Proof. As above, let $K_p$ be a splitting field for the characteristic polynomial of A over $\mathbb {Q}_p$ , and let $v_1,\ldots , v_d$ be an eigenbasis of $K_p^d$ for A. For every $v=\sum \alpha _i v_i \in \mathbb {Q}_p^d$ define $\lVert v\rVert = \max _i\{\lvert \alpha _i\rvert _{p}\}$ . It is straightforward to verify that this defines a norm on $\mathbb {Q}_p^d$ with the desired properties.

Proof. As before, for every $p\in S$ we let $K_p$ be a splitting field for the characteristic polynomial of A over $\mathbb {Q}_p$ , and let $\unicode{x3bb} _{p,1},\ldots ,\unicode{x3bb} _{p,r_p}$ be the eigenvalues of A inside $K_p$ with $\lvert \unicode{x3bb} _{p,i}\rvert _p> 1$ . We claim that such eigenvalues exist (that is, $r_p>0$ for some p). Let $\ell \in \mathbb {N}$ be the least common multiple of the denominators of the coefficients of the characteristic polynomial of A. If $\ell = 1$ then the polynomial has integer coefficients, and since we assume no eigenvalue of A is a root of unity, a well-known theorem of Kronecker implies that $\lvert \unicode{x3bb} \rvert _\infty> 1$ for some eigenvalue (we mention that, in contrast to the case of an integer matrix, a rational matrix may satisfy $E_\infty ^s=E_\infty ^u=0$ ; see, for example, [Reference LindLin82b]) $\unicode{x3bb} \in \mathbb {C}$ . If $\ell> 1$ , then the following stronger assertion is proven in [Reference Lind and WardLW88, pp. 412, 416]:

$$ \begin{align*}\prod_{p\in S\setminus\{\infty\}}\prod_{i=1}^{r_p} \lvert \unicode{x3bb}_{p,i}\rvert_p = \ell,\end{align*} $$

so in particular $r_p> 0$ for some p. For this p, we have $\widetilde {E_p^u}\ne 0$ (equation (5.1)) and hence also $E_p^u\ne 0$ and $E^u\ne 0$ .

Proof. For a subgroup H of $\mathbb {Q}_S^d$ , denote by $H^\bot $ its annihilator in $\widehat {\mathbb {Q}_S^d}$ . By duality theory, it suffices to show that $(E^u + \Delta _S^d)^\bot =0$ . Since $(\Delta _S^d)^\bot = \widehat {X_S^d}$ , we have, as stated in equation (2.2),

(5.8) $$ \begin{align} (\Delta_S^d)^\bot = \bigg\{(-x,x,\ldots,x):x\in \mathbb{Z}\bigg[\frac{1}{S}\bigg]^d\bigg\}\subseteq\mathbb{R}^d\times\prod_{p\in S\setminus\{\infty\}}\mathbb{Q}_p^d \simeq \widehat{\mathbb{Q}_S^d}. \end{align} $$

Moreover, writing $H^{\bot p}$ for the annihilator of a subgroup $H\le \mathbb {Q}_p^d$ in $\widehat {\mathbb {Q}_p^d}$ , we have

(5.9) $$ \begin{align} (E^u)^\bot = \prod_{p\in S} (E_p^u)^{\bot p}. \end{align} $$

Apply Lemma 5.3 to choose some $p\in S$ with $E_p^u\ne 0$ . Then $(E_p^u)^{\bot p}\cap \mathbb {Q}^d$ is an $A^\top $ -invariant subspace of $\mathbb {Q}^d$ , and since the characteristic polynomial of A is irreducible over $\mathbb {Q}$ , it must be either $0$ or $\mathbb {Q}^d$ . The latter case is impossible since $E_p^u\ne 0$ , and therefore $(E_p^u)^{\bot p}\cap \mathbb {Q}^d = 0$ . Combining this with equations (5.8) and (5.9), we conclude that

$$ \begin{align*} (E^u + \Delta_S^d)^\bot = (E^u)^\bot\cap (\Delta_S^d)^\bot = 0, \end{align*} $$

and the proof is complete.

Proof. Let $\varepsilon>0$ . Since $\Delta _S^d$ is cocompact, the last lemma implies that there exists $R>0$ such that $B^u(R) + \Delta _S^d$ is $\varepsilon $ -dense in $\mathbb {Q}_S^d$ . It follows from equation (5.7) (with $n=0$ ) that

$$ \begin{align*}\mathbb{Q}_S^d = B^u(R) + \Delta_S^d + B(\varepsilon) \subseteq B^u(R+\varepsilon) + B^s(\varepsilon) + B^c(\varepsilon) + \Delta_S^d.\end{align*} $$

By equation (5.6), $B^u(R+\varepsilon ) \subseteq A^n {(}B^u(\varepsilon ){)}$ for every sufficiently large $n\in \mathbb {N}$ , and the assertion follows.

