Proof. Let $s>2$ , $n\in \mathbb {N}^*$ and $f_1,f_2\in H^s(\mathbb {T}^2).$ We recall that for $i\in \{1,2\}$ , using the definition (3.8), for $q_1,q_2\in \mathbb {Z}$ ,

(3.11) $$ \begin{align} \vert \widehat{f_i}(q_1,q_2)(1+q_1^2+q_2^2)^{{s}/{2}}\vert \leq C_{f_i}(s). \end{align} $$

For $\xi _1,\xi _2\in \mathbb {Z}$ , let

$$ \begin{align*}g_{(\xi_1,\xi_2)} : x\in\mathbb{T}\mapsto e^{2i\pi\xi_1x}\bigg(\int_{\mathbb{T}} \overline{f_1}(x,y)e^{2i\pi\xi_2 y} \,d\unicode{x3bb}(y) \bigg).\end{align*} $$

By Fourier decomposition, for $\xi _1,\xi _2\in \mathbb {Z}^d$ and for all $x\in \mathbb {T}$ ,

(3.12) $$ \begin{align} g_{(\xi_1,\xi_2)}(x)=\sum_{k\in\mathbb{Z}}\overline{\widehat{f_1}(k,\xi_2)}e^{2i\pi (\xi_1-k)x}. \end{align} $$

By (3.12),

(3.13) $$ \begin{align}\int_{\mathbb{T}^2} \overline{f_1}(f_2\circ T^n)\,d(\mu\otimes\unicode{x3bb}) =\sum_{\xi\in\mathbb{Z}^2}\widehat{f_2}(\xi)\int_{\mathbb{T}} e^{2i\pi n\xi_2x}g_{(\xi_1,\xi_2)}(x)\,d\mu(x) \end{align} $$

and for all $k,\xi _1,\xi _2\in \mathbb {Z}$ ,

(3.14) $$ \begin{align}\widehat{g_{(\xi_1,\xi_2)}}(k)=\widehat{f_1}(\xi_1-k,\xi_2).\end{align} $$

By (3.11) and (3.14),

(3.15) $$ \begin{align}\vert \widehat{g_{(\xi_1,\xi_2)}}(k)\vert (1+k^2)^{{s}/{2}}\leq C_{f_1}\frac{(1+k^2)^{{s}/{2}}}{(1+\xi_2^2+(\xi_1-k)^2)^{{s}/{2}}}.\end{align} $$

Moreover, by (3.13) and (3.14), for $n\in \mathbb {N}$ ,

$$ \begin{align*}\mathbb{E}_{\mu\otimes\unicode{x3bb}}(\mathrm{Cov}_n(f_1,f_2\vert \mathcal{I}))=\sum_{\xi\in\mathbb{Z}\times\mathbb{Z}^*}\widehat{f_2}(\xi) \int_{\mathbb{T}} e^{2i\pi n\xi_2x}g_{(\xi_1,\xi_2)}(x)\,d\mu(x) \end{align*} $$

and

(3.16) $$ \begin{align} \mathbb{E}_{\mu\otimes\unicode{x3bb}}(\mathrm{Cov}_n(f_1,f_2\vert \mathcal{I}))=\sum_{(k,\xi)\in\mathbb{Z}\times \mathbb{Z}\times\mathbb{Z}^*}\widehat{f_2}(\xi)\widehat{f_1}(k,\xi_2)\widehat{\mu}(\xi_1-k+n\xi_2). \end{align} $$

We estimate the sum above, considering three different cases. In the first case, we suppose that $k,\xi _1,\xi _2\in \mathbb {Z}$ and $n\in \mathbb {N}$ are such that

(3.17) $$ \begin{align} \vert \xi_1-k+n\xi_2\vert \leq \frac{n \vert \xi_2\vert}{4}\end{align} $$

and

(3.18) $$ \begin{align} \vert n\xi_2-k\vert \leq \frac{n\vert \xi_2\vert}{2}. \end{align} $$

Therefore,

(3.19) $$ \begin{align} \vert k\vert \geq \frac{n \vert \xi_2\vert}{2},\end{align} $$

(3.20) $$ \begin{align} \vert \xi_1\vert \geq \frac{3}{4}n \vert \xi_2\vert ,\end{align} $$

and

(3.21) $$ \begin{align} \vert \widehat{\mu}(\xi_1-k+n\xi_2)\vert \leq 1.\end{align} $$

Assuming (3.17) and (3.18), by (3.15), (3.19), (3.20), and (3.21),

(3.22) $$ \begin{align} & \sum_{\substack{(k,\xi_1,\xi_2)\in \mathbb{Z}\times\mathbb{Z}\times\mathbb{Z}^*\\ \vert \xi_1-k+n\xi_2\vert \leq ({n \vert \xi_2\vert})/{4}\\ \vert n\xi_2-k\vert \leq ({n \vert \xi_2\vert})/{2}}}\vert \widehat{f_2}(\xi_1,\xi_2)\vert \vert \widehat{f_1}(k,\xi_2) \vert \widehat{\mu}(\xi_1-k+n\xi_2)\vert \nonumber\\ &\qquad \leq \frac{2^{6(s+1)}\zeta(2(s-1))C_{f_1}(s)C_{f_2}(s)}{3^sn^{2(s-1)}}.\end{align} $$

In the second case, we assume (3.17) and

(3.23) $$ \begin{align} \vert n\xi_2-k\vert> \frac{n \vert \xi_2\vert}{2}.\end{align} $$

Thus, we get again (3.21) and

(3.24) $$ \begin{align} \vert \xi_1\vert>\frac{n\vert \xi_2\vert}{4}.\end{align} $$

By (3.21) and (3.11), $\vert \widehat {f_1}(k,\xi _2)\vert \vert \widehat {\mu }(\xi _1-k+n\xi _2)\vert \leq ({C_{f_1}(s)}/{(1+k^2+\xi _2^2)^{{s}/{2}}})$ . Moreover, by (3.11) and (3.24),

(3.25) $$ \begin{align}\vert \widehat{f_2}(\xi_1,\xi_2)\vert \leq \frac{4^sC_{f_2}(s)}{\vert \xi_2\vert^{s}n^{s}}.\end{align} $$

By (3.25) and series-integral criterion with $s>2$ , there exists $C_s>0$ such that

$$ \begin{align*}\sum_{\substack{(k,\xi_1,\xi_2)\in \mathbb{Z}\times\mathbb{Z}\times\mathbb{Z}^*\\ \vert \xi_1-k+n\xi_2\vert \leq ({n\vert \xi_2\vert})/{4}\\ \vert n\xi_2-k\vert> ({n \vert \xi_2\vert })/{2}}}\frac{C_{f_1}(s) \vert \widehat{f_2}(\xi_1,\xi_2)\vert }{(1+k^2+\xi_2^2)^{{s}/{2}}}\leq \frac{C_s4^sC_{f_1}(s)C_{f_2}(s)}{n^{(s-1)}}.\end{align*} $$

Finally, in the third case, we assume

(3.26) $$ \begin{align} \vert \xi_1-k+n\xi_2\vert> \frac{n \vert \xi_2\vert }{4}.\end{align} $$

Here, $\mu $ is a Rajchman measure and $r>0$ its order, by the definition (2.14), there exists $C_\mu>0$ such that for $q\in \mathbb {Z}$ ,

(3.27) $$ \begin{align} \vert \widehat{\mu}(q)\vert \leq\frac{C_\mu}{ \vert q\vert ^r}. \end{align} $$

