 OK, good morning everybody. So today we're going to discuss some applications of the theory that we've developed so far. And the great strength of what you've learned today, the dynamics of flows on the space of lattices in n dimensions, is that you can solve many problems that were out of reach, in particular for dimensions greater or equal to 3. So there are many findings that people have in two dimensional distribution problems or one dimensional problems. And then they usually say, oh, we don't have good continued fraction of the higher dimension. We can't do it. So the space of lattices, you can do it in some cases. And that's when sort of some of these breakthroughs happen. So I want to discuss some of these things today. But I will focus on very simple examples. So I will not discuss the Oppenheim conjecture or quantum unique ergodicity. I will discuss much simpler examples. And I, in fact, will restrict myself, in this lecture, to the two dimensional situation where the results were known before. But just to illustrate to you the techniques in a simple setting. And then, if you like, you can look at the papers or you can try it yourself on how these techniques generalize to the n dimensional case. It will be more or less sort of straightforward for what I'm going to tell you. So the first application I want to talk about is the statistics of fairy fractions. Now fairy fractions are fairy fractions of level Q are all rational numbers. And let's restrict them to the unit interval. So I'm doing reduced fractions, which means that I assume P and Q are integers and they are coprime. So this thing here just means primitive lattice points, i.e. lattice points with GCD of P and Q equal to 1. And level Q means that you look at all the denominators up to Q. And we also exclude 0 here. So that's implicit in this definition. OK. The first fact, classic fact, is that you know how many fairy fractions of level Q there are when Q is large. So asymptotically, the number of elements in FQ goes like Q squared over 2 times zeta at 2, which is more explicitly 3 over pi squared. And we're going to call this function sigma of Q. So this is when Q is large. So that's fact number 1, fact number 2. So we are interested in the distribution mode 1 of this sequence. So here's our unit interval. And here are our fairy points. OK, so here's 1 half. Here's 1 third. Here's 2 thirds, something like this. And then you have all the other guys in here lining up. And you can make your little numerical pictures, if you like. And so the first question is, is the sequence, as Q tends to infinity, uniformly distributed modular 1? And that's the case. So if you take a subinterval of the unit interval, and you count the number of points in there, this is asymptotic to the length of that subinterval times the total number of points. So this always refers to the length or the measure, the back measure of a set in R, as Q tends to infinity. OK, and when I say modular 1 here, don't really have to say modular 1, because this is a subset of the unit interval. But I'm now going to study the fine-scale statistics. And one example of the fine-scale statistics that I mentioned in the very first lecture is the distribution of gaps. So you now look at the gaps between your elements for a fixed Q. So you fix Q, and you get finite number of points. So you get Q minus 1 gaps between your points. But what I would like to do is also consider the gap between the first and the last guy by just identifying 0 and 1. So I'm just going to go around. And then I get exactly Q gaps. OK, so what did I say? Yeah, sigma Q gaps. I'm sorry, sigma Q gaps, exactly. That's the number of points that we have. So I have sigma Q gaps now. And here, because we have sigma Q gaps, the average gap size is 1 over sigma Q. So they're very, very close together. And now we want to simply say, what is the number of gaps that fall into a certain interval? And this interval I have to scale with sigma over Q, because the average gap is sigma over Q. So I have to measure my gaps and units of the average gap. And just to write this a little more formally. So I'm going to identify my unit interval with the circle r modulo z. So this is, again, the same coset that we've seen before at g mod gamma by simply taking a point here and associating with it the coset x plus z. So that's a bijection. If you just take as a fundamental domain here, the unit interval. So now the first statistics that one can consider here, I call it given, I call them test sets, because they will test my distribution. So I have a test set i, which is now a subset of the circle and a subset j of the real line. And now I look at the following statistics. And I call them fine scale statistics, because they measure randomness on this very fine scale of the average gap versus a sort of global statistics where you just look at distribution mod 1. So this here I would call a global statistics, fact number 2, because it measures things on a very large scale. OK, so what do we do? Well, we count how many points are in a small interval. In this interval here, that's, if you like, setting at a point x. So