Proof. Let $0<\delta <\varepsilon $ be sufficiently small and then let $N\in \mathbb {N}$ correspond to $\delta $ as in Corollary 5.5. Next, let $n\in \mathcal {P}$ be sufficiently large. The precise meaning of ‘sufficiently small’ and ‘sufficiently large’ will be clarified in the course of the proof. To begin with, we require that $n\delta \ge N$ . Let $x\in X_S^d$ and choose $x'\in \mathbb {Q}_S^d$ which projects to x. By the choice of N, we can write $x'-A^n x' = v + w$ , where $w\in \Delta _S^d$ and

(5.10) $$ \begin{align} v = v_u + v_s + v_c \in A^{\lfloor n\delta\rfloor} {(}B^u(\delta){)} + B^s(\delta) + B^c(\delta). \end{align} $$

We claim that $(I-A^n)^{-1}v\in B(\varepsilon , n - \lfloor n\varepsilon \rfloor )$ (where I is the identity matrix and $(I-A^n)$ is invertible since no eigenvalue of A is a root of unity; see also equation (5.5)). Let us first indicate how to complete the proof, assuming this holds. Set $y' = (I-A^n)^{-1} w\in \mathbb {Q}_S^d$ . Then $A^n y' = y' - w$ , which means that $y'$ projects to a point of period n in $X_S^d$ . Furthermore,

$$ \begin{align*}x' - y' = (I-A^n)^{-1}v\in B(\varepsilon, n-\lfloor n\varepsilon\rfloor),\end{align*} $$

and hence the projection of $y'$ to $X_S^d$ has the desired properties.

We are left to prove that $(I-A^n)^{-1}v\in B(\varepsilon , n - \lfloor n\varepsilon \rfloor )$ . By equation (5.7) it suffices to show that

(5.11) $$ \begin{align} (I-A^n)^{-1}v_s \in B^s(\varepsilon),\quad (I-A^n)^{-1}v_c \in B^c(\varepsilon),\quad (I-A^n)^{-1}v_u\in A^{-n + \lfloor n\varepsilon\rfloor}B^u(\varepsilon). \end{align} $$

Note that $A^m y \underset {m\to \infty }{\to }0$ uniformly on $B^s(\delta )$ , and hence if $\delta $ is small enough and n is large enough then $(I-A^n)^{-1}v_s \in B^s(\varepsilon )$ . By the same reasoning, $(A^{-n}-I)^{-1}B^u(\delta )\subseteq B^u(\varepsilon )$ . Thus,

$$ \begin{align*}(I-A^n)^{-1}A^{\lfloor n\delta\rfloor}{(}B^u(\delta){)} = A^{-n + \lfloor n\delta\rfloor}(A^{-n}-I)^{-1}{(}B^u(\delta){)}\subseteq A^{-n + \lfloor n\delta\rfloor}{(}B^u(\varepsilon){)},\end{align*} $$

and in particular by equation (5.10) (and also equation (5.6), replacing $n\delta $ with $n\varepsilon $ ), $(I-A^n)^{-1}v_u\in A^{-n + \lfloor n\varepsilon \rfloor }{(}B^u(\varepsilon ){)}$ .

We now proceed to prove the final assertion of equation (5.11), concerning $B^c(\varepsilon )$ . For every $p\in S$ , let $K_p$ be the splitting field of f over $\mathbb {Q}_p$ that appears in the definition of $\mathcal {P}$ (Definition 5.6), and $\unicode{x3bb} _{p,1},\ldots ,\unicode{x3bb} _{p,r_p}\in K_p$ the unimodular roots of f. Assume first that $r_p> 0$ . Since we assume f is irreducible over $\mathbb {Q}$ , it has no multiple roots, and hence we can choose $M_p\in \operatorname {\mathrm {GL}}_{r_p}(K_p)$ that diagonalizes ${A}|_{E^c}$ . Then

$$ \begin{align*}(I-{A^n}|_{E^c})^{-1}=M_p^{-1} \begin{pmatrix} (1-\unicode{x3bb}_{p,1}^n)^{-1} \\ & \ddots \\ & & (1-\unicode{x3bb}_{p,r_p}^n)^{-1} \end{pmatrix}M_p\end{align*} $$

(cf. [Reference Einsiedler and LindenstraussEL22, Proposition 2.2], [Reference BerendBer84]). By the definition of $\mathcal {P}$ ,

$$ \begin{align*}\sup_{\substack{1\le i\le r_p \\ m\in\mathcal{P}}}\lvert (1-\unicode{x3bb}_{p,i}^m)^{-1}\rvert_p<\infty,\end{align*} $$

and thus the collection of matrices $\{(I-{A^m}|_{E^c})^{-1}:m\in \mathcal {P}\}$ is contained in a compact subset of matrices over $K_p$ . Denote $B_p^c(\delta ) = \{z\in E_p^c:\lVert z\rVert _p<\delta \}$ . Then for sufficiently small $\delta $ (independent of n), we have $(I-A^n)^{-1}{(}B_p^c(\delta ){)} \subseteq B_p^c (\varepsilon )$ . If $r_p = 0$ , then $E_p^c = 0$ and the last inclusion holds trivially. Since this inclusion holds for all $p\in S$ , we conclude that $(I-A^n)^{-1}{(}B^c(\delta ){)} \subseteq B^c (\varepsilon )$ , verifying equation (5.11) and completing the proof.