Applying (3.27) to $q=\xi _1-k+n\xi _2$ and by assumption (3.26),

(3.28) $$ \begin{align} \vert \widehat{\mu}(\xi_1-k+n\xi_2)\vert \leq \frac{4^rC_\mu}{(n \vert \xi_2\vert )^{r}}.\end{align} $$

By (3.28), (3.11), and the series-integral criterion, for $s>2$ , there exists a constant $C_s^{\prime }$ such that

(3.29) $$ \begin{align} \sum_{\substack{(k,\xi_1,\xi_2)\in \mathbb{Z}\times\mathbb{Z}\times\mathbb{Z}^*\\ \vert \xi_1-k+n\xi_2\vert> ({n \vert \xi_2\vert })/{4}}}\!\!\!\frac{C_{\!f_1}(s) \vert \widehat{f_2}(\xi_1,\xi_2)\vert \vert \widehat{\mu}(\xi_1{\kern-1pt}-{\kern-1pt}k{\kern-1pt}+{\kern-1pt}n\xi_2)\vert}{(1+k^2+\xi_2^2)^{{s}/{2}}}\!\leq\! \frac{4^r(1{\kern-1pt}+{\kern-1pt}3\zeta({s}/{2}))C_\mu C_s^{\prime}C_{\!f_1}(s)C_{\!f_2}(s)}{n^{r}}. \end{align} $$

Let us pose

$$ \begin{align*}M(r,s)=\sup\bigg\{4^r\bigg(1+\zeta\bigg(\frac{s}{2}\bigg)\bigg)C_\mu C_s^{\prime},4^sC_s,2^{6(s+1)}\zeta(2(s-1))\bigg\}.\end{align*} $$

We get

(3.30) $$ \begin{align} \vert \mathbb{E}_{\mu\otimes\unicode{x3bb}}(\mathrm{Cov}_n(f_1,f_2\vert \mathcal{I}))\vert \leq M(r,s)\bigg(\frac{C_{f_1}C_{f_2}}{n^{\min\{s-1,r\}}}\bigg).\end{align} $$

We now prove the optimality of the exponent. The observable f defined by ${f(x,y)=e^{2i\pi y}}$ belongs to $H^s(\mathbb {T}^2)$ . Moreover,

(3.31) $$ \begin{align} \int_{\mathbb{T}^2} \overline{f}(x,y)f(T^n(x,y))\, d\mu\otimes\unicode{x3bb}(x,y)=\widehat{\mu}(n). \end{align} $$

Thus, for $\gamma <\gamma _s$ , there exists $C>0$ such that for all $n\in \mathbb {N}^*$ ,

(3.32) $$ \begin{align} \vert \widehat{\mu}(n)\vert \leq \frac{C}{n^{\gamma}}. \end{align} $$

By definition of the Rajchman order r, $r\geq \gamma $ . Therefore, $r\geq \gamma _s$ .

Proof. Let $f\in \mathbb {L}^2_\mu (\Omega )$ and $U\in \mathcal{A}$ .

Let us prove the direct assertion. Suppose that $f\circ T=f\,\mu \text {-a.s}$ . Let $(x,y) \in U\times \mathbb {T}^d$ . So,

$$ \begin{align*}f(T(\psi_U^{-1}(x,y)))=f(\psi_U^{-1}(x,y)).\end{align*} $$

Namely, $f(\psi _U^{-1}(x,y+v_U(x)))=f(\psi _U^{-1}(x,y))$ . By Fourier series decomposition:

(3.33) $$ \begin{align}\sum_{k\in\mathbb{Z}^d}\widehat{f\circ \psi_U(x,\cdot)}(k)e^{2i\pi{<} k\vert y+v_U(x){>}}=\sum_{k\in\mathbb{Z}^d}\widehat{f\circ \psi_U(x,\cdot)}(k)e^{2i\pi{<} k\vert y{>}}.\end{align} $$

By uniqueness of the coefficients of a Fourier series in (3.33), for $\mu \text {-a.s.}\,x\in U$ , for all $k\in \mathbb {Z}^d$ ,

(3.34) $$ \begin{align}\widehat{f\circ \psi_U(x,\cdot)}(k)(e^{2i\pi {<} k\vert v_U(x){>}}-1)=0.\end{align} $$

Thus, ${<} k\vert v_U(x){>} \in \mathbb {Z}$ or $\widehat {f\circ \psi _U(x,\cdot )}(k)=0$ .

Let us prove the reciprocal. Suppose that f satisfies for all charts $U\in \mathcal{A}$ , $(\mu _{\pi })_{\vert U}\otimes \unicode{x3bb} \text {-a.s.} \,(x,y)\in U\times \mathbb {T}^d$ ,

(3.35) $$ \begin{align} f(\psi_U^{-1}(x,y))=\sum_{k\in\mathbb{Z}^d,{<} k\vert v_U(x){>}\in\mathbb{Z}}a_k(x)e^{2i\pi{<} k\vert y{>}}. \end{align} $$

Let $z\in \Omega $ . There exists $U\in \mathcal{A}$ such that $z\in \pi ^{-1}(U)$ . We have

$$ \begin{align*}f(T(z))=f(T(\psi_U^{-1}(\psi_U(z))))=f(\psi_U^{-1}(\pi(z),\pi_2\circ \psi_U(z)+v_U(\pi(z)))).\end{align*} $$

Applying (3.35) with the above computation,

$$ \begin{align*}\begin{array}{ll} f(T(z))&= \displaystyle\sum_{k\in\mathbb{Z}^d,{<} k\vert v_U(x){>}\in\mathbb{Z}}a_k(\pi(z))e^{2i\pi{<} k\vert \pi_2\circ\psi_U(z)+v_U(\pi(z)){>}}\\ &=\displaystyle\sum_{k\in\mathbb{Z}^d,{<} k\vert v_U(x){>}\in\mathbb{Z}}a_k( \pi(z))e^{2i\pi{<} k\vert \pi_2\circ\psi_U(z){>}}\\ &=f(z). \end{array} \end{align*} $$

Proof. Let $f\in \mathbb {L}^2_\mu (\Omega )$ . Let us start with the direct implication. Suppose that for all $t\in \mathbb {R}$ ,

$$ \begin{align*}f\circ g_t=f\quad \mu\text{-almost everywhere}.\end{align*} $$

Let $U\in \mathcal{A}$ , $t\in \mathbb {R}$ , and $(x,y)\in U\times \mathbb {T}^d$ . Since

$$ \begin{align*}f(g_t(\psi_U^{-1}(x,y)))=f(\psi_U^{-1}(x,y)),\end{align*} $$

we have

$$ \begin{align*}f(\psi_U^{-1}(x,y+tv_U(x)))=f(\psi_U^{-1}(x,y)).\end{align*} $$

By Fourier series decomposition, this gives

$$ \begin{align*}\sum_{k\in\mathbb{Z}^d}\widehat{f\circ \psi_U(x,\cdot)}(k)e^{2i\pi{<} k\vert y+tv_U(x){>}}=\sum_{k\in\mathbb{Z}^d}\widehat{f\circ \psi_U(x,\cdot)}(k)e^{2i\pi{<} k\vert y{>}}.\end{align*} $$

Hence, by uniqueness of Fourier series coefficients, for all $k\in \mathbb {Z}^d$ ,

$$ \begin{align*}\widehat{f\circ \psi_U(x,\cdot)}(k)(e^{2i\pi t{<} k\vert v_U(x){>}}-1)=0.\end{align*} $$

