1 00:00:18,955 --> 00:00:20,330 YUFEI ZHAO: Last time, we started 2 00:00:20,330 --> 00:00:24,950 discussing the extremal problem for bipartide graphs. 3 00:00:24,950 --> 00:00:28,970 And in particular, we saw the Kovari-Sos-Turan theorem, 4 00:00:28,970 --> 00:00:32,810 which tells us that if you forbid your graph from having 5 00:00:32,810 --> 00:00:36,320 a complete bipartide graph, Kst, then 6 00:00:36,320 --> 00:00:42,670 you have this upper bound on the number of edges in your graph. 7 00:00:42,670 --> 00:00:43,480 So we gave a proof. 8 00:00:43,480 --> 00:00:47,680 It was fairly short, used the double counting argument, 9 00:00:47,680 --> 00:00:49,750 and it give you this bound. 10 00:00:49,750 --> 00:00:53,050 And the next question is how tight is this bound? 11 00:00:53,050 --> 00:00:57,760 Is there a lower bound that is off by, let's say, 12 00:00:57,760 --> 00:01:00,130 at most a constant factor? 13 00:01:00,130 --> 00:01:02,260 And that's a major open problem. 14 00:01:02,260 --> 00:01:07,680 It's a conjecture that this bound is tight 15 00:01:07,680 --> 00:01:09,830 up to constant factors. 16 00:01:09,830 --> 00:01:12,960 But that conjecture is known for only a very small number 17 00:01:12,960 --> 00:01:14,010 of graphs. 18 00:01:14,010 --> 00:01:17,590 And we saw a couple of examples last time. 19 00:01:17,590 --> 00:01:20,400 So last time we saw construction that 20 00:01:20,400 --> 00:01:28,050 shows that for S equals to 2, this bound is tight. 21 00:01:31,670 --> 00:01:36,110 So the extremal number for k through 2 is on the order of n 22 00:01:36,110 --> 00:01:37,670 to the 3/2. 23 00:01:37,670 --> 00:01:42,300 So this theta means I'm hiding constant factors. 24 00:01:42,300 --> 00:01:45,090 And our construction used this polarity graph, 25 00:01:45,090 --> 00:01:47,400 which is essentially the point line incidence 26 00:01:47,400 --> 00:01:49,590 graph of a projective plane. 27 00:01:49,590 --> 00:01:53,700 And a basic algebraic or geometric fact, if you will, 28 00:01:53,700 --> 00:01:57,710 that two lines intersecting at most one point. 29 00:01:57,710 --> 00:02:03,010 We also sketched a construction that 30 00:02:03,010 --> 00:02:08,580 showed that for S equals to 3, this bound is also tight. 31 00:02:11,370 --> 00:02:13,060 And this construction here involved 32 00:02:13,060 --> 00:02:21,070 using spheres, again, in some space over a finite field. 33 00:02:21,070 --> 00:02:22,960 So both these constructions are in some sense 34 00:02:22,960 --> 00:02:24,490 algebraic geometric. 35 00:02:24,490 --> 00:02:26,590 And you can ask, is there a way to extend 36 00:02:26,590 --> 00:02:31,540 these ideas to construct other examples of Kst free graphs 37 00:02:31,540 --> 00:02:36,040 with the right number of edges using some ingredients 38 00:02:36,040 --> 00:02:37,870 from algebraic geometry? 39 00:02:37,870 --> 00:02:40,030 And today, I want to show you two different ways 40 00:02:40,030 --> 00:02:40,660 of doing that. 41 00:02:44,870 --> 00:02:51,220 So the state of the art, which I mentioned last time, 42 00:02:51,220 --> 00:02:55,650 let me remind you what is known about constructions 43 00:02:55,650 --> 00:03:00,240 that achieve the right exponent up there. 44 00:03:00,240 --> 00:03:07,560 So in a series of two papers by [INAUDIBLE] 45 00:03:07,560 --> 00:03:11,460 it is shown that if the constants t and s, 46 00:03:11,460 --> 00:03:15,300 such that t is large enough compared to s, 47 00:03:15,300 --> 00:03:20,530 in particular t is bigger than s minus 1 factorial, 48 00:03:20,530 --> 00:03:27,480 then the extremal number of Kst's is 49 00:03:27,480 --> 00:03:30,990 of the order, same order, as given 50 00:03:30,990 --> 00:03:34,980 in the upper bound of the Kovari-Sos-Turan theorem. 51 00:03:34,980 --> 00:03:40,780 In particular, these range of parameters allows you to do 52 00:03:40,780 --> 00:03:46,660 (2, 2), (3, 3), which we already know how to do. 53 00:03:46,660 --> 00:03:49,400 But the next case is 4, 7. 54 00:03:49,400 --> 00:03:52,570 And it is still open how to do 4, 6. 55 00:03:58,940 --> 00:04:00,833 So I want to show you this construction. 56 00:04:00,833 --> 00:04:02,750 And I will tell you exactly what the graph is. 57 00:04:02,750 --> 00:04:05,090 I'll give you an explicit description 58 00:04:05,090 --> 00:04:09,380 of this graph, which is Kst free and has lots of edges. 59 00:04:11,990 --> 00:04:15,860 And as I mentioned earlier, it is an algebraic construction. 60 00:04:15,860 --> 00:04:22,100 So as before, we start with p prime. 61 00:04:22,100 --> 00:04:30,680 And we will take n to be a p raised to the power s. 62 00:04:30,680 --> 00:04:34,172 And let's restrict s to be integer at least 2. 63 00:04:34,172 --> 00:04:35,880 And, of course, that's same as last time, 64 00:04:35,880 --> 00:04:39,390 if you have other values of n, take a prime close 65 00:04:39,390 --> 00:04:42,490 to the desired value and then take it from there. 66 00:04:45,180 --> 00:04:50,160 To describe the construction, let me remind you, 67 00:04:50,160 --> 00:04:55,200 the norm map, if you have a field extension, in this case, 68 00:04:55,200 --> 00:04:59,150 specifically I am looking at the field extension, 69 00:04:59,150 --> 00:05:08,700 this Fp to the s, I can define a norm map as follows. 70 00:05:08,700 --> 00:05:15,060 Sending x to be the product of all the conjugates, 71 00:05:15,060 --> 00:05:19,740 or the Galois conjugates of x, in this field extension. 72 00:05:19,740 --> 00:05:23,500 So explicitly written out is just that expression, 73 00:05:23,500 --> 00:05:27,240 which I can clean and collect and write it down like this. 74 00:05:32,090 --> 00:05:33,960 So I wrote that the image of this norm map 75 00:05:33,960 --> 00:05:36,310 lies in the base field Fp. 76 00:05:36,310 --> 00:05:37,185 And that is because-- 77 00:05:41,388 --> 00:05:44,280 well, one of the many reasons why this is the case, 78 00:05:44,280 --> 00:05:47,310 is that if you look at-- 79 00:05:47,310 --> 00:05:49,740 so I'll denote this norm map by N-- 80 00:05:49,740 --> 00:05:57,990 if you look at N of x, it raised to power p, 81 00:05:57,990 --> 00:06:00,420 leaves this value unchanged. 82 00:06:00,420 --> 00:06:04,500 And the base field is the field where it 83 00:06:04,500 --> 00:06:07,110 is invariant under power by p. 84 00:06:09,930 --> 00:06:15,550 So here's the graph, which I'll denote the norm 85 00:06:15,550 --> 00:06:20,180 graph with parameters p and s. 86 00:06:20,180 --> 00:06:30,640 So the norm graph will have as vertices just the elements 87 00:06:30,640 --> 00:06:41,690 of this field extension. 88 00:06:41,690 --> 00:06:54,300 And the edges will be the set of all pairs of vertices, not 89 00:06:54,300 --> 00:07:00,180 equal, of course, such that a norm of their sum equals to 1. 90 00:07:04,547 --> 00:07:05,380 So that's the graph. 91 00:07:05,380 --> 00:07:07,912 This is an explicit description of what the vertices 92 00:07:07,912 --> 00:07:08,870 and what the edges are. 93 00:07:11,430 --> 00:07:14,760 So now, we need to verify a couple of things. 94 00:07:14,760 --> 00:07:20,040 One is that this graph has the desired number of edges. 95 00:07:20,040 --> 00:07:21,980 It has lots of edges. 96 00:07:21,980 --> 00:07:26,170 And two is that this graph is Kst free. 97 00:07:26,170 --> 00:07:28,980 So let's do both of those things. 98 00:07:28,980 --> 00:07:33,350 So the first is let's check it has the right number of edges. 99 00:07:33,350 --> 00:07:36,770 So that's a relatively easy task. 100 00:07:36,770 --> 00:07:42,580 What we need to do is to count for every a 101 00:07:42,580 --> 00:07:47,860 how many choices of b are there in this field extension 102 00:07:47,860 --> 00:07:53,650 such that a plus b has norm exactly 1? 103 00:07:53,650 --> 00:07:56,430 And I claim that that number-- 104 00:07:56,430 --> 00:08:01,110 well, so here's a basic algebra fact, 105 00:08:01,110 --> 00:08:07,710 that the number of elements in this field extension with norm 106 00:08:07,710 --> 00:08:15,120 exactly 1 is precisely p to the s minus 1 107 00:08:15,120 --> 00:08:17,850 divided by p to the s. 108 00:08:23,646 --> 00:08:27,300 And this is because really we're looking 109 00:08:27,300 --> 00:08:31,150 in the multiplicative subgroup. 110 00:08:31,150 --> 00:08:36,640 So the multiplicative group in this Fp to the s. 111 00:08:36,640 --> 00:08:38,258 And it has a cyclical-- 112 00:08:42,162 --> 00:08:44,130 so there's a generator-- 113 00:08:44,130 --> 00:08:51,950 the order of the cyclic group is p to the s minus 1. 114 00:08:51,950 --> 00:08:57,930 So you're asking, how many elements 115 00:08:57,930 --> 00:09:08,357 when raised to this power here ends up at the identity? 116 00:09:10,985 --> 00:09:11,860 So that's the answer. 117 00:09:15,970 --> 00:09:17,430 So that's one aspect. 118 00:09:17,430 --> 00:09:27,230 And so as a result, every vertex is adjacent to, well, 119 00:09:27,230 --> 00:09:29,330 how many vertices? 120 00:09:29,330 --> 00:09:33,130 For every given a I need to solve for b. 121 00:09:33,130 --> 00:09:35,270 And basically, this many solutions, 122 00:09:35,270 --> 00:09:38,000 I have to be just slightly careful because I 123 00:09:38,000 --> 00:09:40,430 don't want loops in my graph. 124 00:09:40,430 --> 00:09:42,290 So I may need to subtract 1. 125 00:09:42,290 --> 00:09:50,090 So it's adjacent to at least this number up here minus 1 126 00:09:50,090 --> 00:10:00,740 to account for possible loops, which is pretty large, 127 00:10:00,740 --> 00:10:05,070 p to the s minus 1, which in other words 128 00:10:05,070 --> 00:10:12,140 is n raised to 1 minus 1 over s, that many vertices, 129 00:10:12,140 --> 00:10:14,840 and you see that this gives you the right number of edges. 130 00:10:28,620 --> 00:10:30,820 So this is a graph with lots of edges. 