 Okay, thank you very much for the invitation. Um, yeah, so I'd like to talk about Vanda that numbers. Only I can't advance my slides there we are. So, there's a very famous theorem of Vanda that from the 1920s, which in its two color version states the following. So take integers R and S. The number, which I'm going to call WR comma S in this talk, which has the following property. If I take an integer N that's bigger than that number, WRS, and if I two color, the integers one up to N. So my colors are always going to be blue and red, then either there will be an R term arithmetic progression that's blue, or an S term arithmetic progression that's red. So that's Vanda van theorem from the 1920s. Now it will take WRS to be the smallest number with this property. Obviously, any bigger number would also work. So define WRS to be the smallest number with this property. That's called a Vanda van number. So for the most part, trying to estimate what these are is very much an unsolved problem. And a particularly notorious case is when R is equal to S. So you're looking for either a blue R term progression or a red R term progression. And there the bounds are out by several towers of exponentials. And in fact, until Tim Gowers did his famous work 20 years ago, the best known upper bounds for those numbers were enormous sort of Ackerman type function or certainly very large towers of twos. So these are sort of notorious open questions. I'm going to be interested in what's called an off diagonal Vanda van number, which is where one of the numbers is three so you're looking at blue three term progressions versus red K term progressions. So W3 of K is the smallest number, such that if you color one up to N, if N is bigger than that number, if you color one up to N blue and red, there's either going to be a blue three term progression, or a red K term progression. So what's known about this, the best known upper bound is not too bad. It's a little bit better than E to the K. And that was done very recently by by Thomas Schoen. And that also would follow from the celebrated recent work of Blumen seesask on roster on three term progressions. So that's the best known upper bound at the moment. Here C is some very, very small positive constant, a little bit better than E to the K. The best known low bounds in the literature are much worse than this, basically K squared, in fact slightly worse than K squared K over log K squared by Lee and shoe. And there's quite a lot of numerical calculation of these w three K numbers. In fact, it's a favored sort of environment for people who do things with sat solvers to test what they're doing. So actually, there's numerical evidence, numerical values for w three K up to K equals 20, and then lower bounds which are speculated to be the right numbers for bigger care as well up to about K equals 40. So I'll show you those on the next slide. And a lot of names associated with that endeavor that I've written down there. So let's have a look at this data. This is a table of known values of w three K. And once you get up to K equals 20. These are in italics, which means that they're rigorous lower bands. And I think they definitely conjectured to be optimal. So that is to say actually the correct values up to K equals 30, and maybe a little bit more speculative after that. So these values should be taken with a pinch of salt. And I believe that it's certainly very reasonable to conjecture given this data. In fact, almost the only thing you could conjecture is that w three of K grows like K squared. So I first heard this question from Ron Graham 15 years or so ago. And I think Ron said to me, I think w three of K is roughly K squared, certainly bounded above by constant times K squared. Being very arrogant in those days, even more arrogant. I immediately said that cannot be right. And here's why it can't be right. All we need to do is to take a blue set B that's got no three time progressions. So that's already half the battle color everything else red. And then with luck, that red set won't have any long progressions. Now, if that red set looked like a random set, whatever that means, then we'd expect the longest progression in R to have size about N over the size of B up to logarithmic factors. If instead of progression you you think interval that's a sort of very well known principle if you've got a set B inside one up to N, the gaps are going to be on average of size N over B. So basically there won't be any too much longer than that. And there are many examples, many famous examples of these blue sets B with no three time progressions of size much bigger than square root N. And therefore, if this heuristic that I've put up there had any merit, those red sets would have no progressions longer than about square root N, which would disprove Ron Graham's conjecture. All you need to do is try it on your favorite examples of large blue sets with no three time progressions and here are some the folklore example that consists of numbers with just zeros and ones in ternary, which also came up as an IMO problem actually in 1983. So that's got size bigger than route and it's got size n to the log two over log three, and it's an easy exercise to check that that has no three time progressions. It's much better than that very famous example of their end, based on a slightly earlier example of Salam and Spencer gets better than any power of n less than one so n e to the minus constant root log n. This comes from taking the lattice points on a sphere so you can observe that if you take the points on a sphere that has no three time progression in fact doesn't even have three points on the line. So if you get that down in a suitable way down to the