WEBVTT 00:00:19.023 --> 00:00:21.190 PROFESSOR: We spent the last few lectures developing 00:00:21.190 --> 00:00:23.530 the theory of graph limits. 00:00:23.530 --> 00:00:26.230 And one of the motivations I gave 00:00:26.230 --> 00:00:28.720 at the beginning of the lecture on graph limits 00:00:28.720 --> 00:00:32.830 was that there were certain graph inequalities. 00:00:32.830 --> 00:00:37.580 Specifically, if I tell you that your graph has edge density one 00:00:37.580 --> 00:00:41.840 half, what's the minimum possible C4 density? 00:00:41.840 --> 00:00:43.810 So for those kinds of problems, graph limits 00:00:43.810 --> 00:00:47.890 gives us a very nice language for describing 00:00:47.890 --> 00:00:49.990 what the answer is, and also sometimes 00:00:49.990 --> 00:00:51.830 for solving these problems. 00:00:51.830 --> 00:00:56.620 So today, I want to dive more into these types of problems. 00:00:56.620 --> 00:00:59.470 Specifically, we're going to be talking about homomorphism 00:00:59.470 --> 00:01:00.730 density inequalities. 00:01:06.502 --> 00:01:07.340 Homomorphism. 00:01:17.990 --> 00:01:20.450 So trying to understand what is the relationship 00:01:20.450 --> 00:01:24.900 between possible subgraph densities or homomorphism 00:01:24.900 --> 00:01:28.830 densities within a large graph. 00:01:28.830 --> 00:01:31.180 We've seen these kind of problems in the past. 00:01:31.180 --> 00:01:32.640 So one of the very first theorems 00:01:32.640 --> 00:01:35.105 that we did in this course was Turan's theorem 00:01:35.105 --> 00:01:35.980 and Mantel's theorem. 00:01:41.710 --> 00:01:44.310 So specifically, for Mantel's theorem, 00:01:44.310 --> 00:01:48.930 it tells us something about the possible edge versus triangle 00:01:48.930 --> 00:01:53.730 densities in the graph, which is something 00:01:53.730 --> 00:01:56.940 that I want to spend the first part of today's lecture 00:01:56.940 --> 00:01:57.810 focusing on. 00:01:57.810 --> 00:02:00.130 So what is the possible relationship? 00:02:00.130 --> 00:02:07.650 What are all the possible edge versus triangle densities 00:02:07.650 --> 00:02:10.539 in the graph? 00:02:10.539 --> 00:02:12.630 Mantel's theorem tells us something-- 00:02:12.630 --> 00:02:18.120 namely, that if your edge density exceeds one half, 00:02:18.120 --> 00:02:22.130 then your triangle density cannot be zero. 00:02:22.130 --> 00:02:24.120 So that's what Mantel's theorem tells us. 00:02:24.120 --> 00:02:27.230 And let me write it down like this. 00:02:27.230 --> 00:02:29.150 So the statement I just said, the one 00:02:29.150 --> 00:02:32.690 about Mantel's theorem, and more generally for Turan's theorem 00:02:32.690 --> 00:02:42.110 tells us that if the Kr plus 1 density in W is 0, 00:02:42.110 --> 00:02:52.930 then necessarily the edge density is at most 1 minus 1 00:02:52.930 --> 00:02:53.430 over r. 00:02:58.210 --> 00:03:01.090 So this is what it tells us. 00:03:01.090 --> 00:03:03.130 It gives us some information about what 00:03:03.130 --> 00:03:04.570 are the possible densities. 00:03:04.570 --> 00:03:07.490 But I would like to know more generally, 00:03:07.490 --> 00:03:09.370 or a good complete picture, of what 00:03:09.370 --> 00:03:13.600 is a set of edge versus triangle density inequalities. 00:03:13.600 --> 00:03:17.340 So let me draw a picture that captures 00:03:17.340 --> 00:03:19.870 what we're looking for. 00:03:19.870 --> 00:03:27.950 So here on the x-axis, I have all the possible edge 00:03:27.950 --> 00:03:33.360 densities, and on the vertical axis, 00:03:33.360 --> 00:03:36.700 I have the triangle density. 00:03:36.700 --> 00:03:40.160 And I would like to know what is a set of feasible 00:03:40.160 --> 00:03:42.590 points in this box. 00:03:45.790 --> 00:03:48.730 Mantel's theorem tells us already something-- 00:03:48.730 --> 00:03:54.530 namely, when can you-- so this region, 00:03:54.530 --> 00:03:58.300 the horizontal line at zero extends at most 00:03:58.300 --> 00:04:01.300 until the halfway point. 00:04:01.300 --> 00:04:05.418 Beyond this point, it's not a part of the feasible region. 00:04:05.418 --> 00:04:07.210 So far that's the information that we know. 00:04:09.790 --> 00:04:13.860 Our discussion about graph limits, and in particular-- 00:04:13.860 --> 00:04:16.870 so let me first write down what is the question. 00:04:16.870 --> 00:04:24.870 So if you look at the set of possible edge versus triangle 00:04:24.870 --> 00:04:31.840 densities, so there is this region here. 00:04:36.070 --> 00:04:37.780 What is this region? 00:04:37.780 --> 00:04:41.530 It's a subset of this unit square. 00:04:41.530 --> 00:04:43.270 We would like to understand what is 00:04:43.270 --> 00:04:47.120 the set of all possibilities. 00:04:47.120 --> 00:04:49.910 The compactness of the space of graphons 00:04:49.910 --> 00:04:54.260 tells us that this region is compact. 00:04:54.260 --> 00:05:02.570 So let me call this region D23 for edge versus triangle. 00:05:02.570 --> 00:05:09.050 So D23 is compact because the space of graphons 00:05:09.050 --> 00:05:13.190 is compact under the cut metric, and densities are 00:05:13.190 --> 00:05:16.890 continuous under cut distance. 00:05:16.890 --> 00:05:21.500 So in particular, if you have some limit point 00:05:21.500 --> 00:05:25.610 of some sequence of graphs, that limit point's 00:05:25.610 --> 00:05:29.030 achieved by a corresponding limit graphon. 00:05:29.030 --> 00:05:33.280 So you really have a nice closed region over here. 00:05:33.280 --> 00:05:35.040 So we don't have to-- 00:05:35.040 --> 00:05:36.968 I should be able to tell you the answer. 00:05:36.968 --> 00:05:37.760 This is the region. 00:05:37.760 --> 00:05:39.780 There should not be any additional quantifiers. 00:05:39.780 --> 00:05:41.812 There's no optimizer zero, missing this point 00:05:41.812 --> 00:05:42.770 and missing that point. 00:05:42.770 --> 00:05:43.770 It's a closed region. 00:05:43.770 --> 00:05:46.970 So what is this closed region? 00:05:46.970 --> 00:05:50.450 Equivalently, we can ask the following question. 00:05:50.450 --> 00:05:56.820 Suppose I give you the edge density. 00:05:56.820 --> 00:06:00.080 In other words, look at a particular horizontal place 00:06:00.080 --> 00:06:01.310 in this picture. 00:06:01.310 --> 00:06:12.687 What is the maximum and minimum possible triangle densities? 00:06:21.620 --> 00:06:27.580 So I tell you that the edge density is 0.75. 00:06:27.580 --> 00:06:31.590 What is the upper and lower boundaries of this region? 00:06:31.590 --> 00:06:35.000 I want you to think about why this region is-- 00:06:35.000 --> 00:06:38.090 the vertical cross-section is a line segment. 00:06:38.090 --> 00:06:40.160 You cannot have any hulls. 00:06:40.160 --> 00:06:42.388 So that requires an argument, and I'll 00:06:42.388 --> 00:06:43.430 let you think about that. 00:06:50.270 --> 00:06:52.530 So I want to complete this picture, 00:06:52.530 --> 00:06:53.780 and I'll show you some proofs. 00:06:53.780 --> 00:06:55.945 And at the end of-- 00:06:55.945 --> 00:06:57.570 well, by the middle of today's lecture, 00:06:57.570 --> 00:06:59.695 we'll see a picture of what this region looks like. 00:07:03.820 --> 00:07:04.360 All right. 00:07:04.360 --> 00:07:07.870 First, let me do the easier direction, which 00:07:07.870 --> 00:07:12.370 is to find the upper boundary of this region. 00:07:12.370 --> 00:07:15.360 So what is the maximum possible triangle density 00:07:15.360 --> 00:07:17.160 for a given edge density? 00:07:28.950 --> 00:07:34.680 And the answer-- so it turns out to be what I will-- 00:07:34.680 --> 00:07:37.080 the result I will tell you is a special case 00:07:37.080 --> 00:07:40.012 of what's called Kruskal-Katona. 00:07:46.150 --> 00:07:47.290 Think about it this way. 00:07:47.290 --> 00:07:49.850 Suppose I give you a very large number of vertices 00:07:49.850 --> 00:07:52.060 and I give you some large number of edges, 00:07:52.060 --> 00:07:55.300 and I want you to put the edges into the graph in a way that 00:07:55.300 --> 00:07:59.350 generates as many triangles as possible. 00:07:59.350 --> 00:08:01.120 Intuitively, how should you put the edges 00:08:01.120 --> 00:08:04.380 in to try to make as many triangles as you can? 00:08:08.196 --> 00:08:09.150 AUDIENCE: Clique. 00:08:09.150 --> 00:08:10.460 PROFESSOR: In a clique. 00:08:10.460 --> 00:08:14.920 So you put all the edges as closely together as possible, 00:08:14.920 --> 00:08:16.820 try to form a clique. 00:08:16.820 --> 00:08:24.730 So maximize number of triangles by forming a clique. 00:08:28.370 --> 00:08:29.620 And that is indeed the answer. 00:08:29.620 --> 00:08:33.340 And this is what we'll prove, at least in the graph densities 00:08:33.340 --> 00:08:34.270 version. 00:08:34.270 --> 00:08:38.419 So we will show that the upper boundary is given by the curve 00:08:38.419 --> 00:08:41.440 y equals to x to the 3/2. 00:08:44.420 --> 00:08:46.570 So don't worry about the specific function. 00:08:46.570 --> 00:08:54.170 But what's important is that the upper bound is achieved 00:08:54.170 --> 00:08:55.360 by the following graphon. 00:08:58.700 --> 00:09:01.490 Namely, this graphon corresponding to a clique. 00:09:08.450 --> 00:09:14.780 For this graphon here, the edge density is a squared, 00:09:14.780 --> 00:09:19.980 and the triangle density is a cubed. 00:09:19.980 --> 00:09:22.440 And it turns out this graphon is the best that you 00:09:22.440 --> 00:09:27.170 can do with a given edge density in order to generate 00:09:27.170 --> 00:09:31.500 as many triangles or the most triangle density possible. 