 Okay, thanks for having me. Yeah, so I'm a Fourier analyst and I'm gonna talk today about application of Fourier analysis in number theory. And to be honest, when I first started working in this area, I found the proof so technical that it was hard for me to talk about it even in an analysis seminar. But I've been working on it for the past several years and I'm excited to try to talk about the proof with a broad audience. And well, like Alina said, I'm gonna pause a bunch of times to check if there are questions and also stop me anytime. It's always a good idea, but especially because I'm from a different field from any of you. Okay, so first, let me tell you what is the Vener-Gradoff mean value conjecture? So one way to look at it is it's a conjecture about the number of solutions to a diafontein system. So the system is these equations here. So I have two S numbers and one through N2S and the sum of the j-th powers of the first S is the same as the sum of the j-th powers of the second S. And that's true for a range of J between one and K. And then we ask how many solutions are there? So J sub SK of capital N is the number of integer solutions to this system where the variables run between one and capital N. So the Vener-Gradoff mean value conjecture estimates how many solutions there are up to a fairly small factor. And this is the conjecture. But it's equivalent to a conjecture about exponential subs. So here's the exponential sum version. It's also equal to this integral. Okay, so let me pause to digest all the notation. So E of X is short for E to the two pi I X. And then I have an exponential sum here. And so I add up a little n goes from one to capital N, E of this thing. And it depends on alpha one through alpha K who are living in the K dimensional unit cube. So that's my exponential sum for each trace of alpha I added up and I get some complex number. And I integrate, I take the two S moment I integrate the norm to the two S. And that's also equal to the number of solutions to the diafontein system because of the fundamental ideas of the circle method of number three. Okay. And the conjecture is a conjecture about how big is this integral of this exponential sum? Excuse me, I have a question. Did Vina Granov actually make that conjecture? Yeah, that's a good question. I'm not positive about that, but it seems hard to believe that he didn't think of it as a reasonable conjecture. Yeah, I can't find it anywhere in the literature before I put it in my book. Okay, thank you. Yeah. So Vina Granov did prove this estimate for some values of S and K and he used his estimates for a bunch of problems in number theory, such as giving better estimates for vile sums and giving better estimates for the very wearing problem in high degrees. And the full conjecture was proven in the last decade and by now there are several proofs. Woolly proved it for K equals three. That was the first using something called efficient congruencing and Bergan and Dempeter and I proved it for all K using decoupling and then Woolly also proved it for all K using efficient congruencing. And then more recently, Guo Li, Yang and Zoran Kronich proved it also for all K. They combined all of these ideas and also some new ideas and they made a very concise proof. It feels like a proof from the book, only 10 pages long. Yilin, thank you. Pull on me up. Sorry about that. Is there a conjecture for unlike power systems? Yeah, okay. So this is a system of diathontine equations here and you could imagine changing the system in lots of different ways. Unlike powers might mean that the different powers here might be different from each other. And yeah, so I don't feel an expert. I feel like I could make a naive conjecture using the heuristics that these powers behave sort of randomly. Yeah, but maybe I'm inclined not to go into details. At the end, if there's time, I'll show you another conjecture by Hardy and Littlewood with like powers, but that's open and it tries to give a sense of the boundaries of our methods. Okay, so there are several proofs and my main goal in the talk is to describe some of the ideas of the proof or proofs of the mean value conjecture. Now, one thing that's a little difficult is that all of the proofs involve some really long formulas and computations which are hard to go through in a seminar, especially an online seminar. And what I wanna try to do is to explain the ingredients of those computations and try to limit the long formulas as we talk together. I'm gonna focus on the decoupling proof mostly because I know it the best, but some of the comments that I make apply to all of the proofs. And if there's time, I'll talk about some limitations of the method and some open problems. Okay, so let me make some big picture comments about the proof. So the first comment I wanna make is that decoupling proof is purely analysis. There's nothing that you would call a number theory input. So the kinds of ingredients are things like orthogonality and holders inequality and integration by parts like bread and butter tools of analysis and induction on scales or combining estimates from many scales. We'll talk a lot about that in a minute. And it's perhaps kind of surprising that this theorem can be proven by these kinds of techniques. And historically we built a lot on a 2014 breakthrough by Bergen and Demeter. They proved something that's called the sharp decoupling for the paraboloid. I'm not gonna take the time to state it, but it gives some sharp LP estimates or a moment estimates for some other exponential sums that come up in PDE. And when they did that, and it has this same list of ingredients. And when they did that, I was totally shocked that that conjecture could be proven by using these tools. And even after that, I was surprised that the meat