 So, let me thank the organizers again a little bit more loudly. And so, I will give four lectures, and the topics in each lecture will be quite different. So, if you don't understand something in this lecture, you can come to the next lecture and hopefully, maybe it will be better, or maybe not. So, I'll try to illustrate various techniques that appear when you try to solve counting problems. And I will try to illustrate them in the simplest examples, but try to illustrate the ideas. And the examples that I'll try to work out, I think, contain the main ideas that appear in much more complicated situations. So, in particular, if you have questions about anything that I am explaining, you should not hesitate to stop me as I'm explaining. And if I haven't defined some object that you're not familiar with, I'm happy to define it here. So, yeah, so the goal is for people to try to understand what I'm actually saying. And to begin, I'll start with probably the simplest sort of classical problem, which is called the Gauss-Circle problem, and it's concerned with the following question. You have Z2 in R2, and you'd like to understand how many integer points there are in the bowl of radius R. So, Br is X is less than R, or X is in R2. And the question one would like to understand is the number of integer points in this big disk. And the fact is that, of course, the number of such integral points is proportional to the area. This is the area of, maybe I'll put this here, the area of Br by R squared. If I move this, I think it will be better. Okay, so the area is quadratic, but what I want to discuss in this lecture is how to deal with the error term. So, the easiest error term is just linear, so you can get a linear error. And how do you see this linear error while you put a square of size 1 around each lattice point? And you can see that if you change the radius by, let's say, 2, then the area that you get here is an undercount for the number of lattice points. And if you change the radius by plus 2, the area will be an overcount for the number of lattice points. So, getting this linear term is quite easy. So, this was essentially known to Gauss. And what I want to explain today is how to obtain a stronger error term, which is, so this is due to Sierpiński, and it wasn't done today, but in 100 years ago. So, the error term, so you can improve the error term to R to the 2 thirds. And the conjectured error term is, exponent here is 1 half plus epsilon for any epsilon. But, and there have been some improvements to this 2 thirds, but there hasn't been, the 1 half hasn't been reached. So, as I said, I'll explain how to obtain this error term today. But the techniques that are used are actually the kind of basic steps that will appear, are useful in other general contexts, and hopefully in the next lectures, I'll start connecting these kinds of counting problems with dynamics on homogeneous spaces. And there, you can also get error terms of this type. So, this type of 2 thirds, so there's even a name for this kind of 2 thirds type estimate in other contexts. And the techniques are essentially the same. So, if you understand how to do it in this simple case, you can do it in these more general contexts of semi-simple league groups and so on. Okay, so what is the plan for proving this? So, the plan is essentially the following. So, we're going to use Fourier transforms. So, we're going to take, so we're going to define the function f of r to be this. Sorry, let me introduce some notation. So, define f tilde of r of x to be 1 if x is in vr. And 0 otherwise. So, this is just the indicator function of this big ball. And then, we consider the projection from r2 to r2 mod z2. And we just push forward this function. This is going to be our function that we'll work with, f of r. So, you have this big ball and you push it to this fundamental domain. And you see how it overlaps. And the quantity that we're interested in is we want to understand fr of 0. The value of this function at the point 0. And this is exactly the quantity that we are interested in. So, it's the number of lattice points in the ball of radius r. Okay, so this is the object we're interested in. And we are going to proceed as follows. We're going to estimate the freer coefficients of this function. Then we're going to regularize f. So, what is push forward? So, you just... Okay, maybe let me... I'll define it in one second. I apologize. So, what I mean by push forward is you just take the function and you kind of throw it down. So, if you want to evaluate the function at a point here, you take all the pre-images under the projection map and you sum their values. So, I'll write it formally. Let me just write this. So, forward of f at x is defined to be the sum. So, x is now an equivalence class of points. So, it's the sum of... This is f tilde of x, of x prime. Oops. So, this is x prime. Where x prime is x plus some v and v is in z2. So, you sum over all the values that you see above you. And again, this kind of process where you have your kind of uniformization, you have this universal covering type object and then you have a function defined on that and then you push it to some compact manifold or some finite volume manifold. This is a standard step. Okay. So, are there other questions about the objects I've defined so far? So, I'm trying to explain the plan of how we're going to do the estimates. So, we have this function f of r that we've pushed forward. We're going to estimate the free acquisitions. Then we're going to regularize f. So, what do I mean by regularizing f? Well, f is not really even a differentiable function because of the edges. The function is one here and zero here. So, this