WEBVTT 00:00:19.142 --> 00:00:20.600 PROFESSOR: So we've been discussing 00:00:20.600 --> 00:00:23.900 Szemeredi's regularity level for the past couple of lectures. 00:00:23.900 --> 00:00:25.580 And one of the theorems that we proved 00:00:25.580 --> 00:00:30.410 was Roth's theorem, which tells us how large can a subset of 1 00:00:30.410 --> 00:00:35.220 through N be if it has no 3-term arithmetic progressions. 00:00:35.220 --> 00:00:37.130 And I mentioned at the end of last time 00:00:37.130 --> 00:00:41.780 that we can construct fairly large subsets of 1 00:00:41.780 --> 00:00:44.990 through N without 3-term arithmetic progressions. 00:00:44.990 --> 00:00:48.740 I want to begin today's lecture with showing you that example, 00:00:48.740 --> 00:00:51.140 showing you that construction. 00:00:51.140 --> 00:00:55.250 But first, let's try to do something 00:00:55.250 --> 00:00:59.580 that is somewhat more naive, somewhat more straightforward, 00:00:59.580 --> 00:01:14.240 to try to just construct greedily 3-AP-free sets of 1 00:01:14.240 --> 00:01:15.830 through n. 00:01:15.830 --> 00:01:19.970 And recall from last time, we showed Roth's theorem 00:01:19.970 --> 00:01:27.740 that such a set must have size little o of N. 00:01:27.740 --> 00:01:29.570 So what's one thing you might do? 00:01:29.570 --> 00:01:33.533 Well, you can try to greedily construct such a set. 00:01:33.533 --> 00:01:35.450 It will just make our life a little bit easier 00:01:35.450 --> 00:01:37.740 if we start with 0 instead of 1. 00:01:37.740 --> 00:01:40.310 So you put one element in, and you 00:01:40.310 --> 00:01:42.200 keep putting in the next integer, 00:01:42.200 --> 00:01:46.500 as long as it doesn't create a 3-term arithmetic progression. 00:01:46.500 --> 00:01:47.820 So you keep doing this. 00:01:47.820 --> 00:01:51.140 Well, you can't put 2 in, so let's skip 2. 00:01:51.140 --> 00:01:53.654 So we go to the next one, 3. 00:01:53.654 --> 00:01:55.340 So 4 is OK. 00:01:57.870 --> 00:02:00.180 So we skip 5. 00:02:00.180 --> 00:02:05.260 We have to skip 6 as well because 0, 3, 6, that's a 3-AP. 00:02:05.260 --> 00:02:06.000 So we keep going. 00:02:06.000 --> 00:02:08.904 So what's the next number we can put in? 00:02:08.904 --> 00:02:11.257 AUDIENCE: 9. 00:02:11.257 --> 00:02:11.840 PROFESSOR: Up. 00:02:11.840 --> 00:02:12.610 Go to 9. 00:02:15.980 --> 00:02:19.010 Then the next one is 10. 00:02:19.010 --> 00:02:21.888 What's the next one we can put in? 00:02:21.888 --> 00:02:22.930 So we can play this game. 00:02:25.460 --> 00:02:30.490 Find out what is the next number that you can put in. 00:02:30.490 --> 00:02:32.360 Greedily, if you could put it in, 00:02:32.360 --> 00:02:37.460 put it in in a way that generates a subset of integers 00:02:37.460 --> 00:02:41.530 without 3-term arithmetic progressions. 00:02:41.530 --> 00:02:43.867 So this actually has a name, so this 00:02:43.867 --> 00:02:45.075 is called a Stanley sequence. 00:02:51.540 --> 00:02:55.530 And there's an easier way to see what the sequence is. 00:02:55.530 --> 00:03:01.930 Namely, if you write the sequence in base 3, 00:03:01.930 --> 00:03:09.758 you find that these numbers are 0; 0, 1; 1, 0; 1, 1. 00:03:12.370 --> 00:03:13.060 1, 0, 0. 00:03:13.060 --> 00:03:14.160 1, 0, 1. 00:03:14.160 --> 00:03:14.950 1, 1, 0. 00:03:14.950 --> 00:03:16.980 1, 1, 1. 00:03:16.980 --> 00:03:25.460 So these are just numbers whose base 3 representation 00:03:25.460 --> 00:03:30.050 consists of zeros and ones. 00:03:30.050 --> 00:03:32.600 So I'll leave it to you as an exercise 00:03:32.600 --> 00:03:34.830 to figure out why this is the case, if you generate 00:03:34.830 --> 00:03:38.570 the sequence greedily this way, this is exactly the sequence 00:03:38.570 --> 00:03:41.140 that you obtain. 00:03:41.140 --> 00:03:44.040 But once you know that, it's not too hard to find out 00:03:44.040 --> 00:03:46.300 how many numbers you generate. 00:03:49.240 --> 00:03:57.440 So up to-- suppose N is equal to 3 to the k. 00:03:57.440 --> 00:04:08.500 We get 2 to the k terms, which gives you 00:04:08.500 --> 00:04:12.450 N raised to the power of log base 3 of 2. 00:04:14.852 --> 00:04:17.310 So you can figure out what that number is, but some numbers 00:04:17.310 --> 00:04:18.279 strictly less than 1. 00:04:28.810 --> 00:04:30.540 And actually, for a very long time, 00:04:30.540 --> 00:04:32.830 people thought this was the best construction. 00:04:32.830 --> 00:04:35.620 So this construction was known even before Stanley, 00:04:35.620 --> 00:04:39.630 so it was something that is very natural to come up with. 00:04:39.630 --> 00:04:45.658 And it was somewhat of a surprise when in the '40s, 00:04:45.658 --> 00:04:47.200 it was discovered that you can create 00:04:47.200 --> 00:04:50.380 much larger subsets of the positive integers 00:04:50.380 --> 00:04:52.275 without arithmetic progressions. 00:04:52.275 --> 00:04:54.400 So that's the first thing I want to show you today. 00:04:58.470 --> 00:05:06.180 So these sets were first discovered by Salem and Spencer 00:05:06.180 --> 00:05:08.310 back in the '40s. 00:05:08.310 --> 00:05:17.080 And a few years later, Behrend gave a somewhat improved 00:05:17.080 --> 00:05:19.820 and simplified version of the Salem-Spencer construction. 00:05:19.820 --> 00:05:24.190 So these were all back in the '40s. 00:05:24.190 --> 00:05:26.160 So these days, we usually refer, at least 00:05:26.160 --> 00:05:27.910 in the additive combinatorics community, 00:05:27.910 --> 00:05:30.665 to this construction as Behrend's construction. 00:05:30.665 --> 00:05:33.040 And this, indeed, what I will show you is due to Behrend, 00:05:33.040 --> 00:05:36.150 but somehow, this Salem-Spencer name has been forgotten 00:05:36.150 --> 00:05:38.650 and I just wanted to point out that it 00:05:38.650 --> 00:05:45.640 was Salem and Spencer who first demonstrated that there exists 00:05:45.640 --> 00:05:50.590 a subset of 1 through N that is 3-AP-free 00:05:50.590 --> 00:05:56.710 and has size N to the 1 minus little o 1, that there's 00:05:56.710 --> 00:05:58.330 no power saving. 00:05:58.330 --> 00:06:01.100 So there exists examples with no power saving. 00:06:01.100 --> 00:06:04.240 And this is important because as we 00:06:04.240 --> 00:06:06.150 saw last time, the proof of Roth's theorem-- 00:06:06.150 --> 00:06:08.800 and we basically spent two full lectures 00:06:08.800 --> 00:06:11.020 proving Roth's theorem. 00:06:11.020 --> 00:06:13.010 And it's somewhat involved, right. 00:06:13.010 --> 00:06:15.340 It wasn't this one line of inequality 00:06:15.340 --> 00:06:18.620 you can just use to deduce the result. 00:06:18.620 --> 00:06:21.700 And that is part of the difficulty. 00:06:21.700 --> 00:06:24.640 So having such an example is indication 00:06:24.640 --> 00:06:28.350 that Roth's theorem should not be so easy. 00:06:28.350 --> 00:06:30.130 That's not a rigorous demonstration, 00:06:30.130 --> 00:06:32.910 but it's some indication of the difficulty of Roth's theorem. 00:06:36.430 --> 00:06:39.355 So let's see this construction due to Behrend. 00:06:47.890 --> 00:06:50.190 So let me write down the precise statement. 00:06:50.190 --> 00:06:56.760 There exists a constant, C, such that there exists a subset of 1 00:06:56.760 --> 00:07:07.070 through N that is 3-AP-free and has size at least N times e 00:07:07.070 --> 00:07:17.320 to the minus C root log N. 00:07:17.320 --> 00:07:19.870 Perhaps somewhat amazingly enough, this bound 00:07:19.870 --> 00:07:23.290 is still the current best known. 00:07:23.290 --> 00:07:25.920 We do not know any constructions that 00:07:25.920 --> 00:07:29.040 is essentially better than Behrend's construction 00:07:29.040 --> 00:07:30.870 from the '40s. 00:07:30.870 --> 00:07:33.180 There were some small recent improvements 00:07:33.180 --> 00:07:39.983 that clarified what the C could be, but in this form, 00:07:39.983 --> 00:07:41.650 we do not know anything better than what 00:07:41.650 --> 00:07:43.050 was known back in the '40s. 00:07:43.050 --> 00:07:46.020 And the construction, I will show yow-- 00:07:46.020 --> 00:07:48.387 hopefully you should believe that it's quite simple. 00:07:48.387 --> 00:07:50.220 So I will show you what the construction is. 00:07:50.220 --> 00:07:52.980 It's clever, but it's not complicated. 00:07:52.980 --> 00:07:56.350 And it's a really interesting direction to figure out 00:07:56.350 --> 00:08:00.330 is this really the best construction out there. 00:08:00.330 --> 00:08:01.080 Can you do better? 00:08:04.358 --> 00:08:04.900 So let's see. 00:08:08.590 --> 00:08:12.100 We're going to set some parameters to be decided later, 00:08:12.100 --> 00:08:14.200 m and d. 00:08:14.200 --> 00:08:22.510 Let me consider x to be this discrete box in d dimensions. 