and then it's scaled down by the average gap. And this will be my interval j, scaled down by my average gap. And then we remember we are in the circle, so an x in r mod z. So this will be an interval in the circle, this interval. And now I'm going to throw that interval at random by simply taking the left point of my interval to be random. So x will be random, and how will it be random? Well, it be random in another big interval, i. So x will simply be thrown at this x here, will be thrown at random. I'm not going to draw it, because otherwise you get all confused. And what does random mean? Well, random just means I look at the measure of points x. This is the Lebesgue measure of points x in this interval i, so that I have exactly k points. Just a second, Andreas, give me a chance. Give me a chance. Give me a chance. Oh, yes. So far, I haven't mentioned my fk at all. That's right. So here is my interval. Number of points in here intersected with my sequence, of course. And then this should be equal to k. Is that good? So I take my, and then I want to compare this. So with the total number of things that I'm counting here, and this will simply be the number of points, number of total points in the interval i that fall into, ah, sorry, just the length of the interval. So this is just a normalization to say that I'm actually looking here at the fraction of xs for which the corresponding shifted interval intersects the sequence I'm interested in exactly k points. So I'm going to just call this the, do I have a name for this? Well, this is the local statistics for this point set. And we are now interested in where this goes when q tends to infinity. Another statistics that I'm going to look at is very similar. But instead of throwing the left point of my interval here at random, I'm going to put it onto my sequence. So let me write it down. So I'm now counting the number of points in my sequence, fq in i, so that r plus 1 over sigma q times the interval j plus z intersect with fq is equal to k divided by all the points that I'm counting here. So this would be the number of points, fq intersect with i. So what this means, if I just draw one element here, I have here an r, which is an fq. And I'm just asking, what is the number of elements in my sequence so that there are k points in the interval to the right of it? When I say to the right of it, I'm already thinking of j being an interval of this form, say, 0s. But it doesn't have to be like this, right? Question? Yeah, j is an interval, test sets, intervals. Contained, yes, very good. Yeah, OK, so this is just some intervals. You could take more general test sets as long as they have boundary of measure 0 and everything would work as well. So this is the idea, OK? And so in particular, if you would choose this particular interval here and you would say take k equal to 0, then this would give you exactly the number of elements r that have a gap to the right with no other element in your sequence of distance at least s. So this will be the gap distribution where we showed you the numerics early on, OK? So these are fine scale statistics. They actually can test the randomness of a sequence. So when you just look at the global distributions, mod 1 like this, you can have very regular distributions like a picket fence distributions that are completely non-random, but still are uniformly distributed, mod 1. Now these things pick up randomness. And indeed, if my sequence of numbers would be random in the sense that the elements in fq would be uniformly distributed on the unit interval independently of each other. So if, and they are not, of course, but if fq were independently uniformly distributed on 0, 1, random variables, then pq kij would converge to, as q tends to infinity, the Poisson distribution. And similar, this guy would converge to the Poisson distribution. Now, OK, so this is a triangular array of random variables where for each q, you have sigma q points. And this convergence would hold for the connoisseurs. This would hold almost surely in the appropriate probability space of triangular arrays. So this is the Poisson distribution. And so whenever we get some convergence, we just compare it with the Poisson distribution. And if it's Poisson, which can happen for some sequences, this can be hard to prove, or sometimes you can have. And sometimes you can prove these things. For instance, when your points will be generated, say, by a dynamical system, you look at the first n points of a random initial condition, you look at the first n points, and then you look at these statistics. And you can prove that if the system is hyperbolic, you converge to a Poisson distribution. So Sol and Vienti and others have many beautiful theorems in this direction. But we are interested now in this number theoretic sequence. And I'd like to show you now how the techniques that we've developed can prove this convergence in one step, more or less. What I have to do now is I have to translate this problem into a problem of dynamics on our lovely or beloved space of lattices. And that's going to be one step. And we're going to reprove in two