 Thank you very much. Thank you Elina and thank you the organisers for inviting me to speak here. It's a great honor. I Want to talk to you about random Diofantine equations and some joint work with Pierre LaBoudec and Will Savin So Pierre reminded me actually that we started thinking about this project on June the 2nd in 2017 so that's almost exactly three years ago and we finally put the paper on the archive this morning I've spoken about this a couple of times before but I've We've sort of been constantly trying to squeeze the most that we can out of the method And I think we we've now as it as it as you can see on the archive we put down we put this to rest So Oops, let me just This talk is all about Diofantine equations. So these are just polynomial equations with integer coefficients And we want to understand when we have integer or possibly rational solutions to these equations So i'm going to keep this Talk fairly colloquial um I think to keep the analytic number theorists happy. I Will set in introduce some iterated logarithms. I might change the order of summation at some point But the rest will be at a very sort of gentle pace. I want to start by talking a little bit about the decidability problem for this question I will focus on cubic equations For a little bit and then I'll talk about the main result and and how it's proved using the geometry of numbers So the kind of famous results or question here is Goes back to hill, but in his famous address at the turn of the previous century we was the last century And he asks for an algorithm About whether you can uh given any Diofantine or any polynomial with integer coefficients. Can you decide whether or not there's an integer solution? so This is known when the number of variables is very small. So if a polynomial is in one variable It's also known if the degree is at most two In fact, it was a nice observation of scholar using the non negative non negativity of the squares And in fact, it suffices to always assume that you've got polynomials of degree at most four That's an example there involving the modell curve But we now know this this dream of hill, but is uh can't can't be attained. There's no general algorithm for deciding Whether or not any given Diofantine equation has a solution And in the absence of that, um, we're left with finding, uh, perhaps classes of Diofantine equations for which we can exhibit algorithms But I want to point out that even in the case of very simple Quadratic equations, it's just a quadratic equation in two variables. There remains much that's unknown So there's a a nice paper of jeff lagerius from the 80s Who discussed the the algorithm for deciding the solubility of the negative pel equation? so I mean here you you just look at the continued fraction of d and The equation is soluble if and only if the period um Of the continued fraction expansion is odd And in fact, lagerius proved that this is an np algorithm So if you manage to come up with a polynomial time algorithm, you will in particular Solve this thorny issue of p is equal to np I think it's unknown whether or not this is in np complete, but Just want to illustrate that even in the case of degree two equations, there's much that's unknown So the the kind of main test for solubility that's going to motivate this talk Comes from the most obvious necessary condition for solubility Namely, it should be soluble over all of the completions of q. So it should be soluble over the reels And over all of the pedics or like a modulo every prime power. There should be a solution So we say that a class of polynomials satisfies the hassa principle um, if This necessary condition is also sufficient And this is a nice thing to have When you know that a class of polynomials satisfies the hassa principle because this is in fact a decidable problem So let me discuss it briefly in the case of decidability over the pedics So for this you just want to pick a large prime p such that this polynomial remains irreducible when you reduce it modulo p Then by invoking the lang vei estimates, which is sort of generalization of the The resolution of the Riemann hypothesis for curves um This tells you that the number of fp points on your On your hypersurface or the number fp solutions to this equation Can be approximated by p to the n minus 1 Plus an error term, which is smaller So in particular again if p is large enough that's enough to guarantee that there's a solution over this finite field um in which the The partial derivatives of your polynomial don't all vanish and now you appeal to Hensel's lemma, which really just goes back to to isaac newton, I think And that allows you to lift these fp solutions to get genuine solutions over qp okay, so large primes You automatically get p added to solubility And then for the finitely many and in fact all of these constants can be made Explicit and then for the finitely many small primes you just use a computer Do by hand So in principle this uh, this has a principle is is decidable um nonetheless, it's very uh So they still present dive under equations still present lots of challenges even in relatively benign instances like Quadratic equations as I've