WEBVTT 00:00:18.955 --> 00:00:20.330 YUFEI ZHAO: Last time, we started 00:00:20.330 --> 00:00:24.950 discussing the extremal problem for bipartide graphs. 00:00:24.950 --> 00:00:28.970 And in particular, we saw the Kovari-Sos-Turan theorem, 00:00:28.970 --> 00:00:32.810 which tells us that if you forbid your graph from having 00:00:32.810 --> 00:00:36.320 a complete bipartide graph, Kst, then 00:00:36.320 --> 00:00:42.670 you have this upper bound on the number of edges in your graph. 00:00:42.670 --> 00:00:43.480 So we gave a proof. 00:00:43.480 --> 00:00:47.680 It was fairly short, used the double counting argument, 00:00:47.680 --> 00:00:49.750 and it give you this bound. 00:00:49.750 --> 00:00:53.050 And the next question is how tight is this bound? 00:00:53.050 --> 00:00:57.760 Is there a lower bound that is off by, let's say, 00:00:57.760 --> 00:01:00.130 at most a constant factor? 00:01:00.130 --> 00:01:02.260 And that's a major open problem. 00:01:02.260 --> 00:01:07.680 It's a conjecture that this bound is tight 00:01:07.680 --> 00:01:09.830 up to constant factors. 00:01:09.830 --> 00:01:12.960 But that conjecture is known for only a very small number 00:01:12.960 --> 00:01:14.010 of graphs. 00:01:14.010 --> 00:01:17.590 And we saw a couple of examples last time. 00:01:17.590 --> 00:01:20.400 So last time we saw construction that 00:01:20.400 --> 00:01:28.050 shows that for S equals to 2, this bound is tight. 00:01:31.670 --> 00:01:36.110 So the extremal number for k through 2 is on the order of n 00:01:36.110 --> 00:01:37.670 to the 3/2. 00:01:37.670 --> 00:01:42.300 So this theta means I'm hiding constant factors. 00:01:42.300 --> 00:01:45.090 And our construction used this polarity graph, 00:01:45.090 --> 00:01:47.400 which is essentially the point line incidence 00:01:47.400 --> 00:01:49.590 graph of a projective plane. 00:01:49.590 --> 00:01:53.700 And a basic algebraic or geometric fact, if you will, 00:01:53.700 --> 00:01:57.710 that two lines intersecting at most one point. 00:01:57.710 --> 00:02:03.010 We also sketched a construction that 00:02:03.010 --> 00:02:08.580 showed that for S equals to 3, this bound is also tight. 00:02:11.370 --> 00:02:13.060 And this construction here involved 00:02:13.060 --> 00:02:21.070 using spheres, again, in some space over a finite field. 00:02:21.070 --> 00:02:22.960 So both these constructions are in some sense 00:02:22.960 --> 00:02:24.490 algebraic geometric. 00:02:24.490 --> 00:02:26.590 And you can ask, is there a way to extend 00:02:26.590 --> 00:02:31.540 these ideas to construct other examples of Kst free graphs 00:02:31.540 --> 00:02:36.040 with the right number of edges using some ingredients 00:02:36.040 --> 00:02:37.870 from algebraic geometry? 00:02:37.870 --> 00:02:40.030 And today, I want to show you two different ways 00:02:40.030 --> 00:02:40.660 of doing that. 00:02:44.870 --> 00:02:51.220 So the state of the art, which I mentioned last time, 00:02:51.220 --> 00:02:55.650 let me remind you what is known about constructions 00:02:55.650 --> 00:03:00.240 that achieve the right exponent up there. 00:03:00.240 --> 00:03:07.560 So in a series of two papers by [INAUDIBLE] 00:03:07.560 --> 00:03:11.460 it is shown that if the constants t and s, 00:03:11.460 --> 00:03:15.300 such that t is large enough compared to s, 00:03:15.300 --> 00:03:20.530 in particular t is bigger than s minus 1 factorial, 00:03:20.530 --> 00:03:27.480 then the extremal number of Kst's is 00:03:27.480 --> 00:03:30.990 of the order, same order, as given 00:03:30.990 --> 00:03:34.980 in the upper bound of the Kovari-Sos-Turan theorem. 00:03:34.980 --> 00:03:40.780 In particular, these range of parameters allows you to do 00:03:40.780 --> 00:03:46.660 (2, 2), (3, 3), which we already know how to do. 00:03:46.660 --> 00:03:49.400 But the next case is 4, 7. 00:03:49.400 --> 00:03:52.570 And it is still open how to do 4, 6. 00:03:58.940 --> 00:04:00.833 So I want to show you this construction. 00:04:00.833 --> 00:04:02.750 And I will tell you exactly what the graph is. 00:04:02.750 --> 00:04:05.090 I'll give you an explicit description 00:04:05.090 --> 00:04:09.380 of this graph, which is Kst free and has lots of edges. 00:04:11.990 --> 00:04:15.860 And as I mentioned earlier, it is an algebraic construction. 00:04:15.860 --> 00:04:22.100 So as before, we start with p prime. 00:04:22.100 --> 00:04:30.680 And we will take n to be a p raised to the power s. 00:04:30.680 --> 00:04:34.172 And let's restrict s to be integer at least 2. 00:04:34.172 --> 00:04:35.880 And, of course, that's same as last time, 00:04:35.880 --> 00:04:39.390 if you have other values of n, take a prime close 00:04:39.390 --> 00:04:42.490 to the desired value and then take it from there. 00:04:45.180 --> 00:04:50.160 To describe the construction, let me remind you, 00:04:50.160 --> 00:04:55.200 the norm map, if you have a field extension, in this case, 00:04:55.200 --> 00:04:59.150 specifically I am looking at the field extension, 00:04:59.150 --> 00:05:08.700 this Fp to the s, I can define a norm map as follows. 00:05:08.700 --> 00:05:15.060 Sending x to be the product of all the conjugates, 00:05:15.060 --> 00:05:19.740 or the Galois conjugates of x, in this field extension. 00:05:19.740 --> 00:05:23.500 So explicitly written out is just that expression, 00:05:23.500 --> 00:05:27.240 which I can clean and collect and write it down like this. 00:05:32.090 --> 00:05:33.960 So I wrote that the image of this norm map 00:05:33.960 --> 00:05:36.310 lies in the base field Fp. 00:05:36.310 --> 00:05:37.185 And that is because-- 00:05:41.388 --> 00:05:44.280 well, one of the many reasons why this is the case, 00:05:44.280 --> 00:05:47.310 is that if you look at-- 00:05:47.310 --> 00:05:49.740 so I'll denote this norm map by N-- 00:05:49.740 --> 00:05:57.990 if you look at N of x, it raised to power p, 00:05:57.990 --> 00:06:00.420 leaves this value unchanged. 00:06:00.420 --> 00:06:04.500 And the base field is the field where it 00:06:04.500 --> 00:06:07.110 is invariant under power by p. 00:06:09.930 --> 00:06:15.550 So here's the graph, which I'll denote the norm 00:06:15.550 --> 00:06:20.180 graph with parameters p and s. 00:06:20.180 --> 00:06:30.640 So the norm graph will have as vertices just the elements 00:06:30.640 --> 00:06:41.690 of this field extension. 00:06:41.690 --> 00:06:54.300 And the edges will be the set of all pairs of vertices, not 00:06:54.300 --> 00:07:00.180 equal, of course, such that a norm of their sum equals to 1. 00:07:04.547 --> 00:07:05.380 So that's the graph. 00:07:05.380 --> 00:07:07.912 This is an explicit description of what the vertices 00:07:07.912 --> 00:07:08.870 and what the edges are. 00:07:11.430 --> 00:07:14.760 So now, we need to verify a couple of things. 00:07:14.760 --> 00:07:20.040 One is that this graph has the desired number of edges. 00:07:20.040 --> 00:07:21.980 It has lots of edges. 00:07:21.980 --> 00:07:26.170 And two is that this graph is Kst free. 00:07:26.170 --> 00:07:28.980 So let's do both of those things. 00:07:28.980 --> 00:07:33.350 So the first is let's check it has the right number of edges. 00:07:33.350 --> 00:07:36.770 So that's a relatively easy task. 00:07:36.770 --> 00:07:42.580 What we need to do is to count for every a 00:07:42.580 --> 00:07:47.860 how many choices of b are there in this field extension 00:07:47.860 --> 00:07:53.650 such that a plus b has norm exactly 1? 00:07:53.650 --> 00:07:56.430 And I claim that that number-- 00:07:56.430 --> 00:08:01.110 well, so here's a basic algebra fact, 00:08:01.110 --> 00:08:07.710 that the number of elements in this field extension with norm 00:08:07.710 --> 00:08:15.120 exactly 1 is precisely p to the s minus 1 00:08:15.120 --> 00:08:17.850 divided by p to the s. 00:08:23.646 --> 00:08:27.300 And this is because really we're looking 00:08:27.300 --> 00:08:31.150 in the multiplicative subgroup. 00:08:31.150 --> 00:08:36.640 So the multiplicative group in this Fp to the s. 00:08:36.640 --> 00:08:38.258 And it has a cyclical-- 00:08:42.162 --> 00:08:44.130 so there's a generator-- 00:08:44.130 --> 00:08:51.950 the order of the cyclic group is p to the s minus 1. 00:08:51.950 --> 00:08:57.930 So you're asking, how many elements 00:08:57.930 --> 00:09:08.357 when raised to this power here ends up at the identity? 00:09:10.985 --> 00:09:11.860 So that's the answer. 00:09:15.970 --> 00:09:17.430 So that's one aspect. 00:09:17.430 --> 00:09:27.230 And so as a result, every vertex is adjacent to, well, 00:09:27.230 --> 00:09:29.330 how many vertices? 00:09:29.330 --> 00:09:33.130 For every given a I need to solve for b. 00:09:33.130 --> 00:09:35.270 And basically, this many solutions, 00:09:35.270 --> 00:09:38.000 I have to be just slightly careful because I 00:09:38.000 --> 00:09:40.430 don't want loops in my graph. 00:09:40.430 --> 00:09:42.290 So I may need to subtract 1. 00:09:42.290 --> 00:09:50.090 So it's adjacent to at least this number up here minus 1 00:09:50.090 --> 00:10:00.740 to account for possible loops, which is pretty large, 00:10:00.740 --> 00:10:05.070 p to the s minus 1, which in other words 00:10:05.070 --> 00:10:12.140 is n raised to 1 minus 1 over s, that many vertices, 00:10:12.140 --> 00:10:14.840 and you see that this gives you the right number of edges. 00:10:28.620 --> 00:10:30.820 So this is a graph with lots of edges. 