00:08:22.510 --> 00:08:27.850 So this is the box of lattice points in d dimensions, 1 00:08:27.850 --> 00:08:30.790 through m raised to d. 00:08:30.790 --> 00:08:34.210 And let me consider an intersection 00:08:34.210 --> 00:08:44.640 of this box by a sphere of radius root L. 00:08:44.640 --> 00:08:49.190 So namely, we look at points in this x, 00:08:49.190 --> 00:08:53.550 such that the sum of their squares 00:08:53.550 --> 00:08:59.890 is equal to exactly L. You take a bunch of these spheres, 00:08:59.890 --> 00:09:12.040 they partition your set x, so the smallest possible sum 00:09:12.040 --> 00:09:16.180 of squares, largest possible sum of squares. 00:09:16.180 --> 00:09:22.440 So in particular, there exists some L such that x of L 00:09:22.440 --> 00:09:33.250 is large, just by pigeonhole principle. 00:09:33.250 --> 00:09:36.790 You can probably do this step with a bit more finesse, 00:09:36.790 --> 00:09:40.132 but it's not going to change any of the asymptotics. 00:09:43.366 --> 00:09:46.340 The intuition here-- and we'll come back to this in a second-- 00:09:46.340 --> 00:09:49.950 is that xL lies on a sphere. 00:09:49.950 --> 00:09:51.480 So this is the set of lattice points 00:09:51.480 --> 00:09:53.070 that lies in a given sphere. 00:09:53.070 --> 00:09:57.180 And because you are looking at a sphere, it has no 3-term APs. 00:09:59.580 --> 00:10:01.080 So we're going to use that property, 00:10:01.080 --> 00:10:04.570 but right now it's not yet a subset of the integers. 00:10:04.570 --> 00:10:09.510 So what we're going to do is to take this section of a sphere 00:10:09.510 --> 00:10:14.460 and then project it to one dimension, 00:10:14.460 --> 00:10:17.770 so that it is now going to be a subset of integers. 00:10:17.770 --> 00:10:19.440 And we're going to do this in such a way 00:10:19.440 --> 00:10:25.104 that it does not affect the presence of 3-term APs. 00:10:25.104 --> 00:10:33.090 So let us map x to the integers by base expansion. 00:10:43.290 --> 00:10:47.660 So this is base 2m expansion. 00:10:57.580 --> 00:10:58.080 All right. 00:10:58.080 --> 00:11:01.356 So what's the point of this construction here so far? 00:11:01.356 --> 00:11:05.870 Well, you verified a couple of properties, the first being 00:11:05.870 --> 00:11:13.550 that this construction here is injective, this map. 00:11:13.550 --> 00:11:20.310 So if you call this map phi, this phi is injective. 00:11:20.310 --> 00:11:24.480 Well, it's base expansion, so it's injective. 00:11:24.480 --> 00:11:28.680 But also a somewhat stronger claim 00:11:28.680 --> 00:11:31.230 is that if you have three points-- 00:11:36.630 --> 00:11:40.050 so if you have three points in x, 00:11:40.050 --> 00:11:47.770 that map to a 3-AP in the integers, 00:11:47.770 --> 00:11:58.750 then the three points originally must be a 3-AP in x. 00:12:02.494 --> 00:12:04.020 Again, this is not a hard claim. 00:12:04.020 --> 00:12:05.670 Just think about it. 00:12:05.670 --> 00:12:08.833 Here, because we're using base 2m expansion 00:12:08.833 --> 00:12:10.750 and you're only allowed to use digits up to m, 00:12:10.750 --> 00:12:12.190 you don't have any wrap around effects. 00:12:12.190 --> 00:12:14.482 You don't have any carryovers when you do the addition. 00:12:17.760 --> 00:12:26.800 So combining these two observations, 00:12:26.800 --> 00:12:43.130 we find that the image of xL is a 3-AP-free set off 1 00:12:43.130 --> 00:12:48.430 through N. So what is N? 00:12:48.430 --> 00:12:53.430 I can take N to be, for instance, 2m raised to power d, 00:12:53.430 --> 00:12:57.710 so all the numbers up there are less than this quantity. 00:12:57.710 --> 00:13:04.400 And the size is the size of x sub 00:13:04.400 --> 00:13:11.512 L, which is N to the d divided by d m squared. 00:13:11.512 --> 00:13:13.720 And now you just need to find the appropriate choices 00:13:13.720 --> 00:13:19.430 of the parameters m and d to maximize the size of x sub L. 00:13:19.430 --> 00:13:23.440 And that's an exercise. 00:13:23.440 --> 00:13:29.690 So, for instance, if you take m to be e to the root log 00:13:29.690 --> 00:13:37.880 N and d to be root log N, then you 00:13:37.880 --> 00:13:42.980 find that the size here is the bound that we claim. 00:13:48.570 --> 00:13:54.920 And that finishes the construction of Behrend, 00:13:54.920 --> 00:13:57.960 giving you a fairly large subset of 1 00:13:57.960 --> 00:14:01.480 through N without 3-term arithmetic progressions. 00:14:01.480 --> 00:14:04.640 And the idea here is you look at a higher dimensional 00:14:04.640 --> 00:14:07.490 object, namely, a higher dimensional sphere which 00:14:07.490 --> 00:14:09.380 has this property of being 3-AP-free, 00:14:09.380 --> 00:14:13.185 and then you project it onto the integers. 00:14:13.185 --> 00:14:14.060 Any questions so far? 00:14:20.850 --> 00:14:21.610 OK. 00:14:21.610 --> 00:14:24.050 So we have some proofs, we have some examples. 00:14:24.050 --> 00:14:27.960 So now let's go on to variations of Roth's theorem. 00:14:27.960 --> 00:14:31.570 And I want to show you a higher dimensional 00:14:31.570 --> 00:14:33.750 version of Roth's theorem. 00:14:33.750 --> 00:14:35.500 So I mentioned in the very first lecture-- 00:14:35.500 --> 00:14:38.110 so you have this whole host of theorems and additive 00:14:38.110 --> 00:14:38.880 combinatorics-- 00:14:38.880 --> 00:14:42.880 Roth, Szemeredi, multi-dimensional Szemeredi. 00:14:42.880 --> 00:14:45.640 So this is, in some sense, the simplest example 00:14:45.640 --> 00:14:48.760 of multi-dimensional Szemeredi theorem. 00:14:48.760 --> 00:14:50.770 And this example is known as corners. 00:14:53.330 --> 00:14:54.370 So what's a corner? 00:14:54.370 --> 00:14:58.660 A corner is, well, we're working inside two dimensions, 00:14:58.660 --> 00:15:04.700 so a corner is simply three points, such that the two-- 00:15:04.700 --> 00:15:07.990 so they're positioned like that-- such that these two 00:15:07.990 --> 00:15:11.132 segments, they are parallel to the axes 00:15:11.132 --> 00:15:12.340 and they are the same length. 00:15:12.340 --> 00:15:14.890 So that's, by definition, what a corner is. 00:15:14.890 --> 00:15:16.810 And the question is, if you give me 00:15:16.810 --> 00:15:19.840 a subset of a grid that is corner-free, 00:15:19.840 --> 00:15:21.862 how large can this subset be? 00:15:25.478 --> 00:15:27.170 Here's a theorem. 00:15:27.170 --> 00:15:38.650 So if A is a subset of 1 through N with no corners, 00:15:38.650 --> 00:15:44.320 particular, no three points of the form, x comma y; 00:15:44.320 --> 00:15:48.680 x plus d comma y; and x comma y plus d, 00:15:48.680 --> 00:15:56.780 where d is some positive integer, then the size of A 00:15:56.780 --> 00:15:59.030 has to be little o of N squared. 00:16:02.840 --> 00:16:03.340 Question. 00:16:03.340 --> 00:16:04.965 AUDIENCE: Do we only care about corners 00:16:04.965 --> 00:16:06.260 oriented in that direction? 00:16:06.260 --> 00:16:07.120 PROFESSOR: OK, good question. 00:16:07.120 --> 00:16:08.495 So that's one of the first things 00:16:08.495 --> 00:16:10.150 we will address in this proof. 00:16:10.150 --> 00:16:12.880 So your question was, do we only care about corners oriented 00:16:12.880 --> 00:16:14.510 in the positive direction. 00:16:14.510 --> 00:16:18.300 So you can have a more relaxed version of this problem 00:16:18.300 --> 00:16:22.150 where you allow d to be negative as well. 00:16:22.150 --> 00:16:23.650 The first step in the proof is we'll 00:16:23.650 --> 00:16:26.806 see that that constraint doesn't actually matter. 00:16:26.806 --> 00:16:27.306 Yes. 00:16:32.790 --> 00:16:33.350 OK. 00:16:33.350 --> 00:16:33.850 Great. 00:16:37.360 --> 00:16:38.110 Let's get started. 00:16:40.640 --> 00:16:47.070 So as Michael mentioned, we do have this constraint in this 00:16:47.070 --> 00:16:48.950 over here that d is positive. 00:16:48.950 --> 00:16:52.680 And it's somewhat annoying because if you remember 00:16:52.680 --> 00:16:56.160 in our proof of Roth's theorem, they are positive, negative, 00:16:56.160 --> 00:16:57.810 they don't play a role. 00:16:57.810 --> 00:17:01.820 So let's try to find a way to get rid of this constraint 00:17:01.820 --> 00:17:05.650 so that we are in a more flexible situation. 00:17:05.650 --> 00:17:07.510 So first step is that we'll get rid 00:17:07.510 --> 00:17:14.190 of this d being positive requirement. 00:17:17.480 --> 00:17:20.119 So here's a trick. 00:17:20.119 --> 00:17:24.160 Let's consider the sumset A plus A. So sumset 00:17:24.160 --> 00:17:31.450 here means I'm looking at all pairwise sums as a set. 00:17:31.450 --> 00:17:35.860 So you don't keep track of duplicates, 00:17:35.860 --> 00:17:37.720 so you'll keep it as a set. 00:17:37.720 --> 00:17:40.930 And we're living inside this grid, 00:17:40.930 --> 00:17:43.600 but now somewhat wider in width. 00:17:49.930 --> 00:18:01.670 Then there exists an element in this domain of the sumset. 00:18:01.670 --> 00:18:03.350 By pigeonhole, that's represented 00:18:03.350 --> 00:18:05.900 in many different ways. 00:18:05.900 --> 00:18:11.870 So there exists a z represented as a plus b 00:18:11.870 --> 00:18:19.290 in at least size of A squared divided 00:18:19.290 --> 00:18:26.640 by size of 2N squared different ways, 00:18:26.640 --> 00:18:29.190 just because if you look at how many things come up 00:18:29.190 --> 00:18:34.086 in that representation, just use pigeonhole. 