Proof. Let $\varepsilon>0$ and set $\delta = (1-\rho ^{-1})\varepsilon $ , where $\rho>1$ is as in equation (5.6). Apply Corollary 5.5 to find $M\in \mathbb {N}$ such that for every $m\ge M$ ,

(5.12) $$ \begin{align} \mathbb{Q}_S^d= A^m (B^u(\delta)) + B^s(\delta) + B^c(\delta) + \Delta_S^d. \end{align} $$

Denote by $\pi :\mathbb {Q}_S^d\to X_S^d$ the canonical projection, and for every $x \in X_S^d$ and $n\in \mathbb {N}_0$ consider the subsets of $X_S^d$ defined by

$$ \begin{align*}\mathcal{C}(x,n) = A^n{(}(x+ \pi(B^u(\delta)){)},\quad \mathcal{D}(x,n) = x + \pi(B(\delta,n))\end{align*} $$

(see equation (5.5)). We claim that for every $x,y\in X_S^d$ and every $n,m\in \mathbb {N}$ with $m\ge M$ we have $\mathcal {C}(x,m)\cap \mathcal {D}(y, n)\ne \emptyset $ . Indeed, choose preimages $x',y'\in \mathbb {Q}_S^d$ of $x,y$ . By equation (5.12), there exist $z_s\in B^s(\delta )$ and $z_c\in B^c(\delta )$ such that

$$ \begin{align*}y'-A^mx' + z_s + z_c \in A^m (B^u(\delta)) + \Delta_S^d.\end{align*} $$

Projecting the last equation to $X_S^d$ shows that $y + \pi (z_s + z_c) \in \mathcal {C}(x,m)$ , and equation (5.7) implies that also $y + \pi (z_s + z_c) \in \mathcal {D}(y,n)$ .

Let $\{(x_i; a_i,b_i)\}_{i=1}^r$ be an M-spaced specification in $X_S^d$ . We now define a sequence of points $y_1,\ldots ,y_r\in X_S^d$ by induction on $k\le r$ , such that for every $1\le i \le k$ and $a_i\le n < b_i$ ,

$$ \begin{align*}d_{X_S^d}(A^n x_i, A^n y_k) < \delta S_{k-i}\quad\text{where}\ S_\ell = \sum_{j=0}^{\ell}\rho^{-j}.\end{align*} $$

This will conclude the proof, since $y_r$ will satisfy $d_{X_S^d}(A^n x_i, A^n y_r)< \delta (1-\rho ^{-1})^{-1}=\varepsilon $ . For $k=1$ we can simply take $y_1 = x_1$ . Assume now $y_k$ has been defined. Since $a_{k+1}-b_k \ge M$ , we have

(5.13) $$ \begin{align} \mathcal{C}(A^{b_k}y_k, a_{k+1}-b_k)\cap \mathcal{D}(A^{a_{k+1}}x_{k+1}, b_{k+1}-a_{k+1})\ne\emptyset, \end{align} $$

so we may choose a point $y_{k+1}$ such that $A^{a_{k+1}}y_{k+1}$ lies in the above intersection. The membership in the latter set implies that for every $a_{k+1}\le n < b_{k+1}$ ,

$$ \begin{align*}d_{X_S^d}(A^n x_{k+1}, A^n y_{k+1}) < \delta.\end{align*} $$

Let $1\le i\le k$ and $a_i\le n < b_i$ . Unpacking the definition of the former set in equation (5.13) and applying $A^{-(a_{k+1}-n)}$ to both sides yields

$$ \begin{align*} A^n y_{k+1} \in A^n y_k + A^{-(b_k-n)}\pi(B^u(\delta))\subseteq A^n y_k + \pi(B^u(\rho^{-(b_k-n)}\delta)),\end{align*} $$

where we used equation (5.6) to obtain the last inclusion. Note that $b_k - n \ge b_k-b_i + 1\ge k - i + 1$ , and thus from the induction hypothesis and the last equation we obtain

$$ \begin{align*} d_{X_S^d}(A^nx_i,A^n y_{k+1})&\le d_{X_S^d}(A^n x_i, A^n y_k) + d_{X_S^d}(A^n y_k, A^n y_{k+1}) \\ &< \delta S_{k-i} + \delta\rho^{-(k+1-i)}=\delta S_{k+1-i}, \end{align*} $$

completing the construction of the sequence $(y_k)$ and the proof.

Proof of Proposition 5.7.

Let $\varepsilon>0$ , and we may assume that $\varepsilon <1$ . Let $M\in \mathbb {N}$ be provided by Proposition 5.9 for $\varepsilon /2$ , and $N\in \mathbb {N}$ large enough so that Proposition 5.8 holds for $\varepsilon /2$ and for every $n\in \mathcal {P}$ with $n\ge N$ . Suppose we are given an M-spaced specification $\xi =\{(x_i;a_i,b_i)\}_{i=1}^r$ in $X_S^d$ and some $n\in \mathcal {P}$ such that $n\ge \max \{N,(1+\varepsilon )b_r\}$ . By Proposition 5.9, there exists $x\in X_S^d$ that $\varepsilon /2$ -partially traces $\xi $ . Next we use Proposition 5.8 to find $y\in X_S^d$ of period n such that $d_{X_S^d}(A^ix,A^iy)<\varepsilon /2$ for every $0\le i \le (1-\varepsilon /2)n$ . But $(1-\varepsilon /2)n\ge (1-\varepsilon /2)(1+\varepsilon )b_r\ge b_r$ , and it follows that $\xi $ is $\varepsilon $ -partially traced by y.

