WEBVTT 00:00:17.367 --> 00:00:18.950 YUFEI ZHAO: For the past few lectures, 00:00:18.950 --> 00:00:21.310 we've been discussing the structure of set addition, 00:00:21.310 --> 00:00:25.260 and which culminated in the proof of Freiman's theorem. 00:00:25.260 --> 00:00:28.270 So this was a pretty big and central result 00:00:28.270 --> 00:00:30.640 in additive combinatorics, which gives you 00:00:30.640 --> 00:00:35.500 a complete characterization of sets with small doubling. 00:00:35.500 --> 00:00:38.650 Today, I want to look at a somewhat different issue also 00:00:38.650 --> 00:00:40.780 related to sets of small doubling, 00:00:40.780 --> 00:00:44.860 but this time we want to have a somewhat different 00:00:44.860 --> 00:00:48.760 characterization of what does it mean for a set to have 00:00:48.760 --> 00:00:51.652 lots of additive structure. 00:00:51.652 --> 00:00:53.110 So in today's lecture, we're always 00:00:53.110 --> 00:00:55.968 going to be working in an Abelian group. 00:01:00.450 --> 00:01:02.990 Let me define the following quantity. 00:01:02.990 --> 00:01:09.450 Given sets A and B, we define the additive energy 00:01:09.450 --> 00:01:16.450 between A and B to be denoted by E of A and B. 00:01:16.450 --> 00:01:17.820 So A and B are subgroups. 00:01:17.820 --> 00:01:21.840 They're subsets of this arbitrary Abelian group. 00:01:21.840 --> 00:01:26.250 So E of A and B is defined to be the number of quadruples, a1, 00:01:26.250 --> 00:01:32.820 a2, b1, b2, where a1, a2 are elements of A, and b1, 00:01:32.820 --> 00:01:41.310 b2 are elements of B, such that a1 plus b1 00:01:41.310 --> 00:01:43.680 equals to a2 plus b2. 00:01:48.650 --> 00:01:52.610 So the additive energy is the number 00:01:52.610 --> 00:01:56.510 of quadruples of these elements where you 00:01:56.510 --> 00:01:59.510 have this additive relation. 00:01:59.510 --> 00:02:02.000 And we would like to understand sets 00:02:02.000 --> 00:02:04.790 with large additive energy. 00:02:04.790 --> 00:02:07.460 So, intuitively, if you have lots of solutions 00:02:07.460 --> 00:02:09.680 to this equation in your sets, then the 00:02:09.680 --> 00:02:14.590 sets themselves should have lots of internal additive structure. 00:02:14.590 --> 00:02:17.900 So it's a different way of describing additive structure, 00:02:17.900 --> 00:02:19.400 and we'd like to understand how does 00:02:19.400 --> 00:02:21.800 this way of describing additive structure 00:02:21.800 --> 00:02:26.510 relate to things we've seen before, namely small doubling. 00:02:30.480 --> 00:02:33.840 When you have not two sets but just one set-- 00:02:33.840 --> 00:02:37.170 slightly easier to think about-- 00:02:37.170 --> 00:02:42.690 we just write E of A. I mean E of A comma A. 00:02:42.690 --> 00:02:55.150 And these objects are analogous to 4 cycles in graph theory. 00:02:55.150 --> 00:02:59.020 Because if you about this expression here in a Cayley 00:02:59.020 --> 00:03:02.650 graph, let's say over F2, then this 00:03:02.650 --> 00:03:05.110 is the description of a 4 cycle. 00:03:05.110 --> 00:03:07.240 You go around 4 steps, and you come 00:03:07.240 --> 00:03:09.460 back to where you started from. 00:03:09.460 --> 00:03:13.360 So these objects are the analogs of 4 cycles. 00:03:13.360 --> 00:03:16.720 And we already saw in our discussion of quasi-randomness, 00:03:16.720 --> 00:03:18.970 and also elsewhere, that 4 cycles 00:03:18.970 --> 00:03:22.030 play an important role in graph theory. 00:03:22.030 --> 00:03:24.160 And, likewise, these additive energies 00:03:24.160 --> 00:03:27.190 are going to play an important role in describing sets 00:03:27.190 --> 00:03:28.847 with additive structure. 00:03:33.320 --> 00:03:35.570 Consider the following quantity. 00:03:35.570 --> 00:03:42.530 We're going to let r sub A comma B of x to be the number of ways 00:03:42.530 --> 00:03:45.410 to write x as a plus b. 00:03:49.710 --> 00:03:53.330 So x equals to a plus b. 00:03:56.080 --> 00:03:59.750 So r sub A comma B of x is the number of ways 00:03:59.750 --> 00:04:04.040 I can write x as a plus b, where a comes from big A, little b 00:04:04.040 --> 00:04:09.110 comes from big B. Then, reinterpreting the formula 00:04:09.110 --> 00:04:12.590 up there, we see that the additive energy between two 00:04:12.590 --> 00:04:18.758 sets A and B is simply the sum of the squares of A-- r sub A 00:04:18.758 --> 00:04:26.630 comma B. As x ranges over all elements of the group, 00:04:26.630 --> 00:04:34.580 we only need to take x in the sumset A plus B. 00:04:34.580 --> 00:04:38.780 So the basic question, like when we discussed 00:04:38.780 --> 00:04:41.480 additive combinatorics, in the sense of when we discussed sets 00:04:41.480 --> 00:04:44.750 of small doubling, there we asked, 00:04:44.750 --> 00:04:49.910 if you have a set A of a certain size, how big can a plus a be? 00:04:49.910 --> 00:04:51.350 Here, let's ask the same. 00:04:51.350 --> 00:04:55.490 If I give you set A of a certain size, how big or how small 00:04:55.490 --> 00:05:00.070 can the additive energy of the set be? 00:05:00.070 --> 00:05:02.440 What's the most number of possible number 00:05:02.440 --> 00:05:03.520 of additive quadruples. 00:05:03.520 --> 00:05:07.760 What's the least possible number of additive quadruples? 00:05:07.760 --> 00:05:09.950 There's some trivial bounds, just 00:05:09.950 --> 00:05:12.550 like in the case of sumsets. 00:05:17.950 --> 00:05:19.290 So what are some trivial bounds? 00:05:23.660 --> 00:05:29.330 On one hand, by taking a1 equal to a2, and b2 equal to b2, 00:05:29.330 --> 00:05:33.620 we see that the energy is always at least the square 00:05:33.620 --> 00:05:37.970 of the size of A. On the other hand, 00:05:37.970 --> 00:05:40.300 if I fix three of the four elements, 00:05:40.300 --> 00:05:42.750 then the fourth element is determined. 00:05:42.750 --> 00:05:49.030 So the upper bound is cube of the size of A. 00:05:49.030 --> 00:05:51.430 And you convince yourself that, except up 00:05:51.430 --> 00:05:54.340 to maybe a constant factors, this 00:05:54.340 --> 00:05:58.300 is the best possible general upper and lower bound. 00:05:58.300 --> 00:06:01.510 Similar situation with sumsets, where you have lower bound 00:06:01.510 --> 00:06:04.550 linear, upper bound quadratic. 00:06:04.550 --> 00:06:07.948 Which is the side with additive structure? 00:06:07.948 --> 00:06:10.250 So if you have lots of additive structure, 00:06:10.250 --> 00:06:12.650 you have high energy. 00:06:12.650 --> 00:06:16.930 So this range is when you have lots of additive structure. 00:06:16.930 --> 00:06:19.360 And we would like to understand, what can you 00:06:19.360 --> 00:06:23.290 say about a set with high additive energy? 00:06:27.460 --> 00:06:32.030 Well, what are some examples of sets with high additive energy? 00:06:32.030 --> 00:06:34.450 It turns out that if you have a set that 00:06:34.450 --> 00:06:39.640 has small doubling, then, automatically, 00:06:39.640 --> 00:06:42.010 it implies large additive energy. 00:06:49.030 --> 00:06:54.740 So, in particular, intervals, or GAPs, or a large subset 00:06:54.740 --> 00:06:57.480 of GAPs, or all these examples that we saw-- in fact, 00:06:57.480 --> 00:07:00.590 these are all the examples coming from Freiman's theorem. 00:07:00.590 --> 00:07:01.790 Also, arbitrary groups. 00:07:01.790 --> 00:07:02.840 You can have subgroups. 00:07:02.840 --> 00:07:05.630 And so all of these examples have large additive energy. 00:07:08.490 --> 00:07:10.490 So let me-- I'll you the proof just in a second. 00:07:10.490 --> 00:07:11.740 It's not hard. 00:07:11.740 --> 00:07:14.330 But the real question is, what about the converse? 00:07:14.330 --> 00:07:17.500 So can you see much in the reverse direction? 00:07:17.500 --> 00:07:20.670 But, first, let me show you this claim that small doubling 00:07:20.670 --> 00:07:23.690 implies large additive energy. 00:07:23.690 --> 00:07:28.310 Well, if you have small doubling, if a plus A is size, 00:07:28.310 --> 00:07:33.940 at most, k times the size of A, then 00:07:33.940 --> 00:07:37.030 it turns out the additive energy of A 00:07:37.030 --> 00:07:42.080 is at least the maximum possible, 00:07:42.080 --> 00:07:46.370 which is A cubed divided by k. 00:07:46.370 --> 00:07:49.480 So that's within a constant factor of the maximum. 00:07:49.480 --> 00:07:50.450 It's pretty large. 00:07:50.450 --> 00:07:54.900 If you have small doubling, then large additive energy. 00:07:54.900 --> 00:07:57.580 So let's see the proof. 