discussed and cubic equations. So this is a question. I like a lot which integers are representable as a sum of three cubes So the local condition here the local test is that your integer shouldn't be plus or minus four mod nine And we saw a beautiful talk by drew sutherland in this series A few a few weeks ago It's a it's a sort of very beguiling problem because for some k You get very small solutions. So if you take k to be three um, it's not very hard to work out what the smallest solution is in that case But if you take other values of k such as the number 42 It took an enormous amount of ingenuity and computer power Um to actually write down the smallest solution One might ask about other cubic polynomials Representing integers Um, and in fact you you get a different story sometimes. So Here we're asking about representations of integers as sum of three squares minus um, the product of the variables x y z Again, the local test it's quite easy to work out that Provided k is not three mod four and not plus or minus three mod nine That's enough to guarantee solubility Over all of the pedics But nonetheless This is not necessarily enough to guarantee that there are solutions over the integers And there's a very nice paper by gosh and synaq a few years ago We've actually proved for this particular equation That there are infinitely many counter examples To the house of principle So it'd be interesting to to figure out what's going on in these kinds of cases. I think this is not completely understood So for the rest of the talk, I'm going to talk about uh irreducible homogeneous diaphanetine equations um, so All of them no more all of the monomials have the same degree and this has the effect that talking about uh solubility over the integers Is the same thing as talking about solubility over the rational numbers and uh, the motivation is going to be to ask about Uh, this test this sort of house of principle for random equations um, let me motivate this by a discussion of cubic equations homogeneous cubic equations and i'm going to start by those which define Curves so it was in dimension one so for us a cubic curve is going to be Uh, a homogeneous polynomial of degree three the locus of zeros polynomial of degree three which is homogeneous So one of the mantras in uh diaphanetine geometry is that uh the complex geometry of the variety cut out by the equation the polynomial equation should have a very strong effect on the uh arithmetic of the problem and we see this quite uh distinctly here so um the problem is very easy if you've got a singular cubic curve um, so first of all If your if your curve is irreducible um, you It's not hard to see that you always have a unique singularity um, if there were two singularities two double points Then the line joining the two double points would Hit the cubic curve in four points, uh, which is impossible by pursuit's theorem Assuming the curve is irreducible So it contains a unique singularity and then by uh, let's go to this singularity has to be invariant under the galois group Um, and so this is actually always going to be a rational point Uh on the cubic curve So singular cubic curves always have a rational point in the fact you can use this Uh rational point to parameterize All of the rational points and find that your curve is is really just a copy of p1 Things are much uh rational subtle for non-singular cubic curves and there's a famous example here of selmer central you've seen before Uh, which shows that this local test is not sufficient in general. So it doesn't Are famous at the house of principle Let's move to dimension. Oh, no stick with curves. Um Individual curves difficult to Analyze that's sort of many questions about them um Bargava started seriously looking at um answering these types of questions, uh on average And this will motivate what comes later Uh, so this is a result I like a lot. Um, and he has proved that if you, uh, order all ternary cubic forms Uh by height So all ternary cubic forms over the rational numbers by height Then a positive proportion of these actually have non-trivial solutions over the rational numbers So this is a tour de force which builds on his work with shankar On the average size of three selma groups of elliptic curves And his work with skinner showing that a positive proportion of elliptic curves actually have rank one um, so in the paper he also, uh Postulates a conjecture that I've written it in a in a fairly strange way, but um the the exact proportion of cubic curves which have a rational point Should be one-third times 0.97 It should be the proportion of cubic curves which are soluble over cube So the 0.97 actually comes through joint work with manjul and John Cremona and tom fischer Where they calculate the exact probability that a ternary cubic form Passes all of the local tests. So it's locally soluble everywhere 97 percent of Plain cubic curves are everywhere locally soluble and among these Vagava conjectures that precisely a third of those Actually have rational points. So this third sort of comes out of the method and is related to Sort of rank distribution conjectures of goldfeld cats and sonic Okay, so now I think we're