131 00:10:30,820 --> 00:10:33,170 So that part wasn't so hard. 132 00:10:33,170 --> 00:10:36,040 The next part, it's much trickier, 133 00:10:36,040 --> 00:10:41,060 which is we want to check that this graph has no Kst's. 134 00:10:41,060 --> 00:10:43,100 So previously in our algebraic construction, 135 00:10:43,100 --> 00:10:46,520 we used some geometric facts, such as no two lines 136 00:10:46,520 --> 00:10:51,230 intersect in more than one point to show that there's no k to 2 137 00:10:51,230 --> 00:10:53,120 in the polarity graph. 138 00:10:53,120 --> 00:10:55,550 So there's going to be something like that here. 139 00:10:55,550 --> 00:11:05,410 So the claim is that this construction, this norm graph, 140 00:11:05,410 --> 00:11:12,880 is Ks, s factorial plus 1 free. 141 00:11:15,780 --> 00:11:17,890 So it's not quite the bound I claimed. 142 00:11:17,890 --> 00:11:19,180 So it's a little bit weaker. 143 00:11:19,180 --> 00:11:21,430 But it is in the spirit of what I am claiming, 144 00:11:21,430 --> 00:11:27,060 namely that for t large enough, this graph is Kst free. 145 00:11:27,060 --> 00:11:29,570 So for t a large enough comes constant here 146 00:11:29,570 --> 00:11:33,470 will show s factorial plus 1. 147 00:11:33,470 --> 00:11:41,350 And as a result, it would follow that the extremal number 4s 148 00:11:41,350 --> 00:11:54,060 sub s factorial plus 1 is at least 1/2 149 00:11:54,060 --> 00:11:56,280 minus little 1 of the constant-- 150 00:11:56,280 --> 00:11:59,010 I don't already worry about that much, 151 00:11:59,010 --> 00:12:03,568 but it's on the order of n to the 2 minus 1 over s. 152 00:12:03,568 --> 00:12:06,860 OK, everyone with me? 153 00:12:06,860 --> 00:12:10,490 So we need to verify this graph here has no Kst. 154 00:12:10,490 --> 00:12:11,411 Yes, question? 155 00:12:11,411 --> 00:12:14,638 AUDIENCE: Should that t be an s? 156 00:12:14,638 --> 00:12:16,180 YUFEI ZHAO: Yes, that should be an s. 157 00:12:16,180 --> 00:12:16,680 Thank you. 158 00:12:21,480 --> 00:12:23,152 Any more questions? 159 00:12:23,152 --> 00:12:26,960 AUDIENCE: Should that be s minus 1 factorial? 160 00:12:26,960 --> 00:12:29,060 YUFEI ZHAO: So we will show later 161 00:12:29,060 --> 00:12:32,450 on a better result using s minus 1 factorial, 162 00:12:32,450 --> 00:12:35,420 but for now I'll show you the slightly weaker result, which 163 00:12:35,420 --> 00:12:36,810 is still in the same spirit. 164 00:12:36,810 --> 00:12:37,310 Yep. 165 00:12:37,310 --> 00:12:39,953 AUDIENCE: Is the stronger result using the same graph? 166 00:12:39,953 --> 00:12:41,870 YUFEI ZHAO: We'll change to a different graph. 167 00:12:41,870 --> 00:12:44,412 For the stronger result, we will change to a different graph. 168 00:12:50,000 --> 00:12:55,260 OK, so now let's show that this graph here is Kst free. 169 00:12:55,260 --> 00:12:58,653 And for that claim, we need to invoke an algebraic fact, which 170 00:12:58,653 --> 00:12:59,570 let me write down now. 171 00:13:02,770 --> 00:13:05,750 So suppose we have a field f. 172 00:13:10,280 --> 00:13:14,090 Any field will work with a finite field. 173 00:13:14,090 --> 00:13:15,290 Any field is fine. 174 00:13:15,290 --> 00:13:22,520 And I have a bunch of elements from the field, 175 00:13:22,520 --> 00:13:31,320 such that a sub ij is different from a sub 176 00:13:31,320 --> 00:13:37,120 rj for all i not the same as r. 177 00:13:40,240 --> 00:13:43,200 Then the system of equations-- 178 00:13:47,468 --> 00:13:48,760 and I'll write down the system. 179 00:13:48,760 --> 00:13:50,450 So x minus 1-- 180 00:13:50,450 --> 00:13:59,460 x1 minus a 1, 1, x2 minus a 1, 2, and so on. 181 00:13:59,460 --> 00:14:04,286 xs minus a1x equals to b1. 182 00:14:04,286 --> 00:14:06,070 That's the first equation. 183 00:14:06,070 --> 00:14:09,240 Second equation, x1 minus a 2, 1. 184 00:14:09,240 --> 00:14:13,630 It almost looks like the usual system of linear equations. 185 00:14:13,630 --> 00:14:15,790 But I'm taking products. 186 00:14:21,170 --> 00:14:22,470 And so on. 187 00:14:22,470 --> 00:14:34,910 So the last one being x1 minus a s1 x2 minus a s2, dot dot dot, 188 00:14:34,910 --> 00:14:40,460 xs minus a ss equals to b sub s. 189 00:14:40,460 --> 00:14:54,400 The system has at most s factorial solutions, where 190 00:14:54,400 --> 00:14:56,320 I'm working inside this field. 191 00:14:59,780 --> 00:15:01,370 So that's the claim. 192 00:15:01,370 --> 00:15:03,782 So let me just give you some intuition for this claim. 193 00:15:08,710 --> 00:15:16,840 Suppose the right side vector is 0, all zeroes. 194 00:15:16,840 --> 00:15:19,730 Then I claim that this is trivial. 195 00:15:19,730 --> 00:15:21,280 So what is the saying? 196 00:15:21,280 --> 00:15:28,740 I need to select x1 to be one of the a's from the first row 197 00:15:28,740 --> 00:15:32,110 and x2 be one of the a's from the second row, and so on. 198 00:15:32,110 --> 00:15:34,990 But each column of a is distinct. 199 00:15:34,990 --> 00:15:36,670 So that's the hypothesis. 200 00:15:36,670 --> 00:15:40,660 You have all the a's in the first column are distinct. 201 00:15:40,660 --> 00:15:45,940 So no two of the x's can-- 202 00:15:49,150 --> 00:15:53,350 so I need to set one of the xi's to be one 203 00:15:53,350 --> 00:15:55,330 of the a's from the first row. 204 00:15:55,330 --> 00:16:01,210 But you see that you cannot set x1 to be a 1, 1, 205 00:16:01,210 --> 00:16:05,420 and x1 to be a s1 at the same time. 206 00:16:05,420 --> 00:16:08,450 So the solution just counts permutations, 207 00:16:08,450 --> 00:16:10,339 which is exactly s factorial. 208 00:16:21,590 --> 00:16:24,910 So this algebraic fact plays a key role 209 00:16:24,910 --> 00:16:29,550 in the proof of the theorem that the lower bound 210 00:16:29,550 --> 00:16:34,720 that we're stating up there, if you look at the paper, 211 00:16:34,720 --> 00:16:36,160 they give a proof of this result. 212 00:16:36,160 --> 00:16:37,810 And it's not a long proof. 213 00:16:37,810 --> 00:16:41,980 But it uses some commutative algebra and algebraic geometry. 214 00:16:41,980 --> 00:16:45,490 And usually in a class, if the instructor 215 00:16:45,490 --> 00:16:48,790 doesn't present the proof, it's for one of several reasons. 216 00:16:48,790 --> 00:16:50,950 Maybe the proof is too short. 217 00:16:50,950 --> 00:16:52,360 It doesn't need to be presented. 218 00:16:52,360 --> 00:16:54,700 Maybe it's too long or too difficult. Maybe 219 00:16:54,700 --> 00:16:57,790 it's not instructive to the class. 220 00:16:57,790 --> 00:16:59,920 And the last reason, which is the case here, 221 00:16:59,920 --> 00:17:01,930 is that I don't actually understand the proof. 222 00:17:01,930 --> 00:17:04,780 As in I can follow it line by line, 223 00:17:04,780 --> 00:17:08,079 but I don't understand why it is true. 224 00:17:08,079 --> 00:17:12,640 And if one of you wants to come up with a different proof 225 00:17:12,640 --> 00:17:16,690 or try to explain to me how this seemingly 226 00:17:16,690 --> 00:17:23,829 elementary algebraic fact is proved, I would appreciate it. 227 00:17:23,829 --> 00:17:26,710 For small values of x, you can check it by hand. 228 00:17:26,710 --> 00:17:29,110 So x equals to 2, you're solving a system 229 00:17:29,110 --> 00:17:31,390 of two quadratic equations. 230 00:17:31,390 --> 00:17:32,890 And that you can check by hand. 231 00:17:32,890 --> 00:17:35,050 And three maybe you can do it with some work. 232 00:17:35,050 --> 00:17:40,270 But even with 4 it's not so clear how to do it. 233 00:17:40,270 --> 00:17:47,410 And also, one of the geometric intuition is that if b is o0, 234 00:17:47,410 --> 00:17:50,320 then you have exactly s factorial solutions. 235 00:17:50,320 --> 00:17:52,090 And the geometric intuition is somehow 236 00:17:52,090 --> 00:17:56,290 that if you move b around, then the fiber, 237 00:17:56,290 --> 00:17:59,080 the sides of the fiber, the number of solutions x 238 00:17:59,080 --> 00:18:00,110 can only go down. 239 00:18:00,110 --> 00:18:00,940 It can not go up. 240 00:18:00,940 --> 00:18:05,170 And this corresponds to some algebraic geometry phenomenon. 241 00:18:05,170 --> 00:18:08,440 And that's all I will say about this algebraic fact, which 242 00:18:08,440 --> 00:18:09,670 we'll now use as a black box. 243 00:18:17,330 --> 00:18:19,520 Great. 244 00:18:19,520 --> 00:18:22,790 So now we have that as our algebraic input, 245 00:18:22,790 --> 00:18:27,950 let us show that a norm graph is Kst free. 246 00:18:27,950 --> 00:18:30,080 It's actually not so hard once you 247 00:18:30,080 --> 00:18:32,310 assume that theorem up there. 248 00:18:36,480 --> 00:18:47,400 So let's show at a norm graph is Ks, s factorial plus 1 free. 249 00:18:50,070 --> 00:18:55,170 Well, what does it mean to have a Kst? 250 00:18:55,170 --> 00:18:59,160 It means that if you have distinct vertices, 251 00:18:59,160 --> 00:19:05,330 which then correspond to elements, 252 00:19:05,330 --> 00:19:12,580 s elements, y1 through ys, of this field, 253 00:19:12,580 --> 00:19:17,580 then the common neighbors of these elements 254 00:19:17,580 --> 00:19:28,800 correspond to solutions of this system of equations 255 00:19:28,800 --> 00:19:32,560 where I set all of these values to be 1. 256 00:19:32,560 --> 00:19:35,320 But I can write l exactly what these guys are 257 00:19:35,320 --> 00:19:39,480 because I have this form, that representation there 258 00:19:39,480 --> 00:19:44,690 for the norm map, so I can write it out. 259 00:19:44,690 --> 00:19:48,560 And now remember this fact that when 260 00:19:48,560 --> 00:19:53,080 you are in characteristic p, x plus y raised to the power p, 261 00:19:53,080 --> 00:20:00,500 is the same as x to the p plus yp and characteristic p. 262 00:20:00,500 --> 00:20:04,720 So I can expand the remaining parenthesis like that. 263 00:20:13,340 --> 00:20:16,050 So I want the first line to equal to 1, and so on. 264 00:20:16,050 --> 00:20:19,950 And each of the lines has that equal to 1. 