00:09:31.500 --> 00:09:34.220 In other words, what we'll prove is 00:09:34.220 --> 00:09:40.670 that the triangle density throughout W will be a graphon. 00:09:40.670 --> 00:09:42.980 So W is always a graphon. 00:09:42.980 --> 00:09:44.870 So values between 0 and 1. 00:09:50.230 --> 00:09:52.434 Then you have the following inequality. 00:10:00.840 --> 00:10:03.420 So let's prove it. 00:10:03.420 --> 00:10:05.700 First let me draw you what this shape looks like. 00:10:19.490 --> 00:10:22.760 Because of the relationship between graphs and graph 00:10:22.760 --> 00:10:25.760 limits, any of these inequalities about graph 00:10:25.760 --> 00:10:29.660 limits, about graphons, it's sufficient to prove 00:10:29.660 --> 00:10:32.750 the corresponding inequality for graphs 00:10:32.750 --> 00:10:35.840 because the set of graphs is dense 00:10:35.840 --> 00:10:39.620 within the space of graphons according to the topology-- 00:10:39.620 --> 00:10:41.930 namely, the cut metric that we discussed. 00:10:41.930 --> 00:10:52.100 So it suffices to show the corresponding inequality 00:10:52.100 --> 00:10:54.350 about graphs-- 00:10:54.350 --> 00:10:59.750 namely, that the K3 density in a graph 00:10:59.750 --> 00:11:08.580 is at most the K2 density in a graph raised to the power 3/2. 00:11:08.580 --> 00:11:11.460 So let me belabor this point just a little bit more. 00:11:11.460 --> 00:11:14.850 This inequality is a subset of those inequalities 00:11:14.850 --> 00:11:18.210 up there because graphs sit inside a space of graphons. 00:11:18.210 --> 00:11:22.860 But because they sit inside in a dense subset, 00:11:22.860 --> 00:11:27.100 if you know this inequality and everything is continuous, 00:11:27.100 --> 00:11:29.290 then you know that inequality. 00:11:29.290 --> 00:11:32.240 So these two are equivalent to each other. 00:11:32.240 --> 00:11:36.760 Now, with graphs-- and specifically, 00:11:36.760 --> 00:11:42.850 these counts here, so triangle densities and edge densities-- 00:11:42.850 --> 00:11:46.850 they correspond to counting closed walks in the graph. 00:11:46.850 --> 00:11:53.290 So in particular, if we're interested in the number 00:11:53.290 --> 00:11:56.590 of K3 homomorphisms in a graph, this 00:11:56.590 --> 00:12:01.180 is the same as counting closed walks of length 3. 00:12:01.180 --> 00:12:04.300 And there was important identity we used earlier 00:12:04.300 --> 00:12:07.630 when we were discussing the proof of quasi-random graphs, 00:12:07.630 --> 00:12:11.080 that for counting closed walks you should 00:12:11.080 --> 00:12:14.210 look at the spectral moment. 00:12:14.210 --> 00:12:16.030 So that's a very important tool to look 00:12:16.030 --> 00:12:17.410 at the spectral moment-- 00:12:17.410 --> 00:12:21.850 namely, the third power of the eigenvalues 00:12:21.850 --> 00:12:24.580 of the adjacency matrix of this graph. 00:12:27.430 --> 00:12:41.820 This is the eigenvalues of the adjacency matrix of G. 00:12:41.820 --> 00:12:46.110 I claim that this sum here is upper bounded 00:12:46.110 --> 00:12:52.230 by a corresponding sum of squares raised 00:12:52.230 --> 00:12:56.307 to the power that normalizes. 00:12:56.307 --> 00:12:58.640 The first time I saw this I was a bit confused because I 00:12:58.640 --> 00:12:59.850 remember, power means inequality. 00:12:59.850 --> 00:13:00.990 Shouldn't go the other way. 00:13:00.990 --> 00:13:03.390 But actually, no, this is the correct direction. 00:13:03.390 --> 00:13:05.070 So let me remind you why. 00:13:05.070 --> 00:13:11.930 So if you have a positive t, then claim that-- 00:13:11.930 --> 00:13:17.550 and you have a bunch of non-negative reals, 00:13:17.550 --> 00:13:23.430 then the claim is that this t-th power 00:13:23.430 --> 00:13:33.200 sum is less than or equal to the t-th power of the sum. 00:13:33.200 --> 00:13:35.635 Now, there are several ways to see why this is true. 00:13:35.635 --> 00:13:36.510 You can do induction. 00:13:36.510 --> 00:13:40.320 But let me show you one way which is quite neat. 00:13:40.320 --> 00:13:43.140 Because it's homogeneous in the variables, 00:13:43.140 --> 00:13:56.480 I can assume that the sum is 1, in which case 00:13:56.480 --> 00:14:01.280 the left-hand side is equal to this sum of t-th powers. 00:14:05.540 --> 00:14:12.080 So because I assumed that everything is non-negative, 00:14:12.080 --> 00:14:14.460 all these a's are between 0 and 1. 00:14:14.460 --> 00:14:18.950 So now, this sum is less than or equal to the same sum 00:14:18.950 --> 00:14:24.160 without the t's because you're using it like that. 00:14:24.160 --> 00:14:27.410 And that's equal to 1, which is the right-hand side. 00:14:36.380 --> 00:14:38.430 So this is true. 00:14:38.430 --> 00:14:41.460 And now we have the sum of the squares 00:14:41.460 --> 00:14:44.430 of the eigenvalues, which is also 00:14:44.430 --> 00:14:52.053 a moment of the eigenvalues-- 00:14:52.053 --> 00:14:53.220 namely, corresponding to K2. 00:15:05.850 --> 00:15:08.640 So the same inequality is true for graph homomorphisms. 00:15:08.640 --> 00:15:13.440 And to get to the inequality for densities, 00:15:13.440 --> 00:15:18.120 we just divide by the number of vertices raised 00:15:18.120 --> 00:15:20.880 to the third power from both sides, 00:15:20.880 --> 00:15:23.050 and we get the inequality that we're looking for. 00:15:27.180 --> 00:15:31.010 So that's the proof of the upper bound. 00:15:31.010 --> 00:15:31.910 Any questions? 00:15:37.440 --> 00:15:40.350 There is something that bothers me slightly about this proof. 00:15:40.350 --> 00:15:41.610 Look, it's a correct proof. 00:15:41.610 --> 00:15:43.360 So there is nothing wrong with this proof. 00:15:43.360 --> 00:15:44.235 Everything is kosher. 00:15:44.235 --> 00:15:46.932 Everything is correct. 00:15:46.932 --> 00:15:48.390 You might ask, is there a way to do 00:15:48.390 --> 00:15:51.750 this spectral argument in graphons 00:15:51.750 --> 00:15:53.730 without passing to graphs? 00:15:53.730 --> 00:15:56.940 And yes, you can, because for graphons you 00:15:56.940 --> 00:15:59.370 can also talk about spectrum. 00:15:59.370 --> 00:16:01.440 It turns out to be a compact operator, 00:16:01.440 --> 00:16:03.045 so that spectrum makes sense. 00:16:03.045 --> 00:16:05.670 You have to develop a little bit more theory about the spectrum 00:16:05.670 --> 00:16:07.800 of compact operators, but everything, more or less, 00:16:07.800 --> 00:16:09.470 works exactly the same way. 00:16:09.470 --> 00:16:12.300 It's just easier to talk about graphs. 00:16:12.300 --> 00:16:14.730 But what bothers me about this proof 00:16:14.730 --> 00:16:18.000 is that we started with what I would 00:16:18.000 --> 00:16:21.090 call a physical inequality, meaning that it only 00:16:21.090 --> 00:16:28.290 has to do with the actual edges and subgraph densities. 00:16:28.290 --> 00:16:32.950 But the proof involved going to the spectrum. 00:16:32.950 --> 00:16:35.150 And that bothers me a little bit. 00:16:35.150 --> 00:16:37.680 There's nothing incorrect about it, but somehow in my mind 00:16:37.680 --> 00:16:40.980 a physical inequality deserves a physical proof. 00:16:40.980 --> 00:16:43.780 So I use the word physical in contrast 00:16:43.780 --> 00:16:47.910 to frequency, which is coming from Fourier analysis. 00:16:47.910 --> 00:16:50.770 And that's the next thing we'll do in this course. 00:16:50.770 --> 00:16:52.540 But this proof goes to the spectrum. 00:16:52.540 --> 00:16:54.710 It goes to something beyond the physical domain. 00:16:54.710 --> 00:16:55.210 OK. 00:16:55.210 --> 00:16:55.750 It's neat. 00:16:55.750 --> 00:16:58.420 But I want to show you a different proof that stays 00:16:58.420 --> 00:17:00.540 within the physical domain. 00:17:00.540 --> 00:17:01.582 And this other proof-- 00:17:01.582 --> 00:17:03.790 I mean, it's always nice to see some different proofs 00:17:03.790 --> 00:17:07.282 because you can use it to apply to different situations. 00:17:07.282 --> 00:17:08.740 And there are some situations where 00:17:08.740 --> 00:17:15.730 you might not be able to use this spectral characterization. 00:17:15.730 --> 00:17:20.190 For example, what if your K3 is now K4? 00:17:20.190 --> 00:17:24.260 A similar inequality is true, but this proof doesn't show it, 00:17:24.260 --> 00:17:25.420 at least not directly. 00:17:25.420 --> 00:17:27.819 You have to do a little bit of extra work. 00:17:36.240 --> 00:17:39.420 So let me show you a different proof of the upper bound. 00:17:42.480 --> 00:17:45.282 And we'll prove a slightly stronger statement. 00:17:58.580 --> 00:18:03.820 Namely, that for all not just graphons-- 00:18:03.820 --> 00:18:06.380 it's not so important but for all symmetric measurable 00:18:06.380 --> 00:18:12.135 functions, from the unit square to r, 00:18:12.135 --> 00:18:15.330 one has the following inequality-- 00:18:15.330 --> 00:18:19.890 namely, that the K3 density in W is 00:18:19.890 --> 00:18:26.340 upper bounded by the K2 density of W squared raised 00:18:26.340 --> 00:18:28.700 to power 3/2. 00:18:28.700 --> 00:18:31.730 Here, the square is meant to be a point-wise square. 00:18:40.563 --> 00:18:41.480 So a couple of things. 00:18:41.480 --> 00:18:46.620 If your graph is a graphon or a graph, then-- 00:18:46.620 --> 00:18:48.910 if it's a graph, then it's 0 comma 1 value. 00:18:48.910 --> 00:18:51.990 So taking this point by square doesn't do anything. 00:18:51.990 --> 00:18:54.980 If you're a graphon, you can always 00:18:54.980 --> 00:18:57.380 put one more inequality that replaces it by the thing 00:18:57.380 --> 00:18:59.870 that we're looking for because W is always 00:18:59.870 --> 00:19:02.000 bounded between 0 and 1. 