Proof Suppose first that $E=E(\tau ;1)$ (a.e.). Then every point of E has a unique representative in the fundamental domain $2\pi I_0$ , and the functions

$$\begin{align*}(2\pi)^{-d/2}\,e^{-u(\ell\cdot\xi)}\,\chi_E(\xi), \qquad \ell\in\mathbb Z^d, \end{align*}$$

form an orthonormal basis for $L^2(E)$ , because $q(\xi )\in \mathbb C_u$ commutes with $e^{-u(\ell \cdot \xi )}$ . Thus, any $\widehat f$ supported in E admits the expansion

$$\begin{align*}\widehat f = \sum_{\ell\in\mathbb Z^d} \widehat{T_\ell\psi}\; \langle\,\widehat{T_\ell\psi}\mid \widehat f\,\rangle, \end{align*}$$

so $H_{0,E}\widehat f=\widehat f$ .

Conversely, suppose $H_{0,E}\widehat f=\widehat f$ for all $\widehat f$ supported in E but that $E(\tau ;m)$ has positive measure for some $m\ge 2$ . Let $\widetilde {E(\tau ;m)}\subset 2\pi I_0$ be its canonical representative. Since $m\ge 2$ , there exist distinct $\ell _1,\ell _2\in \mathbb Z^d$ such that the translates $\widetilde {E(\tau ;m)}+2\pi \ell _1$ and $\widetilde {E(\tau ;m)}+2\pi \ell _2$ both lie in E with positive measure.

Choose $\widehat f$ supported on $\widetilde {E(\tau ;m)}+2\pi \ell _1$ and $\widehat g$ supported on $\widetilde {E(\tau ;m)}+2\pi \ell _2$ , both nonzero. By translation orthogonality, one has

$$\begin{align*}\langle\,\widehat{T_\ell\psi}\mid \widehat f\,\rangle \perp \langle\,\widehat{T_\ell\psi}\mid \widehat g\,\rangle \qquad\text{for all }\ell\in\mathbb Z^d, \end{align*}$$

and hence the reproducing operator cannot send $\widehat f+\widehat g$ to itself, contradicting the assumed identity property of $H_{0,E}$ . Thus, no multiplicity $m\ge 2$ can occur on a set of positive measure, and $E=E(\tau ;1)$ (a.e.).

Proof All scalar multipliers appearing below lie in the slice $\mathbb C_u$ , so they commute and integrals are $\mathbb C_u$ -valued.

Fix $f\in L^2(\mathbb R^d,\mathbb H)$ . For each $\ell \in \mathbb Z^d$ write, on the Fourier side,

$$ \begin{align*} \widehat{T_\ell\psi_E}(\xi) &=(2\pi)^{-d/2} e^{-u(\ell\cdot\xi)}\chi_E(\xi) q_E(\xi), \\ \widehat{T_\ell\psi_F}(\xi) &=(2\pi)^{-d/2} e^{-u(\ell\cdot\xi)}\chi_F(\xi) q_F(\xi). \end{align*} $$

Compute the inner product (a $\mathbb C_u$ -scalar):

$$\begin{align*}c_\ell := \big\langle \widehat{T_\ell\psi_E}\mid \widehat{T_\ell\psi_F}\big\rangle = (2\pi)^{-d}\int_{\mathbb R^d} \overline{q_E(\xi)}\,e^{u(\ell\cdot\xi)}\, e^{-u(\ell\cdot\xi)}\,q_F(\xi)\, \chi_E(\xi)\chi_F(\xi)\,d\xi. \end{align*}$$

Since $e^{u(\ell \cdot \xi )}e^{-u(\ell \cdot \xi )}=1$ and $q_E,q_F\in \mathbb C_u$ , the integrand simplifies to a $\mathbb C_u$ -valued function supported on $E\cap F$ . By hypothesis, the periodized representatives of E and F are disjoint in $2\pi I_0$ ; the supports of the periodized exponentials arising from E and F are therefore disjoint modulo $2\pi \mathbb Z^d$ . Hence, the integrand vanishes almost everywhere, and $c_\ell =0$ for every $\ell $ .

