Proof of Revised Theorem 7.1

Given a nilpotent Lie group G, denote by $G=G_1 \trianglerighteq G_2\trianglerighteq \cdots \trianglerighteq G_{s}\trianglerighteq \{1_G\}$ its lower central series. For $k\in {\mathbb {N}}$ , define $H^{(1)}(G),\ldots , H^{(k-1)}(G)$ as

(2.2) $$ \begin{align} H^{(i)}(G):=\{ (g^{\binom{1}{i}}, g^{\binom{2}{i}},\ldots,g^{\binom{k}{i}}): g\in G_i\}\subseteq G^k, \end{align} $$

where $\binom {j}{i}=0$ for $j<i$ , and let $H(G)$ be given by

(2.3) $$ \begin{align} H(G):= H^{(1)}(G) H^{(2)}(G)\cdots H^{(k-1)}(G) G_k^k. \end{align} $$

Also, for a co-compact lattice $\Gamma \subset G$ define $\Delta (G,\Gamma ):=H(G)\cap \Gamma ^k$ . Since $H(G)$ is a rational subgroup of $G^k$ , it follows from [Reference Leibman2, Lemma 1.11] that $\Delta (G,\Gamma )$ is a uniform and discrete subgroup of $H(G)$ . Define the nilmanifold $Y(G,\Gamma ):=H(G)/\Delta (G,\Gamma )$ . Note that we can naturally identify $Y(G,\Gamma )$ with a subnilmanifold of $(G/\Gamma )^k$ .

For $b\in G$ , define $R_b:G/\Gamma \to G/\Gamma $ to be the map $R_b(g\Gamma )=(bg)\Gamma $ and let

(2.4) $$ \begin{align} S_b:= R_b\times R_b^2\times \cdots\times R_b^k. \end{align} $$

For $x=g\Gamma \in G/\Gamma $ define

(2.5) $$ \begin{align} Y_x:=\overline{\{S_b^n(x,x,\ldots, x): n\in\mathbb{Z}\}}\subseteq (G/\Gamma)^k. \end{align} $$

It was shown in [Reference Moreira and Richter3, Proposition 7.5, part (iv)] that for almost every $x=g\Gamma \in G/\Gamma $ the map $R_{g^{-1}}\times \cdots \times R_{g^{-1}}: (G/\Gamma )^k\to (G/\Gamma )^k$ is an isomorphism from the nilsystem $(Y_x, S_a)$ to the nilsystem $(Y(G,\Gamma ),S_{g^{-1}ag})$ .

Suppose now that $X=G/\Gamma $ is the system in the statement of the theorem and let $a\in ~G$ be such that $R=R_a$ . Take $\theta \in [0,1)$ . Our goal is to show that if $\theta \notin \sigma (X,R)$ then ${\theta \notin \sigma (Y_x, S_a)}$ for almost every $x\in X$ . Let us first deal with the case when $\theta $ is irrational.

Observe that $\theta $ is not an eigenvalue of $(X,R_a)$ if and only if the product system $(X, R_a)\times (\mathbb {T},R_{\theta })$ is ergodic, where $R_{\theta }\colon t\mapsto t+\theta $ is rotation by $\theta $ . Notice that ${X\times \mathbb {T}=(G\times \mathbb {R})/(\Gamma \times \mathbb {Z})}$ is a nilmanifold too, and hence $(X, R_a)\times (\mathbb {T},R_{\theta })$ is a nilsystem. In accordance with (2.4) and (2.5) let

$$ \begin{align*} S_{(a,\theta)}=(R_a\times R_\theta)\times(R_a^2\times R_{2\theta})\times\cdots\times(R_a^k\times R_{k\theta}) \end{align*} $$

and

$$ \begin{align*} Y_{(x,t)}:=\overline{\{S_{(a,\theta)}^n((x,t),\ldots, (x,t)): n\in\mathbb{Z}\}}\subseteq (X\times\mathbb{T})^k. \end{align*} $$

As was mentioned above, for almost every $(x,t)=(g\Gamma ,t)\in X\times \mathbb {T}$ , the nilsystem $(Y_{(x,t)},S_{(a,\alpha )})$ is isomorphic to $(Y(G\times \mathbb {R},\Gamma \times \mathbb {Z}),S_{(g^{-1}ag,\theta )})$ .

We claim that $Y(G\times \mathbb {R},\Gamma \times \mathbb {Z})\cong Y(G,\Gamma )\times Y(\mathbb {R},\Gamma )$ . Assuming this claim for now, it follows that

$$ \begin{align*} (Y_{(x,t)},S_{(a,\theta)})&\cong (Y(G\times\mathbb{R},\Gamma\times\mathbb{Z}),S_{(g^{-1}ag,\theta)}) \\ &\cong(Y(G,\Gamma),S_{g^{-1}ag})\times ( Y(\mathbb{R},\mathbb{Z}), S_\theta) \\ &\cong (Y(G,\Gamma),S_{g^{-1}ag})\times (\mathbb{T},R_\theta) \\ &\cong (Y_x,S_{a})\times (\mathbb{T},R_\theta). \end{align*} $$

Recall that any transitive nilsystem is ergodic. Since $(Y_{(x,t)},S_{(a,\theta )})$ is transitive by definition, it follows that it is ergodic, which implies that $(Y_x,S_{a})\times (\mathbb {T},R_\theta )$ is ergodic for almost every $x\in X$ . However, $(Y_x,S_{a})\times (\mathbb {T},R_\theta )$ can only be ergodic if $\theta $ is not in the discrete spectrum of $(Y_x,S_{a})$ , which finishes the proof that $\theta \notin \sigma (Y_x, S_a)$ for almost every $x\in X$ .