00:19:02.000 --> 00:19:04.510 So it's a slightly stronger inequality. 00:19:04.510 --> 00:19:10.550 Let me show it to you by writing down a series of inequalities 00:19:10.550 --> 00:19:14.190 and applying the Cauchy-Schwarz inequality repeatedly. 00:19:14.190 --> 00:19:18.770 So it's, again, an exercise in using Cauchy-Schwartz. 00:19:18.770 --> 00:19:24.890 And we will apply three applications of Cauchy-Schwarz. 00:19:29.210 --> 00:19:31.430 Essentially, three applications-- one 00:19:31.430 --> 00:19:35.640 corresponding to every edge of this triangle. 00:19:35.640 --> 00:19:43.510 So let me begin by writing down the expression in graphons 00:19:43.510 --> 00:19:45.670 corresponding to the K3 density. 00:20:02.590 --> 00:20:07.690 I'm going to apply Cauchy-Schwarz by-- 00:20:07.690 --> 00:20:09.880 so I'm going to apply Cauchy-Schwarz 00:20:09.880 --> 00:20:18.960 to the variable x, holding all the other variables constant. 00:20:18.960 --> 00:20:22.490 So hold dy and dz constant. 00:20:22.490 --> 00:20:23.840 Going to apply to dz. 00:20:23.840 --> 00:20:27.080 You see there are two factors that involve the variable x. 00:20:27.080 --> 00:20:29.990 So apply Cauchy-Schwarz to them, you split each of them 00:20:29.990 --> 00:20:30.560 into an L2. 00:20:37.980 --> 00:20:39.850 So one of these factors become that. 00:20:39.850 --> 00:20:41.850 By the way, all of these are definite integrals. 00:20:41.850 --> 00:20:43.725 I'm just omitting the domain of integrations. 00:20:43.725 --> 00:20:48.640 All the integrals are integrated over from 0 to 1. 00:20:48.640 --> 00:20:50.850 So the second application-- sorry, 00:20:50.850 --> 00:20:56.670 the second factor becomes like that. 00:20:56.670 --> 00:20:59.315 And the third factor is left intact. 00:21:07.700 --> 00:21:09.910 So that's the first application of Cauchy-Schwarz. 00:21:09.910 --> 00:21:13.840 You apply it with respect to dx to these two factors. 00:21:13.840 --> 00:21:15.806 Split them like that. 00:21:15.806 --> 00:21:18.355 AUDIENCE: There's a normalization missing. 00:21:18.355 --> 00:21:19.230 PROFESSOR: Thank you. 00:21:19.230 --> 00:21:20.750 There is a normalization missing. 00:21:25.550 --> 00:21:26.660 OK. 00:21:26.660 --> 00:21:28.630 Guess what the second step is? 00:21:28.630 --> 00:21:31.310 Going to apply Cauchy-Schwarz again, but now 00:21:31.310 --> 00:21:36.430 to dy, to one more variable. 00:21:39.220 --> 00:21:43.140 Cauchy-Schwarz with respect to dy. 00:21:43.140 --> 00:21:46.800 There are two factors now that involve the letter y. 00:21:46.800 --> 00:21:51.570 So I apply Cauchy-Schwarz and I get the following. 00:21:51.570 --> 00:22:01.630 The first factor now just becomes the L2 norm of W. 00:22:01.630 --> 00:22:03.610 The second factor does not involve y, 00:22:03.610 --> 00:22:04.780 so it is left intact. 00:22:12.840 --> 00:22:19.290 And the third factor is again integrated with respect 00:22:19.290 --> 00:22:23.070 to y after taking the square. 00:22:29.080 --> 00:22:31.010 And there's now dz that remains. 00:22:33.630 --> 00:22:35.100 Last step. 00:22:35.100 --> 00:22:37.440 You can guess, you integrate with respect 00:22:37.440 --> 00:22:42.570 to dz and apply Cauchy-Schwarz. 00:22:42.570 --> 00:22:47.420 Apply Cauchy-Schwarz to the last two factors. 00:22:47.420 --> 00:22:50.310 And there, actually, the outside integral goes away. 00:23:15.690 --> 00:23:16.190 OK. 00:23:16.190 --> 00:23:17.690 So you get this product. 00:23:17.690 --> 00:23:24.470 And you see every single term is just the L2 norm of W. 00:23:24.470 --> 00:23:31.170 So you have that, which is the same as what I wrote over here. 00:23:38.430 --> 00:23:42.002 Any questions? 00:23:42.002 --> 00:23:42.986 Yeah. 00:23:42.986 --> 00:23:46.167 AUDIENCE: Where do you use the fact that W is symmetric? 00:23:46.167 --> 00:23:47.250 PROFESSOR: Great question. 00:23:47.250 --> 00:23:50.348 So where do I use the fact that W is symmetric? 00:23:53.150 --> 00:23:56.065 So let's see. 00:23:56.065 --> 00:23:57.690 In some sense, we're not using the fact 00:23:57.690 --> 00:23:59.640 that W is symmetric because there 00:23:59.640 --> 00:24:02.940 is a slightly more general inequality you can write down. 00:24:02.940 --> 00:24:06.810 And actually, the question gives me a good chance 00:24:06.810 --> 00:24:11.610 to do a slight diversion into how this inequality is 00:24:11.610 --> 00:24:13.470 related to Holder's inequality. 00:24:13.470 --> 00:24:15.690 So this is actually one of my favorite inequalities 00:24:15.690 --> 00:24:20.940 for this kind of combinatorial inequalities on graphons. 00:24:20.940 --> 00:24:24.690 So many of you may be familiar with Holder's inequality 00:24:24.690 --> 00:24:25.650 in the following form. 00:24:25.650 --> 00:24:29.220 If I have three functions, if I integrate them, 00:24:29.220 --> 00:24:33.930 then you can upper bound this integral 00:24:33.930 --> 00:24:38.520 by the product of the L3 norms. 00:24:38.520 --> 00:24:42.400 And likewise, if you have more functions. 00:24:42.400 --> 00:24:45.600 So if you apply just this inequality directly, 00:24:45.600 --> 00:24:49.707 you get a weaker estimate. 00:24:49.707 --> 00:24:52.040 So you don't get anything that's quite as strong as what 00:24:52.040 --> 00:24:54.390 you're looking for over there. 00:24:54.390 --> 00:24:58.580 So what happens is that if you know-- 00:24:58.580 --> 00:25:06.620 so if f, g, and h each depends only 00:25:06.620 --> 00:25:18.500 on a subset of the coordinates in the following way 00:25:18.500 --> 00:25:26.950 that f depends on only x and y, g depends only on x and z, 00:25:26.950 --> 00:25:30.450 and h depends only on y and z, then 00:25:30.450 --> 00:25:32.640 if you repeat that proof verbatim 00:25:32.640 --> 00:25:34.430 with three different functions, you 00:25:34.430 --> 00:25:38.100 will find that you can upper bound 00:25:38.100 --> 00:25:45.060 this product, this integral, by the product of the L2 norms. 00:25:50.280 --> 00:25:53.640 So L2 norms are in general less than or equal to the L3 norms. 00:25:53.640 --> 00:25:57.100 So here we're inside a probability measure space. 00:25:57.100 --> 00:26:02.140 So the entire space has volume 1. 00:26:02.140 --> 00:26:03.713 So this is a stronger inequality, 00:26:03.713 --> 00:26:05.880 and this is the inequality that comes up over there. 00:26:09.550 --> 00:26:10.135 Yeah. 00:26:10.135 --> 00:26:12.718 AUDIENCE: Is there an entirely graph theoretic proof of this-- 00:26:12.718 --> 00:26:14.985 say, for graphs instead of graphons-- that doesn't 00:26:14.985 --> 00:26:16.272 involve going to spectrum? 00:26:16.272 --> 00:26:16.980 PROFESSOR: Great. 00:26:16.980 --> 00:26:18.640 So the question is, is there entirely 00:26:18.640 --> 00:26:20.680 graph theoretic proof of this? 00:26:20.680 --> 00:26:26.020 So the reason why I mentioned that this result is a special 00:26:26.020 --> 00:26:27.606 case of Kruskal-Katona-- 00:26:32.850 --> 00:26:35.940 so Kruskal-Katona actually is a stronger result, 00:26:35.940 --> 00:26:43.560 which tells you precisely how you should construct a graph. 00:26:43.560 --> 00:26:50.700 So given exactly m edges, what's the maximum number 00:26:50.700 --> 00:26:52.200 of triangles? 00:26:56.800 --> 00:27:00.040 And the statement that there is actually-- 00:27:00.040 --> 00:27:02.380 it's a very precise result. It tells 00:27:02.380 --> 00:27:08.760 you, for example, if you have K choose 2 edges, 00:27:08.760 --> 00:27:11.580 you have at most K choose 3 triangles. 00:27:14.600 --> 00:27:17.530 It's not just at the density level but exactly. 00:27:17.530 --> 00:27:20.530 And even if the number of edges is not in the form of K 00:27:20.530 --> 00:27:22.830 choose 2, it tells you what to do. 00:27:22.830 --> 00:27:25.140 And actually, the answer is pretty easy to describe. 00:27:25.140 --> 00:27:29.520 It's almost intuitive so if I give you a bunch of matchsticks 00:27:29.520 --> 00:27:32.310 and ask you to construct a graph with as many triangles as you 00:27:32.310 --> 00:27:33.780 can, what should you do? 00:27:33.780 --> 00:27:37.160 You start with one, two, filling a triangle. 00:27:37.160 --> 00:27:40.680 Start filling a triangle. 00:27:40.680 --> 00:27:41.730 Another vertex. 00:27:41.730 --> 00:27:43.100 1, 2, 3, 4. 00:27:43.100 --> 00:27:44.690 You keep going. 00:27:44.690 --> 00:27:47.190 And that's the best way to do it. 00:27:47.190 --> 00:27:49.470 And that's what Kruskal-Katona tells you. 00:27:49.470 --> 00:27:53.410 So that's a more precise version of this inequality. 00:27:53.410 --> 00:27:56.680 And the Kruskal-Katona, the combinatorial version, 00:27:56.680 --> 00:28:00.100 is proved via a combinatorial shifting argument, also known 00:28:00.100 --> 00:28:01.493 as a compression argument. 00:28:01.493 --> 00:28:03.160 Namely, if you start with a given graph, 00:28:03.160 --> 00:28:05.500 there are some transformations you do to that graph 00:28:05.500 --> 00:28:08.050 to push your edges in one direction that 00:28:08.050 --> 00:28:10.840 saves the number of edges exactly the same 00:28:10.840 --> 00:28:13.650 but increases the number of triangles at each step. 00:28:13.650 --> 00:28:17.218 And eventually, you push everything into a clique. 00:28:17.218 --> 00:28:18.760 So it's something you can read about. 00:28:18.760 --> 00:28:22.033 It's a very nice result. Other questions? 00:28:26.110 --> 00:28:30.770 So we've solved the upper bound. 00:28:30.770 --> 00:28:33.340 So from examples and from this upper bound proof, 00:28:33.340 --> 00:28:35.560 we see that it's the upper bound. 