dimensions here in this particular setting a beautiful observation by Hall from 1970, who proved this kind of convergence only for the gap distribution, but that doesn't matter. And the key thing, as I mentioned before, is that we can now generalize this without any further worry, modular technicalities, to fairy sequence in higher dimensional spaces, which I will not discuss. But you see that whatever I do here, nothing is really special to two dimensions, except that the notation is simpler. OK, so how are we going to do this? So let me erase the Poisson distribution. So the first observation is that if q is sufficiently large, will be larger than a certain constant. You'll see in a minute what sufficiently large means. Then we have the following observation that the number of points in our sequence is equal to the number of points in primitive lattice points. And then I apply our unstable Horus cyclic group to it. Just a second, have I skipped something here? Oh, I've skipped something. I do apologize. I do apologize. I've skipped the whole page. We'll get there in a second. So I first need to have some notation here. I'm going to define a certain triangle, which I call CJ. And this is a triangle, obviously, in R2. And it looks like this. So it's the point x and y in R cross the unit interval so that x over y is sigma n times this times j. You can convince yourself this is a triangle. Make a little picture and you'll see what it is for any interval j here. And I remind you, this was what we called the expanding Horus sphere generated by this element. I'm just going to put a minus here. It doesn't matter, right? If you go left or right, notational, it's better. And then we also had our diagonal action. And I'm going to use this matrix here, again, just for notational reasons. And then we have the following lemma. And that's sort of the key in the translation of our problem to the dynamics. For any point pq in R2, we have that if pq is in x plus 1 over sigma qj, and q is strictly positive and less or equal to capital Q. So this is equivalent, the statement, to if and only if pq h minus x dq is in cj, where cj is the cylinder that I wrote down, or not the cylinder, sorry, the triangle that I've just written down and erased immediately. And now I'm going to write it down again. So this is a little exercise that you can do. And I'm sure it's wrong, because I never get these things, right? So you can correct my mistake. It's here, it's here. Thank you very much. Yeah, thank you very much. Is in sigma, let me write it like this, 1 over sigma 1. What is sigma 1? Is it 3 over pi squared or something? That's my way of writing. So you remember I given you sigma q. What was sigma q? Was 3 over pi squared times q squared or something like that? So q is 1 here. Just a constant, right? So it's just a special case of our growth function. Just a scaling constant. What is? I can't hear you. d of q? OK, thank you. Good questions. Right, and now, so if q sufficiently large, and sufficiently large means here so that our scaled interval is actually in the unit interval. So this is a, I'm scaling this interval down here. And so all I want to make sure is that it's now sort of sufficiently small. That's achieved when q is sufficiently large. Then the number that I'm interested in, this is the number of points qj plus z. This is what I had to raise with fq. Is equal to, so this follows from this lemma now, the number of primitive lattice points under the linear transformation h minus of x times dq intersect with that triangle, cj. This is all elementary. All elementary. It's just plain geometry. You can check all these things. But that is the key that makes everything work. And that works in any dimension, of course, right? Nothing here that I've told you is more difficult in higher dimensions. OK, so what does this mean now? Well, this now means that we can write the counting function that we're interested in here. Let's start with this one in the following way. So let's start with pq kij. This can now be written as well. It's the measure of x in i. So that, and now I'm just going to use this alternative description that we, probably you can't read this, yes? Can you read this? Yes or no? I need yes or no? No, OK. Further down, yes? Let me write it here so I have more space. So it's the measure of x in i. So that, I'm just copying what I had there, h minus x dq intersects the cylinder at the triangle, cj. And that that number is equal to k. That's my set. I'm looking at the measure of that divided by i. So we have something that looks very similar to what you've seen yesterday in Andreas' lecture. We have here something that looks like z squared m. m is a matrix and z squared is your square lattice. And Andreas told you about the space of lattices that you get when m runs through all of sl2r. Now here we have the primitive lattice points. But also the primitive lattice points, z hat squared, are invariant under sl2z. And so we now have a space of primitive lattice points, if you like, OK? And that's we're going to discuss a little more in the exercise, that that also makes sense. So the space of lattices