one dimensional integers, then you get a large set for your three time progressions. And there's a variant of that construction due to Julia Wolf and myself in 2007. So there's three things that you can try here that will give you very large blue set. What you need to do then is just to show that the red set the complement of that blue set has no long progressions. So that was the thinking behind my saying to Ron Graham that I thought his question was ridiculous. Unfortunately, I have a question in the audience. So, recently, maybe you want to unmute and just ask away. I just read the question. So there's a connection PD wonder that numbers and Ramsey numbers. That's the question. Well, the definitions are somewhat similar. Ramsey numbers, the most natural Ramsey number that's of a similar flavor to this would be our three K where you're two coloring the complete graph, and you want either a blue triangle or a clique of size K that's red. But other than that, I don't know of any direct connection. Somehow clique is a somewhat quite a considerably less structured object than an arithmetic progression, I would say. There's similar definitions but no direct connection. Yeah, so back to this sort of simple minded attempt to refute Ron, Ron Graham's conjecture. So the first thing to try is the folklore example of the set with zeros and ones in its ternary expansion. And then you immediately see that there might be some difficulties here, because the complement of that said contains enormously long arithmetic progressions in fact it contains everything that's to mod three. That means progressions of linear size. So this is going to be a terrible example for the van der Verden number problem. And unfortunately the Salem Spencer and Baron examples also don't work for somewhat similar reasons. I won't tell you precisely what those examples are but they're all consists of things whose digits are somehow constrained in somewhat small bases. So both have digits restricted to just zero up to D minus one in some base to D minus one. And here D is going to be quite small. So it's something like order log n. So again, such a red, the complement of that set B, which is going to be the red points has very, very long arithmetic progression way bigger than the square return. Okay. So here's a couple of Julia Wolf and me. I want to talk about this in a bit more detail because actually it's the starting point for the construction. I'm going to talk about later. So here we take some integer dimension D and look at the tourist T to the D. And let pi be the natural projection from R to the D to T to the D. So that's actually notation I'm going to keep throughout the talk. And D, consider a Euclidean annulus of radius one quarter, and of this rather small width, enter the minus four over D. That turns out to be a convenient choice. I'll explain a bit more why in a minute project it down to the Torah. So you've got this little annulus sitting inside a tourist in D dimensions. And now we're going to create a coloring from that. So what you do is you pick a random theta vector theta in the tourists and look at the orbit theta two theta three theta and so on. And let the blue points be the set of times N at which that orbit hits the annulus. Okay, so I'm going to put a little picture up. Let's see the complement of that set of points. So here is a picture. So that is supposed to be a tourist. I've drawn a two dimensional one. And the annulus is there quite clearly. And then you're going to take an orbit. I think the one I drew starts from here. Red red red red red. And then when it hits the annulus it's blue red blue and so on and then it sort of wraps around up to here comes down like that. And then you go over here. And then probably down like this. So this this particular segment of an orbit only has two blue points on it. Everything else is red. So that's the construction. And why you'd expect that the number of blue points here, because these are as picked randomly, it's going to be proportional to the volume of that annulus times N. So that's roughly end to the one minus four over D. So as soon as D is at all large, say D is bigger than eight. The number of blue points is going to be bigger than square root N. And again, you can hope that this red set shouldn't contain any long progressions. But before I get to that, I've got to explain to you why I hope that this set B doesn't actually contain any blue three term progressions. Why am I taking this annulus? Well, this is just a sort of slightly padded out version of the fact that a sphere doesn't have three points in a line, just a sort of thickened up version of that statement. So A is this annulus. And then we have the parallelogram law, simple fact from Euclidean geometry. And so if I've got three points in that annulus A, and I'm now thinking in R to the D. And if they're in arithmetic progression, then the common difference of that arithmetic progression has got to be absolutely tiny by this parallelogram law. It's got to be its L2 norm is at most N to the minus two over D. That's in R to the D. But when you push things down to T to the D, because the radius of this tourist is one quarter, you're not going to gain any extra progressions by sort of wraparound effects. So here's a picture of this. The only way you're going to get a blue progression is if it just sort of grazes against the very edge of this tourist like this. And if the step, the common difference is very, very small. So, although inside this annulus because it's an open set, but definitely are collinear points. There are, they've got to be very, very highly bunched together. If there was a three term arithmetic progression and N plus DN plus 2D of blue