00:28:35.560 --> 00:28:41.050 Now let me tell you a fairly general result that 00:28:41.050 --> 00:28:45.180 says something about graph theoretic inequalities 00:28:45.180 --> 00:28:48.590 but for a specific kind of linear inequalities. 00:28:48.590 --> 00:28:50.910 So here's a theorem due to Bollobas. 00:28:55.190 --> 00:29:04.900 I'm interested in an inequality of the form like-- 00:29:04.900 --> 00:29:19.380 so I'm interested in inequality of this form, where 00:29:19.380 --> 00:29:26.920 I have a bunch of real coefficients, 00:29:26.920 --> 00:29:29.170 and I'm looking at a linear combination 00:29:29.170 --> 00:29:32.120 of the clique densities. 00:29:32.120 --> 00:29:36.770 I would like to know if this inequality is true. 00:29:36.770 --> 00:29:39.390 So somebody gives us this inequality, 00:29:39.390 --> 00:29:41.430 whatever the numbers may be. 00:29:41.430 --> 00:29:42.910 You can also have a constant term. 00:29:42.910 --> 00:29:46.000 The constant term corresponds to r equals to 1. 00:29:46.000 --> 00:29:47.730 So the point density. 00:29:47.730 --> 00:29:50.100 That's the constant term. 00:29:50.100 --> 00:29:54.290 And asks you to decide is this inequality true. 00:29:54.290 --> 00:29:55.410 And if so, prove it. 00:29:55.410 --> 00:29:57.343 If not, find a counterexample. 00:30:00.800 --> 00:30:03.260 So the theorem tells you that this is actually not 00:30:03.260 --> 00:30:04.040 hard to do. 00:30:04.040 --> 00:30:09.230 So this inequality holds for all G 00:30:09.230 --> 00:30:17.000 if and only if it holds whenever G is a clique. 00:30:23.043 --> 00:30:24.710 Maybe somebody gives you this inequality 00:30:24.710 --> 00:30:28.350 about-- it's a linear inequality about clique densities. 00:30:28.350 --> 00:30:30.200 Then, to check this inequality, you only 00:30:30.200 --> 00:30:35.160 have to check over all cliques G, 00:30:35.160 --> 00:30:39.300 which is much easier than checking for all graphs. 00:30:39.300 --> 00:30:42.927 For each clique G this is just some specific expression 00:30:42.927 --> 00:30:44.510 you can write down, and you can check. 00:30:51.340 --> 00:30:55.890 So I want to show you the proof of Bollobas' theorem. 00:30:55.890 --> 00:30:59.600 It's a quite nice result. But before that, any 00:30:59.600 --> 00:31:01.477 questions about the statement. 00:31:05.293 --> 00:31:08.640 All right. 00:31:08.640 --> 00:31:11.360 So the reason I say that this is very easy to check 00:31:11.360 --> 00:31:15.030 if I actually give you what the numbers are 00:31:15.030 --> 00:31:20.770 is because this inequality for cliques-- 00:31:20.770 --> 00:31:30.720 so the inequality is equivalent to just the statement 00:31:30.720 --> 00:31:33.950 of inequality that I'm writing down now, 00:31:33.950 --> 00:31:40.710 where I tell you precisely what the r clique 00:31:40.710 --> 00:31:42.520 density is in an n clique. 00:31:42.520 --> 00:31:45.092 Because that's just some combinatorial expression. 00:31:48.217 --> 00:31:49.800 So to check whether this inequality is 00:31:49.800 --> 00:31:51.720 true for all graphs, I just have to check 00:31:51.720 --> 00:31:54.540 the specific inequality for all integers n, which 00:31:54.540 --> 00:31:56.103 is straightforward. 00:31:59.970 --> 00:32:00.780 All right. 00:32:00.780 --> 00:32:04.540 So let's see how to prove that inequality up there. 00:32:04.540 --> 00:32:05.640 And here we're-- 00:32:05.640 --> 00:32:07.290 I mean, we're not going to exactly use 00:32:07.290 --> 00:32:09.668 the theorems about graphons, but it's 00:32:09.668 --> 00:32:10.960 useful to think about graphons. 00:32:19.580 --> 00:32:25.380 So if and only if one of the directions is trivial-- 00:32:25.380 --> 00:32:28.300 so let's get that out of the way first. 00:32:28.300 --> 00:32:32.860 But also-- so the only if is clear. 00:32:32.860 --> 00:32:35.750 So for the if direction, first note 00:32:35.750 --> 00:32:41.860 that this is true for all graphs if and only 00:32:41.860 --> 00:32:48.670 if it is true for all graphons and where I replaced G 00:32:48.670 --> 00:32:53.380 by W. By the general theory of graph limits and whatnot, 00:32:53.380 --> 00:32:56.080 this is true. 00:32:56.080 --> 00:33:00.210 So in particular, there is one class of class 00:33:00.210 --> 00:33:02.460 that I would like to look at-- 00:33:02.460 --> 00:33:08.430 namely, I want to consider the set of node weighted-- 00:33:16.270 --> 00:33:19.900 so I want to consider the set of node weighted simple graphs. 00:33:27.150 --> 00:33:29.020 So node weighted simple graphs, by this 00:33:29.020 --> 00:33:35.960 I mean a graph where some of the edges are present 00:33:35.960 --> 00:33:41.190 and I have a node weight-- 00:33:41.190 --> 00:33:43.010 a weight for each node. 00:33:43.010 --> 00:33:45.210 And to normalize things properly, 00:33:45.210 --> 00:33:50.850 I'm going to assume that the nodes' weights add up to 1. 00:33:50.850 --> 00:33:54.690 Now, you see that each graph like that, you can represented 00:33:54.690 --> 00:34:03.180 by a graphon where-- 00:34:13.300 --> 00:34:15.800 so you can have a graphon. 00:34:15.800 --> 00:34:17.690 So they're not meant to be the same picture, 00:34:17.690 --> 00:34:20.659 but you have some graphon like this, which corresponds 00:34:20.659 --> 00:34:23.210 to a node weighted graph. 00:34:23.210 --> 00:34:25.699 And the set of such node weighted graphs 00:34:25.699 --> 00:34:30.174 is dense in the space of graphons. 00:34:37.420 --> 00:34:40.630 In particular, as far as graph densities are concerned, 00:34:40.630 --> 00:34:44.020 they include all the simple graphs. 00:34:44.020 --> 00:34:45.460 So it suffices-- 00:34:45.460 --> 00:34:49.000 I mean, it's equivalent to-- 00:34:49.000 --> 00:34:51.699 the inequality is equivalent to it 00:34:51.699 --> 00:34:55.650 being true for all node weighted simple graphs. 00:34:55.650 --> 00:35:01.170 But for this space of graphs, suppose 00:35:01.170 --> 00:35:03.585 that the inequality fails. 00:35:06.900 --> 00:35:10.390 Suppose that inequality is false. 00:35:10.390 --> 00:35:19.440 Then there exists a node weighted simple graph. 00:35:22.750 --> 00:35:25.330 I'm going to actually drop the word simple from now on. 00:35:25.330 --> 00:35:44.390 So node weighted graph H, such that f of H being the above sum 00:35:44.390 --> 00:35:46.080 is less than zero. 00:35:48.840 --> 00:35:53.670 And there could be many possibilities for such an H. 00:35:53.670 --> 00:35:54.750 But let me choose. 00:35:54.750 --> 00:35:59.490 So among all the possible H's, let's choose 00:35:59.490 --> 00:36:04.290 one that is minimal in the sense that it has the smallest 00:36:04.290 --> 00:36:06.090 possible number of nodes. 00:36:09.038 --> 00:36:09.955 So with this minimum-- 00:36:23.060 --> 00:36:24.710 has a minimum number of nodes. 00:36:24.710 --> 00:36:40.990 And furthermore, among all H with this number of nodes, 00:36:40.990 --> 00:36:53.230 choose the node weights, which we'll denote by a1 through a n, 00:36:53.230 --> 00:36:54.700 summing to 1. 00:36:54.700 --> 00:37:01.030 Chooses node weights so that this expression, the sum, 00:37:01.030 --> 00:37:02.753 is minimized. 00:37:08.680 --> 00:37:11.117 And by compactness-- and now we're 00:37:11.117 --> 00:37:13.450 not even talking about compact in the space of graphons. 00:37:13.450 --> 00:37:15.075 You have a finite number of parameters. 00:37:15.075 --> 00:37:16.340 It's a continuous function. 00:37:16.340 --> 00:37:19.000 So just by compactness, there exists 00:37:19.000 --> 00:37:21.340 such an H for which the minimum is achieved. 00:37:24.530 --> 00:37:26.155 This is minimizing over integers. 00:37:26.155 --> 00:37:28.840 And here, minimizing over a finite set 00:37:28.840 --> 00:37:32.300 of bounded real numbers. 00:37:32.300 --> 00:37:34.750 So the name of the game now is we have 00:37:34.750 --> 00:37:37.870 this H, which is minimizing. 00:37:37.870 --> 00:37:41.620 And I want to show that H has certain properties. 00:37:41.620 --> 00:37:43.120 If it doesn't have these properties, 00:37:43.120 --> 00:37:45.340 I can decrease those values. 00:38:01.450 --> 00:38:06.250 So let's see what properties this H must 00:38:06.250 --> 00:38:11.150 have if it has the minimum number of nodes and f of H 00:38:11.150 --> 00:38:12.310 is minimum possible. 00:38:16.180 --> 00:38:25.620 So first I claim that all the node weights are positive. 00:38:25.620 --> 00:38:27.480 If not, I can delete that node and decrease 00:38:27.480 --> 00:38:28.400 the number of nodes. 00:38:38.880 --> 00:38:41.970 I would like to claim that H must 00:38:41.970 --> 00:38:54.010 be a complete graph because if some ij is not edge of H-- 00:38:54.010 --> 00:38:56.715 so here i is different from j. 00:38:56.715 --> 00:38:57.590 I do not allow loops. 00:38:57.590 --> 00:38:59.410 It's just simple. 00:38:59.410 --> 00:39:05.412 Then let's think about what this expression f of H should be. 00:39:05.412 --> 00:39:06.870 So I don't want to write this down, 00:39:06.870 --> 00:39:08.520 but I want you to imagine in your head. 00:39:08.520 --> 00:39:15.990 So you have this graphon H. I'm Looking at the clique density. 00:39:15.990 --> 00:39:19.600 It's some polynomial. 00:39:19.600 --> 00:39:23.990 In fact, it's some multilinear-- 00:39:23.990 --> 00:39:27.380 it's some polynomial in these node weights. 00:39:30.700 --> 00:39:34.490 So I want to understand what is the shape 00:39:34.490 --> 00:39:39.010 of this polynomial as a function of the node weights. 00:39:42.130 --> 00:39:53.950 And I observe that it has to be multilinear in-- 00:39:53.950 --> 00:39:58.130 has to be multilinear in particular in 00:39:58.130 --> 00:40:01.470 alpha i and alpha j. 00:40:06.090 --> 00:40:06.915 It's a polynomial. 