265 00:20:31,560 --> 00:20:35,790 How many solutions in x does this system of equations have? 266 00:20:35,790 --> 00:20:39,510 So even if I treat each of x and x to the p 267 00:20:39,510 --> 00:20:42,330 and so on as separate variables, that theorem 268 00:20:42,330 --> 00:20:49,200 appear tells me that there are at most s factorial solutions 269 00:20:49,200 --> 00:20:50,250 in x. 270 00:20:53,310 --> 00:20:57,500 Satisfies all the hypotheses of that theorem up there. 271 00:20:57,500 --> 00:21:04,250 Therefore, the graph is Ks sub s factorial plus 1 free. 272 00:21:04,250 --> 00:21:07,610 You do not have more than s plus 1 different values 273 00:21:07,610 --> 00:21:11,360 of x satisfying this system of equations. 274 00:21:11,360 --> 00:21:15,287 And that's the proof that this norm graph is Kst free. 275 00:21:15,287 --> 00:21:15,870 Yes, question? 276 00:21:15,870 --> 00:21:22,832 AUDIENCE: Why can't powers of like [INAUDIBLE] 277 00:21:22,832 --> 00:21:24,790 YUFEI ZHAO: Sorry, can you repeat the question? 278 00:21:24,790 --> 00:21:29,345 AUDIENCE: Why cannot the powers of the y's be the same? 279 00:21:29,345 --> 00:21:31,970 YUFEI ZHAO: The question is why cannot the powers of the p's be 280 00:21:31,970 --> 00:21:33,360 the same? 281 00:21:33,360 --> 00:21:36,210 So you are asking, down the second column, 282 00:21:36,210 --> 00:21:39,330 let's say, why are all these y's different? 283 00:21:39,330 --> 00:21:42,400 Because you're working inside a field. 284 00:21:42,400 --> 00:21:47,370 And raising to a p in this field is a bijection. 285 00:21:51,370 --> 00:21:54,040 So think about the order of the cyclic group. 286 00:21:54,040 --> 00:21:57,840 It has co-prime to p. 287 00:21:57,840 --> 00:22:00,040 But great question. 288 00:22:00,040 --> 00:22:02,640 Anything else? 289 00:22:02,640 --> 00:22:04,340 OK, so this gives you a construction 290 00:22:04,340 --> 00:22:11,870 that gives you Kst free for t bigger than s plus s factorial. 291 00:22:11,870 --> 00:22:14,630 Now, let me show you how to improve this construction 292 00:22:14,630 --> 00:22:19,280 to do a little bit better, to get s minus 1 factorial. 293 00:22:19,280 --> 00:22:21,080 And the idea is to take a variant 294 00:22:21,080 --> 00:22:24,790 of this norm graph, which we'll call the projective norm graph. 295 00:22:31,670 --> 00:22:34,430 And the projective norm graph will define it 296 00:22:34,430 --> 00:22:39,100 for s at least 3 is rather similar. 297 00:22:39,100 --> 00:22:41,160 But there's a twist. 298 00:22:41,160 --> 00:22:46,410 I have as the vertex set, not just the field extension-- 299 00:22:46,410 --> 00:22:50,490 OK, so now I take field extension, but one level less. 300 00:22:50,490 --> 00:22:54,210 And I take a second coordinate, which consists 301 00:22:54,210 --> 00:22:57,060 of non-zero elements from Fp. 302 00:23:01,340 --> 00:23:09,650 The edges are formed by putting an edge between these two 303 00:23:09,650 --> 00:23:13,730 elements, if and only if, the norm 304 00:23:13,730 --> 00:23:17,070 of the sum of the first coordinates 305 00:23:17,070 --> 00:23:20,340 equals to the product of the second coordinates. 306 00:23:24,480 --> 00:23:27,760 So now, you can run through a similar calculation that 307 00:23:27,760 --> 00:23:29,290 tells you the number of edges. 308 00:23:29,290 --> 00:23:31,570 So first of all, the number of vertices 309 00:23:31,570 --> 00:23:35,440 is p to the s minus 1 times p minus 1, 310 00:23:35,440 --> 00:23:39,600 so basically the same as p to s. 311 00:23:39,600 --> 00:23:49,200 And additionally, every vertex has degree exactly 312 00:23:49,200 --> 00:23:54,690 p raised to s minus 1 minus 1. 313 00:23:54,690 --> 00:24:04,460 And the reason is that if I tell you the values of x little x 314 00:24:04,460 --> 00:24:08,390 and big y, which cannot equal minus x, 315 00:24:08,390 --> 00:24:10,340 or else you will never form an edge, 316 00:24:10,340 --> 00:24:14,408 then they together uniquely determine little y. 317 00:24:18,160 --> 00:24:20,520 So for every value of big X and little x, 318 00:24:20,520 --> 00:24:23,880 I just need to run through all the values of big Y other 319 00:24:23,880 --> 00:24:24,690 than minus x. 320 00:24:29,110 --> 00:24:34,870 So the number of edges then equals to 1/2 times 321 00:24:34,870 --> 00:24:39,120 the number of vertices times the degree of every vertex, which, 322 00:24:39,120 --> 00:24:48,520 as before, is the claimed asymptotic. 323 00:24:48,520 --> 00:24:51,450 And the remaining thing to show is that this projective norm 324 00:24:51,450 --> 00:24:52,730 graph is Kst free. 325 00:25:05,820 --> 00:25:11,385 So it's K sub s, s minus 1 factorial plus 1 free. 326 00:25:16,840 --> 00:25:19,010 It's a similar calculation as the one before, 327 00:25:19,010 --> 00:25:22,240 but we need to take into account the small variant 328 00:25:22,240 --> 00:25:23,520 in the construction. 329 00:25:27,120 --> 00:25:36,520 So suppose we fix s vertices, labeled by this big Y's, little 330 00:25:36,520 --> 00:25:37,020 y's. 331 00:25:43,040 --> 00:25:49,100 And now we need to solve for uppercase X, 332 00:25:49,100 --> 00:26:05,800 lowercase x in this system of equations, 333 00:26:05,800 --> 00:26:09,230 so asking how many different pairs, big X, little 334 00:26:09,230 --> 00:26:13,710 x, can appear as a solution to this system of equations? 335 00:26:13,710 --> 00:26:21,030 Well, first of all, if some pairs of the first coordinates, 336 00:26:21,030 --> 00:26:31,610 big I is equal to big J, then if you have a solution, 337 00:26:31,610 --> 00:26:37,400 then that forces little y to be the same as little j. 338 00:26:37,400 --> 00:26:40,890 And so the y's wouldn't have been distinct to begin with. 339 00:26:40,890 --> 00:26:44,490 So this is not possible. 340 00:26:44,490 --> 00:26:48,690 So all the big Y's are distinct. 341 00:26:55,990 --> 00:27:03,320 Well, now, let's divide these equations 342 00:27:03,320 --> 00:27:04,760 by the final equation. 343 00:27:10,040 --> 00:27:13,040 And we get that the i-th equation 344 00:27:13,040 --> 00:27:26,970 becomes like that, which you can rewrite 345 00:27:26,970 --> 00:27:37,850 by dividing by the coordinate of the norm of big Y i 346 00:27:37,850 --> 00:27:39,800 minus big Y s. 347 00:27:39,800 --> 00:27:43,150 This is non-zero because we just showed that all the big Y 348 00:27:43,150 --> 00:27:45,130 i's are distinct. 349 00:27:45,130 --> 00:27:49,780 If you divide by this norm here and rearrange appropriately, 350 00:27:49,780 --> 00:27:53,860 we find that the equations become like this. 351 00:28:15,870 --> 00:28:17,570 So after doing some rearranging-- so 352 00:28:17,570 --> 00:28:20,540 this is the equation the set of new questions that we get. 353 00:28:20,540 --> 00:28:29,210 And you see that if you use new variables, x prime, 354 00:28:29,210 --> 00:28:33,300 do a substitution, this being x prime, 355 00:28:33,300 --> 00:28:37,020 then it has basically the same form as the one 356 00:28:37,020 --> 00:28:41,200 that we just saw with a different set of constants. 357 00:28:41,200 --> 00:28:44,640 And in particular from what we just saw, 358 00:28:44,640 --> 00:28:49,980 we see that you cannot have more than s minus 1 factorial 359 00:28:49,980 --> 00:28:53,250 solutions in x. 360 00:28:53,250 --> 00:28:56,910 Now, they're s minus 1 equations. 361 00:28:56,910 --> 00:29:00,380 And the field extension working as the x minus 1 362 00:29:00,380 --> 00:29:03,880 field extension. 363 00:29:03,880 --> 00:29:08,690 So we saved an equation by using this projectivization. 364 00:29:08,690 --> 00:29:11,060 And that's it. 365 00:29:11,060 --> 00:29:14,510 So this shows you the claim of constructing 366 00:29:14,510 --> 00:29:20,110 a Kst free graph for t bigger than s minus 1 factorial, which 367 00:29:20,110 --> 00:29:22,700 has the desired number of edges. 368 00:29:22,700 --> 00:29:23,583 Yes, question? 369 00:29:23,583 --> 00:29:26,072 AUDIENCE: Why do you have the if some capital 370 00:29:26,072 --> 00:29:29,040 Y equals capital IJ? 371 00:29:29,040 --> 00:29:35,290 YUFEI ZHAO: OK, so the question is why do I say this part? 372 00:29:35,290 --> 00:29:37,330 So I'm maybe skipping a sentence. 373 00:29:37,330 --> 00:29:39,670 I'm saying, if there is a solution to this system 374 00:29:39,670 --> 00:29:42,640 of equations x, if these vertices have 375 00:29:42,640 --> 00:29:47,770 a common neighbor, then if you have some x satisfying 376 00:29:47,770 --> 00:29:51,970 this system of equations, then having two different big Y's 377 00:29:51,970 --> 00:29:56,640 being the same forces you to have the two smaller y's 378 00:29:56,640 --> 00:29:57,546 being the same. 379 00:29:57,546 --> 00:29:58,340 AUDIENCE: OK. 380 00:29:58,340 --> 00:29:59,090 YUFEI ZHAO: Right. 381 00:29:59,090 --> 00:30:02,380 And then the Y's will have been distinct. 382 00:30:02,380 --> 00:30:04,150 So for them to have some common neighbors, 383 00:30:04,150 --> 00:30:08,000 you better have these big Y's being distinct. 384 00:30:08,000 --> 00:30:10,620 Any more questions? 385 00:30:10,620 --> 00:30:12,033 Great. 386 00:30:12,033 --> 00:30:13,700 So as I mentioned, it is an open problem 387 00:30:13,700 --> 00:30:21,090 to determine whether what is the extremal number for K44, K45, 388 00:30:21,090 --> 00:30:22,880 K46. 389 00:30:22,880 --> 00:30:26,586 And you may ask, well, we have this nice construction-- 390 00:30:26,586 --> 00:30:31,600 it maybe somewhat mysterious because of that, but explicit. 391 00:30:31,600 --> 00:30:33,580 And you can write this graph down. 392 00:30:33,580 --> 00:30:37,180 And you can ask is this graph K46 free? 393 00:30:37,180 --> 00:30:40,480 So do we gain one extra number for free, 394 00:30:40,480 --> 00:30:44,410 maybe because we didn't analyze things properly? 