Proof. Let $c\in \mathbb {N}$ . It suffices to construct an increasing sequence $(n_k)_{k=1}^\infty $ in $\mathbb {N}$ with bounded gaps, such that

(5.14) $$ \begin{align} \inf_{\substack{1\le i\le r \\ k\in\mathbb{N}}}d_G(g_i^{cn_k}, 1)> 0, \end{align} $$

since we can then set $\mathcal {P} = \{cn_k :k\in \mathbb {N}\}$ . For $\delta>0$ denote by $B(\delta )\subseteq G$ the open ball of radius $\delta $ centered at $1$ . As the elements $g_i$ have infinite order, we can choose $\delta> 0$ such that for every $i=1,\ldots ,r$ and $m=1,\ldots ,r+1$ ,

(5.15) $$ \begin{align} B(\delta)\cap g_i^{cm}B(\delta)=\emptyset. \end{align} $$

In particular, for every $n\in \mathbb {N}$ and $i=1,\ldots ,r$ ,

(5.16) $$ \begin{align} {\lvert}\{1\le m \le r+1 : g_i^{cm+cn}\in B(\delta)\}{\rvert} \le 1. \end{align} $$

We construct a sequence $(n_k)$ with bounded gaps inductively so that for every $i=1,\ldots ,r$ , we have $g_i^{cn_k}\notin B(\delta )$ . By equation (5.15) we may take $n_1=1$ . Suppose $n_k$ has been defined. Then, by equation (5.16), for each $i=1,\ldots ,r$ , there exists at most one integer $1 \le m_i \le r + 1$ such that $g_i^{c(n_k+m_i)} \in B(\delta )$ . Hence, we can choose an integer $1 \le m \le r+1$ with $m\ne m_1,\ldots , m_r$ , and set $n_{k+1} = n_k + m$ , completing the construction. It follows that the left-hand side of equation (5.14) is bounded below by $\delta $ .

Proof. As in Definition 5.6, for every $p\in S$ let $K_p$ be a splitting field of f over $\mathbb {Q}_p$ , with the absolute value $\lvert \;\rvert _{p}$ . Denote by $\unicode{x3bb} _{p,1},\ldots ,\unicode{x3bb} _{p,r_p}\in K_p$ the roots of f which belong to the unit group $G_p:= \{x\in K_p: \lvert x\rvert _{p} = 1\}$ . Set $G:= \prod _{p\in S} G_p$ and define a G-invariant metric on G by $d((x_p)_p,(y_p)_p) = \max _p \{\lvert x_p - y_p\rvert _{p}\}$ . Applying the previous lemma to the group G and the images of the elements $\{\unicode{x3bb} _{p,i}:p\in S, i=1,\ldots ,r_p\}$ under the canonical embeddings $G_p \hookrightarrow G$ yields the desired set.

Proof. Fix metrics $d_X$ and $d_Y$ on X and Y, and let $\phi : X\to Y$ be the continuous factor map. Let $\varepsilon>0$ and choose $0<\delta <\varepsilon $ such that for every $x_1,x_2\in X$ ,

$$ \begin{align*}d_X(x_1,x_2)<\delta \Longrightarrow d_Y(\phi(x_1), \phi(x_2))<\varepsilon.\end{align*} $$

Let $N,M\in \mathbb {N}$ be the parameters provided by the partial specification for X, corresponding to $\delta $ (Definition 4.2). Given an M-spaced specification $\xi =\{(y_i;a_i,b_i)\}_{i=1}^r$ in Y, we can choose preimages $x_i\in \phi ^{-1}(y_i)$ and lift $\xi $ to an M-spaced specification $\xi '=\{(x_i;a_i,b_i)\}_{i=1}^r$ in X. Thus, if $\mathcal {P}\ni n\ge \max \{N, (1+\varepsilon )b_r\}$ we can find $x\in X$ of period n which $\delta $ -partially traces $\xi '$ , and $\phi (x)$ will $\varepsilon $ -partially trace $\xi $ .

Proof of Theorem 1.2.

Pick an ergodic abelian group dynamical system $(X',T')$ , of which $(X,T)$ is a continuous factor, and $T'$ -invariant closed subgroups $0=X_n\le \cdots \le X_0 = X'$ as described in Proposition 3.1. For every quotient $(X_k/X_{k+1}, T')$ not isomorphic to a Bernoulli shift, apply case (2) of Proposition 3.1 to select suitable $S_k\subseteq \mathbb {N}\cup \{\infty \}$ and $A_k \in \operatorname {\mathrm {GL}}_{d_k}(\mathbb {Q})$ such that $(X_k/X_{k+1}, T')$ is a continuous factor of the $S_k$ -adic solenoid $(X_{S_k}^{d_k}, A_k)$ . Let $f_k$ denote the characteristic polynomial of $A_k$ , and set $S=\bigcup _k S_k$ and $f=\prod _k f_k$ .

Fix some $c\in \mathbb {N}$ , and apply Corollary 5.11 to find $\mathcal {P}\subseteq c\mathbb {N}$ which is bounded below for f and S. Clearly, $\mathcal {P}$ is also bounded below for every pair $f_k,S_k$ , and hence by Proposition 5.7 together with Lemma 5.12, $(X_k/X_{k+1}, T')$ satisfies partial specification with periods $\mathcal {P}$ . By Example 4.4, any quotient $(X_j/X_{j+1}, T')$ that is isomorphic to a Bernoulli shift also satisfies partial specification with periods $\mathcal {P}$ . Proposition 4.5 then implies that $(X',T')$ satisfies partial specification with periods $\mathcal {P}$ , and a further application of Lemma 5.12 shows that $(X,T)$ does as well. Since c was arbitrary, $(X,T)$ satisfies partial specification.