00:07:57.580 --> 00:07:59.440 So you can often tell how hard a proof 00:07:59.440 --> 00:08:02.320 is by how simple the statement is, although that's not always 00:08:02.320 --> 00:08:05.080 the case, as we've seen with some of our theorems, 00:08:05.080 --> 00:08:07.660 like Plunnecke's inequality. 00:08:07.660 --> 00:08:09.990 But in this case, it turns out to be fairly simple. 00:08:09.990 --> 00:08:20.620 So we see that r sub A comma A is supported on A plus A. 00:08:20.620 --> 00:08:26.250 So we use Cauchy-Schwarz to write-- 00:08:26.250 --> 00:08:33.419 so, first, we write additive energy in terms of the sum 00:08:33.419 --> 00:08:35.460 of the squares of these r's. 00:08:38.260 --> 00:08:46.670 And now, by Cauchy-Schwarz, we find that you can replace 00:08:46.670 --> 00:08:50.510 the sum of the squared r's by the sum of the r's. 00:08:50.510 --> 00:08:55.370 But now the key point here is that we take out 00:08:55.370 --> 00:08:59.090 this factor coming from Cauchy-Schwarz, which is only 00:08:59.090 --> 00:09:03.560 A plus A. So if the support size is small, we gain in this step. 00:09:06.310 --> 00:09:11.110 But what is the sum of r's? 00:09:11.110 --> 00:09:13.210 I mean, r of x is just number of ways 00:09:13.210 --> 00:09:16.320 to write x as little a1 plus little ab-- 00:09:16.320 --> 00:09:17.800 little a2. 00:09:17.800 --> 00:09:23.500 So if I sum over all x, I'm just looking at different two ways-- 00:09:23.500 --> 00:09:28.810 we're just looking at ways of picking an ordered pair from A. 00:09:28.810 --> 00:09:34.210 So this last expression is equal to the size of A 00:09:34.210 --> 00:09:39.980 to power 4 divided by A plus A. And now we 00:09:39.980 --> 00:09:43.430 use that A has small doubling to conclude 00:09:43.430 --> 00:09:47.700 that the final quantity is at least A cubed divided by k. 00:09:53.820 --> 00:09:58.512 So we see small doubling implies large additive energy. 00:09:58.512 --> 00:09:59.720 And this kind of makes sense. 00:09:59.720 --> 00:10:03.830 If your set doesn't expand, then there 00:10:03.830 --> 00:10:08.180 are many collisions of sums. 00:10:08.180 --> 00:10:11.110 And so you must have lots of solutions to that equation 00:10:11.110 --> 00:10:13.220 up there. 00:10:13.220 --> 00:10:15.230 But what about the converse? 00:10:15.230 --> 00:10:18.440 If I give you a set with large additive energy, 00:10:18.440 --> 00:10:21.170 must it necessarily have small doubling? 00:10:24.922 --> 00:10:27.015 Oh. 00:10:27.015 --> 00:10:28.140 Let me show you an example. 00:10:30.760 --> 00:10:38.320 So, well-- so a large additive energy, 00:10:38.320 --> 00:10:45.070 does it imply small doubling? 00:10:47.680 --> 00:10:50.730 So consider the following example, where 00:10:50.730 --> 00:10:53.610 you take a set A which is a combination, 00:10:53.610 --> 00:10:56.970 is a union of a set with small doubling 00:10:56.970 --> 00:11:03.874 plus a bunch of elements without additive structure. 00:11:10.170 --> 00:11:12.230 So I take a set with small doubling 00:11:12.230 --> 00:11:16.940 plus a bunch of elements without additive structure. 00:11:16.940 --> 00:11:21.190 Then it has large additive energy, just coming 00:11:21.190 --> 00:11:25.120 from this interval itself. 00:11:25.120 --> 00:11:31.990 So the energy of A is order N cubed. 00:11:31.990 --> 00:11:34.630 N is the number of elements. 00:11:34.630 --> 00:11:38.320 What about A plus A? 00:11:38.320 --> 00:11:41.650 Well, for A plus A, this part doesn't-- 00:11:41.650 --> 00:11:43.120 that's the part that contributes, 00:11:43.120 --> 00:11:48.270 or the part of this A without additive structure. 00:11:48.270 --> 00:11:52.250 And we see that the size of A plus A 00:11:52.250 --> 00:11:56.830 is quadratic in the size of A. 00:11:56.830 --> 00:12:00.470 So, unfortunately, the converse fails. 00:12:00.470 --> 00:12:05.480 So you can have sets that have large additive energy and also 00:12:05.480 --> 00:12:07.590 large doubling. 00:12:07.590 --> 00:12:11.100 But, you see, the reason why this has large additive energy 00:12:11.100 --> 00:12:14.110 is because there is a very highly structured additively 00:12:14.110 --> 00:12:15.990 structured piece of it. 00:12:15.990 --> 00:12:20.290 And, somehow, we want to forget about this extra garbage. 00:12:20.290 --> 00:12:24.790 And that's part of the reason why the converse is not true. 00:12:24.790 --> 00:12:26.820 So we would like a statement that 00:12:26.820 --> 00:12:30.150 says that if you have large additive energy, then 00:12:30.150 --> 00:12:33.750 it must come from some highly structured piece that 00:12:33.750 --> 00:12:36.380 has small doubling. 00:12:36.380 --> 00:12:38.260 And that is true, and that's the content 00:12:38.260 --> 00:12:40.540 of the Balog-Szemeredi-Gowers theorem, which 00:12:40.540 --> 00:12:43.660 is the main topic today. 00:12:43.660 --> 00:12:51.890 So the Balog-Szemeredi-Gowers theorem says that if you have 00:12:51.890 --> 00:12:52.760 a set-- 00:12:52.760 --> 00:12:56.060 so we're working always in some arbitrary Abelian group. 00:12:56.060 --> 00:13:00.650 If you have a set with large energy, 00:13:00.650 --> 00:13:07.280 then there exists some subset A prime of A such 00:13:07.280 --> 00:13:13.460 that A prime is a fairly large proportion of A. 00:13:13.460 --> 00:13:18.260 And here, by large I mean up to polynomial changes in the error 00:13:18.260 --> 00:13:18.920 parameters. 00:13:22.010 --> 00:13:28.310 So this A prime is such that A prime has small doubling. 00:13:34.340 --> 00:13:36.760 If you have large additive energy, 00:13:36.760 --> 00:13:40.960 then I can pick out a large piece with small doubling 00:13:40.960 --> 00:13:44.200 constant, and I only loose a polynomial 00:13:44.200 --> 00:13:45.775 in the error factors. 00:13:48.200 --> 00:13:50.075 So that's the Balog-Szemeredi-Gowers theorem, 00:13:50.075 --> 00:13:56.420 and it describes this example up here. 00:13:56.420 --> 00:13:58.058 Any questions about the statement? 00:14:01.200 --> 00:14:04.970 So what I will actually show you is a slight variant, actually 00:14:04.970 --> 00:14:08.782 a more general statement, where, instead of having one set, 00:14:08.782 --> 00:14:09.990 we're going to have two sets. 00:14:12.720 --> 00:14:16.880 So here's Balog-Szemeredi-Gowers theorem version 00:14:16.880 --> 00:14:25.460 2, where now we have two sets. 00:14:25.460 --> 00:14:27.050 Again, A and B are-- 00:14:27.050 --> 00:14:28.640 I'm not going to write any-- 00:14:28.640 --> 00:14:29.690 I'm not going to write it in this lecture, 00:14:29.690 --> 00:14:32.210 but A and B are always subsets of some arbitrary Abelian 00:14:32.210 --> 00:14:32.710 group. 00:14:32.710 --> 00:14:34.790 So A and B both have size of, at most, 00:14:34.790 --> 00:14:40.325 n, and the energy between A and B is large. 00:14:44.670 --> 00:14:53.580 Then there exists a subset A prime of A, B prime of B such 00:14:53.580 --> 00:14:59.820 that both A prime and B prime are 00:14:59.820 --> 00:15:05.250 large fractions of their parent set, 00:15:05.250 --> 00:15:14.070 and such that A prime plus B prime is not 00:15:14.070 --> 00:15:15.790 too much bigger than n. 00:15:21.170 --> 00:15:24.150 It's not so obvious why the second version 00:15:24.150 --> 00:15:26.020 implies the first version. 00:15:26.020 --> 00:15:29.030 So you can say, well, take A and B to be the same. 00:15:29.030 --> 00:15:31.580 But then the conclusion gives you 00:15:31.580 --> 00:15:36.200 possibly two different subsets, A prime and B prime. 00:15:36.200 --> 00:15:39.980 But the first version, I only want one subset 00:15:39.980 --> 00:15:43.320 that has small doubling. 00:15:43.320 --> 00:15:45.270 So, fortunately, the second version 00:15:45.270 --> 00:15:47.782 does imply the first version. 00:15:47.782 --> 00:15:48.490 So let's see why. 00:15:54.020 --> 00:15:58.610 The second version implies the first version because, if we-- 00:16:03.090 --> 00:16:06.350 so there's a tool that we introduced 00:16:06.350 --> 00:16:08.630 early on when we discussed Freiman's theorem, 00:16:08.630 --> 00:16:14.390 and this is the Ruzsa triangle inequality. 00:16:14.390 --> 00:16:16.410 So the spirit of Ruzsa triangle inequality 00:16:16.410 --> 00:16:19.680 is it allows you to relate, to sort of go 00:16:19.680 --> 00:16:23.010 back and forth between different sumsets in different sets. 00:16:23.010 --> 00:16:31.250 So by Ruzsa triangle inequality, if we apply the second version 00:16:31.250 --> 00:16:34.535 with A equals to B, then-- 00:16:37.750 --> 00:16:40.290 and we pick out this A prime and B prime, 00:16:40.290 --> 00:16:43.050 then we see that A prime plus A prime 00:16:43.050 --> 00:16:54.370 is, at most, A prime plus B prime squared over B prime. 