Minerva Lectures 2013 - Terence Tao Talk 1: Sets with few ordinary lines

on behalf of the Department of Mathematics at Princeton University it's my pleasure to welcome you to the third Minerva lecture series and before professor Bhargava introduces professor Terry tell I'd like to say a few words about the Minerva lectures so the the Minerva lectures are funded by the Fernholz foundation and and the Fernholz foundation is was graciously and generously set up by louisa and robert Fernholz and and directed by louisa Fernholz and and the fern Holtz also actually endow the Minerva Foundation which has been longtime generous supporters of the math department so it's it's very happy to recognize their generous contributions so it's my great pleasure to to welcome professor module bhargava who will introduce our nervous speaker thanks Dave and I'm going to join Dave in thanking again the Minerva foundation and the fern Holtz's for their generosity in making these Minerva lectures possible their help has has really helped make this department much more active and made a big difference in our activities in the mathematics department so thank you again to them so it's just a great pleasure to introduce my longtime friend and inspiration today's speaker professor Terence Tao of the University of California at Los Angeles so when Dave asked me to do this I was wondering how I could possibly say something about Terry's mathematics in a couple minutes and then I was looking at what I had prepared to say about our first maneuver speaker professor Sarah and realized that what I wrote for him also applied very well to Terry so so here's what I said about professor Sayer so you can see that and you can see my laziness is oh no our speaker today has been a major source of inspiration for scores of people in this room his transformational contributions to so many areas of mathematics are discussed in our department and as prolific writings are read by our department members pretty much every day I won't go into the list of the highest honors and awards that our speaker has received because if I did that would take up the entire hour instead what I want to say is that what impresses me most about the speaker it's kind of amazing it's all applies is that not only is he one of the most prolific mathematicians of our time but that he also has an unwavering commitment to mathematical exposition and to the dissemination of mathematics his writings are crystal clear and super enlightening and he works magic with his chalkboard and blackboard when he lectures anyway that's what I wrote from Professor Sayer and it you can see it applies it applies just perfectly to Terry of course Terry is slightly younger than professor said always it makes it more amazing all these things are really true about him already and as I said tires made transformational contributions to numerous areas of mathematics including analysis combinatorics number theory geometry and the list increases everyday so check back tomorrow and you'll get a feel for this range of topics inherent and Terry's work as you see Terry's lectures this week so find out more on a more personal note Terry is also incredible as a person and a wonderful friend one of my favorite trips as a tourist was when we toured Paris together a few years ago and so we toured the Louvre and visited the Arc de Triomphe and then and then did some nice mathematics at the top of the Eiffel Tower hear that so that's what's what knowing Terry is like and for all of these reasons I can't think of a better person to and our inaugural Minerva year lectures this year with so please join me in welcoming welcoming our third and last Minerva speaker of this year he'll speak on sets with few ordinary lines professor Terence Tao thank you very much for that amazing introduction it's a never think of it a seer before that's really quite not and it's great be back here I was here as a graduate student for four years and it still feels like home I think much has changed even the snow looks the same okay so I'm getting three talks and there aren't fairly different topics but the first two talks are sort of loosely connected and themed there they concern a trend which has happened in the last few years in combinatoric s-- especially instance comet oryx which is a loosely defined subject but basically counting things involving points and lines and circles and and part on your curves and other sort of geometric objects and how they intersect each other so more and more we've been getting a lot of progress on internet problems long-standing problems in instance comet Oryx using techniques both algebraic geometry and also algebraic topology in a sense itself that's surprising in retrospect that these tools are useful since instant as comet Oryx is the study of geometric objects which is defined by your algebra curves so it's not so surprising that the algebraic tools are useful but but but they have really been there's been really dramatic progress in solving problems that were considered almost intractable by other means and now I can attack them by these tools generally speaking if your problems of a finite field then algebraic geometry methods seem to work well and if they're over the reals then a combination algebraic geometry and algebraic topology methods seem to work so there's many examples of this now this two that I won't discuss today but probably the most striking is about 2008 I think Syria was here the instant Institute