# 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.

## Join the Conversation

10 Comments

1. Do you know how to give a nice eigenbasis for this operator? I tried but did not succeed.

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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.

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2. Alexander Smal says:

It looks like sqrt(n) is missing in the left hand side of (2) and (3).

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1. Alexander Smal says:

OK, I see, I was wrong, nothing is missing.

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3. adityaguharoy says:

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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4. 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 (*).)

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1. Dear Gil, glad to see you! $(\star)$ is an equation after the words “crucial property is”. I added the detailed calculation to the place you ask about.

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