00:10:30.820 --> 00:10:33.170 So that part wasn't so hard. 00:10:33.170 --> 00:10:36.040 The next part, it's much trickier, 00:10:36.040 --> 00:10:41.060 which is we want to check that this graph has no Kst's. 00:10:41.060 --> 00:10:43.100 So previously in our algebraic construction, 00:10:43.100 --> 00:10:46.520 we used some geometric facts, such as no two lines 00:10:46.520 --> 00:10:51.230 intersect in more than one point to show that there's no k to 2 00:10:51.230 --> 00:10:53.120 in the polarity graph. 00:10:53.120 --> 00:10:55.550 So there's going to be something like that here. 00:10:55.550 --> 00:11:05.410 So the claim is that this construction, this norm graph, 00:11:05.410 --> 00:11:12.880 is Ks, s factorial plus 1 free. 00:11:15.780 --> 00:11:17.890 So it's not quite the bound I claimed. 00:11:17.890 --> 00:11:19.180 So it's a little bit weaker. 00:11:19.180 --> 00:11:21.430 But it is in the spirit of what I am claiming, 00:11:21.430 --> 00:11:27.060 namely that for t large enough, this graph is Kst free. 00:11:27.060 --> 00:11:29.570 So for t a large enough comes constant here 00:11:29.570 --> 00:11:33.470 will show s factorial plus 1. 00:11:33.470 --> 00:11:41.350 And as a result, it would follow that the extremal number 4s 00:11:41.350 --> 00:11:54.060 sub s factorial plus 1 is at least 1/2 00:11:54.060 --> 00:11:56.280 minus little 1 of the constant-- 00:11:56.280 --> 00:11:59.010 I don't already worry about that much, 00:11:59.010 --> 00:12:03.568 but it's on the order of n to the 2 minus 1 over s. 00:12:03.568 --> 00:12:06.860 OK, everyone with me? 00:12:06.860 --> 00:12:10.490 So we need to verify this graph here has no Kst. 00:12:10.490 --> 00:12:11.411 Yes, question? 00:12:11.411 --> 00:12:14.638 AUDIENCE: Should that t be an s? 00:12:14.638 --> 00:12:16.180 YUFEI ZHAO: Yes, that should be an s. 00:12:16.180 --> 00:12:16.680 Thank you. 00:12:21.480 --> 00:12:23.152 Any more questions? 00:12:23.152 --> 00:12:26.960 AUDIENCE: Should that be s minus 1 factorial? 00:12:26.960 --> 00:12:29.060 YUFEI ZHAO: So we will show later 00:12:29.060 --> 00:12:32.450 on a better result using s minus 1 factorial, 00:12:32.450 --> 00:12:35.420 but for now I'll show you the slightly weaker result, which 00:12:35.420 --> 00:12:36.810 is still in the same spirit. 00:12:36.810 --> 00:12:37.310 Yep. 00:12:37.310 --> 00:12:39.953 AUDIENCE: Is the stronger result using the same graph? 00:12:39.953 --> 00:12:41.870 YUFEI ZHAO: We'll change to a different graph. 00:12:41.870 --> 00:12:44.412 For the stronger result, we will change to a different graph. 00:12:50.000 --> 00:12:55.260 OK, so now let's show that this graph here is Kst free. 00:12:55.260 --> 00:12:58.653 And for that claim, we need to invoke an algebraic fact, which 00:12:58.653 --> 00:12:59.570 let me write down now. 00:13:02.770 --> 00:13:05.750 So suppose we have a field f. 00:13:10.280 --> 00:13:14.090 Any field will work with a finite field. 00:13:14.090 --> 00:13:15.290 Any field is fine. 00:13:15.290 --> 00:13:22.520 And I have a bunch of elements from the field, 00:13:22.520 --> 00:13:31.320 such that a sub ij is different from a sub 00:13:31.320 --> 00:13:37.120 rj for all i not the same as r. 00:13:40.240 --> 00:13:43.200 Then the system of equations-- 00:13:47.468 --> 00:13:48.760 and I'll write down the system. 00:13:48.760 --> 00:13:50.450 So x minus 1-- 00:13:50.450 --> 00:13:59.460 x1 minus a 1, 1, x2 minus a 1, 2, and so on. 00:13:59.460 --> 00:14:04.286 xs minus a1x equals to b1. 00:14:04.286 --> 00:14:06.070 That's the first equation. 00:14:06.070 --> 00:14:09.240 Second equation, x1 minus a 2, 1. 00:14:09.240 --> 00:14:13.630 It almost looks like the usual system of linear equations. 00:14:13.630 --> 00:14:15.790 But I'm taking products. 00:14:21.170 --> 00:14:22.470 And so on. 00:14:22.470 --> 00:14:34.910 So the last one being x1 minus a s1 x2 minus a s2, dot dot dot, 00:14:34.910 --> 00:14:40.460 xs minus a ss equals to b sub s. 00:14:40.460 --> 00:14:54.400 The system has at most s factorial solutions, where 00:14:54.400 --> 00:14:56.320 I'm working inside this field. 00:14:59.780 --> 00:15:01.370 So that's the claim. 00:15:01.370 --> 00:15:03.782 So let me just give you some intuition for this claim. 00:15:08.710 --> 00:15:16.840 Suppose the right side vector is 0, all zeroes. 00:15:16.840 --> 00:15:19.730 Then I claim that this is trivial. 00:15:19.730 --> 00:15:21.280 So what is the saying? 00:15:21.280 --> 00:15:28.740 I need to select x1 to be one of the a's from the first row 00:15:28.740 --> 00:15:32.110 and x2 be one of the a's from the second row, and so on. 00:15:32.110 --> 00:15:34.990 But each column of a is distinct. 00:15:34.990 --> 00:15:36.670 So that's the hypothesis. 00:15:36.670 --> 00:15:40.660 You have all the a's in the first column are distinct. 00:15:40.660 --> 00:15:45.940 So no two of the x's can-- 00:15:49.150 --> 00:15:53.350 so I need to set one of the xi's to be one 00:15:53.350 --> 00:15:55.330 of the a's from the first row. 00:15:55.330 --> 00:16:01.210 But you see that you cannot set x1 to be a 1, 1, 00:16:01.210 --> 00:16:05.420 and x1 to be a s1 at the same time. 00:16:05.420 --> 00:16:08.450 So the solution just counts permutations, 00:16:08.450 --> 00:16:10.339 which is exactly s factorial. 00:16:21.590 --> 00:16:24.910 So this algebraic fact plays a key role 00:16:24.910 --> 00:16:29.550 in the proof of the theorem that the lower bound 00:16:29.550 --> 00:16:34.720 that we're stating up there, if you look at the paper, 00:16:34.720 --> 00:16:36.160 they give a proof of this result. 00:16:36.160 --> 00:16:37.810 And it's not a long proof. 00:16:37.810 --> 00:16:41.980 But it uses some commutative algebra and algebraic geometry. 00:16:41.980 --> 00:16:45.490 And usually in a class, if the instructor 00:16:45.490 --> 00:16:48.790 doesn't present the proof, it's for one of several reasons. 00:16:48.790 --> 00:16:50.950 Maybe the proof is too short. 00:16:50.950 --> 00:16:52.360 It doesn't need to be presented. 00:16:52.360 --> 00:16:54.700 Maybe it's too long or too difficult. Maybe 00:16:54.700 --> 00:16:57.790 it's not instructive to the class. 00:16:57.790 --> 00:16:59.920 And the last reason, which is the case here, 00:16:59.920 --> 00:17:01.930 is that I don't actually understand the proof. 00:17:01.930 --> 00:17:04.780 As in I can follow it line by line, 00:17:04.780 --> 00:17:08.079 but I don't understand why it is true. 00:17:08.079 --> 00:17:12.640 And if one of you wants to come up with a different proof 00:17:12.640 --> 00:17:16.690 or try to explain to me how this seemingly 00:17:16.690 --> 00:17:23.829 elementary algebraic fact is proved, I would appreciate it. 00:17:23.829 --> 00:17:26.710 For small values of x, you can check it by hand. 00:17:26.710 --> 00:17:29.110 So x equals to 2, you're solving a system 00:17:29.110 --> 00:17:31.390 of two quadratic equations. 00:17:31.390 --> 00:17:32.890 And that you can check by hand. 00:17:32.890 --> 00:17:35.050 And three maybe you can do it with some work. 00:17:35.050 --> 00:17:40.270 But even with 4 it's not so clear how to do it. 00:17:40.270 --> 00:17:47.410 And also, one of the geometric intuition is that if b is o0, 00:17:47.410 --> 00:17:50.320 then you have exactly s factorial solutions. 00:17:50.320 --> 00:17:52.090 And the geometric intuition is somehow 00:17:52.090 --> 00:17:56.290 that if you move b around, then the fiber, 00:17:56.290 --> 00:17:59.080 the sides of the fiber, the number of solutions x 00:17:59.080 --> 00:18:00.110 can only go down. 00:18:00.110 --> 00:18:00.940 It can not go up. 00:18:00.940 --> 00:18:05.170 And this corresponds to some algebraic geometry phenomenon. 00:18:05.170 --> 00:18:08.440 And that's all I will say about this algebraic fact, which 00:18:08.440 --> 00:18:09.670 we'll now use as a black box. 00:18:17.330 --> 00:18:19.520 Great. 00:18:19.520 --> 00:18:22.790 So now we have that as our algebraic input, 00:18:22.790 --> 00:18:27.950 let us show that a norm graph is Kst free. 00:18:27.950 --> 00:18:30.080 It's actually not so hard once you 00:18:30.080 --> 00:18:32.310 assume that theorem up there. 00:18:36.480 --> 00:18:47.400 So let's show at a norm graph is Ks, s factorial plus 1 free. 00:18:50.070 --> 00:18:55.170 Well, what does it mean to have a Kst? 00:18:55.170 --> 00:18:59.160 It means that if you have distinct vertices, 00:18:59.160 --> 00:19:05.330 which then correspond to elements, 00:19:05.330 --> 00:19:12.580 s elements, y1 through ys, of this field, 00:19:12.580 --> 00:19:17.580 then the common neighbors of these elements 00:19:17.580 --> 00:19:28.800 correspond to solutions of this system of equations 00:19:28.800 --> 00:19:32.560 where I set all of these values to be 1. 00:19:32.560 --> 00:19:35.320 But I can write l exactly what these guys are 00:19:35.320 --> 00:19:39.480 because I have this form, that representation there 00:19:39.480 --> 00:19:44.690 for the norm map, so I can write it out. 00:19:44.690 --> 00:19:48.560 And now remember this fact that when 00:19:48.560 --> 00:19:53.080 you are in characteristic p, x plus y raised to the power p, 00:19:53.080 --> 00:20:00.500 is the same as x to the p plus yp and characteristic p. 00:20:00.500 --> 00:20:04.720 So I can expand the remaining parenthesis like that. 00:20:13.340 --> 00:20:16.050 So I want the first line to equal to 1, and so on. 00:20:16.050 --> 00:20:19.950 And each of the lines has that equal to 1. 