is actually the same as the space of primitive lattices. They are both bijective with g mod gamma, where g is sl2r and gamma is sl2z. So it looks like this. So now, OK, well then, my claim now is that I can write this as an integral of the form that you had yesterday. Some function that lives on g mod gamma, where we integrate over the horizontal cycle. And then we flow with a diagonal flow. So Andreas would have written this as a t. And of course, we can do this. And then t would be nothing but. So e to the t in the setting would be nothing but q squared. OK, so this is just identical to a t. And Andreas was going backwards in time. I'm going forwards in time, but that's why I'm now using h minus. So we also know that these guys will become equidistributed. We called it n minus. I'm so sorry. And I called it h everywhere. I'm so sorry. OK, guys, so you know that n minus was defined as g times n minus. How do we do this? Like this? Yeah, Andreas? OK, so I'm now confusing matrices and maps and all of that. So forgive me. And we had a x here. We had a x here. And I would have an x here or a t or an s or whatever we had. And now I'm using a slightly different notation. I should really use this notation. But anyway, I put a minus in here so I can use a new notation. Forget these technicalities. The point is we average over a horror cycle. We push it by the diagonal flow. And we know that this thing becomes equidistributed. So we know that this would become equidistributed, which means that this converges to f of g times r measure as q tends to infinity, which is the same as t tends to infinity, same thing. But what is the function f that we're using here? That's the question. So here we have our statistics. And what I'm saying is that equidistribution of horror cycles gives you the limit. The limit's there. Nothing more to prove. So in one step, you've solved the problem of Hall on the distribution. One step, just apply the equidistribution theorem. Just do a little bit of simple geometry and then you're done. But there is a little technical subtlety here that becomes apparent when I write down what the function f is that we need to put in here. Because obviously, that is a very specific function f to get my statistics. So what is this function? The function f of g that I'm using in my integral here that will give me pq is a characteristic function. So it's 1 if the number of points, of primitive points transformed by g in my triangle is equal to k and 0 if not. That's the function. So it's a characteristic function. This function is not continuous. So let me say this function is, so this is a function. You can convince yourself that this function is well defined as a function. So g, again, is SL2r and gamma is SL2z. So that's something you can think about or we can do it in an exercise. That's well defined on this quotient. And it's well defined because z hat squared is also invariant under SL2z, as I mentioned earlier. It's bounded. That's good, but not continuous. And in Andrea's equidistribution theorems, he always assumed continuity. And this is the little technical subtlety that we have. And the way you can get around this, and we will discuss this later in the course, is that as long as you can show that the discontinuities here are contained in a nice set of measure 0. So in other words, the boundary of the set whose characteristic function you look at here has to have measure 0 with respect to the limiting measure that we have here. And that's the harm measure. So if you can show this, then you can approximate such characteristic functions by continuous functions. And then you're fine. And that's the only thing you still need to prove. So now let me just state the consequence of our observation here. And that is theorem 1. So modulo this little technical detail is that the limit as q tends to infinity of p, q, k, i, j is equal to the measure, the harm measure, of points in g mod gamma so that the number of primitive lettuce points g intersect with cj is equal to k. And that's that. From this, you can derive all other local statistics, for instance, the whole distribution. Now what I had planned is to also tell you how to get to the whole distribution in a very quick way. And maybe I'm going to do that. What do you think, Andreas? Shall I go straight to the three steps, or shall I do the whole distribution? How do you see the whole distribution directly? Well, let me continue. Because I think it's so nice and explicit that you get a bit more appreciation that this stuff is not just abstract nonsense. So I want to now discuss these statistics here, which is closer to the gap distribution, even though knowing this kind of thing will imply knowing the gap distribution by some probabilistic arguments. But let's discuss this one directly. The strategy is the same. The only interesting thing here is that we now need a new equidistribution theorem on g mod gamma. And this looks in the following way. So I'm now considering pq0. And instead of the equidistribution theorem that Andreas had on the board yesterday morning, I need the following. For any bounded continuous function on