value conjecture could be proven using these tools. Okay, so that's the first comment. It's gonna be an analysis proof. And the tool I wanna highlight, as in red now, is induction on scales, combining estimates from many scales. Kevin asks why I was surprised. Let me come to the moment. Let me try to come back to that later as we talk about the proof. All right, so the induction on scales is crucial and it's crucial to all of the proofs, even going back to Vino Gradoff's work. They all use this in some way. And one of my main goals today, my main goal is to explain what we mean when we say induction on scales and why it helps in this problem. Okay, the third comment I wanna make about the proof is that it's pretty visual or geometric. And so I wanna try to draw pictures to convey some of the things. And so at the very beginning, we're gonna choose a coordinate system that makes the pictures nice. It actually makes the algebra a little bit worse. So it's not the one number theorists usually use, but it makes the pictures nicer. And it's the one that I'm used to. Okay, and to keep the discussion simple, I'm gonna focus on just K equals two, which means that the pictures will be two dimensional. And we won't prove the full thing. We'll prove a weaker estimate, but the discussion will illustrate the tools, which is really what I wanna talk to you about, the tools that go into the proof. So those are some big picture comments. And now I wanna formulate this weaker estimate. So we'll write down a precise theorem and then we'll talk about its proof. So we're gonna look at an exponential sum. There's a variable X, which is in R2. So it has components X1 and X2. And here's our exponential sum. It's the sum little n goes from one to big n of the complex exponential with this one here. So this is a complex exponential with a frequency. And the frequency is a vector. The first component is little n over big n. The second component of the frequency vector is little n squared over big n squared. And if you draw a picture of all of those frequency vectors in the plane, you'll see that they all lie on the parabola that the second component is the first component squared. And that shape of the parabola is important in the analysis approach to this problem. So here's our exponential sum that we're gonna be studying. And I'm gonna write Q sub S for a square of side length S. And Q sub S of X means it's centered at X. And the mean value conjecture it translates into a conjecture for the integral of F to the sum power over a big square. And so the power, the interesting power here turns out to be six and the interesting size of the square turns out to be big n squared. So that's the kind of thing we'd like to estimate. And to start to give a sense for it, I wanna try to visualize this function on this square. So let's try to make a picture. So here's our function. What can we see about it by hand? Well, if we plug in X equals zero, then each of these is just one. So F of zero is n. And that's the biggest that F could ever be by the triangle inequality. But F doesn't usually have norm n by orthogonality. If I take the integral of F squared over a big square, then that's less than, it's easiest for me to remember this if I average. So I divide by the area of the big square. And then orthogonality will tell me that this is bounded by capital N because there are n terms here in their orthogonality. And that means that the size of F of X should be not much bigger than square root of N for most of the points X. So there's a, there's 1.0 where the function's really big, but at most points it's much smaller. Okay, another thing we can see about it is that the function is periodic in the X one variable with period capital N. So it's not only big at zero, it's big at N comma zero to N comma zero and so on. So all of those red dots. And another thing that maybe not quite as easy to see is that if you move X just a little bit just within a unit square, then this function generally doesn't change very much. So this isn't a completely precise statement, but the norm of F of X is roughly constant on each unit square. So it's not just big at the point zero, it's big on like a unit ball or unit square around zero. So it's big on all of these unit squares. Okay, so this is a very rough picture, but we've located some places where the function is big and we know that at most places it's much smaller than that. Okay, so that's starting to orient ourselves. And now let's think about what we might like to prove. So we're gonna, eventually one would like to estimate an integral of the norm of F to some power, but I like to visualize the super level sets. So U lambda of F is the set of points where the norm of F is bigger than lambda. And the norm of a set will mean its measure, which is its area in this case. Okay, so if we wanna say the function is not that big, we can state that in terms of moments, but we can also state it in terms of the size of super level sets. So here's a baby version of the mean value conjecture. If you look at the points in this square where F is bigger than N over 10, so almost as big as it could be by the triangle inequality, the mean value conjecture implies that the area of that set should be at most N to the one plus epsilon. So here epsilon could be anything bigger than zero and C epsilon depends on epsilon. Okay, so let's compare with our picture. So we already know that all of these unit balls have F bigger than N over 10. So they're all in this set and they already have area N. And so the conjecture says that that is basically the size of this set. There might be somewhere a few other places, but the order of magnitude of the size of the set is just these red dots. So that's our baby version of the mean value conjecture that I'm gonna describe to