The sensitivity conjecture, induced subgraphs of cubes, and Clifford algebras

TL;DR

This paper presents a new proof of the Sensitivity Conjecture using Clifford algebras, showing that large induced subgraphs of hypercubes have high maximum degree, and extends the result to weighted cases.

Contribution

It introduces a novel Clifford algebra-based approach to prove the Sensitivity Conjecture and provides a weighted version of the theorem.

Findings

Induced subgraphs with over half the vertices have maximum degree at least √n

The proof employs Clifford algebras in a natural way

A weighted version of the main result is established

Abstract

We give another version of Huang's proof that an induced subgraph of the n-dimensional cube graph containing over half the vertices has maximal degree at least $\sqrt{n}$, which implies the Sensitivity Conjecture. This argument uses Clifford algebras of positive definite signature in a natural way. We also prove a weighted version of the result.

Equations

$$v^2=B(v,v)=Q(v)\text{or equivalently}vw+wv=2B(v,w)$$

$$e_i^2=1\text{and}{e}_i{e}_j=-e_j{e}_i.$$

$$\left(\sqrt{n}+S\right)S=\sqrt{n}\left(\sqrt{n}+S\right)\text{and}\left(-\sqrt{n}+S\right)S=-\sqrt{n}\left(-\sqrt{n}+S\right)$$

$$\mathrm{Cl}(n)=\mathrm{Cl}(n){\alpha }_{+}\oplus \mathrm{Cl}(n){\alpha }_{-}=\ker \left(M_S-\sqrt{n}\right)\oplus \ker \left(M_S+\sqrt{n}\right).$$

$$\sqrt{n}\left|x_{v_0}\right|\le \sum _{w\in N(v_0)}\left|x_w\right|\le \sum _{w\in N(v_0)}\left|x_{v_0}\right|=\left|N(v_0)\right|\left|x_{v_0}\right|.$$

$$\sqrt{a_1^2+\cdots +a_n^2}\left|x_{v_0}\right|=\left|\sum _{i=1}^n{\epsilon }_i{a}_i{x}_{v_0(i)}\right|\le \sum _{i=1}^n{a}_i\left|x_{v_0(i)}\right|\le \left|x_{v_0}\right|\sum _{v_0(i)\in W}{a}_i$$

Full text

1 Introduction

In [2], Huang proves a theorem about induced subgraphs of hypercube graphs, and uses it to prove the Sensitivity Conjecture. The main result of that paper is as follows. Let $n\geq 1$ be a integer, and let $Q^{n}$ be the $n$-dimensional (hyper)cube graph, with $2^{n}$ vertices given by $\{0,1\}^{n}$ and edges connecting points which differ at exactly one coordinate. For a graph $G$, let $\Delta(G)$ denote the maximum degree of its vertices.

Theorem 1.1**.**

Let $H$ be a subgraph of $Q^{n}$ induced by a subset of $(2^{n-1}+1)$ vertices. Then $\Delta(H)\geq\sqrt{n}$.

Huang’s proof uses a sequence of matrices $A_{n}$, which have size $2^{n}\times 2^{n}$. In this short paper we observe that these matrices are the matrices of multiplication by a certain natural element $S$ in positive definite Clifford algebras. From this we are able to reformulate the proof in terms of Clifford algebras.

We note that Karasev in [3] has related Huang’s matrices $A_{n}$ to exterior algebras; these are special cases of Clifford algebras when the quadratic form is zero. By using a nontrivial quadratic form for our Clifford algebras we are able to see a close connection between the multiplicative structure of the Clifford algebras, and the combinatorics of the cube graph and its subgraphs.

Our approach generalises to give a “weighted” version of the theorem as follows. Let $a_{1},\ldots,a_{n}$ be non-negative real numbers. Each edge of $Q^{n}$ joins points whose coordinates differ in one place; for a vertex $v$ denote by $v(i)$ the unique vertex which differs from $v$ only in the $i$th place, so that the vertices adjacent to $v$ in $Q^{n}$ are precisely $v(1),\ldots,v(n)$. Let edges joining points whose coordinates differ in the $i$th place (i.e. vertices of the form $v$ and $v(i)$) have weight $a_{i}$. Then for a vertex $v$ of a subgraph $H$ of $Q^{n}$, its weighted degree is the sum of the weights on adjacent edges. Denote by $\Delta_{a}(H)$ the maximum weighted degree of the vertices of $H$.