00:16:54.370 --> 00:16:56.290 Well, actually, this uses the-- 00:16:56.290 --> 00:16:58.420 vice versa it uses a slightly stronger version 00:16:58.420 --> 00:17:01.910 that we had to use Plunnecke-Ruzsa key lemma 00:17:01.910 --> 00:17:02.650 to prove. 00:17:02.650 --> 00:17:04.089 But you can come up-- 00:17:04.089 --> 00:17:06.730 I mean, if you don't care about the precise loss 00:17:06.730 --> 00:17:09.040 in the polynomial factors, you can also 00:17:09.040 --> 00:17:10.780 use the basic Ruzsa triangle inequality 00:17:10.780 --> 00:17:13.270 to deduce a similar statement. 00:17:13.270 --> 00:17:14.920 This is easier to deduce. 00:17:14.920 --> 00:17:16.300 So you have that. 00:17:16.300 --> 00:17:19.990 And now, the second version tells you 00:17:19.990 --> 00:17:24.150 that the numerator is, at most, poly kn, 00:17:24.150 --> 00:17:30.340 and the denominator is, at most-- at least, 00:17:30.340 --> 00:17:33.170 n divided by poly k. 00:17:33.170 --> 00:17:40.440 Remember, over here, to get this hypothesis, we automatically 00:17:40.440 --> 00:17:45.830 have that the size of A and B are not 00:17:45.830 --> 00:17:47.580 two much smaller than n. 00:17:47.580 --> 00:17:50.700 Or else this cannot be true. 00:17:50.700 --> 00:17:58.200 So putting all these estimates together, we get that. 00:17:58.200 --> 00:18:02.705 So these two versions, they are equivalent to each other. 00:18:02.705 --> 00:18:04.080 Second version implies the first. 00:18:04.080 --> 00:18:06.240 The second one is stronger. 00:18:06.240 --> 00:18:08.292 The first one is slightly more useful. 00:18:08.292 --> 00:18:09.750 They're not necessarily equivalent, 00:18:09.750 --> 00:18:13.190 but the second one is stronger. 00:18:13.190 --> 00:18:16.370 Any questions? 00:18:16.370 --> 00:18:17.847 All right. 00:18:17.847 --> 00:18:19.680 So this is a Balog-Szemeredi-Gowers theorem. 00:18:19.680 --> 00:18:21.570 So the content of today's lecture 00:18:21.570 --> 00:18:24.930 is to show you how to prove this theorem. 00:18:24.930 --> 00:18:26.940 A remark about the naming of this theorem. 00:18:26.940 --> 00:18:29.100 So you might notice that these three letters do not 00:18:29.100 --> 00:18:31.738 coming in alphabetical order. 00:18:31.738 --> 00:18:33.780 And the reason is that this theorem was initially 00:18:33.780 --> 00:18:37.380 approved by Balog and Szemeredi, but using 00:18:37.380 --> 00:18:40.890 a more involved method that didn't 00:18:40.890 --> 00:18:43.470 give polynomial high bounds. 00:18:43.470 --> 00:18:47.190 And Gowers, in his proof of Szemeredi's theorem, 00:18:47.190 --> 00:18:49.530 his new proof of Szemeredi's theorem with good bounds, 00:18:49.530 --> 00:18:51.030 he required-- 00:18:51.030 --> 00:18:52.470 well, he looked into this theorem 00:18:52.470 --> 00:18:54.930 and gave a new proof that resulted 00:18:54.930 --> 00:18:56.990 in this polynomial type bounds. 00:18:56.990 --> 00:18:59.700 And it is that idea that we're going to see today. 00:19:09.567 --> 00:19:11.650 So this course is called graph theory and additive 00:19:11.650 --> 00:19:12.850 combinatorics. 00:19:12.850 --> 00:19:15.490 And the last two topics of this course-- 00:19:15.490 --> 00:19:17.680 today being Balog-Szemeredi-Gowers, 00:19:17.680 --> 00:19:20.740 and tomorrow we're going to see sum-product problem-- 00:19:20.740 --> 00:19:23.500 are both great examples of problems 00:19:23.500 --> 00:19:26.740 in additive combinatorics where tools from graph theory 00:19:26.740 --> 00:19:29.960 play an important role in their solutions. 00:19:29.960 --> 00:19:33.910 So it's a nice combination of the subject where we see 00:19:33.910 --> 00:19:36.350 both topics at the same time. 00:19:36.350 --> 00:19:39.190 So I want to show you the proof of Balog-Szemeredi-Gowers, 00:19:39.190 --> 00:19:41.890 and the proof goes via a graph analog. 00:19:41.890 --> 00:19:44.860 So I'm going to state for you a graphical version 00:19:44.860 --> 00:19:48.960 of the Balog-Szemeredi-Gowers theorem. 00:19:48.960 --> 00:19:50.520 And it goes like this. 00:19:50.520 --> 00:20:02.860 If G is a bipartite graph between vertex sets A and B-- 00:20:02.860 --> 00:20:06.140 and here A and B are still subsets of the Abelian group-- 00:20:17.740 --> 00:20:24.910 we define this restricted sumset, A plus sub G of B, 00:20:24.910 --> 00:20:33.820 to be the set of sums where I'm only 00:20:33.820 --> 00:20:40.256 taking sums across edges in g. 00:20:44.540 --> 00:20:47.500 So, in particular, if G is the complete bipartite graph, 00:20:47.500 --> 00:20:50.100 then this is the usual sumset. 00:20:50.100 --> 00:20:54.390 But now I may allow G to be a subset 00:20:54.390 --> 00:20:56.220 of the complete bipartite graph. 00:20:56.220 --> 00:20:59.430 So only taking some but not all of the-- 00:20:59.430 --> 00:21:01.920 only taking-- yes, some of this sums but not all of them. 00:21:04.600 --> 00:21:12.940 The graphical version of Balog-Szemeredi-Gowers 00:21:12.940 --> 00:21:23.290 says that if you have A and B be subsets of an Abelian group, 00:21:23.290 --> 00:21:27.610 both having size, at most, n, and G 00:21:27.610 --> 00:21:35.770 is a bipartite graph between A and B, 00:21:35.770 --> 00:21:42.460 such that G has lots of edges, has at least n squared 00:21:42.460 --> 00:21:43.570 over k edges. 00:21:47.290 --> 00:21:56.090 If the restricted sumset between A and B is small-- 00:21:56.090 --> 00:22:02.470 So here we're not looking at all the sums but a large fraction 00:22:02.470 --> 00:22:04.540 of the possible pairwise sums. 00:22:04.540 --> 00:22:07.040 If that sumset has small size, this 00:22:07.040 --> 00:22:10.210 is kind of like a restricted doubling constant. 00:22:10.210 --> 00:22:16.090 Then there exists A prime, subset 00:22:16.090 --> 00:22:26.530 of A, B prime, subset of B, with A prime and B prime 00:22:26.530 --> 00:22:32.830 both fairly large fractions of their parent set, 00:22:32.830 --> 00:22:36.970 and such that the unrestricted sumset between A prime and B 00:22:36.970 --> 00:22:40.270 prime is not too large. 00:22:48.020 --> 00:22:50.390 So let me say it again. 00:22:50.390 --> 00:22:52.760 So we have a fairly dense-- 00:22:52.760 --> 00:22:55.180 so a constant fraction edge density, 00:22:55.180 --> 00:22:59.480 a fairly dense bipartite graph between A and B. A and B 00:22:59.480 --> 00:23:02.660 are subsets of the Abelian group. 00:23:02.660 --> 00:23:08.700 Then-- and such that the restricted sumset is small. 00:23:08.700 --> 00:23:14.840 Then I can restrict A and B to subsets, fairly large subsets, 00:23:14.840 --> 00:23:19.070 so that the complete sumset between the subsets A prime 00:23:19.070 --> 00:23:21.170 and B prime is small. 00:23:26.180 --> 00:23:29.720 Let me show you why the graphical version of BSG 00:23:29.720 --> 00:23:33.270 implies the version of BSG I stated up there. 00:23:50.630 --> 00:23:54.511 But, so why do we care about this graphical version? 00:23:54.511 --> 00:23:59.530 Well, suppose we-- so we have all of these hypotheses. 00:23:59.530 --> 00:24:08.030 Let's write-- so we have all of those hypotheses up there. 00:24:08.030 --> 00:24:11.938 So let's write r to be r sub A comma 00:24:11.938 --> 00:24:16.930 B, so I don't have to carry the subscripts all around. 00:24:16.930 --> 00:24:17.920 What do you think-- 00:24:17.920 --> 00:24:20.760 so I start with A and B up there, 00:24:20.760 --> 00:24:24.660 and I need to construct that graph G. 00:24:24.660 --> 00:24:26.340 So what should we choose as our graph? 00:24:30.940 --> 00:24:34.460 Let's consider the popular sums. 00:24:40.370 --> 00:24:44.900 So the popular sums are going to be elements 00:24:44.900 --> 00:24:50.390 in the complete sumset such that it 00:24:50.390 --> 00:24:54.530 is represented as a sum in many different ways. 00:25:02.760 --> 00:25:07.840 And we're going to take edges that correspond 00:25:07.840 --> 00:25:12.760 to these popular sums. 00:25:12.760 --> 00:25:26.670 So let's consider bipartite graph G such 00:25:26.670 --> 00:25:39.770 that A comma B is an edge if and only A plus B is a popular sum. 00:25:46.900 --> 00:25:50.390 So let's verify some of the hypotheses. 