Ergodic translation action of the ring of integers

Statement

Fix $\omega\in K_\infty^d$ . The maps

S_b([x])=[x+b\omega]\qquad(b\in\mathcal O_K)

form a Haar-preserving action of the additive group $(\mathcal O_K,+)$ on $(K_\infty/\mathcal O_K)^d$ . This action is ergodic if and only if

k\cdot\omega\notin\mathcal O_K\qquad(0\ne k\in\mathcal O_K^d).

Equivalently, $1,\omega_1,\ldots,\omega_d$ are linearly independent over the diagonally embedded field $K$ .

Proof using characters

A character indexed by $a\in(\mathcal O_K^\vee)^d$ is fixed by every $S_b$ exactly when

\operatorname{Tr}_\infty(b(a\cdot\omega))\in\mathbb Z \quad\text{for every }b\in\mathcal O_K,

or $a\cdot\omega\in\mathcal O_K^\vee$ . The latter equivalence also holds for elements initially in $K_\infty$ : expansion in the trace-dual basis forces integral coefficients.

If a nonzero such $a$ exists, choose $N$ clearing the codifferent and put $k=Na$ . Then $k\in\mathcal O_K^d\setminus\{0\}$ and $k\cdot\omega\in\mathcal O_K$ . Conversely any such $k$ is already a codifferent frequency with codifferent pairing. Fourier completeness proves the criterion. Clearing rational denominators in a $K$ -linear relation proves the final equivalence.

Scope

These are translations indexed by the additive group, not multiplication maps $[x]\mapsto[bx]$ . When $K=\mathbb Q$ , this action is iteration of one translation. For higher-degree fields it involves several commuting generators.