points, then theta times D, where theta is this randomly chosen rotation on the tourists would have to have very, very small norm where this notation here means that all of the coordinates, Modulo one, that's small. So all of the coordinates are at most N to the minus two over D. And if you fix D, and remember that thesis chosen randomly, the probability of this happening is absolutely minute. It's actually enter the minus two. So we've got a theta times D sits inside some box of width, enter the minus two over D the volume of that box is enter the minus two. And then you can just take a union bound over all of the possible choices of D of which there are most N. And you see that the chance that there's any blue three term progression at all is absolutely minute. So this is why annual I are a good thing to choose. They're basically just slightly thickened up spheres, spheres have no three term progressions. So I feel I have very, very few three term progression. So that's why it's a sensible construction to consider. So what about red progressions. So the red points are the complement of the blue points. Let's just recap some of the notation quickly. So I've got this random rotation theta on the tourists. Let me write theta one up to theta D for its coordinates. And then an actual small piece of number theory, I can apply Dirichlet's theorem. The most simple thing that one learns about time approximation. So there's a non zero D less than square root n, such that the fractional part of theta one times D is at most one over squares. Now, theta is a randomly chosen vector. You certainly so theta one is also random. And you expect the sequence theta one to three to one three to one and so on to equal distribute very, very rapidly module one. And so certainly before too long you'll find an n naught such that these are one times n naught is close to a half. But then if I take the progression that starts at n naught and has common differences D that I found using Dirichlet's theorem. Well that's going to stay near a half for a long time. So one times n naught plus LD, here L's an integer will be a half plus something that's on the order of L over root m. So this is a picture of what happens. So this progression starts over here, it's x coordinate if you like it one coordinate is kind of close to a half. And then it just stays there for a very long time sort of wraps around and around and around, but it never hits this annulus. So it's it's first coordinate just is stuck near a half. And that happens all the way up to L being close to squaries event. So unfortunately, this construction by Julia Wolf and myself doesn't work either. So you can use this construction to get a large set that's free of three 10 progressions color it blue. But by Dirichlet's theorem. You're still going to be left with a red progression whose length is is basically squaries event. So at this point, I have to admit that I don't know how to disprove Ron Graham's conjecture it's possible that w three of K is bounded by constant K square. And in fact, I went ahead and conjecture this myself will be reiterated his conjecture in print. So everything so far is consistent with the fact that this off diagonal van der bedden number, the smallest n such that if you two color one up to end, you're guaranteed to have either blue three 10 progression or red K 10 progression. That could be quadratic in K. However, it turns out that that's not the case so this is the new theorem that I want to talk about today. So actually w three of K does grow quicker than any polynomial in K. And in fact it grows like basically K to the log K to the one third. So in particular it grows faster than any polynomial in K. So, let me talk a little bit about how that construction works. And I showed you three possible constructions there was the one based on the digits in ternary. And then there was one based on the baron construction also really really didn't work. And then there was the one based on my work with Julia Wolf which did a little bit better in that the red progressions at least weren't linear in size they looked as if they were down to square root and inside that didn't prove that there were no longer progressions, but the ones that I found only had size square 10. So what was the problem with that construction that prevented you rolling out progressions of length through 10. The problem was that I took this annulus of radius of quarter pushed it down to the tourists. The problem was that the complement of that annulus contained cosets of co dimension one sub to right. In fact, in that case it just just the sub tourist T one equals a half of coset T one equals a half. And then you could find orbits theta P where P is a progression of length square root n using Dirichlet's theorem you can find an orbit of that length, which concentrates really near such a sub tourist. And so that progression is guaranteed to be read. So the first new idea is that is that taking one annulus perhaps we can take a union of several annual I am slightly smaller radius which are called row. And perhaps if you choose those premises judiciously, you can block sub to right so if you take that union of annulus annual I. There's going to be no co dimension one sub tourist in the complement. And hopefully there'll also still be no blue three term progressions. So that's the, that's the basic idea is take more than one annulus instead of just one. So let's take a picture of this. These are only lie of radius row where row is something like one 12 in this picture. And the way I've drawn them. There's no obvious way of sort of threading a co dimension one sub tourist through them without somehow help however you try and do that. You're