you. I'm gonna do a proof sketch. So I wanna make another comment about the proof of decoupling, which is that it is robust in a certain sense. So well, maybe before getting to that, I should mention that the theorem I've actually stated here is not very new and not very difficult from a number theory perspective. It has a couple of different short proofs that use number theory and that are fairly elementary. But so the proof we're gonna give the main point is that it's going to illustrate the ideas that go into the whole proof of the mean value conjecture. But another comment about it is that it's robust if we perturb these frequencies a little bit. So here's the actual parabola in black and along it are the actual frequencies that appear in the actual function F. But suppose instead of this parabola, I moved it, I perturbed it a little bit. So I have a smooth function that's close to the function omega one squared, but not exactly. And along it I put some dots that are kind of close to the actual frequency, the original frequencies, but not exactly. And I make an exponential sum where I use these orange frequencies instead of the black one. This theorem would still be true in that generality. And then I don't think it has such a short proof. I think if we do that generalization, the proof I'm going to show you is approximately the shortest proof that I'm aware of. And also I think in that generality that it is pretty new, that it probably wasn't known before we're gonna do it. Okay. So maybe just put together all the notation that we have and the statement, the theorem that we're going to try to prove and give you a moment to make sure it's all clear. So Q sub s is a cube of side length s. U lambda of f is the set where norm of f is bigger than lambda. Norm of a set denotes its measure. This is the exponential sum we're going to talk about. And we'd like to prove that the set where it's as big as it could be up to a constant factor intersected with this big square is not much bigger than the obvious places where it was big that we saw before. Okay, so any questions or comments about the setup? So what I wanna do now is I'm gonna introduce one tool at a time from the toolbox of decoupling. And each time we add a tool, we'll see how close we can get to our goal. So the first tool is orthogonality. So these complex exponential functions are orthogonal to each other. And in fact, if you take a cube of side length n, then these functions are orthogonal on that cube. So therefore I can estimate the integral on a cube of side length n of the norm of f square. So it's less than the size of the cube times n. You write it this way. So the average value of this on the cube is less than n. Okay, and I can use that to estimate the size of the set where f is big. And what I'll see is that it's at most a constant times n for any one of these cubes. The area of the whole cube is n squared and this is smaller. Okay, so I made a picture that shows the information that we have so far. And this red set shows the way u n over 10 of f might look given the things we figured out so far. And all we know so far is that in each one of these cubes of side length n, there's at most that much of it. And it could be arranged, who knows how. Okay, so in this picture that we'll look at a lot, there are a few scales. The big square is the square of side length n squared, which is what we ultimately wanna understand. And each of these smaller squares has side length n labeled here. Okay, so for each smaller square, I have an estimate. I just put those all together. I get an estimate for the bigger square and I get a bounded n cubed. So that's a bound, but it's significantly bigger than our goal. And that's just orthogonality. Okay, so how can we do any better than orthogonality? We have to introduce a few different tools to give our next improvement. Maybe I should say there are multiple ways of beating orthogonality, but I'm just, I'm going to show you one that comes from Fourier analysis. Okay, so the next thing, the next tool I wanna introduce is we won't only look at this whole sum, we'll also look at pieces of the sum. So suppose I have an interval inside of the numbers from one to n, then f sub i of x would be, I just sum up the little n's in that interval of my complex x-menages. And if I take one up to n and I partition it into a bunch of intervals, say of the same length l, then my whole sum is the sum of all the f sub i of x. So if there's one up to n and I'm gonna divide it up into some pieces, i one is here and i two is here. So actually all of the proofs of the mean value conjecture involve looking at pieces like this for many different i. So Vinogrado's proof involves looking at pieces like this for many different i. And well, you've already seen, I'm no expert on the history. That's the first example that I'm aware of where people looked at a lot of these sub pieces that's part of their work. Okay, so just these are important character in the story. We don't yet have a new estimate, but let me make a comment about them. So f of x is the sum of the f sub i of x's. And if the norm of f is going to be big, then many of these norms need to be big. So a little more precisely, the biggest that f could possibly be is n and the biggest that f i could possibly be is l, which is the number of the length of this interval, the number of terms in the sum. And the number of different intervals is n over l. So if the norm of f is almost as big as it could be, say n over 10, the only way that could happen is that many of these f i's are almost as big as they could be. So quantitatively you can check that the norm of f i should be at least l over 20 and that should happen for at least one 20th of all the different i's. Okay, so if we wanna understand the places where f