00:20:31.560 --> 00:20:35.790 How many solutions in x does this system of equations have? 00:20:35.790 --> 00:20:39.510 So even if I treat each of x and x to the p 00:20:39.510 --> 00:20:42.330 and so on as separate variables, that theorem 00:20:42.330 --> 00:20:49.200 appear tells me that there are at most s factorial solutions 00:20:49.200 --> 00:20:50.250 in x. 00:20:53.310 --> 00:20:57.500 Satisfies all the hypotheses of that theorem up there. 00:20:57.500 --> 00:21:04.250 Therefore, the graph is Ks sub s factorial plus 1 free. 00:21:04.250 --> 00:21:07.610 You do not have more than s plus 1 different values 00:21:07.610 --> 00:21:11.360 of x satisfying this system of equations. 00:21:11.360 --> 00:21:15.287 And that's the proof that this norm graph is Kst free. 00:21:15.287 --> 00:21:15.870 Yes, question? 00:21:15.870 --> 00:21:22.832 AUDIENCE: Why can't powers of like [INAUDIBLE] 00:21:22.832 --> 00:21:24.790 YUFEI ZHAO: Sorry, can you repeat the question? 00:21:24.790 --> 00:21:29.345 AUDIENCE: Why cannot the powers of the y's be the same? 00:21:29.345 --> 00:21:31.970 YUFEI ZHAO: The question is why cannot the powers of the p's be 00:21:31.970 --> 00:21:33.360 the same? 00:21:33.360 --> 00:21:36.210 So you are asking, down the second column, 00:21:36.210 --> 00:21:39.330 let's say, why are all these y's different? 00:21:39.330 --> 00:21:42.400 Because you're working inside a field. 00:21:42.400 --> 00:21:47.370 And raising to a p in this field is a bijection. 00:21:51.370 --> 00:21:54.040 So think about the order of the cyclic group. 00:21:54.040 --> 00:21:57.840 It has co-prime to p. 00:21:57.840 --> 00:22:00.040 But great question. 00:22:00.040 --> 00:22:02.640 Anything else? 00:22:02.640 --> 00:22:04.340 OK, so this gives you a construction 00:22:04.340 --> 00:22:11.870 that gives you Kst free for t bigger than s plus s factorial. 00:22:11.870 --> 00:22:14.630 Now, let me show you how to improve this construction 00:22:14.630 --> 00:22:19.280 to do a little bit better, to get s minus 1 factorial. 00:22:19.280 --> 00:22:21.080 And the idea is to take a variant 00:22:21.080 --> 00:22:24.790 of this norm graph, which we'll call the projective norm graph. 00:22:31.670 --> 00:22:34.430 And the projective norm graph will define it 00:22:34.430 --> 00:22:39.100 for s at least 3 is rather similar. 00:22:39.100 --> 00:22:41.160 But there's a twist. 00:22:41.160 --> 00:22:46.410 I have as the vertex set, not just the field extension-- 00:22:46.410 --> 00:22:50.490 OK, so now I take field extension, but one level less. 00:22:50.490 --> 00:22:54.210 And I take a second coordinate, which consists 00:22:54.210 --> 00:22:57.060 of non-zero elements from Fp. 00:23:01.340 --> 00:23:09.650 The edges are formed by putting an edge between these two 00:23:09.650 --> 00:23:13.730 elements, if and only if, the norm 00:23:13.730 --> 00:23:17.070 of the sum of the first coordinates 00:23:17.070 --> 00:23:20.340 equals to the product of the second coordinates. 00:23:24.480 --> 00:23:27.760 So now, you can run through a similar calculation that 00:23:27.760 --> 00:23:29.290 tells you the number of edges. 00:23:29.290 --> 00:23:31.570 So first of all, the number of vertices 00:23:31.570 --> 00:23:35.440 is p to the s minus 1 times p minus 1, 00:23:35.440 --> 00:23:39.600 so basically the same as p to s. 00:23:39.600 --> 00:23:49.200 And additionally, every vertex has degree exactly 00:23:49.200 --> 00:23:54.690 p raised to s minus 1 minus 1. 00:23:54.690 --> 00:24:04.460 And the reason is that if I tell you the values of x little x 00:24:04.460 --> 00:24:08.390 and big y, which cannot equal minus x, 00:24:08.390 --> 00:24:10.340 or else you will never form an edge, 00:24:10.340 --> 00:24:14.408 then they together uniquely determine little y. 00:24:18.160 --> 00:24:20.520 So for every value of big X and little x, 00:24:20.520 --> 00:24:23.880 I just need to run through all the values of big Y other 00:24:23.880 --> 00:24:24.690 than minus x. 00:24:29.110 --> 00:24:34.870 So the number of edges then equals to 1/2 times 00:24:34.870 --> 00:24:39.120 the number of vertices times the degree of every vertex, which, 00:24:39.120 --> 00:24:48.520 as before, is the claimed asymptotic. 00:24:48.520 --> 00:24:51.450 And the remaining thing to show is that this projective norm 00:24:51.450 --> 00:24:52.730 graph is Kst free. 00:25:05.820 --> 00:25:11.385 So it's K sub s, s minus 1 factorial plus 1 free. 00:25:16.840 --> 00:25:19.010 It's a similar calculation as the one before, 00:25:19.010 --> 00:25:22.240 but we need to take into account the small variant 00:25:22.240 --> 00:25:23.520 in the construction. 00:25:27.120 --> 00:25:36.520 So suppose we fix s vertices, labeled by this big Y's, little 00:25:36.520 --> 00:25:37.020 y's. 00:25:43.040 --> 00:25:49.100 And now we need to solve for uppercase X, 00:25:49.100 --> 00:26:05.800 lowercase x in this system of equations, 00:26:05.800 --> 00:26:09.230 so asking how many different pairs, big X, little 00:26:09.230 --> 00:26:13.710 x, can appear as a solution to this system of equations? 00:26:13.710 --> 00:26:21.030 Well, first of all, if some pairs of the first coordinates, 00:26:21.030 --> 00:26:31.610 big I is equal to big J, then if you have a solution, 00:26:31.610 --> 00:26:37.400 then that forces little y to be the same as little j. 00:26:37.400 --> 00:26:40.890 And so the y's wouldn't have been distinct to begin with. 00:26:40.890 --> 00:26:44.490 So this is not possible. 00:26:44.490 --> 00:26:48.690 So all the big Y's are distinct. 00:26:55.990 --> 00:27:03.320 Well, now, let's divide these equations 00:27:03.320 --> 00:27:04.760 by the final equation. 00:27:10.040 --> 00:27:13.040 And we get that the i-th equation 00:27:13.040 --> 00:27:26.970 becomes like that, which you can rewrite 00:27:26.970 --> 00:27:37.850 by dividing by the coordinate of the norm of big Y i 00:27:37.850 --> 00:27:39.800 minus big Y s. 00:27:39.800 --> 00:27:43.150 This is non-zero because we just showed that all the big Y 00:27:43.150 --> 00:27:45.130 i's are distinct. 00:27:45.130 --> 00:27:49.780 If you divide by this norm here and rearrange appropriately, 00:27:49.780 --> 00:27:53.860 we find that the equations become like this. 00:28:15.870 --> 00:28:17.570 So after doing some rearranging-- so 00:28:17.570 --> 00:28:20.540 this is the equation the set of new questions that we get. 00:28:20.540 --> 00:28:29.210 And you see that if you use new variables, x prime, 00:28:29.210 --> 00:28:33.300 do a substitution, this being x prime, 00:28:33.300 --> 00:28:37.020 then it has basically the same form as the one 00:28:37.020 --> 00:28:41.200 that we just saw with a different set of constants. 00:28:41.200 --> 00:28:44.640 And in particular from what we just saw, 00:28:44.640 --> 00:28:49.980 we see that you cannot have more than s minus 1 factorial 00:28:49.980 --> 00:28:53.250 solutions in x. 00:28:53.250 --> 00:28:56.910 Now, they're s minus 1 equations. 00:28:56.910 --> 00:29:00.380 And the field extension working as the x minus 1 00:29:00.380 --> 00:29:03.880 field extension. 00:29:03.880 --> 00:29:08.690 So we saved an equation by using this projectivization. 00:29:08.690 --> 00:29:11.060 And that's it. 00:29:11.060 --> 00:29:14.510 So this shows you the claim of constructing 00:29:14.510 --> 00:29:20.110 a Kst free graph for t bigger than s minus 1 factorial, which 00:29:20.110 --> 00:29:22.700 has the desired number of edges. 00:29:22.700 --> 00:29:23.583 Yes, question? 00:29:23.583 --> 00:29:26.072 AUDIENCE: Why do you have the if some capital 00:29:26.072 --> 00:29:29.040 Y equals capital IJ? 00:29:29.040 --> 00:29:35.290 YUFEI ZHAO: OK, so the question is why do I say this part? 00:29:35.290 --> 00:29:37.330 So I'm maybe skipping a sentence. 00:29:37.330 --> 00:29:39.670 I'm saying, if there is a solution to this system 00:29:39.670 --> 00:29:42.640 of equations x, if these vertices have 00:29:42.640 --> 00:29:47.770 a common neighbor, then if you have some x satisfying 00:29:47.770 --> 00:29:51.970 this system of equations, then having two different big Y's 00:29:51.970 --> 00:29:56.640 being the same forces you to have the two smaller y's 00:29:56.640 --> 00:29:57.546 being the same. 00:29:57.546 --> 00:29:58.340 AUDIENCE: OK. 00:29:58.340 --> 00:29:59.090 YUFEI ZHAO: Right. 00:29:59.090 --> 00:30:02.380 And then the Y's will have been distinct. 00:30:02.380 --> 00:30:04.150 So for them to have some common neighbors, 00:30:04.150 --> 00:30:08.000 you better have these big Y's being distinct. 00:30:08.000 --> 00:30:10.620 Any more questions? 00:30:10.620 --> 00:30:12.033 Great. 00:30:12.033 --> 00:30:13.700 So as I mentioned, it is an open problem 00:30:13.700 --> 00:30:21.090 to determine whether what is the extremal number for K44, K45, 00:30:21.090 --> 00:30:22.880 K46. 00:30:22.880 --> 00:30:26.586 And you may ask, well, we have this nice construction-- 00:30:26.586 --> 00:30:31.600 it maybe somewhat mysterious because of that, but explicit. 00:30:31.600 --> 00:30:33.580 And you can write this graph down. 00:30:33.580 --> 00:30:37.180 And you can ask is this graph K46 free? 00:30:37.180 --> 00:30:40.480 So do we gain one extra number for free, 00:30:40.480 --> 00:30:44.410 maybe because we didn't analyze things properly? 