's this 4 by 4 matrix with whole number entries and similarly with those you can easily check that if I square any of these matrices you'll get the identity the determinant for those of you know what that is any of these is minus 1 the inverse well from this of this matrix is also a whole number a matrix with whole numbers so I'm going to take these matrices and I'm going to look at all expressions in them or what is called the group generated by these 4 by 4 matrices this group we call the epillonian group it's the symmetry group of this problem it's the group which captures that feature that any time I don't know whether to choose in epillonius's theorem which of those two circles I don't know which to choose that's the sign of a group that's what the Goa theory is about this problem so concretely we're just going to look at all matrices that you get you will never leave the 4 by 4 matrices of determinant 1 that's all minus 1 that's all this big notation is over there and I get a matrix group a subgroup of the 4 by 4 integer matrices that means any element here can be multiplied with any other element and you get a new element in there and the inverse is also in there this group is the key thing in the whole problem and I'm going to explain to you why understanding the difantine properties of epillonian packing is difficult precisely because this group is a new kind of beast so it is a symmetry group of the packing alright now I'm going to just reformulate what I've been saying in words if you take your 4 vector so this is a point whose coordinates are all integers in 4 dimensions so it's in 4 a 4 tuple of curvatures remember you think of our starting guy which was minus 11 and those other 3 numbers and you take this epillonian group and you multiply you fix A and you take all elements in the group and you multiply each element in the group times A you'll get a new 4 vector and if you follow roughly what I was saying you will find that these 4 tuples that you get of whole numbers consist precisely of all circles of circles in that packing which are mutually tangent but of course I don't know which elements I get here in z4 this is called an orbit of a group and that's a very subtle question maybe it's even undecidable you might guess and if you generalize the problem a little bit that is the case as to who's in the orbit or even who's in the group we only have generators for the group the generators are as 1 as 2 as 3 as 4 and the group had another description but that's what I'm going to show you is not the case number one since all these vectors we're producing here are 4 tuples of mutually tangent circles they must obey Descartes theorem so they must all these integers we'll ever get will all lie on f of A equals 0 that was Descartes theorem now f is a quadratic form and 0 set of a quadratic form by the law of inertia this has signature 3 1 for those who know what I'm talking about this is quadratic form the 0 set is a cone so these points all lie in a cone in 4 dimensional space and they all whole number points you might ask if you're getting all points this way well let's look a little further let me take those s1 s2 s3 s4 those transformations f of sj times x is f of x that's not this surprising because these s's cone back into itself it's probably the case that they take me they don't change the value of f I hope most people know what the rotations in 3 dimensions are motions which keep you at a fixed distance from the origin so you rotate about an axis that would be the rotation group of x1 squared plus x2 squared plus x3 squared here we have a more complicated quadratic form it's the Descartes form it's which is 3 1 the rotations the transformations 4 by 4 transformations which preserve f meaning f of x s sx is f of x for any point x are called orthogonal transformations for this form and the group is called the orthogonal group and clearly whatever our apollonian group is it's a subgroup of the orthogonal group that's all I'm saying here so I will denote all 4 by 4 matrices which preserve this quadratic form by O of f all 4 by 4 matrices whose entries are whole numbers that's a much more frightening thing but they are many of them I'll call O f of z these are the 4 by 4 matrices which preserve the form and are integral and our group I've just told you is a subgroup of that if it were all of that we would be in the realms of ordinary number theory so I'm going to spend giving you what standard questions are but that's not the case this group happens to be very small in there it's infinite index so it's small in a very explicit way let me not define the index I'll just say it's thin I just want to give you a feel the mathematicians know what I'm talking about but on the other hand it's not too small this is why we'll be able to say anything it's a risky dense an algebraus could not tell this group from that group by polynomial equations so the risky topology is a way of telling is a set of equations a space defined by set of equations and if you look at the smallest set of equations which contain all these points it will be the same as the smallest set of equations which contain all those points which is just the orthogonal group itself so don't worry too much about this but you've got to take two things away from this that one this group has got something good about it in terms of geometry algebraic geometry and it's got something terrible about it that it's not something that we ever study in number theory because it's infinite index the theory of quadratic forms for this whole group this group is a well studied group goes back to Hilbert I'm going to tell you what Hilbert's 11th problem is is well understood thanks to many many people including Myler who studied these kind of groups a famous Myler compactness theorem is all about these kind of groups there's a very beautiful compactness principle so this is the theory that I don't want to talk about because it doesn't address our question here but just to show you how hard our question is let's just remind ourselves a typical problem of this nature so Hilbert's 11th problem concerns solvability of quadratic equations in whole