When k is such that $\widehat {f\circ \psi _U(x,\cdot )}(k)\ne 0$ , we get that for all $t\in \mathbb {R}, t{<} k\vert v_U(x){>} \in \mathbb {Z}$ . So, ${<} k\vert v_U(x){>} =0$ and then $k\in \{v_U(x)\}^\perp $ . Hence,

$$ \begin{align*} \begin{gathered} (\mu_{\pi})_{\vert U}\otimes\unicode{x3bb}\text{-a.e.} \,(x,y)\in U\times\mathbb{T}^d, \\ f(\psi_U^{-1}(x,y)) =\sum_{k\in\mathbb{Z}^d\cap\{v_U(x)\}^\perp}\widehat{f\circ \psi_U(x,\cdot)}(k)e^{2i\pi{<} k\vert y{>}}.\end{gathered}\end{align*} $$

Let us see the reciprocal implication. Let $t\in \mathbb {R}$ . Suppose that f satisfies for all charts $U\in \mathcal{A}$ ,

$$ \begin{align*}(\mu_{\pi})_{\vert U}\otimes\unicode{x3bb}\text{-a.e.} \,(x,y)\in U\times\mathbb{T}^d{,}\quad f(\psi_U^{-1}(x,y))=\sum_{k\in\mathbb{Z}^d\cap\{v_U(x)\}^\perp}a_k(x)e^{2i\pi{<} k\vert y{>}}.\end{align*} $$

Let $z\in \Omega $ and take $ U\in \mathcal{A}$ such that $z\in \pi ^{-1}(U)$ . We have

$$ \begin{align*} f(g_t(z))=f(g_t(\psi_U^{-1}(\psi_U(z))))=f(\psi_U^{-1}(\pi(z),\pi_2\circ \psi_U(z)+tv_U(\pi(z)))).\end{align*} $$

Thus,

$$ \begin{align*} \begin{aligned} f(g_t(z))&=\sum_{k\in\mathbb{Z}^d\cap\{v_U(\pi(z))\}^\perp}a_k(\pi(z))e^{2i\pi{<} k\vert \pi_2\circ\psi_U(z)+tv_U(\pi(z)){>}} \\ &=\sum_{k\in\mathbb{Z}^d\cap\{v_U(\pi(z))\}^\perp}a_k(\pi(z))e^{2i\pi{<} k\vert \pi_2\circ\psi_U(z){>}} \end{aligned} \end{align*} $$

by orthogonality. Hence,

$$ \begin{align*}f(g_t(z))=f(\psi_U^{-1}(\psi_U(z)))=f(z).\\[-37pt] \end{align*} $$

Proof. We prove the direct inclusion. Let $A\in \mathcal{I}$ . Let $U\in \mathcal{A}$ and set $C=\psi _U(A\cap \pi ^{-1}(U))$ . By Proposition 4.6,

$$ \begin{align*}\mathbb{1}_C(x,y)=\sum_{k\in \{v_U(x)\}^\perp\cap\mathbb{Z}^d}a_k(x)e^{2i\pi{<} k\vert y{>}}\,(\mu_{\pi})_{\vert U}\otimes \unicode{x3bb}\text{-a.s.},\end{align*} $$

which shows that $C\in \mathcal{I}_U$ . Hence, $A\cap \pi ^{-1}(U)\in \psi _U^{-1}(\mathcal{I}_U)$ and, finally,

$$ \begin{align*}A= \bigcup_{U\in\mathcal{A}} A\cap\pi^{-1}(U) .\end{align*} $$

Now, we prove the reciprocal inclusion.

Let $A\in \mathcal{B}(\Omega )$ such that $\text { there exists } (B_U)_{U\in \mathcal{A}}\in {{\prod _{U\in \mathcal{A}}}} \psi _U^{-1}(\mathcal{I}_U),\ A= {\bigcup _{U\in \mathcal{A}}} B_U$ . Then, for all $t\in \mathbb {R}$ ,

$$ \begin{align*}g_t^{-1}(A)= {\bigcup_{U\in\mathcal{A}}} g_t^{-1}(B_U) .\end{align*} $$

Let $U\in \mathcal{A}$ . By bijectivity of $\psi _U$ , $\psi _U(B_U)\in \mathcal{I}_U$ . Hence,

$$ \begin{align*} \begin{gathered} (\mu_{\pi})_{\vert U}\otimes\unicode{x3bb}\text{-a.e.}\,(x,y)\in U\times\mathbb{T}^d, \\ \mathbb{1}_{\psi_U(B_U)}(x,y) =\sum_{k\in \{v_U(x)\}^\perp\cap\mathbb{Z}^d}\widehat{\mathbb{1}_{\psi_U(B_U)}(x,\cdot)}(k)e^{2i\pi{<} k\vert y{>}}, \end{gathered}\end{align*} $$

which proves that $A\in \mathcal{I}$ by Proposition 4.6.

Proof. Let $U\in \mathcal{A}$ , $\xi \in \mathbb {Z}^d\setminus \{0\}$ . Let $\gamma>r((m_{\xi ,U})_{\vert \mathbb {R}^*})$ . Consider

$$ \begin{align*} f : z\in\Omega \mapsto e^{2i\pi {<}\xi\vert (\pi_2\circ\psi_{U}^{-1})(z){>}}\mathbb{1}_{\pi^{-1}(U)}(z)\in \mathcal{C}^\infty(\pi^{-1}(U))\cap\mathbb{L}^2_\mu(\Omega). \end{align*} $$

We get

$$ \begin{align*} \int_{\Omega} \overline{f}(f\circ g_t)\, d\mu = \int_{U} e^{2i\pi t{<}\xi\vert v_{U}(x){>}}\, d(\mu_\pi)(x) = \int_{\mathbb{R}} e^{2i\pi tz} \,dm_{\xi,U}(z) .\end{align*} $$

By optimality of $r((m_{\xi ,U})_{\vert \mathbb {R}^*})$ , for all $C>0$ , for all $T>0$ , there exists $t>T$ such that

$$ \begin{align*} \bigg\vert \int_{\mathbb{R}} e^{2i\pi tz} \,d(m_{\xi,U})_{\vert \mathbb{R}^*} \bigg\vert>\frac{C}{t^\gamma}.\end{align*} $$

The result follows by definition of conditional correlations.