00:30:44.410 --> 00:30:46.580 And it turns out that's not the case. 00:30:46.580 --> 00:30:50.200 So there was a very recent paper just released last month 00:30:50.200 --> 00:30:54.850 showing that this graph here for s equals to 4 actually does 00:30:54.850 --> 00:30:56.710 contain some K46's. 00:31:00.010 --> 00:31:05.140 So if you want to prove a corresponding lower bound 00:31:05.140 --> 00:31:10.750 for K46, you better come up with a different construction. 00:31:10.750 --> 00:31:13.030 And that's I think an interesting direction 00:31:13.030 --> 00:31:13.680 to explore. 00:31:17.190 --> 00:31:18.300 Any questions? 00:31:18.300 --> 00:31:18.800 Yes. 00:31:18.800 --> 00:31:21.245 AUDIENCE: Do we know of any similar results 00:31:21.245 --> 00:31:23.423 about this construction not working for larger s? 00:31:23.423 --> 00:31:24.840 YUFEI ZHAO: The question is, do we 00:31:24.840 --> 00:31:30.810 know any similar result of about does this graph contain 00:31:30.810 --> 00:31:36.060 Kst for other values of s and t less 00:31:36.060 --> 00:31:38.070 than the claimed threshold? 00:31:38.070 --> 00:31:39.250 It is unclear. 00:31:39.250 --> 00:31:42.630 So the paper that was uploaded, it 00:31:42.630 --> 00:31:45.486 doesn't address the issue of s bigger than 4. 00:31:45.486 --> 00:31:45.986 Yeah. 00:31:45.986 --> 00:31:50.870 AUDIENCE: Why Fp to the power s? 00:31:50.870 --> 00:31:53.530 YUFEI ZHAO: So question is why Fp to the power of s? 00:31:53.530 --> 00:31:55.690 So let's go back to the norm graph construction. 00:31:55.690 --> 00:31:58.660 So where do we use Fp to the power of s? 00:31:58.660 --> 00:32:06.010 Well, certainly we needed it to have the right edge count. 00:32:06.010 --> 00:32:08.360 So that comes up in the edge count. 00:32:08.360 --> 00:32:12.500 And also in the norm expression, you 00:32:12.500 --> 00:32:14.840 have the correct number of factors. 00:32:20.960 --> 00:32:22.690 So I encourage you to try for if you 00:32:22.690 --> 00:32:24.880 use a smaller or bigger value of s, 00:32:24.880 --> 00:32:27.460 you either don't get something which is Kst free 00:32:27.460 --> 00:32:31.130 or you have the wrong number of edges. 00:32:31.130 --> 00:32:34.830 Any more questions? 00:32:34.830 --> 00:32:37.650 So later, I will show you a different construction 00:32:37.650 --> 00:32:41.850 of Kst free graphs, again, for t large compared to s 00:32:41.850 --> 00:32:44.790 that will not do as well as this one. 00:32:44.790 --> 00:32:47.130 But it is a genuinely different construction. 00:32:47.130 --> 00:32:50.340 And it uses the idea of randomized algebraic 00:32:50.340 --> 00:32:54.210 construction, which is something that actually was only 00:32:54.210 --> 00:32:55.600 developed a few years ago. 00:32:55.600 --> 00:32:57.947 It's a very recent development. 00:32:57.947 --> 00:32:58.780 And it's quite nice. 00:32:58.780 --> 00:33:01.260 So it combines some of the things we've talked about 00:33:01.260 --> 00:33:03.860 with constructing using random graphs to construct 00:33:03.860 --> 00:33:06.730 H free graphs on one hand, and on the other hand, 00:33:06.730 --> 00:33:08.250 some of the algebraic ideas. 00:33:08.250 --> 00:33:11.250 In particular, we're not going to use that theorem up there, 00:33:11.250 --> 00:33:14.777 but we'll use some other algebraic geometry fact. 00:33:14.777 --> 00:33:16.110 OK, so let's take a quick break. 00:33:19.500 --> 00:33:23.820 So what I want to discuss now is a relatively new idea 00:33:23.820 --> 00:33:28.740 called a randomized algebraic construction, which 00:33:28.740 --> 00:33:33.930 combines some ideas from both the randomized construction 00:33:33.930 --> 00:33:36.170 and also the algebraic construction that we just saw. 00:33:39.970 --> 00:33:43.620 So this idea is due to Boris Bukh just a few years ago. 00:33:47.660 --> 00:33:53.900 And the goal is to give an alternative construction 00:33:53.900 --> 00:34:09.620 of a Kst free graph with lots of edges 00:34:09.620 --> 00:34:19.610 provided that as before, t is much larger compared to s. 00:34:23.659 --> 00:34:26.050 So this band here will not be as good as the one 00:34:26.050 --> 00:34:27.159 that we just saw. 00:34:27.159 --> 00:34:29.280 And I will even tell you what it is. 00:34:29.280 --> 00:34:30.520 But it's some constant. 00:34:30.520 --> 00:34:32.650 So for every s there is some t, such 00:34:32.650 --> 00:34:33.870 that this construction works. 00:34:37.630 --> 00:34:44.239 As before, we working inside some finite field geometry. 00:34:44.239 --> 00:34:47.062 So let's start with a, a prime power. 00:34:47.062 --> 00:34:48.520 You can think of prime if you like. 00:34:48.520 --> 00:34:49.810 It doesn't make so much difference. 00:34:49.810 --> 00:34:51.550 So we're working inside a finite field. 00:34:54.389 --> 00:34:59.180 And let's assume s is fixed and at least 4. 00:35:03.513 --> 00:35:04.930 Let me write down some parameters. 00:35:04.930 --> 00:35:06.320 Don't worry about them for now. 00:35:06.320 --> 00:35:08.730 Just think of them as sufficiently large constants. 00:35:08.730 --> 00:35:11.470 So d is this quantity here. 00:35:11.470 --> 00:35:15.080 So we'll come back to it later when it comes up. 00:35:15.080 --> 00:35:19.000 OK, so what's the idea? 00:35:19.000 --> 00:35:21.430 When we looked at the randomized construction, 00:35:21.430 --> 00:35:22.690 we took a random graph. 00:35:22.690 --> 00:35:24.970 We took an Erdos Renyi graph. 00:35:24.970 --> 00:35:27.280 Every edge appeared independently. 00:35:27.280 --> 00:35:30.010 And saw that it has lots of edges, 00:35:30.010 --> 00:35:33.730 if you choose p property and not so many copies of h. 00:35:33.730 --> 00:35:36.160 So you can remove all the copies of h 00:35:36.160 --> 00:35:41.410 to get a graph with lots of edges that is h free. 00:35:41.410 --> 00:35:43.310 What we're going to do now is instead 00:35:43.310 --> 00:35:45.380 of taking the edges randomly, we're 00:35:45.380 --> 00:35:48.205 going to take a random polynomial. 00:35:54.030 --> 00:36:07.690 F will be a random polynomial chosen uniformly 00:36:07.690 --> 00:36:19.720 among all polynomials with-- 00:36:19.720 --> 00:36:24.510 so I wrote uppercase Y and uppercase X and Y. Actually, X 00:36:24.510 --> 00:36:26.340 and Y, they're not single variants. 00:36:26.340 --> 00:36:39.280 They are-- so each of them is a vector of s variables. 00:36:41.870 --> 00:36:46.550 So in other words, x1 through xs are the variables 00:36:46.550 --> 00:36:47.570 in the polynomial. 00:36:47.570 --> 00:36:51.800 And then y1 through ys. 00:36:51.800 --> 00:36:54.058 So it's a 2s variable polynomial. 00:36:58.180 --> 00:37:06.050 So among all polynomials with degree, at most d-- 00:37:06.050 --> 00:37:07.790 d being the number up here-- 00:37:07.790 --> 00:37:13.840 in each of x and y sets of variables. 00:37:13.840 --> 00:37:16.700 So you look at it as xs variables, 00:37:16.700 --> 00:37:20.750 each monominal has their exponents sum 00:37:20.750 --> 00:37:24.170 to our most d, and likewise with each monomial for the y 00:37:24.170 --> 00:37:25.790 variables. 00:37:25.790 --> 00:37:29.260 So this is the random object. 00:37:29.260 --> 00:37:32.300 It's a random polynomial in 2s variables. 00:37:32.300 --> 00:37:33.450 And the degree is bounded. 00:37:33.450 --> 00:37:35.990 So you only have a finite number of possibilities, 00:37:35.990 --> 00:37:38.120 and I choose one of them uniformly at random. 00:37:44.750 --> 00:37:47.030 And now, what's my graph? 00:37:47.030 --> 00:37:49.120 We're going to construct a bipartide graph 00:37:49.120 --> 00:37:52.810 G. The bipartideness, it's not so crucial. 00:37:52.810 --> 00:37:54.690 But it'll make our life somewhat easier. 00:37:58.870 --> 00:38:00.330 So it's a bipartide graph. 00:38:00.330 --> 00:38:03.820 So it hs two vertex parts, which I will label left and right, 00:38:03.820 --> 00:38:09.320 L and R. And they are both the s dimensional vector 00:38:09.320 --> 00:38:11.600 space over f cube. 00:38:11.600 --> 00:38:18.530 And we'll put an edge between two vertices, 00:38:18.530 --> 00:38:21.200 if and only if that polynomial up there 00:38:21.200 --> 00:38:27.960 f evaluates to 0 on these values. 00:38:27.960 --> 00:38:29.200 That's the graph. 00:38:29.200 --> 00:38:32.090 So I give you a random polynomial f. 00:38:32.090 --> 00:38:38.720 And then you put in edges according to when f vanishes. 00:38:38.720 --> 00:38:44.420 So f, if you view the bipartide graph as a subset of Fq 00:38:44.420 --> 00:38:49.210 to s cross F to the s, then this is the zero set. 00:38:49.210 --> 00:38:52.970 The edge set is the zero set. 00:38:52.970 --> 00:38:56.480 Just like in random graphs, with the construction 00:38:56.480 --> 00:38:58.950 with random graphs, we'll need to show a couple of things. 00:38:58.950 --> 00:39:02.330 One is that has lots of edges, which will not be hard to show. 00:39:02.330 --> 00:39:06.050 And second, that it will have typically a small number 00:39:06.050 --> 00:39:09.650 of copies of Kst. 00:39:09.650 --> 00:39:13.000 And that will have some ingredients which 00:39:13.000 --> 00:39:16.660 are similar to the random graphs case we saw before, 00:39:16.660 --> 00:39:22.820 but it will have some new ideas coming from algebraic geometry. 00:39:22.820 --> 00:39:27.360 First, let's show that this graph has lots of edges. 