going to run into one of these annual life. So there's no obvious way through. And then the construction once again is going to be, I take an orbit theta two theta three theta. I look at where it hits this union of annual I, and when it hits them, I color blue. And when it doesn't hit them, I color red. Okay. So, I first of all need to justify that I can that there's a chance of doing this in such a way that I block. Co dimension one subject. So in other words, there are enough of these annual I that I can't thread a sub tourist through them without hitting them. But they're not so many of them that there's guaranteed to be a three 10 progression for trivial reasons. So you've got to make sure that you can balance the parameters to achieve these two somewhat contradictory aims. And I'll just tell you, it's probably the most mathematical part of the talk. We're going to in too much detail. Try and explain to you why that can be done. So here's how one should choose the parameters you fix, fix a dimension D, which is going to be the dimension of this tourist. The radius of those annual I will be D to the minus four there's quite a lot of flexibility there but it should basically be a power of D one over a moderate power of D. So we're going to look at. First of all just one annulus a of radius this row in Euclidean space. And now push it down to the tourists. And now I'm going to pick a number of random translates at that inside the tourists. And the right number of random translates to take a game there's quite a lot of flexibility but we'll take D to the D of them. And then we're going to consider D to the D of these little annual I of radius row. So all the annual I congruent, and we translate them around D to the D times, and we'll call that set S. And then the construction which I drew on the previous slide, pick a random theta in T to the D, and to find the blue points to be the set of return times of this orbit theta two theta three theta and so on. And then we'll call that set S, which is a union of annual I, and the red points to be the complement of the blue points. So that's the construction. Okay. So why can I hope that there are no blue three term progressions. Well this is essentially a fact from to do with coding theory intuitions really. You can pack a lot in high dimensions you can pack a lot of these points in, while still keeping them highly separated. And actually I need them to be a bit better than highly separated I need the sort of difference of them to be highly separated like this. But I claim that with high probability. So the three term progression of these D to the D to the D centers of these annual I are not only not in three 10 progression but they're very far from being in three 10 progression so they're separated by 10 row. And that means that if you put a little annulus around each one. Well the only way that you could end up with a three 10 progression is if it's completely inside just one of those annual you can't have a three 10 progression that's across different. Okay, so that's the first claim. So the progression in the set S has got to lie entirely in just one of those annually, it can't sit across different time. And therefore, by the parallelogram law as I talked about before, it's got to have very very small common difference. So that will eventually lead to that being with high probability, no blue three term progressions. So as I said this is reminiscent of simple sort of packing bounds that lead to the simplest bounds in sphere packing really. I want I two and I three. If they're not all equal, then this XI one minus two XI two plus XI three is uniformly distributed on the tourists. Therefore the chance that it's sub norm is at most 10 row is very small. It's at most, it's 20 row to the D. You can work through the parameters carefully and check that the probability of this happening for any one of the m cubed choices of the indices I one I two and I three is minuscule. So that's why there are no blue three 10 progressions. What about the red progressions. So I need to make sure that this union of annual I blocks subtory so I need to make sure that there isn't a co-dimension one sub torrents in the complement of those. So for the blue progressions. I couldn't afford for there to be too many centers XI. And this is pulling in the opposite direction. So here, I need there to be lots of the centers XI to make sure that there's a chance of blocking subtory. So it doesn't even mean mathematically to block subtory. But what it really means is that if you take a character from additive character from T to the D to our mod Z, then it's subjective on that set S. So in particular, Kai could be the character that just takes the first coordinate. So we're saying that there's no fixed. If you look at a sub to co-dimension one subtory is with just a fixed value of the first coordinate. Intersex S. So S has blocked that particular co-dimension one sub torrents. Very rough sketch of the proof. Well there's only a bounded number of these size I need to worry about because as soon as say XI one is large, it will sort of stretch out the annulus to the point where it automatically covers everything. Just one of these annual I pie a it has radius row. If this character has got a large coefficient, it will sort of stretch out the annulus and its projection mod one will be everything. So there's only a sort of bounded number of them I need to worry about. And then I just need to check that the centers X1 up to XM under such a character are dense mod one. And then I will, the rest of the annulus will push forward to cover everything. And this is what's called a bin packing problem. So essentially the problem is for each fixed