Proof. Many periodic specification-like properties are known to imply this assertion, with the details of the proof adapted to the specific situation; see, for example, [Reference MarcusMar80, Reference SigmundSig70] and [Reference Gelfert and KwietniakGK18, Theorem 1.1(1)]. Our case is no different, and the argument proceeds in essentially the same way. For this reason we only sketch the proof and omit the technical details.

Fix a metric d on $\mathcal {M}_T(X)$ . Let $\mu \in \mathcal {M}_T(X)$ and $\varepsilon>0$ . Then there exist ergodic measures $\mu _1,\ldots ,\mu _k\in \mathcal {M}_T(X)$ and a convex combination $\sum _{i=1}^k t_i\mu _i$ such that $d(\mu , \sum t_i\mu _i)<\varepsilon $ . For $i=1,\ldots , k$ , let $x_i\in X$ be a $\mu _i$ -generic point. Pick $N\in \mathbb {N}$ such that for every i and $n\ge N$ , $d(\mu _i,\mathfrak {m}_{T,x_i,n})<\varepsilon $ . Consider the orbit segments $x_i,\ldots ,T^{n_i}x_i$ , where $n_i\ge N$ and the ratio ${n_i}/(n_1+\cdots +n_k)$ approximates its weight $t_i$ . Using these segments, we construct a suitably spaced specification, and apply the partial specification of $(X,T)$ to obtain a periodic point $x\in X$ whose orbit approximates them, except on a negligible set. The T-invariant measure supported on the finite orbit of x then approximates $\mu $ .

Proof. Consider the group $G = \operatorname {\mathrm {SL}}_d(\mathbb {R})\ltimes \mathbb {R}^d$ (with the standard action of $\operatorname {\mathrm {SL}}_d(\mathbb {R})$ on $\mathbb {R}^d$ ) and its lattice $\Gamma = \operatorname {\mathrm {SL}}_d(\mathbb {Z})\ltimes \mathbb {Z}^d$ . We adopt the convention that for $(A,x), (B,y)\in G$ , multiplication is given by $(A,x)(B,y)=(AB,x+Ay)$ , and denote the identity matrix by $1$ . The map $\phi : \mathbb {T}^d \to G/\Gamma $ defined for $x\in \mathbb {R}^d$ by $\phi (x+\mathbb {Z}^d) = (1,x)\Gamma $ is a homeomorphism on its image

$$ \begin{align*}\phi(\mathbb{T}^d) = \{(1, x)\Gamma :x \in \mathbb{R}^d\} \subseteq G/\Gamma.\end{align*} $$

Moreover, for $A\in \operatorname {\mathrm {SL}}_d(\mathbb {Z})$ and $x\in \mathbb {R}^d$ we have

$$ \begin{align*}(A,0)\phi(x+\mathbb{Z}^d) = (A, Ax)\Gamma = (1, Ax)(A,0)\Gamma = \phi(Ax+\mathbb{Z}^d),\end{align*} $$

so $\phi $ is $\operatorname {\mathrm {SL}}_d(\mathbb {Z})$ -equivariant. In particular, given an ergodic measure $\mu \in \mathcal {M}_{U}(\mathbb {T}^d)$ , we obtain an ergodic U-invariant measure $\nu := \phi _*\mu $ on $\phi (\mathbb {T}^d)$ . By Shah’s theorem [Reference ShahSha98, Theorem 1.1], which generalizes Ratner’s classical measure classification theorem, there exist a subgroup $S< G$ and $x\in \mathbb {R}^d$ such that $\nu $ is the S-invariant measure supported on the orbit $S(1,x)$ . Identifying $\mathbb {R}^d$ with its image in G, we define

$$ \begin{align*} \widetilde{H} = \{g\in \mathbb{R}^d : g_*\nu = \nu\}.\end{align*} $$

$\mathbb {R}^d$ is a normal subgroup of G, and it follows that S normalizes $\widetilde {H}$ . Replace S with the subgroup of G generated by S and $\widetilde {H}$ ; then $\widetilde {H} = S\cap \mathbb {R}^d$ is a normal subgroup of S. Next, note that for every $y\in \mathbb {R}^d$ and non-integer matrix $A \in \operatorname {\mathrm {SL}}_d(\mathbb {R})$ , $(A,y)\phi (\mathbb {T}^d)$ is disjoint from $\phi (\mathbb {T}^d)$ , and hence $S \subseteq \operatorname {\mathrm {SL}}_d(\mathbb {Z})\ltimes \mathbb {R}^d$ . Consequently, $S/\widetilde {H}$ can be identified with a subgroup of $\operatorname {\mathrm {SL}}_d(\mathbb {Z})$ .