395 00:30:44,410 --> 00:30:46,580 And it turns out that's not the case. 396 00:30:46,580 --> 00:30:50,200 So there was a very recent paper just released last month 397 00:30:50,200 --> 00:30:54,850 showing that this graph here for s equals to 4 actually does 398 00:30:54,850 --> 00:30:56,710 contain some K46's. 399 00:31:00,010 --> 00:31:05,140 So if you want to prove a corresponding lower bound 400 00:31:05,140 --> 00:31:10,750 for K46, you better come up with a different construction. 401 00:31:10,750 --> 00:31:13,030 And that's I think an interesting direction 402 00:31:13,030 --> 00:31:13,680 to explore. 403 00:31:17,190 --> 00:31:18,300 Any questions? 404 00:31:18,300 --> 00:31:18,800 Yes. 405 00:31:18,800 --> 00:31:21,245 AUDIENCE: Do we know of any similar results 406 00:31:21,245 --> 00:31:23,423 about this construction not working for larger s? 407 00:31:23,423 --> 00:31:24,840 YUFEI ZHAO: The question is, do we 408 00:31:24,840 --> 00:31:30,810 know any similar result of about does this graph contain 409 00:31:30,810 --> 00:31:36,060 Kst for other values of s and t less 410 00:31:36,060 --> 00:31:38,070 than the claimed threshold? 411 00:31:38,070 --> 00:31:39,250 It is unclear. 412 00:31:39,250 --> 00:31:42,630 So the paper that was uploaded, it 413 00:31:42,630 --> 00:31:45,486 doesn't address the issue of s bigger than 4. 414 00:31:45,486 --> 00:31:45,986 Yeah. 415 00:31:45,986 --> 00:31:50,870 AUDIENCE: Why Fp to the power s? 416 00:31:50,870 --> 00:31:53,530 YUFEI ZHAO: So question is why Fp to the power of s? 417 00:31:53,530 --> 00:31:55,690 So let's go back to the norm graph construction. 418 00:31:55,690 --> 00:31:58,660 So where do we use Fp to the power of s? 419 00:31:58,660 --> 00:32:06,010 Well, certainly we needed it to have the right edge count. 420 00:32:06,010 --> 00:32:08,360 So that comes up in the edge count. 421 00:32:08,360 --> 00:32:12,500 And also in the norm expression, you 422 00:32:12,500 --> 00:32:14,840 have the correct number of factors. 423 00:32:20,960 --> 00:32:22,690 So I encourage you to try for if you 424 00:32:22,690 --> 00:32:24,880 use a smaller or bigger value of s, 425 00:32:24,880 --> 00:32:27,460 you either don't get something which is Kst free 426 00:32:27,460 --> 00:32:31,130 or you have the wrong number of edges. 427 00:32:31,130 --> 00:32:34,830 Any more questions? 428 00:32:34,830 --> 00:32:37,650 So later, I will show you a different construction 429 00:32:37,650 --> 00:32:41,850 of Kst free graphs, again, for t large compared to s 430 00:32:41,850 --> 00:32:44,790 that will not do as well as this one. 431 00:32:44,790 --> 00:32:47,130 But it is a genuinely different construction. 432 00:32:47,130 --> 00:32:50,340 And it uses the idea of randomized algebraic 433 00:32:50,340 --> 00:32:54,210 construction, which is something that actually was only 434 00:32:54,210 --> 00:32:55,600 developed a few years ago. 435 00:32:55,600 --> 00:32:57,947 It's a very recent development. 436 00:32:57,947 --> 00:32:58,780 And it's quite nice. 437 00:32:58,780 --> 00:33:01,260 So it combines some of the things we've talked about 438 00:33:01,260 --> 00:33:03,860 with constructing using random graphs to construct 439 00:33:03,860 --> 00:33:06,730 H free graphs on one hand, and on the other hand, 440 00:33:06,730 --> 00:33:08,250 some of the algebraic ideas. 441 00:33:08,250 --> 00:33:11,250 In particular, we're not going to use that theorem up there, 442 00:33:11,250 --> 00:33:14,777 but we'll use some other algebraic geometry fact. 443 00:33:14,777 --> 00:33:16,110 OK, so let's take a quick break. 444 00:33:19,500 --> 00:33:23,820 So what I want to discuss now is a relatively new idea 445 00:33:23,820 --> 00:33:28,740 called a randomized algebraic construction, which 446 00:33:28,740 --> 00:33:33,930 combines some ideas from both the randomized construction 447 00:33:33,930 --> 00:33:36,170 and also the algebraic construction that we just saw. 448 00:33:39,970 --> 00:33:43,620 So this idea is due to Boris Bukh just a few years ago. 449 00:33:47,660 --> 00:33:53,900 And the goal is to give an alternative construction 450 00:33:53,900 --> 00:34:09,620 of a Kst free graph with lots of edges 451 00:34:09,620 --> 00:34:19,610 provided that as before, t is much larger compared to s. 452 00:34:23,659 --> 00:34:26,050 So this band here will not be as good as the one 453 00:34:26,050 --> 00:34:27,159 that we just saw. 454 00:34:27,159 --> 00:34:29,280 And I will even tell you what it is. 455 00:34:29,280 --> 00:34:30,520 But it's some constant. 456 00:34:30,520 --> 00:34:32,650 So for every s there is some t, such 457 00:34:32,650 --> 00:34:33,870 that this construction works. 458 00:34:37,630 --> 00:34:44,239 As before, we working inside some finite field geometry. 459 00:34:44,239 --> 00:34:47,062 So let's start with a, a prime power. 460 00:34:47,062 --> 00:34:48,520 You can think of prime if you like. 461 00:34:48,520 --> 00:34:49,810 It doesn't make so much difference. 462 00:34:49,810 --> 00:34:51,550 So we're working inside a finite field. 463 00:34:54,389 --> 00:34:59,180 And let's assume s is fixed and at least 4. 464 00:35:03,513 --> 00:35:04,930 Let me write down some parameters. 465 00:35:04,930 --> 00:35:06,320 Don't worry about them for now. 466 00:35:06,320 --> 00:35:08,730 Just think of them as sufficiently large constants. 467 00:35:08,730 --> 00:35:11,470 So d is this quantity here. 468 00:35:11,470 --> 00:35:15,080 So we'll come back to it later when it comes up. 469 00:35:15,080 --> 00:35:19,000 OK, so what's the idea? 470 00:35:19,000 --> 00:35:21,430 When we looked at the randomized construction, 471 00:35:21,430 --> 00:35:22,690 we took a random graph. 472 00:35:22,690 --> 00:35:24,970 We took an Erdos Renyi graph. 473 00:35:24,970 --> 00:35:27,280 Every edge appeared independently. 474 00:35:27,280 --> 00:35:30,010 And saw that it has lots of edges, 475 00:35:30,010 --> 00:35:33,730 if you choose p property and not so many copies of h. 476 00:35:33,730 --> 00:35:36,160 So you can remove all the copies of h 477 00:35:36,160 --> 00:35:41,410 to get a graph with lots of edges that is h free. 478 00:35:41,410 --> 00:35:43,310 What we're going to do now is instead 479 00:35:43,310 --> 00:35:45,380 of taking the edges randomly, we're 480 00:35:45,380 --> 00:35:48,205 going to take a random polynomial. 481 00:35:54,030 --> 00:36:07,690 F will be a random polynomial chosen uniformly 482 00:36:07,690 --> 00:36:19,720 among all polynomials with-- 483 00:36:19,720 --> 00:36:24,510 so I wrote uppercase Y and uppercase X and Y. Actually, X 484 00:36:24,510 --> 00:36:26,340 and Y, they're not single variants. 485 00:36:26,340 --> 00:36:39,280 They are-- so each of them is a vector of s variables. 486 00:36:41,870 --> 00:36:46,550 So in other words, x1 through xs are the variables 487 00:36:46,550 --> 00:36:47,570 in the polynomial. 488 00:36:47,570 --> 00:36:51,800 And then y1 through ys. 489 00:36:51,800 --> 00:36:54,058 So it's a 2s variable polynomial. 490 00:36:58,180 --> 00:37:06,050 So among all polynomials with degree, at most d-- 491 00:37:06,050 --> 00:37:07,790 d being the number up here-- 492 00:37:07,790 --> 00:37:13,840 in each of x and y sets of variables. 493 00:37:13,840 --> 00:37:16,700 So you look at it as xs variables, 494 00:37:16,700 --> 00:37:20,750 each monominal has their exponents sum 495 00:37:20,750 --> 00:37:24,170 to our most d, and likewise with each monomial for the y 496 00:37:24,170 --> 00:37:25,790 variables. 497 00:37:25,790 --> 00:37:29,260 So this is the random object. 498 00:37:29,260 --> 00:37:32,300 It's a random polynomial in 2s variables. 499 00:37:32,300 --> 00:37:33,450 And the degree is bounded. 500 00:37:33,450 --> 00:37:35,990 So you only have a finite number of possibilities, 501 00:37:35,990 --> 00:37:38,120 and I choose one of them uniformly at random. 502 00:37:44,750 --> 00:37:47,030 And now, what's my graph? 503 00:37:47,030 --> 00:37:49,120 We're going to construct a bipartide graph 504 00:37:49,120 --> 00:37:52,810 G. The bipartideness, it's not so crucial. 505 00:37:52,810 --> 00:37:54,690 But it'll make our life somewhat easier. 506 00:37:58,870 --> 00:38:00,330 So it's a bipartide graph. 507 00:38:00,330 --> 00:38:03,820 So it hs two vertex parts, which I will label left and right, 508 00:38:03,820 --> 00:38:09,320 L and R. And they are both the s dimensional vector 509 00:38:09,320 --> 00:38:11,600 space over f cube. 510 00:38:11,600 --> 00:38:18,530 And we'll put an edge between two vertices, 511 00:38:18,530 --> 00:38:21,200 if and only if that polynomial up there 512 00:38:21,200 --> 00:38:27,960 f evaluates to 0 on these values. 513 00:38:27,960 --> 00:38:29,200 That's the graph. 514 00:38:29,200 --> 00:38:32,090 So I give you a random polynomial f. 515 00:38:32,090 --> 00:38:38,720 And then you put in edges according to when f vanishes. 516 00:38:38,720 --> 00:38:44,420 So f, if you view the bipartide graph as a subset of Fq 517 00:38:44,420 --> 00:38:49,210 to s cross F to the s, then this is the zero set. 518 00:38:49,210 --> 00:38:52,970 The edge set is the zero set. 519 00:38:52,970 --> 00:38:56,480 Just like in random graphs, with the construction 520 00:38:56,480 --> 00:38:58,950 with random graphs, we'll need to show a couple of things. 521 00:38:58,950 --> 00:39:02,330 One is that has lots of edges, which will not be hard to show. 522 00:39:02,330 --> 00:39:06,050 And second, that it will have typically a small number 523 00:39:06,050 --> 00:39:09,650 of copies of Kst. 524 00:39:09,650 --> 00:39:13,000 And that will have some ingredients which 525 00:39:13,000 --> 00:39:16,660 are similar to the random graphs case we saw before, 526 00:39:16,660 --> 00:39:22,820 but it will have some new ideas coming from algebraic geometry. 527 00:39:22,820 --> 00:39:27,360 First, let's show that this graph has lots of edges. 