approved the final kakeya conjecture using towards my algebraic geometry which was a big surprise and then figures later another nice outcome it's by Larry Goosen and that's Cass who almost proved the edge distance problem almost solve even the smallest number of distances you can form of endpoints in the plane they go to bouches within a square root of logarithm of optimal using tools from both algebraic geometry and Ontrack topology so I won't talk about these two but I'm going to talk about two other results of this flavor where methods from algebraic geometry and algebraic topology are being used to prove things in kinetics today I'll be talking about a fairly low tech application of algebraic geometry and our direct apology the direct apology will come in Oilers formula and out of a geometry is the classical 19th century algebraic geometry of cubic curves saker's of degree n was 3 so not very advanced the next lecture will have a slightly more advanced algebraic geometry 20th century SGA coming in but this is this is this small classical necessary joint work of Ben green okay so this is a talk about about lines appointing lies in the plane so it's not solve a definition so suppose it's a piece set of end points so let P be a set of n points on the plane final set of points okay so then you can you can start connecting these points by lines and you make some definitions we see that a carriage fly line and r2 in the plane that is instant or - that contains exactly K points from P ok so this is a three which line is a 2 which line support and then with it n sub K the number of K which lines now this is only an interesting concept for cave because an equal to of course you have infinitely many one which in zero which lines ok this is it's the two which signs in higher we check which are finite and so the question is we're going to be Wow and when K equals two so if you pick endpoints generically you expect most that your two-inch lines with no higher so we will call it ordinary line so a two inch line it's also called an ordinary life okay so this is very classical topic ordinary lines and one of the first films in the subject is a very famous one called sophistical I theorem so this was posed as an exercise as a problem by Sylvester in 1893 the first published proof actually appears in 1944 like a lie so so the claim is is so the theorem is that there's always at least one ordinary line except in the in one exceptional case with all the points if oil points lie collinear then of course you don't get any ordinary lines once you get a few more points so if P is it the points in P P then do exist laundry line so n2 is at least one okay so this is a great theorem maybe just to point out that this seems like highly trivial it's important that you live over the reals if you replace the rules by other fields this is not true so this is not true okay so for example if you work over a finite field and my PSA you could just say P to be everybody take the entire finite field as long as peas at least for me then it's easy to see there's no ordinary line because every line is P which a less less trivial example if you work in the complexes complex plane then there is an example for the has configuration which has nine points not collinear and no one realize so the way it works is that in situ what you do is is that you you make an elliptic curve and in situ of the picket booklet or I if you squinted it the right way and and what you do is you take you take the free torsion points so it's a group you take the key you take the points of order three in your lip recurve and those are nine points III cause III and because three points in there it could add up to zero if they're collinear you can quickly see that that the that every line that is to these point hits hits a third in fact this has the John which I think of the final play so that's that's that's an example where it fails so if you want to prove this theorem you must use something about the real line which the finite fields in the complex people don't have so you have to do something slightly non-trivial now the standard proof is a beautiful proof that you thought well you probably all seen it before but only give it again just because it's so pretty so the standard proof the sophistical eye theorem is that what you do is that you consider lines which are at least two rich and points which so every time you have a rich line because the points are not collinear must be at least one other point not on on that line and then so that that point is is some distance away from your rich fine so for all pairs of which lines and points you find the pair that minimizes the distance okay so there's a finite number points so you can eventually find a point P and a line L which minimizes this distance okay and so you've got a point and you've got your eye which is really close but not but not been on the line and if this line is too rich if this line is ordinary then you're done then your only line if it's not ordinary then there's at least three points on this line so two of them are on one side of this Bob is perpendicular or maybe maybe one is on them actually on the pinnacle it doesn't really matter but two of them will be or will be say on on the right side of this perpendicular and then by it is easy to see geometrically then at this point is closer to this line then this point is closer to this line this is a contradiction okay will be like this assessment infinite descent so this is