is not a smooth function and things don't work so well for non-smooth functions. And then we're going to... After we regularize, we're going to estimate again what happens and we'll see that something good will come out of it. Okay. So, this is the plan. So, now we fail. So, the first step is to estimate the free acquisitions. So, I have to remind you what the free transform is. So, if you have a function, the free transform is just the integral. So, we're here t2 of f of x e to the negative 2 pi i cx dx. So, we're going to work with free transforms either the torus or r2. So, c is in here z2. But if you take c to be in r2, then you're integrating over r2 and x can also live on r2. Okay. So, there's this basic free theory and the kind of formula we want to use is free inversion which tells you that you can recover the value of a function if you know it's free i series. So, it says that f of x is the sum of over xc and z2 of f hat of xc times e to the 2 pi i. So, this is the free inversion formula but the problem is that it's not true in general. It doesn't hold for any function. So, this holds if f is, for instance, smooth. So, if f is a smooth function, if this formula is true, our function is not smooth. So, we'll have to deal with this problem later. So, what is the main technique for dealing with non-smooth with functions free transform? So, I'm going to give you an exercise which I'll essentially solve later. I'll tell you the main idea but the exercise is the following. So, if f, if, let me call it eta, is ck, so has k derivatives, then you have an estimate for the free coefficients. So, if you take the norm, the absolute value of the free coefficient, then, so I'll explain in a second, plus an eta ck times c to the minus k. Okay, so what does this say? It says that, so let me explain the notation first. So, I'm going to use this less than or equal sign, so a less than or equal b, and here we have x, if exists a constant c which could depend on x, such that a is less than or equal to c times b. So, what is this exercise? It says that if a function is pretty smooth, if it has k derivatives, then the free coefficients decay up to some constant which depends on the k we're talking about and the norm of the k derivative, they decay like a power, and the power is the power here in the derivative. So, this is a standard fact, and as I said, I'll essentially give you the idea of the proof of this. Okay, so I said we have to compute the free coefficient of our function, so we have r of c by definition. So, if you unfold that projection, it's going to be the integral over rn of this after all the r of e to the minus 2 pi i c x dx, which is simply the integral over the bowl of radius r of e to the negative 2 pi i c x dx. So, the function is 1 on this bowl and 0 outside, and also notice that, so here we have the dot product, right? We have the dot product of two vectors and the bowl is symmetric, so we can rotate. So, rotating, so this integral is rotation invariant, so it means that we can take c to be 1 0, sorry, not 1 0, but absolute value of c 0. So, we can rotate c to our free coefficient to whatever value we like, and we can normalize it to be this. So, let me draw a picture of what integral we're computing. We have this sphere and we've normalized our direction c to point in this direction, and you see this function does not depend on, so let me write this here. So, c times x is c absolute value of c times x 1, so if x is x 1, x 2. So, it doesn't depend on the second coordinate, so we can integrate out this second coordinate. And how large is it? Well, if this is x, and this is r, then this is square root of r squared minus x squared. So, this integral is just going to be the integral from minus r, so we have minus r here and r here, then go from minus r to r of e to the minus 2 pi i c x, or absolute value of c x, times square root of r squared minus x squared dx. This is the integral, and so we're going to factor out the r squared here. So, I'm just going to do a manipulation to normalize it. Can people see here? No, okay. Can you follow what I'm saying, or is it okay? I'm working with general what? Yes, r2, yes. Everything is in r2. Thank you. Okay, so let me write the integral. So, we have an integral from minus r to r of e to the minus 2 pi i c x. So, I lost the factor of 2 there, but it's not too important. 2 r squared minus x squared to the 1 half dx. So, I'm going to write this as integral from minus r to r, e to the minus 2 pi i c r x over r, times 1 minus x over r squared to the 1 half dx over r, and r squared in front. Okay, so this is just algebra. Hopefully, it is correct. So, now if you call the new variable x over r, so this just becomes the normalized integral, integral from negative 1 to 1, e to the negative 2 pi i, your absolute value of c times r, times x dx. Oh, sorry, I forgot the main thing, that's 1 minus x to the 1 half dx. So, we have this integral. So, let me try to draw a picture of what we're looking at. We have something that looks like this. Okay, there's a 2 that I keep forgetting. So, we have this function from negative 1 to 1, and we're integrating it against this sign, which oscillates more and more and more. So, we're trying to integrate something like this, and we want to say that there's a lot of consolation, because you see this integral, this sign is oscillating. So, we want to say that as this number gets larger, then we have lots of consolation. And this was the type of problem that is also illustrated in that exercise. So, if you want to say that it's a function of smooth, then you have some consolation. So, let me make the claim that this integral, so I'm going to just call this c times