numbers that's it called Hilbert's 11th problem was only solved recently the final straw being put by Piottsky, Shapiro, Kogdel and myself I want to remind you of a very simple equation so that you at least see the relation what Hilbert was asking compared to the problem we'll be asking here so Hilbert's 11th problem is the following is the generalization of the following problem that goes back to Gauss and Legendre and I like this because it's very, very involved this looks simple but it's quite deep which numbers are some of three squares now that has got no deficiencies or packings that's just an equation in whole numbers that's what most of number theory, much of number theory about is a theory of diathantine equations can we solve this equation so this mankind pondered very early on which numbers are some of three squares these are everything to be whole numbers the one condition that you quickly see is n better be not negative because squares are not negative so there's an obstruction there but that's, if you think a little more you'll see another obstruction no number which gives remainder 7 when divided by 8 is a sum of 3 squared why? because if I look at division by 8, I look at only I do all my operations in the whole numbers I only look at the remainder when divided by 8 then my remainder is 0, 1, 2, 3 all the way up to 7 there are only finitely many 8 possibilities and if I square any of those numbers I'll get 0, 4, 1 those are the only possibilities for the squares and if I add any 3 of those you just look at the possibilities you'll never get 7 so any number which gives remainder 7 when divided by 8 is not a sum of 3 squares there is an obstruction and similarly any number which is 4 to the 8 times 8b plus 7 is also not a sum of 3 squares so there are these obvious reasons that a number is not a sum of 3 squares that anybody who thinks about it quickly comes up with the hard and the deep theorem is once you can't find any more reasons that it's not a sum of 3 squares maybe it is a sum of 3 squares and that's one of the great theorems it's a theorem of Gauss there's a big debate who first proved this well he knew the theorem but he didn't quite have a complete proof Gauss has a proof it's an entry in his diary around 1800 he starts out Eureka he was trying to prove this for a long time he carries on in this thing at the end of the entry he says and this is fantastic I proved this and at the end he says by the way my fifth daughter was born today a minor event in this context of this theorem and it's a good example of a local to global principle meaning that this is the best of all possible worlds we have reasons that a number the equation can't be solved and the minute we overcome we pass this condition we are solvable there are very few equations which obey this local to global principle the quadratic is 1 the question we are asking about the Apollonian packing is the same kind of question it was born by making this packing and it's not an arithmetic group it's not the full orthogonal group OFZ it's infinite index let me just amplify this point a little bit so again I go back to that cone the set of all x such as f of x is 0 this is a cone in four dimensions the Apollonian group or this group OFZ which are the four by four matrices which preserve the quadratic form of which all a's all elements of a are in there preserves a little more so I define v prime primitives to be the points whose coordinates have no common factor the greatest common divisor of a 1, a2, a3, a4 is 1 we start with such a point and we apply any of those Apollonian moves you will not destroy the fact that the greatest common factor is 1 and what's interesting is if you take this big group OFZ the one that's friendly in the sense that mankind knows about it not that it's easy, this is a great achievement of many people to understand Hilbert's 11th problem but if you take the orbit of any point a which is primitive under this you will get all the points so that's why that problem has another description the orbit there is exactly all points which satisfy an equation while the set of circles in an Apollonian packing are the set of points which you get by keep on making these moves I don't have another description of those points so if you want to work out what's there you have to develop a new theory and that's why the theory the modern theory the theory that developed after Hilbert is not good enough for the simple question that Soddy asked I won't go into it but I've coined these thin groups because they were infinite index whatever there was they were deficient they were born this way and they come up everywhere now now that we start looking they come up in monodrome in Calabi yows a yow was here recently selling his book maybe you look there you'll see Calabi yows they're thin groups there amazingly there are some of these groups you can't even tell who's in the group that's the undecidability properties in any event what's happened in the last six years is a great development of theory with many contributors making that theory not as well understood as the theory of following Hilbert but enough to answer some questions and now I'm going to tell you what other questions we can answer for the Apollonian group so far I haven't told you anything about the Apollonian group except what Soddy found that their whole number alright so now let me go through and that's the rest of the lecture I'm going to explain to you what we can show let me amplify that point that this is just an example and a very beautiful example with beautiful pictures of what is a very general theory that has come out which allows you to deal with these kind of situations so the first question if you're going to try and look for numbers with some properties if you want to count prime numbers and you erasthostanese or somebody like that if you want to produce prime numbers there first thing is you need to know how many numbers there are so if I give you a big number x and I ask you how many numbers are less than x well more or less x x could be any real number if x is a whole number it's the floor function of x or something I don't know if the