moving to dimension two um so Cubic curves were Homogeneous cubic polynomials in three variables cubic surfaces are homogenous cubic polynomials in four variables and again, we have this dichotomy between singular and non-singular surfaces so The singular ones are easier Provided you stay away from cones. I think cones over genus one curves will be no easier than looking at the arithmetic of genus one curves, but So the geometry of singular cubic surfaces is very classical. You always have at most Well, assuming that it's uh, it's it's a normal cubic surface. You always have at most four singularities um, it was proved over 50 years ago that the hassa principle always holds for these cubic surfaces I've illustrated a lot of my favorites. This is the Kaley cubic surface It has the maximum number of singularities and it has a bunch of Rational points on this on this cubic surface So the non-singular cubic surfaces are much more Recalcitrant We we know that the the hassa principle can fail for these cubic surfaces But we expect and in fact this is a conjecture of um uh sans souk and Collier to lend from the 1980s That the only way that it can fail is if the Brouwer group associated to the cubic surface is non-trivial Whereas for singular cubic surfaces, we've had some success at even Understanding the distribution of rational points on those cubic surfaces non-singular cubic surfaces are much harder I've 3d printed. I don't see my video, but I've 3d printed a cubic surface Here, this is the klepsch cubic surface and all the pink dots correspond to rational points on these on this cubic surface uh, we have Very little idea about how they're distributed. I mean we have lots of conjectures But So I wanted to Point out that Yes, we'll sort of underline the point that understanding individual non-singular cubic surfaces Um is difficult Um, so this is one of the strongest results in this direction that we have Um, I'm focusing attention on diagonal cubic surfaces. So I assume that it just has these four Cubic monomials and they're all cube free and swedish and dia the turn of the millennium was able to show that these cubic surfaces admit rational points assuming that They pass all of the local tests for solubility plus some extra explicit conditions on the coefficients, which I'm not going to record here um So that's a very strong result But as with many of these results in this area It's highly conditional. So he has to assume that the Tatech-Averevich group Is finite for any elliptic curve. That's a big open problem. Um, so in the 80s Collier to Lenkinevsky and Sonsuk made a detailed study of diagonal cubic surfaces And they explored what the Brown group or the brown man in obstruction has to say about solubility for these surfaces And so the calculation one of the calculations they did was to show that This obstruction is empty. So there is no obstruction to The hassa principle if Well, whenever you have a single prime p Which divides One of the coefficients to the first power to Pianic valuation one and doesn't divide any of the other coefficients so one consequence of that is that Failures of the hassa principle in this particular family should be exceedingly rare um, if you think about it You can only possibly have a brown man in obstruction If whenever there's a prime dividing one of the coefficients Either its square divides the coefficient Or it divides another one of the other coefficients And that places a very strong multiplicative constraint on the coefficients. That's not very hard to show that Well, assuming this big conjecture that the brown group explains failures of the hassa principle that the proportion of failures should be Decay like one over b squared um Nonetheless, uh, so this is a sort of conditional result about cubic surfaces, but it is still possible to prove Unconditional results about The existence of rational points on cubic surfaces And in fact, you're able to prove Examples of families of diagonal cubic surfaces for which we always have a rational point um, so There's a result by heath brown and morose Which states that if you have the coefficients one to a b And you assume that a is congruent to plus or minus b mod nine Then this cubic surface always has rational points So this this is really a A A sort of high point of marrying sort of heavy analytic machinery with Uh, sort of work on Hegana points in elliptic curves So there's a result by satchel who shows that The Cubic equation x naught cube plus two x one cubed equal to p for a prime p congruent to two mod nine Always has a rational point So if you're able to produce prime values of your binary cubic form In the right congruence class Then you're able to deduce the existence of rational points on this surface So i'm sort of inspired by this and by the works of uh bargava To think about a similar family. So here i'm looking at a family of cubic surfaces Which are of the shape binary cubic form equal to a binary cubic form in disjoint variables And uh and one can show that uh in this family Um a positive proportion of them actually have rational points Basically by reducing it to a problem about curves in a fairly obvious way so In the same