00:18:34.086 --> 00:18:50.033 And now let's take A prime to be A intersect z minus A. So 00:18:50.033 --> 00:18:50.950 what's happening here? 00:18:50.950 --> 00:18:55.870 So you have this set A. And basically what 00:18:55.870 --> 00:18:58.660 I want to do is look at the-- 00:18:58.660 --> 00:19:00.130 so suppose A looks like that. 00:19:00.130 --> 00:19:04.900 So look at minus A and then shift minus A 00:19:04.900 --> 00:19:10.420 over so that they intersect in as many elements as you can. 00:19:10.420 --> 00:19:12.400 And by pigeonhole, I can guarantee 00:19:12.400 --> 00:19:14.820 that their intersection is fairly large. 00:19:21.130 --> 00:19:24.570 So because the size of A prime, it's 00:19:24.570 --> 00:19:27.600 essentially the number of ways that z can be represented 00:19:27.600 --> 00:19:31.450 in the aforementioned manner. 00:19:31.450 --> 00:19:44.303 So it suffices to show that A prime is little o of N squared. 00:19:44.303 --> 00:19:45.970 If you show that, then you automatically 00:19:45.970 --> 00:19:48.920 show A is little o of N squared. 00:19:48.920 --> 00:19:51.820 But now A prime is symmetric. 00:19:51.820 --> 00:19:54.460 A prime is symmetric around 0 over 2. 00:19:59.290 --> 00:20:05.370 So this is centrally symmetric about z over 2, 00:20:05.370 --> 00:20:09.450 meaning that A prime equals to z minus A prime. 00:20:12.280 --> 00:20:13.900 And so you see now we've gotten rid 00:20:13.900 --> 00:20:19.090 of this d positive requirement because if A prime had 00:20:19.090 --> 00:20:23.390 a positive corner, then it has a negative corner and vice versa. 00:20:27.647 --> 00:20:35.520 So no corner in A prime with d positive 00:20:35.520 --> 00:20:44.520 implies that no corner with d negative. 00:20:44.520 --> 00:20:46.170 So now let's forget about A and A prime 00:20:46.170 --> 00:20:50.640 and just replace A by A prime and forget 00:20:50.640 --> 00:20:53.890 about this d positive condition. 00:20:56.650 --> 00:21:03.143 So let's forget about this part, but I do want d not equal to 0. 00:21:03.143 --> 00:21:05.310 Otherwise, you have trivial corners, and, of course, 00:21:05.310 --> 00:21:07.310 you always have trivial corners. 00:21:07.310 --> 00:21:07.810 All right. 00:21:15.120 --> 00:21:20.080 So let's remember how the proof of Roth's theorem went. 00:21:20.080 --> 00:21:22.610 And this relates to the very first lecture where 00:21:22.610 --> 00:21:26.480 I showed you this connection between additive combinatorics 00:21:26.480 --> 00:21:29.440 on one hand and graph theory on the other hand, 00:21:29.440 --> 00:21:31.930 or you take some arithmetic pattern 00:21:31.930 --> 00:21:34.840 and you try to encode it in a graph 00:21:34.840 --> 00:21:37.240 so that the patterns in your arithmetic 00:21:37.240 --> 00:21:43.160 set correspond to patterns in the graph. 00:21:43.160 --> 00:21:45.840 So we're going to do the same thing here. 00:21:45.840 --> 00:21:50.000 So we're going to encode the subset of the grid 00:21:50.000 --> 00:21:53.990 as a tripartite graph in such a way 00:21:53.990 --> 00:21:58.520 that the corners correspond to triangles in the graph. 00:21:58.520 --> 00:22:03.365 So I'll show you how to do this and it will be fairly simple. 00:22:03.365 --> 00:22:07.557 And in general, sometimes it takes a little bit of ingenuity 00:22:07.557 --> 00:22:09.890 to figure out what is the right way to set up the graph. 00:22:09.890 --> 00:22:12.920 So one of an upcoming homework problem 00:22:12.920 --> 00:22:14.630 will be for you to figure out how 00:22:14.630 --> 00:22:18.710 to set up a corresponding graph when the pattern is not 00:22:18.710 --> 00:22:20.690 a corner, but a square. 00:22:20.690 --> 00:22:24.260 If I add one extra point up there, how would you do that? 00:22:27.470 --> 00:22:29.690 So what does this graph look like? 00:22:29.690 --> 00:22:41.784 Let's build a tripartite graph where-- 00:22:41.784 --> 00:22:43.570 so this should be somewhat reminiscent 00:22:43.570 --> 00:22:48.120 of the proof of Roth's theorem-- 00:22:48.120 --> 00:22:56.240 where I give you three sets, x, y, and z; and x and y 00:22:56.240 --> 00:22:59.600 are both going to have N elements 00:22:59.600 --> 00:23:03.355 and z is now going to have 2N elements. 00:23:08.720 --> 00:23:13.000 x is supposed to enumerate or index 00:23:13.000 --> 00:23:19.680 all the vertical lines in your grid. 00:23:19.680 --> 00:23:30.130 So you have this grid of N by N. So x has size 4. 00:23:30.130 --> 00:23:33.160 Each vertex in x corresponds to a vertical line. 00:23:38.450 --> 00:23:48.350 y corresponds to the horizontal lines and z corresponds 00:23:48.350 --> 00:23:52.220 to these negatively sloped diagonal lines-- 00:23:52.220 --> 00:23:55.713 slope minus 1 lines. 00:23:55.713 --> 00:23:57.380 And of course you should only take lines 00:23:57.380 --> 00:24:01.250 that affect your N by N grid. 00:24:01.250 --> 00:24:03.440 So that's why there are this many of them 00:24:03.440 --> 00:24:06.080 for each direction. 00:24:06.080 --> 00:24:07.760 So what's the graph? 00:24:07.760 --> 00:24:21.685 I join two vertices if their corresponding lines 00:24:21.685 --> 00:24:39.055 meet in a point of A. So I might have-- 00:24:45.320 --> 00:24:50.060 this may be A, a point of A. So I 00:24:50.060 --> 00:24:53.570 put an edge between those two lines-- 00:24:53.570 --> 00:24:56.300 one for x, one for z-- 00:24:56.300 --> 00:25:02.075 because their intersection lies in A. And more explicitly-- 00:25:02.075 --> 00:25:03.450 I mean, that is pretty explicit-- 00:25:03.450 --> 00:25:06.000 and alternatively, you could also 00:25:06.000 --> 00:25:12.465 write this graph by telling us how to put 00:25:12.465 --> 00:25:15.540 in the edge between x and z. 00:25:15.540 --> 00:25:22.280 Namely, you do this if x comma z minus x lies in A; 00:25:22.280 --> 00:25:28.960 you put an edge between x and y if x comma y lies in A; 00:25:28.960 --> 00:25:31.710 and likewise, you put in the final edge 00:25:31.710 --> 00:25:37.680 if z minus y comma y lies in A. 00:25:37.680 --> 00:25:39.760 So two equivalent descriptions of the same graph. 00:25:44.050 --> 00:25:46.900 Any questions? 00:25:46.900 --> 00:25:47.498 All right. 00:25:47.498 --> 00:25:49.540 So the rest of the proof is more or less the same 00:25:49.540 --> 00:25:52.940 as the proof of Roth's theorem we saw last time. 00:25:52.940 --> 00:25:55.930 So we need to figure out how many edges are there. 00:26:00.710 --> 00:26:05.840 Well, every element of A gives you three edges. 00:26:10.510 --> 00:26:12.990 So that's one of the edges corresponding to that element. 00:26:12.990 --> 00:26:17.060 The other two pairs with two other edges. 00:26:17.060 --> 00:26:24.530 And most importantly, the triangles in this graph, 00:26:24.530 --> 00:26:28.435 I claim, correspond to corners. 00:26:31.890 --> 00:26:33.680 So what is a triangle? 00:26:33.680 --> 00:26:36.790 A triangle corresponds to a horizontal line, 00:26:36.790 --> 00:26:40.660 a vertical line, and a slope 1 minus line 00:26:40.660 --> 00:26:44.520 that pairwise intersect in elements of A. 00:26:44.520 --> 00:26:47.730 If you look at their intersections, that's a corner. 00:26:47.730 --> 00:26:51.550 And conversely, if you have a corner, you build a triangle. 00:26:51.550 --> 00:26:59.040 And because your graph, your set is corner-free, well, 00:26:59.040 --> 00:27:00.390 you don't have any triangles. 00:27:00.390 --> 00:27:02.340 Actually, no, that's not true. 00:27:02.340 --> 00:27:05.670 So like we saw in Roth's theorem, 00:27:05.670 --> 00:27:07.860 you have some triangles, but corresponding 00:27:07.860 --> 00:27:10.410 to trivial corners. 00:27:10.410 --> 00:27:18.880 So triangles correspond to trivial corners 00:27:18.880 --> 00:27:24.160 because your set A is corner-free. 00:27:24.160 --> 00:27:28.840 Trivial corners, meaning the same point with d close to 0, 00:27:28.840 --> 00:27:34.770 so it's not a genuine corner like that. 00:27:34.770 --> 00:27:39.050 And in particular, in this graph, 00:27:39.050 --> 00:27:46.515 every edge is in a unique triangle. 00:27:55.412 --> 00:27:56.870 And so we're in the same situation. 00:27:56.870 --> 00:28:02.900 By the corollary of triangle removal lemma, 00:28:02.900 --> 00:28:11.350 we find that the number of edges must 00:28:11.350 --> 00:28:17.060 be little o of the number of vertices. 00:28:20.930 --> 00:28:23.400 The number of edges must be subquadratic, 00:28:23.400 --> 00:28:27.380 so it must be little o of N squared. 00:28:27.380 --> 00:28:30.360 And so that implies that the size of A 00:28:30.360 --> 00:28:32.310 is little o of N squared. 00:28:35.256 --> 00:28:39.300 So that proves the Corners theorem. 00:28:39.300 --> 00:28:40.381 Any questions? 00:28:44.530 --> 00:28:46.270 So once you set up this graph, the rest 00:28:46.270 --> 00:28:47.540 is the same as Roth's theorem. 00:28:47.540 --> 00:28:49.840 So this connection between graph theory and additive 00:28:49.840 --> 00:28:52.210 combinatorics, well, you need to figure out 00:28:52.210 --> 00:28:53.390 how to set up this graph. 