We are now in a position to apply Lemma 6.2, stated below, which asserts that $\nu $ is supported on finitely many $\widetilde {H}$ -orbits. Thus, applying $\phi ^{-1}$ shows that the original measure $\mu $ on $\mathbb {T}^d$ is supported on finitely many cosets of $H'$ , where $H'$ is the closure of the projection of $\widetilde {H}$ to $\mathbb {T}^d$ . In addition, the $\widetilde {H}$ -invariance of $\nu $ implies that the restriction of $\mu $ to each of those cosets is $H'$ -invariant, and therefore must be a translate of the Haar measure on $H'$ . Finally, we define H to be the identity connected component of $H'$ . $\mu $ remains a translate of Haar measure on each of the resulting finitely many cosets of H. That the cosets can be written as $a+H,\ldots ,U^{m-1} a+H$ follows from the fact that U acts transitively on them, as $\mu $ is ergodic. Finally, note that H must be a proper subgroup of $\mathbb {T}^d$ since the Haar measure on $\mathbb {T}^d$ is not U-ergodic.

Proof. S acts on the orbit space $H\setminus X$ : for $s\in S$ and an orbit $Hx$ we have $sH.x=Hs.x$ . The transitivity of the S-action on X implies that this action is transitive as well. Therefore, by the S-invariance of $\nu $ , all H-orbits must have the same measure. Since $S/H$ is countable, we can choose $s_1,s_2,\ldots \in S$ and $x\in X$ such that $X = S.x = \bigsqcup _n Hs_n.x$ , where the union is either finite or countably infinite. Then

$$ \begin{align*}1 = \sum_n \nu(Hs_n.x) = \sum_n \nu(Hs_1.x).\end{align*} $$

This equation is valid only when the sum contains finitely many positive terms, from which the claim follows.

Proof. By Proposition 6.1, $\operatorname {\mathrm {supp}}(\mu )= \bigcup _{i=0}^{m-1}U^i a + H$ for some $a\in \mathbb {T}^d$ and a closed connected subgroup $H\le \mathbb {T}^d$ . We begin by fixing $n\in \mathbb {N}$ coprime to m and showing that $\mu $ is $U^n$ -ergodic. The ergodic decomposition $\mu =\int \nu \; d\unicode{x3bb} (\nu )$ with respect to $U^n$ yields a representation of $\mu $ as an average of U-invariant measures $\mu = \int ({1}/{n})\sum _{i=0}^{n-1}U_*^i\nu \; d\unicode{x3bb} (\nu )$ . But since $\mu $ is U-ergodic, there exists a $U^n$ -ergodic measure $\nu \in \mathcal {M}_{U^n}(\mathbb {T}^d)$ such that $\mu =({1}/{n})\sum _{i=0}^{n-1}U_*^i\nu $ . Applying Proposition 6.1 to $U^n$ and $\nu $ , we can write $\operatorname {\mathrm {supp}}(\nu )= \bigcup _{i=0}^{m'-1}U^{ni} a' + H'$ , with $m'\in \mathbb {N}$ , $a'\in \mathbb {T}^d$ and $H'\le \mathbb {T}^d$ being a closed connected subgroup. The equality of the supports of $\mu $ and $({1}/{n})\sum _{i=0}^{n-1}U_*^i\nu $ implies that $a' + H'\subseteq \bigcup _i U^i a + H$ . The connectedness of $H'$ then implies that $H'$ is contained in a single coset of H, which must be H; thus, $H'\subseteq H$ . An analogous argument gives $H\subseteq H'$ , and hence $H=H'$ .

It follows that $\operatorname {\mathrm {supp}}(\nu )$ consists of a subset of the H-cosets comprising $\operatorname {\mathrm {supp}}(\mu )$ . But since $\mu $ is U-ergodic and n is coprime to m, $U^n$ acts transitively on the collection of cosets $\{U^i a + H\}_{i=0}^{m-1}$ , and hence $\operatorname {\mathrm {supp}}(\nu )=\operatorname {\mathrm {supp}}(\mu )$ . As both measures are given by the translate of the Haar measure to each coset, we conclude that $\mu =\nu $ , and $\mu $ is $U^n$ -ergodic.

We now prove that $(\operatorname {\mathrm {supp}}(\mu ), {U}|_{\operatorname {\mathrm {supp}}(\mu )})$ is uniquely ergodic. First let us assume that $\operatorname {\mathrm {supp}}(\mu )= a + H$ is connected. Note that

$$ \begin{align*}a + H = U(a + H) = Ua + UH,\end{align*} $$

and since the subgroups $H, UH$ differ by a translation, they must be equal, namely $UH = H$ and $Ua - a \in H$ . Consider the affine transformation $T:H\to H$ defined by $T(h) = Uh + Ua - a$ . Observe that the translation map $\ell _a(x)=x+a$ satisfies $\ell _a \circ T = U\circ \ell _a$ on H, and it takes the Haar measure on H to $\mu $ , so the unique ergodicity of ${U}|_{a+H}$ will follow once we show T is uniquely ergodic. As H is closed and connected, the dynamical system $(H,T)$ is isomorphic to a torus $\mathbb {T}^k$ for some $k\le d$ equipped with an affine transformation whose linear part is unipotent. Moreover, this system is ergodic with respect to Haar measure on $\mathbb {T}^k$ as it is conjugate to the ergodic system $(\mathbb {T}^d,U,\mu )$ . The study of unique ergodicity for affine transformations of the torus was carried out by Hahn; in particular, [Reference HahnHah63, Theorems 6 and 18] show that an ergodic affine transformation with a unipotent linear part is uniquely ergodic.