00:39:27.360 --> 00:39:34.530 And that's a simple calculation, because for every pair 00:39:34.530 --> 00:39:44.620 of points of vertices, I claim that the probability-- 00:39:44.620 --> 00:39:46.940 so here f is the random object-- 00:39:46.940 --> 00:39:53.790 the probability that f evaluates to 0 on this pair is exactly q. 00:39:53.790 --> 00:39:58.970 So exactly 1 over q, 1 over the size of the field. 00:39:58.970 --> 00:40:01.930 So this is not too hard. 00:40:01.930 --> 00:40:06.850 And the reason is that the distribution of f 00:40:06.850 --> 00:40:32.210 is identical to if you are an extra random constant on f, 00:40:32.210 --> 00:40:34.070 chosen uniformly at random. 00:40:34.070 --> 00:40:35.840 So I took a random polynomial. 00:40:35.840 --> 00:40:37.430 I shifted by a random constant. 00:40:37.430 --> 00:40:39.320 It's still uniformly random polynomial, 00:40:39.320 --> 00:40:41.180 according to that distribution. 00:40:41.180 --> 00:40:46.670 But now, you see that this, whatever f 00:40:46.670 --> 00:40:49.810 evaluated to, if I shift by a random constant, 00:40:49.810 --> 00:40:53.700 you will end up with a uniform distribution. 00:40:53.700 --> 00:41:03.640 So it tells you that that guy up there is uniform distribution 00:41:03.640 --> 00:41:12.600 on every fixed point u, v. So in particular, 00:41:12.600 --> 00:41:18.220 it hits 0 with probability exactly 1 over q. 00:41:18.220 --> 00:41:25.310 And as a result, the number of edges of g in the expectation 00:41:25.310 --> 00:41:31.620 is exactly n squared over q, where 00:41:31.620 --> 00:41:34.270 n is the number of vertices. 00:41:34.270 --> 00:41:37.410 So n is actually not the number of vertices, 00:41:37.410 --> 00:41:43.655 but the size of each vertex part, namely q to the s. 00:41:43.655 --> 00:41:48.770 So you see that it gives you the right number of edges, so n 00:41:48.770 --> 00:41:51.290 to the 2 minus 1 over s. 00:41:51.290 --> 00:41:53.280 So we have the right number of edges. 00:41:53.280 --> 00:41:58.280 And now, we want to show that this graph here typically does 00:41:58.280 --> 00:42:01.690 not have too many copies of Kst's. 00:42:01.690 --> 00:42:03.500 It might have some copies of Kst's. 00:42:03.500 --> 00:42:05.620 Somehow that is unavoidable. 00:42:05.620 --> 00:42:09.650 Just as in the random case, you do have some copies of Kst's. 00:42:09.650 --> 00:42:11.360 But if they are not too many copies, 00:42:11.360 --> 00:42:14.300 I can remove them and obtain a Kst free graph. 00:42:17.830 --> 00:42:20.070 OK, so what is the intuition? 00:42:20.070 --> 00:42:22.590 How does it compare to the case when 00:42:22.590 --> 00:42:27.260 you have a genuine Erdos-Renyi random graph? 00:42:27.260 --> 00:42:32.603 Well, what is the expected number of common neighbors? 00:42:36.050 --> 00:42:43.120 So if you fix some u with let's say 00:42:43.120 --> 00:42:49.600 on the left side with exactly s vertices, 00:42:49.600 --> 00:43:02.300 I want to understand, how many common neighbors does u have? 00:43:09.845 --> 00:43:11.220 But because the common neighbors, 00:43:11.220 --> 00:43:14.070 if he has too many common neighbors, then that's a Kst. 00:43:16.820 --> 00:43:20.070 It is not hard to calculate the expectation 00:43:20.070 --> 00:43:22.760 of this quantity, both in the random graph case 00:43:22.760 --> 00:43:24.780 as well as in this case. 00:43:24.780 --> 00:43:27.780 And you can calculate if you pretend every edge occurs 00:43:27.780 --> 00:43:32.570 independently, the expected number of common neighbors 00:43:32.570 --> 00:43:39.210 is exactly n to the q to the minus s. 00:43:39.210 --> 00:43:42.870 So there are s elements of u, which is exactly 1. 00:43:45.980 --> 00:43:50.600 And you know that for a binomial distribution with expectation 1 00:43:50.600 --> 00:43:53.720 and a large number of variables, the distribution 00:43:53.720 --> 00:43:55.574 is approximately Poissonian. 00:44:03.802 --> 00:44:11.640 Ah, but that's in the case when it's independently distributed, 00:44:11.640 --> 00:44:14.760 which is the case in the case of GMP. 00:44:14.760 --> 00:44:17.760 But it turns out for the algebraic setting 00:44:17.760 --> 00:44:22.240 we're doing here, things don't behave independently. 00:44:22.240 --> 00:44:26.170 It's not that you're doing coin flops for every possible edge. 00:44:26.170 --> 00:44:28.800 We're doing some randomized algebraic construction. 00:44:28.800 --> 00:44:31.110 And for algebraic geometry reasons, 00:44:31.110 --> 00:44:33.670 you will see that the distribution is very much not 00:44:33.670 --> 00:44:35.660 like Poissonian. 00:44:35.660 --> 00:44:40.040 It will turn out that either the number of common neighbors 00:44:40.040 --> 00:44:43.700 is bounded or it is very large. 00:44:46.810 --> 00:44:53.620 And that means that we can show using some Markov inequality 00:44:53.620 --> 00:44:57.130 that the probability that it is very large is quite small. 00:44:57.130 --> 00:45:02.300 So typically, it will not have many common neighbors? 00:45:02.300 --> 00:45:06.740 And that's the intuition, and so let's work out this intuition. 00:45:06.740 --> 00:45:07.620 Any questions so far? 00:45:12.220 --> 00:45:13.650 So how do we do this calculation? 00:45:13.650 --> 00:45:16.070 So first, let's start with something that's 00:45:16.070 --> 00:45:17.240 actually fairly elementary. 00:45:17.240 --> 00:45:25.060 So suppose you have some parameters, r and s. 00:45:25.060 --> 00:45:27.160 So I think of r and s-- 00:45:27.160 --> 00:45:29.980 so I have some parameters r and s. 00:45:29.980 --> 00:45:34.160 And thick of them as constants. 00:45:34.160 --> 00:45:36.160 Have some restrictions, but don't worry too much 00:45:36.160 --> 00:45:38.630 about them. 00:45:38.630 --> 00:45:47.600 Suppose I have two bounded subsets of the finite field, 00:45:47.600 --> 00:45:54.590 where you have size s and v has size r. 00:45:54.590 --> 00:45:58.310 Then the claim is that the probability 00:45:58.310 --> 00:46:09.240 that f vanishes on the Cartesian product of u and v. OK, 00:46:09.240 --> 00:46:12.840 so what do you expect it to be? 00:46:12.840 --> 00:46:17.310 So I have s r elements, and I want f, this random polynomial, 00:46:17.310 --> 00:46:19.240 to vanish on the entire product. 00:46:19.240 --> 00:46:23.370 Well, if s, its value behaved independently for every point, 00:46:23.370 --> 00:46:26.030 you should expect that the probability is exactly 00:46:26.030 --> 00:46:30.060 q to the power minus s r. 00:46:30.060 --> 00:46:32.710 And it turns out that is the case. 00:46:32.710 --> 00:46:33.480 So this is true. 00:46:33.480 --> 00:46:35.297 This is an exact statement. 00:46:38.570 --> 00:46:39.730 OK, so why is this true? 00:46:39.730 --> 00:46:42.100 So this is in some sense a generalization 00:46:42.100 --> 00:46:44.250 of this claim over here. 00:46:44.250 --> 00:46:46.040 And you have to do a little bit more work. 00:46:46.040 --> 00:46:48.520 But it's not too difficult. 00:46:48.520 --> 00:46:55.080 So let's first consider this lemma 00:46:55.080 --> 00:47:11.950 in a somewhat simpler case, where all the first coordinates 00:47:11.950 --> 00:47:14.080 of x are distinct-- 00:47:14.080 --> 00:47:16.370 of u are distinct. 00:47:19.440 --> 00:47:23.720 And all the first coordinates of v are distinct. 00:47:23.720 --> 00:47:25.660 Suppose u and v have that form. 00:47:25.660 --> 00:47:29.140 So I write down the list of points for u 00:47:29.140 --> 00:47:31.270 and first coordinates are all distinct. 00:47:33.990 --> 00:47:39.000 What I want to do is to give you a random shift, 00:47:39.000 --> 00:47:43.470 to do a uniform random shift. 00:47:43.470 --> 00:47:47.280 And I will shift it by a polynomial g, which 00:47:47.280 --> 00:47:48.960 is a bivariate polynomial. 00:47:48.960 --> 00:47:50.600 So these are not vectors. 00:47:50.600 --> 00:47:55.180 They are just single variables. 00:47:55.180 --> 00:48:03.770 And I look at all possible sum of monomials, 00:48:03.770 --> 00:48:12.560 where the degree in i is less than s, and the degree in j 00:48:12.560 --> 00:48:13.880 is less than r. 00:48:16.640 --> 00:48:24.360 And these a's are chosen uniformly independently 00:48:24.360 --> 00:48:32.330 at random from the ground field fq. 00:48:34.910 --> 00:48:39.800 And as before, we see that f and f plus g 00:48:39.800 --> 00:48:45.890 have the same distribution, the same probability distribution. 00:48:45.890 --> 00:48:49.660 And so all it remains to show that is whatever f comes out 00:48:49.660 --> 00:48:53.530 to be, if I tack on this extra random g, 00:48:53.530 --> 00:48:56.680 it creates a uniform distribution on the values 00:48:56.680 --> 00:49:04.270 on the entire u cross v. But, actually, see, 00:49:04.270 --> 00:49:09.350 I have sr choices exactly for these coefficients. 00:49:09.350 --> 00:49:14.970 And I have sr values that I'm trying to control. 00:49:14.970 --> 00:49:17.420 So really it's a counting problem. 00:49:17.420 --> 00:49:27.210 And it suffices to show a bijection, namely 00:49:27.210 --> 00:49:37.400 that for every possible vector of values, 00:49:37.400 --> 00:49:40.630 there exists a choice of coefficients, 00:49:40.630 --> 00:49:50.230 as above, such that g evaluates to the prescribed values 00:49:50.230 --> 00:49:52.544 with the given coefficients. 00:50:02.710 --> 00:50:07.390 And that uniformity will follow just because you have the exact 00:50:07.390 --> 00:50:10.410 if you just do a counting. 