psi, these psi applied to X1 up to psi applied to XM. They're like balls that are being packed in bins. The number of bins is one over one over the size of each bin is row. The basic point is that I just got way more balls than I have. And so the probability that I've got at least one ball in every bin is overwhelmingly high. So, and then I can just sum over all of the possibilities for XI which is row to the minus D. And again do a calculation and this calculation wins by absolutely miles because I basically got an E to the minus D to the D here versus just a D to the D. And so that's how it all wins by a large margin. So it turns out these two contradictory aims. So choosing few enough annual I that you don't create three time progressions, but enough annual I that you block all co-dimension one sub toray. They can be achieved simultaneously by the right choice of parameters. Now, I've been talking about this co-dimension one sub toray. They came about because I applied Dirichlet's theorem. Now you can also apply Dirichlet's theorem with more than one variable. And then you have to start worrying about co-dimension two and three and four sub toray and so on. But actually with the right choice of parameters, you can block a whole load of those all at once. So the construction I just described it gives you a blue red coloring of one up to one with no three time progression. And there's no obvious way of making a red progression of length square root n. So where obvious I mean you cannot use Dirichlet's theorem to make a red progression of length square root n. Because I blocked all of the co-dimension one sub toy. And as I said just now, by making sure that higher co-dimension sub toy are blocked. You can also prevent more higher number of variables application of Dirichlet's theorem from creating progressions of length n to the one third n to the one quarter and so on. So this gives you a construction that at least has a chance of having no blue three time progressions and no long red three time progressions. But of course it's nothing like a proof that there are no such red progressions. All I've done is I made a construction which defeats somehow the most obvious way of producing long red progressions that might very well be other long red progressions. Okay, so let's talk a bit about how you can hope to show that in fact there are no such things. So it's very natural to fix a value of r and focus on rolling out red progressions of length n to the one over r. So maybe I just as an aside. So I said that to block progressions of length root n you need to look at co-dimension one sub toy. But then you can apply Dirichlet's theorem with two variables, potentially to find red progressions of length n to the one third. And there's no, there's nothing in between these two things you either apply Dirichlet's theorem with one number, or with two, but nothing in between. And this I think is probably behind that data that I showed you right at the start of the talk, where you see quadratic behavior. So what I sort of expect is going to happen is that you see quadratic behavior up to a point, and then it will jump to something more like cubic behavior as sort of small numbers phenomena go away and it starts being more efficient to look at constructions like this, where you're blocking Torah. And then there might be another phase change later on. Anyway, for this reason it's very natural to introduce a parameter r and try and study progressions of length n to the one over r. So anything about what D is. So I'm going to perform that union of annual construction that I showed you with a specific value of D that's tailored to the R that I'm interested in. So you can think of our being three if you like. So D will be some constant times r squared. And to perform the annual construction with a randomly chosen theta. So Dp a progression of length x inside one up to n, and I want to show that it's not all red doesn't all get colored red. So in other words with high probability it contains a blue point. So how are we going to do that. Well I'll do it in two stages. First of all, use the first half of that progression. Not to hit an annulus that's too much to do all in one go, but to get inside one of those annual. So we'll first show that if you follow the orbit theta P. Then, by the time you followed half of that orbit, you've gone inside a little ball of radius row over 10 remember rose is the radius of the annual I. So you're, you're right inside one of these annual I, and just for simplicity of notation let me suppose that the center of this annulus is the origin zero. So you're in the middle of one of these annual I and then the natural thing to do is to show that you sort of have to hit the annulus on your way out as it were. So we're going to use the remaining portion of this progression P to actually hit this annulus. So we're going to use the X, the projected version of the set of points X, whose our two norm is between row minus 10 to the minus four of D and row. Okay. So these are two quite different flavors of problem the first one is what I call a core scale equal distribution results, you've got an orbit, a linear orbit on the tourists, and you want to show it hits some ball. So you've got the numbers of reasonably large size, and you can apply the kind of techniques you'd expect to apply in this sort of equal distribution problem so for your base techniques exponential some techniques to show that under suitable conditions. You do get inside one of these balls. And then the second stages is a much finer scale result. The annulus is very, very thin. You cannot just go and apply the core scale for your techniques to try and locate a point inside one of