00:28:53.390 --> 00:28:57.760 And then sometimes it's clear, but sometimes you 00:28:57.760 --> 00:29:00.760 have to really think hard about how to do this. 00:29:04.000 --> 00:29:05.080 All right. 00:29:05.080 --> 00:29:08.680 What does corner have to do with Roth's theorem, 00:29:08.680 --> 00:29:12.415 other than that they have very similar looking proofs? 00:29:12.415 --> 00:29:14.290 Well, actually, you can deduce Roth's theorem 00:29:14.290 --> 00:29:15.550 from the Corners theorem. 00:29:18.990 --> 00:29:24.070 And to show you this precisely, so let me use r sub 3 of N 00:29:24.070 --> 00:29:37.950 to denote the size of the largest 3-AP-free set of 1 00:29:37.950 --> 00:29:41.520 through N. 00:29:41.520 --> 00:29:44.790 So this notation, the r sub 3 is actually fairly standard. 00:29:44.790 --> 00:29:47.443 The next one's not so standard, but let's just do that. 00:29:47.443 --> 00:29:49.110 So that's not an L, but that's a corner. 00:29:51.690 --> 00:29:58.410 So that is the size of the largest corner-free subset 00:29:58.410 --> 00:30:02.940 of N squared. 00:30:02.940 --> 00:30:06.890 So we gave bounds for both quantities, 00:30:06.890 --> 00:30:11.780 but they are actually related to each other through this fairly 00:30:11.780 --> 00:30:14.300 simple proposition that if you have 00:30:14.300 --> 00:30:16.370 an upper bound for corners, then you 00:30:16.370 --> 00:30:20.497 have an upper bound for Roth's theorem. 00:30:25.270 --> 00:30:39.430 Indeed, given A a 3-AP-free subset of 1 through N, 00:30:39.430 --> 00:30:42.720 let me build for you a corner-free subset 00:30:42.720 --> 00:30:47.430 of the grid that is a fairly large subset of that grid. 00:30:50.220 --> 00:31:04.220 I can form B by setting it to be the set of pairs 00:31:04.220 --> 00:31:09.930 inside this grid whose difference, x minus y, 00:31:09.930 --> 00:31:16.530 lies in A. So what does this look like? 00:31:23.910 --> 00:31:28.450 This is the grid of size 2N. 00:31:28.450 --> 00:31:37.590 And if I start with A that is 3-AP-free, what I can do-- 00:31:37.590 --> 00:31:38.880 so over here-- 00:31:38.880 --> 00:31:52.260 is look at the lines like that, putting all of those points. 00:31:52.260 --> 00:31:54.060 And you see that this set of points 00:31:54.060 --> 00:32:00.590 should not have any corners because if they have corners, 00:32:00.590 --> 00:32:02.920 then the corners would project down to a 3-AP. 00:32:06.740 --> 00:32:08.386 So B is corner-free. 00:32:16.970 --> 00:32:19.680 To recap, if we have upper bound on corners, 00:32:19.680 --> 00:32:22.630 we have upper bound on Roth's theorem. 00:32:22.630 --> 00:32:25.425 But you also know that if you have lower 00:32:25.425 --> 00:32:26.800 bound on Roth's theorem, then you 00:32:26.800 --> 00:32:28.420 have lower bound on corners. 00:32:28.420 --> 00:32:31.210 So the Behrend construction we saw at the beginning of today 00:32:31.210 --> 00:32:34.540 extends to, you know, through exactly this way to a fairly 00:32:34.540 --> 00:32:35.980 large corner-free subset. 00:32:35.980 --> 00:32:38.920 And that's more or less the best thing that we know how to do. 00:32:38.920 --> 00:32:41.760 In fact, there aren't that many constructions 00:32:41.760 --> 00:32:43.990 in additive combinatorics that's known. 00:32:43.990 --> 00:32:45.430 Almost everything that I know how 00:32:45.430 --> 00:32:47.830 to construct that's fairly large come 00:32:47.830 --> 00:32:50.160 from Behrend's construction or some variant 00:32:50.160 --> 00:32:51.850 of Behrend's construction. 00:32:51.850 --> 00:32:53.170 So it looks pretty simple. 00:32:53.170 --> 00:32:55.060 It comes from the '40s, yet we don't really 00:32:55.060 --> 00:32:59.110 have too many new ideas besides playing and massaging 00:32:59.110 --> 00:33:00.520 Behrend's construction. 00:33:04.360 --> 00:33:07.180 Let me tell you what is the best known upper bound 00:33:07.180 --> 00:33:08.680 on the Corners theorem. 00:33:14.440 --> 00:33:15.440 This is due to Shkredov. 00:33:18.430 --> 00:33:22.185 So the proof using triangle removal lemma, 00:33:22.185 --> 00:33:25.110 it goes to Szemeredi's regularity lemma. 00:33:25.110 --> 00:33:26.760 It gives you pretty horrible bounds, 00:33:26.760 --> 00:33:31.070 but using Fourier analytic methods-- 00:33:31.070 --> 00:33:33.010 so you see if you have upper bound on Roth, 00:33:33.010 --> 00:33:35.200 it doesn't give you an upper bound on corner, 00:33:35.200 --> 00:33:37.700 so you need to do something extra. 00:33:37.700 --> 00:33:43.330 And so the best known bound so far is of the form, 00:33:43.330 --> 00:33:47.825 N squared divided by polylog log N, 00:33:47.825 --> 00:33:50.305 so log log N raised to some small constant, 00:33:50.305 --> 00:33:53.748 C. Any questions? 00:34:02.050 --> 00:34:04.730 Last time, we discussed the triangle counting lemma 00:34:04.730 --> 00:34:06.070 and the triangle removal lemma. 00:34:08.870 --> 00:34:11.020 Well, it shouldn't be a surprise to you 00:34:11.020 --> 00:34:12.480 that if we can do it for triangles, 00:34:12.480 --> 00:34:15.929 we may be able to do it for other subgraphs. 00:34:15.929 --> 00:34:17.790 So that's the next thing I want to discuss-- 00:34:17.790 --> 00:34:22.380 how to generalize the techniques and results 00:34:22.380 --> 00:34:25.500 that we obtain for triangles to other graphs, 00:34:25.500 --> 00:34:27.090 and what are some of the implications 00:34:27.090 --> 00:34:30.000 if you combine it with Szemeredi's regularity lemma. 00:34:49.179 --> 00:34:56.236 So let's generalize the triangle counting lemma. 00:35:09.370 --> 00:35:11.650 So the strategy for the triangle counting lemma, 00:35:11.650 --> 00:35:15.110 let me remind you, was that we embedded the vertices one 00:35:15.110 --> 00:35:15.610 by one. 00:35:18.280 --> 00:35:20.890 Putting a vertex and a typical vertex here 00:35:20.890 --> 00:35:27.610 should have many neighbors to both vertex sets. 00:35:27.610 --> 00:35:32.980 So these two guys should have sizes typically roughly 00:35:32.980 --> 00:35:36.730 the same as a fraction of them corresponding 00:35:36.730 --> 00:35:38.920 to the edge density between the vertex parts. 00:35:41.610 --> 00:35:43.490 And if they're not too small, then 00:35:43.490 --> 00:35:46.670 from the epsilon regularity of these two sets, 00:35:46.670 --> 00:35:50.250 you can deduce the number of edges between them. 00:35:50.250 --> 00:35:54.210 So that was the strategy for the triangle counting lemma. 00:35:54.210 --> 00:35:58.580 So you can try to extend the same strategy for other graphs. 00:35:58.580 --> 00:36:01.410 So let me show you how this would have been done, 00:36:01.410 --> 00:36:04.520 but I don't want to give too many details because it 00:36:04.520 --> 00:36:07.420 does get somewhat hairy if you try to execute it. 00:36:11.300 --> 00:36:18.730 So the first strategy is to embed the vertices of H one 00:36:18.730 --> 00:36:19.230 at a time. 00:36:24.030 --> 00:36:29.100 So my H now is going to be, let's say, a K4. 00:36:32.770 --> 00:36:38.190 And I wish to embed this H in this setting 00:36:38.190 --> 00:36:41.940 where you have these four parts in the regularity partition, 00:36:41.940 --> 00:36:45.570 and they are pairwise epsilon regular with edge densities 00:36:45.570 --> 00:36:49.700 that are not too small. 00:36:49.700 --> 00:36:52.600 Well, what you can try to do, mimicking the strategy 00:36:52.600 --> 00:36:58.570 over there, is to first find a typical image 00:36:58.570 --> 00:36:59.875 for the top vertex. 00:37:04.760 --> 00:37:07.920 And a typical image over here, minus 00:37:07.920 --> 00:37:14.460 some small bad exceptions, will have many neighbors 00:37:14.460 --> 00:37:15.600 to each of the three parts. 00:37:19.160 --> 00:37:25.340 Next, I need to figure out where this vertex can go. 00:37:25.340 --> 00:37:31.020 I'm going to embed this vertex somewhere here. 00:37:31.020 --> 00:37:34.440 Again, a typical place for this vertex, 00:37:34.440 --> 00:37:37.440 modulo some small fraction, which I'm going to throw away. 00:37:37.440 --> 00:37:40.380 So now you see you need somewhat stronger hypotheses 00:37:40.380 --> 00:37:43.260 on the epsilon regularity, but they are still 00:37:43.260 --> 00:37:45.720 all polynomial dependent, so you just 00:37:45.720 --> 00:37:47.830 have to choose your parameters correctly. 00:37:47.830 --> 00:37:52.120 So this typical green vertex should have 00:37:52.120 --> 00:37:56.940 lots of neighbors over here. 00:37:59.740 --> 00:38:01.450 So you just keep embedding. 00:38:01.450 --> 00:38:03.280 So it's almost this greedy strategy. 00:38:03.280 --> 00:38:06.190 You keep embedding each vertex, but you 00:38:06.190 --> 00:38:10.930 have to guarantee that there are still lots of options left. 00:38:16.690 --> 00:38:20.170 So embed vertices one at a time. 00:38:20.170 --> 00:38:34.050 And I want to embed each vertex so that the yet to be embedded 00:38:34.050 --> 00:38:47.550 vertices have many choices left. 00:39:01.690 --> 00:39:05.490 So epsilon regularity guarantees that you can always do this. 00:39:05.490 --> 00:39:08.290 And you do this all the way until the end 00:39:08.290 --> 00:39:11.490 and you arrive at some statement. 