528 00:39:27,360 --> 00:39:34,530 And that's a simple calculation, because for every pair 529 00:39:34,530 --> 00:39:44,620 of points of vertices, I claim that the probability-- 530 00:39:44,620 --> 00:39:46,940 so here f is the random object-- 531 00:39:46,940 --> 00:39:53,790 the probability that f evaluates to 0 on this pair is exactly q. 532 00:39:53,790 --> 00:39:58,970 So exactly 1 over q, 1 over the size of the field. 533 00:39:58,970 --> 00:40:01,930 So this is not too hard. 534 00:40:01,930 --> 00:40:06,850 And the reason is that the distribution of f 535 00:40:06,850 --> 00:40:32,210 is identical to if you are an extra random constant on f, 536 00:40:32,210 --> 00:40:34,070 chosen uniformly at random. 537 00:40:34,070 --> 00:40:35,840 So I took a random polynomial. 538 00:40:35,840 --> 00:40:37,430 I shifted by a random constant. 539 00:40:37,430 --> 00:40:39,320 It's still uniformly random polynomial, 540 00:40:39,320 --> 00:40:41,180 according to that distribution. 541 00:40:41,180 --> 00:40:46,670 But now, you see that this, whatever f 542 00:40:46,670 --> 00:40:49,810 evaluated to, if I shift by a random constant, 543 00:40:49,810 --> 00:40:53,700 you will end up with a uniform distribution. 544 00:40:53,700 --> 00:41:03,640 So it tells you that that guy up there is uniform distribution 545 00:41:03,640 --> 00:41:12,600 on every fixed point u, v. So in particular, 546 00:41:12,600 --> 00:41:18,220 it hits 0 with probability exactly 1 over q. 547 00:41:18,220 --> 00:41:25,310 And as a result, the number of edges of g in the expectation 548 00:41:25,310 --> 00:41:31,620 is exactly n squared over q, where 549 00:41:31,620 --> 00:41:34,270 n is the number of vertices. 550 00:41:34,270 --> 00:41:37,410 So n is actually not the number of vertices, 551 00:41:37,410 --> 00:41:43,655 but the size of each vertex part, namely q to the s. 552 00:41:43,655 --> 00:41:48,770 So you see that it gives you the right number of edges, so n 553 00:41:48,770 --> 00:41:51,290 to the 2 minus 1 over s. 554 00:41:51,290 --> 00:41:53,280 So we have the right number of edges. 555 00:41:53,280 --> 00:41:58,280 And now, we want to show that this graph here typically does 556 00:41:58,280 --> 00:42:01,690 not have too many copies of Kst's. 557 00:42:01,690 --> 00:42:03,500 It might have some copies of Kst's. 558 00:42:03,500 --> 00:42:05,620 Somehow that is unavoidable. 559 00:42:05,620 --> 00:42:09,650 Just as in the random case, you do have some copies of Kst's. 560 00:42:09,650 --> 00:42:11,360 But if they are not too many copies, 561 00:42:11,360 --> 00:42:14,300 I can remove them and obtain a Kst free graph. 562 00:42:17,830 --> 00:42:20,070 OK, so what is the intuition? 563 00:42:20,070 --> 00:42:22,590 How does it compare to the case when 564 00:42:22,590 --> 00:42:27,260 you have a genuine Erdos-Renyi random graph? 565 00:42:27,260 --> 00:42:32,603 Well, what is the expected number of common neighbors? 566 00:42:36,050 --> 00:42:43,120 So if you fix some u with let's say 567 00:42:43,120 --> 00:42:49,600 on the left side with exactly s vertices, 568 00:42:49,600 --> 00:43:02,300 I want to understand, how many common neighbors does u have? 569 00:43:09,845 --> 00:43:11,220 But because the common neighbors, 570 00:43:11,220 --> 00:43:14,070 if he has too many common neighbors, then that's a Kst. 571 00:43:16,820 --> 00:43:20,070 It is not hard to calculate the expectation 572 00:43:20,070 --> 00:43:22,760 of this quantity, both in the random graph case 573 00:43:22,760 --> 00:43:24,780 as well as in this case. 574 00:43:24,780 --> 00:43:27,780 And you can calculate if you pretend every edge occurs 575 00:43:27,780 --> 00:43:32,570 independently, the expected number of common neighbors 576 00:43:32,570 --> 00:43:39,210 is exactly n to the q to the minus s. 577 00:43:39,210 --> 00:43:42,870 So there are s elements of u, which is exactly 1. 578 00:43:45,980 --> 00:43:50,600 And you know that for a binomial distribution with expectation 1 579 00:43:50,600 --> 00:43:53,720 and a large number of variables, the distribution 580 00:43:53,720 --> 00:43:55,574 is approximately Poissonian. 581 00:44:03,802 --> 00:44:11,640 Ah, but that's in the case when it's independently distributed, 582 00:44:11,640 --> 00:44:14,760 which is the case in the case of GMP. 583 00:44:14,760 --> 00:44:17,760 But it turns out for the algebraic setting 584 00:44:17,760 --> 00:44:22,240 we're doing here, things don't behave independently. 585 00:44:22,240 --> 00:44:26,170 It's not that you're doing coin flops for every possible edge. 586 00:44:26,170 --> 00:44:28,800 We're doing some randomized algebraic construction. 587 00:44:28,800 --> 00:44:31,110 And for algebraic geometry reasons, 588 00:44:31,110 --> 00:44:33,670 you will see that the distribution is very much not 589 00:44:33,670 --> 00:44:35,660 like Poissonian. 590 00:44:35,660 --> 00:44:40,040 It will turn out that either the number of common neighbors 591 00:44:40,040 --> 00:44:43,700 is bounded or it is very large. 592 00:44:46,810 --> 00:44:53,620 And that means that we can show using some Markov inequality 593 00:44:53,620 --> 00:44:57,130 that the probability that it is very large is quite small. 594 00:44:57,130 --> 00:45:02,300 So typically, it will not have many common neighbors? 595 00:45:02,300 --> 00:45:06,740 And that's the intuition, and so let's work out this intuition. 596 00:45:06,740 --> 00:45:07,620 Any questions so far? 597 00:45:12,220 --> 00:45:13,650 So how do we do this calculation? 598 00:45:13,650 --> 00:45:16,070 So first, let's start with something that's 599 00:45:16,070 --> 00:45:17,240 actually fairly elementary. 600 00:45:17,240 --> 00:45:25,060 So suppose you have some parameters, r and s. 601 00:45:25,060 --> 00:45:27,160 So I think of r and s-- 602 00:45:27,160 --> 00:45:29,980 so I have some parameters r and s. 603 00:45:29,980 --> 00:45:34,160 And thick of them as constants. 604 00:45:34,160 --> 00:45:36,160 Have some restrictions, but don't worry too much 605 00:45:36,160 --> 00:45:38,630 about them. 606 00:45:38,630 --> 00:45:47,600 Suppose I have two bounded subsets of the finite field, 607 00:45:47,600 --> 00:45:54,590 where you have size s and v has size r. 608 00:45:54,590 --> 00:45:58,310 Then the claim is that the probability 609 00:45:58,310 --> 00:46:09,240 that f vanishes on the Cartesian product of u and v. OK, 610 00:46:09,240 --> 00:46:12,840 so what do you expect it to be? 611 00:46:12,840 --> 00:46:17,310 So I have s r elements, and I want f, this random polynomial, 612 00:46:17,310 --> 00:46:19,240 to vanish on the entire product. 613 00:46:19,240 --> 00:46:23,370 Well, if s, its value behaved independently for every point, 614 00:46:23,370 --> 00:46:26,030 you should expect that the probability is exactly 615 00:46:26,030 --> 00:46:30,060 q to the power minus s r. 616 00:46:30,060 --> 00:46:32,710 And it turns out that is the case. 617 00:46:32,710 --> 00:46:33,480 So this is true. 618 00:46:33,480 --> 00:46:35,297 This is an exact statement. 619 00:46:38,570 --> 00:46:39,730 OK, so why is this true? 620 00:46:39,730 --> 00:46:42,100 So this is in some sense a generalization 621 00:46:42,100 --> 00:46:44,250 of this claim over here. 622 00:46:44,250 --> 00:46:46,040 And you have to do a little bit more work. 623 00:46:46,040 --> 00:46:48,520 But it's not too difficult. 624 00:46:48,520 --> 00:46:55,080 So let's first consider this lemma 625 00:46:55,080 --> 00:47:11,950 in a somewhat simpler case, where all the first coordinates 626 00:47:11,950 --> 00:47:14,080 of x are distinct-- 627 00:47:14,080 --> 00:47:16,370 of u are distinct. 628 00:47:19,440 --> 00:47:23,720 And all the first coordinates of v are distinct. 629 00:47:23,720 --> 00:47:25,660 Suppose u and v have that form. 630 00:47:25,660 --> 00:47:29,140 So I write down the list of points for u 631 00:47:29,140 --> 00:47:31,270 and first coordinates are all distinct. 632 00:47:33,990 --> 00:47:39,000 What I want to do is to give you a random shift, 633 00:47:39,000 --> 00:47:43,470 to do a uniform random shift. 634 00:47:43,470 --> 00:47:47,280 And I will shift it by a polynomial g, which 635 00:47:47,280 --> 00:47:48,960 is a bivariate polynomial. 636 00:47:48,960 --> 00:47:50,600 So these are not vectors. 637 00:47:50,600 --> 00:47:55,180 They are just single variables. 638 00:47:55,180 --> 00:48:03,770 And I look at all possible sum of monomials, 639 00:48:03,770 --> 00:48:12,560 where the degree in i is less than s, and the degree in j 640 00:48:12,560 --> 00:48:13,880 is less than r. 641 00:48:16,640 --> 00:48:24,360 And these a's are chosen uniformly independently 642 00:48:24,360 --> 00:48:32,330 at random from the ground field fq. 643 00:48:34,910 --> 00:48:39,800 And as before, we see that f and f plus g 644 00:48:39,800 --> 00:48:45,890 have the same distribution, the same probability distribution. 645 00:48:45,890 --> 00:48:49,660 And so all it remains to show that is whatever f comes out 646 00:48:49,660 --> 00:48:53,530 to be, if I tack on this extra random g, 647 00:48:53,530 --> 00:48:56,680 it creates a uniform distribution on the values 648 00:48:56,680 --> 00:49:04,270 on the entire u cross v. But, actually, see, 649 00:49:04,270 --> 00:49:09,350 I have sr choices exactly for these coefficients. 650 00:49:09,350 --> 00:49:14,970 And I have sr values that I'm trying to control. 651 00:49:14,970 --> 00:49:17,420 So really it's a counting problem. 652 00:49:17,420 --> 00:49:27,210 And it suffices to show a bijection, namely 653 00:49:27,210 --> 00:49:37,400 that for every possible vector of values, 654 00:49:37,400 --> 00:49:40,630 there exists a choice of coefficients, 655 00:49:40,630 --> 00:49:50,230 as above, such that g evaluates to the prescribed values 656 00:49:50,230 --> 00:49:52,544 with the given coefficients. 657 00:50:02,710 --> 00:50:07,390 And that uniformity will follow just because you have the exact 658 00:50:07,390 --> 00:50:10,410 if you just do a counting. 