00:40:06.915 --> 00:40:07.790 That should be clear. 00:40:07.790 --> 00:40:10.940 It is multilinear because, well, you have-- 00:40:16.810 --> 00:40:20.070 why is it multilinear? 00:40:20.070 --> 00:40:24.180 Why do I not have alpha i squared? 00:40:33.910 --> 00:40:35.266 Either of you. 00:40:35.266 --> 00:40:38.580 AUDIENCE: It says the 0 is not [INAUDIBLE].. 00:40:38.580 --> 00:40:41.270 PROFESSOR: So we're forbidding-- 00:40:41.270 --> 00:40:45.350 so here's alpha 1, alpha 2, alpha 3, alpha 4, alpha 1, 00:40:45.350 --> 00:40:48.630 alpha 2, alpha 3, alpha 4. 00:40:48.630 --> 00:40:52.877 So understand what the triangle density-- 00:40:52.877 --> 00:40:54.460 if you write down the triangle density 00:40:54.460 --> 00:40:58.990 as an expression in terms of the parameters, 00:40:58.990 --> 00:41:01.420 think about what comes out, what it looks like. 00:41:01.420 --> 00:41:06.340 And they essentially consist of you choosing a subgraph, 00:41:06.340 --> 00:41:10.260 which you cannot have repeats. 00:41:10.260 --> 00:41:11.460 So it's multilinear. 00:41:11.460 --> 00:41:14.690 So it's multilinear in particular in alpha 00:41:14.690 --> 00:41:17.440 i and alpha j. 00:41:17.440 --> 00:41:31.010 So no term has the product alpha i alpha j in it 00:41:31.010 --> 00:41:33.950 because ij is not an edge. 00:41:36.720 --> 00:41:41.820 So here's where we're really using that we're only 00:41:41.820 --> 00:41:44.580 considering clique densities. 00:41:52.380 --> 00:41:56.670 So the theorem is completely false without the assumption 00:41:56.670 --> 00:41:57.660 of clique densities. 00:41:57.660 --> 00:42:02.573 If we have a general inequality, general linear inequality, then 00:42:02.573 --> 00:42:03.990 the statement is completely false. 00:42:06.800 --> 00:42:11.240 So it's multilinear. 00:42:11.240 --> 00:42:14.360 So if we now fix all the other variables 00:42:14.360 --> 00:42:17.480 and just think about how to optimize, 00:42:17.480 --> 00:42:23.530 how to minimize f of H by tweaking alpha i and alpha j, 00:42:23.530 --> 00:42:27.350 well, it's linear, so you should minimize it 00:42:27.350 --> 00:42:29.620 by setting one of them to be zero. 00:42:32.190 --> 00:42:36.100 And that would then decrease the number of nodes. 00:42:36.100 --> 00:42:49.410 So can shift alpha i and alpha j while preserving alpha i 00:42:49.410 --> 00:42:52.755 plus alpha j and not changing-- 00:42:55.545 --> 00:43:05.670 so not increasing f of H. And then 00:43:05.670 --> 00:43:13.310 we get either alpha i to go to zero 00:43:13.310 --> 00:43:16.340 or alpha j to go to zero, in which case 00:43:16.340 --> 00:43:23.030 we decrease the number of nodes, thereby contradicting 00:43:23.030 --> 00:43:24.808 the minimality assumption. 00:43:30.180 --> 00:43:36.290 So this argument here then tells you that H must be a clique. 00:43:36.290 --> 00:43:42.110 So hence, H is complete. 00:43:46.470 --> 00:43:53.020 And if H is complete, then as a polynomial in these alphas, 00:43:53.020 --> 00:43:55.660 what should f look like? 00:43:55.660 --> 00:43:58.810 Well, it has to be symmetric with respect 00:43:58.810 --> 00:44:01.270 to all these alphas. 00:44:01.270 --> 00:44:05.750 So in particular, it has to be-- 00:44:05.750 --> 00:44:13.440 so since H is complete, we see that, in fact, now you 00:44:13.440 --> 00:44:15.720 can write down exactly what f of H 00:44:15.720 --> 00:44:20.000 is in terms of the parameters described in the problem. 00:44:20.000 --> 00:44:26.400 Namely, it's Cr times r factorial times Sr, 00:44:26.400 --> 00:44:34.150 where Sr is a symmetric polynomial where you look at-- 00:44:34.150 --> 00:44:40.290 you choose r of the terms, r of these alphas 00:44:40.290 --> 00:44:42.790 for each term in this sum. 00:44:42.790 --> 00:44:46.980 It's just elementary symmetric polynomial. 00:44:46.980 --> 00:44:50.080 And I would like to know, given such a polynomial, 00:44:50.080 --> 00:44:54.810 how to minimize this number by choosing the alphas. 00:44:54.810 --> 00:44:56.940 But if you think about what happens 00:44:56.940 --> 00:45:01.350 if you fix again everything but two of the alphas-- 00:45:01.350 --> 00:45:11.310 so by fixing all of, let's say, alpha 3 to alpha n, 00:45:11.310 --> 00:45:12.750 we find that-- 00:45:12.750 --> 00:45:20.170 so as a function in just alpha 1 and alpha 2, f of H 00:45:20.170 --> 00:45:21.550 has the following form. 00:45:32.860 --> 00:45:35.475 And because it's symmetric, these two B's 00:45:35.475 --> 00:45:36.750 are actually the same. 00:45:40.770 --> 00:45:47.390 So if we now vary alpha 1 and alpha 2 but fixing everything 00:45:47.390 --> 00:45:53.870 else, because alpha 1 plus alpha 2 is constant, 00:45:53.870 --> 00:45:57.230 I can even get rid of this linear part. 00:46:00.850 --> 00:46:03.050 So that linear part is fixed as a constant. 00:46:07.020 --> 00:46:09.810 I want to minimize this expression with alpha 1 plus 00:46:09.810 --> 00:46:11.790 alpha 2, how it's fixed. 00:46:11.790 --> 00:46:13.770 So there are two possibilities depending on 00:46:13.770 --> 00:46:19.170 whether C is positive or negative or, I guess, 0. 00:46:19.170 --> 00:46:20.370 So now you're here. 00:46:20.370 --> 00:46:35.250 So depending if C is positive or negative, 00:46:35.250 --> 00:46:43.400 it's minimized by either the two alphas equal to each other 00:46:43.400 --> 00:46:52.340 or one of the two of alphas should be zero. 00:46:52.340 --> 00:46:55.520 The latter cannot occur because we assume minimality. 00:46:55.520 --> 00:46:58.150 So the first must occur. 00:46:58.150 --> 00:47:01.880 And hence, by symmetry, if you apply the same argument 00:47:01.880 --> 00:47:04.550 to all the other alphas, all the alphas 00:47:04.550 --> 00:47:09.190 are equal to each other, which means 00:47:09.190 --> 00:47:11.190 that H is a simple clique. 00:47:15.490 --> 00:47:17.150 It's basically an unweighted clique. 00:47:20.060 --> 00:47:22.780 So in other words, if this inequality 00:47:22.780 --> 00:47:28.360 fails for some H, some node weighted H, 00:47:28.360 --> 00:47:33.740 then it must fail for a simple clique H. 00:47:33.740 --> 00:47:36.630 And that's the claim above. 00:47:36.630 --> 00:47:37.130 Yeah? 00:47:37.130 --> 00:47:38.630 AUDIENCE: So in the statement, there 00:47:38.630 --> 00:47:41.137 are two n's, are those two n's different n's then? 00:47:41.137 --> 00:47:41.720 PROFESSOR: OK. 00:47:41.720 --> 00:47:42.230 Question. 00:47:42.230 --> 00:47:43.210 There are two n's. 00:47:43.210 --> 00:47:44.670 Yeah. 00:47:44.670 --> 00:47:45.380 Thank you. 00:47:45.380 --> 00:47:47.450 So these are two-- 00:47:50.460 --> 00:47:50.960 yeah. 00:47:50.960 --> 00:47:53.122 So these are two different n's. 00:48:01.554 --> 00:48:03.042 Great. 00:48:03.042 --> 00:48:04.034 yeah. 00:48:04.034 --> 00:48:05.540 AUDIENCE: I have a question. 00:48:05.540 --> 00:48:10.130 Which is the node weight such that f of H [INAUDIBLE]?? 00:48:10.130 --> 00:48:11.590 PROFESSOR: Question is, why can we 00:48:11.590 --> 00:48:17.590 assume that you can choose H so that f of H is minimized? 00:48:17.590 --> 00:48:18.590 Its because once-- 00:48:18.590 --> 00:48:19.090 OK. 00:48:19.090 --> 00:48:20.830 So you agreed the first thing you 00:48:20.830 --> 00:48:24.187 can minimize because the number of nodes is a positive integer. 00:48:24.187 --> 00:48:25.770 So if there's a counterexample, choose 00:48:25.770 --> 00:48:27.590 the minimum counterexample. 00:48:27.590 --> 00:48:32.215 Now, you fixed that number of vertices, and the number of-- 00:48:32.215 --> 00:48:33.970 then this is an optimization problem. 00:48:33.970 --> 00:48:36.040 It's minimizing continuous function 00:48:36.040 --> 00:48:38.200 with a finite number of variables. 00:48:38.200 --> 00:48:42.600 So it has a minimum just by compactness 00:48:42.600 --> 00:48:44.095 of a continuous function. 00:48:44.095 --> 00:48:45.220 So I choose that minimizer. 00:48:50.370 --> 00:48:53.380 Any more questions? 00:48:53.380 --> 00:48:56.950 So we have this rather general looking theorem. 00:48:56.950 --> 00:48:59.283 So in the second part of today's lecture, 00:48:59.283 --> 00:49:00.700 after taking a short break, I want 00:49:00.700 --> 00:49:02.800 to discuss what are some of the consequences 00:49:02.800 --> 00:49:06.960 and also variations of that statement up there. 00:49:06.960 --> 00:49:10.240 And I want to also show you what the rest of this picture 00:49:10.240 --> 00:49:10.750 looks like. 00:49:15.030 --> 00:49:18.390 So let's continue to deduce some consequences of this theorem 00:49:18.390 --> 00:49:20.190 up there that tells us that it is 00:49:20.190 --> 00:49:22.650 pretty easy to decide linear inequalities 00:49:22.650 --> 00:49:25.830 between clique densities. 00:49:25.830 --> 00:49:31.380 Namely, to decide it, just check the inequalities on cliques. 00:49:31.380 --> 00:49:44.590 So as a corollary for each n-- 00:49:44.590 --> 00:49:47.890 yes, for each n, the extremal points-- 00:49:55.200 --> 00:50:09.440 so the extremal points of the convex hull of this set 00:50:09.440 --> 00:50:29.140 where I record the clique densities overall graphons W. 00:50:29.140 --> 00:50:32.610 So think about this set as the higher dimensional 00:50:32.610 --> 00:50:35.910 generalization of that picture I drew up there. 00:50:35.910 --> 00:50:39.810 But no previously we had n equals to 3, 00:50:39.810 --> 00:50:42.480 and we're still interested in n equals to 3. 00:50:42.480 --> 00:50:51.330 But in general, you have this set sitting in this box. 00:50:51.330 --> 00:50:55.220 And so it's some set. 00:50:55.220 --> 00:50:59.240 And if I take the convex hull of the set, what 00:50:59.240 --> 00:51:03.620 that theorem tells us-- and it requires maybe one bit 00:51:03.620 --> 00:51:04.890 of extra computation. 