floor includes x or not you always miss by one ok so there are roughly x integers less than x and if you were to do a sieve which is the kind of thing we may want to do we want to find all prime numbers up to x then you would remove all the numbers divisible by 2 and that's roughly half of x remove all the numbers divisible by 3 it's roughly one third of x add all the numbers divisible by 6 and then you keep this going and then when you left you're just left with primes that's fine if you're running it on a computer but if you actually want to count the number of primes it's extremely difficult that's called the sieve and it's very difficult from that procedure to actually do the counting you have to do something much more difficult so can we count here this is the following let's look at all the circles in the packing for which their curvatures are less than a big number x you should think here of x is a very big number and x is going to infinity how many circles are there in the very first picture you saw there whose curvature is less than x this doesn't, this has nothing to do with diaffantine this could be any packet David Boyd many years ago was able to answer the first step in this he showed that this number grows like x to a power which is some weird fractal dimension in other words he showed log of this number divided by log x when x goes to infinity tends to a number that exists there and that's the number to a few decimal places we don't know this number exactly but it is important for the rest of this lecture that this number is bigger than one it's actually the house of dimension of something called the apollonian gasket he showed that too and his arguments were all elementary in Euclidean geometry estimating inductively the new circles from the old circles using formulae one of the things that one wanted to understand is that maybe there's a refinement of this, an asymptotic law and there is and this is one of the first theorems that has emerged from this new theory and it's a theorem of Alex Kontorovich and he owe from two years ago it's just appeared in jams probably one of the better journals but not the Princeton Journal I have to sell the journal I'm an editor it's going down the hill down the tube I'll tell you outside later some stories so anyway this is a beautiful theorem and it gives you the refinement that the number is in fact there's another constant such that the number of circles whose curvature is less than x is about b times x to the delta to show you that probably you'll never get this elementarily is you might, I can't prove that but you would have to rediscover something rather striking this b is basically determined by a base vibrating mode on a hyperbolic three-dimensional infinite volume manifold scattering theory is needed here so it is quite remarkable that you can use these beautiful theories from spectral theory and geometry to do this and if you gave an elementary proof of this you would have to rediscover this base eigenfunction in another language it's not impossible, but it would be very interesting so we do know the number and this is very important as we advance to try to understand the arithmetic so let's turn to the arithmetic now this will be used freely the question you should be asking and we have alluded to all the time which numbers appear there are there some congruence are there some arithmetic obstruction just like I said that if you take a number and you want to write it as sum of three squares you can't if it gives remainder seven when divided by eight now those numbers were too small but I expect you all notice that the numbers in the coverages there satisfy some congruences I'll now tell you what they are and the interesting theorem here is we know exactly what these congruences are let me remind you what a congruence is I've already been talking about it but just to review we do when I say mod Q I will always look at all integers or any expression that you see there in arithmetic when you divide by Q Q is any number greater than or equal to one so then there are only finitely many possibilities and being finite we can manipulate much more easily and understand that problem and so that's the local theory and we want to first understand what's happening locally and then hopefully globally when we look for the integers themselves that's the motto of modern numbers here so for P naught I can't expect you to have noticed but those are exactly the congruences every curvature you saw you could certainly see it from the first four that I had there but you generate all those numbers and they always satisfy they always one of these numbers and mod 24 is one of these numbers and this is the only arithmetic such obstruction that's not obvious and that was proved by Elena Fuchs in her thesis last year that's the initial one that's I'm lecturing around those three coins but in fact there are only two possibilities universally there's only one other possibility so she answers it for any integral epilonian packing so P could be replaced this is for P naught that this is true for all of those is more or less obvious you start off with your four matrices and then you got S1, S2, S3, S4 and if S1, S2, S3, S4 you look at what they do mod 24 this is now a finite little diagram and you can see what you trace out and you see you only get these numbers the more difficult thing is to see there's no obstruction there's no congruent subtle thing going on modulo some other big prime and I should say that that's where the risky density comes in and a very beautiful theorem of a man called Weisfeller who tragically disappeared in South America many years ago, unexplained alright now why are we so interested in counting suppose I want to know I'm going to tell you a conjectured answer and something we know towards it which numbers are curvatures there well the first thing is remember the number of curvatures up to X is about X to the 1.3 that was the theorem of Kontorovich and O but the number of integers up to X is only X so roughly if God is making them at random each integer is going to be hit roughly X to the 0.3 times but not every integer is there because