Now if $\operatorname {\mathrm {supp}}(\mu )$ is not connected, we still have $\operatorname {\mathrm {supp}}(\mu )=\bigcup _{i=0}^{m-1}E_i$ , where $E_i:= U^i a + H$ . Since $\mu $ is ergodic, the induced transformation $U^m$ on each coset $E_i$ is ergodic with respect to ${1}/({\mu (E_i)}){\mu }|_{E_i}$ (see [Reference Einsiedler and WardEW11, Lemma 2.43]). Thus, by what we have just proved, this is the unique $U^m$ -invariant measure on $E_i$ , and it follows that $\mu $ is the unique U-invariant measure on $\operatorname {\mathrm {supp}}(\mu )$ .

Finally, if n is coprime to m, then $\mu $ is $U^n$ -ergodic by the first part of the proof. Now the second part, applied to $U^n$ , shows that ${U^n}|_{\operatorname {\mathrm {supp}}(\mu )}$ is uniquely ergodic.

Proof. First, let us assume that U is in Jordan normal form. Since $X/X^\circ $ is finite, say of order $\ell \in \mathbb {N}$ (where $X^\circ $ denotes the identity component of X), we have $\ell x \in X^\circ $ . Choosing $y\in X^\circ $ with $\ell x = \ell y$ , we see that $x - y$ admits a representative in $({1}/{\ell })\mathbb {Z}^d$ . Thus, the connected component of x in X is given by

$$ \begin{align*}x + X^\circ = (x-y) + X^\circ = v_0 + \operatorname{\mathrm{Span}}_{\mathbb{R}} \{v_1,\ldots,v_m\} + \mathbb{Z}^d\end{align*} $$

for some $m\le d$ , $v_0\in ({1}/{\ell })\mathbb {Z}^d$ and $v_1,\ldots ,v_m\in \mathbb {Z}^d$ .

Choose sequences of integers $(r_{i,n})_{n=1}^\infty $ , for $i=1,\ldots ,m$ , such that the sequence $(\widetilde {x}_n)\subseteq \mathbb {R}^d$ , defined by

$$ \begin{align*}\widetilde{x}_n := v_0 + \sum_i \frac{r_{i,n}}{n}v_i,\end{align*} $$

satisfies $x_n:= \widetilde {x}_n+\mathbb {Z}^d \underset {n\to \infty }{\longrightarrow }x$ . We claim that $x_n$ is of period $p_n:= (d!\ell )n$ . To see that, first observe that for every integer $0\le k <d$ , we have $({p_n}/{k!})\widetilde {x}_n\in \mathbb {Z}^d$ . Now since U is in Jordan normal form, every entry of $U^{p_n}\widetilde {x}_n$ is of the form

$$ \begin{align*}U^{p_n}(\widetilde{x}_n)[i] = \widetilde{x}_n[i] + \binom{p_n}{1}\widetilde{x}_n[i+1] + \cdots + \binom{p_n}{k}\widetilde{x}_n[i+k]\end{align*} $$

for some $0\le k < d$ . Thus, $U^{p_n}(\widetilde {x}_n) - \widetilde {x}_n \in \mathbb {Z}^d$ , confirming that $U^{p_n} x_n = x_n$ .

For a general unipotent matrix U, let J denote its Jordan normal form, and let Q be an integer matrix such that $U=QJQ^{-1}$ (noting that $Q^{-1}$ may not be an integer matrix). Define $Y = \{y\in \mathbb {T}^d:Qy\in X\}$ and choose some $y\in Y$ such that $Qy = x$ (which is possible since Q induces a surjective endomorphism of $\mathbb {T}^d$ ). Apply the previous part of the proof to find a sequence $(y_n)$ in Y which converges to y and such that $J^{p_n} y_n = y_n$ . Then the sequence $(Qy_n)$ has the required properties.

Proof. By Proposition 6.1, $\operatorname {\mathrm {supp}}(\mu ) = \bigcup _{i=0}^{m-1} U^i a + H$ for some $a\in \mathbb {T}^d$ and a closed connected subgroup $H\le \mathbb {T}^d$ . If H is finite then so is $\operatorname {\mathrm {supp}}(\mu )$ . Then for every $n\in \mathbb {N}$ we may take $x_n$ to be any point in $\operatorname {\mathrm {supp}}(\mu )$ ; by ergodicity, the U-invariant measure supported on its orbit is exactly $\mu $ .

Otherwise, $a+H$ contains non-torsion elements, and by replacing a if necessary, we may assume that a itself is non-torsion. Thus, we can apply Lemma 6.4 to obtain $c\in \mathbb {N}$ and a sequence $(x_n)$ contained in the subgroup $\overline {\langle a\rangle }$ such that $x_n$ is of period $cn$ and $x_n\underset {n\to \infty }{\longrightarrow }a$ .

Moreover, since $a+H$ is $U^m$ -invariant and $x_n \in \overline {\langle a \rangle }$ , every coset $x_n + H$ is $U^m$ -invariant as well. Consequently, $\mathfrak {m}_{U, x_n, cn}$ is a U-invariant measure whose support is contained inside $\bigcup _{i=0}^{m-1}U^i x_n +H$ , and combined with the fact that $x_n \to a$ , it implies that every accumulation point of the sequence $(\mathfrak {m}_{U, x_n, cn})_{n}$ is supported on a subset of $\operatorname {\mathrm {supp}}(\mu )$ . Now by Proposition 6.3 (and the compactness of $\mathcal {M}_U(\mathbb {T}^d)$ ), $\mathfrak {m}_{U, x_n, cn} \to \mu $ .