Proof. Here, M is compact Hausdorff, so the number of critical points for each function ${<} \xi \vert v{>} $ is finite. Let $(a_k)_{k\in {[\![} 1,m{]\!]} }$ be the family of critical point of ${<} \xi \vert v(\cdot ){>} $ with respective orders $l_k-1\geq 1$ , $1\leq m<l_k,$

$$ \begin{align*} {<}\xi\vert v^{(m)}(a_k){>}=0\quad\text{and}\quad {<}\xi\vert v^{(l_k)}(a_k){>}\ne 0.\end{align*} $$

Let $(U_k)_{k\in {[\![} 1,m{]\!]} }$ be charts such that for all $k\in {[\![} 1,m{]\!]} $ , $a_k\in U_k$ . We have

$$ \begin{align*} {<}\xi\vert v(\varphi_{U_k}^{-1}(x)){>}&={<}\xi\vert v(a_k){>}+\frac{(x-\varphi_{U_k}(a_k))^{l_k}}{l_k!}{<}\xi\vert v^{(l_k)}(a_k){>}\\ &\quad+(x-\varphi_{U_k}(a_k))^{l_k} h(x-\varphi_{U_k}(a_k)) \end{align*} $$

with $h : \mathbb {R}\to \mathbb {R}\, \mathcal{C}^1$ such that $h(x)\xrightarrow [x\to 0]{}h(0)=0.$ Let us pose for $k\in {[\![} 1,m{]\!]} ,$

$$ \begin{align*}w_k : x\in\mathbb{R} \mapsto (x-\varphi_{U_k}(a_k))\bigg(\frac{{<}\xi\vert v^{(l_k)}(a_k){>}}{l_k!}+h(x-\varphi_{U_k}(a_k))\bigg)^{{1}/{l_k}}.\end{align*} $$

Therefore,

(4.4) $$ \begin{align}w_k^{\prime} : x\in\mathbb{R}\mapsto \bigg(\frac{{<}\xi\vert v^{(l_k)}(a_k){>}}{l_k!}+h(x-\varphi_{U_k}(a_k))\bigg)^{{1}/{l_k}}+(x-\varphi_{U_k}(a_k))\kappa(x-\varphi_{U_k}(a_k))\end{align} $$

with $\kappa _k(x)={h^{\prime }(x)}/{l_k({<} \xi \vert v^{(l)}(a_k){>} +h(x))^{({l_k-1})/{l_k}}}$ . Applying (4.4) to $x=\varphi _{U_k}(a_k)$ ,

$$ \begin{align*}w_k^{\prime}(\varphi_{U_k}(a_k))=\bigg(\frac{{<}\xi\vert v^{(l_k)}(a){>}}{l_k!}\bigg)^{{1}/{l_k}}\ne 0.\end{align*} $$

We get a local inverse of $w_k$ on a neighborhood $W_k$ of $0$ . Let $(V_k)_{k\in {[\![} 1,m+m^{\prime }{]\!]} }$ be a family of charts on M such that, for all $k\in {[\![} 1,m{]\!]} $ ,

(4.5) $$ \begin{align} (V_k = w_k^{-1}(W_k)\subset U_k \text{ and } \text{ for all } j\in {[\![} m+1,m^{\prime}{]\!]},a_k\not\in V_k).\end{align} $$

Let $(\psi _k)_{k\in {[\![} 1,m+m^{\prime }{]\!]} }$ be a partition of unity subordinated to $(V_k)_{k\in {[\![} 1,m+m^{\prime }{]\!]} }$ . We get immediately that for $k\in {[\![} 1,m{]\!]} $ , $\psi _k(a_k)=1$ . By inverse function theorem, for $k\in {[\![} 1,m{]\!]} $ , we get that on $W_k$ , for all $x\in W_k$ ,

(4.6) $$ \begin{align}{<}\xi\vert v(\varphi_{U_k}^{-1}(w_k^{-1}(x))){>}={<}\xi\vert v^{(l_k)}(a_k){>}+x^{l_k}.\end{align} $$

Writing $\check {\unicode{x3bb} }$ as the push forward measure by charts of Lebesgue measure on the manifold M, for $k\in {[\![} 1,m{]\!]} $ , we get

(4.7) $$ \begin{align} & \int_{\varphi_{U_k}^{-1}(w^{-1}(W_k))} e^{2i\pi t{<}\xi\vert v(x){>}}\psi_k(x) \,d\check{\unicode{x3bb}}(x) \nonumber\\ &\quad =e^{2i\pi t{<}\xi\vert v(\varphi_{W_k}^{-1}(w^{-1}(a_k))){>}}\int_{W_k} e^{2i\pi tx^{l_k}}\check{\psi_k}(x)\, d\unicode{x3bb}(x) \end{align} $$

with

$$ \begin{align*}\check{\psi_k} : x\in W_k\mapsto \psi(\varphi_{U_k}^{-1}(w_k^{-1}(x)))J(\varphi_{U_k}^{-1})(w_k^{-1}(x))J(w_k^{-1})(x)\end{align*} $$

and $J(f)(x)$ denotes the Jacobian of the function f at the point x. We know that $\check {\psi _k}$ is subordinated to $(V_j)_{j\in {[\![} 1,m+m^{\prime }{]\!]} }$ . Moreover, by (4.5), the support of $\check {\psi _k}$ is included in $U_k$ . Thus, by (4.7), for all $t\ne 0$ ,

$$ \begin{align*} & \int_{\varphi_{U_k}^{-1}(w_k^{-1}(W_k))} e^{2i\pi t{<}\xi\vert v(x){>}}\psi(x)\, d\check{\unicode{x3bb}}(x) \\ &\quad =e^{2i\pi t{<}\xi\vert v(\varphi_{U_k}^{-1}(w_k^{-1}(a_k))){>}} \dfrac{1}{t^{{1}/{l_k}}} \int_{\mathbb{R}} e^{2i\pi x^{l_k}}\check{\psi_k}\bigg(\dfrac{x}{t^{{1}/{l_k}}}\bigg)\, d\unicode{x3bb}(x). \end{align*} $$

Integrating by parts $u(x)v^{\prime }(x)$ with $u(x)=({\check {\psi _k}({x}/{t^{{1}/{l_k}}})}/({2il_k\pi x^{l_k-1}}))$ and $v(x)=e^{2i\pi x^{l_k}}-1$ ,

$$ \begin{align*} \begin{aligned} &\int_{\mathbb{R}} e^{2i\pi x^{l_k}}\check{\psi_k}\bigg(\frac{x}{t^{{1}/{l_k}}}\bigg) \,d\unicode{x3bb}(x)\\ &\quad =\frac{1}{2i\pi l_k}\bigg( \int_{\mathbb{R}} \frac{(l_k-1)(e^{2i\pi x^{l_k}}-1)}{x^{l_k}}\check{\psi_k}\bigg(\frac{x}{t^{{1}/{l_k}}}\bigg)\, d\unicode{x3bb}(x) \\ &\qquad -\frac{1}{t^{{1}/{l_k}}} \int_{\mathbb{R}} \frac{(e^{2i\pi x^{l_k}}-1)}{x^{l_k-1}}\check{\psi_k}^{\prime}\bigg(\frac{x}{t^{{1}/{l_k}}}\bigg)\, d\unicode{x3bb}(x) \bigg). \end{aligned} \end{align*} $$

We know that for all $\alpha>0,$

(4.8) $$ \begin{align}x\in \mathbb{R}\mapsto {\check{\psi_k}^{\prime}(x)}{(1+ \vert x\vert^\alpha)},x\in \mathbb{R}\mapsto {\check{\psi_k}(x)}{(1+ \vert x\vert^\alpha)}\text{ are bounded by a constant } C>0.\end{align} $$

Using the development at $x=0$ of $x\in \mathbb {R}\mapsto ({(e^{2i\pi x^{l_k}}-1)}/{x^{l_k-1}})$ , we get that

(4.9) $$ \begin{align}x\in\mathbb{R}\mapsto {\frac{(e^{2i\pi x^{l_k}}-1)}{x^{l_k-1}}}{(1+ \vert x\vert^{l_k-1})}\quad\text{is bounded by a constant } C_2>0.\end{align} $$

For all $t\ne 0$ such that $ \vert t\vert>1$ , by (4.8), for all $\alpha>0$ ,

(4.10) $$ \begin{align} x\in \mathbb{R}\mapsto {\check{\psi_k}^{\prime}\bigg(\frac{x}{t^{{1}/{l_k}}}\bigg)}\frac{(1+ \vert x\vert ^\alpha)}{t^{{1}/{l_k}}}\quad\text{is bounded by a constant } C_3>0. \end{align} $$