659 00:50:10,410 --> 00:50:13,663 And the one-dimensional version of this claim, 660 00:50:13,663 --> 00:50:15,080 so let's think about what that is. 661 00:50:15,080 --> 00:50:19,000 So if I have, let's say, three points on the line 662 00:50:19,000 --> 00:50:22,330 and a degree 2 polynomial, what I am saying 663 00:50:22,330 --> 00:50:26,860 is that if you give me the values you want on these three 664 00:50:26,860 --> 00:50:31,090 points, I can produce for you a unique polynomial that 665 00:50:31,090 --> 00:50:35,460 evaluates to the prescribed values on these three points. 666 00:50:35,460 --> 00:50:39,250 And that you should all know as Lagrange interpolation. 667 00:50:39,250 --> 00:50:42,720 So it tells you exactly how to do that. 668 00:50:42,720 --> 00:50:45,390 And that works for many reasons. 669 00:50:45,390 --> 00:50:51,240 One of them is that the random indeterminate is invertible. 670 00:50:51,240 --> 00:50:54,250 Here, we have multi-variables. 671 00:50:54,250 --> 00:50:58,290 So let's do Lagrange interpolation twice, once 672 00:50:58,290 --> 00:51:01,920 for each variable. 673 00:51:01,920 --> 00:51:16,030 So we'll apply Lagrange interpolation twice. 674 00:51:16,030 --> 00:51:23,360 So the first time, we'll see that for all values of u, 675 00:51:23,360 --> 00:51:29,890 there exists a single variate polynomial 676 00:51:29,890 --> 00:51:38,850 in the y variable with degree at most r minus 1 that 677 00:51:38,850 --> 00:51:46,940 evaluates to the correct values on the fixed little u. 678 00:51:54,280 --> 00:51:56,090 So do it for one variable at a time. 679 00:51:56,090 --> 00:51:59,600 For fixed u, do Lagrange interpolation 680 00:51:59,600 --> 00:52:02,200 on the y variable. 681 00:52:02,200 --> 00:52:06,490 And now, once we have those things there, 682 00:52:06,490 --> 00:52:16,580 viewing the g that we want to find as a polynomial whose 683 00:52:16,580 --> 00:52:22,840 coefficients are polynomials in the x variables, but is itself 684 00:52:22,840 --> 00:52:30,570 is a polynomial in the y variable, 685 00:52:30,570 --> 00:52:34,500 we find that again using Lagrange interpolation, 686 00:52:34,500 --> 00:52:40,310 there exists these values for these coefficients 687 00:52:40,310 --> 00:52:51,510 here, such that each coefficient of if you plug 688 00:52:51,510 --> 00:53:04,650 in the first entry into little u agrees with the coefficients 689 00:53:04,650 --> 00:53:09,460 of the g that we just found. 690 00:53:09,460 --> 00:53:11,540 And that should be the case for every little u. 691 00:53:14,910 --> 00:53:17,320 So once you find these polynomials 692 00:53:17,320 --> 00:53:20,280 and now you have a bonafide polynomial in g. 693 00:53:20,280 --> 00:53:23,610 And that's the claim above. 694 00:53:23,610 --> 00:53:26,990 So using Lagrange interpolation twice, once for each variable. 695 00:53:26,990 --> 00:53:31,370 So this is-- if you're confused, just think about it. 696 00:53:31,370 --> 00:53:34,590 There is nothing deep here. 697 00:53:34,590 --> 00:53:37,890 So that finishes the claim in the case 698 00:53:37,890 --> 00:53:42,160 when the first coordinates are all distinct. 699 00:53:42,160 --> 00:53:45,180 So we use that fact crucially in doing this Lagrange 700 00:53:45,180 --> 00:53:47,200 interpolation. 701 00:53:47,200 --> 00:53:52,350 Now, for general u and v, where we 702 00:53:52,350 --> 00:53:53,910 don't have this assumption of having 703 00:53:53,910 --> 00:53:57,090 distinct first coordinates, well, 704 00:53:57,090 --> 00:54:00,840 let's make them to have distinct first coordinates 705 00:54:00,840 --> 00:54:04,350 by considering a random linear transformation, 706 00:54:04,350 --> 00:54:07,980 so using a probabilistic method. 707 00:54:07,980 --> 00:54:20,880 So we suffice as to find invertible linear maps, p 708 00:54:20,880 --> 00:54:34,170 and s, on this vector space, such that TU and SV have 709 00:54:34,170 --> 00:54:35,712 the above properties. 710 00:54:40,910 --> 00:54:43,150 So let me show you how to do it for u. 711 00:54:43,150 --> 00:54:45,170 So I need to find you a invertible linear 712 00:54:45,170 --> 00:54:48,060 transformation t. 713 00:54:48,060 --> 00:54:50,170 Well, it's just the first coordinate that matters. 714 00:54:50,170 --> 00:54:55,910 So it suffices to find just a linear map 715 00:54:55,910 --> 00:55:03,600 corresponding to the first coordinate that 716 00:55:03,600 --> 00:55:05,886 is injective of u. 717 00:55:08,840 --> 00:55:12,020 Whatever linear map you have, even if it's zero, that's fine, 718 00:55:12,020 --> 00:55:18,845 I can extend it to the remaining coordinates. 719 00:55:18,845 --> 00:55:20,220 Actually, if it's zero, then it's 720 00:55:20,220 --> 00:55:21,585 not going to be injective of u. 721 00:55:21,585 --> 00:55:24,990 So it better not be zero. 722 00:55:24,990 --> 00:55:27,720 OK, well, let's find this map randomly. 723 00:55:27,720 --> 00:55:40,520 So pick t uniform via random among all linear maps. 724 00:55:46,310 --> 00:55:48,310 And I want to understand what is the probability 725 00:55:48,310 --> 00:55:53,440 of collision, bad event, if two elements of u 726 00:55:53,440 --> 00:55:56,460 end up getting mapped to the same point. 727 00:55:56,460 --> 00:55:58,710 Well, that's not too hard. 728 00:55:58,710 --> 00:56:07,150 So for every distinct pair of points in fq to the s 729 00:56:07,150 --> 00:56:23,100 the probability that they collide, 730 00:56:23,100 --> 00:56:24,600 think about why this is true. 731 00:56:24,600 --> 00:56:27,212 It's exactly 1 over q. 732 00:56:30,990 --> 00:56:33,750 If x and x prime, they're differing at least one 733 00:56:33,750 --> 00:56:35,920 coordinate, then even just along that coordinate, 734 00:56:35,920 --> 00:56:38,650 I can make them distinct. 735 00:56:38,650 --> 00:56:41,470 So this is the case for every pair. 736 00:56:41,470 --> 00:56:47,170 So now by union bound, the probability 737 00:56:47,170 --> 00:56:55,900 that t1 is injective on u is at least 1 minus the size 738 00:56:55,900 --> 00:57:01,370 if u choose 2 times 1 over q. 739 00:57:01,370 --> 00:57:07,150 And that's why we chose q to be large enough. 740 00:57:07,150 --> 00:57:13,540 So q is at least r squared, s So this number here is positive. 741 00:57:13,540 --> 00:57:16,730 So such a t exists. 742 00:57:16,730 --> 00:57:21,350 And so we can transform this u and v, two configurations where 743 00:57:21,350 --> 00:57:23,120 the first coordinates are all distinct, 744 00:57:23,120 --> 00:57:26,790 and then run the argument as before. 745 00:57:26,790 --> 00:57:27,892 OK, great. 746 00:57:30,660 --> 00:57:36,780 So what we've shown so far is that if you look at these Ksr 747 00:57:36,780 --> 00:57:41,620 structures, they appear with probability exactly-- 748 00:57:41,620 --> 00:57:44,580 well, with expectation exactly what 749 00:57:44,580 --> 00:57:49,320 you might expect as in independent random case. 750 00:57:49,320 --> 00:57:51,210 But what we really want to understand 751 00:57:51,210 --> 00:57:55,610 is the distribution of the number of common neighbors. 752 00:57:55,610 --> 00:58:00,050 In particular, we want to upper bound the probability 753 00:58:00,050 --> 00:58:02,360 that there are too many common neighbors. 754 00:58:02,360 --> 00:58:08,160 We want to understand some kind of tail probabilities. 755 00:58:08,160 --> 00:58:11,600 And to do that, one way to do tail probabilities 756 00:58:11,600 --> 00:58:13,155 is to consider moments. 757 00:58:13,155 --> 00:58:15,622 Yes, question? 758 00:58:15,622 --> 00:58:21,920 AUDIENCE: [INAUDIBLE] 759 00:58:21,920 --> 00:58:25,485 YUFEI ZHAO: Sorry, can you repeat the question? 760 00:58:25,485 --> 00:58:27,900 AUDIENCE: How do you have the equality right 761 00:58:27,900 --> 00:58:29,832 before the lemma? 762 00:58:29,832 --> 00:58:32,790 YUFEI ZHAO: Question, how do I have the equality right 763 00:58:32,790 --> 00:58:34,800 before the lemma? 764 00:58:34,800 --> 00:58:43,470 So there, I'm actually saying for the Erdos-Renyi random 765 00:58:43,470 --> 00:58:44,860 graph case-- 766 00:58:44,860 --> 00:58:47,820 that's the case-- in Erdos-Renyi random graph, 767 00:58:47,820 --> 00:58:52,680 each edge, if you have the same edge probability, is 1 over q. 768 00:58:52,680 --> 00:58:55,932 And then that's the number of common neighbors 769 00:58:55,932 --> 00:58:56,640 you would expect. 770 00:59:00,200 --> 00:59:05,150 So that's a heuristic for the Erdos-Renyi random graph case. 771 00:59:05,150 --> 00:59:07,220 OK, so now let's try to understand 772 00:59:07,220 --> 00:59:10,680 the distribution of the number of common neighbors. 773 00:59:10,680 --> 00:59:19,850 So let's fix a u subset of Fp to the s with exactly s elements. 774 00:59:19,850 --> 00:59:22,790 And I want to understand how many common neighbors it has. 775 00:59:22,790 --> 00:59:36,330 So let's consider the number of common neighbors of u 776 00:59:36,330 --> 00:59:40,140 and the d-th moment of this random variable. 777 00:59:40,140 --> 00:59:44,470 So this is a common way to do upper tail bounds. 778 00:59:44,470 --> 00:59:47,590 And one way to analyze such moments 779 00:59:47,590 --> 00:59:51,670 is to decompose this count as a sum of indicator 780 00:59:51,670 --> 00:59:53,000 random variables. 781 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 782 01:00:03,670 --> 01:00:08,770 is 0 for all with ou and big U. In the words, 783 01:00:08,770 --> 01:00:10,730 it's a common neighbor. 784 01:00:10,730 --> 01:00:16,737 It's 1 if v is a common neighbor for u, and 0 otherwise. 785 01:00:20,220 --> 01:00:22,410 So then the number of common neighbors 786 01:00:22,410 --> 01:00:31,200 would simply be the sum of this indicator as v ranges 787 01:00:31,200 --> 01:00:33,080 over the entire vertex set. 788 01:00:36,590 --> 01:00:39,170 And I can expand the sum. 789 01:00:39,170 --> 01:00:41,780 Then all of these are standard things 790 01:00:41,780 --> 01:00:43,925 to do when you're trying to calculate moments. 