Theorem 1.2**.**

Let $H$ be a subgraph of $Q^{n}$ induced by a subset of $(2^{n-1}+1)$ vertices, and let the weights $a_{1},\ldots,a_{n}$ be any non-negative real numbers. Then $\Delta_{a}(H)\geq\sqrt{a_{1}^{2}+\cdots+a_{n}^{2}}$.

Setting $a_{1}=\cdots=a_{n}=1$ in theorem 1.2 recovers theorem 1.1.

A connection between Huang’s proof and Clifford algebras has also been given by Tao [6].

After the first version of this paper was posted, the author was informed that essentially identical observations (including theorems 1.1 and 1.2) had independently been made by T. Mrowka some weeks earlier, although they had not been been published [5].

Acknowledgments.

The author is supported by Australian Research Council grant DP160103085.

2 Clifford algebras

2.1 Background

We recall some well-known notions about Clifford algebras, which can be found in any standard text or introductory article on the subject (e.g. [1, 4, 7]). Let $V$ be a real vector space, equipped with a symmetric bilinear form $B\colon V\times V\longrightarrow\mathbb{R}$, or equivalently, a quadratic form $Q\colon V\longrightarrow\mathbb{R}$, related by $Q(v)=B(v,v)$ and $B(v,w)=\frac{1}{2}\left(Q(v+w)-Q(v)-Q(w)\right)$. The Clifford algebra $\operatorname{Cl}(V,Q)$ is the associative algebra freely generated by $V$, subject to the relations

[TABLE]

for all $v,w\in V$. Alternatively, $\operatorname{Cl}(V,Q)$ is given by the tensor algebra of $V$, modulo the ideal generated by elements of the form $v\otimes v-Q(v)$.

We are interested here in positive definite $B$ and $Q$. If $V$ has dimension $n$, and $e_{1},\ldots,e_{n}$ form an orthonormal basis of $V$ with respect to $B$, then the relations imply that for all $1\leq i,j\leq n$,

[TABLE]

As a vector space, $\operatorname{Cl}(V,Q)$ has dimension $2^{n}$ and a basis is given by $e_{i_{1}}\cdots e_{i_{k}}$, for each sequence $1\leq i_{1}<\cdots<i_{k}\leq n$. This includes the empty sequence, whose corresponding basis element is the identity $1$. In the positive definite case $\operatorname{Cl}(V,Q)$ only depends on the dimension $n$ and we write $\operatorname{Cl}(n)$ for $\operatorname{Cl}(V,Q)$. It is known that $\operatorname{Cl}(n)$ is given as a direct sum of one or two matrix algebras, over the real numbers, complex numbers or quaternions, depending on $n$ modulo $8$.

The involution $\phi\colon V\longrightarrow V$ given by $\phi(v)=-v$ extends to an involution of $\operatorname{Cl}(n)$, which we also denote $\phi$, such that for each $1\leq i_{1}<\cdots<i_{k}\leq n$, $\phi(e_{i_{1}}\cdots e_{i_{k}})=(-1)^{k}e_{i_{1}}\cdots e_{i_{k}}$.

2.2 Clifford algebras and cubes

We can identify the basis element $e_{i_{1}}\cdots e_{i_{k}}$ (where $1\leq i_{1}<\cdots<i_{k}\leq n$) of for $\operatorname{Cl}(n)$ with the vertex of $Q^{n}$ with ones in positions $i_{1},\ldots,i_{k}$ and zeroes elsewhere. Thus we write $e_{v}$ for the basis element corresponding to the vertex $v\in\{0,1\}^{n}$. A general element $x\in\operatorname{Cl}(n)$ can be written uniquely as $x=\sum_{v}x_{v}e_{v}$, where the sum is over vertices $v\in\{0,1\}^{n}$ of $Q^{n}$ and each $x_{v}\in\mathbb{R}$.

Observe that multiplying a basis element $e_{v}$ by $e_{i}$, for $1\leq i\leq n$ sends $e_{v}$ to $\pm e_{v(i)}$. In other words, multiplication by $e_{i}$, up to sign, permutes basis elements by translating them along edges of $Q^{n}$ in the $i$th direction.