is really big, we can say, ah, those are places where each f i, not each, but almost each f i is really big. Okay, so how does that help us? Next, we have to think a little bit about the f i's one at a time and what they're like. So tool number three is something about the shape of the f i. Okay, so let's think about f i. So we take the n's in the interval i and we look at all these frequencies. So if this is my interval i, then these are the frequencies that appear in f i. And notice that they all lie in a small rectangular box. The length of this box is about the length of, from here to here, so that's l over n, because these are the integers over n, I have l of them. And then this is much smaller, and this looks like a little parabola in here. So this length is about that length squared. So this length is like l over n, all quantity squared. So they lie in a small box. And the fact that those frequencies lie in a small box tells us something about what f i looks like. So here's how it works. So the frequencies of i lie in this small box. Then I tile R2 by rectangles that are dual to the small box. And I'll explain more what that means in a minute. So what you see here is a tiling by rectangles. And then heuristically, the norm of f i is roughly constant on each rectangle in the tile. So to clarify what that means, maybe on this rectangle here, the norm of f i is around 100. And on this rectangle here, the norm of f i might be quite different. So maybe the norm of f i here is around just 10. And on this rectangle here, maybe the norm of f i is around 1,000. Okay, so if I take a point in this rectangle and a point in that rectangle, f i might have very different sizes. But if I take two points in the same rectangle, the size of f i should be at least usually pretty simple. So that's explaining, so I didn't make a precise statement and that's because the precise statement is messier and a little kind of unpleasant. But what we all do is we first imagine this statement, which is not quite true when we do our proofs and then we go back and we fill in the precise statement and we roll up our sleeves and do a little work, make sure everything is okay. Okay, so that's explaining what it means, roughly constant on each rectangle in the tiling. And now that may explain to you more carefully, what is this tiling? Okay, so the frequencies lie in this box here that has dimensions L over N by L squared over N squared. The dimensions of the tiles over here are the reciprocals. So N over L and N squared over L squared. And the orientation of the tiling is that the long axis here is at the same angle as the short axis here. So this long axis has the same angle as that short axis and its length is the reciprocal of the length of the short axis and that's the tiling. Okay, I put some pause marks in my presentation to remind myself at some key places. So I think that's a good one. This is a very Fourier analyst kind of idea, I think. Anyway, so let me pause and see if there are questions or comments. Okay, so let me recap what we've done so far and put it together. So our first tool is orthogonality. These exponentials are orthogonal to each other on each cube of side length N or square side length N. And then my second tool is that we're going to consider partial sums F sub i for various intervals inside of here, say at length L. And our third tool is the shape of F sub i that the norm of F sub i is roughly constant on the rectangles from this tiling that we saw on the last page. So let's try to combine all these tools and see what we can say about F sub i. An interesting trace of the length of the interval is the square root of N, let's try that. And then let's try to understand the set where F sub i is almost as big as it could be. The biggest it could be is L. It's going to look at the set where it's at least L over 12. Okay, so our first tool orthogonality, we can use that to estimate the integral of F i squared on any cube. So, so tool one, what it tells us about is the integral over Q N of F i squared. But, and then once we have a bound for this integral, then it gives us a bound for the set where F is big. And if you work it out, you'll see that the bound is N to the three halves. Okay, but the third tool, so that tells us about the measure of this set. And the third tool tells us about the shape of the set. It tells us that this set is organized into rectangles that have dimension N to the half by N. Notice that the area of N to the half by N rectangle is N to the three halves. So, by the area bound, this set actually, this set here actually only has a couple of rectangles in it. So, so this set is organized into rectangles and there are order of one rectangles in each square Q N. Okay, let me draw you a picture. So, F sub i looks basically like this. So, F sub i is in this light blue color here. It's our use of L over 20 of F sub i. The places where F sub i is almost as big as L that's in blue. And in each one of these squares, there's at most one blue rectangle. So it may, it looks like this. There may be less than one. I mean, the set even could be empty, but the biggest it could be looks like this. So that combines our three tools and then you pause to see if that makes sense so far. So we just thought really carefully about F sub i for one i and F is given by adding up the F sub i for many different i's. So now let's think what happens when we put all of the different i's into the game. And our third tool is transversality. It says that for different i, these rectangles that we're drawing are oriented in different directions. So here's i one, i two, i three and so on. And if you look at the set where i one is big, I've drawn that in purple. It's organized into rectangles at most one per small square and the rectangles are perpendicular