791 01:00:56,150 --> 01:01:06,240 OK, so I can bring this expectation inside 792 01:01:06,240 --> 01:01:10,170 and try to understand what is the expectation of this object 793 01:01:10,170 --> 01:01:10,670 inside. 794 01:01:14,100 --> 01:01:17,420 Well, if all the v's are distinct, 795 01:01:17,420 --> 01:01:22,045 then this is simply the expected number of Kst's. 796 01:01:30,090 --> 01:01:32,090 But the v's might not be distinct. 797 01:01:32,090 --> 01:01:34,220 So we need to be a little bit careful. 798 01:01:34,220 --> 01:01:36,290 But that's not too hard to handle. 799 01:01:36,290 --> 01:01:39,440 So let me write M sub r to be the number 800 01:01:39,440 --> 01:01:45,640 of subjective functions from d element set to an r elements 801 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. 802 01:01:53,100 --> 01:01:58,390 Then, let's consider how many distinct v values are there. 803 01:01:58,390 --> 01:02:03,820 If there are distinct v values, then they 804 01:02:03,820 --> 01:02:07,570 take on that many possible values. 805 01:02:07,570 --> 01:02:10,540 And Mr for the number of subjections, 806 01:02:10,540 --> 01:02:16,240 and for each possible r, the exact number 807 01:02:16,240 --> 01:02:18,640 for the exact value of this expectation 808 01:02:18,640 --> 01:02:20,380 is q to the minus rs. 809 01:02:20,380 --> 01:02:22,210 And that's exactly what we showed. 810 01:02:22,210 --> 01:02:25,970 So this comes from the lemma just now. 811 01:02:29,770 --> 01:02:31,710 But we chose-- I mean, look, this 812 01:02:31,710 --> 01:02:34,420 is that binomial coefficient. 813 01:02:34,420 --> 01:02:35,670 And you have this number here. 814 01:02:35,670 --> 01:02:37,530 So they multiply it to at most 1. 815 01:02:42,280 --> 01:02:45,580 So we have this number, this quantity there, 816 01:02:45,580 --> 01:02:47,090 which I think of as a constant. 817 01:02:47,090 --> 01:02:49,520 So this is a constant. 818 01:02:49,520 --> 01:02:51,590 So the d-th moment is bounded. 819 01:02:57,900 --> 01:03:02,260 And one way to get tail bounds once you have the moments 820 01:03:02,260 --> 01:03:07,286 is that we can use Markov's inequality. 821 01:03:07,286 --> 01:03:12,850 It tells us that the number of common neighbors of u, 822 01:03:12,850 --> 01:03:18,460 the probability that u has too many common neighbors, more 823 01:03:18,460 --> 01:03:22,590 than lambda common neighbors. 824 01:03:22,590 --> 01:03:29,690 I can rewrite this inequality here 825 01:03:29,690 --> 01:03:34,730 by raising both sides to the power d. 826 01:03:34,730 --> 01:03:42,350 And then, using Markov inequality, take expectation. 827 01:03:51,950 --> 01:03:54,010 So all of these are standard techniques 828 01:03:54,010 --> 01:03:56,740 for upper tail estimation. 829 01:03:56,740 --> 01:03:59,800 You want to understand the upper tails on random variable, 830 01:03:59,800 --> 01:04:03,670 understand its moments and use Markov on its moments. 831 01:04:06,710 --> 01:04:12,920 But now, we know we have some bound for the d-th moment, 832 01:04:12,920 --> 01:04:14,600 right? 833 01:04:14,600 --> 01:04:16,880 Which is M as we just showed. 834 01:04:16,880 --> 01:04:18,560 So there is this bound here. 835 01:04:21,320 --> 01:04:23,420 So far you can run the same argument 836 01:04:23,420 --> 01:04:25,100 in the random graphs case. 837 01:04:25,100 --> 01:04:27,550 And you wouldn't really do much different. 838 01:04:27,550 --> 01:04:29,300 I mean everything is more or less the same 839 01:04:29,300 --> 01:04:31,010 what I've said so far, although we 840 01:04:31,010 --> 01:04:33,140 had to do a special calculation algebraically that 841 01:04:33,140 --> 01:04:34,290 didn't really make sense-- 842 01:04:34,290 --> 01:04:37,910 I mean, that you have to show some kind of near independence. 843 01:04:37,910 --> 01:04:38,705 Question? 844 01:04:38,705 --> 01:04:43,455 AUDIENCE: Is that less than or equal to, right? 845 01:04:43,455 --> 01:04:44,393 The Markov inequality. 846 01:04:44,393 --> 01:04:45,850 YUFEI ZHAO: Ah, thank you. 847 01:04:45,850 --> 01:04:47,470 So this is less than or equal to. 848 01:04:47,470 --> 01:04:50,030 Thank you. 849 01:04:50,030 --> 01:04:52,950 But now is where the algebra comes in, 850 01:04:52,950 --> 01:04:56,180 so the algebraic geometry nature of this argument comes in. 851 01:04:56,180 --> 01:04:58,940 It turns out that this quantity here-- 852 01:05:02,990 --> 01:05:05,630 previously, we said, at least heuristically, 853 01:05:05,630 --> 01:05:08,720 in the random graphs case, it behaves like a qua Poisson 854 01:05:08,720 --> 01:05:10,140 random variable. 855 01:05:10,140 --> 01:05:13,900 So it's fairly uniform in a Poisson sense. 856 01:05:13,900 --> 01:05:15,940 It turns out because of the algebraic nature 857 01:05:15,940 --> 01:05:17,860 of the construction, this random variable 858 01:05:17,860 --> 01:05:20,020 behaves nothing like a Poisson. 859 01:05:20,020 --> 01:05:31,590 So it turns out it's highly constrained due to reasons 860 01:05:31,590 --> 01:05:32,790 using algebraic geometry. 861 01:05:32,790 --> 01:05:35,400 And I'll tell you exactly why. 862 01:05:35,400 --> 01:05:37,320 So that the number of common neighbors 863 01:05:37,320 --> 01:05:40,620 is either very small or very large. 864 01:05:40,620 --> 01:05:50,550 And here is the claim that for every s and d, 865 01:05:50,550 --> 01:05:54,660 there exists some c, such that if I 866 01:05:54,660 --> 01:06:07,040 have a bunch of polynomials on fq to the s of degree 867 01:06:07,040 --> 01:06:14,260 o at most d, then if you look at the number of zeros, 868 01:06:14,260 --> 01:06:29,860 common zeros, of the f's, how many common zeros can you have? 869 01:06:29,860 --> 01:06:32,230 It turns out it cannot just be some arbitrary number. 870 01:06:32,230 --> 01:06:42,400 So this set has size either bounded at most c. 871 01:06:42,400 --> 01:06:48,600 Or it is at least q minus something very small. 872 01:06:51,520 --> 01:06:52,990 And I'll explain just a little bit 873 01:06:52,990 --> 01:06:55,790 why this is the case, although I will not give a proof. 874 01:06:55,790 --> 01:06:58,930 So either somehow you are working 875 01:06:58,930 --> 01:07:06,890 in a zero dimensional case, or if this algebraic variety that 876 01:07:06,890 --> 01:07:10,670 comes with it has positive dimension, 877 01:07:10,670 --> 01:07:12,600 then you should have a lot more points. 878 01:07:26,360 --> 01:07:29,300 And the reason for this dichotomy 879 01:07:29,300 --> 01:07:33,740 has to do with how many points are there 880 01:07:33,740 --> 01:07:39,060 on the algebraic variety over a finite field? 881 01:07:39,060 --> 01:07:41,362 So I will not give a proof, although if you 882 01:07:41,362 --> 01:07:43,570 look on the course website for a link to a reference, 883 01:07:43,570 --> 01:07:45,180 that does have a proof. 884 01:07:45,180 --> 01:07:47,670 But I will tell you what the key algebraic geometric 885 01:07:47,670 --> 01:07:50,950 input is to that claim up here. 886 01:07:50,950 --> 01:07:54,910 And this is a important and famous theorem 887 01:07:54,910 --> 01:07:56,330 called a Lang Weil bound. 888 01:08:01,180 --> 01:08:04,970 So the Lang Weil bound tells you that if you 889 01:08:04,970 --> 01:08:13,250 have an algebraic variety, v. And for now, it is important 890 01:08:13,250 --> 01:08:19,760 in order to say this properly to work in the algebraic closure, 891 01:08:19,760 --> 01:08:27,060 so Fq bar as the algebraic closure, 892 01:08:27,060 --> 01:08:29,560 it's the smallest field extension where I can 893 01:08:29,560 --> 01:08:33,510 solve all polynomial equations. 894 01:08:33,510 --> 01:08:41,211 Then, the variety cut out by a set of polynomials-- 895 01:08:45,450 --> 01:08:46,710 so v is this variety. 896 01:08:46,710 --> 01:08:53,420 So if it is irreducible, it cannot be written as a union 897 01:08:53,420 --> 01:08:57,300 of finite number of smaller varieties, 898 01:08:57,300 --> 01:09:02,490 irreducible over Fq bar. 899 01:09:02,490 --> 01:09:09,570 And all of these polynomials have degree bounded. 900 01:09:12,240 --> 01:09:18,220 Then the question is, if I take these polynomials 901 01:09:18,220 --> 01:09:22,830 and I look at how many Fq points does it have? 902 01:09:22,830 --> 01:09:25,390 So in other words, now I leave the field. 903 01:09:25,390 --> 01:09:29,649 I come back down to Earth, to the base field and ask, 904 01:09:29,649 --> 01:09:33,220 what's the number of solutions where 905 01:09:33,220 --> 01:09:34,460 the coordinates are in Fq? 906 01:09:39,040 --> 01:09:40,790 OK, so how many points do we expect? 907 01:09:40,790 --> 01:09:43,020 Well, the simplest example of an algebraic variety 908 01:09:43,020 --> 01:09:45,189 is that of a subspace. 909 01:09:45,189 --> 01:09:48,640 If you have a d-dimensional subspace over Fq, 910 01:09:48,640 --> 01:09:51,850 you have q to the d points exactly. 911 01:09:51,850 --> 01:09:57,460 So you expect something like q raised to the dimension 912 01:09:57,460 --> 01:09:59,010 of the variety. 913 01:09:59,010 --> 01:10:01,930 Now the dimensions is actually a somewhat subtle concept. 914 01:10:01,930 --> 01:10:02,750 I won't define. 915 01:10:02,750 --> 01:10:06,070 But there are many definitions in algebraic geometry. 916 01:10:06,070 --> 01:10:08,380 It turns out it's not always exactly as nice 917 01:10:08,380 --> 01:10:10,360 as in the case of linear subspaces. 918 01:10:10,360 --> 01:10:13,630 But the Lang Weil bound tells us that is not too far off. 919 01:10:13,630 --> 01:10:19,330 You have a deviation that is at most 920 01:10:19,330 --> 01:10:22,360 on the order of q 1 over root q, where 921 01:10:22,360 --> 01:10:25,060 there are some hidden constants depending 922 01:10:25,060 --> 01:10:28,690 on the description of your variety in terms 923 01:10:28,690 --> 01:10:30,940 of the degrees of the polynomial with the dimension 924 01:10:30,940 --> 01:10:32,410 and the number of polynomials. 