spirit of the heath brown rose paper It's sufficient here to just study the the cubic curves in p2 which look like binary cubic form equal to a cube And then you can exploit um a connection with These j invariant zero modell curves And one finds that A soluble cubic curves of this special shape were given discriminant actually correspond to The image of This e mod phi e in the phi selma group of this modell curves This curve has a as a sort of obvious three-eyes oxygenate And then you can appeal to uh work of um chris and lee who actually are able to show unconditionally that a positive proportion of these Modell curves have positive rank Okay So let me stick with cubics for a little bit longer. Um and Under the uh philosophy that any mathematics talk should always contain a proof Um, let me Present the following result. So I want to introduce some notation which will be used later as well. So curly f Of a with subscript d and n This is the space of all homogeneous degree d polynomials in n plus one variables with integer coefficients and whose coefficients all have modulus of most a so a here is a Parameter that will be tending to infinity So the theorem it's about cubic surfaces Is that when you order these cubic surfaces by height? You actually get that a positive proportion of them are soluble over the rational numbers Okay, so that sounds like quite an impressive result, but it follows very easily from uh from bargamas result about cubic curves that I mentioned earlier So we're interested in estimating the number of Cubic polynomial polynomials cubic forms in four variables which have a rational Solution I can lower down that by just counting the number of cubic polynomials in four variables Which have a rational solution in which the last variable is zero The last coordinate is zero Um, so there are 10 monomials in uh cubic surfaces which have which involve The x3 variable um, so I can easily so I essentially have a A free sum over 10 of the the coefficients and I get A lower bound of the shape greater than greater than a to the 10 Now times the number of cubic polynomials In only three variables which have a rational point I have a rational solution and that's Uh, exactly the content of bargamas result and he showed that there was a positive proportion Which just means that there's a lower bound of the shape of constant times a to the 10 there are 10 monomials in any cubic Uh polynomial in three three variables As of that give you the result um Now let me get on to uh talking about uh the the main result That we've been working on so again, it's the same same notation as before curly f is the space of all of these Homogeneous degree d polynomials in n plus one variables with integer coefficients and height at most a So then the result is that provided that we're in the so-called fanno range. So the number n um It should be at least d And we have to exclude the case of cubic surfaces the case d and n being three and that's Okay, it's unfortunate, but I have to exclude the case of cubic surfaces then the proportion of um polynomials which Have a rational point Tends to a constant c sub dn As a goes to infinity and this constant has a nice description Uh, if you're outside the setting of plain conics It admits a description as a product of local densities and a positive product of local densities And if you're in the setting of plain conics, this constant is zero So that was already well known. That's a nice result of ser from the 90s that a random plain conic Doesn't have any rational points So I won't define these densities here, but uh sigma p uh, for example, is the probability that um A homogeneous degree d polynomial in n plus one variables Has a qp point So another way of saying this Is that Again, if you make the same assumptions of n is at least d We're not looking at cubic surfaces anymore When you order these all by height a hundred percent of them Satisfy the hassa principle Okay, so while Previously we saw some instances of counter examples to the hassa principle This is at least confirming that these counter example counter examples are rare For this range of n and d So that this This question or this result goes back to a question or to a paper A poon and a voller can this influence me a lot? I think this was just when I was finishing my phd. It was a conference one of my first conferences at to aim about rational points and This was one of the things that was discussed there. So they conject made this conjecture that it should be true For all n greater than or equal to d including cubic surfaces And they also had a nice argument which showed that It follows from a very general conjecture of collier to len Uh, so if you've not seen that conjecture before that states that If you have any rationally connected a variety The Brown and an obstruction should be the only obstruction to the hassa principle So if you take that on faith Actually, it tells you something much stronger in the setting of phano hyperservices because one can show that if n is at least four And x happens to be non-singular which certainly happens a hundred percent of the time Then this brown group is trivial. So therefore According to