to i one. So they're vertical. Dido i two that we already talked about and dido i three. So if the set where F i three is big is drawn in green, it's organized into rectangles and the green rectangles are perpendicular to the green rectangle here in frequency space. So the different colored rectangles are pointing in all different directions. And that's tool number four. Why does that matter? Remember that if the norm of F is big, then the norm of F i has to be big for many i's. So in particular, if X is in u n over 10 of F, then X should lie in a rectangle of almost every color. So the only place where this set could be in my picture is like there, there, there. I don't know if I missed one, but that's it. So that gives us actually a lot of new information about this set. So, so what we see is that if we look at the set, u n over 10 of F and we intersect it with one of these medium cubes, q n, it actually is in a much smaller red circle or red square of side length and to the half. So the set that we're interested in is it's contained in the red circles and there's at most one red circle in each of these squares. That's some important new information about this set from the Fourier analysis approach to the mean value conjunction. So we're gonna unwind on the next slide what quantitative estimate we can get out of this. Yeah, okay, great. So the question is, is transversality a consequence of the shape of omega? And is that why it still works when it's perturbed? Yeah, great question and the answer is yes. So the reason we have transversality is that the orientation of these rectangles are changing as we move along the curve. In other words, it's because the parabola is curved and that feature is robust. So if you replace the parabola by a slightly different curve, it will still have that feature that the tangent space is changing. Yeah, great. Any other questions or comments? Okay, so what do we get from this? We know that un over 10 intersected with a medium square of side length n actually is contained in one order of one, much smaller squares. And then we ask ourselves, well, if we look at un over 10 inside this smaller square, how big is it? That's a very similar question and we can actually repeat the same argument. I won't do the details, but by the same reasoning, if you look at un over 10 of f inside of this n to the half square, it's contained in just order of one, even smaller squares of size n to the quarter. And if you keep iterating that, it gives a nice really quite strong sounding estimate that if you look at un over 10 of f intersected a medium square, so inside of here, it actually has area n to the epsilon. It's in almost constant number of unit squares. So by using transversality, by repeating this argument inside the red square and doing that many times, we get this bound and this is an optimum bound because it can happen that un over 10 of f intersected with one of these squares. It could have area one. So this is essentially best possible. And then if we wanna look at the size of un over 10 of f in the bigger square, we can add up each smaller square and there are n squared of them, so we get this bound. So I need to make a historical remark about where this argument came from that basically is Jacob's question. This transversality comes from the curvature of the parabola and the circle of ideas goes back in Fourier analysis to Stein's work in the 60s. He began a program to investigate the connection between the curvature of this set in frequency space and what the function f looks like in x space. And these ideas were developed by many people and the argument that I just sketched is due to Tom Wolf and to Bennett, Carberry and Tao. And I'm gonna call it the orthogonality transversality argument because those are the two main ingredients in it. And that gave us an improvement over just orthogonality. So taking stock, just orthogonality, we bounded the measure of our set by n cubed and with orthogonality and transversality together, we got down to n to the two plus epsilon. And that was the best estimate that was available in harmonic analysis before Brigand and Demerick. Okay, so I phrased this a little bit carefully. This goal was known a long time ago. I don't actually know who to attribute to it, but probably, you know, but before Wienegroff for this exact function by using number theoretic arguments about these exact phases. But if we do, if we wanted to make it more robust than if we perturbed these frequencies, then this was the best estimate that was known before Brigand and Demeter, period. And I was surprised that it could be beaten by harmonic analysis. People knew that it could be beaten by using number theory, by using the special structure, number theoretic structure of these frequencies, but it didn't seem clear that it could be beaten using harmonic analysis. So now I can come back to Kevin's question. Why was I surprised? Okay, so the situation that we are worried about here is that there's one red dot in each of these squares. And that would correspond to this. Now, if you pick any one of these squares, it's very plausible that there's a red dot there. Let me say this more carefully. If you take the function and you add some weights here, some coefficients, C and which have norm at most one, then it's easy to choose those coefficients so that the function F has norm one at any single desired point. So it could easily happen that there's a red dot in any one of these squares, but we have to prove that if there are red dots in some of the squares, then there can't be red dots in other squares. If something sort of unfortunate happens in this square, it somehow prevents something unfortunate happening in that square. And it didn't look to me like the harmonic analysis argument could say anything