Proof. By the unique ergodicity, it suffices to show that every accumulation point of $(\mathfrak {m}_{T^K, z_n, p_n})$ is supported on a subset of $\operatorname {\mathrm {supp}}(\mu )$ . Let $x\in X\setminus \operatorname {\mathrm {supp}}(\mu )$ and choose an open ball B containing x with $\overline {B}\subseteq X\setminus \operatorname {\mathrm {supp}}(\mu )$ , so that $\mu (\overline {B})=0$ . Since $2p_nK\ge q_n$ for every sufficiently large n, we obtain from the Portmanteau theorem

$$ \begin{align*}\mathfrak{m}_{T^K, z_n, p_n}(B) \le 2K\cdot\mathfrak{m}_{T,z_n,q_n}(B) \to 2K\cdot\mu(B) = 0.\end{align*} $$

Applying the Portmanteau theorem once again, if $\nu $ is any accumulation point of $(\mathfrak {m}_{T^K, z_n, p_n})_n$ then $\nu (B)=0$ . Hence, $x\notin \operatorname {\mathrm {supp}}(\nu )$ , and the proof is complete.

Proof. By Proposition 6.1, $\operatorname {\mathrm {supp}}(\mu )= \bigcup _{i=0}^{m-1} U^i a + H$ , where $a\in \mathbb {T}^d$ , $H\le \mathbb {T}^d$ is a closed connected proper subgroup and the restriction of $\mu $ to each coset is the translate of the Haar measure on H. Take an open neighborhood $\operatorname {\mathrm {supp}}(\mu )\subseteq O\subseteq \mathbb {T}^d$ sufficiently small so that there exists a partition of boxes $X_1,\ldots , X_{r}$ of O that satisfies the following properties.

1. (1) The diameter of each $X_j$ is less than $\varepsilon $ .

2. (2) No side of $X_j$ is contained in $\operatorname {\mathrm {supp}}(\mu )$ .

By item (2), the intersection of $\operatorname {\mathrm {supp}}(\mu )$ with any side of $X_j$ is lower-dimensional and hence $\mu (\partial X_j) = 0$ . Thus, the Portmanteau theorem implies that if $\mu _n\to \mu $ then $\mu _n(X_j)\to \mu (X_j)$ and $\mu _n(O) \to 1$ .

By Proposition 6.5 we can choose $c\in \mathbb {N}$ (depending only on U and $\mu $ ) and a sequence $(z_n)$ in $\mathbb {T}^d$ with $z_n$ of period $cn$ such that $\mathfrak {m}_{U,z_n,cn}\to \mu $ . The system $(\operatorname {\mathrm {supp}}(\mu ), {U^K}|_{\operatorname {\mathrm {supp}}(\mu )})$ is uniquely ergodic by Proposition 6.3, and hence we can apply Lemma 6.6 to conclude that $\mathfrak {m}_{U^K,z_n,\lfloor cn/K\rfloor }\to \mu $ as well. Therefore, fixing a sufficiently large N and putting $z:= z_N$ , $q:= cN$ and $p:= \lfloor q/K\rfloor $ , we have for every $j=1,\ldots ,r$ ,

$$ \begin{align*} {\lvert} \mathfrak{m}_{U^K,z,p}(X_j) - \mu(X_j) {\rvert} < \frac{\varepsilon}{8r}. \end{align*} $$

Since x is $\mu $ -generic, a similar argument implies that $\mathfrak {m}_{U^K,x,\lfloor cn/K\rfloor }\to \mu $ , and enlarging N if necessary, we have for every $j=1,\ldots ,r$ ,

$$ \begin{align*} {\lvert} \mathfrak{m}_{U^K,x,p}(X_j) - \mu(X_j) {\rvert} < \frac{\varepsilon}{8r}. \end{align*} $$

It follows from the last two equations that for every $j=1,\ldots ,r$ ,

$$ \begin{align*} {\lvert} \lvert\{0\le i < p: (U^K)^i x \in X_j\}\rvert - \lvert\{0\le i < p: (U^K)^i z \in X_j\}\rvert{\rvert} < \frac{\varepsilon p}{4r}. \end{align*} $$

Moreover, by enlarging N if necessary, we may assume that $\mathfrak {m}_{U^K,x,p}(O)> 1-\varepsilon /4$ . Thus, we can choose a permutation $\pi $ of $\{0,\ldots ,q-1\}\cap K\mathbb {N}_0$ such that $U^{iK} z$ and $U^{\pi (iK)}x$ belong to the same $X_j$ for more than

$$ \begin{align*}\bigg(1-\frac{\varepsilon}{4}\bigg)p - r\frac{\varepsilon p}{4r} = \bigg(1-\frac{\varepsilon}{2}\bigg)p\end{align*} $$

values of i, and hence for every such i, $d_{\mathbb {T}^d}(U^{iK} z,U^{\pi (iK)}x) < \varepsilon $ . Enlarging N once again if necessary, we have $(1-\varepsilon /2)p> (1-\varepsilon )q/K$ .