00:39:11.490 --> 00:39:13.540 And depending on how you do this, 00:39:13.540 --> 00:39:15.340 the exact formulation of the statement 00:39:15.340 --> 00:39:16.798 will be somewhat different, but let 00:39:16.798 --> 00:39:19.120 me give you one statement which we will not prove, 00:39:19.120 --> 00:39:22.060 but you can deduce using this strategy. 00:39:22.060 --> 00:39:25.720 And this is known as a graph embedding lemma. 00:39:25.720 --> 00:39:28.420 And again, as I mentioned when I started 00:39:28.420 --> 00:39:29.855 discussing Szemeredi's regularity 00:39:29.855 --> 00:39:34.870 lemma, the exact statements, they are fairly robust 00:39:34.870 --> 00:39:39.520 and they are not as important as the spirit of the ideas. 00:39:39.520 --> 00:39:41.380 So if you have some application in mind, 00:39:41.380 --> 00:39:44.170 you might have to go into the proof, tweak a thing 00:39:44.170 --> 00:39:47.210 here and there, but you get what you want. 00:39:47.210 --> 00:39:50.240 So the graph embedding lemma says, for example, 00:39:50.240 --> 00:39:59.020 that if H is bipartite, and with maximum degree and most delta-- 00:39:59.020 --> 00:40:03.120 so the H maximum degrees at most delta-- 00:40:03.120 --> 00:40:11.930 and suppose you have vertex sets V1 through Vr, such 00:40:11.930 --> 00:40:21.970 that each vertex set is not too small; 00:40:21.970 --> 00:40:29.620 and if these vertex sets are pairwise epsilon 00:40:29.620 --> 00:40:36.550 regular and the density is not too small. 00:40:40.890 --> 00:40:44.440 So here I'm assuming some lower bound 00:40:44.440 --> 00:40:50.220 on the density which depends on your epsilon. 00:40:50.220 --> 00:41:02.910 Then the conclusion is that G contains a copy of H. 00:41:02.910 --> 00:41:06.090 I just want to give a few remarks on the statement 00:41:06.090 --> 00:41:06.900 of this theorem. 00:41:06.900 --> 00:41:10.000 Again, we will not discuss the proof. 00:41:10.000 --> 00:41:14.250 So what is this hypothesis on H being 00:41:14.250 --> 00:41:16.200 r-partite have to do with anything? 00:41:16.200 --> 00:41:19.620 So here, as an example, when r equals to 4, instead 00:41:19.620 --> 00:41:24.870 of this K4, maybe I also care about that graph over there. 00:41:29.520 --> 00:41:31.520 Well, maybe some more vertices, some more edges, 00:41:31.520 --> 00:41:32.840 but if it's 4-partite. 00:41:32.840 --> 00:41:37.640 And the point is that I want to embed the vertices in such 00:41:37.640 --> 00:41:42.290 a way that that top vertex goes to the part 00:41:42.290 --> 00:41:45.160 that it's supposed to go into. 00:41:45.160 --> 00:41:47.660 So I'm embedding this configuration 00:41:47.660 --> 00:41:52.400 in the same way that corresponds to the proper coloring of H. 00:41:52.400 --> 00:41:56.840 So if you do this, there's enough room still 00:41:56.840 --> 00:42:01.580 to go, as long as you have some lower bound on the edge 00:42:01.580 --> 00:42:06.240 density between the individual parts, 00:42:06.240 --> 00:42:10.170 and you only depend really on not the number of edges 00:42:10.170 --> 00:42:13.740 of H, but the maximum degree. 00:42:13.740 --> 00:42:18.640 Because if you look at how many times each vertex, 00:42:18.640 --> 00:42:22.170 its possibilities can be shrunk, it's in most delta times. 00:42:25.290 --> 00:42:26.993 Now, this graph embedding lemma-- 00:42:26.993 --> 00:42:28.410 so I give you this statement here, 00:42:28.410 --> 00:42:29.920 but it's a fairly robust statement. 00:42:29.920 --> 00:42:32.800 And if you want to get, for example, not just a single copy 00:42:32.800 --> 00:42:35.505 of H, if you want to get many copies of H, 00:42:35.505 --> 00:42:36.880 you can tweak the hypotheses, you 00:42:36.880 --> 00:42:38.920 can tweak the proofs somewhat to get you 00:42:38.920 --> 00:42:42.045 what you want, again, following what we did for the triangle 00:42:42.045 --> 00:42:42.670 counting lemma. 00:42:45.350 --> 00:42:46.670 Question. 00:42:46.670 --> 00:42:50.226 AUDIENCE: Is the bound on the H edge density between partitions 00:42:50.226 --> 00:42:51.040 correct? 00:42:51.040 --> 00:42:53.770 So if the maximum degree increases, 00:42:53.770 --> 00:42:55.240 the lower bound decreases? 00:42:55.240 --> 00:42:58.981 PROFESSOR: If maximum degree increases, this number goes up. 00:42:58.981 --> 00:42:59.963 AUDIENCE: Oh, OK. 00:43:08.810 --> 00:43:11.300 PROFESSOR: I want to show you a different way 00:43:11.300 --> 00:43:15.710 to do counting that does not go through this embedding vertices 00:43:15.710 --> 00:43:18.680 one by one, but instead we will try 00:43:18.680 --> 00:43:22.250 to analyze what happens if you take out an edge of H one 00:43:22.250 --> 00:43:24.190 by one. 00:43:24.190 --> 00:43:27.010 And that's an alternative approach which I like more. 00:43:27.010 --> 00:43:29.270 It's somewhat less intuitive if you are not 00:43:29.270 --> 00:43:31.880 used to thinking about it, but the execution 00:43:31.880 --> 00:43:33.350 works out to be much cleaner. 00:43:33.350 --> 00:43:37.460 And it's also in line with some of the techniques 00:43:37.460 --> 00:43:41.398 that we'll see later on when we discuss graph limits. 00:43:41.398 --> 00:43:42.440 Let's take a quick break. 00:43:45.620 --> 00:43:46.630 Any questions so far? 00:43:54.100 --> 00:43:58.880 We will see a second strategy for proving a graph counting 00:43:58.880 --> 00:44:00.650 lemma. 00:44:00.650 --> 00:44:06.100 And the second strategy is more analytic in nature, 00:44:06.100 --> 00:44:15.740 which is to embed, well, to analytically somehow we'll 00:44:15.740 --> 00:44:19.670 analyze what happens when we pick out one edge of H 00:44:19.670 --> 00:44:20.292 at a time. 00:44:26.635 --> 00:44:28.260 So let me give you the statement first. 00:44:38.050 --> 00:44:41.070 So the graph counting lemma says that if you 00:44:41.070 --> 00:44:51.030 have a graph H with vertex set elements of 1 through k-- 00:44:51.030 --> 00:44:55.110 I also have an epsilon parameter and I 00:44:55.110 --> 00:45:04.580 have a graph G and subsets V1 through Vk of G, 00:45:04.580 --> 00:45:11.900 such that Vi Vj is epsilon regular 00:45:11.900 --> 00:45:18.370 whenever ij is an edge of H. 00:45:18.370 --> 00:45:21.130 So here, the setup is slightly different from the picture 00:45:21.130 --> 00:45:22.910 I drew up there. 00:45:22.910 --> 00:45:28.770 So what's going on here is that you have, 00:45:28.770 --> 00:45:35.390 let's say, H being this graph. 00:45:35.390 --> 00:45:38.990 And suppose you are in a situation where 00:45:38.990 --> 00:45:41.180 I know that some of these-- 00:45:47.700 --> 00:45:53.370 so I know that five of these pairs are epsilon regular, 00:45:53.370 --> 00:45:58.200 and what I really want to do is embed this H 00:45:58.200 --> 00:45:59.933 into this configuration. 00:46:06.090 --> 00:46:08.920 So 1, 2; V1, V2; and so on. 00:46:11.800 --> 00:46:15.010 And I want to know how many ways can you embed this way. 00:46:18.690 --> 00:46:21.240 The conclusion is that the number 00:46:21.240 --> 00:46:36.900 of tuples, the number of such embeddings, such that Vi, Vj 00:46:36.900 --> 00:46:45.130 is an edge of G for all ij being the edge of H-- 00:46:45.130 --> 00:46:47.040 so exactly as showing that picture, 00:46:47.040 --> 00:46:52.390 the number of such embeddings, the little v's, is 00:46:52.390 --> 00:46:57.040 within a small error, which is the number of edges 00:46:57.040 --> 00:47:05.660 of H epsilon times the total, the product of these vertex 00:47:05.660 --> 00:47:16.810 set sizes of this number, which is what you would predict 00:47:16.810 --> 00:47:21.400 the number of embeddings to be if all of your bipartite graphs 00:47:21.400 --> 00:47:23.100 were actually random. 00:47:36.350 --> 00:47:38.530 So like in the triangle counting lemma, 00:47:38.530 --> 00:47:42.520 if you look at the edge densities in this configuration 00:47:42.520 --> 00:47:46.420 and predict how many copies of H you would get, 00:47:46.420 --> 00:47:48.400 that's the number you should write down. 00:47:48.400 --> 00:47:51.730 And this counting lemma tells you 00:47:51.730 --> 00:47:55.955 that the truth is not so far from the prediction. 00:48:04.510 --> 00:48:05.679 Any questions? 00:48:14.170 --> 00:48:16.000 So we will prove the graph counting 00:48:16.000 --> 00:48:21.430 lemma in an analytic manner. 