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Rado's Graph has no Quantum Symmetry

Husam Ismaeel

TL;DR

This paper proves that the Rado graph $R$ admits no quantum symmetry by showing that every quantum automorphism yields commuting generator projections, eliminating nonclassical quantum automorphisms. The argument leverages the graph's extension property via carefully defined projections $p_\alpha$, $q_\beta$ and partitions $V_{\alpha\beta}$ to force commutativity of the quantum permutation entries. Consequently, the automorphism algebra is classical, and $R$ has only trivial (classical) symmetries in the quantum sense. The result advances the understanding of quantum symmetries in countable homogeneous graphs and connects graph-theoretic extension properties to operator-algebraic rigidity.

Abstract

We prove that Rado's graph admits no quantum symmetries.

Rado's Graph has no Quantum Symmetry

TL;DR

This paper proves that the Rado graph admits no quantum symmetry by showing that every quantum automorphism yields commuting generator projections, eliminating nonclassical quantum automorphisms. The argument leverages the graph's extension property via carefully defined projections , and partitions to force commutativity of the quantum permutation entries. Consequently, the automorphism algebra is classical, and has only trivial (classical) symmetries in the quantum sense. The result advances the understanding of quantum symmetries in countable homogeneous graphs and connects graph-theoretic extension properties to operator-algebraic rigidity.

Abstract

We prove that Rado's graph admits no quantum symmetries.

Paper Structure

This paper contains 3 sections, 2 theorems, 16 equations.

Table of Contents

  1. Introduction
  2. Preliminaries
  3. Main Result

Key Result

Lemma 3.1

Let $(H,u)$ be a quantum automorphism of $R$, let $P_0, P_1, Q_0, Q_1$ be finite subsets of $V_R$ such that $(P_0 \cup Q_0) \cap(P_1 \cup Q_1) = \varnothing$, and let $y, t \in V_R$ be two distinct vertices. Consider the projections Then for any $v \in p_1(H)$ there exist vectors $v_0 \in p_0^\perp(H) \cap q_0^\perp(H), v_1 \in p_0^\perp(H) \cap q_1^\perp(H)$ such that $v = v_0 + v_1$ and $\lan

Theorems & Definitions (6)

  • Definition 2.1
  • Definition 2.2
  • Lemma 3.1
  • proof
  • Theorem 3.1
  • proof