this is the standard proof of the silver cycle active now you know it's very short very beautiful and it very much uses the properties of the real light you know to minimize distance and so forth and to talk left and right us very much something that only the wheels can do well they go all over the rationals or things in the wheels okay okay but that that proof is somehow it's too clever it doesn't seem to extend to do other things other than prove the sophistical a theorem unfortunately but it is very pretty but anyway once you see this theorem if you if you do instance Comet Oryx is their natural oskins okay so then the question is what the mineral value and tuefel fixed in within points not volcania okay so if you have n points not all online so Vesta classes there's at least one ordinary line but questions will have a few ordering lines can you make of course it can make a lot of ordered me lines if you take your pick if you take a look at the end point be generic you should get n choose two ordering lines you can get every line pretty much where two points ordinary but the interest questions that's what's the minimal number okay so so little there's some basic counter examples so Watsky which I can't spell for some reason okay there is hardscape gave the following car example so first comment this problem is set in the affine plane but without loss of generality an exactly a little bit more natural to put in the projective plane because this question is invariant under projective transformations that preserves lines and so forth and if you're on the projective plane which is the plane with the line infinity you can apply a projective transformation move all the points on the line if India off one by infinity so without loss of generality we can work in the projective plane that that's a bit more technically convenience it turns out so there's a standard example that suggests how how low you can go if n is even what you can do is you take the unit circle and you can take n over two equidistant points and so I get n over 2 which unity if you will and then you pick with all these chords and then at the line infinity so draw over here you take all these chords and use where they hit and there are n over two add up there n over two places where they where these chords can can can reach I guess you all something clear the tangents as well and so the place n over two sort of four roots of unity at infinity and if you do the computation these this so together they if they from end points in the projective plane we have only an allergic or relax of course most most of the cords here will hit a third point infinity the only lines are ordinary I want tangent to one of the points at this atmosphere I create an over to ordinary lines so given when n is even you can create and over to ordinary lines when n is odd you can eat you can't quite do this but you can add or remove a point on this line or the circle or fill in the origin and you do the computation you can get a similar example but basically 3/4 in ordinary lines okay so this is this is a configuration book with very few ordinary lines you can't get slightly fewer than this in for very small configurations if so they do if you take a triangle and its midpoints and a centroid this is a configuration with a 7 which for seven points and when you free one realize ok these three lines are the ordinary lines and that's slightly less than these bounds here and there's a more complicated configuration which I won't draw which is N equals 13 and 7 would be lines that's not right six or nine eyes which is also slightly wrote in these bouts and so the conjecture was a visa sharp so in particular once you get more than 13 and you give n points not all collinear then the number warden Elijah pissed n over 2 where n is even and 3 and over 4 okay these configurations should be the best there's a general principle in this subject that so of algebraic configurations things Excel come from some some very symmetric algebraic object tend to give you the minimal configurations although actually proving that's often a challenge so this this this was the conjecture there was a so there's a lot of partial progress on this conjecture so as I said so best eagle-eye gives you that n over 2 is at least one in what year in 1940 my uncle gave another proof was this mystical I threw him that she was in pack n 2 is at least three and bigger than something and bigger like three now this men although class has a great improvement but but his proof is much more robust than the Foo Fighters gave and that's that's important and then there's some other work ok so in 51 my skin so the entry must roll like at least the root of N in 58 care Nosa sure the short of linear bound for the first time which is 3 over 7 which exactly matches this first configuration that I said here and then until recently the best bound was in 73 I seem in soya sugar into it then 6 in with it if n is at least 7 ok so getting closer number 2 and again that bound is tight because of this this other example that I didn't I didn't draw okay so what Benin I did once we prove this through this conjecture for a large enough get that too bad for four points specially large into the please go to be even okay and yeah okay so we can actually prove this interview at least and large enough which in principle leaves the rest of the conjecture is a finite check and working in large enough force is 10 to the 10 to the 10 so we can't actually set up a problem