00:51:04.890 --> 00:51:09.980 But what it tells us is that the extremal points are precisely 00:51:09.980 --> 00:51:23.600 the points given by W equals to Km for all m equal to 1. 00:51:23.600 --> 00:51:27.950 So evaluate, find what this point is for each m, 00:51:27.950 --> 00:51:31.250 and you have a bunch of points. 00:51:31.250 --> 00:51:34.520 And those are the convex hull. 00:51:34.520 --> 00:51:37.250 So I'll illustrate by drawing what the points are 00:51:37.250 --> 00:51:38.840 for the picture over there. 00:51:38.840 --> 00:51:42.170 But it essentially follows from Bollobas' theorem 00:51:42.170 --> 00:51:44.390 with one extra bit of computation 00:51:44.390 --> 00:51:47.340 to make sure that all of these are actually 00:51:47.340 --> 00:51:49.100 extremal points of the convex hull. 00:51:49.100 --> 00:51:51.770 None of them is contained in the convex hull 00:51:51.770 --> 00:51:54.750 of the other points. 00:51:54.750 --> 00:51:59.998 So for example, we can also deduce very easily 00:51:59.998 --> 00:52:00.665 Turan's theorem. 00:52:05.840 --> 00:52:08.350 So what does Turan's theorem tell us? 00:52:08.350 --> 00:52:20.180 It tells us that if the r plus 1 clique density is zero, 00:52:20.180 --> 00:52:32.930 then the K2 density is at most 1 minus r. 00:52:32.930 --> 00:52:36.245 So why does Turan's theorem follow from the above claims? 00:52:41.350 --> 00:52:44.740 It should follow because all the data here has 00:52:44.740 --> 00:52:46.360 to do with clique densities. 00:52:46.360 --> 00:52:47.860 And everything we saw so far says 00:52:47.860 --> 00:52:50.110 that if you just want to understand 00:52:50.110 --> 00:52:52.630 linear inequalities between clique densities, 00:52:52.630 --> 00:52:53.320 it's super easy. 00:53:01.600 --> 00:53:04.372 Maybe I'll draw the picture for triangles, 00:53:04.372 --> 00:53:05.830 and then you'll see what it's like. 00:53:08.700 --> 00:53:12.060 So the corollary tells us for this picture, 00:53:12.060 --> 00:53:15.540 corresponding to n equals to 3, what 00:53:15.540 --> 00:53:19.670 the points, the extreme point of the convex hull are. 00:53:19.670 --> 00:53:22.100 So let me let me draw these points for you. 00:53:22.100 --> 00:53:26.340 So one of these points is this 1/2 comma 0. 00:53:26.340 --> 00:53:29.180 So that corresponds to Mantel's theorem. 00:53:32.400 --> 00:53:36.680 Now, if you go to the other values of m, 00:53:36.680 --> 00:53:39.290 you find that those points-- 00:53:39.290 --> 00:53:41.930 so the extreme points-- 00:53:41.930 --> 00:53:50.200 they are of the form m minus 1 divided by m, m minus 1 m 00:53:50.200 --> 00:53:57.370 minus 2 divided by m squared for positive integers m. 00:54:00.396 --> 00:54:03.760 So for m equals to 2, that's the point that we just drew. 00:54:03.760 --> 00:54:04.630 And the next point-- 00:54:09.270 --> 00:54:12.110 so next two points, one third and one fourth, 00:54:12.110 --> 00:54:14.690 they are at, if you plug it in-- 00:54:17.850 --> 00:54:18.350 thank you. 00:54:18.350 --> 00:54:21.760 2/3 and 3/4. 00:54:21.760 --> 00:54:25.720 They correspond to 2/9 and 3/8. 00:54:28.780 --> 00:54:30.940 So let me show you where these points are. 00:54:30.940 --> 00:54:38.740 So they are at here and over there. 00:54:38.740 --> 00:54:41.860 And you have this sequence of points going up. 00:54:51.170 --> 00:54:53.110 So this is the convex hull. 00:54:53.110 --> 00:54:55.900 And from that information, you should already 00:54:55.900 --> 00:54:58.710 be able to deduce Mantel's theorem 00:54:58.710 --> 00:55:03.870 because this right half is not part of this convex hull. 00:55:03.870 --> 00:55:05.670 So that's what Mantel's theorem. 00:55:05.670 --> 00:55:09.900 And similarly, the deduction to Turan's theorem 00:55:09.900 --> 00:55:11.490 also follows by a similar logic. 00:55:13.810 --> 00:55:14.310 OK. 00:55:14.310 --> 00:55:17.110 So you have this sequence of points. 00:55:17.110 --> 00:55:21.430 Now, it happens that all of these points lie on a curve. 00:55:21.430 --> 00:55:26.190 So let me try to draw what this extra curve is. 00:55:30.140 --> 00:55:34.966 So there is some curve, like that. 00:55:38.438 --> 00:55:40.230 So there's some curve like that. 00:55:42.780 --> 00:55:51.750 The equation of this curve happens to be x 2x minus 1. 00:55:51.750 --> 00:55:53.940 And because the regions is contained 00:55:53.940 --> 00:55:56.010 in the convex hull, the yellow points, 00:55:56.010 --> 00:56:01.720 it certainly lies above this convex red curve. 00:56:01.720 --> 00:56:03.160 You've seen this red curve before. 00:56:06.470 --> 00:56:06.970 From where? 00:56:19.010 --> 00:56:20.010 So what is that saying? 00:56:20.010 --> 00:56:23.740 It's saying that if your edge density is beyond 00:56:23.740 --> 00:56:27.010 above one half, then you have some lower bound 00:56:27.010 --> 00:56:29.730 on the triangle density. 00:56:29.730 --> 00:56:31.560 Where have we seen this before? 00:56:31.560 --> 00:56:32.658 Problem set one. 00:56:32.658 --> 00:56:34.450 There was a problem on problem set one that 00:56:34.450 --> 00:56:36.890 says exactly this inequality. 00:56:36.890 --> 00:56:38.660 So go back and compare what it did. 00:56:43.580 --> 00:56:46.710 But of course, the convex hull result 00:56:46.710 --> 00:56:48.410 tells you even a little bit more-- 00:56:48.410 --> 00:56:51.950 namely, that you can draw line segments 00:56:51.950 --> 00:56:54.770 between these convex hull points. 00:56:59.660 --> 00:57:01.850 So you have some polygonal reason that 00:57:01.850 --> 00:57:03.230 lower bounds the actual region. 00:57:06.312 --> 00:57:07.520 So what is the actual region? 00:57:07.520 --> 00:57:09.390 So leaving you in suspense. 00:57:09.390 --> 00:57:11.990 So let me tell you what the actual region is now. 00:57:11.990 --> 00:57:14.680 So it turns out to be actually-- it's beautiful 00:57:14.680 --> 00:57:19.043 and it's quite deep, that the region is now 00:57:19.043 --> 00:57:19.960 completely understood. 00:57:19.960 --> 00:57:22.600 And it's a fairly recent result. It's only about 10 years ago 00:57:22.600 --> 00:57:31.250 roughly that there are some concave curves. 00:57:31.250 --> 00:57:39.120 The sequence of scallops going up to the top right corner. 00:57:39.120 --> 00:57:42.930 And this is now understood to be the complete region 00:57:42.930 --> 00:57:44.580 between these lower and upper curves. 00:57:47.510 --> 00:57:50.680 So this is the complete set of feasible regions for edge 00:57:50.680 --> 00:57:54.310 versus triangle densities. 00:57:54.310 --> 00:57:57.100 So this lower curve is a difficult result 00:57:57.100 --> 00:57:57.870 due to Razborov. 00:58:02.730 --> 00:58:07.600 And I want to give you a statement what this curve is. 00:58:07.600 --> 00:58:12.760 And Razborov came up with this machinery, this technique, 00:58:12.760 --> 00:58:14.770 known by the name of flag algebra. 00:58:14.770 --> 00:58:16.930 So actually, he came up with this name. 00:58:22.390 --> 00:58:24.390 So I won't really tell you what flag algebra is, 00:58:24.390 --> 00:58:26.670 but it's kind of a computerized way of doing 00:58:26.670 --> 00:58:28.440 Cauchy-Schwarz inequalities. 00:58:28.440 --> 00:58:33.930 So many of our proofs for this graph through inequalities, 00:58:33.930 --> 00:58:35.850 they go through some kind of Cauchy-Schwarz 00:58:35.850 --> 00:58:38.130 or sum of squares equivalently. 00:58:38.130 --> 00:58:42.033 But there are some very large or difficult inequalities 00:58:42.033 --> 00:58:43.200 you can also prove this way. 00:58:43.200 --> 00:58:45.840 But it may be difficult to find exactly what is 00:58:45.840 --> 00:58:48.630 the actual inequality-- the chain of Cauchy-Schwarz 00:58:48.630 --> 00:58:50.970 or the sum of squares that you should write down. 00:58:50.970 --> 00:58:56.550 So this machinery, flag algebra, is a language, 00:58:56.550 --> 00:58:59.910 is a framework for setting up those sum of squares 00:58:59.910 --> 00:59:02.220 inequalities in the context of proving 00:59:02.220 --> 00:59:04.710 graph theoretic inequalities. 00:59:04.710 --> 00:59:07.120 So it can be used in many different ways. 00:59:07.120 --> 00:59:11.760 And notably, a lot of people have used serious computer 00:59:11.760 --> 00:59:13.180 computations. 00:59:13.180 --> 00:59:14.820 If I want to prove something is true, 00:59:14.820 --> 00:59:19.050 I plug it into what's called a semidefinite program that 00:59:19.050 --> 00:59:23.190 allows me to decide what kinds of Cauchy-Schwarz inequalities 00:59:23.190 --> 00:59:26.680 I should be applying to derive the result I want to prove. 00:59:26.680 --> 00:59:28.310 So that's what flag algebra roughly is. 00:59:32.130 --> 00:59:35.948 So what Razborov proved is the following. 00:59:42.430 --> 00:59:46.030 So Razborov's theorem, which is drawn up there-- 00:59:46.030 --> 00:59:47.830 that's the lower curve-- 00:59:47.830 --> 00:59:50.820 is that for fixed-- 00:59:50.820 --> 01:00:02.410 so for a fixed value of edge densities, 01:00:02.410 --> 01:00:15.670 if it lies between two specific points, drawn above, 01:00:15.670 --> 01:00:23.820 the minimum value of triangle density 01:00:23.820 --> 01:00:26.530 with a fixed value of edge density 01:00:26.530 --> 01:00:30.164 is attained via the following construction. 01:00:33.960 --> 01:00:37.920 It's attained by the step function of the graphon 01:00:37.920 --> 01:00:48.340 corresponding to a K clique. 