g mod gamma, so again, I'm now looking at a test function f, evaluated along a horror cycle. But instead of integrating over a horror cycle, I take point measures. Namely, I just sum over the points of my fairy sequence in my interval i. And I divide by the total number of points. So that's an average. And I take the limit as q tends to infinity. So the picture somehow is that we are on the unit tangent bundle of the modular surface. So every point is a point in the direction. We have some horror cycle piece here. And these are the directions. These are the, this is the horror cycle corresponding to this interval i. And now I'm pushing this with my diagonal action. But now rather than integrating over this piece, I'm just summing over points. And as Andreas mentioned yesterday in general, this will not converge. So if I'm just summing over some points here, in general, this will not converge to a limiting measure. But here we are so lucky. And this indeed does converge to a limiting measure. And this looks as follows. It's very, very nice and simple to go from 0 to r. And to go from 1 to infinity, 0 to 1. And oh, I'm sorry. It's again nx times square root y, 1 over square root y, 0, 0. dx dy over y squared. That's the answer. So you see, we are here averaging over points on a curve in a three-dimensional space, our unit tangent bundle. But the limit measure is sitting on a two-dimensional sub-surface, which is exactly, I'm drawing it, it's this thing here, with all vectors pointing upwards, if you think of it as a sub. So it's this rectangle in the hyperbolic plane. And to tell you how this looks like in the unit tangent bundle, I have to tell you at each point what is the unit vector. And it's pointing upwards at every point. So this defines a two-dimensional sub-surface with boundary, this is the boundary, in the three-dimensional unit tangent bundle, i.e. in SL2R, modulo SL2Z. OK. And so this, yeah, this you can simply think of in our identification with the unit tangent bundle as point x plus iy and angle 0. Now this is a theorem. I proved in arbitrary dimensions in a paper in 2010. So this is for SLDR, where I proved a limit distribution for Frobenius numbers in this way. And this distribution was only known for two-dimensional numbers and not in higher dimensions. That's where this really made sense. Here the proof is very, very simple in two dimensions. So I'm just asking you to swallow that and believe that equidistribution result. It has the same flavor as equidistribution of horror cycles, just that now we now get equidistributed on a smaller sub-surface. And I'm happy to discuss this with you in detail how this works. In two dimensions, it's sort of half a page maybe to prove that. And now we are going to apply this philosophy in exactly the same way as we arrived to theorem 1 for the PQ statistics. Because now you see you can go through exactly the same argument as before, but now for P0, you can now put a 0 everywhere. And then instead of the measure, you have the sum or you have the number of points here of R in FQ intersect I. And then you have the points R here. And then here you need to divide by the number of points, FQ intersect I, and so on. You go through this. So now instead of this, you get your sum over R divided by the normalization. And now you get, instead, you get the other limit. You get the limit star, which I've written down here. And so the theorem 3 will follow exactly along the same lines, except that the limit distribution that we have is now not given by this formula, but by the following formula. And the fact that the limit measure is so much simpler will help you in the explicit computation of the whole distribution. So let me write it down. And I'm not going to write it down here. I'm going to write it down somewhere else. P0, yes, now it's P0. And now let me write the limit distribution here. So instead of R measure, I now have to use this measure. And so the nicest way of writing this is as an integral from over Y. And then the measure, so I'm splitting this integration up. I first do the X integration. So I have the measure of points X, so that the number of points squared 1X01, right? N, this N here, remember, is 1X01. Let me just write it explicitly. Intersect with a cylinder Cj is equal to K. And then I integrate over Y. And now you can see that to compute this measure here doesn't look too hard, because you are looking at the measure of points in the unit interval for which a certain condition is satisfied. And you can integrate that. You can do it, OK? And it's done in a short paper that I've written that's on the reading list that has multidimensional fairy sequences on there. And we'll do it for the case K equal to 0. I'll do it there. And J is equal to 0S, or actually 0S. Absolutely. I'm sorry, yes. Completely correct. And did I do it right there? Yes. Yeah, absolutely. So you can do this computation. You calculate this. And then you have the integral over Y left, which you can do. And what you find then is that in this special case, you get the following. So I'm going to introduce an auxiliary variable, high squared over 3 times