but asymptotically at least we do have we do have a result okay so all right and and this you'll see the algebraic geometry come in in the prove you already saw just a hint of it with this elliptic curve construction in fact you know that if you look carefully this is also a cubic curve here the circle and why it's also a cubic curve is the cube occurs all over the place it's you know it's not initially obvious because it's a question about about lines where the Q becomes come from and that comes from a very classical theorem in in our geometry of the KB bacharach doing so talk about that later but let me just okay so before I go on to talk our other students proven let me otherwise else there's a there's another was out which was which was very closely related to to this one which is another question Sylvester where you don't know where into Tory n3 the maxi okay so this is called the orchard planted problem so given any points in the thinest in the place hub in space how many three in a row is can you make it what's the most more funerals you can make go to fix ten now here you don't need to make the end points non collinear because Colonia is not not a very good choice to make it to make plaza for universe and you want to maximize because you've minimized very easy you need to configure with no funerals okay but you want to maximize in the term of universe so there's a trivial about so if you double count every pair of points in your configuration must lie on some cabbage line and every Kalish line has K choose two pairs of points so you do have this bound which tells you an upper bound and threes at most one third of n choose two okay so you do you do have you can't have much more than N squared of six of in fact you can have close to N squared over six triples but you can come very close to that and again using an ultra curve so forth for the current example if you if you let say see the curve you know something like y squared equals x plus ax B to curve then as you spawn on the elliptical points on the picker form a group there's Pullman and additive group and three points on the curve sum to zero if if if they learn the right physical unit if Antonio so what you can do is that you get in a lip to curve you can take subgroup no 2n and we got to curve it either as a group it's either the torus or two copies of tourists so you can always find you can we find a group of order n for any n and then so there's lots of triples in a you know in a sub we water and there's some to 0 and you can just compute it and you find that in this configuration the number of few which lines you can you can get is n minus 3 over 6 plus 1 plus N squared over 6 minus n over 2 also bounded error so you can see that elliptic curves come very close to saturating the trip you're bound almost every line that goes through two points goes to a third there's only a few points at don't and those are the lies are tangent to to one point and hit note you can modify this example just by a little bit you can also look at a coset all of them open up the curve you can translate it by an element whose Triple R is isn't is in this is back in innocence in this intercept so there's like two other Co sets of every subgroup which which also give you this construction there's also a degenerate case where you take what's called an AK nodal a cubic curve it looks a bit like this so it's y squared times y squared equals x squared X cube minus X it's a good example where this loop here has shrunk to a point and now this is this part is still it's still as a group structure and you can take a regular sample point ten points over here and you can still get the plus is a point infinity and so you can still generate the same example in this degenerate case okay so this is this is a construction that that almost saturates the trivial bound and we were able to prove that this sexy the best possible for n large enough it's pretty large this is this bound is the best bound ok the elliptic curve example cannot be improved okay so so this seems a little strange because this is you know I mean the extremal configurations I had this cubic structure and the original problem just has these points in lines so the way we prove this these two theorems they're both consequences of another theorem actually I guess I have the second here okay so the first observation is that is that if there's bound we're not true if you had more if yet even more free which lines than the elliptic curve count examples then there'd be so many finish lines that there's not enough room for the two lines then you can show that the two ish lines can't go much bigger than some some constant over N okay and that just follows from from some computing but with this distribute identity here so if you had a counter example to this theorem then you have a set with very few ordinary lines and just a bigger inordinate lines which is which is which is not expected you know if you take endpoint in general you get N squared or me lines this is I said very few ordinary lines and similarly if you have a connect to this theorem event E was less than this and of course clearly and do also the would be bigger event so our approach to these two problems was not to look at the actual conclusions but just to study those configurations that are that have few ordinary lines and this is the this is a type of strategy that's proven very useful in common topics recently that if you want to suit your size study some statistic which