With $c_\ell \equiv 0$ , we have for all f

$$\begin{align*}\sum_{\ell\in\mathbb Z^d}\big\langle f\mid \widehat{T_\ell\psi_E}\big\rangle \widehat{T_\ell\psi_F} = \sum_{\ell\in\mathbb Z^d} \widehat{T_\ell\psi_F}\, \underbrace{\big\langle \widehat{T_\ell\psi_E}\mid f\big\rangle}_{\in\mathbb C_u} = \sum_{\ell\in\mathbb Z^d} \widehat{T_\ell\psi_F}\,d_\ell, \end{align*}$$

where each scalar $d_\ell =\langle \widehat {T_\ell \psi _E}\mid f\rangle $ is $\mathbb C_u$ -valued. But since the families have disjoint periodized supports, the sum of these rank-one terms is orthogonally supported on disjoint measurable sets and therefore equals the zero function in $L^2$ . This completes the proof.

Proof We work on the Fourier side and use the identity

$$\begin{align*}\widehat{D_A^k T_\ell\psi}(\xi) = |\det A|^{-k/2}\,(2\pi)^{-d/2}\,e^{-u(\ell\cdot\xi)}\, \chi_{(A')^k E}(\xi)\,q((A')^{-k}\xi), \end{align*}$$

which follows from the dilation and translation formulas and the form of $\widehat \psi $ . Note that $q((A')^{-k}\xi )\in \mathbb C_u$ a.e. by the slice hypothesis, and $e^{-u(\ell \cdot \xi )}\in \mathbb C_u$ , so any scalar coefficients lie in $\mathbb C_u$ and commute.

Decompose

$$ \begin{align*}E(\tau;m)=\bigsqcup\limits_{j=1}^{m} E^{(j)}(\tau;m)\end{align*} $$

and observe that

$$\begin{align*}(A')^k E(\tau;m)=\bigsqcup\limits_{j=1}^{m} (A')^k E^{(j)}(\tau;m)\; (a.e.).\end{align*}$$

It suffices to prove the claimed representation on each supporting piece $(A')^k E^{(j)} (\tau ;m)$ . Fix $j,$ and for $\xi \in (A')^k E^{(j)}(\tau ;m),$ write $\eta :=(A')^{-k}\xi \in E^{(j)}(\tau ;m)\subset E$ . The inner product coefficient for a translation index $\ell $ equals

$$ \begin{align*} c_{k,\ell}(\widehat f) &=\big\langle \widehat{D_A^k T_\ell\psi}\mid \widehat f\big\rangle\\ &= |\det A|^{-k/2}(2\pi)^{-d/2} \int_{\mathbb R^d} \overline{q((A')^{-k}\zeta)} \,e^{u(\ell\cdot\zeta)}\,\chi_{(A')^k E}(\zeta)\,\widehat f(\zeta)\,d\zeta, \end{align*} $$

a $\mathbb C_u$ -valued scalar depending (right-linearly) on $\widehat f$ . Because the support of the integrand is contained in the union of the translated copies of $(A')^k\widetilde {E(\tau ;m)}$ and the partition of $E(\tau ;m)$ into the $E^{(j)}$ pieces separates these copies, the integral naturally organizes into a periodization over the lattice $2\pi (A')^{k}\mathbb Z^d$ (i.e., translates of the fundamental domain $(A')^k 2\pi I_0$ ). Equivalently, for $\xi \in (A')^k E^{(j)}(\tau ;m)$ , one can rewrite the rank-one summand as

$$ \begin{align*} &\widehat{D_A^k T_\ell\psi}(\xi)\,c_{k,\ell}(\widehat f)\\ &= \Big[\,|\det A|^{-k/2}(2\pi)^{-d/2} e^{-u(\ell\cdot\xi)} q((A')^{-k}\xi)\,\Big] \,c_{k,\ell}(\widehat f)\,\chi_{(A')^k E^{(j)}(\tau;m)}(\xi). \end{align*} $$

Summing over $\ell \in \mathbb Z^d$ , the dependence on $\ell $ in the bracketed factor produces the usual Fourier series (periodization) representation of the restriction of $H_{k,E}$ to the piece $(A')^k E^{(j)}(\tau ;m)$ . Collecting the scalar coefficients into a single $(2\pi (A')^{k}\mathbb Z^d)$ -periodic function yields the claimed formula: on $(A')^k E^{(j)}(\tau ;m)$ ,

$$\begin{align*}H_{k,E(\tau;m)}\widehat f(\xi) = F_{k,m}^{(j)}(\widehat f)(\xi)\,\chi_{(A')^k E^{(j)}(\tau;m)}(\xi), \end{align*}$$

where $F_{k,m}^{(j)}(\widehat f)$ is precisely the periodization of the component of $\widehat f$ associated with the jth translation fiber. Since each scalar factor and each periodization are $\mathbb C_u$ -valued, the functions $F_{k,m}^{(j)}$ take values in $\mathbb C_u$ and depend right-linearly on $\widehat f$ . Combining the $j=1,\dots ,m$ pieces gives the full decomposition, completing the proof.