00:25:50.390 --> 00:25:53.110 So we're going to assume graph BSG, 00:25:53.110 --> 00:25:57.340 and let's verify the hypothesis in graph BSG. 00:25:57.340 --> 00:25:59.500 On one hand, because each element of S 00:25:59.500 --> 00:26:05.170 is a popular sum, if we consider its multiplicity, 00:26:05.170 --> 00:26:13.750 we find that the size of S multiplied by n over 2k, lower 00:26:13.750 --> 00:26:19.245 bound be size of A times the size of B. 00:26:19.245 --> 00:26:26.750 So if you think about all the different pairs in A and B, 00:26:26.750 --> 00:26:31.780 each sum here, each popular sum, contributes this many times 00:26:31.780 --> 00:26:36.330 to this A cross B. 00:26:36.330 --> 00:26:41.370 So, as a result, because size of A and size of B 00:26:41.370 --> 00:26:44.880 are both, at most, n, we find that the size of S 00:26:44.880 --> 00:26:46.180 is, at most, 2kn. 00:26:49.382 --> 00:26:51.780 And if you think about what G is, 00:26:51.780 --> 00:27:00.840 then this implies also that the restricted sumset of A and B 00:27:00.840 --> 00:27:02.310 across this graph G-- 00:27:02.310 --> 00:27:04.080 which only requires the popular sums. 00:27:04.080 --> 00:27:10.718 So the restricted sumset is precisely the popular sums. 00:27:10.718 --> 00:27:13.660 So restricted sumset is not too large. 00:27:18.930 --> 00:27:19.900 OK, good. 00:27:19.900 --> 00:27:24.020 So we got one of the conditions, that the restricted sumset 00:27:24.020 --> 00:27:25.910 is not too large. 00:27:25.910 --> 00:27:30.150 And now we want to show that this graph has lots of edges. 00:27:30.150 --> 00:27:31.360 It has lots of edges. 00:27:36.210 --> 00:27:39.120 And here's where we would need to use the hypothesis that, 00:27:39.120 --> 00:27:44.166 between A and B, originally there is large additive energy. 00:27:44.166 --> 00:27:49.980 And the point here is that these unpopular sums cannot 00:27:49.980 --> 00:27:55.140 contribute very much to the additive energy in total, 00:27:55.140 --> 00:27:58.240 because each one of them is unpopular. 00:27:58.240 --> 00:28:01.960 So the dominant contributions to the additive energy 00:28:01.960 --> 00:28:05.280 are going to come from the popular sums, 00:28:05.280 --> 00:28:08.910 and we're going to use that to show that G has lots of edges. 00:28:12.660 --> 00:28:16.980 So let's lower bound the number of edges of G by first showing 00:28:16.980 --> 00:28:18.820 that-- 00:28:18.820 --> 00:28:35.030 so we'll show that the unpopular sums contribute very little 00:28:35.030 --> 00:28:42.210 to the additive energy between A and B. Indeed, 00:28:42.210 --> 00:28:49.860 the sums of the squares of the r's, if for x 00:28:49.860 --> 00:28:58.130 not in popular sums, is upper bounded by-- 00:28:58.130 --> 00:29:01.170 well, claim that it is upper bounded 00:29:01.170 --> 00:29:09.910 by the following quantity, that n over 2k times n squared. 00:29:14.520 --> 00:29:19.470 Because I can take out one factor r, 00:29:19.470 --> 00:29:24.180 upper bound by this number, just by definition, 00:29:24.180 --> 00:29:27.820 and the sums of the r's is n squared. 00:29:32.540 --> 00:29:39.150 So you have this additive energy between A and B. 00:29:39.150 --> 00:29:41.190 I know that it is large by hypothesis. 00:29:45.940 --> 00:29:48.310 Whereas, I also know that I can write it 00:29:48.310 --> 00:29:52.570 as a sum of the squares of the r's, which 00:29:52.570 --> 00:30:00.550 I can break into the popular contributions 00:30:00.550 --> 00:30:02.530 and the unpopular contributions. 00:30:05.533 --> 00:30:06.950 And, hopefully, this should all be 00:30:06.950 --> 00:30:09.470 somewhat reminiscent of basically all these proofs 00:30:09.470 --> 00:30:11.200 that we did so far in this course, 00:30:11.200 --> 00:30:14.750 where we separate a sum into the dominant terms 00:30:14.750 --> 00:30:16.510 and the minor terms. 00:30:16.510 --> 00:30:20.010 This came up in Fourier analysis in particular. 00:30:20.010 --> 00:30:24.320 So we do this splitting, and we upper 00:30:24.320 --> 00:30:28.820 bound the unpopular contributions by the estimate 00:30:28.820 --> 00:30:29.890 from just now. 00:30:36.810 --> 00:30:40.800 So, as a result, bringing this small error term, 00:30:40.800 --> 00:30:44.610 it doesn't cancel much of the energy. 00:30:44.610 --> 00:30:52.350 So we still have a lower bound on the sum of the squares 00:30:52.350 --> 00:30:56.590 of the r's in the popular sums. 00:31:00.010 --> 00:31:04.240 But I can also give a fairly trivial upper bound to a single 00:31:04.240 --> 00:31:08.050 r, namely it cannot be bigger than n. 00:31:16.220 --> 00:31:23.860 And so the number of edges of G-- 00:31:23.860 --> 00:31:27.480 so what's the number of edges of G? 00:31:27.480 --> 00:31:28.260 Look at that. 00:31:28.260 --> 00:31:33.470 Each x here contributes rx many edges. 00:31:33.470 --> 00:31:36.750 So the number of edges of G is simply the sums of these rx's. 00:31:41.310 --> 00:31:42.690 Which is quite large. 00:31:49.740 --> 00:31:56.070 So the hypothesis of graph BSG are satisfied. 00:31:56.070 --> 00:31:59.850 And so we can use the conclusion of graph BSG, which 00:31:59.850 --> 00:32:02.730 is the conclusion that we're looking for in BSG. 00:32:11.520 --> 00:32:12.532 Any questions? 00:32:17.095 --> 00:32:17.595 Good. 00:32:17.595 --> 00:32:19.860 So the remaining task is to prove 00:32:19.860 --> 00:32:23.160 the graphical version of BSG. 00:32:23.160 --> 00:32:26.040 So let's take a quick break, and when 00:32:26.040 --> 00:32:30.030 we come back we'll focus on this theorem, 00:32:30.030 --> 00:32:35.140 and it has some nice graph theoretic arguments. 00:32:35.140 --> 00:32:37.430 OK, let's continue. 00:32:37.430 --> 00:32:42.230 We've reduced the proof of the Balog-Szemeredi-Gowers theorem 00:32:42.230 --> 00:32:44.540 to the following graphical result. 00:32:44.540 --> 00:32:46.170 Well, it's not just graphical, right? 00:32:46.170 --> 00:32:49.370 Still-- we're still inside some an Abelian group, 00:32:49.370 --> 00:32:52.570 still looking at some set in some Abelian group, 00:32:52.570 --> 00:32:57.140 but, certainly, now it has a graph attached to it. 00:32:57.140 --> 00:33:01.410 Let me show this theorem through several steps. 00:33:01.410 --> 00:33:04.700 First, something called a path of length 2 lemma. 00:33:15.502 --> 00:33:17.860 So the path of length 2 lemma, the statement 00:33:17.860 --> 00:33:21.340 is that you start with a graph G which 00:33:21.340 --> 00:33:27.130 is a bipartite graph between vertex sets A and B. 00:33:27.130 --> 00:33:29.050 And now A and B no longer need-- 00:33:29.050 --> 00:33:30.100 they're just sets. 00:33:30.100 --> 00:33:31.150 They're just vertex sets. 00:33:31.150 --> 00:33:34.550 We're not going to have sums. 00:33:34.550 --> 00:33:38.175 And the number of edges is at least a constant fraction 00:33:38.175 --> 00:33:39.175 of the maximum possible. 00:33:45.570 --> 00:33:48.620 Then the conclusion is that there 00:33:48.620 --> 00:33:55.385 exists some U, a subset of A, such that U is fairly large. 00:33:59.650 --> 00:34:10.199 And between most pairs of elements of U-- 00:34:10.199 --> 00:34:24.880 so between 1 minus epsilon fraction of pairs of U-- 00:34:24.880 --> 00:34:30.840 there are lots of common neighbors. 00:34:30.840 --> 00:34:36.650 So at least epsilon delta squared 00:34:36.650 --> 00:34:46.230 B over 2 common neighbors. 00:34:46.230 --> 00:34:58.550 So you start with this bipartite graph A and B. Lots of edges. 00:34:58.550 --> 00:35:01.520 And we would like to show that there 00:35:01.520 --> 00:35:08.840 exists a pretty large subset U such that between most pairs-- 00:35:08.840 --> 00:35:11.150 all but an epsilon fraction-- 00:35:11.150 --> 00:35:12.980 of ordered pairs-- they could be the same, 00:35:12.980 --> 00:35:15.350 but it doesn't really matter-- 00:35:15.350 --> 00:35:22.600 the number of paths of length 2 between these two vertices 00:35:22.600 --> 00:35:24.920 is quite large. 00:35:24.920 --> 00:35:28.430 So they have lots of common neighbors. 00:35:28.430 --> 00:35:30.440 Where have we seen something like this before? 00:35:30.440 --> 00:35:30.890 There's a question? 00:35:30.890 --> 00:35:32.694 AUDIENCE: Is there a [INAUDIBLE] epsilon? 00:35:32.694 --> 00:35:33.600 YUFEI ZHAO: Ah, yes. 00:35:33.600 --> 00:35:37.156 So for every epsilon and every delta. 00:35:37.156 --> 00:35:43.634 So let epsilon, delta be parameters. 00:35:48.080 --> 00:35:51.860 Where have we seen something like this before? 00:35:51.860 --> 00:35:54.620 So in a bipartite graph with lots of edges, 00:35:54.620 --> 00:35:59.180 I want to find a large subset of one of the parts 00:35:59.180 --> 00:36:02.270 so that every pair of elements, or almost 00:36:02.270 --> 00:36:05.842 every pair of elements, have lots of common neighbors. 00:36:11.254 --> 00:36:12.238 Yes. 00:36:12.238 --> 00:36:13.570 AUDIENCE: [INAUDIBLE]. 