helpful about that because the very first thing we did is we started to look at the F sub i's and to focus attention on what's going on in each of these squares. And so it seemed like we'd have to do something very different to prove an estimate that relates what's happening in this square to what's happening in this square. Okay. But they did and it turned out it didn't involve a really radically different idea. It involved this one extra trick of induction on scales which we come to next. Tool number five, the part of the talk, induction on scales. Okay. So here's the big idea. We discussed some estimates for F sub i and for the set where F sub i is big. Those estimates played a really big role in our discussion so far. And if you pause to reflect on this sub problem just focus on this sub problem. How big is that? It's kind of similar to our original problem. Because who's F sub i? F sub i is an exponential sum that you get by just taking some of the terms in this original sum. And we're asking about where it's big. It's a cousin of our original problem. And in fact, it's a super close cousin. For each interval i of length l, you can make a change of variables, just a linear change of variables that turns this function into our original function except with only l terms instead of n terms. So in fact, all the things that we did so far, we can apply them to estimate u sub l over 20 of F sub i. In the transversality orthogonality argument we had to estimate the set u sub l over 20 of F sub i. And we estimated it only by using orthogonality. But instead we can estimate it using the whole story that we told with all four tools. And it gives a stronger thought. That's the new idea. So let's think again about this F sub i with the same length n to the half. So we used orthogonality and that's what we did the first time and we said, okay, we can bounce. How much of that set is in a square side length n for each square? That was the one we, this is the tool we already used. But we can also use induction on scales. And in other words, we can use all the arguments we've already made and apply them to this function. Maybe for convenience after doing a change of variables to make it look just like our original function but with a smaller value of n. And then we do formally, we might do induction on n. When we do that, it gives a bound for the area of UL over 20 in the big square and it's a stronger bound. It's new information. So if you put these together, then it tells us that this set has at most one rectangle per Qn. But also its total area is quite small. So there's not a rectangle in every Qn. There's only a rectangle in a small fraction of Qn. And that gives us an advantage over the information that we've had before. Okay, so let me take another pause. The main idea is roughly, an important idea in the proof is roughly on this slide that by induction we can find out more about this function f sub i than we knew before. Yeah, I haven't had a question. So when you say change of variable, which variables are you changing? I guess all of them. But the x's are the n's. Right, so you change the x's. So you can think about it this way. Do a linear change of variables in the x's. So y is L of x, where L is a matrix. Oh, okay, okay. So then what happens? Then we can write, so I'll call it g i of y is equal to f i of L inverse y, which is x. And then write this g i as an exponential sum and see what happens. And you'll see that. The reason I asked is because in the classical methods the change of variables occur with the n's, not the x's, right? Because you use this underlying property that the system is a translation invariant and as well as being scaled invariant. And that doesn't seem to occur here, which surprises me. Because that's a fundamental property of the Vinogradov system. Right, right. I think it's the same thing. So I think it's just a question of my habits. I think we do the change of variables on the n first and then see what happens to the n's. Yeah, yeah, yeah, yeah, okay. Actually, let me write something more precise. Oh, no, it's awkward in my coordinate system to write this more precise. Yeah, okay, yeah. Let me repeat what Bob said for everybody because it's important. If we go back to this slide number one. Sorry about that. That's okay. And we look at this system of equations, the system of diaphanetine equations. It is translation invariant, which means that if you replace n sub i by n sub i plus t for all of the different i's then the new system has the same number of solutions as the old system. And that translates, that symmetry translates into a symmetry of the f sub i's. And it tells us how to relate the moments of f sub i to the moments of f. So there's a nice symmetry there, which is equivalent in slightly different language to the symmetry that I talked about. Thank you. So that's what each f i looks like. Locally, it looks the same as before, but we know there are many fewer rectangles total. Now what happens when you put all the f i's together? Well, for each i, there's a few rectangles where f i is big and there's at most one per square and there's not too many of them total. And that's all that we know. The worst thing that could happen is that they could all conspire to be in the same squares. It could happen that the purple rectangles are in different squares from the blue rectangles. That would be even better for us. This picture is the worst case. And then there are places that lie in a rectangle of every color, and that's where u n over 10 of f lives. And you can see that we have a better estimate than before because we have a red circle in only a few of the squares instead of a red circle in every square. So we should get some improvement. And what's not so obvious, if you have to do a computation to see what