How about the Fourier expansion? Suppose we choose and according to this distribution. That is, we choose a random permutation, and then choose initial segments and independently with binomial probabilities on and . What is the expected value of ?

To make the calculation friendlier, I’m going to work out something else that I hope will turn out to be the same. It’s more like Ryan’s p-q approach. To pick a random point I’ll pick a number , binomially distributed with parameters and 1/2, and I’ll then set and choose a point with the measure . Damn, that doesn’t work, because the probability that I end up choosing the empty set is strictly greater than (because there’s a chance I’ll pick it if takes a typical value round 1/2, and also a probability of taking ). OK, it’s believable that any average of binomial distributions that ends up with mean is guaranteed to be less concentrated than the binomial distribution itself, so probably this idea was doomed.

OK I don’t at the moment see a conceptual argument for calculating the Fourier expression. Perhaps it’s a question of a brute-force calculation that one hopes will cancel a lot. But it would be disappointing if the expectation of above were not zero when .

]]>I want to think some more about Ryan’s observation that the permutations trick can be used with the uniform measure. In particular, I want to understand from a non-permutations point of view what the resulting measure is on pairs .

So let’s fix two sets of sizes and , respectively. The probability that a random permutation gives both and as initial segments is and the probability that we choose those two initial segments (if we binomially choose a random and a random without conditioning on their order) is So the probability that we choose the sets and is which equals which equals

Therefore, the probability of choosing conditional on having chosen is If and were independent, this probability would of course be , so the extra weight is … hmm, I’m not really getting anywhere here.

]]>Terry, I’ve tried to understand your argument properly but there are a few points where I’ve had difficulty working out what you mean. I think they all stem from one main point, which comes in the last paragraph where you say “by hypothesis”. I’m not sure I know what the hypothesis is. You established that it was impossible to get a substantial density *increase* on a local 1-set, but here you seem to need that you can’t have a substantial density *decrease*. Have I misunderstood something? I hope so. (I’m actually writing this offline in my car so by the time I get to post it perhaps this matter will have been cleared up.)

We said that two edges span the same cube if substituting all 1-s and 2-s by a wildcard we get the same 0,3,* sequence. Similarly we can say that every edge, e, spans a particular subcube; substitute all 1-s and 2-s with wildcards to get a 0,3,* sequence. Any edge, which differs from e in the * positions only, is in the subcube spanned by e. Our moderate goal here is to show that if there are no three edges of the form (v1,v2), (w1,w2), and (v1,w2) in G(A,B) that (v1,v2) and (w1,w2) span the same subcube – in which case we were done – then there is a subcube spanned by many edges containing many other edges (see in post 845). The vertices of the spanning edges provide us the large empty subgraph.

We will follow a simple algorithm. At first consider the set of edges spanning the whole cube. Those are edges without 0 or 3. If there are at least such edges, then stop. If there are less than such edges, then select the densest n-1 dimensional subcube. (Such subcubes are spanned by edges having exactly one 0 and no 3-s and by edges with one 3 and no 0-s.) If the densest subcube is spanned by at least edges then stop, otherwise select the densest n-2 dimensional subcube. If we repeat the algorithm and , then it should terminate in a c-dense d-dimensional subcube spanned by at least edges. By the initial assumption there is no edge connecting the end-vertices of the spanning edges. It gives us the huge empty bipartite graph, . I didn’t do the actual calculations, but it seems correct to me.

Here we show that a density increment argument implies the Moser-type theorem in 846. It isn’t new that a density increment would give us what we are looking for, the new feature is that it might be easier to prove density increment in a graph with a huge empty subgraph. For practical reasons we switch to a bipartite settings. Consider a bipartite graph G(A.B), between two copies of the n-dimensional cube, . G(A,B) has vertices and, say, edges. There is a natural one to one mapping between the set of edges and a pointset of . The two vertices of an edge are 0-1 sequences. From the two sequences we get one as follows. The i-th position of the new sequence (0,1,2, or 3) is given by where x is the number (0 or 1) in the i-th position of the vertex in A and y is the number in the i-th position of the vertex in B. We say that two edges span the same cube if substituting all 1-s and 2-s by a wildcard, *, we get the same sequence. Observe that if (v1,v2) and (w1,w2), two edges of G(A,B), span the same cube and (v1,w2) is in G(A,B), then we are done, there are three points formed by taking a string with one or more wildcards and in it and replacing the first type wildcards by 1,2 and 3, and the second type wildcards by 4,3, and 2, respectively. (For example 2312 gives 123412, 223312, and 323212.) Equivalently, it will give a corner on the Cartesian product of an n-dimensional Hilbert cube. I will continue, but I would like to see if the typing is correct so far or not.