00:36:13.570 --> 00:36:15.070 YUFEI ZHAO: Dependent random choice. 00:36:15.070 --> 00:36:17.150 So in the very first chapter of this course, 00:36:17.150 --> 00:36:19.340 when we did extremal graph theory 00:36:19.340 --> 00:36:21.530 forbidding bipartite subgraphs, there 00:36:21.530 --> 00:36:26.900 was a technique for proving the extremal number, upper bounds, 00:36:26.900 --> 00:36:29.540 for bipartite graphs of bounded degree. 00:36:29.540 --> 00:36:32.810 And there we used something called dependent random choice 00:36:32.810 --> 00:36:35.860 that had a conclusion that was very similar flavor. 00:36:35.860 --> 00:36:39.848 So there, we had every pair-- so a fairly large, but not as 00:36:39.848 --> 00:36:41.390 large as this-- a fairly large subset 00:36:41.390 --> 00:36:45.640 where every pair of elements had lots of common neighbors. 00:36:45.640 --> 00:36:48.230 For every couple, every k couple of vertices, 00:36:48.230 --> 00:36:50.330 have lots of common neighbors. 00:36:50.330 --> 00:36:51.470 So it's very similar. 00:36:51.470 --> 00:36:53.960 In fact, it's the same type of technique 00:36:53.960 --> 00:36:56.480 that we'll use to prove this lemma over here. 00:37:00.390 --> 00:37:05.030 So who remembers how dependent random choice goes? 00:37:05.030 --> 00:37:09.120 So the idea is that we are going to choose U 00:37:09.120 --> 00:37:11.200 not uniformly at random. 00:37:11.200 --> 00:37:12.872 So that's not going to work. 00:37:12.872 --> 00:37:15.860 Going to choose it in a dependent random way. 00:37:15.860 --> 00:37:19.630 So I want elements of U to have lots of common neighbors, 00:37:19.630 --> 00:37:20.720 typically. 00:37:20.720 --> 00:37:24.950 So one way to guarantee this is to choose U to be 00:37:24.950 --> 00:37:28.550 a neighborhood from the right. 00:37:28.550 --> 00:37:33.640 So pick a random element in B and choose 00:37:33.640 --> 00:37:37.300 U to be its neighborhood. 00:37:37.300 --> 00:37:39.230 So let's do that. 00:37:39.230 --> 00:37:41.210 So we're going to use dependent random choice. 00:37:47.505 --> 00:37:49.380 See, everything in the course comes together. 00:37:56.000 --> 00:38:04.480 So let's pick v an element of B uniformly at random. 00:38:09.580 --> 00:38:15.440 And let U be the neighborhood v. So, first of all, 00:38:15.440 --> 00:38:19.100 by linearity of expectations, the size of U 00:38:19.100 --> 00:38:27.600 is at least delta of A. So because the average degree 00:38:27.600 --> 00:38:32.472 from the right from B is at least delta A just based 00:38:32.472 --> 00:38:33.430 on the number of edges. 00:38:36.550 --> 00:38:43.560 If you have two vertices a and a prime in A 00:38:43.560 --> 00:39:04.220 with a small number of common neighbors, then the size of-- 00:39:04.220 --> 00:39:07.020 so sorry. 00:39:07.020 --> 00:39:10.520 Let me-- I skipped ahead a bit. 00:39:10.520 --> 00:39:15.170 So if a and a prime have a small number of common neighbors, 00:39:15.170 --> 00:39:22.730 then the probability that a and a prime both lie in U 00:39:22.730 --> 00:39:25.580 should be quite small. 00:39:25.580 --> 00:39:28.670 Because if they both had-- 00:39:28.670 --> 00:39:33.040 if a and a prime have a small number of common neighbors, 00:39:33.040 --> 00:39:36.863 in order for a and a prime to be included in this U, 00:39:36.863 --> 00:39:37.780 you must have chosen-- 00:39:41.550 --> 00:39:44.920 so suppose this were their common neighbor. 00:39:44.920 --> 00:39:49.550 Then in order that a and a prime be contained in U, 00:39:49.550 --> 00:39:54.470 it must have chosen this v to be inside the common neighborhood 00:39:54.470 --> 00:39:55.300 of a and a prime. 00:39:57.840 --> 00:39:59.940 Which is unlikely if a and a prime 00:39:59.940 --> 00:40:02.940 had a small number of common neighbors. 00:40:02.940 --> 00:40:08.600 So this probability is, at most, epsilon delta squared over 2. 00:40:12.460 --> 00:40:15.520 Just think about how U is constructed. 00:40:15.520 --> 00:40:26.740 So if we let x be the number of a and a primes in U cross U 00:40:26.740 --> 00:40:34.800 with, at most, epsilon delta squared over 2 times B 00:40:34.800 --> 00:40:42.900 common neighbors, then, by linearity of expectations, 00:40:42.900 --> 00:40:46.300 the expectation of x is-- 00:40:46.300 --> 00:40:54.420 well, by summing up all of these probabilities of a and a prime, 00:40:54.420 --> 00:40:56.890 both being in U-- 00:40:56.890 --> 00:41:02.010 so this is, at most, epsilon delta squared 00:41:02.010 --> 00:41:04.740 over 2 times size of A squared. 00:41:08.250 --> 00:41:12.030 So, typically, at least in expectation, 00:41:12.030 --> 00:41:16.260 you do not expect very many pairs of elements in U 00:41:16.260 --> 00:41:20.510 with few common neighbors. 00:41:20.510 --> 00:41:22.940 But we can also turn such an estimate 00:41:22.940 --> 00:41:24.830 into a specific instance. 00:41:28.015 --> 00:41:33.840 And the way to do this is to consider the quantity size of U 00:41:33.840 --> 00:41:39.280 squared minus x over epsilon. 00:41:39.280 --> 00:41:43.070 Well, first of all, we can lower bound this quantity, 00:41:43.070 --> 00:41:47.630 because the size of second moment of U 00:41:47.630 --> 00:41:53.030 is at least the first moment of U squared. 00:41:53.030 --> 00:42:02.450 And we also know that the size of x in expectation 00:42:02.450 --> 00:42:04.830 is not very large. 00:42:04.830 --> 00:42:07.700 So the whole expression can be lower bounded 00:42:07.700 --> 00:42:15.910 by delta squared over 2 times the size of A squared. 00:42:25.630 --> 00:42:26.935 So this is epsilon, sorry. 00:42:30.120 --> 00:42:34.880 Therefore, there is some concrete instance 00:42:34.880 --> 00:42:39.110 of this randomness resulting in some specific U such 00:42:39.110 --> 00:42:41.180 that this inequality holds. 00:42:41.180 --> 00:42:54.330 So there exists some U such that this inequality holds. 00:42:54.330 --> 00:43:03.310 And, in particular, we find that the size of U is at least-- 00:43:03.310 --> 00:43:05.110 just forget about this minus term-- 00:43:05.110 --> 00:43:08.950 is at least that right-hand side, square root. 00:43:08.950 --> 00:43:11.500 So, in particular, the size of U is at least 00:43:11.500 --> 00:43:14.415 delta over 2 times the size of A. 00:43:14.415 --> 00:43:18.307 And, just looking at the left-hand side, which 00:43:18.307 --> 00:43:20.890 must be a non-negative quantity because the right-hand side is 00:43:20.890 --> 00:43:26.800 non-negative, we find that x is, at most, an epsilon 00:43:26.800 --> 00:43:29.786 fraction of U squared. 00:43:34.480 --> 00:43:38.730 So putting these together, we arrive 00:43:38.730 --> 00:43:42.280 at the path of length 2 lemma. 00:43:42.280 --> 00:43:43.830 So let me go through it again. 00:43:43.830 --> 00:43:46.100 So this is the dependent random choice method, 00:43:46.100 --> 00:43:50.480 where we're going to-- we want to find this U, 00:43:50.480 --> 00:43:52.430 where most pairs of vertices in U 00:43:52.430 --> 00:43:55.920 have lots of common neighbors. 00:43:55.920 --> 00:43:58.640 So we start from the right side. 00:43:58.640 --> 00:44:02.480 We start from B, pick a uniform random vertex, which 00:44:02.480 --> 00:44:08.598 you call v, and let U be the neighborhood of v. 00:44:08.598 --> 00:44:11.140 And I claim that this U, typically, should 00:44:11.140 --> 00:44:13.170 have the desired property. 00:44:13.170 --> 00:44:18.160 And the reason is that, if you have a pair of vertices 00:44:18.160 --> 00:44:24.030 on the left that do not have many common neighbors, 00:44:24.030 --> 00:44:27.360 then I claim it is highly unlikely that these two 00:44:27.360 --> 00:44:33.360 vertices both appear in U. Because for them to both appear 00:44:33.360 --> 00:44:38.310 in U, your v have been selected inside the common neighborhood 00:44:38.310 --> 00:44:42.390 of a and a prime, which is unlikely if a and a prime 00:44:42.390 --> 00:44:46.550 have few common neighbors. 00:44:46.550 --> 00:44:50.050 So, as a result, the expected number 00:44:50.050 --> 00:44:58.338 of pairs in U with small number of common neighbors is small. 00:44:58.338 --> 00:45:00.130 And, already, that's a very good indication 00:45:00.130 --> 00:45:01.020 that we're on the right track. 00:45:01.020 --> 00:45:02.670 And, to finish things off, we look 00:45:02.670 --> 00:45:07.830 at this expression, which we can lower bound by convexity. 00:45:07.830 --> 00:45:10.890 And we know the size of U in expectation is large. 00:45:10.890 --> 00:45:13.470 And, also, the size of x, that we just saw, 00:45:13.470 --> 00:45:17.260 is small in expectation. 