our parameter. So, integral from negative 1 to 1 e to the minus 2 pi i. So, I'm going to get rid of the minus, because the minus is not relevant. kx, 1 minus x squared to the 1 half dx, equal to O of k to the 3 halves, negative 3 halves. So, k is some positive, large positive number, a real number. So, assuming this claim, so assuming the claim we found our first estimate, so we get that the free coefficient, the r free coefficient. So, what is our k? Our k is c times r, and we had an r squared here. So, we had r squared, and then r times c to the negative 3 halves. So, we get that this is less than or equal up to some constants. So, less than or equal to what? To r to the power 1 half times absolute value of c to the power negative 3 halves. Okay, so let me write this first estimate. So, we'll have it in the future. f r hat of c, oops, less than or equal to r to the 1 half, e to the minus 3 halves. So, I want to, okay, so let me make a few comments about this estimate. So, the claim here which implies what we want is, so if you look in standard textbooks on this material, they'll tell you lots of things about Bessel functions and lots of words which somehow essentially mean this. So, you have an oscillator integral that cancels out a lot, and you can in fact have much more precise asymptotics. You can not just bound it, but you can say it's equal to something plus some error, but this is the only thing you will need, and you should know that these kinds of, these things are called oscillator integrals, and in this particular problem, something that comes up a lot is Bessel functions. So, you get an estimate of this type. Let me remark the following. This estimate is kind of weak. So, if you look at this formula, this free inversion formula, if you try to apply this estimate, you're going to get that the function is not even summable. Okay, so this series does not even converge if you put absolute values with that estimate, and that estimate is essentially accurate. So, now we're going to prove the claim. So, we want to estimate this integral. So, we have negative from negative one to one of, where one half e to the two pi i kx dx. So, the way to estimate all of these integrals is the same, namely, you integrate by parts. So, integration by parts will give you that it's the integral from negative one to one. So, what do you do? You differentiate this, and you have to integrate this. So, you get negative x one minus x squared to the negative one half, and here you have e to the two pi i kx over two pi i k dx. This is just integration by parts, but you've gained something. You've gained this one over k factor. Okay, so you've gained. So, this is the key point. You've gained already a k to the minus one in this decay, but this is still not too good. So, I want to emphasize one thing, which, so when you have integration by parts, you have a boundary term. And here we didn't have a boundary term because the function ventures at the boundary. Here, this function already blows up at the boundary. So, you cannot integrate by parts again. You cannot integrate. So, this function one minus, so negative x one minus x squared to the negative one half at plus or minus one is equal to infinity. So, you can't integrate by parts again. So, if you had a smooth function, you could do the same trick, and you would do the exercise. You could do it as many times as you want, and you would get as high a power of k as you ever wanted. So, here you can't do this. So, what can you do? What you can do is the following. So, you have your minus one one, and what, here zero. So, you will fix an alpha. So, fix alpha, which, so, and consider, and you break up the interval, interval into x is less than one minus alpha. So, you set this parameter to be alpha, and absolute value of x is bigger than y minus alpha less than or equal to alpha. So, let me call this interval a, and this interval b. So, now you have to play a little game, because you cannot integrate by parts. You have to do what is possible. So, you can integrate by parts on this interval. And here, so, one thing I didn't mention is that this integral is bounded, right? This function, even though it's infinity, it's integrable at the edges. Okay? So, since this function is integrable at the edges, this integral is finite. So, if we could just estimate this integral by absolute value, you would get a k to the minus one. But we want a k to the minus three-halves, so we're going to do this trick. So, on interval a, so, on interval a, use integration by parts again, and you'll get a boundary term. And on b, you can just estimate by absolute values. So, you can say it again? On b, yeah. So, b is this interval. So, this is b, this is a. Yeah? So, the absolute value of this term, where's the typo? Is this better? Thank you. Is that the only complaint? Okay. So, the point is this. On this interval, we can gain something by using integration by parts. On this interval, we cannot gain anything, so we just estimate by the absolute value and we'll have to figure out what alpha is. So, because I want to actually finish this proof today, I'll leave the actual integrations by, you know, I think it's clear what you have to do and probably the effect of the coffee is wearing out. So, I'll just tell you what the estimates are. So, okay. So, what is it? Yeah. So, on b, so, okay. So, on b you get, so, the integral is o of alpha to the one-half and on a, on a it's, the integral is o of one over k alpha to the minus one-half. So, you see that this kind of makes sense in that if you make alpha to be very small, the boundary, the b term goes to zero, but this thing blows up as you would want it. And so, you have to optimize and