the conjecture of collier to len the hassa principle should always hold Actually the light of that is possibly interesting to comment on the error term in this theorem um I suppose if if So it follows from this argument based on uh collier to len's conjecture that That this should be true For all n greater than or equal to d Uh, certainly with an error term, which is polynomial polynomial decay in a So we're not able to obtain polynomial decay in a in this result that we do at least get some explicit power of a logarithm Um I guess as I was preparing these slides it occurred to be of me interesting to think about What one might expect in a slightly more general context. So if you give yourself a family If you have some smooth Projective variety z with a morphism down to some projective space One might ask about what's the probability that a random fiber satisfies the hassa principle So I think it's really only makes sense if you put some conditions on the family, so I think Assuming that Z itself is everywhere locally soluble seems to be a fairly basic assumption I guess you would want the fibers to be Rationally connected perhaps But it'd be interesting to To think about what one might expect here um So at least the local solubility condition Or result here has been worked out in joint work with martin brighton dan lochran So if you assume that over every co-dimension one point in such a family Um The fiber is irreducible Then you can at least show that sort of using a To sort of cilvecadal type machinery That a positive proportion of the fibers are everywhere locally soluble So I don't know is this condition enough to guarantee that a 100% satisfy the hassa principle Okay, um, maybe I should discuss a little bit about n less than d Okay, so that's completely off the cards Poon and nvolok also conjectured that If you're in the world of general type varieties 0% of them should have a solution of the cube So interestingly here It's only true that a positive proportion of them are everywhere locally soluble So we expect failures of the hassa principle to dominate But even writing down a single example where the hassa principle fails Is very very difficult We have some Results in this direction, but they're all conditional on things like abc conjecture On the area lang time things Um, one could also ask about special families Um, so if the family becomes too special, then our methods Uh, unfortunately break down, but there is some prior work in this direction. Um So, uh, yerg brood and rinoditman Uh, considered the case of diagonal hypersurfaces and under a slightly stronger condition on the relationship between n and d So ending speed now is 3d plus 1 They're able to show that 100% of them satisfy the hassa principle Okay, so actually we're able to prove something slightly stronger than what I've revealed so far We can say something about the existence of small solutions So let me write double lines f To be the height of a polynomial f so the maximum modulus of the coefficients of the polynomial f So there's a Uh, a classical result of castles about quadratic forms that states that if you have any non-singular isotropic quadratic form Then it's least integer solution it's a nice least non-trivial integer solution Always has euclidean norm at most of constant times the height of q to the n plus two Sorry n over two So that's quite Entry so this gives a completely different sort of Solubility test to the hassa principle This is telling you that you know, I mean if you wanted to know whether such a thing was soluble You just have to run through all possible integer vectors up to this search range And uh, figure out whether you have a solution or not. There's been work on improving the implied constant by several authors And in fact, this is known to be best possible. So knesa or nasa That's his name Wrote down an example of a quadratic form Which had large height and whose smallest solution had large height Matching this up a bound of castles Um, however, if you look at the the equation he writes or the polynomial he writes down It's a very special shape. And in fact, it actually has discrete discriminant one determinant one So it's very atypical So you might expect that on average, this isn't the true The true behavior and this is sort of what we also managed to prove So again the same assumptions on n and d n is at least d We're not looking at cubic surfaces. Unfortunately And so when you order Everywhere locally soluble degree d integer forms in n plus one variables Then a hundred percent of these have a solution in which the The norm of this vector the cleaning length of this vector is bounded by the height To the one over n plus one minus d times something that goes to infinity slowly Yes, I promise there'll be some iterated logs. I think that's the only point at which they appear So you can see that in the case d equals to two This is a much much smaller upper bound than what we're seeing in castle's result Okay So actually this chimes with work of Yuri Chinko Ells and Hans and Yarnall so they also postulated the idea that The point of least height on a fano variety should be something like one over the Tamagawa measure