00:45:17.260 --> 00:45:19.890 So you have this inequality over here. 00:45:19.890 --> 00:45:21.690 And because there's an expectation, 00:45:21.690 --> 00:45:25.560 it implies that there's some specific instance such that, 00:45:25.560 --> 00:45:28.800 without the expectation, the inequality holds. 00:45:28.800 --> 00:45:30.570 So take that specific instance. 00:45:30.570 --> 00:45:34.740 We obtain some U such that this inequality is true, 00:45:34.740 --> 00:45:37.800 which simultaneously implies that U is large 00:45:37.800 --> 00:45:40.910 and x, the number of bad pairs, is small. 00:45:43.796 --> 00:45:47.850 So that was dependent random choice. 00:45:47.850 --> 00:45:48.953 Any questions? 00:45:51.731 --> 00:45:54.046 All right. 00:45:54.046 --> 00:45:56.250 So that was the path of length 2 lemma. 00:45:56.250 --> 00:45:57.900 So it tells us I can take a large set 00:45:57.900 --> 00:46:02.590 with lots of paths of length 2 between most pairs of vertices. 00:46:02.590 --> 00:46:07.888 Let's upgrade this lemma to a path of length 3 lemma. 00:46:18.850 --> 00:46:20.640 So, in the path of length 3 lemma, 00:46:20.640 --> 00:46:27.378 we start with a bipartite graph, as before, between A and B. 00:46:27.378 --> 00:46:33.970 So G is a bipartite between A and B. 00:46:33.970 --> 00:46:39.230 And, as before, we have a lot of edges between A and B. 00:46:39.230 --> 00:46:42.690 It's the delta fraction of all possible edges. 00:46:42.690 --> 00:46:50.840 Then the conclusion is that there exists A prime in A and B 00:46:50.840 --> 00:46:57.270 prime subset of B such that A prime and B 00:46:57.270 --> 00:47:01.560 prime are both large fractions of their parent set. 00:47:08.070 --> 00:47:15.820 And now, the-- and, furthermore, every pair 00:47:15.820 --> 00:47:24.390 between A prime and B prime is joined 00:47:24.390 --> 00:47:28.895 by many paths of length 3. 00:47:36.820 --> 00:47:39.300 So a path of length 3 means there's 3 edges. 00:47:42.610 --> 00:47:50.130 And, here, this eta is basically the original error term 00:47:50.130 --> 00:47:51.690 up to a polynomial change. 00:48:00.270 --> 00:48:05.020 So starting with this bipartite graph that's fairly dense, 00:48:05.020 --> 00:48:08.500 the lemma tells us that we can find 00:48:08.500 --> 00:48:13.870 some large A prime and large B prime so 00:48:13.870 --> 00:48:17.440 that between every vertex in A prime and every vertex in B 00:48:17.440 --> 00:48:21.960 prime, there are lots of paths of length 3 between them. 00:48:28.530 --> 00:48:29.263 Every time. 00:48:33.500 --> 00:48:37.215 So we should think about all of these constants as-- 00:48:37.215 --> 00:48:39.830 plus you only make polynomial changes in the constants, 00:48:39.830 --> 00:48:42.290 we're happy. 00:48:42.290 --> 00:48:46.560 Here, eta is a polynomial change in the delta. 00:48:46.560 --> 00:48:49.130 There's a convention which I like which is not universal, 00:48:49.130 --> 00:48:51.860 but it's often solved, unlike this convention. 00:48:51.860 --> 00:48:53.930 It's the difference between the little c 00:48:53.930 --> 00:48:56.630 and the big C is that a little c is better 00:48:56.630 --> 00:48:59.440 if you make it smaller, and a big C is better-- 00:48:59.440 --> 00:49:02.420 I mean, it's better in the sense that if this 00:49:02.420 --> 00:49:04.970 is true for little c and big C, and you 00:49:04.970 --> 00:49:10.050 make little c smaller and big C bigger, then it is still true. 00:49:10.050 --> 00:49:12.570 So big C is a sufficiently large constant, 00:49:12.570 --> 00:49:15.436 and little c is a sufficiently small constant. 00:49:15.436 --> 00:49:16.422 Just a-- 00:49:30.740 --> 00:49:36.650 So let's see the path of length 3 lemma, see it's proof. 00:49:36.650 --> 00:49:39.460 We're going to use the path of length 2 lemma, 00:49:39.460 --> 00:49:42.070 but we need a bit of preparation first. 00:49:42.070 --> 00:49:46.930 So the proof has some nice ideas, but it's also-- 00:49:46.930 --> 00:49:50.740 some parts of it are slightly tedious, so bear with me. 00:49:50.740 --> 00:49:58.690 So we're going to construct a chain of subsets A-- 00:49:58.690 --> 00:50:02.717 inside A. So A1, A2, A3. 00:50:02.717 --> 00:50:04.800 And this is just because there's a few cleaning up 00:50:04.800 --> 00:50:08.100 steps that need to be done. 00:50:08.100 --> 00:50:19.880 Let's call two vertices in A friendly 00:50:19.880 --> 00:50:23.980 if they have lots of common neighbors. 00:50:23.980 --> 00:50:25.430 And, precisely, we're going to say 00:50:25.430 --> 00:50:28.750 they're friendly if they have more than delta 00:50:28.750 --> 00:50:34.770 squared over 80 times the size of B common neighbors. 00:50:41.770 --> 00:50:46.590 Let me construct this sequence of subsets as follows. 00:50:46.590 --> 00:50:53.870 First, let A1 be all the vertices in A 00:50:53.870 --> 00:50:58.200 with degree not too small. 00:50:58.200 --> 00:51:02.500 So this is in preparation. 00:51:02.500 --> 00:51:05.950 So it will make our life quite a bit easier later on. 00:51:05.950 --> 00:51:09.100 Let's just trim all the really small degree vertices 00:51:09.100 --> 00:51:11.850 so that we don't have to think about them. 00:51:11.850 --> 00:51:15.870 So you trim all the small degree vertices. 00:51:15.870 --> 00:51:20.420 And think about how many edges you trim. 00:51:20.420 --> 00:51:25.700 You cannot trim so many edges, because each time you trim such 00:51:25.700 --> 00:51:30.100 a vertex, you only get rid of a small number of edges. 00:51:30.100 --> 00:51:34.300 So, in the end, at least half of the original set of edges 00:51:34.300 --> 00:51:36.690 must remain. 00:51:36.690 --> 00:51:43.320 And, as a result, the size of A1 is at least 00:51:43.320 --> 00:51:50.590 a delta over 2 fraction of the original vertex set. 00:51:50.590 --> 00:51:53.050 Otherwise, you could not have contained half 00:51:53.050 --> 00:51:57.460 of the original set of edges. 00:51:57.460 --> 00:51:59.380 So this is the first trimming step. 00:52:02.180 --> 00:52:07.390 So we got rid of some edges, but we got rid of fewer than half 00:52:07.390 --> 00:52:10.940 of the original edges. 00:52:10.940 --> 00:52:15.720 And because now you have a minimum degree on A1, 00:52:15.720 --> 00:52:18.670 the number of edges between A1 and B 00:52:18.670 --> 00:52:22.660 is quite large, still quite large. 00:52:22.660 --> 00:52:27.040 So think about passing down to A1 now. 00:52:27.040 --> 00:52:31.530 In the second step, we are going to apply the path of length 2 00:52:31.530 --> 00:52:34.480 lemma to this A1. 00:52:34.480 --> 00:52:41.880 So A2 is going to be constructed from-- 00:52:41.880 --> 00:52:50.170 so using the path of length 2 lemma, 00:52:50.170 --> 00:52:56.620 specifically with parameter epsilon being delta over 10. 00:52:56.620 --> 00:52:59.440 Although, remember, now the density of the graph 00:52:59.440 --> 00:53:01.760 went from delta to delta over 2. 00:53:01.760 --> 00:53:04.245 Again, if you don't care about the specific numbers, 00:53:04.245 --> 00:53:05.620 they're all polynomials in delta. 00:53:05.620 --> 00:53:06.703 So don't worry about them. 00:53:06.703 --> 00:53:08.590 Everything's poly delta. 00:53:08.590 --> 00:53:11.860 So we're going to apply the path of length 2 lemma 00:53:11.860 --> 00:53:16.240 to find this subset A2. 00:53:16.240 --> 00:53:25.660 And it has the property that A2 is quite large, 00:53:25.660 --> 00:53:45.320 and all but a small fraction of pairs in A2 are friendly. 00:53:54.580 --> 00:53:59.540 So we passed down to, first, trimming small degree vertices, 00:53:59.540 --> 00:54:02.120 and then passed down further to A2, 00:54:02.120 --> 00:54:06.020 where all but a small fraction of elements in A2, 00:54:06.020 --> 00:54:08.563 or all but a small fraction of the pairs 00:54:08.563 --> 00:54:10.230 are friendly to each other, meaning they 00:54:10.230 --> 00:54:11.770 have lots of common neighbors. 00:54:15.020 --> 00:54:16.870 And now let's look at the other side. 00:54:16.870 --> 00:54:21.610 Let's look at B. So we're in this situation 00:54:21.610 --> 00:54:24.520 now where you have-- 00:54:27.760 --> 00:54:31.390 so we're now in a situation where you've passed down 00:54:31.390 --> 00:54:42.630 to A2 and in B, where, because of what we did initially, 00:54:42.630 --> 00:54:47.410 every vertex in here have large degree. 