these. So you've got to do something different. And I'm going to talk mostly about stage two in the rest of the talk. So what kind of a problem is this. So let's suppose I'm at zero. So that's what stage one has given me. And now I've got this orbit. And this is actually that the common difference of P is one so I'm really just looking at the progression 1234 song. And I want to show that one of the points and these are one dots up to empty to D lies inside this annulus. And so the rough kind of flavor of this problem is showing that a certain quadratic form. So the variation here means the absolute value of the fractional parts it's the norm on the circle lies very close to row squared. So the basic flavor is I've got something like a quadratic form and I want to show that it takes a value in a very short interval. But of course the, the obvious problem is this isn't a quadratic form. It's a kind of bracket quadratic form. So that's going to obviously make things a lot nastier to deal with. So the problem is to show a certain quadratic form where quadratic form means bracket quadratic form takes at least one value in a very small interval. So the first step is to take this nasty bracket quadratic form and turn it into something that's really an actual quadratic form. And now we use some more sophisticated geometry of numbers that beyond Dirichlet's theorem. So basically Minkowski's second theorem. This is a fairly well-trained path in, in attitude combinatorics and geometry of numbers. The set of n for which all of the fractional parts of these n theaters are small, let's say at most 110, has a multi-dimensional structure. So this is a well-known fact from geometry of numbers. So more precisely it means that if you look inside the set of such n, it contains a very large multi-dimensional progression. It's a kind of grid based on points n1 up to md plus one with side lengths capital Li. And the product of those side lengths capital L1 times up to capital LD plus one is comparable to x, which is the total number of points and consideration. So I really most a large number of the points and for which this is true, which all their fractional parts at most attempt are captured by this multi-dimensional structure. So this goes into misprint. That's really what it is. This is somehow an artifact of the fact that we're working over the integers in additive combinatorics. This sort of thing is usually done in finite cyclic groups modular prime. And there is D instead of D plus one. Okay, so you can put a multi-dimensional structure on things. So here's roughly what it's going to look like you can imagine this progression P. It's going to look multi-dimensional. By passing to this new basis that geometry of numbers gives me. Here it's two dimensional. That's wrong. It should be three dimensional, but a three dimensional drawing looks a total mess. And when it hits this annulus, I've colored things blue. So that's the sort of situation I have. So what's the task in this new basis, which geometry of numbers has given me. I have a genuine quadratic form, no brackets anymore in D plus one variables. And I want to show that this quadratic form hits a user specified, so a sort of given short interval. Now representative case just to get one's head around the sort of nature of this problem is that all of the lengths L one up to LD plus one of this generalized progression multi-dimensional progression. They're all the same. So in the generic case, you have to worry about widely differing lengths as well, but just it's a reasonable representative case. So those lengths would all be about X to the one over D plus one. The coefficients of that quadratic turn out to be fairly small. So in particular the values of that quadratic are bounded in size. And then the length of the interval that I'm aiming for which is enter the minus four over D. So the parameters. There's probably too many to remember here but I chose D and X was like enter the one over R. The bottom line is that the length of the interval that I'm trying to hit with this quadratic form is about L to the minus four R. So remember here R was. I'm trying to rule out read progressions of length and to the one over R. And I said if you like you can think about our being three. Of course this is L to the minus 12. The basic problem is I've got a fixed quadratic form in some large number of variables. I want to show that it hits some interval of length L to the minus 12. As the variables in that quadratic form range up to L. So you can forget the rest of the construction and all of this picture now let me just restate what the problem that I've reduced things to is. So R is an integer. I said you can think about RS three. D is some large constant times R squared. Take a quadratic form on Z to the D plus one with coefficients of size one upon L squared on the box out of the power D plus one. This Q is bounded in size. And the problem is just to show that it takes at least one value on any given interval of length L to the minus four R. Okay, so that's the nature of the problem. It should be true by miles because the number of different values of Q is L to the D. But D is big constant times R squared. And so you've got all of these many, many values of Q. It's bounded in size so by the pigeonhole principle that the sort of average gap between values is is more like L to the minus D. And D is massively bigger than four R. So it should be true heuristically by miles. So that's the naive heuristic can you take so many values that it really you feel like it ought to be the case that it hits every interval of length L to the minus four R with masses of room to spare. It's a bit too naive. And the reason is that there's an