00:50:10.410 --> 00:50:13.663 And the one-dimensional version of this claim, 00:50:13.663 --> 00:50:15.080 so let's think about what that is. 00:50:15.080 --> 00:50:19.000 So if I have, let's say, three points on the line 00:50:19.000 --> 00:50:22.330 and a degree 2 polynomial, what I am saying 00:50:22.330 --> 00:50:26.860 is that if you give me the values you want on these three 00:50:26.860 --> 00:50:31.090 points, I can produce for you a unique polynomial that 00:50:31.090 --> 00:50:35.460 evaluates to the prescribed values on these three points. 00:50:35.460 --> 00:50:39.250 And that you should all know as Lagrange interpolation. 00:50:39.250 --> 00:50:42.720 So it tells you exactly how to do that. 00:50:42.720 --> 00:50:45.390 And that works for many reasons. 00:50:45.390 --> 00:50:51.240 One of them is that the random indeterminate is invertible. 00:50:51.240 --> 00:50:54.250 Here, we have multi-variables. 00:50:54.250 --> 00:50:58.290 So let's do Lagrange interpolation twice, once 00:50:58.290 --> 00:51:01.920 for each variable. 00:51:01.920 --> 00:51:16.030 So we'll apply Lagrange interpolation twice. 00:51:16.030 --> 00:51:23.360 So the first time, we'll see that for all values of u, 00:51:23.360 --> 00:51:29.890 there exists a single variate polynomial 00:51:29.890 --> 00:51:38.850 in the y variable with degree at most r minus 1 that 00:51:38.850 --> 00:51:46.940 evaluates to the correct values on the fixed little u. 00:51:54.280 --> 00:51:56.090 So do it for one variable at a time. 00:51:56.090 --> 00:51:59.600 For fixed u, do Lagrange interpolation 00:51:59.600 --> 00:52:02.200 on the y variable. 00:52:02.200 --> 00:52:06.490 And now, once we have those things there, 00:52:06.490 --> 00:52:16.580 viewing the g that we want to find as a polynomial whose 00:52:16.580 --> 00:52:22.840 coefficients are polynomials in the x variables, but is itself 00:52:22.840 --> 00:52:30.570 is a polynomial in the y variable, 00:52:30.570 --> 00:52:34.500 we find that again using Lagrange interpolation, 00:52:34.500 --> 00:52:40.310 there exists these values for these coefficients 00:52:40.310 --> 00:52:51.510 here, such that each coefficient of if you plug 00:52:51.510 --> 00:53:04.650 in the first entry into little u agrees with the coefficients 00:53:04.650 --> 00:53:09.460 of the g that we just found. 00:53:09.460 --> 00:53:11.540 And that should be the case for every little u. 00:53:14.910 --> 00:53:17.320 So once you find these polynomials 00:53:17.320 --> 00:53:20.280 and now you have a bonafide polynomial in g. 00:53:20.280 --> 00:53:23.610 And that's the claim above. 00:53:23.610 --> 00:53:26.990 So using Lagrange interpolation twice, once for each variable. 00:53:26.990 --> 00:53:31.370 So this is-- if you're confused, just think about it. 00:53:31.370 --> 00:53:34.590 There is nothing deep here. 00:53:34.590 --> 00:53:37.890 So that finishes the claim in the case 00:53:37.890 --> 00:53:42.160 when the first coordinates are all distinct. 00:53:42.160 --> 00:53:45.180 So we use that fact crucially in doing this Lagrange 00:53:45.180 --> 00:53:47.200 interpolation. 00:53:47.200 --> 00:53:52.350 Now, for general u and v, where we 00:53:52.350 --> 00:53:53.910 don't have this assumption of having 00:53:53.910 --> 00:53:57.090 distinct first coordinates, well, 00:53:57.090 --> 00:54:00.840 let's make them to have distinct first coordinates 00:54:00.840 --> 00:54:04.350 by considering a random linear transformation, 00:54:04.350 --> 00:54:07.980 so using a probabilistic method. 00:54:07.980 --> 00:54:20.880 So we suffice as to find invertible linear maps, p 00:54:20.880 --> 00:54:34.170 and s, on this vector space, such that TU and SV have 00:54:34.170 --> 00:54:35.712 the above properties. 00:54:40.910 --> 00:54:43.150 So let me show you how to do it for u. 00:54:43.150 --> 00:54:45.170 So I need to find you a invertible linear 00:54:45.170 --> 00:54:48.060 transformation t. 00:54:48.060 --> 00:54:50.170 Well, it's just the first coordinate that matters. 00:54:50.170 --> 00:54:55.910 So it suffices to find just a linear map 00:54:55.910 --> 00:55:03.600 corresponding to the first coordinate that 00:55:03.600 --> 00:55:05.886 is injective of u. 00:55:08.840 --> 00:55:12.020 Whatever linear map you have, even if it's zero, that's fine, 00:55:12.020 --> 00:55:18.845 I can extend it to the remaining coordinates. 00:55:18.845 --> 00:55:20.220 Actually, if it's zero, then it's 00:55:20.220 --> 00:55:21.585 not going to be injective of u. 00:55:21.585 --> 00:55:24.990 So it better not be zero. 00:55:24.990 --> 00:55:27.720 OK, well, let's find this map randomly. 00:55:27.720 --> 00:55:40.520 So pick t uniform via random among all linear maps. 00:55:46.310 --> 00:55:48.310 And I want to understand what is the probability 00:55:48.310 --> 00:55:53.440 of collision, bad event, if two elements of u 00:55:53.440 --> 00:55:56.460 end up getting mapped to the same point. 00:55:56.460 --> 00:55:58.710 Well, that's not too hard. 00:55:58.710 --> 00:56:07.150 So for every distinct pair of points in fq to the s 00:56:07.150 --> 00:56:23.100 the probability that they collide, 00:56:23.100 --> 00:56:24.600 think about why this is true. 00:56:24.600 --> 00:56:27.212 It's exactly 1 over q. 00:56:30.990 --> 00:56:33.750 If x and x prime, they're differing at least one 00:56:33.750 --> 00:56:35.920 coordinate, then even just along that coordinate, 00:56:35.920 --> 00:56:38.650 I can make them distinct. 00:56:38.650 --> 00:56:41.470 So this is the case for every pair. 00:56:41.470 --> 00:56:47.170 So now by union bound, the probability 00:56:47.170 --> 00:56:55.900 that t1 is injective on u is at least 1 minus the size 00:56:55.900 --> 00:57:01.370 if u choose 2 times 1 over q. 00:57:01.370 --> 00:57:07.150 And that's why we chose q to be large enough. 00:57:07.150 --> 00:57:13.540 So q is at least r squared, s So this number here is positive. 00:57:13.540 --> 00:57:16.730 So such a t exists. 00:57:16.730 --> 00:57:21.350 And so we can transform this u and v, two configurations where 00:57:21.350 --> 00:57:23.120 the first coordinates are all distinct, 00:57:23.120 --> 00:57:26.790 and then run the argument as before. 00:57:26.790 --> 00:57:27.892 OK, great. 00:57:30.660 --> 00:57:36.780 So what we've shown so far is that if you look at these Ksr 00:57:36.780 --> 00:57:41.620 structures, they appear with probability exactly-- 00:57:41.620 --> 00:57:44.580 well, with expectation exactly what 00:57:44.580 --> 00:57:49.320 you might expect as in independent random case. 00:57:49.320 --> 00:57:51.210 But what we really want to understand 00:57:51.210 --> 00:57:55.610 is the distribution of the number of common neighbors. 00:57:55.610 --> 00:58:00.050 In particular, we want to upper bound the probability 00:58:00.050 --> 00:58:02.360 that there are too many common neighbors. 00:58:02.360 --> 00:58:08.160 We want to understand some kind of tail probabilities. 00:58:08.160 --> 00:58:11.600 And to do that, one way to do tail probabilities 00:58:11.600 --> 00:58:13.155 is to consider moments. 00:58:13.155 --> 00:58:15.622 Yes, question? 00:58:15.622 --> 00:58:21.920 AUDIENCE: [INAUDIBLE] 00:58:21.920 --> 00:58:25.485 YUFEI ZHAO: Sorry, can you repeat the question? 00:58:25.485 --> 00:58:27.900 AUDIENCE: How do you have the equality right 00:58:27.900 --> 00:58:29.832 before the lemma? 00:58:29.832 --> 00:58:32.790 YUFEI ZHAO: Question, how do I have the equality right 00:58:32.790 --> 00:58:34.800 before the lemma? 00:58:34.800 --> 00:58:43.470 So there, I'm actually saying for the Erdos-Renyi random 00:58:43.470 --> 00:58:44.860 graph case-- 00:58:44.860 --> 00:58:47.820 that's the case-- in Erdos-Renyi random graph, 00:58:47.820 --> 00:58:52.680 each edge, if you have the same edge probability, is 1 over q. 00:58:52.680 --> 00:58:55.932 And then that's the number of common neighbors 00:58:55.932 --> 00:58:56.640 you would expect. 00:59:00.200 --> 00:59:05.150 So that's a heuristic for the Erdos-Renyi random graph case. 00:59:05.150 --> 00:59:07.220 OK, so now let's try to understand 00:59:07.220 --> 00:59:10.680 the distribution of the number of common neighbors. 00:59:10.680 --> 00:59:19.850 So let's fix a u subset of Fp to the s with exactly s elements. 00:59:19.850 --> 00:59:22.790 And I want to understand how many common neighbors it has. 00:59:22.790 --> 00:59:36.330 So let's consider the number of common neighbors of u 00:59:36.330 --> 00:59:40.140 and the d-th moment of this random variable. 00:59:40.140 --> 00:59:44.470 So this is a common way to do upper tail bounds. 00:59:44.470 --> 00:59:47.590 And one way to analyze such moments 00:59:47.590 --> 00:59:51.670 is to decompose this count as a sum of indicator 00:59:51.670 --> 00:59:53.000 random variables. 00:59:53.000 --> 01:00:03.670 So let me write I of v to be the quantity which is 1 if f of uv 01:00:03.670 --> 01:00:08.770 is 0 for all with ou and big U. In the words, 01:00:08.770 --> 01:00:10.730 it's a common neighbor. 01:00:10.730 --> 01:00:16.737 It's 1 if v is a common neighbor for u, and 0 otherwise. 01:00:20.220 --> 01:00:22.410 So then the number of common neighbors 01:00:22.410 --> 01:00:31.200 would simply be the sum of this indicator as v ranges 01:00:31.200 --> 01:00:33.080 over the entire vertex set. 01:00:36.590 --> 01:00:39.170 And I can expand the sum. 01:00:39.170 --> 01:00:41.780 Then all of these are standard things 01:00:41.780 --> 01:00:43.925 to do when you're trying to calculate moments. 