01:00:48.340 --> 01:00:51.160 So a complete graph on K vertices 01:00:51.160 --> 01:01:03.450 with node weights alpha 1 through alpha K summing to 1, 01:01:03.450 --> 01:01:07.680 and such that the first K minus 1 of the node weights 01:01:07.680 --> 01:01:09.300 are equal. 01:01:09.300 --> 01:01:18.970 And the last one is Smaller 01:01:18.970 --> 01:01:19.690 All right. 01:01:19.690 --> 01:01:24.610 And the point here is that if you are given a specific edge 01:01:24.610 --> 01:01:27.490 weight, edge density, then there is 01:01:27.490 --> 01:01:33.580 a unique choice of these alphas that achieve that edge density. 01:01:33.580 --> 01:01:37.420 And that is the graphon you should use that minimizes 01:01:37.420 --> 01:01:38.800 the triangle density-- 01:01:38.800 --> 01:01:40.998 describes the lower curve. 01:01:40.998 --> 01:01:43.540 So you can write down specific equations for the lower curve, 01:01:43.540 --> 01:01:44.510 but it's not so important. 01:01:44.510 --> 01:01:46.052 This is a more important description. 01:01:46.052 --> 01:01:48.070 These are the graphs that come out. 01:01:48.070 --> 01:01:51.430 And what is something that is actually quite-- 01:01:51.430 --> 01:01:53.680 I mean, why you should suspect this theorem is 01:01:53.680 --> 01:01:57.220 difficult is that unlike Turan's theorem-- so Turan's theorem, 01:01:57.220 --> 01:02:00.580 which corresponds to all those discrete points. 01:02:00.580 --> 01:02:04.370 In Turan's theorem, the minimizer is unique. 01:02:04.370 --> 01:02:07.950 I tell you the number-- 01:02:07.950 --> 01:02:10.680 I tell you that the edge density is 2/3, 01:02:10.680 --> 01:02:13.220 and I want you to minimize the number of triangles. 01:02:17.190 --> 01:02:19.250 Not from Turan's theorem, but it turns out 01:02:19.250 --> 01:02:22.790 that this extremal point is unique. 01:02:22.790 --> 01:02:26.750 Essentially corresponds to a complete three partite graph. 01:02:26.750 --> 01:02:30.080 But for the intermediate values, the constructions 01:02:30.080 --> 01:02:31.920 are not unique. 01:02:31.920 --> 01:02:43.560 So unless the K2 density is exactly of this form, 01:02:43.560 --> 01:02:49.760 the minimizer is not unique. 01:02:52.770 --> 01:02:54.680 And the reason why it is not unique 01:02:54.680 --> 01:03:03.130 is that you can replace-- 01:03:03.130 --> 01:03:04.790 so what's going on here? 01:03:04.790 --> 01:03:06.400 So you have this graphon. 01:03:11.040 --> 01:03:14.270 Alpha 1, alpha 2, alpha 3. 01:03:14.270 --> 01:03:34.600 I can replace this graphon here by any triangle free graphon 01:03:34.600 --> 01:03:35.830 of the same edge density. 01:03:41.240 --> 01:03:43.670 And there are lots and lots of them. 01:03:43.670 --> 01:03:46.550 And the non-uniqueness of the minimizer 01:03:46.550 --> 01:03:54.370 makes this minimization problem much more difficult. 01:03:54.370 --> 01:03:56.580 So Razborov proved this result for edge 01:03:56.580 --> 01:03:58.320 versus triangle densities. 01:03:58.320 --> 01:04:04.290 And this program was later completed to K4, 01:04:04.290 --> 01:04:07.800 and more generally, to Kr So K4 is 01:04:07.800 --> 01:04:17.060 due to a result of Nikiforov, and the Kr result of Reiher 01:04:17.060 --> 01:04:18.200 So a similar picture. 01:04:18.200 --> 01:04:19.850 It's more or less that picture up 01:04:19.850 --> 01:04:22.840 there but with the actual numbers shifted. 01:04:22.840 --> 01:04:26.380 Instead of edge versus triangle, it is now edge versus Kr. 01:04:29.740 --> 01:04:32.818 I should say that it's worth-- so this is a picture that I 01:04:32.818 --> 01:04:35.110 drew up there, and this is roughly the picture that you 01:04:35.110 --> 01:04:36.520 see in textbooks-- 01:04:36.520 --> 01:04:38.730 how they draw these scallops. 01:04:38.730 --> 01:04:42.970 I once plotted what this picture looks like in Mathematica, 01:04:42.970 --> 01:04:46.090 just to see for myself where the actual graph is. 01:04:46.090 --> 01:04:48.340 And it doesn't actually look like that. 01:04:48.340 --> 01:04:50.500 The concaveness is very subtle. 01:04:50.500 --> 01:04:53.170 If you draw it on a computer, they look like straight lines. 01:04:53.170 --> 01:04:55.320 So in some sense, that's a cartoon. 01:04:55.320 --> 01:04:59.170 So the concaveness is caricatured. 01:04:59.170 --> 01:05:02.563 So it's not actually as concave as it is drawn, 01:05:02.563 --> 01:05:04.480 but I think it's a good illustration of what's 01:05:04.480 --> 01:05:05.355 happening in reality. 01:05:09.700 --> 01:05:12.830 Questions? 01:05:12.830 --> 01:05:18.040 So on one hand, every polynomial graph inequality-- 01:05:18.040 --> 01:05:20.840 so what do I mean by a polynomial graph inequality? 01:05:20.840 --> 01:05:27.270 So something like-- suppose I have 01:05:27.270 --> 01:05:28.920 some inequality of this form. 01:05:38.890 --> 01:05:40.890 And I want to know, is this true? 01:05:43.420 --> 01:05:44.850 It turns out that I don't actually 01:05:44.850 --> 01:05:49.380 need these squares in some sense because I can always 01:05:49.380 --> 01:05:58.600 replace them by what happens if you take disjoint unions. 01:05:58.600 --> 01:06:03.100 So all I'm trying to say is that every polynomial graph 01:06:03.100 --> 01:06:09.670 inequality can be written as a linear graph 01:06:09.670 --> 01:06:11.980 inequality of densities. 01:06:18.710 --> 01:06:21.170 But nevertheless, this still captures a very large class 01:06:21.170 --> 01:06:22.890 of graph inequalities. 01:06:22.890 --> 01:06:25.850 And if I just give you some arbitrary one that is not 01:06:25.850 --> 01:06:27.380 of that form, it can be often very 01:06:27.380 --> 01:06:30.000 difficult to decide whether it is true or not. 01:06:30.000 --> 01:06:31.250 So over here it's not so hard. 01:06:31.250 --> 01:06:32.625 You just plug it in, and then you 01:06:32.625 --> 01:06:35.690 can decide whether it is true. 01:06:35.690 --> 01:06:39.080 I mean, it turns out to decide whether this inequality is 01:06:39.080 --> 01:06:42.110 true, it's really a polynomial. 01:06:42.110 --> 01:06:44.600 And then you just check. 01:06:44.600 --> 01:06:46.200 It's not too hard to do. 01:06:46.200 --> 01:06:50.890 But in general, suppose I give you an inequality of this form. 01:06:50.890 --> 01:06:54.610 So some generalized version of a linear inequality, like that. 01:06:54.610 --> 01:07:00.000 It's even decidable if the inequality holds. 01:07:00.000 --> 01:07:04.110 Decidable in the sense of Turing halting problem. 01:07:04.110 --> 01:07:07.350 So is there some computer program 01:07:07.350 --> 01:07:10.110 give you this inequality is true? 01:07:10.110 --> 01:07:12.120 I wonder, can you write a computer program 01:07:12.120 --> 01:07:16.140 that decides the truthfulness? 01:07:16.140 --> 01:07:17.072 It turns out-- OK. 01:07:17.072 --> 01:07:18.780 So before telling you what the answer is, 01:07:18.780 --> 01:07:22.500 let me just put it in some context. 01:07:22.500 --> 01:07:24.452 What about more classical questions before we 01:07:24.452 --> 01:07:25.410 jump into graph theory? 01:07:25.410 --> 01:07:37.790 If I give you some polynomial p over the real numbers 01:07:37.790 --> 01:07:46.500 and I want to check is that true-- 01:07:46.500 --> 01:07:49.900 so this is not too hard. 01:07:49.900 --> 01:07:51.010 So this is not too hard. 01:07:51.010 --> 01:08:02.220 But what if you have multivariate for all real? 01:08:06.440 --> 01:08:08.360 Does anyone know the answer? 01:08:08.360 --> 01:08:09.320 Is this decidable? 01:08:14.250 --> 01:08:15.720 So as you can imagine, these things 01:08:15.720 --> 01:08:18.810 were studied pretty classically. 01:08:18.810 --> 01:08:22.950 And so it turns out that every first word 01:08:22.950 --> 01:08:25.500 or theory over the real numbers is decidable. 01:08:25.500 --> 01:08:27.939 So this is a result of Tarski. 01:08:27.939 --> 01:08:31.979 In particular, such questions are decidable. 01:08:31.979 --> 01:08:34.946 And in fact, there is a very nice characterization of-- 01:08:34.946 --> 01:08:39.569 so there's a result called Artin's theorem 01:08:39.569 --> 01:08:41.850 that tells you that every such polynomial, 01:08:41.850 --> 01:08:45.000 if it is non-negative, then if and only 01:08:45.000 --> 01:08:47.729 if, it can be written as a sum of squares 01:08:47.729 --> 01:08:50.250 of rational functions. 01:08:50.250 --> 01:08:52.170 So there's a very nice characterization 01:08:52.170 --> 01:08:57.080 of positiveness of polynomials over the reals. 01:09:00.109 --> 01:09:03.410 But now I change the question and I ask, 01:09:03.410 --> 01:09:04.970 what about over the integers? 01:09:09.350 --> 01:09:11.950 So if I give you a polynomial, is it always non-negative 01:09:11.950 --> 01:09:16.760 if I have integer entries? 01:09:16.760 --> 01:09:18.078 Is this decidable? 01:09:21.214 --> 01:09:24.910 So turns out, this is not decidable. 01:09:24.910 --> 01:09:25.859 And this is related. 01:09:25.859 --> 01:09:29.560 So it's more or less the same as the undecidability 01:09:29.560 --> 01:09:42.670 of Diophantine equations, which is also known 01:09:42.670 --> 01:09:44.080 as Hilbert's tenth problem. 01:09:50.660 --> 01:09:54.770 So there is no computer program where we give you a Diophantine 01:09:54.770 --> 01:09:58.370 equation and solves the question or even tells you 01:09:58.370 --> 01:10:02.180 whether the equation has a solution. 01:10:02.180 --> 01:10:05.223 And this is part of what makes number theory, 01:10:05.223 --> 01:10:06.890 makes Diophantine equations interesting. 01:10:06.890 --> 01:10:11.630 So it's undecidable, but we talk about it. 01:10:15.360 --> 01:10:19.620 So undecidability is a famous result due to Matiyasevich. 