00:48:21.430 --> 00:48:24.130 It helps to-- so it will be convenient 00:48:24.130 --> 00:48:28.090 for me to rephrase the result just a little bit 00:48:28.090 --> 00:48:29.940 in this probabilistic form. 00:48:29.940 --> 00:48:34.870 So it has the equivalent to show that if you 00:48:34.870 --> 00:48:42.325 have uniformly randomly chosen vertices, little v1 and big V1, 00:48:42.325 --> 00:48:45.770 and so on-- 00:48:45.770 --> 00:48:50.700 so little vk and big VK. 00:48:50.700 --> 00:48:53.770 So they're independent, uniformly, and random, 00:48:53.770 --> 00:48:59.000 then the probability-- 00:48:59.000 --> 00:49:02.830 so basically, I am putting down a potential image 00:49:02.830 --> 00:49:06.225 for each vertex of H and asking what's 00:49:06.225 --> 00:49:07.600 the probability that you actually 00:49:07.600 --> 00:49:08.980 have an embedding of H. 00:49:08.980 --> 00:49:12.070 So the probability that little vi, little vj 00:49:12.070 --> 00:49:18.910 is an actual edge of G for all ij being an edge of H, 00:49:18.910 --> 00:49:24.150 this number here, we're saying that it differs 00:49:24.150 --> 00:49:28.500 from the prediction, which is simply multiplying all the edge 00:49:28.500 --> 00:49:29.670 densities together. 00:49:36.070 --> 00:49:39.710 So the difference between the actual and the predicted values 00:49:39.710 --> 00:49:43.117 is fairly small. 00:49:43.117 --> 00:49:45.450 So I haven't done anything, just rephrasing the problem. 00:49:45.450 --> 00:49:48.104 Instead of counting, now we're looking at probabilities. 00:49:53.050 --> 00:49:56.920 As I mentioned, we'll take out one edge at a time. 00:49:56.920 --> 00:50:01.510 So relabeling if necessary, let's assume 00:50:01.510 --> 00:50:09.870 that 1, 2 is an edge of H. So now we 00:50:09.870 --> 00:50:18.080 will show the following plane, so I'll denote star. 00:50:18.080 --> 00:50:31.380 That, if you look at this quantity over here 00:50:31.380 --> 00:50:40.700 compared to if you take out just the edge density between V1 00:50:40.700 --> 00:50:47.670 and V2, but now you put in a similar quantity 00:50:47.670 --> 00:50:55.820 where I'm considering all of the edges of H, except for 1, 2. 00:51:11.750 --> 00:51:14.930 I claim that this difference is, at most, epsilon. 00:51:24.390 --> 00:51:26.070 So you can think of this quantity 00:51:26.070 --> 00:51:32.340 here as the same quantity as the green one, except not on H, 00:51:32.340 --> 00:51:34.990 but on H minus the edge 1, 2. 00:51:43.350 --> 00:51:50.190 To show this star claim, let us couple the two random processes 00:51:50.190 --> 00:51:51.860 choosing these little vi's. 00:52:07.990 --> 00:52:11.710 By that, I mean here you have the random little vi's and here 00:52:11.710 --> 00:52:14.590 you have the random little vi's, but use the same little vi's 00:52:14.590 --> 00:52:17.430 in both probabilities. 00:52:17.430 --> 00:52:21.720 So in both, there are two different random events, 00:52:21.720 --> 00:52:24.430 but you use the same little vi's, 00:52:24.430 --> 00:52:35.310 it suffices to show this inequality star with-- 00:52:37.900 --> 00:52:39.770 so what's this process, this random process? 00:52:39.770 --> 00:52:42.830 You pick V1, you pick V2, you pick V3, all of them 00:52:42.830 --> 00:52:45.690 independently uniformly at random. 00:52:45.690 --> 00:52:50.520 But if I show you the inequality under a further constraint 00:52:50.520 --> 00:52:55.540 of arbitrary V3 through VN, then that's even better. 00:52:58.540 --> 00:53:10.510 So with V3 through Vk, fixed arbitrary, 00:53:10.510 --> 00:53:18.420 and only little v1 and little v2 random. 00:53:27.028 --> 00:53:29.570 So you can phrase this in terms of conditional probabilities, 00:53:29.570 --> 00:53:31.450 if you like. 00:53:31.450 --> 00:53:33.950 You're comparing these two probabilities. 00:53:33.950 --> 00:53:38.660 Now I fix the V3's through Vk's, and I just 00:53:38.660 --> 00:53:40.850 let V1 and V2 be random. 00:53:40.850 --> 00:53:47.720 And if condition on V3 through Vk, you have this inequality, 00:53:47.720 --> 00:53:51.080 then letting V3 through Vk go, letting 00:53:51.080 --> 00:53:53.750 them be random as well, by triangle inequality, 00:53:53.750 --> 00:53:57.260 you obtain the bound that we're looking for. 00:54:02.060 --> 00:54:03.340 Any questions about this step? 00:54:06.220 --> 00:54:08.630 If you're confused about what's happening here, 00:54:08.630 --> 00:54:11.840 another way to bypass all the probability language 00:54:11.840 --> 00:54:14.720 is to go back to the counting language. 00:54:14.720 --> 00:54:17.260 So in other words, we're trying to count embeddings. 00:54:17.260 --> 00:54:23.590 And I'm asking if you arbitrarily fix V3 through Vk, 00:54:23.590 --> 00:54:26.600 how many different choices are there to V1 and V2 00:54:26.600 --> 00:54:27.850 in the two different settings. 00:54:32.600 --> 00:54:34.420 OK. 00:54:34.420 --> 00:54:47.950 So let A1 be the set of places where little v1 can go, 00:54:47.950 --> 00:54:51.880 if you already knew what V3 through Vk should be. 00:55:05.980 --> 00:55:06.510 OK. 00:55:06.510 --> 00:55:12.600 So you look at all the neighbors of V1, neighbors of 1 and H, 00:55:12.600 --> 00:55:14.650 except for 2. 00:55:14.650 --> 00:55:19.250 And I want to make sure that V1, Vi, as i 00:55:19.250 --> 00:55:24.820 ranges over all such neighbors, is indeed a valid H in G. 00:55:24.820 --> 00:55:26.410 I'll draw a picture in a second. 00:55:26.410 --> 00:55:36.870 And A2, likewise, is the same quantity, 00:55:36.870 --> 00:55:38.310 but with 2 instead of 1. 00:55:47.200 --> 00:55:51.640 So for example, if you're trying to embed a K4, what's 00:55:51.640 --> 00:56:00.260 happening here is that you have this V1, this V2, and somebody 00:56:00.260 --> 00:56:03.130 already arbitrarily fixed the locations 00:56:03.130 --> 00:56:06.120 where V3 and V4 are embedded. 00:56:06.120 --> 00:56:10.500 And you're asking, well, how many different choices are now 00:56:10.500 --> 00:56:12.960 left for little v1. 00:56:12.960 --> 00:56:21.830 It's the common neighborhood of V3 and V4. 00:56:21.830 --> 00:56:24.730 So that's for A1. 00:56:24.730 --> 00:56:32.150 And V3 and V4, also their common neighborhood in B2 is A2. 00:56:38.550 --> 00:56:39.050 OK. 00:56:39.050 --> 00:56:42.680 So with that notation, what is it that we're 00:56:42.680 --> 00:56:46.060 trying to show over here? 00:56:46.060 --> 00:56:48.590 So if you rewrite this inequality with the V3's 00:56:48.590 --> 00:56:56.880 through Vk's fixed, you find that what we're trying to show 00:56:56.880 --> 00:56:59.116 is the following. 00:57:03.880 --> 00:57:10.680 So we claim-- and this claim implies the star, 00:57:10.680 --> 00:57:16.040 that if you look at what the first term should be-- 00:57:16.040 --> 00:57:21.820 so this is the number of edges between A1 and A2 00:57:21.820 --> 00:57:26.590 as a fraction of the product of V1 and V2. 00:57:33.120 --> 00:57:36.990 And then the second guy here is if you use 00:57:36.990 --> 00:57:41.890 the prediction d of V1 and V2. 00:57:53.230 --> 00:57:56.620 So this is each of them, each of these two factors, 00:57:56.620 --> 00:58:01.420 there's a probability that little v1 lies in A1, 00:58:01.420 --> 00:58:03.790 little v2 lies in A2, and then you 00:58:03.790 --> 00:58:09.030 tack on this extra constant, namely, this constant here. 00:58:09.030 --> 00:58:11.580 So we're trying to show that this difference is small. 00:58:11.580 --> 00:58:14.430 And the claim is that this difference is, indeed, 00:58:14.430 --> 00:58:15.900 always small. 00:58:15.900 --> 00:58:22.800 There's always, at most, epsilon for every A1 in V1 and A2 00:58:22.800 --> 00:58:23.400 in V2. 00:58:29.970 --> 00:58:33.950 And here, in particular, so this statement 00:58:33.950 --> 00:58:35.990 looks somewhat like the definition of epsilon 00:58:35.990 --> 00:58:38.150 regularity, but there's no restrictions 00:58:38.150 --> 00:58:42.710 on the sizes of A1 and A2, and they don't have to be big. 00:58:46.690 --> 00:58:48.670 And as you can imagine, this statement, 00:58:48.670 --> 00:58:51.250 we're not really using all that much. 00:58:51.250 --> 00:58:56.170 All we're assuming is the epsilon regularity 00:58:56.170 --> 00:58:58.150 between B1 and B2. 00:58:58.150 --> 00:59:01.000 So we will deduce this inequality 00:59:01.000 --> 00:59:07.535 from the hypotheses of epsilon regularity between B1 and B2. 00:59:07.535 --> 00:59:08.160 So let's check. 00:59:11.730 --> 00:59:19.950 So we know that B1 and B2 is epsilon regular by hypotheses. 00:59:19.950 --> 00:59:30.770 So if either A1 or A2 is too small, 00:59:30.770 --> 00:59:40.950 so if A1 is too small or A2 is too small, 00:59:40.950 --> 00:59:50.920 then we see that both of these terms 00:59:50.920 --> 00:59:52.780 here are, at most, epsilon. 01:00:00.040 --> 01:00:02.770 So if the A's are too small, then neither of these terms 01:00:02.770 --> 01:00:03.730 can be too large. 01:00:03.730 --> 01:00:05.920 Here it's bounded by-- 01:00:05.920 --> 01:00:08.640 if you took the product of A1 and 01:00:08.640 --> 01:00:11.743 A2 and likewise over there. 01:00:11.743 --> 01:00:13.660 So in this case, their difference is, at most, 01:00:13.660 --> 01:00:14.910 epsilon, and we're good to go. 01:00:17.570 --> 01:00:24.750 Otherwise, if A1 and A2 are both at least an epsilon fraction 01:00:24.750 --> 01:00:34.220 of their [INAUDIBLE] sets, then we find that-- 01:00:37.200 --> 01:00:38.330 so what happens? 01:00:38.330 --> 01:00:45.480 So here, so by the hypothesis of epsilon regularity, 01:00:45.480 --> 01:00:52.990 we find that d of V1 and V2 differs 01:00:52.990 --> 01:00:56.500 from the number of edges between A1 and A2, 01:00:56.500 --> 01:00:59.860 divided by the product of their sizes. 