Low-level variant of Huang’s argument for sensitivity conjecture

Consider the set \Omega_n of all 2^n words t_1t_2\ldots t_n,\, t_i\in \{0,1\}, over the alphabet \{0,1\} (the vertices of Boolean cube). We naturally multiply the words by concatenation: if x\in \Omega_n, y\in \Omega_k, then xy\in \Omega_{n+k}. For i=1,2,\ldots,n and x\in \Omega_n denote by T_ix the word which differs from T_ix only in i-th position. Clearly the operators T_i,i=1,\ldots,n are involutive and mutually commute. We further use the same notations T_i for different values of n (namely, both for n and n-1.) We say that x,y are joined by an edge if y=T_ix for some i=1,\ldots,n. This graph (the skeleton of the Boolean cube) is bipartite by the chessboard coloring. In particular, it contains an independent subset of size 2^{n-1}.

The recent sensational result by Hao Huang https://arxiv.org/pdf/1907.00847 is that if we take a subset A\subset \Omega_n, |A|=2^{n-1}+1, the induced subgraph on this subset already contains a vertex of degree at least \sqrt{n}. That is sharp by the earlier example of F. R. K. Chung, Z. Füredi, R. L. Graham and P. Seymour https://faculty.math.illinois.edu/~z-furedi/PUBS/furedi_chung_graham_seymour_cube.pdf

Everybody is very excited since this estimate settles the so called sensitivity conjecture, and this is not what I am going to tell about (cause of complete ignorance.)

The aim of this post is to give an explicit exposition of Huang’s argument without using any theorems but the following: the linear system of m equations and m+1 variables has a non-trivial solution.

Following Huang, we take a not-everywhere-zero function f:\Omega_n\to \mathbb{R} which vanishes outside A and satisfies some (other) 2^{n-1} linear relations. Such a function exists by the aforementioned theorem. For describing these relations, we define w_i(x)=(-1)^{t_1+\ldots+t_{i-1}} for x=t_1\ldots t_k\in \Omega_{k},i=1,\ldots,k. Clearly w_i(x)=w_i(T_ix) and the straightforward but crucial property is

\displaystyle w_i(x)w_j(x)w_j(T_ix)w_i(T_jx)=-1\,\,\text{for}\,i\ne j. \quad \quad\quad (\star)

Now we are ready to formulate the 2^{n-1} relations on f, they are the following:

\displaystyle \sqrt{n}f(y)=\sum_{i=1}^{n}w_i(y)f(T_i y),\,\,\text{for all}\, y\in \Omega_{n}.\quad (1)

You may say that the number of relations is 2^{n}, but actually the relations (1) for y starting from 0, say y=0x,x\in \Omega_{n-1}, which read as:

\displaystyle f(1x)=\sqrt{n}f(0x)-\sum_{i=1}^{n-1}w_i(x)f(0T_ix),x\in \Omega_{n-1},\quad (2)

imply relations (1) for y=1x,x\in \Omega_{n-1}:

\displaystyle f(0x)=\sqrt{n}f(1x)+\sum_{i=1}^{n-1}w_i(x)f(1T_ix),x\in \Omega_{n-1}.\quad (3)

This reduces to (\star) after we express f(1x),f(1T_ix) via the values f(0T_iT_jx). Indeed, Right Hand Side of (3) reads as

\sqrt{n}(\sqrt{n}f(0x)-\sum_{i=1}^{n-1}w_i(x)f(0T_ix))+\sum_{i=1}^{n-1}w_i(x)(\sqrt{n}f(0T_ix)-\sum_{j=1}^{n-1}w_j(T_ix)f(0T_jT_ix)).

the coefficient of f(0x) in this expression equals \sqrt{n}\cdot \sqrt{n}-\sum_{i=1}^{n-1} w_i(x)w_i(T_ix)=1. The coefficient of f(0T_ix) is -\sqrt{n}w_i(x)+\sqrt{n}w_i(x)=0 for all i=1,\ldots,n-1. Finally for i\ne j the coefficient of f(0T_iT_jx) equals -w_i(x)w_j(T_ix)-w_j(x)w_i(T_jx)=0 by (\star).

Now if y\in \Omega_n is chosen so that |f(y)| is maximal, we have y\in A and looking at the right hand side we conclude that at least \sqrt{n} words of the form T_i y must belong to the support of f, thus to A. The proof is complete.

Of course the reason why 2^{n-1} equations (2) imply the other 2^{n-1} equations (3) is that the corresponding to (1) operator f(y)\mapsto \sum_{i=1}^{n}w_i(y)f(T_i y) has an eigensubspace of dimension 2^{n-1} for the eigenvalue \sqrt{n}. Or viceversa, if you prefer. I cannot say that I understand well where does this operator come from.

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10 Comments

    1. Well, you may partition the cube onto two parallel facets, take your favourite basis for functions on the facet (for example, one value is 1 and others are 0), and the values on the other facet are determined uniquely by (2). I do not know whether it is nice, but at least it is quite explicit and consists of functions with small support. In general, if the operator satisfies an equation like X^2-nI=(X-\sqrt{n}I)(X+\sqrt{n}I)=0, the kernel of X-\sqrt{n}I is the same as the image of X+\sqrt{n}I, that guy has a basis consisting of the columns of the corresponding matrix.

      Liked by 1 person

  1. The description put by you is really interesting (tasty I would like to say). I did not know a word about the topic prior to reading this post. But I could still understand some parts of the post quite clearly. That’s powerful explanation sir. It’s really good.

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  2. Dear Fedor, the sentence, “This reduces to (*) after we express f(1x),f(1T_ix) via the values f(0T_iT_jx)” is not clear.
    can you add a few sentences of explenation to make the post self contained. (Also, what is (*).)

    Like

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