we know that you have to satisfy these arithmetic obstructions so you build in the arithmetic and then you expect that you hit everything you're supposed to hit and it's bad news, you don't but the best that's because very small numbers which satisfy the congruence are not hit so then you hope maybe something becomes true when X gets large and that's where I'm aiming but before you do that you try on so much simpler question which is how many different numbers not telling me which one X without multiplicity so how many distinct integers up to X are the curvatures of circles in the Apollonian packing this was a basic conjecture in the paper not of Saudi but of the Garia center and this too was solved recently by Borghain and Fuchs showing that their conjecture was a positive proportion of all numbers are hit and that's proved and that uses all sorts of more standard rules from number theory, not this really modern theory and you can find the basic idea outlined in a letter of mine so if you go to my name at Princeton they made a very important improvement on what I was doing getting where I had some slightly weaker result the basic idea is an ad hoc idea I would say coupled with standard number see my philosophy in this lecture is anything that's standard I won't talk about much but the real question here that I don't know the answer to is there local to global principle like in Hilbert's 11th problem and I'm sure there is it's just we don't have the tools to prove it that is, is the set of integers which you see in this Apollonian packing exactly all integers which give one of those remainders when divided by 24 except so that's not true is that true except for finitely many exceptions is it true that if we know the exceptions we just find them galore I'm going to show you a picture now but if you go far enough because you have many opportunities to hit each number about x to the 0.3 according to that count maybe you do eventually hit every number and that would be a local to global principle that would be a beautiful answer to this you would actually have a description of exactly who are curvatures so that we don't know but we have some nice experiments here by Fuchs and Sandin to Princeton students I think I'm running a bit slow so I won't go into it but let me tell you that the so look here whatever this says here it says there are 536 exceptions between 10 to the 8 4 times 10 to the 8 and 5 times 10 to the 8 so you might say how could I say with any with a straight face that this conjecture is true when there are 538 exceptions out when we've gone up to 10 to the 8 we have some theoretical ways of understanding why this is catching up slowly like yeah there are no exceptions so I am quite confident that this local to global principle is true and it's much more difficult than Hilbert's 11 problem because the group A is deficient in the way I've described alright one can prove other things this is what I had set out initially are there infinitely many circles whose curvatures are prime for example if you went if the original picture up here I'm asking now for some memory you'll find in the middle that they actually even twin prime there are two circles next to each other both of which curvatures are prime the answer is yes there are infinitely many circles to a prime and there better still there are infinitely many circles pairs of circles for prime and that proof is an ad hoc proof it does use all the modern tools of half dimensional serves of a reduction but it's true and it's something that is true for any integral apollonian packing one of the reasons I like to put this down I make this joke always it's a bad joke but I've got to make it I've always wanted to prove the twin prime conjecture so I prove the twin prime conjecture there are infinitely many twin prime in this picture where I define what I mean by twin prime the twin prime conjecture it's a much simpler problem that it's natural it's debatable but any I think anybody here who's a young person I don't see too many but who's supposed to come to these lectures and saw a list of primes which can't help but asking are there infinitely many primes whose distance from each other is two it's not a natural question it's a very famous story about Vinogrado the guy who invented many of the tools to do things with primes he was the head of the Steklov Institute a nasty man and one of the young men the young guy is different in each story but the story seems accurate enough he says he was a boss why is our boss so famous he proved every large odd number to some of three primes what does that mean I haven't thought about primes since high school but I thought primes not add them it's not a natural thing to ask if there are infinitely many primes in this picture but I can't resist so that's the nature of the beast we like these kind of questions yeah oh yeah twins alright I'm near the end here you might ask is there prime number theorem so we are a little short of proving that and I'll just end with that the prime number theorem is a theorem about the ordinary prime numbers the number of integers less than x is x the number of primes less than x how many prime numbers are there less than x there's no formula that's useful for giving primes but we can try to count them by all sorts of ingenious ideas and one of the great achievements at the turn of the century 1900 just before 1896 due to Audemars and de la Vallée Pousson was to prove that the number of primes less than x is roughly x divided by log x it's called the prime number theorem and it was one of the high points of mathematics at the time the techniques were quite stunning in using things from analysis and zeta functions so you could ask here how many primes are there in p or p0 any one of those whose curvatures are less than x and such that ac is prime at this point thanks to my other the theorem I mentioned there that I was able to show this does go to infinity there are infinitely many circles which are prime but I didn't produce the right number of them to make this true so you could ask is there a prime number theorem is it true that pi p of x is asymptotic to the total