S to the minus 1. And I get the whole distribution. So just recall, if I choose exactly this interval, this gives me the probability that the nearest neighbor to an element in my sequence is distance at least as a way. That's exactly what this probability gives you here, because you don't want to have a point in that interval. That's what you're conditioning on. Yeah, is that clear? OK, and so here is the whole distribution. So this now follows. And I hope you can believe that now from a computation. And it's a very nice explicit formula that was calculated by Hall in 1970 using a completely different, well, completely different is maybe overdoing it. But he didn't use this viewpoint, because if he had, he could have solved the problem in any dimension. Saying that, I have to admit that in higher dimensions, we'd have no explicit formulas for these kind of things. We only know tail estimates of those distributions. And that has to do with the fact that the geometry in higher dimensions is just so much more complicated. If you remember, even the Ziegels sets are already quite involved. OK, very nice. So that was about the first application, the local statistics of various points. And I hope I'll illustrate it to you that once we understand how to translate the original number theoretic problem into a problem of equidistribution on G mod gamma, it's just one technical step to get the limit. And we get a formula for the limit in terms of how measure or in terms of this measure. Now, in the remaining 10 minutes, I will talk about another beautiful classical problem. And that's the three-step theorem or Steinhaus conjecture. Oh, you are very, very good. You are very good. Looks bad, right? A one-half under the square root. No, where is it? Ah, OK. Where is it? It's a one-half. I have a one-half here, but it might be wrong. It's very possible that you're right, but I have it in my notes like that and maybe my notes are wrong. Have a look online in my paper. And then you can let us know by the end of the lecture and I can correct it. All right. So this is the second application, B3 gap theorem, also known as a Steinhaus conjecture, which was proved shortly after it was formulated in the 50s by a whole host of people, including Vera Sauss and others. And it's a very, very beautiful observation. So again, this is about gaps. Now we look at, again, the unit interval viewed as a circle with opposite points identified. And now we look at the sequence k alpha, mod 1. So we take the fractional parts of k alpha, where k runs through the integers. And the observation is that for every n, if you look at your sequence, so how will they look like while they will be equally spaced, OK. And then they'll wrap around and so on. Of course, one of our favorite dynamical systems. You look at the gaps that you can find. Do the same as before. The three-step theorem says that there are at most three distinct gaps that can appear here. And this statement has not only been proved immediately by four different people independently or something like that, but has been rediscovered in the last 50 years, again by probably many people and many different settings. And there's a paper, I think, by Krikoryan and Blank, I think, in a dynamical reformulation of this in terms of trajectories of the continuous translation on the torus, which all sort of falls into a similar argument. And there's a very nice paper by Slater on this problem, which is on the reading list. So have a look at this paper. And now I'm going to give you a new proof, if you want, new in using the space of lattices, where again this theorem will become more or less geometrically obvious. And we haven't done anything with this. People always ask for higher-dimensional generalizations of the Steinhaus problem. And there are many papers on this as well. And I think more or less there's probably already enough said, but what I'm going to show you gives a nice perspective on this problem, if I want to show you. Who wants me to show you? OK, 10. Oh, oh, OK. So, OK, let's start in the last sort of six minutes and then we'll see where we get to. Everything I say will already be suggestive of the answer. So I can stop as soon as people start collapsing. So we are interested in the gaps here of this sequence. So let us look at the gap of psi k to the right. So I'm writing it like this. I'm not writing psi k and psi k plus 1. Why? Because the psi k's are not ordered. And I really want to have the gap to the nearest psi, the guide to the right to psi k, which is not necessarily psi k plus 1. Because it might be a guy who's just wrapped around and has come closer. All right, so here's the formula. It's the minimum over this set. Gap should be positive. These guys are integers. L runs from 1 to n. And it shouldn't be equal to the k, the guy that you're looking at. So that's the formula for the gap to the right of psi k or to the right of k alpha, if you like. And now you can, of course, rewrite this as the minimum of m alpha. I'm just doing a substitution here where I replace l minus k by m. And you see I'm already writing it in a way using notation of