integrality problem. So Q could, if we were unlucky, be basically rational. So it could have all of its coefficients integers over some fixed Q. So Q here, maybe just a little bit bigger than L squared, just to make sure that this is possible. If this happened then the values of Q would obviously be separated by one over Q the values of little Q the quadratic form is separated by one over big Q. And if this big Q is moderate in size, maybe L cubed or something. Well then the values of this quadratic form are going to be very separated and there's not really any reason to imagine that you'll hit any particular interval of length out to the minus four R. So as I've stated, it has a negative answer, because potentially, I mean, Q could be zero, but there are, there are other worst things that could happen, it could be close to a rational form essentially. So how are we going to get out of this? Well the idea is to use randomness. Q is not some fixed just fixed form that's out there, it depends on the choice of this random rotation theta. So there's a chance that I can show because theta is random. There's maybe a chance that I can show that almost surely Q is not integral in any vague way. Unfortunately, that that seems to be just impossible. I have no idea how to really control the behavior of Q as a function of beta. So the second key idea of the paper is to introduce an enormous amount of randomness elsewhere in the problem. So you can replace these spherical annuli by ellipsoidal annuli where the eccentricity of the ellipse is chosen randomly. And you indeed mentions so the eccentricity of an ellipse is a kind of the configuration spaces d squared dimensional essentially. So you've got this enormous parameter space of random ellipsoids. Instead of taking the identity matrix you perturb it by a little random E. And you define your annulus to be this. So it's an ellipsoidal annulus. All of the arguments that I mentioned earlier to get to do with the parallelogram law and so on they all apply just almost exactly the same with these ellipsoidal annuli. So nothing's been lost there. But you've now introduced an enormous amount of randomness into the problem. And this quadratic form Q that I want to show has small gaps depends on this random E. And if you work hard you can actually show that as he ranges overall choices of eccentricity of this ellipse. Roughly speaking, those quadratic forms range uniformly over all quadratic forms with bounded coefficients. So there's a lot of technical detail that I've suppressed there. But that's the basic point. So I certainly don't know how to show anything like that just using the randomness of theta and spherical annuli. But if you introduce the random ellipsoidal annuli, you get a far richer space of quadratic forms. So by doing this I've reduced the problem I had before about showing that a specific quadratic form. It hits every interval to the same problem but for a random quadratic form essentially. So here's the theorem that we need to prove. B is just some exponent. It's related to that R that I mentioned before but it's not too important. And I claim the following proposition. So suppose that S is bigger than B squared. And let me take lengths L1 up to LS all of size roughly L. And we'll do some quadratic form by choosing the tuple of its coefficients Aij uniformly at random. And we'll do it so that those coefficients are sampled uniformly from minus 100 up to 100. With the exception of the diagonal terms which I want to be strictly positive. So they're going to be sampled just from 10 up to 100. So write down that random quadratic form. Then the claim is that it comes very, very near to any fixed value that you that you want. So here is the write down the event that the values of this quadratic form on the grid with side lengths L1 up to LS. And let Sigma be the event that that discrete set of points comes within L to the minus B of every point between say 0.9 and 1.1. So you want to you want to get a value of this quadratic form that's very near one say so you can actually get it dense in this interval. So that won't happen all the time because Q could be it could be rational. The theorem is that it happens outside of a very small exceptional set. So, in fact, the probability that this doesn't happen is a really small power of a really big power of one over L so L to the minus BS essentially. Now this really is quite a small exceptional probability and that's actually important. So this is the main proposition. Okay, so what goes into the proof of that. Well many number theorists when faced with a problem like this I think would turn straight to the hardy little wood method which is indeed what I did. So you, you've got a quadratic form and you want to try and show that it. It attains values on an interval, the hardy little wood method or more particularly the Davenport harbour invariance of this is really well set up to do that. So it's technically a little bit demanding you need to put in smooth weight functions everywhere. But that is, is basically method and it's some, it's actually a little bit easier than the most standard variants of the hardy little wood method because there's only one major arc about the point zero. There's no major arcs about other rational points in this in this very. So I won't say any more about the details of that but I want to tell you just to finish a couple of things about some slightly interesting ingredients that go into that application. So to make it work. You need a little bit of random matrix theory first of all which was quite surprising to me. And then to get this very tiny exceptional probability. There's something that I call a combinatorial amplification argument. So