01:00:56.150 --> 01:01:06.240 OK, so I can bring this expectation inside 01:01:06.240 --> 01:01:10.170 and try to understand what is the expectation of this object 01:01:10.170 --> 01:01:10.670 inside. 01:01:14.100 --> 01:01:17.420 Well, if all the v's are distinct, 01:01:17.420 --> 01:01:22.045 then this is simply the expected number of Kst's. 01:01:30.090 --> 01:01:32.090 But the v's might not be distinct. 01:01:32.090 --> 01:01:34.220 So we need to be a little bit careful. 01:01:34.220 --> 01:01:36.290 But that's not too hard to handle. 01:01:36.290 --> 01:01:39.440 So let me write M sub r to be the number 01:01:39.440 --> 01:01:45.640 of subjective functions from d element set to an r elements 01:01:45.640 --> 01:01:53.100 set, an M to be sum of these M sub r's for r up to d. 01:01:53.100 --> 01:01:58.390 Then, let's consider how many distinct v values are there. 01:01:58.390 --> 01:02:03.820 If there are distinct v values, then they 01:02:03.820 --> 01:02:07.570 take on that many possible values. 01:02:07.570 --> 01:02:10.540 And Mr for the number of subjections, 01:02:10.540 --> 01:02:16.240 and for each possible r, the exact number 01:02:16.240 --> 01:02:18.640 for the exact value of this expectation 01:02:18.640 --> 01:02:20.380 is q to the minus rs. 01:02:20.380 --> 01:02:22.210 And that's exactly what we showed. 01:02:22.210 --> 01:02:25.970 So this comes from the lemma just now. 01:02:29.770 --> 01:02:31.710 But we chose-- I mean, look, this 01:02:31.710 --> 01:02:34.420 is that binomial coefficient. 01:02:34.420 --> 01:02:35.670 And you have this number here. 01:02:35.670 --> 01:02:37.530 So they multiply it to at most 1. 01:02:42.280 --> 01:02:45.580 So we have this number, this quantity there, 01:02:45.580 --> 01:02:47.090 which I think of as a constant. 01:02:47.090 --> 01:02:49.520 So this is a constant. 01:02:49.520 --> 01:02:51.590 So the d-th moment is bounded. 01:02:57.900 --> 01:03:02.260 And one way to get tail bounds once you have the moments 01:03:02.260 --> 01:03:07.286 is that we can use Markov's inequality. 01:03:07.286 --> 01:03:12.850 It tells us that the number of common neighbors of u, 01:03:12.850 --> 01:03:18.460 the probability that u has too many common neighbors, more 01:03:18.460 --> 01:03:22.590 than lambda common neighbors. 01:03:22.590 --> 01:03:29.690 I can rewrite this inequality here 01:03:29.690 --> 01:03:34.730 by raising both sides to the power d. 01:03:34.730 --> 01:03:42.350 And then, using Markov inequality, take expectation. 01:03:51.950 --> 01:03:54.010 So all of these are standard techniques 01:03:54.010 --> 01:03:56.740 for upper tail estimation. 01:03:56.740 --> 01:03:59.800 You want to understand the upper tails on random variable, 01:03:59.800 --> 01:04:03.670 understand its moments and use Markov on its moments. 01:04:06.710 --> 01:04:12.920 But now, we know we have some bound for the d-th moment, 01:04:12.920 --> 01:04:14.600 right? 01:04:14.600 --> 01:04:16.880 Which is M as we just showed. 01:04:16.880 --> 01:04:18.560 So there is this bound here. 01:04:21.320 --> 01:04:23.420 So far you can run the same argument 01:04:23.420 --> 01:04:25.100 in the random graphs case. 01:04:25.100 --> 01:04:27.550 And you wouldn't really do much different. 01:04:27.550 --> 01:04:29.300 I mean everything is more or less the same 01:04:29.300 --> 01:04:31.010 what I've said so far, although we 01:04:31.010 --> 01:04:33.140 had to do a special calculation algebraically that 01:04:33.140 --> 01:04:34.290 didn't really make sense-- 01:04:34.290 --> 01:04:37.910 I mean, that you have to show some kind of near independence. 01:04:37.910 --> 01:04:38.705 Question? 01:04:38.705 --> 01:04:43.455 AUDIENCE: Is that less than or equal to, right? 01:04:43.455 --> 01:04:44.393 The Markov inequality. 01:04:44.393 --> 01:04:45.850 YUFEI ZHAO: Ah, thank you. 01:04:45.850 --> 01:04:47.470 So this is less than or equal to. 01:04:47.470 --> 01:04:50.030 Thank you. 01:04:50.030 --> 01:04:52.950 But now is where the algebra comes in, 01:04:52.950 --> 01:04:56.180 so the algebraic geometry nature of this argument comes in. 01:04:56.180 --> 01:04:58.940 It turns out that this quantity here-- 01:05:02.990 --> 01:05:05.630 previously, we said, at least heuristically, 01:05:05.630 --> 01:05:08.720 in the random graphs case, it behaves like a qua Poisson 01:05:08.720 --> 01:05:10.140 random variable. 01:05:10.140 --> 01:05:13.900 So it's fairly uniform in a Poisson sense. 01:05:13.900 --> 01:05:15.940 It turns out because of the algebraic nature 01:05:15.940 --> 01:05:17.860 of the construction, this random variable 01:05:17.860 --> 01:05:20.020 behaves nothing like a Poisson. 01:05:20.020 --> 01:05:31.590 So it turns out it's highly constrained due to reasons 01:05:31.590 --> 01:05:32.790 using algebraic geometry. 01:05:32.790 --> 01:05:35.400 And I'll tell you exactly why. 01:05:35.400 --> 01:05:37.320 So that the number of common neighbors 01:05:37.320 --> 01:05:40.620 is either very small or very large. 01:05:40.620 --> 01:05:50.550 And here is the claim that for every s and d, 01:05:50.550 --> 01:05:54.660 there exists some c, such that if I 01:05:54.660 --> 01:06:07.040 have a bunch of polynomials on fq to the s of degree 01:06:07.040 --> 01:06:14.260 o at most d, then if you look at the number of zeros, 01:06:14.260 --> 01:06:29.860 common zeros, of the f's, how many common zeros can you have? 01:06:29.860 --> 01:06:32.230 It turns out it cannot just be some arbitrary number. 01:06:32.230 --> 01:06:42.400 So this set has size either bounded at most c. 01:06:42.400 --> 01:06:48.600 Or it is at least q minus something very small. 01:06:51.520 --> 01:06:52.990 And I'll explain just a little bit 01:06:52.990 --> 01:06:55.790 why this is the case, although I will not give a proof. 01:06:55.790 --> 01:06:58.930 So either somehow you are working 01:06:58.930 --> 01:07:06.890 in a zero dimensional case, or if this algebraic variety that 01:07:06.890 --> 01:07:10.670 comes with it has positive dimension, 01:07:10.670 --> 01:07:12.600 then you should have a lot more points. 01:07:26.360 --> 01:07:29.300 And the reason for this dichotomy 01:07:29.300 --> 01:07:33.740 has to do with how many points are there 01:07:33.740 --> 01:07:39.060 on the algebraic variety over a finite field? 01:07:39.060 --> 01:07:41.362 So I will not give a proof, although if you 01:07:41.362 --> 01:07:43.570 look on the course website for a link to a reference, 01:07:43.570 --> 01:07:45.180 that does have a proof. 01:07:45.180 --> 01:07:47.670 But I will tell you what the key algebraic geometric 01:07:47.670 --> 01:07:50.950 input is to that claim up here. 01:07:50.950 --> 01:07:54.910 And this is a important and famous theorem 01:07:54.910 --> 01:07:56.330 called a Lang Weil bound. 01:08:01.180 --> 01:08:04.970 So the Lang Weil bound tells you that if you 01:08:04.970 --> 01:08:13.250 have an algebraic variety, v. And for now, it is important 01:08:13.250 --> 01:08:19.760 in order to say this properly to work in the algebraic closure, 01:08:19.760 --> 01:08:27.060 so Fq bar as the algebraic closure, 01:08:27.060 --> 01:08:29.560 it's the smallest field extension where I can 01:08:29.560 --> 01:08:33.510 solve all polynomial equations. 01:08:33.510 --> 01:08:41.211 Then, the variety cut out by a set of polynomials-- 01:08:45.450 --> 01:08:46.710 so v is this variety. 01:08:46.710 --> 01:08:53.420 So if it is irreducible, it cannot be written as a union 01:08:53.420 --> 01:08:57.300 of finite number of smaller varieties, 01:08:57.300 --> 01:09:02.490 irreducible over Fq bar. 01:09:02.490 --> 01:09:09.570 And all of these polynomials have degree bounded. 01:09:12.240 --> 01:09:18.220 Then the question is, if I take these polynomials 01:09:18.220 --> 01:09:22.830 and I look at how many Fq points does it have? 01:09:22.830 --> 01:09:25.390 So in other words, now I leave the field. 01:09:25.390 --> 01:09:29.649 I come back down to Earth, to the base field and ask, 01:09:29.649 --> 01:09:33.220 what's the number of solutions where 01:09:33.220 --> 01:09:34.460 the coordinates are in Fq? 01:09:39.040 --> 01:09:40.790 OK, so how many points do we expect? 01:09:40.790 --> 01:09:43.020 Well, the simplest example of an algebraic variety 01:09:43.020 --> 01:09:45.189 is that of a subspace. 01:09:45.189 --> 01:09:48.640 If you have a d-dimensional subspace over Fq, 01:09:48.640 --> 01:09:51.850 you have q to the d points exactly. 01:09:51.850 --> 01:09:57.460 So you expect something like q raised to the dimension 01:09:57.460 --> 01:09:59.010 of the variety. 01:09:59.010 --> 01:10:01.930 Now the dimensions is actually a somewhat subtle concept. 01:10:01.930 --> 01:10:02.750 I won't define. 01:10:02.750 --> 01:10:06.070 But there are many definitions in algebraic geometry. 01:10:06.070 --> 01:10:08.380 It turns out it's not always exactly as nice 01:10:08.380 --> 01:10:10.360 as in the case of linear subspaces. 01:10:10.360 --> 01:10:13.630 But the Lang Weil bound tells us that is not too far off. 01:10:13.630 --> 01:10:19.330 You have a deviation that is at most 01:10:19.330 --> 01:10:22.360 on the order of q 1 over root q, where 01:10:22.360 --> 01:10:25.060 there are some hidden constants depending 01:10:25.060 --> 01:10:28.690 on the description of your variety in terms 01:10:28.690 --> 01:10:30.940 of the degrees of the polynomial with the dimension 01:10:30.940 --> 01:10:32.410 and the number of polynomials. 