Proof ${(1)\Rightarrow (2)}$ . Suppose the uniform partial-sum bound holds: there is $C>0$ such that for every K and all $f\in L^2(\mathbb R^d,\mathbb H)$ , $\left \|\sum \limits _{k=-K}^K H_{k,E}f\right \|_2\le C\|f\|_2$ . If (2) fails, then either the translation multiplicity $\tau _E$ or the dilation multiplicity $\#_E$ is unbounded on a set of positive measure. In that case, one may choose a nonzero f supported inside a region of arbitrarily large multiplicity so that the left-hand side $\sum \limits _\ell |\langle \phi _{k,\ell }|f\rangle |^2$ (for a suitable fixed scale k) becomes arbitrarily large while $\|f\|_2$ remains fixed, contradicting the assumed uniform bound. (This is the standard contradiction used in the scalar theory; the same construction applies here because each coefficient $\langle \phi _{k,\ell }|f\rangle $ is $\mathbb C_u$ -valued and hence its squared modulus is an ordinary nonnegative real number.) Thus, (2) must hold.

${(2)\Rightarrow (1)}$ . Assume (2) holds: there is $M\in \mathbb N$ such that both translation and dilation multiplicities are a.e. $\le M$ . By Lemma 4.5, each scale operator $H_{k,E}$ decomposes into at most M periodized components supported on the disjoint pieces $(A')^k E^{(j)}(\tau ;m)$ (for $j=1,\dots ,m$ with $m\le M$ ). On each such piece, the operator is the periodization (over the lattice $2\pi (A')^k\mathbb Z^d$ ) of a portion of the function f. The classical scalar periodization/Parseval estimate (cf. Dai–Diao–Gu–Han (2003)) yields, for each fixed scale, the scalar bound

$$\begin{align*}\|H_{k,E} f\|_2^2 \le M^{5}\,\| f\cdot\chi_{(A')^k E}\|_2^2. \end{align*}$$

(That is, at scale $k,$ the squared norm is controlled by $M^5$ times the local energy on $(A')^kE$ .)

To pass from scalar to quaternionic functions, write the right-quaternionic function f in a fixed slice decomposition $f=f_1+f_2 v$ , where $f_1$ and $f_2$ take values in the slice $\mathbb C_u$ and v is a quaternion orthogonal to $\mathbb C_u$ . Because all scalar multipliers and inner products appearing in the periodization arguments lie in $\mathbb C_u$ , the scalar estimate applies to each component $f_1,f_2$ separately; summing yields the factor $2$ and hence, for each fixed $k,$

$$\begin{align*}\|H_{k,E} f\|_2^2 \le 2\,M^{5}\,\| f\cdot\chi_{(A')^k E}\|_2^2. \end{align*}$$

Finally, combine the scale bounds. Using the orthogonality/disjointness structure of the dilated pieces (and the periodization orthogonality supplied by Lemmas 4.4 and 4.5), the contributions from different scales can be organized so that their squared norms add without a growing factor in K. Concretely, one obtains, for every integer $K\ge 0$ ,

$$\begin{align*}\Big\| \sum_{k=-K}^K H_{k,E} f\Big\|_2^2 \le 2\,M^5 \sum_{k=-K}^K \| f\cdot\chi_{(A')^k E}\|_2^2 \le 2\,M^5\|f\|_2^2, \end{align*}$$

where the last inequality uses that the sets $(A')^k E$ (over all integers k) cover each frequency point at most $\#_E(\xi )\le M$ times and hence the sum of their indicators is a.e. bounded by M, which is already accounted for in the factor $M^5$ . Thus, $\left \|\sum \limits _{k=-K}^K H_{k,E}\right \|\le \sqrt {2}\,M^{5/2}$ uniformly in K, proving (1).