number np of x the number of integers divided by log x roughly the density of primes is one of a log which is why that should be right and you might even predict the constant and we understand enough heuristics to actually predict this whole story and even check it numerically that's what that picture is you don't need to look at it there's no question it's true with an exact constant the constant is a special value of an L function so we know everything about what's supposed to be true but we can't prove this prime number theorem because we really haven't understood and it's very difficult to understand accurately enough how many had to do this counting accurately enough nevertheless you can give an upper bound so some sub argument coupled with something that's what I want to end with is giving us an upper bound which is at least saying that this upper bound is correct with the wrong constant in other words this is supposed to be roughly asymptotic meaning constant times this over that with a God given constant there this theorem says that the number is at most another constant four times that when x is large so that's already quite remarkable information and I want to mention that particular theorem because it uses the modern machinery that I've been alluding to the theory of thin groups is something called the affine sieve and spectral gaps and even Alex Gumbard happens to be here by coincidence I guess he came because of it he's one of the founders of this together with Morgana myself it's become a big industry Terry Tao who gave his lectures to four years ago also contributed to this it's a theory which allows you to do this kind of mathematics with a thin group and this is a good example of the use of that theory I think I'm coming to the end I said I don't want to give the proof of Descartes theorem it's really not from the book it's not pleasant let me stop there thank you very much well I think we have time for a few questions we won't spend too long because there is a we'll be refreshing this outside but please have a question on too so there are let's call it a tiny bit yes yes yes there's a hold right the five authors look for three dimensions and they are even once in n dimensions they are slews of generalizations and I should give a reference what I promised Andrew is when I finish all these lectures and he posts them on the web it will be off each lecture to complete set of references you can follow these things to see where they are and that's one of the references I would get in fact if you start with any cocks do you know what I mean by that okay the onset they are higher dimensions of generalizations they are a little harder to draw the tools that apply here apply there very much in the same way not every aspect but a month so I hope I was making clear that this is just the juiciest example of this and in some sense it's not the most indicative if you take a sort of monogram here a more complicated situation now situations where we have no idea how to decide who is in orbit because here you can see we are getting smaller and smaller so if we miss somebody we know we will never find it later but imagine I don't know that I am getting smaller and smaller then you can be in a position where you actually can ask this question and not even be able to determine where that specific number is there never mind a universal state anyway there are higher dimensions of generalizations and they certainly already are in the works of the carriers they wrote 5 or 6 papers yeah at the beginning you mentioned that you said this was related to this work and you wanted to remove it so what did you mean what I meant was that one of the key players in the geometry of numbers he developed the foundations of the geometry of numbers the study of integral quadratic forms which is all emerged after Hilbert's problem Hilbert just asked a simple question solve a quadratic equation in several variables with a whole number, the answer is to be a whole number everybody's whole numbers that's a very basic question he asked two questions, one about whole numbers and one about fractions the fraction part is much easier because with the work of Hesse it's very famous Hesse principle and he's completely understood how to solve quadratic equation with a number of fractions but the whole number part is much much harder and that took much longer to solve and the tools that I used I haven't said there automotive forms, the three groups modularization all these rely at the very beginning on something from the geometry of numbers a theory which was started by Kotsky to answer questions about quadratic forms like the second and Mahler was a key player so what I meant was he was a player in this standard and a very important theory mainstream theory which doesn't apply to this problem which is why I'm very interested in it because the group polynomial symmetry A is not that group O, F, Z but where I F, Z this would be translatable to a standard problem not necessarily solve but at least the standard question and the diving theory for quadratic equation but because A is born deficient it doesn't fit in there and so Mahler's work does not mean the same in fact Mahler's work shows that the some quotient space is compact or at least finite volume and the whole defining part of this is very big, it's infinite volume very flat in there, it's very hard to work arithmetic so I'm not saying Mahler could get a page anyway he was working on this a long time ago he's one of the great founders of the geometry of numbers the most many things there are other arithmetic kind of sequences not like this you might view this as a funny construction of new numbers from old numbers each time it's a non-commuted it's not a very big non-commuted group making new numbers from old numbers there are recurrent sequences which are simpler definitions of sequences and where you've asked about patterns in the sequence and he played a very big role in Mahler's Mahler's theory he worked in geometry of numbers on standards have any other questions in that case we can move to refreshments but before we do that please join me in thanking him for the sound