Sort of coincides with that Okay So let me talk a bit about our approach to this so Studying the question of which polynomials actually have solutions is quite quite hard to do and things are smoothed out a lot By instead studying the average behavior of the associated counting function So We could define for a polynomial f we could define n of b To just be the total number of primitive integer vectors in n plus one variables Which are solutions to the diophantine equation f equal to zero and which have norm at most b So this is certainly a counting function that has proved Uh my bread and butter for many years and I've thought about it in many different contexts um, and there have been uh Numerous incarnations of what one might expect for these kinds of counting functions um So here's a conjecture and sort of inspired by Work of many people um If we take The fanno range so n is at least d um and f Is one of these polynomials of degree d and n plus one variables I'm going to assume that it's non singular. Um, and that It uh, it has pick art number one. So it's pick art group. It's just said Then one might expect that there's a positive constant overnight. I need to assume that it has Solutions, otherwise this problem is not very interesting So one might expect that there's a positive constant such that this counting function Behaves asymptotically Like this constant times b to the n plus one minus d So there's a fairly easy Heuristic argument for explaining the exponent n plus one minus d I think in your wildest dreams, you might hope that As in several other instances in analytic number theory that you actually have a square root error term So there's a paper of swedain hendaya where he Conjectures this for cubic surfaces, for example I'm I'm cheating here slightly because there's a There are some issues around dealing with sharp cutoff functions. So there's a paper by damara schindler who Demonstrates that you can get sort of intermediate lower order terms um appearing in asymptotic formulae like this, but I think the consensus is that if you Uh, if you move away from sharp cutoff functions to something smoothly weighted uh by some Smooth weighted compact support say then those uh lower order terms go away So maybe this is a reasonable conjecture to hope for In fact, it's probably work. It'd be interesting probably very very difficult to Think about so, I mean, this is a lot to think about this in the context of Whether or not this Error term if you normalize the error term appearing in this asymptotic formula In the spirit of zeb's talk a couple of weeks ago where he talked about A conjecture of Blair and Dyson about sums of three squares Be very interesting to See whether there's any evidence for this normalized counting function to have a limiting distribution The normalized error term to have a limiting distribution Maybe a Gaussian distribution why not? Um, but at least our Our work is consistent with this conjecture So we basically have two per hours is going to infinity We're doing there's the a which is the height of the polynomial And then there's the b which is the uh, if you like the height of the rational solution to the equation Um, so we have to take a in a certain range with respect to b and all of the work is concerned with dealing with the case where a is close to the Left hand equality inequality So indeed that's the situation where you expect there to be the fewest number of rational points in fact So The basic process or the basic point of view is to try and So start by constructing a convenient localized counting function Which i'm just going to denote by n lock of b And this counting function has got to be Rich enough this localized counting function has got to be rich enough that it approximates our expectation for what the leading behavior should be in the Mannon conjecture But easy enough that we can actually work with it So that took a bit of doing to get that right um, but you can essentially think of it as a A truncated singular series And a truncated singular integral coming from the circle method So the first step and the most important step is to study the mean square of the difference between the counting function and the localized counting function As we average over all of these polynomials homogenous polynomials at degree d in n plus one variables um, and then the result that we are able to prove is uh Is that basically we get a square root error term on average whenever we're dealing with um This range that n is at least d And now we have to exclude the case of plain conics and plain cubics Sorry, cubic surfaces um Yes, I think in earlier iterations of of of talks i've given about this, um Uh, we've been at well i'll say a bit more about this in a minute, but we've been stuck at n being at least d plus two And handling the case n is equal to d It becomes harder and harder as the as as n and d gets smaller Anyway, the upshot of this result is that it is rare For our counting function to not be well approximated by a localized counting function now Going hand in hand with that is um an analysis of the localized counting function itself So we want to show that it is rare for the localized