Thus, by (4.8), there exists $C_4>0$ such that for all $x\in \mathbb {R}$ , for all $t\ne 0$ ,

(4.11) $$ \begin{align} \bigg\vert \frac{(l_k-1)(e^{2i\pi x^{l_k}}-1)}{x^{l_k}}\check{\psi_k}\bigg(\frac{x}{t^{{1}/{l_k}}}\bigg)\bigg\vert{(1+ \vert x\vert^{l_k})}\leq C_4\Vert \psi\Vert _\infty.\end{align} $$

By Lebesgue convergence theorem,(4.9), (4.10), and (4.11),

$$ \begin{align*} \int_{\mathbb{R}} e^{2i\pi x^{l_k}}\check{\psi_k}\bigg(\frac{x}{t^{{1}/{l_k}}}\bigg)\, d\unicode{x3bb}(x)\xrightarrow[t\to\pm\infty]{}\frac{\check{\psi_k}(0)}{2i\pi l_k} \int_{\mathbb{R}} \frac{(l_k-1)(e^{2i\pi x^{l_k}}-1)}{x^{l_k}}\, d\unicode{x3bb}(x).\end{align*} $$

For $k\in {[\![} m+1,m+m^{\prime }{]\!]} ,$ by Green’s theorem, we get

$$ \begin{align*}\int_{\varphi_{U_k}^{-1}(V_k)} e^{2i\pi t{<}\xi\vert v(x){>}}\psi_k(x) \,d\check{\unicode{x3bb}}(x) \in O_{\pm\infty}\bigg(\frac{1}{t}\bigg).\end{align*} $$

Thus,

$$ \begin{align*} & \int_{M} e^{2i\pi t{<}\xi\vert v(x){>}} \,d\check{\unicode{x3bb}}(x) \\ &\quad =\sum_{k=1}^m\frac{1}{t^{{1}/{l_k}}}\bigg(J_{\varphi_{U_k}^{-1}}(\varphi_{U_k}(a_k))\bigg(\frac{l_k!}{{<}\xi\vert v^{(l_k)}(a_k){>}}\bigg)^{{1}/{l_k}}I_{l_k}+o_{\pm\infty}(1)\bigg)+O_{\pm\infty}\bigg(\frac{1}{t}\bigg)\end{align*} $$

with $I_{l} :=({1}/({2i\pi l}))\int _{\mathbb {R}} (({e^{2i\pi x^{l}}-1})/{x^{l}})\, d\unicode{x3bb} (x).$ A simple analysis shows that $I_l$ does not vanish. (For l even, $I_l\in (-\infty ,0)$ trivially, while for l odd, we have $I_l\in i(0,\infty )$ , using a decomposition of the integral on intervals $[2n,2n+2]$ .)

By hypothesis, there exists just one critical a point with maximal order $\ell $ ; hence,

$$ \begin{align*}\int_{M} e^{2i\pi t{<}\xi\vert v(x){>}}\, d\check{\unicode{x3bb}}(x)= \frac{1}{t^{{1}/{\ell}}}\bigg(J_{\varphi_{U_j}^{-1}}(\varphi_{U_j}(a))\bigg(\frac{\ell!}{{<}\xi\vert v^{(\ell)}(a){>}}\bigg)^{{1}/{\ell}}I_\ell+o_{\pm\infty}(1)\bigg).\end{align*} $$

Thus,

$$ \begin{align*}r(\unicode{x3bb}_{{<}\xi\vert v(\cdot){>}})=\frac{1}{\ell}.\\[-41pt]\end{align*} $$

Proof. In the proof of Lemma 4.15, you can take $f_1=e^{2i\pi {<} \xi \vert v(\cdot ){>} }$ and $f_2 = \psi _j$ with j the index of the chart containing the critical point a with a maximal order $\ell $ .

Proof. Let $\xi \ne 0_{\mathbb {Z}^d}$ . Applying the stationary phase theorem on respective charts $(U,\varphi _U)$ of $\mathcal{A}$ containing at most just one critical point $a\in U$ and test functions $\psi $ on U, we get

$$ \begin{align*} \begin{aligned} \int_{M} e^{2i\pi t{<}\xi\vert v(x){>}}\psi(x) \,d\unicode{x3bb}(x) &= \int_{\mathbb{R}^n} e^{2i\pi t{<}\xi\vert v(\varphi_U^{-1}(x)){>}}\psi(\varphi_U^{-1}(x))J(\varphi_U^{-1})(x) \,d\unicode{x3bb}(x)\\ &=\frac{\psi(\varphi_U^{-1}(a))e^{i({\pi}/{4})(\sum_{\unicode{x3bb}\in \sigma(\mathrm{Hess}(a))}\mathrm{sgn}(\unicode{x3bb}))}}{t^{{n}/{2}}\sqrt{\vert \det(\mathrm{Hess}(a))\vert}}+O_{+\infty}\bigg(\frac{1}{t^n}\bigg). \end{aligned} \end{align*} $$

By partition of unity, we obtain

$$ \begin{align*}{\text{ there exists } C\in\mathbb{R}{,}\quad \int_{M} e^{2i\pi{<}\xi\vert v(x){>}}\, d\unicode{x3bb}(x) =\frac{C}{t^{{n}/{2}}}+O_{+\infty}\bigg(\frac{1}{t^n}\bigg)}.\\[-41pt]\end{align*} $$

Proof. Let $s>\tfrac 12$ and $(f_1,f_2)\in (H^{s,0}(\mathbb {R}\times \mathbb {T}))^2$ . Then, by Cauchy–Schwarz inequality and Parseval,

(4.12) $$ \begin{align} \vert \mathbb{E}_{\mu\otimes\unicode{x3bb}}(\mathrm{Cov}_t(f_1,f_2\vert \mathcal{I}))\vert &=\bigg\vert\!\sum_{k\in\mathbb{Z}^*}\int_{\mathbb{R}^2} \overline{\widehat{f_1}}(u,k)\widehat{f_2}(v,k)\widehat{\mu}(kt-(u-v)) \,d\unicode{x3bb}(u,v)\bigg\vert \nonumber\\ &\leq S(t)\Vert f_1\Vert_{H^{s,0}(\mathbb{R}\times\mathbb{T})}\Vert f_2\Vert_{H^{s,0}(\mathbb{R}\times\mathbb{T})} \end{align} $$

with

$$ \begin{align*}S(t) := \sup_{k\in \mathbb{Z}^*} C_\mu\bigg(\int_{\mathbb{R}^2} \frac{1}{(1+\vert kt-(u-v)\vert)^{2r(\mu)}h^{2s}(u)h^{2s}(v)} \,d\unicode{x3bb}(u,v)\bigg)^{{1}/{2}}.\end{align*} $$

We can consider $k{\kern-1pt}={\kern-1pt}1$ considering $kt$ instead of t. Let $D_1(t) {\kern-1pt}:={\kern-1pt} \{(u,v){\kern-1pt}\in{\kern-1pt} \mathbb {R}^2 : \vert t-(u-v)\vert \leq 2t \text { and } \vert v\vert \leq ({t}/{4})\}$ , $D_2(t) := \{(u,v)\in \mathbb {R}^2 : \vert t-(u-v)\vert \leq 2t\text { and } \vert v\vert>({t}/{4})\}$ , and $D_3(t) :=\{(u,v)\in \mathbb {R}^2 : \vert t-(u-v)\vert> 2t\}$ .

