The Independence Number of the Orthogonality Graph in Dimension $2^k$

TL;DR

This paper determines the independence number of the orthogonality graph on hypercubes of dimension 2^k, resolving a long-standing question related to quantum information theory.

Contribution

It extends Frankl's rank argument to compute the independence number for hypercubes of dimension 2^k, answering a question posed in 2001.

Findings

Exact independence number for the orthogonality graph in dimension 2^k

Extension of Frankl's method to new hypercube dimensions

Resolution of a problem in quantum information theory

Abstract

We determine the independence number of the orthogonality graph on $2^k$-dimensional hypercubes. This answers a question by Galliard from 2001 which is motivated by a problem in quantum information theory. Our method is a modification of a rank argument due to Frankl who showed the analogous result for $4p^k$-dimensional hypercubes, where $p$ is an odd prime.

Equations

$$a_n:=4\sum _{i=0}^{n/4-1}\left(\genfrac{}{}{0ex}{}{n-1}{i}\right),$$

$$K_j(i)=K_j(i;n):=\sum _{h=0}^j(-1{)}^h(\genfrac{}{}{0ex}{}{i}{h}\left)\right(\genfrac{}{}{0ex}{}{n-i}{j-h}).$$

$$\{2s:s=0,1,\dots ,2^{k-1}-1,s\ne 2^{k-2}\}.$$

$$\phi (\xi )=\left(\genfrac{}{}{0ex}{}{\xi /2-1}{2^{k-2}-1}\right)\in \mathbb{R}[\xi ],$$

$$\phi (\xi )=\sum _{i=0}^{2^{k-2}-1}{c}_i{K}_i(\xi ).$$

$$X=\sum _{j=0}^{2^k-1}\phi (j)A_j\in \mathcal{A}.$$

$$X=2^{2^k-1}\sum _{i=0}^{2^{k-2}-1}{c}_i{E}_i.$$

$$|C^{\prime }|$$

Full text

The first author is supported by a postdoctoral fellowship of the Research Foundation — Flanders (FWO)

The second author is supported by JSPS KAKENHI Grant Number JP17K05156.

1. Introduction

The orthogonality graph $\Omega_{n}$ has the elements of $\{-1,1\}^{n}$ as vertices, and two vertices are adjacent if they are orthogonal, in other words, if their Hamming distance is $n/2$. The graph $\Omega_{n}$ occurs naturally when comparing classical and quantum communication [3]. In particular, for $n=2^{k}$ the cost of simulating a specific quantum entanglement on $k$ qubits can be reduced to determining the chromatic number $\chi(\Omega_{n})$ of $\Omega_{n}$ [2, 9]. The graph $\Omega_{n}$ is edgeless if $n$ is odd, and is bipartite if $n\equiv 2\pmod{4}$. For $n\equiv 0\pmod{4}$, Frankl [7] and Galliard [9] constructed an independent set of $\Omega_{n}$ of size

[TABLE]

and Galliard [9] asked in 2001 if this is the independence number $\alpha(\Omega_{n})$ of $\Omega_{n}$ when $n=2^{k}$, $k\geqslant 2$. Newman [15] and, according to [8, p. 275, Remark], Frankl conjectured that this holds whenever $n\equiv 0\pmod{4}$. See also [4]. Frankl [7] already showed the conjecture in 1986 for all $n=4p^{k}$ for $k\geqslant 1$, where $p$ is an odd prime. De Klerk and Pasechnik [13] proved the conjecture for $n=16$, i.e., that $\alpha(\Omega_{16})=2304$, using Schrijver’s semidefinite programming bound [16]. Furthermore, Frankl and Rödl [8] showed that $\alpha(\Omega_{n})<1.99^{n}$ if $n\equiv 0\pmod{4}$. In this note, we apply Frankl’s method from [7] to show the following:

Theorem**.**

Let $n=2^{k}$ for some $k\geqslant 2$. Then $\alpha(\Omega_{n})=a_{n}$.

Together with the discussion in [9, Section 5.5], that is using $\chi(\Omega_{n})\geqslant 2^{n}/\alpha(\Omega_{n})$, our result implies an explicit version of Theorem 4 in [2]. Finding such an explicit result is one motivation for Galliard’s work. See also [10, 12].

2. Proof of the Theorem

Let $A_{j}$ be the [math]-$1$-matrix indexed by the vertices of the hypercube $Q_{n}=\{-1,1\}^{n}$ with $(A_{j})_{xy}=1$ if $x$ and $y$ have Hamming distance $j$. The matrices $A_{j}$ have $n+1$ common eigenspaces $V_{0},V_{1},\dots,V_{n}$, and in the usual ordering of the eigenspaces the eigenvalue of $A_{j}$ with respect to $V_{i}$ is given by the Krawtchouk polynomial (see [5, Theorem 4.2])

[TABLE]

It is known that the orthogonal projection matrix $E_{i}$ onto $V_{i}$ has the entry $(E_{i})_{xy}=2^{-n}K_{i}(j)$ if $x$ and $y$ are at Hamming distance $j$ [5, Theorem 4.2], so that we have in particular $\operatorname{rank}E_{i}=\operatorname{trace}E_{i}=K_{i}(0)=\binom{n}{i}$. The $(n+1)$-dimensional matrix algebra spanned by $A_{0}=I,A_{1},\dots,A_{n}$ is called the Bose–Mesner algebra of $Q_{n}$.

Assume now that $n=2^{k}$, $k\geqslant 3$. (The result is trivial if $k=2$.) Let $C$ be an independent set of $\Omega_{2^{k}}$, and let $C_{\mathrm{even}}^{\pm},C_{\mathrm{odd}}^{\pm}\subseteq\{-1,1\}^{2^{k}-1}$ be as in [7]: $C_{\mathrm{even}}^{+}$ is given by taking all the even-weight elements of $C$ that end with $+1$, followed by truncating at the last coordinate, and the other three are analogous. Let $C^{\prime}$ be one of these four families. Then the Hamming distances in $C^{\prime}$ are even and unequal to $2^{k-1}$, so they lie in the following set:

[TABLE]

Below we work with the Bose–Mesner algebra $\mathscr{A}$ of $Q_{2^{k}-1}$. For every $M\in\mathscr{A}$, let $\overline{M}$ denote the principal submatrix corresponding to $C^{\prime}$. Consider the polynomial

[TABLE]

and expand it in terms of the Krawtchouk polynomials $K_{i}(\xi)=K_{i}(\xi;2^{k}-1)$:

[TABLE]

Let

[TABLE]

On the one hand, observe that $\overline{X}$ has only integral entries in view of (1), and an easy application of Lucas’ theorem (cf. [6]) shows moreover that $\overline{X}\equiv\overline{I}\pmod{2}$. In particular, $\overline{X}$ is invertible. On the other hand, from (2) we have

[TABLE]

It follows that

[TABLE]

As $|C|=|C_{\mathrm{even}}^{+}|+|C_{\mathrm{even}}^{-}|+|C_{\mathrm{odd}}^{+}|+|C_{\mathrm{odd}}^{-}|$, the theorem follows.

3. Future Work

Schrijver’s semidefinite programming bound has been extended to hierarchies of upper bounds; see, e.g., [1, 14]. In view of [13], it is interesting to investigate if these bounds in turn prove the conjecture for other values of $n$. One of the referees pointed out to us that using next level in the hierarchy, see [11], yields the correct bound of $a_{24}=178208$ for the case $n=24$.

Problem**.**

Prove the conjecture for $n=40$, which is the first open case.

Acknowledgements

We thank the anonymous referee for solving the case $n=24$.