2.3 A Clifford element for counting degrees

Consider the element $S=e_{1}+\cdots+e_{n}\in\operatorname{Cl}(n)$. Observe that $S^{2}=n$.

For each vertex $v$ then $e_{v}S$ is a signed sum of $e_{w}$ over the vertices $w$ adjacent to $v$ in $Q^{n}$, as is $Se_{V}$ (but in general with different signs). That is, $e_{v}S=\sum_{i=1}^{n}\varepsilon_{i}e_{v(i)}$ where each $\varepsilon_{i}=\pm 1$.

Similarly, if $x\in\operatorname{Cl}(n)$ is a sum of basis elements $x=\sum_{w\in W}e_{w}$, over some subset $W\subseteq\{0,1\}^{n}$, then the coefficient of $e_{v}$ in $xS$ is given by a sum of $\pm 1$s, one for each vertex of $W$ adjacent to $v$. Thus the coefficient of $e_{v}$ in $xS$ is bounded above by the degree of $v$ in the subgraph of $Q^{n}$ induced by $W$. In this way, multiplication by $S$ can be used to bound degrees of vertices.

We therefore consider the map $M_{S}\colon\operatorname{Cl}(n)\longrightarrow\operatorname{Cl}(n)$ given by multiplication on the right by $S$, i.e. $M_{S}(v)=vS$. Observe that

[TABLE]

so $\sqrt{n}+S$ and $-\sqrt{n}+S$ are eigenvectors of $M_{S}$ with eigenvalues $\sqrt{n}$ and $-\sqrt{n}$ respectively. For convenience write $\alpha_{+}=\sqrt{n}+S$ and $\alpha_{-}=-\sqrt{n}+S$.

Indeed then the principal left ideals $\operatorname{Cl}(n)\alpha_{+}$ and $\operatorname{Cl}(n)\alpha_{-}$ are contained in the $\sqrt{n}$ and $-\sqrt{n}$ eigenspaces respectively. Hence their intersection is zero, but as $\alpha_{+}-\alpha_{-}=2\sqrt{n}$, they span $\operatorname{Cl}(n)$. Hence they are the entire eigenspaces and we have

[TABLE]

Since $\alpha_{+}\alpha_{-}=\alpha_{-}\alpha_{+}=0$, this is in fact a product of rings. Indeed, we have $\alpha_{+}^{2}=2\sqrt{n}\alpha_{+}$ and $\alpha_{-}^{2}=-2\sqrt{n}\alpha_{-}$, so $\frac{1}{2\sqrt{n}}\alpha_{+}$ and $\frac{1}{2\sqrt{n}}\alpha_{-}$ are complementary orthogonal idempotents. One can check that the matrix of $M_{S}$ with respect to the lexicographically ordered basis is the matrix $A_{n}$ of [2].

The involution $\phi$ takes $\operatorname{Cl}(n)\alpha_{+}$ to $\operatorname{Cl}(n)\alpha_{-}$ and vice versa, hence both have dimension $2^{n-1}$ as vector spaces.

More generally, if we take real numbers $a_{1},\ldots,a_{n}$ we can define $S_{a}=a_{1}e_{1}+\cdots+a_{n}e_{n}$. If $x=\sum_{w\in W}e_{w}$, over some subset $W\subseteq\{0,1\}^{n}$, then the coefficient of $e_{v}$ in $xS$ is given by a sum of $\pm a_{i}$ terms, one for each vertex of $W$ adjacent to $v$. That is, the coefficient of $e_{v}$ in $xS$ is $\sum_{i}\varepsilon_{i}a_{i}$, where each $\varepsilon_{i}=\pm 1$, and the sum is over $1\leq i\leq n$ such that $v(i)\in W$. This coefficient is bounded above by the weighted degree of $v$ in the subgraph of $Q^{n}$ induced by $w$. We have $S_{a}^{2}=a_{1}^{2}+\cdots+a_{n}^{2}$; multiplication $M_{S_{a}}$ by $S_{a}$ on the right has eigenvectors $\alpha_{\pm}=\pm\sqrt{a_{1}^{2}+\cdots+a_{n}^{2}}+S_{a}$ with eigenvalues $\pm\sqrt{a_{1}^{2}+\cdots+a_{n}^{2}}$. If not all $a_{i}$ are zero, we obtain a direct sum of rings $\operatorname{Cl}(n)\alpha_{+}\oplus\operatorname{Cl}(n)\alpha_{-}=\ker\left(M_{S_{a}}-\sqrt{a_{1}^{2}+\cdots+a_{n}^{2}}\right)\oplus\ker\left(M_{S_{a}}+\sqrt{a_{1}^{2}+\cdots+a_{n}^{2}}\right)$.