We get that for $(u,v)\in D_1(t)$ ,

(4.13) $$ \begin{align} \vert u\vert\geq \frac{3t}{4}\quad \text{and} \quad h^{2s}(u)\geq \frac{16^s}{9^st^{2s}} \end{align} $$

and for $(u,v)\in D_2(t)$ ,

(4.14) $$ \begin{align} h^{2s}(v)> \frac{16^s}{t^{2s}}. \end{align} $$

Using that $s>\tfrac 12$ , for all $\varepsilon \in\ ]0,{1}/{2s}[$ ,

(4.15) $$ \begin{align} \int_{\mathbb{R}} h^{-2s\varepsilon}(u) \,d\unicode{x3bb}(u) <+\infty \end{align} $$

and by (4.13) and (4.14), and for

(4.16) $$ \begin{align} & \int_{D_1(t)} \frac{1}{(1+ \vert t-(u-v)\vert)^{2r(\mu)}h^{2s}(u)h^{2s}(v)} \,d\unicode{x3bb}(u,v) \nonumber\\ &\quad\leq\frac{16^{s(1-\varepsilon)}}{{(3t)}^{2s(1-\varepsilon)}}\int_{\mathbb{R}^2} h^{-2s\varepsilon}(u)h^{-2s}(v)\,d\unicode{x3bb}(u,v), \end{align} $$

(4.17) $$ \begin{align} & \int_{D_2(t)} \frac{1}{(1+\vert t-(u-v)\vert)^{2r(\mu)}h^{2s}(u)h^{2s}(v)} \,d\unicode{x3bb}(u,v) \nonumber\\ &\quad \leq\frac{16^{s(1-\varepsilon)}}{t^{2s(1-\varepsilon)}}\int_{\mathbb{R}^2} h^{-2s}(u)h^{-2s\varepsilon}(v) \,d\unicode{x3bb}(u,v). \end{align} $$

Therefore, by definition of $D_3(t)$ , (4.16), and (4.17),

(4.18) $$ \begin{align} \int_{\mathbb{R}^2}\! \frac{1}{(1+\vert t-(u-v)\vert)^{2r(\mu)}h^{2s}(u)h^{2s}(v)} \,d\unicode{x3bb}(u,v) \leq \frac{D_{s,\varepsilon}^2}{t^{2q_\varepsilon}}\int_{\mathbb{R}^2}\! h^{-2s}(u)h^{-2s\varepsilon}(v) \,d\unicode{x3bb}(u,v) \end{align} $$

with $D_{s,\varepsilon } = \max {\{{1}/({2^{r(\mu )}}),{4}^{s(1-\varepsilon )}\sqrt {2}\}}$ and $q_\varepsilon =\min {\{r(\mu ),s(1-\varepsilon )\}}$ . Thus,

(4.19) $$ \begin{align}\text{ for all } \varepsilon \in\bigg]0,\frac{1}{2s}\bigg[{,} \quad S(t)\leq \frac{C_{\mu}C_{s,\varepsilon}}{t^{q_\varepsilon}}\bigg(\int_{\mathbb{R}} h^{-2s}(u)\, d\unicode{x3bb}(u) \int_{\mathbb{R}} h^{-2s\varepsilon}(v)\, d\unicode{x3bb}(v)\bigg)^{{1}/{2}}.\end{align} $$

By (4.12), (4.15), and (4.19),

$$ \begin{align*}\vert \mathbb{E}_{\mu\otimes\unicode{x3bb}}(\mathrm{Cov}_t(f_1,f_2\vert \mathcal{I}))\vert \leq\frac{C_{s,\varepsilon}}{t^{q_\varepsilon}} \Vert f_1\Vert_{H^{s,0}(\mathbb{R}\times\mathbb{T})}\Vert f_2\Vert_{H^{s,0}(\mathbb{R}\times\mathbb{T})}.\end{align*} $$

Moreover, for $\gamma>0$ , the convergence order on $H^{s,0}(\mathbb {R}{\kern-1pt}\times{\kern-1pt} \mathbb {T})$ , we get ${\min {\{s{\kern-1pt}-{\kern-1pt}\tfrac 12,r(\mu )\}}{\kern-1pt}\leq{\kern-1pt} \gamma }$ and when $\mathrm{supp}(\mu )$ is compact, $\gamma \leq r(\mu )$ .

Proof. Let non-zero $\xi \in \mathbb {Z}^R$ . Let $U\in \mathcal{A}$ and $(f_1,f_2)\in (\mathbb {L}^2_{\mu ^{\prime }_{\vert U}\otimes \mathcal{H}}(U\times G))^2$ . We have the following expression for $b_t(f_1,f_2)$ :

$$ \begin{align*} &=\int_{U\times G} \overline{f_1}(x,\exp(tv_U(x))y)f_2(x,y) \,d\mu^{\prime}_{\vert U}\otimes\mathcal{H}(x,y)\\ &=\int_{U} \int_{G} \overline{f_1}(x,\exp(tv_U(x))y)f_2(x,y)\, d\mathcal{H}(y)\, d\mu^{\prime}_{ \vert U }(x)\\ &=\sum_{k=1}^l\int_{U} \bigg(\int_{h(v_U(x))W_k} \overline{f_1}(x,\check{\varphi^{v_U(x)}}\circ\phi_t^{v_U(x)}\circ\psi_k^{v_U(x)}(y))\\ &\quad\qquad\qquad\times f_2(x,\check{\varphi^{v_U(x)}}\circ\psi_k^{v_U(x)}(y)) d\mathcal{H}(y)\bigg) \,d\mu^{\prime}_{\vert U}(x)\\ &=\sum_{k=1}^l\int_{U} \bigg(\int_{\mathrm{Lie}(T)\times h(v_U(x))W_k} \overline{f_1}(x,\check{\varphi^{v_U(x)}}\circ\phi_t^{v_U(x)}(\theta,r)) \\ &\qquad\qquad\quad \times f_2(x,\check{\varphi^{v_U(x)}}(\theta,r)) dm_{(x,k)}(\theta,r)\bigg) \,d\mu^{\prime}_{\vert U}(x). \end{align*} $$

We have completed the first step by placing ourselves in the Lie algebra of the maximal torus T, which is abelian, so

$$ \begin{align*} \check{\varphi}\circ\phi_t^{v_U(x)}(\theta,r){\kern-1pt}={\kern-1pt}\check{\varphi}(\theta {\kern-1pt}+{\kern-1pt} tv_U(x),r){\kern-1pt}={\kern-1pt}\varphi(\exp(\theta {\kern-1pt}+{\kern-1pt} tv_U(x)),r){\kern-1pt}={\kern-1pt}\varphi(\exp(tv_U(x))\exp(\theta),r). \end{align*} $$