lattices. So people used to do here, usually used rational approximations to alpha and continued fractions to do that. And now I can rewrite it in the following way. So I forgot one thing here. And then m has to be non-zero. Now you can think about that the minimum is actually achieved for such points where you can replace this condition that m is non-zero by both m and n being non-zero, because for the complement or the difference between those sets, the minimum will not be achieved. Little exercise if you can do this. So what I've done here is I've replaced a stronger condition by a weaker condition, removing less points, set is bigger. But the points that I'm admitting here in this larger set will never achieve the minimum. And then you can write this in the language of lattices, where the lattice has a special form. The lattice is of the form z squared m, where m is the matrix 1 alpha 0 1 1 over n 0 0 n. You can check that. And right, so we now have a formulation in terms of the gap size of the k-th element. k appears still here in terms of a special lattice. So now let's say, OK, now this is a special m. Let's extend it to general m, because we love working on the space of all lattices. So let me define this function. The minimum over the second component of the lattice vector that I'm looking at, always excluding 0. And let me, so I've introduced this new variable, t here, so that you get Sn, no, skn by simply setting, making these choices, OK? So again, this is the same trick as before. We have a problem, a discrete problem. We now introduce a function on the space of lattices. That this is a well-defined function on the space of lattices you have to check. You have to check that the minimum exists, and you have to check that this function is actually well defined on g mod gamma, which again it will be because gamma trivially on z squared stabilizes that squared. And now the three-gap theorem can be formulated in terms of this function. And the proof of that theorem is geometrically very, very apparent. So theorem, and that's the three-gap theorem, formulated now in terms of this function f. For every m in g fixed, fm, this function here is piecewise, constant, and takes at most three values. That implies the three-gap theorem because of this formula. And it's, of course, a little bit more general, because I'm not just making a statement over the rationales in t, but for all t. And the proof of this is one minute. It's one drawing, basically. I'm making it one minute. Took me about, well, you have to be, of course, careful. And I'm not careful. So when I send it to Andreas, Andreas is careful. And then you have to forget the right, you shouldn't forget some exceptional cases. But the picture is now clear. What is this function? So here we have a picture of R2. That's a picture of R2. Here's the origin. And now we look at this set. What is this set? Well, y has to be positive, right? This is the x-axis, and this is the y-axis, as usual. And x has to be between minus t and 1 minus t, where t ranges from 0 to 1. So we are ranging over this sort of, we're looking at points in here. So this is minus t, and this is 1 minus t. And this thing is sort of moving around. And what do you need to prove here is that the minimum of all lattice points, as you vary t, can only take three different, at most three distinct values. So how does the lattice look like in here? So here's the origin. So suppose here's your first lattice point R, with a minimum second component. There will be such a point. It could lie here, and ignore that for now. Now if this is the minimum, then of course it's also primitive lattice point. There can't be any point on that line. So there can't be a R over 2, can't be a lattice vector. Because I've assumed that the second component, R2, is the smallest of all lattice points in the big white domain A here. So now let's take the second one. Let's call it S. S has to be above here. And what you can also show is actually if R is on this side, if R1 is positive, then S1 has to be negative. So this is the second lattice vector. And again, it has to be primitive because what did I do wrong? I can't hear you. No, no, no, no. Forget the pink region for now. Forget the pink region. I'm just talking about the white region. Just taking the smallest and the second smallest in the white region. OK? So now this is primitive. This is primitive. A little thought shows that then these have to form a basis for your lattice. Because why? Well, there can't be any point in this rectangle. Because if there was, it would have a smaller second component than S. But if there's no point in this closed rectangle here, there also can't be a point you can convince yourself in the parallelogram. So if you have a parallelogram that's empty of lattice points, it means that the points at the vertices have to form a basis of your lattice. So here would be the point R plus S. And then you can work out that the only points that can contribute to this minimum are S, R, and R plus S. So the second components of R plus S, R, and S are exactly the three values that you can get. End of proof. And I thank you.