3 Proofs of theorems

Let $W\subset\{0,1\}^{n}$ be a set of $2^{n-1}+1$ vertices of $Q^{n}$, and let $H$ be the subgraph of $Q^{n}$ induced by $W$. For a vertex $v\in W$, denote by $N(v)$ its neighbours in $H$, i.e. those vertices in $W$ adjacent to $v$. Thus $|N(v)|$ is the degree of $v$ in $W$, and in theorem 1.1 we want to show some $|N(v)|\geq\sqrt{n}$.

Let $C_{H}\subset\operatorname{Cl}(n)$ be the vector subspace spanned by $e_{w}$, over all $w\in W$. In other words, $C_{H}=\bigoplus_{w\in W}\mathbb{R}e_{w}$.

Proof of theorem 1.1.

As $C_{H}$ has dimension $2^{n-1}+1$, it has nontrivial intersection with $\operatorname{Cl}(n)\alpha_{+}=\ker(M_{S}-\sqrt{n})$. Let $0\neq x=\sum_{w\in W}x_{w}e_{w}$ lie in the intersection. Then $x(S-\sqrt{n})=0$, so $xS=\sqrt{n}x$.

Now in $xS$, for any $v\in\{0,1\}^{n}$, the coefficient of $e_{v}$ is given by a sum $\sum_{w\in N(v)}\varepsilon_{w}x_{w}$, where each $\varepsilon_{w}=\pm 1$, which is at most $\sum_{w\in N(v)}\left|x_{w}\right|$ in absolute value. On the other hand, in $\sqrt{n}\;x$ the coefficient of $e_{v}$ is of course $\sqrt{n}\;x_{v}$. So $\sqrt{n}\;x_{v}=\sum_{w\in N(v)}\varepsilon_{w}x_{w}$ and hence $\sqrt{n}\left|x_{v}\right|\leq\sum_{w\in N(v)}\left|x_{w}\right|$.

This implies that some vertex of $H$ has degree at least $\sqrt{n}$. Indeed, let $v_{0}$ be a vertex such that the coefficient $x_{v}$ is largest in absolute value, i.e. $|x_{v_{0}}|\geq|x_{v}|$ for all $v\in\{0,1\}^{n}$. Then we have

[TABLE]

Thus $\left|N(v_{0})\right|\geq\sqrt{n}$, i.e. $v_{0}$ has degree at least $\sqrt{n}$, so $\Delta(H)\geq\sqrt{n}$. ∎

Proof of theorem 1.2.

If all $a_{i}$ are zero the result is immediate, so assume $a_{1}^{2}+\cdots+a_{n}^{2}>0$. We again have a nontrivial intersection of $C_{H}$ with $\operatorname{Cl}(n)\alpha_{+}=\ker\left(M_{S_{a}}-\sqrt{a_{1}^{2}+\cdots+a_{n}^{2}}\right)$, and we take $0\neq x=\sum_{w\in W}x_{w}e_{w}$ in the intersection. We have $xS_{a}=\sqrt{a_{1}^{2}+\cdots+a_{n}^{2}}\;x$, and comparing coefficients of $e_{v}$ gives $\sqrt{a_{1}^{2}+\cdots+a_{n}^{2}}\;x_{v}=\sum_{i=1}^{n}\varepsilon_{i}a_{i}x_{v(i)}$; here each $\varepsilon_{i}=\pm 1$ and the nonzero terms in this sum correspond precisely to the neighbours of $v$ in the vertex set $W$ of $H$. Taking $v_{0}$ such that $x_{v_{0}}$ is largest in absolute value we then have

[TABLE]

Now $\sum_{v_{0}(i)\in W}a_{i}$ is the weighted degree of $v_{0}$ in $W$, and we have shown it is at least $\sqrt{a_{1}^{2}+\cdots+a_{n}^{2}}$. ∎