01:10:32.410 --> 01:10:36.240 But the point is that the number of points on this variety 01:10:36.240 --> 01:10:40.300 is basically should be around the same as the model 01:10:40.300 --> 01:10:41.665 case, namely that of a subspace. 01:10:44.540 --> 01:10:46.640 And that brings us some intuition 01:10:46.640 --> 01:10:48.230 to why this lemma is true. 01:10:48.230 --> 01:10:50.822 So you have those polynomials up there. 01:10:50.822 --> 01:10:52.280 So there are some subtle points one 01:10:52.280 --> 01:10:54.200 needs to verify about your disability, 01:10:54.200 --> 01:10:56.990 but the punchline is that either you 01:10:56.990 --> 01:10:59.467 are in the zero-dimensional case, in which case 01:10:59.467 --> 01:11:01.300 you have something like [INAUDIBLE] theorem, 01:11:01.300 --> 01:11:05.590 and that tells you that a number of solutions is bounded. 01:11:05.590 --> 01:11:08.085 Or you're in the positive dimensional case, in which case 01:11:08.085 --> 01:11:09.460 the Lang Weil term tells you, you 01:11:09.460 --> 01:11:11.440 must have lots of solutions. 01:11:11.440 --> 01:11:13.180 And there is no middle ground. 01:11:16.170 --> 01:11:20.636 And now, we're ready to finish off Boris Bukh's construction. 01:11:26.100 --> 01:11:35.580 So we see that applying this lemma up there with this, what 01:11:35.580 --> 01:11:37.690 should be my polynomials be? 01:11:37.690 --> 01:11:46.080 I'm going to use my polynomials f sub u of y to be-- 01:11:46.080 --> 01:11:49.300 well, I have a random polynomial up there. 01:11:49.300 --> 01:11:53.640 So I'm trying to find common neighbors. 01:11:53.640 --> 01:11:56.010 I'm trying to find common solutions. 01:11:59.970 --> 01:12:08.450 So these are my polynomial as u ranges over big U. 01:12:08.450 --> 01:12:17.840 So for q large enough, we find that the number 01:12:17.840 --> 01:12:24.320 of common neighbors of u, the probability that it is bigger 01:12:24.320 --> 01:12:30.600 than c, where c is supplied by that lemma, 01:12:30.600 --> 01:12:35.000 is equal to the probability that a number of common neighbors 01:12:35.000 --> 01:12:41.430 of u exceeds q over 2, where q over w is this quantity 01:12:41.430 --> 01:12:42.780 rounded-- 01:12:42.780 --> 01:12:45.410 smaller than that quantity up there. 01:12:45.410 --> 01:12:47.220 So if it has more than c solutions, 01:12:47.220 --> 01:12:49.170 that automatically has a lot of solutions. 01:12:52.230 --> 01:12:55.890 And now, we can apply the Markov inequality up there, 01:12:55.890 --> 01:13:00.330 the tail bound on the moments to deduce that this probability is 01:13:00.330 --> 01:13:05.400 at most M divided by q over 2 raised to the power d. 01:13:08.580 --> 01:13:10.560 And the moral of it is that we should 01:13:10.560 --> 01:13:12.900 have a very small number-- 01:13:12.900 --> 01:13:16.030 I mean, this should occur with very small probability. 01:13:16.030 --> 01:13:29.140 So let's call u being a subset of v bad if u has r elements, u 01:13:29.140 --> 01:13:32.780 is contained entirely on the left side or the right side 01:13:32.780 --> 01:13:35.480 of the original bipartition. 01:13:35.480 --> 01:13:41.240 And most importantly, u has a lot, namely more than c, 01:13:41.240 --> 01:13:49.110 common neighbors in j. 01:13:53.086 --> 01:13:56.650 So how many bad sets do we expect? 01:13:56.650 --> 01:13:58.940 Basically, a very small number. 01:13:58.940 --> 01:14:07.320 So the number of bad sets, bad use, 01:14:07.320 --> 01:14:12.180 is upper bounded by-- well, for each choice of s elements, 01:14:12.180 --> 01:14:14.130 the probability that something is bad 01:14:14.130 --> 01:14:22.810 is this quantity up here, which we chose d to be a large 01:14:22.810 --> 01:14:24.700 enough constant depending on s. 01:14:24.700 --> 01:14:27.880 If you look at the choice of d up there, 01:14:27.880 --> 01:14:32.350 see that this quantity here is quite a bit smaller 01:14:32.350 --> 01:14:34.540 than the number of vertices. 01:14:40.730 --> 01:14:52.070 And now, the last step is the same as our randomized 01:14:52.070 --> 01:14:55.970 construction using Erdos-Renyi random graphs, 01:14:55.970 --> 01:14:59.180 where we remove-- 01:14:59.180 --> 01:15:00.880 well, it's almost the same, but now 01:15:00.880 --> 01:15:06.640 we remove one vertex from every bad set. 01:15:17.360 --> 01:15:22.520 And we get some graph g prime. 01:15:22.520 --> 01:15:26.540 And we just need to check that g prime has lots of edges. 01:15:26.540 --> 01:15:29.000 We know got rid of all the bad sets. 01:15:29.000 --> 01:15:38.180 So g prime is now K sub sc plus 1 free. 01:15:42.390 --> 01:15:46.110 We got rid of all possibilities for s points having 01:15:46.110 --> 01:15:49.040 more than c common neighbors. 01:15:49.040 --> 01:15:52.720 Now, we just need to check that g prime has lots of edges. 01:15:52.720 --> 01:15:56.870 Well, the expected number of edges in g prime 01:15:56.870 --> 01:16:01.180 is at least the x-- 01:16:01.180 --> 01:16:09.750 what we removed one vertex for every bad u. 01:16:09.750 --> 01:16:15.210 And each bad u carries with it at most n edges, 01:16:15.210 --> 01:16:19.034 because there are only n edge on each side of the bipartition. 01:16:26.950 --> 01:16:29.810 And, well, the number of edges of g 01:16:29.810 --> 01:16:33.530 has expectations exactly n squared over q. 01:16:36.360 --> 01:16:46.780 And the number of bad u's we saw up there is not very large. 01:16:46.780 --> 01:16:51.910 So in particular, this quantity, the second term 01:16:51.910 --> 01:16:54.890 is dominated by the first term. 01:16:54.890 --> 01:17:01.640 And so we obtain the claimed number of edges. 01:17:09.300 --> 01:17:16.550 Also, the graph has at most 2n vertices. 01:17:16.550 --> 01:17:18.630 So we may have gotten got rid of some. 01:17:18.630 --> 01:17:21.300 But actually, fewer vertices, the better. 01:17:21.300 --> 01:17:28.790 At most, 2n vertices, and it is a case of s-- 01:17:28.790 --> 01:17:31.190 So it's Kst free for t large enough. 01:17:34.320 --> 01:17:36.510 So this gives you another construction 01:17:36.510 --> 01:17:39.180 of Kst free graphs. 01:17:39.180 --> 01:17:41.180 And so today, we saw two different constructions 01:17:41.180 --> 01:17:45.020 of Kst free graphs for constants s and t, but in both cases, 01:17:45.020 --> 01:17:47.690 t is substantially larger than s. 01:17:47.690 --> 01:17:50.270 But the most important thing is that they both match 01:17:50.270 --> 01:17:54.380 the Kovari-Sos-Turan bound. 01:17:54.380 --> 01:17:57.840 So it gives you some evidence that maybe 01:17:57.840 --> 01:18:01.290 the Kovari-Sos-Turan conjecture, the theorem, 01:18:01.290 --> 01:18:04.710 is tight up to at most a constant factor, although that 01:18:04.710 --> 01:18:06.570 is a major open problem. 01:18:06.570 --> 01:18:11.430 And it remains a very difficult it seems open problem, but one 01:18:11.430 --> 01:18:15.360 that is of central importance in extremal graph theory 01:18:15.360 --> 01:18:20.500 to try to come up perhaps with other constructions 01:18:20.500 --> 01:18:22.200 that can do better. 01:18:22.200 --> 01:18:25.050 Maybe they will have some algebraic input. 01:18:25.050 --> 01:18:28.120 But maybe they will have some input from other ideas. 01:18:28.120 --> 01:18:29.220 We do not know. 01:18:29.220 --> 01:18:30.026 Question? 01:18:30.026 --> 01:18:32.406 AUDIENCE: So is this q defined? 01:18:32.406 --> 01:18:34.429 Because I remember q as a prime power, 01:18:34.429 --> 01:18:35.750 but it doesn't say there. 01:18:35.750 --> 01:18:37.510 YUFEI ZHAO: So question is, is q defined? 01:18:37.510 --> 01:18:40.900 So just like in the proofs of polarity graphs and what not-- 01:18:40.900 --> 01:18:43.750 so you have some n. 01:18:43.750 --> 01:18:47.140 You rounded down to the nearest prime powers. 01:18:47.140 --> 01:18:48.370 So s is a constant. 01:18:48.370 --> 01:18:52.150 So n is basically q to the s. 01:18:52.150 --> 01:18:56.200 So take large n, round it down to the nearest prime power. 01:18:56.200 --> 01:18:57.712 q could be a prime, for instance. 01:18:57.712 --> 01:18:58.545 It could be a prime. 01:18:58.545 --> 01:19:00.140 It could be a prime power. 01:19:00.140 --> 01:19:01.300 Think q to be a prime. 01:19:04.050 --> 01:19:07.420 So I'm saying for every q, there is a construction. 01:19:07.420 --> 01:19:12.790 And for every n, you can round down to the nearest q to the s 01:19:12.790 --> 01:19:17.400 and then run this construction. 01:19:17.400 --> 01:19:19.950 Any more questions? 01:19:19.950 --> 01:19:20.450 Great. 01:19:20.450 --> 01:19:24.230 So next time I will begin by telling you a few more things 01:19:24.230 --> 01:19:27.770 about why people really like this construction 01:19:27.770 --> 01:19:31.198 and some conjectures that were solved using this idea 01:19:31.198 --> 01:19:32.990 and some conjectures that still remain open 01:19:32.990 --> 01:19:34.650 along the same lines. 01:19:34.650 --> 01:19:39.530 And we'll also go beyond Kst, so other bipartide graphs 01:19:39.530 --> 01:19:41.060 and show you how to do upper bounds 01:19:41.060 --> 01:19:43.390 for those bipartide graphs.