01:00:59.860 --> 01:01:03.040 So that's just d of A1 and A2. 01:01:03.040 --> 01:01:16.490 So this difference is, at most, epsilon, which then implies 01:01:16.490 --> 01:01:17.890 the inequality up there. 01:01:21.640 --> 01:01:24.160 So here we're using that the size of A 01:01:24.160 --> 01:01:35.670 is, at most, the size of V. 01:01:35.670 --> 01:01:38.130 So we have this claim. 01:01:38.130 --> 01:01:42.720 And that claim proves this inequality in star. 01:01:42.720 --> 01:01:44.850 And basically, what we've done is 01:01:44.850 --> 01:01:48.630 we showed that if you took out a single edge, 01:01:48.630 --> 01:01:53.190 you change the desired quantity by, at most, an epsilon, 01:01:53.190 --> 01:01:54.540 essentially. 01:01:54.540 --> 01:02:00.030 So now you do this for every edge of H. Alternatively, 01:02:00.030 --> 01:02:06.730 you can do induction on the number of edges of H. 01:02:06.730 --> 01:02:09.360 So to complete a proof of the counting 01:02:09.360 --> 01:02:24.980 lemma, so we do induction on the number of edges of H. 01:02:24.980 --> 01:02:30.520 And when H has exactly one edge, well, that's pretty easy. 01:02:30.520 --> 01:02:34.210 But now if you have more edges, well, you 01:02:34.210 --> 01:02:37.480 apply induction hypothesis to the graph, 01:02:37.480 --> 01:02:41.410 which is H minus the edge 1, 2. 01:02:44.330 --> 01:02:58.880 And you find that this quantity here 01:02:58.880 --> 01:03:14.850 differs from the predicted quantity 01:03:14.850 --> 01:03:20.550 by the number of edges of H minus 1 times epsilon. 01:03:24.760 --> 01:03:27.310 In other words, you run this prove that we just 01:03:27.310 --> 01:03:28.850 did one edge at a time. 01:03:28.850 --> 01:03:30.610 So each time you take out an edge, 01:03:30.610 --> 01:03:32.350 you use epsilon regularity to show 01:03:32.350 --> 01:03:34.660 that the effect of taking that edge out from H 01:03:34.660 --> 01:03:38.890 does not have too big of an effect on the actual number 01:03:38.890 --> 01:03:40.570 of embeddings. 01:03:40.570 --> 01:03:42.940 Do this for one edge at a time, and eventually you 01:03:42.940 --> 01:03:44.745 prove the graph counting lemma. 01:03:49.110 --> 01:03:52.320 So this is one of those proofs which 01:03:52.320 --> 01:03:57.840 may be less intuitive compared to the one I showed earlier, 01:03:57.840 --> 01:04:01.120 in the sense that there's not as nice of a story you can tell 01:04:01.120 --> 01:04:04.180 about putting in one vertex at a time. 01:04:04.180 --> 01:04:07.560 But on the other hand, if you were to carry out this proof 01:04:07.560 --> 01:04:11.010 to bound each time how big the sets have to be, 01:04:11.010 --> 01:04:12.900 it gets much hairier over here. 01:04:12.900 --> 01:04:17.070 And here, the execution is much cleaner, but maybe less 01:04:17.070 --> 01:04:19.650 intuitive, unless you're willing to be 01:04:19.650 --> 01:04:21.760 comfortable with these calculations. 01:04:21.760 --> 01:04:23.010 And it's really not so bad. 01:04:25.810 --> 01:04:29.700 And the strength of these two results are somewhat different. 01:04:29.700 --> 01:04:33.500 So again, it's not so much the exact statements that matter, 01:04:33.500 --> 01:04:36.830 but the spirit of these statements, which 01:04:36.830 --> 01:04:40.550 is that if you have a bunch of epsilon regular pairs, 01:04:40.550 --> 01:04:45.800 then you can embed and kind of pretending that everything 01:04:45.800 --> 01:04:47.290 behaved roughly like random. 01:04:50.890 --> 01:04:52.266 Any questions? 01:04:59.410 --> 01:05:02.740 So now we have Szemeredi's graph regularity lemma, 01:05:02.740 --> 01:05:06.730 we have the graph counting lemma, embedding lemmas, 01:05:06.730 --> 01:05:11.620 we can use it to derive some additional applications that 01:05:11.620 --> 01:05:13.540 don't just involve triangles. 01:05:13.540 --> 01:05:15.940 So when we only had a triangle counting lemma, 01:05:15.940 --> 01:05:18.700 we can only do the triangle removal lemma, 01:05:18.700 --> 01:05:22.330 but now we can do other removal lemmas. 01:05:22.330 --> 01:05:25.930 So in particular, there's the graph removal lemma, 01:05:25.930 --> 01:05:28.490 which generalizes the triangle removal lemma. 01:05:38.690 --> 01:05:40.870 So in the graph removal lemma, the statement 01:05:40.870 --> 01:05:46.040 is that for every H and epsilon, there 01:05:46.040 --> 01:05:55.040 exists a delta, such that every N vertex 01:05:55.040 --> 01:06:05.450 graph with fewer than delta N to the vertex 01:06:05.450 --> 01:06:10.980 of H number of copies of H-- 01:06:10.980 --> 01:06:15.190 so it has very few copies of H-- 01:06:15.190 --> 01:06:29.920 such graph can be made H-free by removing 01:06:29.920 --> 01:06:33.350 a fairly small number of edges. 01:06:33.350 --> 01:06:37.190 All right, so same statement as the triangle removal lemma, 01:06:37.190 --> 01:06:41.250 except now for general graph H. 01:06:41.250 --> 01:06:44.460 And as you expect, the proof is more or less the same 01:06:44.460 --> 01:06:48.140 as that of a triangle removal lemma, 01:06:48.140 --> 01:06:51.470 once we have the H counting lemma. 01:06:51.470 --> 01:06:53.880 So let me remind you how this goes. 01:06:53.880 --> 01:07:01.410 So it's really the same proof as triangle removal, 01:07:01.410 --> 01:07:05.130 where there was this recipe for applying the regularity 01:07:05.130 --> 01:07:07.020 lemma from last time. 01:07:07.020 --> 01:07:08.220 So what is it? 01:07:08.220 --> 01:07:14.980 You first-- so what's the first step when you do regularity? 01:07:14.980 --> 01:07:15.950 You partition. 01:07:15.950 --> 01:07:17.271 So let's do partition. 01:07:20.240 --> 01:07:20.740 OK. 01:07:20.740 --> 01:07:24.020 So apply the regularity lemma to do partitioning. 01:07:24.020 --> 01:07:26.780 And what's the second step? 01:07:26.780 --> 01:07:28.470 So we clean this graph. 01:07:28.470 --> 01:07:30.870 And you do the same cleaning procedure 01:07:30.870 --> 01:07:33.720 as in the triangle removal lemma, 01:07:33.720 --> 01:07:36.210 except maybe you have to adjust the parameters somewhat, so 01:07:36.210 --> 01:07:44.610 remove edges, remove low density pairs, 01:07:44.610 --> 01:07:50.200 irregular pairs, and small vertex sets. 01:07:54.810 --> 01:07:58.070 And the last step-- 01:07:58.070 --> 01:07:59.810 OK, so what do we do now? 01:07:59.810 --> 01:08:00.310 OK. 01:08:00.310 --> 01:08:03.020 So you can count. 01:08:03.020 --> 01:08:11.350 So if there were any H left, then the counting lemma 01:08:11.350 --> 01:08:14.030 shows you that you must have lots of copies of H left. 01:08:26.600 --> 01:08:28.705 So now let me show you how to use the strategy. 01:08:31.600 --> 01:08:35.104 Now that we have this general graph counting lemma, 01:08:35.104 --> 01:08:38.050 we'll prove the Erdos-Stone-Simonovits theorem, 01:08:38.050 --> 01:08:40.500 which we omitted, the proof that we 01:08:40.500 --> 01:08:42.250 omitted from the first part of the course. 01:08:46.020 --> 01:08:54.330 So remind you, the Erdos-Stone-Simonovits theorem 01:08:54.330 --> 01:09:01.020 says that if you have a graph H, then the extremal number of H 01:09:01.020 --> 01:09:06.910 is equal to this quantity which depends only 01:09:06.910 --> 01:09:16.569 on the chromatic number of H. 01:09:16.569 --> 01:09:19.840 The lower bound comes from taking the Turan graph. 01:09:25.390 --> 01:09:28.140 If you take the Turan graph, you get this lower bound, 01:09:28.140 --> 01:09:32.149 so it's really the upper bound that we need to think about. 01:09:34.930 --> 01:09:35.913 All right. 01:09:35.913 --> 01:09:37.080 So what's the strategy here? 01:09:40.760 --> 01:09:42.580 The statement really is that if you 01:09:42.580 --> 01:09:50.229 have a graph G that's N vertex whose number of edges 01:09:50.229 --> 01:09:58.230 is at least that much-- 01:09:58.230 --> 01:10:02.720 OK, so I fixed an epsilon bigger than zero, 01:10:02.720 --> 01:10:04.370 fixed a positive epsilon. 01:10:04.370 --> 01:10:06.760 So the claim, what we're trying to show with 01:10:06.760 --> 01:10:09.500 Erdos-Stone-Simonovits, is that if you have a graph G with too 01:10:09.500 --> 01:10:12.880 many edges-- too many meaning this many edges-- 01:10:12.880 --> 01:10:19.750 then G contains a copy of H if N is sufficiently large. 01:10:27.400 --> 01:10:27.900 OK. 01:10:27.900 --> 01:10:30.240 So let's use the regularity method, 01:10:30.240 --> 01:10:33.000 so applying this three-step recipe. 01:10:33.000 --> 01:10:34.702 First, we partition. 01:10:40.840 --> 01:10:50.370 So partition the vertex set of G into m pieces, 01:10:50.370 --> 01:10:54.870 and in such a way that it is eta regular-- 01:10:54.870 --> 01:10:57.840 and for some parameter eta that we'll decide later. 01:11:14.560 --> 01:11:16.360 The second step is cleaning. 01:11:23.510 --> 01:11:25.920 The cleaning step, again, it's the same kind 01:11:25.920 --> 01:11:27.820 of cleaning as we've done before. 