is generally very large but but occasionally very small trying to classify those sets for which there's some sort of statistic measuring structure is very small often you can get a very good classification and if you get a good enough classification you can then solve a lot of problems relating to that regime where in this case the regime is we have few ordinary lines so our main theorem which implies the other two is calories is actually a classification of sets a few ordinary lines so our main theorem I was structured it's basically as follows that if if an especially large and and a few of me lines let's say less than KN and take and our phone can go off the log log in to a small absolute constant and it's because there's log logs this is why we need these ten to the ten to ten sort of things unfortunately in our zero but for K which which is a synthetic beam is allowed to be somewhat large if you have a set of a few opening lines then after adding or deleting adding and deleting if you go k points and applying it a projective transformation your set P is one of the configurations previously mentioned so the configurations previously mentioned are well there's one case where all the points I want to lie that's that's one configuration with with very few ordinary lines in this case none other configuration was this circle and the line infinity and over two points on each and then the third configuration there's two other configurations one coming from Finance of course of elliptic curve and the cosets two other Co sets and finite subgroups of this act nor given curve okay and these turn out to be basically the only configurations that have fewer new lives if you can only do a few number of things to them you can protectively transfer them around it's like you can turn the circle between the lips for example or some other chronic and you can take the third these cubic curves to not be in wise transform you can happen your cubic curve and you can add or remove a few points but but every time you add or delete a point you tend to create about n or new lines when you add or subtract a point of the visit so after adding or tweeting about big or K points you start getting too many to me ordinary life but this is basically a complete level asymptotically this is a complete classification of sets work with you or me lines and then once you have this the other results are basically just somewhat tedious case checks to show that among all configurations in this class the minimizes our or what we said okay so oops okay so this is our main theorem so explain how we prove it and where the algebraic geometry and the outbreak topology comes in so let me first recall milk that's proof that there's at least three ordinary lines so it's also very nice because it's not quite as short as the other purpose of our circle I that I gave but still it's like a might better fit in one board is very very slick so the way this works is right okay so you have a bunch of end points in the plane and then you have always whipped wine so you take both all the which sides you can find and do all of them okay and so you have a configuration of points and lines and this is in projective plane and then what you do is that you do a lies projective duality and you turn all the points into the dual lines in turn all the lines the dual points okay so just using protective duality replacing every every Lionel point with basically the formal complement you get a dual configuration and and in the dual configuration K which lines get turned into K which points or equivalently points a blob of degree hey if you think of this as a graph or a subdivision of the projective plane into these polygonal regions with with line segments okay which fly gets to lies into a point which has degree 2 K okay so you have this dual configuration and what Melchior did was that he just applied Euler's formula to this to this configuration so you know the first few minutes back topology so it's Euler's formula which is predicted plane is we might see because one although strictly speaking if you need the points not be all Cluny although it did you know it happens but as long as not all collinear there's enough lines and points that this is actually a valid formula so you have this formula and then on the other hand every edge is consider teaching faces and every face since then to at least three edges okay and again here yet again at the user fact that you know all collinear so if you just use these facts to put them together and you double count it's it's not hard okay so did use tuples okay didn't quite fit on one board pretty soon your engine and about the following identity okay all this formula after just a little bit of manipulation comes up ends up looking like this NK is no more cabbage and k which lines which in dual dual license number K which points and M s is the number of phases of si with s sides both there's at least three three sides and the point is that this is positive this non negative because s is at least three this is not negative this is not negative except when when n is two and so you can put the end and tune the other side okay and you get the N two is at least three okay so once so you rewrite all those formula in assert in a certain way that emphasizes the positivity of certain things and then you just extract out the one term that's not actually positive and you get you know cube out so this is this is this is what milk you did simply be all right now over