it doesn't drop straight out of the hardy little wood method you first apply the hardy little wood method and then you do some some kind of an amplification to get a very, very strong error term. So just one slide on each of those two things to finish off. So first the random matrix theory. I needed a lemma like this, not for four by four matrices and need it for n by m matrices, but four by four was the biggest I could go. So random symmetric matrix that's constrained to have ones on the diagonal, all of the other matrices are between all of the other entries are random minus one up to one. Then the claim is that the probability that the determinant of such a matrix isn't most delta is bounded by a constant times delta. Okay. So there are two things here. The first issue is I've got ones down the diagonal which is a bit nonstandard. So I also need this for n by n matrices, and I need it again with just Delta and I need to make sure that the dependence on n isn't absolutely terrible. And I also need it not just with ones on the diagonal but with other fixed diagonal entries but let's let's gloss over that. So the two ideas in the proof so first of all get rid of this annoying feature that the diagonal entries have to be fixed. The idea for that is what I call amplifying the diagonal. So I think this is a fairly standard sort of procedure. If you did have a big set of matrices like this, or with small determinant, then you could make a big set of another, an even bigger set of matrices just by multiplying through by diagonal matrices on the left and right. So diagonal matrix D multiplied through on the left and multiplied through on the right by D, that will still have small determinant. But now it won't have fixed diagonal entries that look more like just an arbitrary symmetric random matrix. So if the problem is if there were many of these matrices with small determinant, there'd be many symmetric matrices. So that gets rid of the fixed diagonal entries issue. So that's not the issue of how to prove that a symmetric random matrix with uniform entries doesn't have large determinant. And I thought this would be something that just immediately you could look up in the literature that I wasn't able to find it. Well, one thing you can do is you can compare this uniform random matrix model to one of the standard models like the GOE ensemble. So this is the random matrix ensemble with symmetric matrices with Gaussian entries. And there there's an explicit formula for the joint eigenvalue density. And of course the determinant being at most delta is just the same as the product of the eigenvalues being at most delta. And what you have then is just a computation to do with that joint eigenvalue. And then you can kind of embed this uniform random matrix model inside the the GOE model in some fairly simple way. So I think probably the reason this isn't in the literature as far as I know is that usually in random matrix theory you don't consider this this particular limit you'd be more interested in the large end limit. Whereas for me and is fixed like for in this picture. And I'm interested in the sort of small determinant probability. Okay, so just the last couple of minutes I want to tell you something about this combinatorial amplification trick so just give a very rough character of how that works. So let's suppose I've got forms in seven variables could take quadratic forms in seven variables. I want to fix the diagonal terms, a one one up to a seven seven and choose the off diagonal terms a i j independently at random. I want to show that the quadratic form with these coefficients doesn't have any large gaps. And what I can do is I can restrict to various subsets of the coefficients. So I've got a Q is the quadratic quadratic form in seven variables. And I've written down here a green form which comes from setting t t one t three t four t five all equal to zero. So just considering t two t six and t seven as variables. I've got a red form where I've set t one t four t six and t seven to be zero. And then a blue form I've restricted t one t two t three and t six to be zero so these are various different restrictions of Q. So if one of these restrictions has small gaps, then the big quadratic form Q in seven variables also has small gaps it's a hereditary property. So I just need one of those events to hold one of the events that either the green form or the red form or the blue form has no large gaps. So these range over a discrete set of points in a box. But these forms those, those conditions are independent. And the reason they're independent is if you draw those off diagonal coefficients. Well they are disjoint triangles in this. This copy of the complete graph on seven verse seat. So actually this is a well known picture this is a decomposition of the final. This is a plane basically into lines. So it's a perfect decomposition of K seven complete graph on seven verses into disjoint triangles. But now this this sort of shows why it was important to be able to fix diagonal terms, because the diagonal terms are definitely not independent so a to two appears here and here. So these triangles that edge just joins and not vertex destroy. So only get this independence if you're allowed to fix the diagonal terms and only vary the off diagonal terms. So that's why it was necessary to deal with that weird random matrix problem on the previous slide. Okay, so if just one of those events holds, then the form Q in seven variables is dense on 0.9 up to 1.1. So you can massively improve the exceptional properties of the probability that the green form is not dense is something you just get to multiply. Okay, so that's all I want to say. Thank you for listening.