00:54:47.410 --> 00:54:53.830 So there's this minimum degree condition 00:54:53.830 --> 00:54:57.190 from every vertex on the left. 00:54:57.190 --> 00:55:00.250 So the average degree is still very high. 00:55:02.960 --> 00:55:07.020 As a result, the average degree from B 00:55:07.020 --> 00:55:09.280 is going to be quite high. 00:55:09.280 --> 00:55:13.720 So let's focus on the B side and pick out vertices in B 00:55:13.720 --> 00:55:16.570 that have high degree. 00:55:16.570 --> 00:55:23.390 So let's B1 denote vertices in B such 00:55:23.390 --> 00:55:30.530 that the degree from B to A2 is at least half 00:55:30.530 --> 00:55:33.390 of what you expect based on average degree. 00:55:37.850 --> 00:55:41.390 And, as before, the same logic as the A1 step. 00:55:41.390 --> 00:55:52.760 We see that B1 has large size, is a large fraction of B. 00:55:52.760 --> 00:55:57.410 And now we pass down to this B1 set. 00:56:04.214 --> 00:56:17.970 Now, finally, let's consider A3 to be vertices in A2 00:56:17.970 --> 00:56:21.490 where a is friendly. 00:56:21.490 --> 00:56:28.650 So vertices a in A2 such that a is friendly to at least 1 00:56:28.650 --> 00:56:31.210 over delta over-- 00:56:31.210 --> 00:56:38.740 so 1 minus delta over 5 fraction of A2. 00:56:42.590 --> 00:56:50.430 So we saw that, in A2, most pairs of vertices are friendly. 00:56:50.430 --> 00:57:00.000 So most, meaning all but a delta over 10 fraction. 00:57:00.000 --> 00:57:05.770 So if we consider vertices which are 00:57:05.770 --> 00:57:10.740 unfriendly to many other vertices in A2, 00:57:10.740 --> 00:57:13.560 there aren't so many of them. 00:57:13.560 --> 00:57:16.440 If there were many of them, you couldn't have had that. 00:57:16.440 --> 00:57:18.850 So that's why I constructed this set 00:57:18.850 --> 00:57:23.540 A3 consisting of elements in A2 that 00:57:23.540 --> 00:57:27.110 are friendly to many elements. 00:57:27.110 --> 00:57:32.990 And the size of A3 is at least half of that of A2. 00:57:40.974 --> 00:57:50.510 So we have this A3 inside. 00:57:50.510 --> 00:57:51.696 All right. 00:57:51.696 --> 00:57:57.235 And now I claim that we can take A3 and B as our final sets, 00:57:57.235 --> 00:58:01.510 and that between every vertex in A3 and every vertex in B1, 00:58:01.510 --> 00:58:04.990 I claim there must be lots of paths of length 3. 00:58:04.990 --> 00:58:07.420 But, first, let's check their sizes. 00:58:07.420 --> 00:58:09.820 I mean, the sizes all should be OK, because we never 00:58:09.820 --> 00:58:11.960 lost too much at each step. 00:58:11.960 --> 00:58:13.740 If you only care about polynomial factors, 00:58:13.740 --> 00:58:15.990 well, you already see that we never lost anything more 00:58:15.990 --> 00:58:17.590 than a polynomial factor. 00:58:17.590 --> 00:58:20.510 But just to be precise, the size of A3 is at least-- 00:58:20.510 --> 00:58:24.320 so if you count up the factor lost at each step, 00:58:24.320 --> 00:58:29.230 so it's 1/2 delta over 4 delta over 2. 00:58:29.230 --> 00:58:34.480 So it's at least delta squared over 16 fraction 00:58:34.480 --> 00:58:36.870 of the original set A. 00:58:36.870 --> 00:58:44.320 And now, if we consider a comma b 00:58:44.320 --> 00:58:49.650 to be an arbitrary pair in A3 cross B1, 00:58:49.650 --> 00:58:53.140 I claim that there must be many paths. 00:58:53.140 --> 00:58:59.090 Because by using-- so what properties do we know? 00:58:59.090 --> 00:59:05.920 We know that b is adjacent to a large fraction. 00:59:05.920 --> 00:59:10.300 So here large means at least delta over 4-- so bounded 00:59:10.300 --> 00:59:16.080 below-- a large fraction of A2. 00:59:16.080 --> 00:59:16.580 Yes. 00:59:16.580 --> 00:59:17.302 So I apologize. 00:59:17.302 --> 00:59:19.260 When I say the word large, depending on context 00:59:19.260 --> 00:59:22.640 it can mean bigger than delta, or it could mean at least 1 00:59:22.640 --> 00:59:23.370 minus delta. 00:59:23.370 --> 00:59:25.380 So you look at what I write down. 00:59:25.380 --> 00:59:31.070 So b is adjacent to at least delta over 4 fraction of A2. 00:59:31.070 --> 00:59:39.300 At the same time, we know that a is friendly to at least 1 00:59:39.300 --> 00:59:43.940 minus delta over 5 fraction of A2. 00:59:49.000 --> 00:59:54.070 So these two sets, they must overlap by at least a delta 00:59:54.070 --> 00:59:55.060 over 20 fraction. 01:00:00.351 --> 01:00:05.260 So let's take a vertex b. 01:00:05.260 --> 01:00:13.100 So you-- so it's adjacent to many vertices here. 01:00:13.100 --> 01:00:17.526 And if you look at a vertex in A, 01:00:17.526 --> 01:00:21.240 it's friendly to a large fraction. 01:00:21.240 --> 01:00:25.570 So, in particular, it's friendly to all these elements 01:00:25.570 --> 01:00:26.070 over here. 01:00:28.760 --> 01:00:34.840 So, to finish off, what does it mean for a-- 01:00:34.840 --> 01:00:37.600 this is-- this vertex is a. 01:00:37.600 --> 01:00:38.810 This vertex is b. 01:00:38.810 --> 01:00:40.970 What does it mean for a to be friendly to all 01:00:40.970 --> 01:00:42.830 of these shaded elements? 01:00:42.830 --> 01:00:47.510 It means that there are lots of paths from a 01:00:47.510 --> 01:00:51.930 to each of these elements. 01:00:51.930 --> 01:00:55.974 And then you can finish off the paths going back to b. 01:00:55.974 --> 01:00:56.750 Yes. 01:00:56.750 --> 01:01:00.092 AUDIENCE: The shaded stuff is allowed to be outside of A3? 01:01:00.092 --> 01:01:01.800 YUFEI ZHAO: No. the shaded-- the question 01:01:01.800 --> 01:01:04.040 is, is the shaded stuff allowed to be outside of A3? 01:01:04.040 --> 01:01:04.540 No. 01:01:04.540 --> 01:01:06.870 The shaded things are inside A3. 01:01:06.870 --> 01:01:10.900 So we're looking at intersections within A3. 01:01:14.550 --> 01:01:15.090 No, sorry. 01:01:17.055 --> 01:01:18.180 Actually, no, you're right. 01:01:18.180 --> 01:01:20.530 So the shaded things can be outside A3. 01:01:20.530 --> 01:01:22.450 So shaded things can be outside A3. 01:01:22.450 --> 01:01:23.130 I apologize. 01:01:23.130 --> 01:01:25.650 So everything now is in A2. 01:01:28.800 --> 01:01:35.100 So b is adjacent to a large fraction of A2. 01:01:35.100 --> 01:01:43.800 And a here is friendly to some part of the neighbors of b. 01:01:43.800 --> 01:01:48.920 So you can complete paths like that. 01:01:52.750 --> 01:01:53.250 Yes. 01:01:53.250 --> 01:01:54.880 So only the starting and ending points 01:01:54.880 --> 01:01:56.410 have to be in A prime and B prime. 01:01:56.410 --> 01:01:58.970 Everything else, they can go outside of the A prime and B 01:01:58.970 --> 01:02:00.666 prime. 01:02:00.666 --> 01:02:03.970 Yes, thank you. 01:02:03.970 --> 01:02:22.516 So the number of paths from a to B to A2 back to b is-- 01:02:22.516 --> 01:02:24.960 let's see if I can stay within B1-- 01:02:24.960 --> 01:02:26.010 so is at least-- 01:02:34.110 --> 01:02:34.610 yes. 01:02:34.610 --> 01:02:36.530 So it's-- sorry. 01:02:36.530 --> 01:02:44.070 This is B. So it's at least delta over 20 times A2 times 01:02:44.070 --> 01:02:49.680 delta over delta squared over 80 times B. So 01:02:49.680 --> 01:02:52.470 if you don't care about polynomial factors in delta, 01:02:52.470 --> 01:02:55.000 then you see that-- 01:02:58.080 --> 01:03:00.140 the point is there's a large fraction of-- 01:03:02.418 --> 01:03:03.460 there are a lot of paths. 01:03:03.460 --> 01:03:07.000 So there are a lot of paths between each little a and each 01:03:07.000 --> 01:03:09.670 little b by the construction we've done. 01:03:15.190 --> 01:03:16.630 So let me just do a recap. 01:03:16.630 --> 01:03:19.990 So there were quite a few details in this proof, 01:03:19.990 --> 01:03:22.300 and some of them have to do with cleaning up. 01:03:22.300 --> 01:03:24.640 Because it's not so nice to work with graphs 01:03:24.640 --> 01:03:26.850 that just have large average degree. 01:03:26.850 --> 01:03:28.480 It's much nicer to work with graphs 01:03:28.480 --> 01:03:29.920 with large minimum degree. 01:03:29.920 --> 01:03:33.580 So there are a couple of steps here to take care of vertices 01:03:33.580 --> 01:03:34.990 with small degrees. 01:03:34.990 --> 01:03:39.290 So we started with, between A and B, lots of edges. 01:03:39.290 --> 01:03:42.410 And we trim vertices from A with small degree. 01:03:42.410 --> 01:03:45.250 So we get A1. 01:03:45.250 --> 01:03:48.970 And then we apply the path of length 2 lemma to get A2. 01:03:48.970 --> 01:03:52.838 So inside A2, most pairs of vertices 01:03:52.838 --> 01:03:54.630 have lots of common neighbors, but not all. 01:03:57.510 --> 01:04:01.860 We then go back to B to get B1, which 01:04:01.860 --> 01:04:04.940 has large minimum degree to A2. 01:04:07.650 --> 01:04:12.030 And then A3 looks at vertices in A 01:04:12.030 --> 01:04:16.490 with many friendly companions in A2. 