Proof (i) $\Rightarrow $ (ii). The definition of $H_{E,q}$ gives

$$\begin{align*}H_{E,q}f=\sum_{k\in\mathbb Z}H_{k,E}f, \qquad H_{k,E}f=\sum_{\ell\in\mathbb Z^d} \widehat{D_A^kT_\ell\psi}\, \langle \widehat{D_A^kT_\ell\psi}\mid \widehat f\rangle. \end{align*}$$

Taking $L^2$ -norms and using Plancherel,

$$\begin{align*}\|H_{E,q}f\|_2^2 = \sum_{k\in\mathbb Z}\sum_{\ell\in\mathbb Z^d} |\langle D_A^kT_\ell\psi|f\rangle|^2. \end{align*}$$

Boundedness of $H_{E,q}$ gives the desired inequality with $C=\|H_{E,q}\|^2$ . Hence, (ii) holds.

(ii) $\Rightarrow $ (iii). Assume (iii) fails. Then at least one multiplicity function is unbounded: either $m_\tau (\xi )$ or $m_{\#}(\xi )$ is infinite or arbitrarily large on a set of positive measure.

Case 1: Unbounded translation multiplicity. Choose $S\subset E$ with positive measure such that $m_\tau (\xi )\ge m$ on S for arbitrarily large integers m. For each such m, select m disjoint translation fibers $S+2\pi \ell _j\subset E$ . Define a function $\widehat f$ supported on the union of these fibers with equal $L^2$ -mass on each piece. Then each $\langle \widehat {T_{\ell _j}\psi }\mid \widehat f\rangle $ contributes the same nonzero value, and the left-hand side of (ii) contains at least m equal nonzero terms. Letting $m\to \infty $ contradicts the boundedness asserted by (ii).

Case 2: Unbounded dilation multiplicity. The argument is identical, using dilates $(A')^{n_j}S$ and applying scale atoms $D_A^{n_j}T_\ell \psi $ . Again one produces arbitrarily many nonzero coefficients, contradicting (ii).

Thus, both multiplicities must be bounded a.e., proving (iii).

(iii) $\Rightarrow $ (i). Here, we use Lemma 4.6 in full. Condition (iii) gives a global multiplicity bound M such that

$$\begin{align*}m_\tau(\xi)\le M,\qquad m_{\#}(\xi)\le M\quad\text{a.e.} \end{align*}$$

Then, Lemma 4.6 asserts that, for every integer K and every f,

$$\begin{align*}\Big\|\sum_{k=-K}^{K} H_{k,E}\, f\Big\|_2^2 \;\le\; 2M^{5}\,\|f\|_2^2. \end{align*}$$

The Fourier-side sum $\sum _{k\in \mathbb Z} H_{k,E}f$ converges in $L^2$ because the partial sums form a Cauchy sequence:

$$\begin{align*}\Big\|\sum_{k=-K}^{K}H_{k,E}f-\sum_{k=-L}^{L}H_{k,E}f\Big\|_2^2 \le 2M^{5}\big\|f\cdot\chi_{\;\bigcup_{|k|>\min(K,L)}(A')^{k}E}\big\|_2^2 \longrightarrow 0, \end{align*}$$

as $K,L\to \infty $ , since the union of dilates exhausted by $(A')^k E$ has finite overlap number M. Therefore, the full series converges to a bounded operator with

$$\begin{align*}\|H_{E,q}\|\;\le\;\sqrt{2}\,M^{5/2}. \end{align*}$$

This proves (i).

Proof We identify the frame operator of the affine system with the global reproducing operator $H_{E,q}$ developed in Lemmas 4.3–4.6. Throughout, all frequency-side scalar factors lie in $\mathbb {C}_u$ , so commutativity holds and the arguments from the complex setting apply without modification at the level of scalar algebra.

(ii) $\Rightarrow $ (i). Assume that E is translation-simple and that the dilates $\{(A')^nE\}_{n\in \mathbb {Z}}$ form an a.e.-disjoint tiling of $\mathbb {R}^d$ .