is the improvement, but it goes all the way to the goal. It proves the theorem that I stated. So I want to make a couple of comments on this. One comment that had me confused into coupling for a long time is how these two bounds work together. I thought initially like, if this bound is better, we should just use it. And if that bound is better, we should just use it. Why are we using both of them? The answer must be that neither bound is better. They are complementary to each other and they tell us different things. So this bound here gives us a better estimate for the total area where fi is big than the one we would get from using this bound on each QF. However, this bound tells us a better estimate on a small square like this. And that helps to control how the rectangles are packed. So suppose for a moment that we had a nice bound of this form, but we didn't know any, we didn't have to use local orthogonality. So that would tell us, we would know how many blue rectangles there are, but we wouldn't know anything about where they are. And then the worst case scenario would be that they were organized, packed closely together. And that would be bad for us because if it happens in every color, oh, that was the wrong direction. If it happens in every color, they can now manage to overlap each other quite a bit. Because of being packed together. So these guys have complementary jobs. This tells us how many blue rectangles there are, but this forces the blue rectangles to be spread out. And that's important because once they're spread out, it's hard for them to interact a lot with the other color rectangles. I should do a pause here. So we've gotten to kind of our goal we've talked about and we should stop soon, so that's a good thing. So we talked about, yeah, okay, great. So Tom has a question. Was the main insight of Bergang Demeter to combine local orthogonality transversality with induction on scales? Yeah, I would say yes. Some form of this induction on scales was around in Vino Gradoff's work, but they didn't understand that, they were originally working on a different for analysis problem that nobody knew that it was related to Vino Gradoff's mean value conjecture or Vino Gradoff's work. And then within harmonic analysis, there was an induction on scales and kind of the spirit that Tom Wolfe did and that was their inspiration. But yeah, putting that in with the orthogonality transversality and seeing that it gives a sharp estimate of was their main new idea. Yeah, other questions or comments? Right, yes. Right, so this kind of induction on scales was in efficient congruencing argument in 2010 and is it fair to say that it was in Vino Gradoff's work? I had thought so, or something of roughly this waiver. Yes, but the inducted use of the many scales was, I think for the first time in the 2010 paper. Vino Gradoff is sort of an induction on scales, but somehow you lose variables each time you do this. So it's sort of one use of scales in each iteration. Okay, so Kevin's asking what was missing in some of these earlier arguments, which proved beautiful estimates, but that were not completely sharp moment estimates for every moment. So maybe I'll do a little, yeah, so I think this is the slide that helps to address that. So why does the induction help so much? I'm gonna cycle back to Kevin's question and do the best I can with it and other people might have better things to say. But why does the induction help? If you look at the orthogonality transversality argument, the first thing, the first scale for I that we picked at that we used was L equals N to the half. And we took advantage of transversality of some rectangles that corresponded to this scale. And then we did iterate. So we zoomed in on a smaller ball and we repeated. So if you unwind that, there are some intervals of other lengths that appear. N to the three quarters and N to the seven eighths and so on. So there are many lengths that appear, but still it's a rather small subset of all the possible lengths. But if you look at the inductive argument and you unwind the induction, then it involves many other scales, many other choices for L and in some sense a dense set of choices. So if you look at all the lengths that appear and you look at their logarithms base N, so it'll be between zero and one. If you unwind everything in the decoupling proof, you'll get a dense set of numbers from zero to one. So every scale is being considered. And after you unwind the induction, I believe that every proof of the mean value conjecture will use a dense set of scales for I and will use some type of transversality at every scale. And so for this second statement, transversality might not, it might not mean exactly what it meant in this talk, but some other effect that broadly could be considered to be some kind of transversality. Okay. So then coming to Kevin's question, well, if you consider arguments of this type, you have to figure out sort of exactly what type of transversality would we like to use and exactly how are we gonna put these different scales together? And depending on how one does that, one doesn't always get sharp estimates. So in this proof, it turns out that this can be arranged in a way that the estimates turn out to be sharp. And there maybe is sort of a matter of craft there, exactly how do we craft the estimates to make that work? But that was what was missing. And I think it was maybe surprising that it's possible to do that for the whole mean value conjecture, or maybe not. So yeah, so this was what I wanted to say about how induction gets into the story and helps a lot with these other arguments and the features that I see being in common among the different proofs. And looking at the time, I think it's probably a good time to stop here. So thank you for having me.