it's clear. So, if you do the optimization, you get that the best choice is to take alpha to be k to the, was it negative one-half? Yes. Negative one-half and you'll get another factor of one-half which will, so you'll get that power saving and this power saving is essentially the key point where you get this minus three-thirds and three-halves and this is the, what will eventually give the right error term. If you want in the exercise session I can actually do these integrals a little bit more slowly. Okay. So, next, we, sorry? Yeah, yeah, yeah, sorry, minus one, you're right, yeah, yeah. Alpha to the one-half is k to the minus one-half, that's right. Okay. So, the next step is to do this smoothing. So, as I said, this estimate is not even good enough to sum the series. So, what you have to do is you have to smooth out the function. So, smoothing, you have, so you'll pick a bump function, so you'll take eta on R2 and B1. So, just take a function which is of integral one and you define eta epsilon of x to be one over epsilon squared eta of x over epsilon. So, this is a bump function which is getting narrower and taller and the normalization is such that the integral of N, that the integral of this gadget is one. So, now you'll convolve these, what we're going to do is we're going to smooth out the edges of our function. Sorry, I'm going to erase this side of the board so that I don't have to write on wet again, so we're going to smooth out the rough function fr on which we have our estimates and then we're going to estimate the new free coefficients of the new function and hopefully this will give us the result. So, the new function is going to be f of r epsilon is going to be defined to be fr convolved with eta epsilon. So, what is the convolution of two functions at x is the integral of f of x minus y g of y dy over, it doesn't matter here, it could be Rn, Tn, your favorite group. So, this is the convolution. So, what does the convolution do geometrically? Let me just draw the picture here. So, you have your function, this is Br, and convolution, better, okay, I'll draw the picture here. Okay, so the convolution does the following, it takes your function and it puts, so it takes this bump function and it moves it around and it puts it. So, you integrate your function, this bump function against this Br and since this has norm one, you see inside it's going to be always, this integral is always going to be one, so it's not changing the function. It's only smoothing it out at the edges. So, okay, so we have, let me write it again, of r epsilon as f of r convolved with eta epsilon. Now, eta epsilon is a smooth function, so eta epsilon of k of c, of a freia coefficient c is less than or equal to any constant absolute value, any k, let's say, eta ck, oh no, sorry, let me, this. So, you get an estimate of the following type for times epsilon to the minus k. So, you get an estimate for the freia coefficients of eta epsilon in terms of the freia, in terms of the ck norm of eta, which is a universal constant, eta is fixed, and then the epsilon gets in here and then you have the c, which gives you lots of dk, where as the epsilon, see the epsilon is small and it's bad for you because it comes with a negative power, the c is good for you, comes with a positive power. So, and we're going to also use two, so let me write this estimate here, because this estimate is a little bit, okay, so this is something that depends on k, and then c times epsilon to the minus k. And we're going to also need a very trivial estimate, which is that this is less than one, and this is true because the, this function has integral one, so this is a trivial estimate, but it's good in the regime, you see this number can be bigger than one when c is very large, this number is, this is a very good estimate, when c is small, this is not a very good estimate. So, we're going to use these two regimes for this estimate, and we're almost there. So, I need one other fact, which I forgot to mention, so f r of zero, f r hat, the first free coefficient is just pi r squared, because what is f hat r of zero, so we don't need to estimate this, this is the main term, this integral of a b r of just one, this is just pi r squared, this is the main term. So, now we can apply free inversion to our regularized function, so f r epsilon, we can write this as f r, sorry, we can write this, I forgot the main thing. Okay, let me write like this, f hat r epsilon c times eta, no, times e to the 2 pi i c x. So, we need to compute the free coefficients of this new function, but these free coefficients are actually very easy to compute, so again, this is an exercise that we can do in the exercise session, that if you convolve two functions and you want to compute the free coefficient of their convolution, then you just take the product of the free coefficients. So, free coefficients behave very nicely under convolution, so the free coefficients here are basically the product of these two guys and we have the following estimates, so we have the estimate on this guy, this kind of decay, and on this guy, we have this kind of decay. So, we have to estimate, we have to see what kind of information this gives us, and as I pointed out earlier, they're essentially the free coefficients of eta epsilon, they depend, we have two estimates for them, and the product of c times epsilon is one, less than one and bigger than one. So, let me write this, so f hat r epsilon of c is f hat r c times eta hat epsilon of c, so we're going to estimate each of these guys individually. For this one, we have a single estimate, we have the estimate when it's less than one and bigger than one. Okay, and finally, the zero of