]]>With Ron Graham we proved a colouring version of 846. For any colouring of there is a monochromatic triple like in 846. The paper is available at

The reasonable bound for the colouring version shows that this problem might be easier to work with than with DHJ k=3. In the next post I will try to prove that 845 implies 846, that is, an “arithmetic density increment” in a graph with a huge hole implies a Moser type statement.

(Sorry about the number of posts, but I don’t want to write long ones as even the shorter ones are full with typos)

844.2. Ah, sorry if you said this already in an earlier comment, Terry; I’m still catching up.

]]>845.2 … and apparently I don’t know how to generate a thread.

Still going backwards, let me state the theorem which would follow from 845: Every dense subset of contains three elements, formed by taking a string with one or more wildcards and in it and replacing the first type wildcards by 1,2 and 3, and the second type wildcards by 4,3, and 2, respectively. For example 123412,

223312, and 323212 is such triple.

oops, I didn’t give my name for the previous note

]]>Tim, Re:842. In case you are considering corners in cubes, let me mention a few things here. Reading some recent blogs, I thought that you are close to prove a density increment result which would imply the corner result. I write the statement first and then I will argue that it proves a Moser type result.

Given an set (or an n-dimensional Hilbert cube if you wish) with elements. Consider the elements as vertices of a graph. The graph is c-dense, i.e. the number of edges is . The graph contains a huge hole, there is an empty subgraph. Prove that then it contains a d-dimensional subcube on the vertices spanning at least edges. might depend on c and but not on n, and d grows with n (arbitrary slowly). Now I will try to post the next part as a new thread.

Ryan, that’s right; being 20-insensitive, we can invoke DHJ(2.5) and get large subspaces in , which of course is a healthy density increment.

]]>What if a random simple set in of density , meaning that its is a random set of density under the -biased distribution?

How should we try to find a combinatorial subspace to increment on?

(Perhaps we shouldn’t; perhaps instead we should exploit the fact that this is extremely 23-insensitive.)

]]>One reason I said the comment in #840 is tiny is that I’m finally starting to catch up the fact (well known to the rest of you, I think) that if you don’t mind density increment arguments (and we certainly don’t!) then the underlying probability measure is quite unimportant. Or rather, you can pass freely between any “reasonable” measures, using the arguments sketched here on the wiki.

In fact, I think “equal-slices” is a good “base measure” to always return to. It has the nice feature that if you have equal-slice density , by density-increment arguments you can assume that you have density very nearly under *almost all* product measures.

I have a very non-mathematically busy day today, so I’ve got time for just one brief post. First, let me say that I’m quite excited by Terry’s 837.2. I haven’t fully digested it yet but I have partially digested it enough to see that it is very definitely of the sort of flavour I was hoping for. It’s frustrating not to be able to spend a few hours today playing around with the argument. And then tomorrow and Tuesday I have two heavy teaching days.

So instead, let me throw out a small thought/question. Recall that in the dictionary I gave in 829, long arithmetic progressions go to combinatorial subspaces in . Therefore, if the general plan of modelling an argument on Ajtai-Szemerédi works, it will have to give a new proof of the corners theorem that is similar to their proof but avoids the use of Szemerédi’s theorem. As I’ve already explained, that sounds discouraging at first, but I think it may in fact not be. But I’ve subsequently realized that these thoughts lead inexorably to a suggestion that Jozsef made, way back in comment 2 (!), that one should look at a generalization of the corners problem in which you don’t have long APs in the two sets of which you take the Cartesian product. Maybe the time is ripe for having a go at that problem, and in particular seeing whether the Ajtai-Szemerédi approach can be pushed through for it.