01:04:20.200 --> 01:04:24.100 And A3 is large, and I claim that between every vertex in A3 01:04:24.100 --> 01:04:28.120 and every vertex in B, you have many paths of length 3. 01:04:28.120 --> 01:04:32.340 Because if you start with a vertex in A3, 01:04:32.340 --> 01:04:35.430 it has many friendly companions. 01:04:35.430 --> 01:04:41.640 So many here means at least 1 minus delta over 5 fraction. 01:04:41.640 --> 01:04:49.470 Whereas every vertex in B1 has lots of neighbors in A2, 01:04:49.470 --> 01:04:53.430 where lots means at least delta over 4. 01:04:53.430 --> 01:04:56.610 So there's necessarily an overlap 01:04:56.610 --> 01:04:59.230 of at least delta over 20. 01:04:59.230 --> 01:05:01.750 And for that overlap, we can create 01:05:01.750 --> 01:05:07.510 lots of paths going through this overlap from A 01:05:07.510 --> 01:05:12.947 to B. Any questions? 01:05:16.220 --> 01:05:16.780 OK, great. 01:05:16.780 --> 01:05:21.940 So let's put everything together to prove the graphical version 01:05:21.940 --> 01:05:23.704 of Balog-Szemeredi-Gowers. 01:05:31.040 --> 01:05:32.730 So we'll prove the graphical version 01:05:32.730 --> 01:05:34.452 of Balog-Szemeredi-Gowers. 01:05:42.660 --> 01:05:46.920 So by-- so, first, note that the hypothesis 01:05:46.920 --> 01:05:49.450 of Balog-Szemeredi-Gowers already 01:05:49.450 --> 01:05:53.730 implies that the size of A and the size of B 01:05:53.730 --> 01:05:55.650 are not too small. 01:05:58.910 --> 01:06:03.287 Because, otherwise, you couldn't have had n squared over k edges 01:06:03.287 --> 01:06:03.870 to begin with. 01:06:08.060 --> 01:06:16.610 So by the path of length 3 lemma, 01:06:16.610 --> 01:06:20.375 there exists A prime in A and B prime 01:06:20.375 --> 01:06:24.980 in B with the following properties. 01:06:24.980 --> 01:06:29.240 That A prime has a large fraction of-- 01:06:32.306 --> 01:06:36.110 so A prime and B prime are both large in size. 01:06:40.460 --> 01:06:46.665 And for all vertices a in A prime 01:06:46.665 --> 01:06:54.090 and vertices b in B prime, there are 01:06:54.090 --> 01:06:59.040 lots of paths of length 3 between these vertices. 01:06:59.040 --> 01:07:05.010 So there are at least k to the minus little o1-- 01:07:05.010 --> 01:07:10.950 to the minus big O1 times n squared 01:07:10.950 --> 01:07:18.050 pairs of intermediate vertices a1, 01:07:18.050 --> 01:07:37.760 b1 in A cross B, such that a b1 a1 b is a path in G. 01:07:37.760 --> 01:07:41.690 So let me draw the situation for you. 01:07:48.920 --> 01:08:00.210 So we have A and B. And so inside A and B, 01:08:00.210 --> 01:08:08.100 we have this fairly large A prime 01:08:08.100 --> 01:08:14.026 and B prime, such that for every little a 01:08:14.026 --> 01:08:23.470 and little b, there are many paths like that 01:08:23.470 --> 01:08:26.890 going to b1 and a2. 01:08:30.439 --> 01:08:44.270 Let me set-- so let me set x to be a plus b1, that sum, 01:08:44.270 --> 01:08:52.240 y to be a1 plus b1, and z to be a1 plus b. 01:09:04.460 --> 01:09:21.380 So now notice that we can write this a plus b in at least k 01:09:21.380 --> 01:09:27.350 to the minus big O1 times n squared ways 01:09:27.350 --> 01:09:39.410 as x minus y plus z by following this path, where x, y, 01:09:39.410 --> 01:09:45.500 and z all lie in the restricted sumset, 01:09:45.500 --> 01:09:49.350 because that's how the restricted sumset is defined. 01:09:49.350 --> 01:09:54.229 So if you have an edge, then the sum of the elements 01:09:54.229 --> 01:09:57.400 across on the two ends, by definition, 01:09:57.400 --> 01:10:00.970 lies in the restricted sumset. 01:10:00.970 --> 01:10:02.900 So the path of length 3 lemma tells us 01:10:02.900 --> 01:10:06.320 that every pair a and b, their sum 01:10:06.320 --> 01:10:12.290 can be written in many different ways as this combination. 01:10:12.290 --> 01:10:21.360 As a result, we see that A prime plus B prime-- 01:10:21.360 --> 01:10:26.477 so this sum, if we consider sum along with its multiplicity-- 01:10:31.450 --> 01:10:36.760 so now we're really looking at all the different sums as well 01:10:36.760 --> 01:10:43.890 as ways of writing the sum as this combination-- 01:10:43.890 --> 01:10:58.060 we see that it is bounded above by the restricted sumset raised 01:10:58.060 --> 01:10:58.920 to the third power. 01:11:11.295 --> 01:11:13.740 Because each of these choices, x, y, and z, 01:11:13.740 --> 01:11:16.330 they come from the restricted sumset. 01:11:16.330 --> 01:11:19.150 But the hypothesis of Balog-Szemeredi-Gowers, 01:11:19.150 --> 01:11:21.880 the graphical version, is that the restricted sumset 01:11:21.880 --> 01:11:24.630 is small in size. 01:11:24.630 --> 01:11:32.430 So we can now upper bound the restricted sumset 01:11:32.430 --> 01:11:36.580 by, basically, the-- 01:11:36.580 --> 01:11:41.840 within a constant, within a factor of the maximum possible. 01:11:41.840 --> 01:11:47.640 And now we are done, because we have deduced 01:11:47.640 --> 01:11:52.770 that the complete sumset between A prime and B prime 01:11:52.770 --> 01:12:02.320 is, at most, a constant factor with change 01:12:02.320 --> 01:12:04.310 in constant by a polynomial. 01:12:04.310 --> 01:12:06.795 So a constant factor more than the maximum possible. 01:12:11.000 --> 01:12:14.490 So it's, at mostly, k to the big O1 poly k times n. 01:12:17.070 --> 01:12:20.310 So that proves the graphical version 01:12:20.310 --> 01:12:22.890 of Balog-Szemeredi-Gowers. 01:12:22.890 --> 01:12:26.070 And because we showed earlier that the graphical version 01:12:26.070 --> 01:12:28.560 of Balog-Szemeredi-Gowers implies Balog-Szemeredi-Gowers, 01:12:28.560 --> 01:12:32.170 this shows the Balog-Szemeredi-Gowers theorem. 01:12:32.170 --> 01:12:35.880 So let me recap some of the ideas we saw today. 01:12:35.880 --> 01:12:38.260 And so the whole point of Balog-Szemeredi-Gowers 01:12:38.260 --> 01:12:42.070 and all of these related lemmas and theorems and variations 01:12:42.070 --> 01:12:47.050 is that you start with something that has 01:12:47.050 --> 01:12:49.600 a lot of additive structure. 01:12:49.600 --> 01:12:54.440 Well, after we passed down to graphs just a lot of edges. 01:12:54.440 --> 01:12:57.700 So you start with a situation where 01:12:57.700 --> 01:13:02.325 you have kind of 1% goodness. 01:13:02.325 --> 01:13:03.700 And you want to show that you can 01:13:03.700 --> 01:13:07.960 restrict to fairly large subsets, 01:13:07.960 --> 01:13:10.610 so that you have perfection. 01:13:10.610 --> 01:13:14.510 So you have complete goodness between these two sets. 01:13:14.510 --> 01:13:17.240 And this is what's going on in both the graphical version 01:13:17.240 --> 01:13:18.920 and the additive version. 01:13:18.920 --> 01:13:21.560 So back to the graph path of length 3 lemma. 01:13:21.560 --> 01:13:25.820 So we were able to boost the path of length 2 lemma, which 01:13:25.820 --> 01:13:31.400 tells us something about 99% of the pairs having lots 01:13:31.400 --> 01:13:37.160 of common neighbors, to 100% of the pairs having 01:13:37.160 --> 01:13:41.010 lots of path of length 3. 01:13:41.010 --> 01:13:43.380 And in the additive setting, we saw that 01:13:43.380 --> 01:13:47.880 by starting with a situation where the hypothesis is 01:13:47.880 --> 01:13:51.210 somewhat patchy, so like a 1% type hypothesis, 01:13:51.210 --> 01:13:54.750 we can pass down to fairly large sets, where 01:13:54.750 --> 01:13:58.590 the complete sumset, starting with just the restricted sumset 01:13:58.590 --> 01:14:01.110 being small, can pass down to large sets 01:14:01.110 --> 01:14:03.540 where the complete sumset is small. 01:14:03.540 --> 01:14:05.730 And this is an important principle, that, often, 01:14:05.730 --> 01:14:11.210 when we have some typicality by an appropriate argument-- 01:14:11.210 --> 01:14:14.300 and, here, it's not at all a trivial argument. 01:14:14.300 --> 01:14:15.890 So there's some cleverness involved, 01:14:15.890 --> 01:14:18.120 that by doing some kind of argument, 01:14:18.120 --> 01:14:21.950 we may be able to pass down to some fairly large set 01:14:21.950 --> 01:14:25.130 where it's not typically good, but everything's 01:14:25.130 --> 01:14:27.020 perfectly good. 01:14:27.020 --> 01:14:31.820 That's the spirit here of the Balog-Szemeredi-Gowers theorem. 01:14:31.820 --> 01:14:35.010 So, next time, for the last lecture of this course, 01:14:35.010 --> 01:14:38.600 I will tell you about the sum-product problem, 01:14:38.600 --> 01:14:40.802 where the-- 01:14:40.802 --> 01:14:44.500 there are also some graph-- very nice graph theoretic inputs.