counting function Um to be smaller than we expect it to be Okay, now the one issue here is that actually we know that for a positive proportion Of polynomials of degree d and n plus one variables this localized counting function vanishes So it certainly can be small And it's here that we really need to work under the assumption that we're only restricting attention to Polynomials which pass all of the local tests for solubility So we're looking at so this curly f lock Is now the restricted set of everywhere locally soluble polynomials Of degree d and n plus one variables in height a So this estimate Tells you that it's rare For this localized counting function to be smaller than we expect it to be And yet still everywhere locally soluble So i'm not going to talk about the proof of this but it really goes back to what we saw on the second or third slide involving Hensel's lemma so Hensel's lemma told us that you know when you're in a situation that the periodic valuations of the partial derivatives are small Then you can lift points you can lift points to get many periodic points on your hypersurface So the guts of the argument here is to sort of show that It's rare that you have polynomials which are locally soluble and yet and yet have All of their partial derivatives featuring large prime powers Okay, I wanted to focus a bit on the first step because this is the bit that involves the geometry of numbers I've just Recalled what the first step is here is a second moment bound and the hardest range Is as you might expect when a is roughly the same size as b to the n plus one minus d So that's the case in which the leading constant In the expected asymptotic formula for nf of b is Roughly one Yeah, please excuse me. We have a question or a comment from seferutnik Oh Please ask your question Tim So you have this two parameter family Ideally you would like to make a Very small relative to b right because you're secretly trying to To capture what happens for an individual hypersurface right But if I understand correctly you are forced to take a quite large which is as you say What you call the hardest range. That's the lower bound So you can't And then you have this upper bound which seems less important because You find this incorrectly. You want a to be small as possible But if you give up on that on that upper bound, could you make a smaller than this power of d? I get I'm not sure. Uh, I think Yes Sorry, can I interject Well, thank you. You're trying to prove like Wait, no, I may refuse myself You're talking about making a very very very small Or b very Ideally you want if I understand the philosophy you really are trying Ideally you would want to do this for an individual Yeah, a equal one. Yeah, ideally, but of course we we can't so your average and so you try to average over a few possible Hyper surfaces as you can which means make a as small as you can while still making it grow with b That's a philosophy. I would imagine Yeah, but your force So you have this constraint that a is essentially larger than what Tim calls the hardest range here B to the n plus one minus D So what is the obstacle to making a smaller? We don't know how to do it. Okay. That's a good one We're going to be reversing some order of summation and so Then Like we're trying we're going to try to be getting cancellation in the sum over a that beats some kind of trivial bound somewhere else Yeah, and you'll probably got a flamin for it on the next slide actually. Okay. Okay. Thank you um Thank you for the question so opening up this This second moment and here comes the interchanging of the order of summation as promised We yeah, the hardest term comes from looking at the average of This counting function squared and so we open that up And then we bring the interchange the order so we're summing over now vectors x and y primitive Uh, size at most b and we're now left with counting these polynomials Which pass through these two points Now that's a much easier problem Um, after all what are what is the space of polynomials of degree d and n plus one variables? It's nothing other than a than a vector space of coefficients A very big vector space of coefficients The fact it has n choose n plus d coordinates for a polynomial homogeneous polynomial of degree d and n plus one variables and One can rewrite that problem as counting integers integer vectors of modulus at most a Which satisfy these two linear equations and the coefficients of the linear equations are just coming from the Deith Veronaise embedding of the two Points, if you like in pn So in principle, this is much easier. We're dealing now just with Counting solutions to linear equations And in fact, that's a lattice point counting problem So let me write capital n for this binomial coefficient And assuming that we we don't have proportional vectors x and y This set this set of coefficient vectors which satisfy these two linear equations Is actually a lattice When it actually has rank n minus two we're cutting up two conditions And it's determinant is roughly the products of the Euclidean norms of these two coefficient vectors new x and new y So the veronaise embedding takes you to degree d