here so but if you stare at this fizz follow this this is this is a nice robust identity that if you know that there's few ordinary lines then that forces since everything is non-negative that forces so one consequence of milk use bound is able and if n2 is small then then most in case have to be have to be at 2 or 3 so it's in fact in fact the most most points in the dual configuration after 3 6 because the points of higher degree will country too much over here and pore spaces have perhaps after three sides so all those formula tells you a lot of structure on your configuration and so on the protective duel aside what starts happening is that your your garage your drawing of all these points and lines must start looking at least comment oily like most big chunks of triangular of our track magnetic signal that is okay this is basically the only configuration in which every point has says as six degrees six in every side every face has has to be three now the so it won't completely be it be the triangle that is it just locally because because n2 is is isn't it's not entirely zero there's a few places where it's it's a quiz because n2 is is it's a little bigger than 3 there is something with some transitional regions where you transition from one executive grid to another executive grid and so forth but the duo configuration must look something like this and that tells you a lot about about the original configuration see if a particular moment you have a shape in the dual configuration that looks like this this dual eyes is to a configuration that looks something like this okay so every time you see a picture in the dual configuration that looks like a hexagon you get in the original configuration a collection of nine points sense and and six line Philip I guess it turns out and and and now we can use a very classical theorem in in in our geometry called the Katie Bacharach review that says that if you have a configuration like this nine points and six lines which are incident in this way and if you can find a cubic curve any curve a degree three that passes through eight of these points has no choice I must also pass through tonight okay so this is the que diga back to him it's a very nice application Pazuzu so it's not like typical to prove you may be more familiar with the special cases of this if you're cubic curve is just three lines this is corner Pappas's theorem yeah three lines and now we go to lines first of all two lines and then you draw three points on each line and you take you take the three points that incident then who's also collinear it's a special case there's also Pascal's theorem which is which is the also the special case when you keep it curve is a conic and I'm like it's a very similar theorem so this is the candy bacharach theorem so so if you use this what you start seeing is that you can pick you pick eight points here you can always find out whose one conic going through eight points just from the new algebra this is a context have like ten degrees of freedom in the plane ten or nine nine maybe okay but it's certainly more than eight okay so given eight points you can find a conic okay cubic going through eight points and then if you have a big trap and hexagonal grid in the dual structure you can keep you can keep applying katieb a crack over and over again and this this cubic curve is forced to pick up more and more points of your of your configuration so what you can show from this is that if into example size we go VIN using the Keely Bacharach theorem you can show that your configuration can be covered by Google of one cubic curves okay so this is already a pretty good structure theorem is it's a non-trivial structure theorem that this is a finite number of cubic curves so fun like this and then be okay conic in a line in the movie three lines over here and between them is this whole mess of our cubic curves cover your whole space now this is well this was the first step but this is still quite far from my four theorem you can you can mess around a bit further with Boo Zeus theorem it's it's it's hard for these these so cubic curve in an organic conic curve they can intersect in too many points and you can rule out a lot of configurations if you have at least two curves that are not lines you're in trouble this is the you don't get a consistent inconsistent decomposition so if you mess run reproduce tomb long enough what you can show that in fact one of three things happen so it's P is either going to be covered by a Bandhan um of lines or tonic and a bottom of lines or you do spoke cubic and turns out that these are actually the idea you can you can eliminate a lot of the complexity of all this and so then you're so then you in a case where out there plus maybe a bigger one of the points okay this is this one for some trash in this last case so yeah this is getting closer to two to the main theorem the main remaining task is strangely enough you get the linear case where bazooms are not so useful so this is some problem with if you already know your points lie on a bunch of lines then you need then you need to generate a lot of ordering lines if you want something will also inform that if you have say K lines and P has so at least epsilon end points on each line okay then you should be able to get a lot of all ordinary lives back with you given a by N squared of them so this if there's a key supplement that you need that if you have you give K lines each of which has a lot of points in it