It remains to show that $Y(G\times \mathbb {R},\Gamma \times \mathbb {Z})\cong Y(G,\Gamma )\times Y(\mathbb {R},\Gamma )$ . Note that ${{H^{(i)}(\mathbb {R})=\{0\}^k}}$ for all $i\geq 2$ , so that $H(\mathbb {R})=\{(t,2t,\ldots ,kt):t\in \mathbb {R}\}$ . More generally, for any G we have $H^{(i)}(G\times \mathbb {R})=H^{(i)}(G)\times \{0\}^k$ whenever $i\geq 2$ . This implies that

$$ \begin{align*}H(G\times\mathbb{R})=H(G)\times H(\mathbb{R}).\end{align*} $$

Finally, since

$$ \begin{align*} \Delta(G\times\mathbb{R},\Gamma\times\mathbb{Z}) &= (H(G)\times H(\mathbb{R}))\cap(\Gamma^k\times\mathbb{Z}^k) \\&= H(G)\cap\Gamma^k\times H(\mathbb{R})\cap\mathbb{Z}^k \\&= \Delta(G,\Gamma)\times\Delta(\mathbb{R},\mathbb{Z}), \end{align*} $$

the claim $Y(G\times \mathbb {R},\Gamma \times \mathbb {Z})\cong Y(G,\Gamma )\times Y(\mathbb {R},\Gamma )$ follows.

Lastly, we deal with the case when $\theta =p/q\in (0,1)$ is rational. Recall that $S_a= R_a\times R_a^2\times \cdots \times R_a^k$ and $Y_x:=\overline {\{S_a^n(x,x,\ldots , x): n\in \mathbb {Z}\}}$ and that

(2.6) $$ \begin{align} (Y_x, S_a)\cong (Y(G,\Gamma), S_{g^{-1}ag}) \end{align} $$

for all $x=g\Gamma \in X'$ , where $X'$ is some full measure subset of X. Observe that (2.6) implies

(2.7) $$ \begin{align} (Y_x, S_a^q)\cong (Y(G,\Gamma), S_{g^{-1}ag}^q), \end{align} $$

for all $x=g\Gamma \in X'$ . Then define

$$ \begin{align*} Y_x^{(q)}:=\overline{\{S_a^{qn}(x,x,\ldots, x): n\in\mathbb{Z}\}}=\overline{\{S_{a^q}^{n}(x,x,\ldots, x): n\in\mathbb{Z}\}}. \end{align*} $$

Since X is connected and $(X, R_a)$ is ergodic, the nilsystem $(X, R_{a}^q)$ is ergodic. This implies that there exists a full measure set $X"\subset X$ such that for all $x=g\Gamma \in X"$ we have

(2.8) $$ \begin{align} (Y_x^{(q)}, S_a^q)\cong (Y(G,\Gamma), S_{g^{-1}ag}^q). \end{align} $$

Combining (2.7) and (2.8), we see that for any $x\in X'\cap X"$ we have

$$ \begin{align*} (Y_x, S_a^q)\cong (Y_x^{(q)}, S_a^q). \end{align*} $$

Since $(Y_x^{(q)}, S_a^q)$ is transitive by definition, it must be ergodic, and thus it follows that for all $x\in X'\cap X"$ the system $(Y_x, S_a^q)$ is ergodic. We conclude that $\theta =p/q$ is not an eigenvalue of $(Y_x, S_a^q)$ and this finishes the proof.

Proof. Given $\theta \in \sigma (X,R)$ , let $f\in L^2(X)$ be an eigenfunction of the system $(X,R)$ with eigenvalue $\theta $ . Since the function $\tilde f\in L^2(Y_{X^\Delta })$ defined by $\tilde f(x_1,\ldots ,x_k)=f(x_1)$ is an eigenfunction for the system $(Y_{X^\Delta },S)$ with eigenvalue $\theta $ , it follows that $\sigma (X,R)\subseteq \sigma (Y_{X^\Delta },S)$ .

Next we prove the converse inclusion. Let $\nu $ be the Haar measure of the nilmanifold $Y_{X^\Delta }$ and let $\nu _x$ be the Haar measure of the nilmanifold $Y_x$ defined by (2.1). Observe that the sets $Y_x$ are precisely the atoms of the invariant $\sigma $ -algebra of the system $(Y_{X^\Delta },S)$ . Therefore, the measures $\nu _x$ form the ergodic decomposition of $\nu $ .

Let $\theta \in \sigma (Y_{X^\Delta },S)$ and let $f\in L^2(Y_{X^\Delta },\nu )$ be an eigenfunction with eigenvalue $\theta $ , that is, for almost every $y\in Y_{X^\Delta }$ we have $Sf(y)=e(\theta )f(y)$ . Since f cannot be $0 \nu $ -almost everywhere, there exists a positive measure set of $x\in X$ for which the restriction of f to the system $(Y_x,\nu _x,S)$ is not the zero function. But for any such x, the restriction of f to the system $(Y_x,\nu _x,S)$ is an eigenfunction with eigenvalue $\theta $ . This implies that $\theta \in \sigma (Y_{X^\Delta },S)$ for all such x. Finally, by invoking Revised Theorem 7.1, we conclude that $\theta \in \sigma (X,R)$ , finishing the proof.