We are now at the second step placing ourselves in the sections of the Lie group G by T. So,

$$ \begin{align*}b_t(f_1,f_2)=\sum_{k=1}^l\int_{U} I(k,U,x)\, d\mu^{\prime}_{\vert U}(x), \end{align*} $$

where the integrand $I(k,U,x)$ is equal to

$$ \begin{align*} &\int_{\mathrm{Lie}(T)\times h(v_U(x))W_k} \overline{f_1}(x,\varphi(\exp(tv_U(x))\chi_{v_U(x)}^{-1}(\chi_{v_U(x)}(e^\theta)),r)) \\ &\quad\times f_2(x,\varphi(e^\theta),r))\ dm_{(x,k)}(\theta,r)\\ &=\int_{\mathbb{T}^{R}\times h(v_U(x))W_k} \overline{f_1}(x,\varphi(\exp(tv_U(x))\chi_{v_U(x)}^{-1}(z),r))f_2(x,\varphi(\chi_{v_U(x)}^{-1}(z),r)) \,dm^{\prime}_{(x,k)}(z,r). \end{align*} $$

We pass here in the torus isomorphic to T. Since

$$ \begin{align*}\exp(tv_U(x))\chi_{v_U(x)}^{-1}(z)=\chi_{v_U(x)}^{-1}(\chi_{v_U(x)}(\exp(tv_U(x)))+z)\end{align*} $$

and

$$ \begin{align*}\chi_{v_U(x)}(\exp(tv_U(x)))+z=z+tD\chi_{v_U(x)}(e)=z+tw_U(x),\end{align*} $$

we get

$$ \begin{align*} &I(k,U,x)\\ &=\int_{\mathbb{T}^{R}\times h(v_U(x))W_k} \overline{f_1}(x,\varphi(\chi_{v_U(x)}^{-1}(z+tw_U(x)),r))f_2(x,\varphi(\chi_{v_U(x)}^{-1}(z),r))\, dm^{\prime}_{(x,k)}(\theta,r). \end{align*} $$

This proves (6.11).

Proof of Theorem 6.2.

Let us begin with the direct implication. Suppose that the dynamical system $(\Omega ,\mu ,(g_t)_{t\in \mathbb {R}})$ exhibits Keplerian shear. Let $U\in \mathcal{A}$ and let

$$ \begin{align*}\xi\in\mathbb{Z}^R\setminus\{0\}.\end{align*} $$

Let $(f_1,f_2)\in (\mathbb {L}^2_{\mu ^{\prime }_{\vert U} \otimes \mathcal{H}}(U\times G))^2$ such that

$$ \begin{align*}\check{f_1} : (x,z,r) \mapsto e^{2i\pi{<}\xi\vert z{>}}\end{align*} $$

and

$$ \begin{align*}\check{f_2} : (x,z,r) \mapsto \mathbb{1}_{{<}\xi\vert w_U(\cdot){>}\ne 0}(x)e^{2i\pi{<}\xi\vert z{>}}\end{align*} $$

taking the notation of Lemma 6.3. Then,

$$ \begin{align*}\int_{\mathbb{R}} e^{-2i\pi tz}\, d\nu^{\xi,U}(z)=\int_{U} e^{-2i\pi t{<}\xi\vert w_U(x){>}}g(x) \,d\mu^{\prime}(x)\end{align*} $$

with

$$ \begin{align*}g : x\in U\mapsto \mathbb{1}_{{<}\xi\vert w_U(\cdot){>}\ne 0}(x).\end{align*} $$

Thus, by Lemma 6.3,

$$ \begin{align*}\int_{U} e^{-2i\pi t{<}\xi\vert w_U(x){>}}g(x)\, d\mu^{\prime}_{\vert U }(x)= \int_{U\times G} \overline{f_1}(x,\exp(tv_U(x))y)f_2(x,y) \,d \mu^{\prime}_{\vert U}\otimes\mathcal{H}(x,y).\end{align*} $$

By the Keplerian shear property, the right-hand term converges and as in (4.1), the limit is zero. So,

$$ \begin{align*}{\int_{\mathbb{R}} e^{-2i\pi tz} \,d\nu^{\xi,U}(z)\xrightarrow[t\to\pm\infty]{}0}.\end{align*} $$

Let us continue with the reciprocal implication. Let R be the dimension of the maximal torus of G.

Consider $k\in {[\![} 1,l{]\!]} $ and define the measure $\omega ^k$ on $U\times \mathbb {T}^R\times W_k$ by

$$ \begin{align*} &\text{ for all } A\in \mathcal{B}(U\times \mathbb{T}^R\times h(v_U(x))W_k), \\ &\quad \omega^k(A) =\int_{U} \int_{\mathbb{T}^R\times h(v_U(x))W_k} \mathbb{1}_A(x,y,z) \,dm^{\prime}_{(x,k)}(y,z)\, d\mu^{\prime}_{\vert U}(x). \end{align*} $$

Let $U\in \mathcal{A}$ and $(\xi _1,\xi _2)\in (\mathbb {Z}^R)^2$ . Let $(f_1,f_2)\in (\mathbb {L}^2_{\mu ^{\prime }_{\vert U }\otimes \mathcal{H}}(U\times G))^2$ such that,

$$ \begin{align*}\check{f_1} : (x,z,r) \mapsto a_1(x,r)e^{2i\pi{<}\xi_1\vert z{>}}\end{align*} $$

and

$$ \begin{align*}\check{f_2} : (x,z,r) \mapsto a_2(x,r)e^{2i\pi{<}\xi_2\vert z{>}} \end{align*} $$

with $a_1$ and $a_2$ square summable functions in the appropriate Lebesgue space. If ${m\neq m'}$ , the scalar product between $f_1,f_2$ is zero. Therefore, we assume that $m=m'$ . So, by Lemma 6.3,

(6.12) $$ \begin{align}\int_{U\times G} \overline{f_1}(x,\exp(tv_U(x))y)f_2(x,y)\, d\mu^{\prime}_{ \vert U }\otimes\mathcal{H}(x,y)=\int_{U} e^{-2i\pi t{<}\xi_1\vert w_U(x){>}}g(x) \,d\mu^{\prime}_{\vert U }(x)\end{align} $$

with

$$ \begin{align*}g : x\in U\mapsto \sum_{k=1}^l\bigg(\int_{h(v_U(x))W_k} \overline{a_1}(x,r)a_2(x,r)e^{2i\pi{<}\xi_2-\xi_1\vert z {>}} \,dm^{\prime}_{(x,k)}(z,r)\bigg).\end{align*} $$

By hypothesis, when $\xi _1\ne 0$ ,

$$ \begin{align*}\int_{\mathbb{R}} e^{-2i\pi tz}\, d\nu^{\xi_1,U}(z)\xrightarrow[t\to\pm\infty]{}0.\end{align*} $$

Then, by Lemma 2.4, we have that $\exp (2i\pi t\cdot )$ converge weakly- $*$ in $\mathbb {L}^{\infty }_{\nu ^{\xi _1,U}}(\mathbb {R})$ to $0$ . Thus,

$$ \begin{align*} & \int_{U} e^{-2i\pi t{<}\xi_1\vert w_U(x){>}}g(x)\, d \mu^{\prime}_{\vert U } (x) \\ &\quad \xrightarrow[t\to\pm\infty]{} \int_{\{{<}\xi_1\vert w_U(\cdot){>}=0\}\times G} \overline{f_1}(x,y)f_2(x,y) \,d\mu^{\prime}_{\vert U}\otimes \mathcal{H}(x,y). \end{align*} $$

Then, remembering (6.12), the integral

$$ \begin{align*} \int_{U\times G} \overline{f_1}(x,\exp(tv_U(x))y)f_2(x,y) \,d \mu^{\prime}_{\vert U }\otimes\mathcal{H}(x,y) \end{align*} $$

converges as $t\rightarrow \pm \infty $ towards

$$ \begin{align*} \int_{\{{<}\xi_1\vert w_U(\cdot){>}=0\}\times G} \overline{f_1}(x,y)f_2(x,y) \,d\mu^{\prime}_{\vert U}\otimes \mathcal{H}(x,y). \end{align*} $$