then it's really hard to make this configuration in such a way that there is no that there's no buzzer jewelry lines and to do this to limit in this case we have to use all we have to use control symmetrical products so I'm able to talk a little bit about this the challenge we sort through two main cases so this um there's a parallel case and then there's a non parlo case like this so the concurrent case with the three lines meet here they meet infinity and and here they don't meet at all so whether they meet at different points the see these are also digital generation of elliptic curves and this is a degeneration where we survey you the goo becomes the additive group of the wheels in this here here's where the group becomes of the more politically for the wheels so over here the problem is that there is sort of a counterexample to to this statement if you're allowed to have an infinite number of points if you're having you take a new kinetic progression by spacing one spacing when I think spacing half here if you have say six three equally spaced lines and if you take an infinite I think progression here here and here with here having twice the spacing half the spacing then then there's no ordering lines every every every line that goes through these points goes to a third so this would be a counter example and if there were any finite subgroups of the reals you could you could cook up a similar example using you using that finite subgroup to to to to to to ruin this lemma but of course there are no fan acceptors are the wheels but yeah yeah yeah they did you need a robust version of a statement there's not even anything that looks like a finite set over the reals this is I'm a data collector box comes in similarly I didn't prepare the notes here but if you work out if this is a single line X ago y equals 0 y equals 1 and x equals 1 you can work out when when three points are collinear when when 0 comma X you for my a 1 comma B and C comma 0 collinear and I think it sounds like CSB - a B or so there's there's some multiplicative relationship between C a and B which which I don't know which I haven't quite remembered right now but this has a structure of optimum the multiplicative reals each of these each of these these lines here and if the molecular veal has had a large final subgroup you would again have a counter example to to the statement and you also need a version of a statement that the more pickle if you also don't have don't have large more positive large minor subgroups or even anything that resembles a large panel subgroup and that's that's more attitude motox then there's more complicated configurations where you have both power lines or concurrent lines and non concurrent lines and then you have to use some called asymptotic phenomenon but you can and you can you can rule that out and then you so you can finally get the point where all the point that you each was in one line or on a conic in one line and on a Hooters for cubic and then from there it's not already so close to the main theorem that you can basically fashions up by hand I'll just point out this there's one was one of theone that were users very cute we actually use that's what's on offer putting that Rubenstein who is actually which yeah Kiki can Galois theory of all things so one one tool that we actually use is the following let you get a unit circle say and you take the entries unity and you draw all the chords connecting F is unity then there'll be a few points that will be something so there'll be some points that lie on a lot of chords so kind of the origin will lie on a lot of chords but it's just turns out if you're not the origin any interior point if any interior point on just stop you rise in at most seven course I can was unity it's a very beautiful paper actually it says it's a nice guy wife everything followed by not so nice computer verification but it the idea this is very nice you can rephrase this as like two chords like like three three chords are concurrent if and only if I think is a certain combination of twelve which you need assemble to zero you can see you can start asking what to cast are classifying when that happens so we actually use this theorem they didn't state it but there's there's a version was for exterior points every point in exterior also lies in at most some constant normal chords we all for our purposes we just need a constant so we use that and then we have a theorem like this for elliptic curves well okay so I given it if you have a if you take a sec water n on elliptic curve and you ask them but we don't know if this theorem of the Rubinstein type that any point not on the curve should lie in at most some constant number of of of cords coming connecting elements in this final subgroup we can't prove that unfortunately but we could we can limit the number of points which lie on a huge number like on on like a constant times n the number of points which are which are incident to - you know like n of 100 quarts of as type the way we have a bandage which is which is constant with this at most a bigger one points up to half that type so we have a weaker version rubenstein theorem for the two curves and using those we can we can we can prune out a lot of remaining cases and get down to this final figure which actually then gives all the applications okay maybe that's a good place to stop you

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Length: 50min 13sec (3013 seconds)

Published: Fri May 24 2013