01:10:24.790 --> 01:10:28.870 So what about graph theoretic inequalities? 01:10:28.870 --> 01:10:45.555 So is a graph homomorphism inequality decidable? 01:10:51.477 --> 01:10:53.310 I mean, the question you should ask yourself 01:10:53.310 --> 01:10:55.290 is, which one is it closer to? 01:10:55.290 --> 01:10:59.280 Is it closer to deciding the positiveness of polynomials 01:10:59.280 --> 01:11:03.430 over reals or over integers? 01:11:03.430 --> 01:11:07.000 On one hand, you might think that it 01:11:07.000 --> 01:11:10.238 is more similar to the question of polynomials over real. 01:11:10.238 --> 01:11:12.280 So first of all, why it's similar to polynomials, 01:11:12.280 --> 01:11:16.090 I hope that's at least intuitively-- 01:11:16.090 --> 01:11:17.830 nothing's a proof, but intuitively it 01:11:17.830 --> 01:11:19.800 feels somewhat similar to polynomials. 01:11:19.800 --> 01:11:22.140 And all of these guys you can write down 01:11:22.140 --> 01:11:23.630 as polynomial-like quantities. 01:11:23.630 --> 01:11:28.633 And we saw this earlier in the proof of Bollobas' theorem. 01:11:28.633 --> 01:11:30.300 So you might think it's similar to reals 01:11:30.300 --> 01:11:32.610 because, well, for graphons, you can 01:11:32.610 --> 01:11:34.380 take arbitrary real weights. 01:11:34.380 --> 01:11:37.150 So it feels like the reals. 01:11:37.150 --> 01:11:46.530 So it turns out, due to a theorem of Hatami and Norine, 01:11:46.530 --> 01:11:50.250 that the answer is no. 01:11:50.250 --> 01:11:53.940 It is not decidable. 01:11:53.940 --> 01:11:59.380 And roughly the reason has to do with this picture. 01:11:59.380 --> 01:12:04.720 Even though the space of graphons is not discrete, 01:12:04.720 --> 01:12:07.840 it's a very continuous object, even if you just 01:12:07.840 --> 01:12:10.120 look at this picture here, you have 01:12:10.120 --> 01:12:14.750 a bunch of discrete points along this scallop. 01:12:14.750 --> 01:12:18.580 So here's a potential strategy for proving 01:12:18.580 --> 01:12:24.190 the undecidability of graph homomorphism inequalities. 01:12:24.190 --> 01:12:29.310 I start by just restricting myself to this curve. 01:12:29.310 --> 01:12:32.892 I restrict myself to the red curve. 01:12:32.892 --> 01:12:34.850 If you restrict yourself to the red curve, than 01:12:34.850 --> 01:12:36.620 the set of possibilities-- 01:12:36.620 --> 01:12:39.845 it's now a discrete set, which is like the positive integers. 01:12:43.670 --> 01:12:46.600 And now I start with-- 01:12:46.600 --> 01:12:51.070 I can reduce the problem to the problem of decidability 01:12:51.070 --> 01:12:54.510 of integer inequalities. 01:12:54.510 --> 01:12:57.220 I start with an integer inequality. 01:12:57.220 --> 01:13:03.480 I convert it to an inequality about points on this red curve. 01:13:05.990 --> 01:13:14.170 And that turns into a corresponding graph inequality, 01:13:14.170 --> 01:13:16.231 which must then be undecidable. 01:13:19.020 --> 01:13:22.970 So this undecidability result is related to the discreteness 01:13:22.970 --> 01:13:25.270 of points on this red curve. 01:13:29.990 --> 01:13:32.370 So general undecidability results are interesting. 01:13:32.370 --> 01:13:35.220 But often, we're interested in specific problems. 01:13:35.220 --> 01:13:38.970 So I give you some specific inequality and ask, is it true? 01:13:38.970 --> 01:13:42.720 And there are a lot of interesting open problems 01:13:42.720 --> 01:13:43.680 of that type. 01:13:43.680 --> 01:13:45.990 My favorite one, and also a very important problem 01:13:45.990 --> 01:13:48.170 in extremal graph theory, is known 01:13:48.170 --> 01:13:50.026 as Sidorenko's conjecture. 01:13:58.657 --> 01:14:00.740 So the main cause conjecture-- it's a conjecture-- 01:14:00.740 --> 01:14:11.500 says that if H is bipartite, then the H density in G or W 01:14:11.500 --> 01:14:14.890 is at least the edge density raised 01:14:14.890 --> 01:14:21.060 to the power of the number of edges of H. 01:14:21.060 --> 01:14:27.820 So we saw one example of this inequality 01:14:27.820 --> 01:14:29.990 when H is the fourth cycle. 01:14:29.990 --> 01:14:31.630 So when we discussed quasi-randomness 01:14:31.630 --> 01:14:33.945 we saw that this is true. 01:14:33.945 --> 01:14:35.820 And in the homework, you'll have a few more-- 01:14:35.820 --> 01:14:37.445 so in the next problem homework, you'll 01:14:37.445 --> 01:14:39.040 have a few more examples where you're 01:14:39.040 --> 01:14:41.790 asked to show this inequality. 01:14:41.790 --> 01:14:42.940 It is open. 01:14:42.940 --> 01:14:44.660 We don't know any counterexamples. 01:14:44.660 --> 01:14:49.240 And the first open example, it's known as something 01:14:49.240 --> 01:14:50.260 called a Mobius strip. 01:14:55.240 --> 01:14:59.920 So the Mobius strip graph, which is 01:14:59.920 --> 01:15:04.150 a fancy name for the graph consisting of taking a K55 01:15:04.150 --> 01:15:05.500 and removing a 10 cycle. 01:15:16.390 --> 01:15:17.570 So that's the graph. 01:15:17.570 --> 01:15:20.560 It is open whether this inequality holds for that graph 01:15:20.560 --> 01:15:22.010 there. 01:15:22.010 --> 01:15:24.187 And this is something of great interest. 01:15:24.187 --> 01:15:26.020 So if you can make progress on this problem, 01:15:26.020 --> 01:15:27.415 people will be very excited. 01:15:31.750 --> 01:15:37.072 Now, why is this called a Mobius strip? 01:15:37.072 --> 01:15:38.530 This took me a while to figure out. 01:15:38.530 --> 01:15:40.363 So there are many different interpretations. 01:15:40.363 --> 01:15:42.610 I think the reason why it's called a Mobius strip is 01:15:42.610 --> 01:15:47.170 that if you think about the usual simplicial complex 01:15:47.170 --> 01:15:47.980 for a Mobius strip. 01:15:50.950 --> 01:15:56.110 And then this is the face vertex incidence bipartite graph. 01:15:56.110 --> 01:15:58.570 So five vertices, one for each face. 01:15:58.570 --> 01:16:02.530 Five vertices, one for each vertex. 01:16:02.530 --> 01:16:05.890 And if you draw the incident structure, that's the graph. 01:16:05.890 --> 01:16:10.150 I'm not sure if this topological formulation will 01:16:10.150 --> 01:16:15.305 help you improving Sidorenko's conjecture or disprove it, 01:16:15.305 --> 01:16:17.680 but certainly that that's why it's called a Mobius strip. 01:16:17.680 --> 01:16:19.222 And there are some people believe who 01:16:19.222 --> 01:16:22.170 believe that it may be false. 01:16:22.170 --> 01:16:23.920 So it's still open. 01:16:23.920 --> 01:16:25.068 It's still open. 01:16:27.820 --> 01:16:30.130 The one last thing I want to mention 01:16:30.130 --> 01:16:35.330 is that even though the inequality written up there 01:16:35.330 --> 01:16:38.630 in general is undecidable, if you only 01:16:38.630 --> 01:16:41.180 want to know whether this inequality is true 01:16:41.180 --> 01:16:45.830 up to an epsilon error, then it has decidable. 01:16:45.830 --> 01:16:49.850 In fact, there is an algorithm that I can tell you. 01:16:49.850 --> 01:16:58.010 So there exists an algorithm that decides, 01:16:58.010 --> 01:17:04.568 for every epsilon, that resides-- 01:17:04.568 --> 01:17:06.860 so I just want to know whether that inequality is true. 01:17:06.860 --> 01:17:10.310 But I allow an epsilon error, meaning 01:17:10.310 --> 01:17:27.860 it decides correctly this inequality is true 01:17:27.860 --> 01:17:36.530 up to an epsilon error for all G or outputs a G such 01:17:36.530 --> 01:17:45.616 that the sum here is negative. 01:17:49.430 --> 01:17:52.930 So up to an epsilon of error, I can give you an algorithm. 01:17:52.930 --> 01:17:56.230 And the algorithm follows-- 01:17:56.230 --> 01:17:59.800 I mean, it's not too hard to describe. 01:17:59.800 --> 01:18:02.500 Basically, the idea is that if I take 01:18:02.500 --> 01:18:07.010 an epsilon regular partition, then all the data 01:18:07.010 --> 01:18:10.090 about edge densities can be encoded 01:18:10.090 --> 01:18:12.340 in the epsilon regular partition. 01:18:12.340 --> 01:18:17.590 So apply even the weak regularity lemma is enough. 01:18:22.190 --> 01:18:32.150 And then we can test the bounded number of possibilities 01:18:32.150 --> 01:18:37.820 with some fixed number of parts. 01:18:41.600 --> 01:18:47.270 And by the counting lemma, you lose some epsilon error 01:18:47.270 --> 01:18:51.270 if I check over all weighted graphs 01:18:51.270 --> 01:18:54.860 on some bounded number of parts whose edge weights are 01:18:54.860 --> 01:18:59.330 multiples of epsilon, let's say, whether this is true. 01:18:59.330 --> 01:19:02.090 If it's true, then it is true with this epsilon. 01:19:02.090 --> 01:19:03.560 If it is false, then I can already 01:19:03.560 --> 01:19:06.860 output a counterexample. 01:19:06.860 --> 01:19:08.870 So there is only finitely many possibilities 01:19:08.870 --> 01:19:11.090 as a result of weak regularity lemma. 01:19:11.090 --> 01:19:15.870 And therefore, this version here is decidable. 01:19:15.870 --> 01:19:20.190 So today, we saw many different graph theoretic inequalities 01:19:20.190 --> 01:19:21.520 and some general results. 01:19:21.520 --> 01:19:24.780 And there are lots of open problems about graph 01:19:24.780 --> 01:19:27.210 homomorphism inequalities. 01:19:27.210 --> 01:19:31.055 So this concludes roughly the extremal graph theory 01:19:31.055 --> 01:19:32.020 section of this course. 01:19:32.020 --> 01:19:33.810 So starting from next lecture, we'll 01:19:33.810 --> 01:19:36.130 be looking at Roth's theorem. 01:19:36.130 --> 01:19:37.710 So looking at the Fourier analytic 01:19:37.710 --> 01:19:40.550 proof of Roth's theorem.