01:11:27.820 --> 01:11:37.910 So let's remove an edge from Vi cross Vj, 01:11:37.910 --> 01:11:43.180 if any of the following hold-- 01:11:43.180 --> 01:11:52.630 if Vi Vj is not A to regular, if the density is 01:11:52.630 --> 01:12:05.000 too small, or either the two sets is too small. 01:12:09.880 --> 01:12:12.900 So same cleaning as before. 01:12:12.900 --> 01:12:20.500 And we can check that the number of edges removed 01:12:20.500 --> 01:12:23.980 is not too small. 01:12:23.980 --> 01:12:25.540 So in the first case, so again, it's 01:12:25.540 --> 01:12:27.360 the same calculation as last time. 01:12:27.360 --> 01:12:29.440 In the first case, the number of edges removed 01:12:29.440 --> 01:12:32.250 is, at most, eta N squared. 01:12:32.250 --> 01:12:41.620 And we'll choose eta to be less than epsilon over 8, 01:12:41.620 --> 01:12:43.753 although it will be actually significantly smaller, 01:12:43.753 --> 01:12:44.920 as you will see in a second. 01:12:47.490 --> 01:12:51.820 In the second step, same as what happened in the triangle 01:12:51.820 --> 01:12:56.830 removal stage, the number of edges 01:12:56.830 --> 01:13:01.700 removed in the second type is, at most, that amount, 01:13:01.700 --> 01:13:04.010 still very small number. 01:13:04.010 --> 01:13:11.440 And the third one here is also a very small number. 01:13:11.440 --> 01:13:15.620 So the third type, they start with one of these sets, 01:13:15.620 --> 01:13:18.580 the m possible-- 01:13:18.580 --> 01:13:20.980 so it's a very small number. 01:13:20.980 --> 01:13:28.675 And so the total is, at most, an epsilon over 2 N squared. 01:13:31.960 --> 01:13:33.465 So maybe I want epsilon over-- 01:13:36.140 --> 01:13:42.650 so let's say epsilon over 2 N squared number of edges. 01:13:42.650 --> 01:13:46.690 And I would like that to be strictly bigger than. 01:13:50.500 --> 01:14:04.640 So now after removing these edges from G, 01:14:04.640 --> 01:14:11.340 we have this G prime, which has strictly more than 1 minus 1 01:14:11.340 --> 01:14:13.330 over r. 01:14:13.330 --> 01:14:18.950 Yeah, 1 minus 1 over r times N squared over 2 edges. 01:14:23.785 --> 01:14:24.660 So now what do we do? 01:14:28.170 --> 01:14:31.980 So we knew from Turan's theorem that if your graph has strictly 01:14:31.980 --> 01:14:38.270 more than this number of edges, you must have a K sub r plus 1. 01:14:38.270 --> 01:14:41.700 So even after deleting all these edges, 01:14:41.700 --> 01:14:45.000 G still has lots of edges left, in particular, 01:14:45.000 --> 01:14:51.560 Turan's theorem implies that G prime contains 01:14:51.560 --> 01:14:56.270 a clique on r plus 1 vertices. 01:14:56.270 --> 01:15:04.063 So here I should say that r is chromatic number of H minus 1. 01:15:08.010 --> 01:15:11.993 So I find one copy of this clique, 01:15:11.993 --> 01:15:13.410 but what does that copy look like? 01:15:17.080 --> 01:15:19.120 I find this one copy of a clique. 01:15:23.310 --> 01:15:26.650 Let's say r equals to 4. 01:15:26.650 --> 01:15:30.400 And the point, now, is that the counting lemma will allow 01:15:30.400 --> 01:15:44.450 me to amplify that clique into H. 01:15:44.450 --> 01:15:50.190 So it will allow me to amplify this clique into a copy of H. 01:15:50.190 --> 01:16:03.230 So, for example, if H were this graph over here, 01:16:03.230 --> 01:16:07.590 so then you would find a copy of H in G, which is what we want. 01:16:07.590 --> 01:16:09.090 So why does the counting lemma allow 01:16:09.090 --> 01:16:12.000 you to do this amplification? 01:16:12.000 --> 01:16:15.500 So it's this point, the ideas are all there, 01:16:15.500 --> 01:16:18.360 but there's a slight wrinkle in the calculations. 01:16:18.360 --> 01:16:20.220 I mean, there's, in the executions, 01:16:20.220 --> 01:16:23.070 I just want to point out, just in case 01:16:23.070 --> 01:16:26.610 some of the vertices of H end up in the same vertex in G. 01:16:26.610 --> 01:16:28.780 But that turns out not to be an issue. 01:16:28.780 --> 01:16:38.020 So by counting lemma, the number of homomorphisms 01:16:38.020 --> 01:16:43.480 from H to G prime, where I'm really only considering 01:16:43.480 --> 01:16:47.200 homomorphisms that map each vertex of H 01:16:47.200 --> 01:16:51.380 to its assigned part. 01:16:51.380 --> 01:16:57.920 It's at least this quantity where 01:16:57.920 --> 01:17:04.130 I'm looking at the predicted density of such homomorphisms, 01:17:04.130 --> 01:17:08.480 and all of these edge densities are at least epsilon over 8. 01:17:08.480 --> 01:17:13.820 So it's at least that amount minus a small error that 01:17:13.820 --> 01:17:15.110 comes from the counting lemma. 01:17:19.720 --> 01:17:23.720 And, well, all of the vertex parts 01:17:23.720 --> 01:17:27.875 are quite large, so all of the vertex parts 01:17:27.875 --> 01:17:35.000 are of size like that. 01:17:38.055 --> 01:17:40.180 So that's the result of the counting lemma combined 01:17:40.180 --> 01:17:43.030 with information about the densities of the parts 01:17:43.030 --> 01:17:48.210 and the sizes of the parts that came out of cleaning. 01:17:48.210 --> 01:18:00.240 So setting eta to be an appropriate value, 01:18:00.240 --> 01:18:08.660 we see that for sufficiently large N, 01:18:08.660 --> 01:18:14.400 this quantity here, is on the order of N 01:18:14.400 --> 01:18:19.590 to the number of vertices of H. But I'm only 01:18:19.590 --> 01:18:21.780 counting homomorphisms, and so it 01:18:21.780 --> 01:18:24.720 could be that some of the vertices of H 01:18:24.720 --> 01:18:29.130 end up in the same vertex of G. And those would not 01:18:29.130 --> 01:18:31.110 be genuine subgraphs, so I shouldn't 01:18:31.110 --> 01:18:32.820 consider those as subgraphs. 01:18:32.820 --> 01:18:35.110 Because otherwise, if you were to allow those, 01:18:35.110 --> 01:18:38.550 then if you found this K4, then you 01:18:38.550 --> 01:18:41.040 found all four chromatic graphs. 01:18:41.040 --> 01:18:44.400 So you shouldn't consider copies that are degenerate, 01:18:44.400 --> 01:18:50.080 but that's OK because the number of maps from the vertex set 01:18:50.080 --> 01:18:55.290 of H to the vertex set of G that are non-injective 01:18:55.290 --> 01:19:00.420 is of a lower order. 01:19:00.420 --> 01:19:02.833 The number of non-injective maps from the vertex set, 01:19:02.833 --> 01:19:04.500 well, you have to pick two vertices of H 01:19:04.500 --> 01:19:07.050 to map to the same vertex, and then the number of choices, 01:19:07.050 --> 01:19:10.750 you have one order less. 01:19:10.750 --> 01:19:14.307 So there are negligible fraction of these homomorphisms. 01:19:16.990 --> 01:19:19.310 And the conclusion, then, is that G prime 01:19:19.310 --> 01:19:27.050 contains a copy of H, which is what we're looking for. 01:19:27.050 --> 01:19:29.850 If G prime contains copy of H, then G contains a copy of H, 01:19:29.850 --> 01:19:32.460 and that proves the Erdos-Stone-Simonovits theorem. 01:19:35.340 --> 01:19:37.950 You get a bit more out of this proof. 01:19:37.950 --> 01:19:44.780 So you see that not only does G contain one copy of H, 01:19:44.780 --> 01:19:46.700 but the counting lemma actually shows you 01:19:46.700 --> 01:19:50.120 it contains many copies of H. 01:19:50.120 --> 01:19:53.360 And this is a phenomenon known as supersaturation, which 01:19:53.360 --> 01:19:55.340 you already saw in the first problem set, 01:19:55.340 --> 01:19:59.360 that often when you are beyond a certain threshold, 01:19:59.360 --> 01:20:04.160 an extremal threshold, you don't just gain one extra copy, 01:20:04.160 --> 01:20:07.570 but you often gain many copies. 01:20:07.570 --> 01:20:11.790 And you see this in this proof here. 01:20:11.790 --> 01:20:15.990 So to summarize, we've seen this proof 01:20:15.990 --> 01:20:19.230 of Erdos-Stone-Simonovits, which comes 01:20:19.230 --> 01:20:22.470 from applying regularity and then finding 01:20:22.470 --> 01:20:25.980 a single copy of a clique from Turan's theorem, 01:20:25.980 --> 01:20:27.840 and then using counting lemma to boost 01:20:27.840 --> 01:20:35.030 that copy from Turan's theorem into an actual copy of H. 01:20:35.030 --> 01:20:38.220 So in the second homework, one of the problems 01:20:38.220 --> 01:20:41.790 is to come up with a different proof of Erdos-Stone-Simonovits 01:20:41.790 --> 01:20:46.920 that is more similar to the proof of Kovari-Sos-Turan, 01:20:46.920 --> 01:20:49.400 more through double-counting like arguments. 01:20:49.400 --> 01:20:51.150 And that is closer in spirit, although not 01:20:51.150 --> 01:20:56.610 exactly the same as the original proof in Erdos-Stone. 01:20:56.610 --> 01:21:00.290 So this regularity proof, I think it's more conceptual. 01:21:00.290 --> 01:21:02.540 You get to see how to do this boosting, 01:21:02.540 --> 01:21:04.250 but it gets a terrible bound. 01:21:04.250 --> 01:21:06.350 And the other proof that you see in the homework 01:21:06.350 --> 01:21:08.270 gives you a much more reasonable bound 01:21:08.270 --> 01:21:13.070 on the dependence between how N grows 01:21:13.070 --> 01:21:17.710 versus how quickly this little o has to go to zero.