coefficient is just what, it didn't change, it's pi r squared, f hat r, sorry, now we have, oops, so now we're actually using the free inversion, so now we want to evaluate our function at zero, so we see that it's pi r squared plus the thing we want to estimate, some c different from zero, f hat r epsilon of c. So again, we divide this sum into two terms, we have a term a where c times epsilon is less than or equal to one and b when epsilon times c is bigger than or equal to one, depending on which estimate is better. And I want to, again, so you see that there's the same game that we played in the previous estimate, we have two positions, two cutoffs, and I haven't told you what epsilon is, so we're going to choose epsilon at the end, not now, and we'll optimize in terms of what we will get. So hopefully, so this is defined on the torus, so this function is so small that you can view it either on the torus or on r2, it doesn't really matter. So everything now is on the torus, but if you want, you can do everything on r2, but you'll just get integrals instead of sums. It's not, the issues here are the exponents, not the summation versus integration. Okay, so let me get the right estimates. Okay, so we have, okay, so summation a is, we have some zero less than c less than one over epsilon, and here we have r to the one half times c to the minus three halves. You see, so I'm using just the free coefficient that's less than or equal to one, the second term in this product. So this is a product, and the second term is just less than or equal to one. So this term is r to the one half times epsilon to the minus one half, and for integral b, so here, so the integral b is going to give us a little bit more savings, so one over epsilon is less than or equal to c. So we're going to have r to the one half, c to the minus three halves, then c to the minus k, then epsilon to the minus k. So you see now the epsilon comes in, and this is less than or equal to, so if you do all the estimates, it comes out to r to the one half times epsilon to the minus one half. So they're actually of the same order of magnitude, as it should have been in some sense, but the point is, okay, so what is the upshot? We're almost there. So the upshot is that our term is our, so we found the following fact, so we found that f r epsilon at zero is pi r squared plus o of r to the one half times epsilon to the minus one half. So it would seem that we can set epsilon to be, let's say, one, and we're getting the r to the one half, which would seem like it's too good, but the problem is that you see our function is a little bit smoothed out, so we have here b r minus epsilon, we have b r plus epsilon, and we have b r. So our function f of r epsilon really is an approximate count. So we have that f of r plus epsilon at zero, so let me get the exact inequalities because, okay, so the claim is that there are two claims. The first one is that f of r at zero, so remember this is an un-regularized function, this is bigger than or equal to f of r minus epsilon, epsilon at zero. So what does this say? You see here we're doing the pure count and here we have the regularized count and the support of the function which is in b r, so we're convolving this b r minus epsilon and we're smoothing it out and its support is contained in here. So this function is everywhere bigger than or equal to this and so this thing as we got here, so this is pi r squared, sorry, pi r minus epsilon squared plus o of r to the one half times epsilon to the minus one half and so this is pi r squared plus two pi r, sorry, minus to pi r epsilon plus this o to the r to the one half times epsilon to the minus one half plus lower order terms. So now again we have these two terms so we want to balance these two terms out so again you see we haven't fixed epsilon yet so we want these terms to be of the same order so that we get some information so we want again r epsilon to be approximately the same as r to the one half times epsilon to the minus one half and so what does this give us? You have that r to the one half is approximately epsilon to minus three halves so then epsilon is r to the minus one third. So if you plug it into here you get that f r of zero is bigger than or equal to pi r squared plus something that's so epsilon is r to the minus one third this is this term so you have r to the one half plus one sixth which is o of r to the two thirds and for the lower bound you play the same trick but you'll have that f r zero is less than or equal to f of r plus epsilon epsilon and you'll get the same type of estimate. Okay so we found so let me just finish and just say that we found this estimate the number of points, integral points involved radius r this is pi r squared plus o of r to the two thirds so before finishing I just want to emphasize what were the main ideas that made this proof work so first we used the spectral decomposition which is something that will have a general semi-simple e-group and this is going to work more generally and so we're going to be able to decompose our functions if you want to count in terms of this Fourier series what you want to do then is you want to smooth out your counting function because things usually don't work so well for these rough counting functions when you just cut off at straight r and then you want to have some parameters you have the smoothing parameter and then maybe some spectral parameters and you want to optimize them so that you get some kind of power savings when you've played your games right so I don't think I'll do any other kinds of epsilons and deltas in the next lectures but I think hopefully you got the idea and probably the simplest example so thanks