monomials. So this Determinant is roughly The norm of x to the d times the norm of y to the d So now we look up our favorite result about lattice point counting In the literature. There are many I guess i'm in austria now, so I turn to Wolfgang Schmidt and He tells us that the number of lattice points in this box Grows like the volume a to the n minus two times a constant Divided by the cone volume of this lattice and the error term is very explicit. It's just of the shape one plus A sum involving smaller exponents of a and the successive minima of this lattice lambda So remember the case successive minimum of of a lattice is just the the minimum radius of a ball containing k linearly independent lattice points okay So that's the kind of result we want to apply. I've written it again at the top of the slide here And again, we're we're dealing with this hard range Where a is roughly of size b to the n plus one minus d So the leading term the main term in in schmidt's result We we can see that that's bigger than a to the n minus two divided by the co volume We saw that the co volume was like x to the the norm of x to the d times the norm of y to the d X and y are both typically of size b So the denominator there we would expect to be typically of size b to the 2d So that's the sort of lower bound for this main term And I mean we're trying to We're doing a sort of dispersion method. So We need asymptotic formulae for each of the terms that comes out of squaring Opening up the square in the In the second moment So we need to know precisely when This main term is actually bigger than the error term so One thing we can do is observe that We are working with integer lattices lattices which lie inside z n And in particular all of the vectors in the lattice are integer vectors and they all have Euclidean length at least one So we can certainly trivially take the lower bound lambda one is greater than or equal to one Lambda two is greater than or equal to one And in that way we see that this error term is bounded about by a to the n minus three So if I restrict attention to this Particularly awkward range where a is of size b to the n plus one minus d and you Compare when that main term dominates the error term You find that you need n plus one minus d to be bigger than 2d So you need n to be at least 3d for this argument to work So that's similar to the 3d that we saw in the result by Bruden and and deidman and I think this also shows why Why you can't expect to do very well unless your a is sort of sufficiently large in terms of b Okay, well, we know rather a lot about successive minima of lattices is a very classical topic So the largest successive minima is always bounded above by the product of the successive minima product of the successive minima by An estimate of minkowski is bounded above and below by absolute constants or constants depending on the rank By the determinant of the lattice and we've seen already that this determinant is roughly x to the d y to the d So in what I've described at the top of the The slide the very worst case for us is when this largest successive minima is basically the co-volume Of the lattice and all of the other successive minima of one So if we can improve that situation if we can get better control over the largest successive minima We will afford your right push up the size of the smallest successive minima and And get in principle a sharper upper bound for this error term Which will allow us to handle ultimately a better range of n and d So I've deliberately not left any time to describe the proof of the following result But this is the sort of key technical lemma that underpins the paper essentially And this states that you're provided that you're dealing with linearly independent vectors integer vectors Then you can always bound this successive minima the largest successive point in fact all of the successive minima By the product of the norm of x and norm of y divided by something which is a Which divided by something which is certainly at least one, but could in principle be larger So this is true for any of these lattices Um, and I think in our earlier versions. We didn't have the denominator in this upper bound for the largest successive minima But with this denominator you now Get into the the range and in fact proving the sorry proving that upper bound with one in the denominator Is relatively easy. You just uh, you can actually construct by hand n minus two linearly independent vectors inside this lattice which have norm at most The normal of x times the norm of y But getting this improvement here is is more complicated. I'm not going to describe it here and then it leads you to exploit the fact that we are In fact, actually just trying to understand the size of this successive minima on average as we average over x and y And in fact, you can gain leverage by showing that there are um There are relatively few x and y for which say d three Of x y is very small And then when you carry this all out and I should and then squeeze as much as you can out of it. You ultimately get to the sort of statements of the main result that I would mention But I think about a time. Thanks very much