]]>1. Terry I agree that with threading we need shorter threads, if you’ll excuse the dual use of the word “thread”. Let’s indeed go for 850.

2. The other thing is that I’ve been meaning to say for ages that I loved your notion that the wiki could be thought of as an online journal of density Hales-Jewett studies, and the discussions as weekly conferences.

]]>840.1 Ryan, I’m not sure how tiny that comment is, and must think about it. Maybe it will turn out that putting the binomial distribution on maximal chains is always more convenient than putting equal-slices measure on the cube. And clearly this same trick works for . This feels to me like a potentially useful further technique to add to our armoury, and when I get the chance I think I’ll add something about it to the wiki.

I agree that the Fourier calculation looks more or less compulsory. And something that’s rather nice about this measure is that it combines the uniform distribution on with a very natural non-uniform distribution on combinatorial lines, so it provides a potential answer to a question we were trying to answer way back in the Varnavides thread. Indeed, another exercise we should do is prove a Varnavides-type version of DHJ(3) for this measure on the combinatorial lines (given DHJ(3) itself).

Oh no wait a moment — it’s not obvious how to generalize to DHJ(3) because we don’t have an analogue of maximal chains. This is something else to think about.

]]>Let be a monotone Boolean function. Let be the integral of the influence of the k-th variable w.r.t. the probability measure . ( is called the “Shapley value” of .) Let be a fixed small real number and let T be the difference between p-q where p is the value for which the probability that is 1 is and q is the value for which the probability that is 1 is . (T is called the length of the “threshold interval for .)

It is known that if all then T is small as well. The known bound is . The proof relies on connecting the influences for different of s. It looks that one could be eliminated and that a more careful understanding of the relation between influences and other properties of for different values of may be useful.

]]>This is very tiny comment. I just wanted to point out that one can prove the “correct” density-Sperner Theorem under the uniform distribution, in the same way Tim proves it under the equal-slices distribution.

Assume under the uniform distribution. Pick a random chain from (0, 0, …, 0) up to (1, 1, …, 1) and then choose Binomial, independently. Define , the th string in the chain, and . Then , as is uniformly distributed.

So we have a distribution on pairs of strings such that and both have the uniform distribution, and such that they form a nondegenerate “line” with probability at least .

PS: If someone wants to express here in terms of Fourier coefficients, I’d be very happy to see it.

]]>One side effect of the threaded system is that we hit 100 comments long before the number assigned to the comments increases by 100; we’re at 838 now but the thread is already longer than most of the other threads. Perhaps we may wish to renew this thread well before 899 (e.g. at 850)?

]]>I semi-checked to myself that one can prove Sperner by increment-free, purely Fourier arguments. One uses:

. the structure theorem (#821).

. Mossel’s “Lemma 6.2” to handle the and parts.

. the triangle inequality argument (#815) to handle the parts.

However, this argument is still quite unsatisfactory to me. For one, it requires . For another, it requires selecting the parameter called in #800 (equivalently, in Terry’s #826) *after* using the Structure Theorem. And most “wrongly”, this parameter must be .

In particular, the following theorem is true (I’m 99% sure, at least; proof by “pushing shadows around with Kruskal-Katona”) — however, I don’t think we have *any* Fourier-based proof at all:

Theorem: Let have density . Fix , , where is, say, . (Or set if you like.) Then , where I’m using the notation from Tim’s wiki entry.

]]>838.2

Ryan, I guess one should specify that f should be bounded within [-1,1], and “basically” means “up to o(1) errors, and with small compared with “.

In your example h should also be bounded between [-1,1]. In that case, flipping 0’s to 1’s will only have an effect of increasing f by or so on average.

The reason for the monotone decrease is because of the absorption formula . Since is basically a contraction for (though it does cease to contract for large , of course), we see that .

I guess the “local” picture (small ) is looking a bit different from the “global” picture. Locally, the operation of flipping 0s to 1s is a measure-preserving operation; globally, of course, it isn’t. (Tragedy of the commons!)

]]>838.1 The nonparsing formula there is . *Now fixed — Tim*

Terry, could you clarify on your statement therein, “the energies … are basically monotone decreasing in “?

Suppose is of the form , where is a nonnegative increasing function. Doesn’t it seem as though will be *increasing* in ?