925 01:10:32,410 --> 01:10:36,240 But the point is that the number of points on this variety 926 01:10:36,240 --> 01:10:40,300 is basically should be around the same as the model 927 01:10:40,300 --> 01:10:41,665 case, namely that of a subspace. 928 01:10:44,540 --> 01:10:46,640 And that brings us some intuition 929 01:10:46,640 --> 01:10:48,230 to why this lemma is true. 930 01:10:48,230 --> 01:10:50,822 So you have those polynomials up there. 931 01:10:50,822 --> 01:10:52,280 So there are some subtle points one 932 01:10:52,280 --> 01:10:54,200 needs to verify about your disability, 933 01:10:54,200 --> 01:10:56,990 but the punchline is that either you 934 01:10:56,990 --> 01:10:59,467 are in the zero-dimensional case, in which case 935 01:10:59,467 --> 01:11:01,300 you have something like [INAUDIBLE] theorem, 936 01:11:01,300 --> 01:11:05,590 and that tells you that a number of solutions is bounded. 937 01:11:05,590 --> 01:11:08,085 Or you're in the positive dimensional case, in which case 938 01:11:08,085 --> 01:11:09,460 the Lang Weil term tells you, you 939 01:11:09,460 --> 01:11:11,440 must have lots of solutions. 940 01:11:11,440 --> 01:11:13,180 And there is no middle ground. 941 01:11:16,170 --> 01:11:20,636 And now, we're ready to finish off Boris Bukh's construction. 942 01:11:26,100 --> 01:11:35,580 So we see that applying this lemma up there with this, what 943 01:11:35,580 --> 01:11:37,690 should be my polynomials be? 944 01:11:37,690 --> 01:11:46,080 I'm going to use my polynomials f sub u of y to be-- 945 01:11:46,080 --> 01:11:49,300 well, I have a random polynomial up there. 946 01:11:49,300 --> 01:11:53,640 So I'm trying to find common neighbors. 947 01:11:53,640 --> 01:11:56,010 I'm trying to find common solutions. 948 01:11:59,970 --> 01:12:08,450 So these are my polynomial as u ranges over big U. 949 01:12:08,450 --> 01:12:17,840 So for q large enough, we find that the number 950 01:12:17,840 --> 01:12:24,320 of common neighbors of u, the probability that it is bigger 951 01:12:24,320 --> 01:12:30,600 than c, where c is supplied by that lemma, 952 01:12:30,600 --> 01:12:35,000 is equal to the probability that a number of common neighbors 953 01:12:35,000 --> 01:12:41,430 of u exceeds q over 2, where q over w is this quantity 954 01:12:41,430 --> 01:12:42,780 rounded-- 955 01:12:42,780 --> 01:12:45,410 smaller than that quantity up there. 956 01:12:45,410 --> 01:12:47,220 So if it has more than c solutions, 957 01:12:47,220 --> 01:12:49,170 that automatically has a lot of solutions. 958 01:12:52,230 --> 01:12:55,890 And now, we can apply the Markov inequality up there, 959 01:12:55,890 --> 01:13:00,330 the tail bound on the moments to deduce that this probability is 960 01:13:00,330 --> 01:13:05,400 at most M divided by q over 2 raised to the power d. 961 01:13:08,580 --> 01:13:10,560 And the moral of it is that we should 962 01:13:10,560 --> 01:13:12,900 have a very small number-- 963 01:13:12,900 --> 01:13:16,030 I mean, this should occur with very small probability. 964 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 965 01:13:29,140 --> 01:13:32,780 is contained entirely on the left side or the right side 966 01:13:32,780 --> 01:13:35,480 of the original bipartition. 967 01:13:35,480 --> 01:13:41,240 And most importantly, u has a lot, namely more than c, 968 01:13:41,240 --> 01:13:49,110 common neighbors in j. 969 01:13:53,086 --> 01:13:56,650 So how many bad sets do we expect? 970 01:13:56,650 --> 01:13:58,940 Basically, a very small number. 971 01:13:58,940 --> 01:14:07,320 So the number of bad sets, bad use, 972 01:14:07,320 --> 01:14:12,180 is upper bounded by-- well, for each choice of s elements, 973 01:14:12,180 --> 01:14:14,130 the probability that something is bad 974 01:14:14,130 --> 01:14:22,810 is this quantity up here, which we chose d to be a large 975 01:14:22,810 --> 01:14:24,700 enough constant depending on s. 976 01:14:24,700 --> 01:14:27,880 If you look at the choice of d up there, 977 01:14:27,880 --> 01:14:32,350 see that this quantity here is quite a bit smaller 978 01:14:32,350 --> 01:14:34,540 than the number of vertices. 979 01:14:40,730 --> 01:14:52,070 And now, the last step is the same as our randomized 980 01:14:52,070 --> 01:14:55,970 construction using Erdos-Renyi random graphs, 981 01:14:55,970 --> 01:14:59,180 where we remove-- 982 01:14:59,180 --> 01:15:00,880 well, it's almost the same, but now 983 01:15:00,880 --> 01:15:06,640 we remove one vertex from every bad set. 984 01:15:17,360 --> 01:15:22,520 And we get some graph g prime. 985 01:15:22,520 --> 01:15:26,540 And we just need to check that g prime has lots of edges. 986 01:15:26,540 --> 01:15:29,000 We know got rid of all the bad sets. 987 01:15:29,000 --> 01:15:38,180 So g prime is now K sub sc plus 1 free. 988 01:15:42,390 --> 01:15:46,110 We got rid of all possibilities for s points having 989 01:15:46,110 --> 01:15:49,040 more than c common neighbors. 990 01:15:49,040 --> 01:15:52,720 Now, we just need to check that g prime has lots of edges. 991 01:15:52,720 --> 01:15:56,870 Well, the expected number of edges in g prime 992 01:15:56,870 --> 01:16:01,180 is at least the x-- 993 01:16:01,180 --> 01:16:09,750 what we removed one vertex for every bad u. 994 01:16:09,750 --> 01:16:15,210 And each bad u carries with it at most n edges, 995 01:16:15,210 --> 01:16:19,034 because there are only n edge on each side of the bipartition. 996 01:16:26,950 --> 01:16:29,810 And, well, the number of edges of g 997 01:16:29,810 --> 01:16:33,530 has expectations exactly n squared over q. 998 01:16:36,360 --> 01:16:46,780 And the number of bad u's we saw up there is not very large. 999 01:16:46,780 --> 01:16:51,910 So in particular, this quantity, the second term 1000 01:16:51,910 --> 01:16:54,890 is dominated by the first term. 1001 01:16:54,890 --> 01:17:01,640 And so we obtain the claimed number of edges. 1002 01:17:09,300 --> 01:17:16,550 Also, the graph has at most 2n vertices. 1003 01:17:16,550 --> 01:17:18,630 So we may have gotten got rid of some. 1004 01:17:18,630 --> 01:17:21,300 But actually, fewer vertices, the better. 1005 01:17:21,300 --> 01:17:28,790 At most, 2n vertices, and it is a case of s-- 1006 01:17:28,790 --> 01:17:31,190 So it's Kst free for t large enough. 1007 01:17:34,320 --> 01:17:36,510 So this gives you another construction 1008 01:17:36,510 --> 01:17:39,180 of Kst free graphs. 1009 01:17:39,180 --> 01:17:41,180 And so today, we saw two different constructions 1010 01:17:41,180 --> 01:17:45,020 of Kst free graphs for constants s and t, but in both cases, 1011 01:17:45,020 --> 01:17:47,690 t is substantially larger than s. 1012 01:17:47,690 --> 01:17:50,270 But the most important thing is that they both match 1013 01:17:50,270 --> 01:17:54,380 the Kovari-Sos-Turan bound. 1014 01:17:54,380 --> 01:17:57,840 So it gives you some evidence that maybe 1015 01:17:57,840 --> 01:18:01,290 the Kovari-Sos-Turan conjecture, the theorem, 1016 01:18:01,290 --> 01:18:04,710 is tight up to at most a constant factor, although that 1017 01:18:04,710 --> 01:18:06,570 is a major open problem. 1018 01:18:06,570 --> 01:18:11,430 And it remains a very difficult it seems open problem, but one 1019 01:18:11,430 --> 01:18:15,360 that is of central importance in extremal graph theory 1020 01:18:15,360 --> 01:18:20,500 to try to come up perhaps with other constructions 1021 01:18:20,500 --> 01:18:22,200 that can do better. 1022 01:18:22,200 --> 01:18:25,050 Maybe they will have some algebraic input. 1023 01:18:25,050 --> 01:18:28,120 But maybe they will have some input from other ideas. 1024 01:18:28,120 --> 01:18:29,220 We do not know. 1025 01:18:29,220 --> 01:18:30,026 Question? 1026 01:18:30,026 --> 01:18:32,406 AUDIENCE: So is this q defined? 1027 01:18:32,406 --> 01:18:34,429 Because I remember q as a prime power, 1028 01:18:34,429 --> 01:18:35,750 but it doesn't say there. 1029 01:18:35,750 --> 01:18:37,510 YUFEI ZHAO: So question is, is q defined? 1030 01:18:37,510 --> 01:18:40,900 So just like in the proofs of polarity graphs and what not-- 1031 01:18:40,900 --> 01:18:43,750 so you have some n. 1032 01:18:43,750 --> 01:18:47,140 You rounded down to the nearest prime powers. 1033 01:18:47,140 --> 01:18:48,370 So s is a constant. 1034 01:18:48,370 --> 01:18:52,150 So n is basically q to the s. 1035 01:18:52,150 --> 01:18:56,200 So take large n, round it down to the nearest prime power. 1036 01:18:56,200 --> 01:18:57,712 q could be a prime, for instance. 1037 01:18:57,712 --> 01:18:58,545 It could be a prime. 1038 01:18:58,545 --> 01:19:00,140 It could be a prime power. 1039 01:19:00,140 --> 01:19:01,300 Think q to be a prime. 1040 01:19:04,050 --> 01:19:07,420 So I'm saying for every q, there is a construction. 1041 01:19:07,420 --> 01:19:12,790 And for every n, you can round down to the nearest q to the s 1042 01:19:12,790 --> 01:19:17,400 and then run this construction. 1043 01:19:17,400 --> 01:19:19,950 Any more questions? 1044 01:19:19,950 --> 01:19:20,450 Great. 1045 01:19:20,450 --> 01:19:24,230 So next time I will begin by telling you a few more things 1046 01:19:24,230 --> 01:19:27,770 about why people really like this construction 1047 01:19:27,770 --> 01:19:31,198 and some conjectures that were solved using this idea 1048 01:19:31,198 --> 01:19:32,990 and some conjectures that still remain open 1049 01:19:32,990 --> 01:19:34,650 along the same lines. 1050 01:19:34,650 --> 01:19:39,530 And we'll also go beyond Kst, so other bipartide graphs 1051 01:19:39,530 --> 01:19:41,060 and show you how to do upper bounds 1052 01:19:41,060 --> 01:19:43,390 for those bipartide graphs.