Translation simplicity implies, by Lemma 4.3, that the family $\{\widehat {T_\ell \psi }\}_{\ell \in \mathbb {Z}^d}$ forms an orthonormal basis of $L^2(E;\mathbb {C}_u)$ (after the usual slice-space identification). Hence, for any $\widehat {g}$ supported in E, we have a Parseval identity:

(4.1) $$ \begin{align} \widehat{g} = \sum_{\ell\in\mathbb{Z}^d} \widehat{T_\ell\psi}\, \langle \widehat{T_\ell\psi} \mid \widehat{g} \rangle, \qquad \|\widehat{g}\|_2^2 = \sum_{\ell\in\mathbb{Z}^d} |\langle \widehat{T_\ell\psi} \mid \widehat{g} \rangle|^2. \end{align} $$

Next, since $\mathbb {R}^d=\bigsqcup \limits _{n\in \mathbb {Z}} (A')^nE$ a.e., every $\widehat {f}\in L^2$ decomposes orthogonally as

(4.2) $$ \begin{align} \widehat{f} = \sum_{n\in\mathbb{Z}} \widehat{f}_n, \qquad \widehat{f}_n := \widehat{f}\,\chi_{(A')^nE}. \end{align} $$

Using Lemma 4.5, each piece $\widehat {f}_n$ is unitarily equivalent (via the slice-QFT dilation map) to a function supported in E, and the corresponding coefficients $\langle f, D_A^nT_\ell \psi \rangle $ are exactly the Fourier coefficients appearing in the Parseval expansion (3.1). Therefore,

(4.3) $$ \begin{align} \sum_{\ell\in\mathbb{Z}^d} |\langle f, D_A^nT_\ell\psi\rangle|^2 = \|\widehat{f}_n\|_2^2. \end{align} $$

Summing (3.3) over $n\in \mathbb {Z}$ and using the orthogonality in (3.2), we obtain a global Parseval identity:

$$\begin{align*}\sum_{n\in\mathbb{Z}}\sum_{\ell\in\mathbb{Z}^d} |\langle f| D_A^nT_\ell\psi\rangle|^2 = \|\widehat{f}\|_2^2 = \|f\|_2^2. \end{align*}$$

Thus, $\mathcal {A}(\psi )$ is a tight frame with frame bound $B=1$ .

(i) $\Rightarrow $ (ii). Assume now that $\mathcal {A}(\psi )$ is tight with frame bound B. Its frame operator S satisfies

(4.4) $$ \begin{align} Sf = Bf, \qquad f\in L^2(\mathbb{R}^d,\mathbb{H}). \end{align} $$

Taking Fourier transforms and using Plancherel,

(4.5) $$ \begin{align} \widehat{Sf}(\xi) = B\,\widehat{f}(\xi). \end{align} $$

On the other hand, by construction and Lemma 4.6, the operator S coincides with the global reproducing operator $H_{E,q}$ , whose Fourier multiplier is determined solely by the translation and dilation multiplicities of E. More precisely,

(4.6) $$ \begin{align} \widehat{Sf}(\xi) = m(\xi)\,\widehat{f}(\xi), \end{align} $$

where $m(\xi )$ counts, with unit weight, the number of scales n such that $\xi \in (A')^nE$ (this follows from the Parseval identities of the translation slices combined with the disjointness properties encoded in Lemma 4.4). Comparing (3.5) and (3.6) for all f yields

(4.7) $$ \begin{align} m(\xi) = B \quad\text{for a.e.\ } \xi\in\mathbb{R}^d. \end{align} $$

We now determine $m(\xi )$ . If E fails to be translation-simple on a set of positive measure, Lemma 4.3 shows that the scale-zero reproducing operator cannot act as the identity on its support; this contradicts the tight identity (3.4). Hence,

(4.8) $$ \begin{align} E = E(\tau;1)\quad\text{a.e.} \end{align} $$

Next, if $\xi $ lies in two distinct dilates $(A')^{n_1}E$ and $(A')^{n_2}E$ with $n_1\neq n_2$ , then ${m(\xi )\ge 2}$ , contradicting (3.7) unless $B\ge 2$ . But testing (3.4) on f supported in a single dilate $(A')^nE$ and using the scale-wise expansion of Lemma 4.5 forces $B=1$ . Therefore, every $\xi $ belongs to exactly one dilate:

(4.9) $$ \begin{align} \#\{\,n\in\mathbb{Z}: \xi\in(A')^nE\,\} = 1 \quad\text{a.e.,} \end{align} $$

which asserts that $\{(A')^nE: n\in \mathbb {Z}\}$ forms an a.e.-disjoint tiling of $\mathbb {R}^d$ . This establishes the second part of (ii).