By totality obtained by the Fourier series expansion, this gives that for all functions $f_1,f_2\in \mathbb {L}^2_{\mu ^{\prime }_{\vert U}\otimes \mathcal{H}}(U\times G)$ , the integral

$$ \begin{align*} \int_{U\times G} \overline{f_1}(x,\exp(tv_U(x))y)f_2(x,y)\, d \mu^{\prime}_{\vert U }\otimes\mathcal{H}(x,y) \end{align*} $$

converges as $t\to \pm \infty $ towards

(6.13) $$ \begin{align} \int_{U\times G} \overline{E_{\mu^{\prime}_{\vert U}\otimes\mathcal{H}}(f_1\vert \mathcal{I}_U)}E_{\mu^{\prime}_{\vert U}\otimes\mathcal{H}}(f_2\vert \mathcal{I}_U) \,d\mu^{\prime}_{\vert U}\otimes\mathcal{H}. \end{align} $$

Proof. Let $(e_i)$ be the canonical basis of $\mathbb {R}^d$ and denote by $\mathcal{C}_i$ the cone of vectors $h\neq 0$ such that the angle between h and $e_i$ is less than or equal to $\pi /8$ .

Let $a\in \mathbb {R}^*$ , and set $A=\{f=a\}$ and $A^{\prime }$ the set of Lebesgue density points of A, that is, $\unicode{x3bb} (A\cap B(x,r))/\unicode{x3bb} (B(x,r))\to 1$ as $r\to 0$ for each $x\in A^{\prime }$ .

Suppose that $x\in A^{\prime }$ . For each $i=1,\ldots ,d$ , there exists a sequence $h_n^i\in \mathcal{C}_i$ such that $x+h_n^i\in A$ and $h_n^i\to 0$ . Without loss of generality, we may assume that $h_n^i/\|h_n^i\|$ converges to some $h^i\in \mathcal{C}_i$ . Since $f(x+h_n^i)=a=f(x)$ , a first-order expansion shows that $d_xf(h^i)=0$ . Since the $h_i$ form a basis of $\mathbb {R}^d$ , this implies that $d_xf=0$ .

By assumption, we get that $A^{\prime }$ , and therefore A, has zero measure.

Proof of Theorem 6.4.

Let $U\in \mathcal{A}$ . Let us show that the measures $\nu ^{(\xi ,U)}$ have the Rajchman property.

Let $\xi \in \mathbb {Z}^d \setminus \{0\}$ . By Radon–Nikodym,

$$ \begin{align*}\int_{U} e^{-2i\pi t{<}\xi\vert w_U(x){>}} \,d\mu^{\prime}_{\vert U }(x)=\int_{U} e^{-2i\pi t{<}\xi\vert w_U(x){>}}\frac{d\mu^{\prime}_{\vert U}}{d\unicode{x3bb}}(x) \,d\unicode{x3bb}(x).\end{align*} $$

However,

$$ \begin{align*}\mu\bigg({\bigcup_{\xi\in \mathbb{Z}^d \setminus \{0\}}} \{x\in U : d{<}\xi\vert w_U(x){>}=0\}\setminus\{x\in U : {<}\xi\vert w_U(x){>}=0\}\bigg)=0.\end{align*} $$

Thus, by Lemma 6.5,

$$ \begin{align*}\text{for all } a\in\mathbb{R}^*{,}\quad \mu\bigg({\bigcup_{\xi\in \mathbb{Z}^d \setminus \{0\}}} \{x\in U : {<}\xi\vert w_U(x){>}=a\}\bigg)=0.\end{align*} $$

By local normal form of submersions, we can choose a countable family $(\varphi _k, V_k)_{k\in \mathbb {N}}$ of open charts on M which are pairwise disjoints, cover $U\setminus ({\bigcup _{\xi \in \mathbb {Z}^d \setminus \{0\}}} \{x\in U : d{<} \xi \vert w_U(x){>} =0\})$ up to a Lebesgue-negligible set, and for all $k\in \mathbb {N}$ and all $x\in \varphi _k(V_k)$ ,

$$ \begin{align*}{<} \xi\vert w_U(\varphi^{-1}_k(x)){>}=x_1.\end{align*} $$

Therefore, for $k\in \mathbb {N}$ ,

$$ \begin{align*}\int_{V_k} e^{-2i\pi t{<}\xi\vert w_U(x){>}}\frac{d\mu^{\prime}_{\vert U}}{d\unicode{x3bb}}(x)\, d\unicode{x3bb}(x)=\int_{\varphi_k(V_k)} e^{-2i\pi tz}\frac{d\mu^{\prime}_{\vert U}}{d\unicode{x3bb}}(\varphi_k^{-1}(z))\vert\!\det(J_{\varphi_k^{-1}})(z)\vert\, d\unicode{x3bb}(z).\end{align*} $$

However, we have

$$ \begin{align*} & \int_{U} e^{-2i\pi t{<}\xi\vert w_U(x){>}}\frac{d\mu^{\prime}_{\vert U}}{d\unicode{x3bb}}(x) \,d\unicode{x3bb}(x) \\ &\quad = \int_{{<}\xi\vert w_U{>}=0} \frac{d\mu^{\prime}_{\vert U}}{d\unicode{x3bb}}(x) \,d\unicode{x3bb}(x) +\sum_{k\in\mathbb{N}}\int_{V_k} e^{-2i\pi t{<}\xi\vert w_U(x){>}}\frac{d\mu^{\prime}_{\vert U}}{d\unicode{x3bb}}(x) \,d\unicode{x3bb}(x). \end{align*} $$

Thus, by Riemann–Lebesgue,

$$ \begin{align*}\int_{U} e^{-2i\pi t{<}\xi\vert w_U(x){>}}\frac{d\mu^{\prime}_{\vert U}}{d\unicode{x3bb}}(x)\, d\unicode{x3bb}(x) \xrightarrow[t\to+\infty]{} \int_{{<}\xi\vert w_U{>}=0} \frac{d\mu^{\prime}_{\vert U}}{d\unicode{x3bb}}(x) \,d\unicode{x3bb}(x).\end{align*} $$

Therefore,

$$ \begin{align*} \int_{U} e^{-2i\pi t{<}\xi\vert w_U(x){>}} \,d \mu^{\prime}_{\vert U }(x)\xrightarrow[t\to+\infty]{}\mu^{\prime}_{\vert U}(\{{<}\xi\vert w_U{>}=0\}).\end{align*} $$

By Theorem 6.2, we get that $(\Omega ,\mu ,(g_t)_{t\in \mathbb {R}})$ exhibits Keplerian shear.

Proof. Let $t\in\ ]-1,1[$ and $v\in Li e(G)\setminus \{0\}$ . Thus, $\bar h(v+tv)=h ({1+t}/{(1+t)\Vert v\Vert }v)=\bar h(v).$ Therefore,

$$ \begin{align*}D\bar h(v)(v)=0.\\[-37pt]\end{align*} $$

Proof. Let $w\in W$ such that $S_\xi (\psi ^{-1}(w))\ne 0$ . By Lemma 6.7, for $u\in \mathbb {R}^d$ , we get

$$ \begin{align*} & D(S_\xi\circ \psi^{-1})(w)(u)\\ &\quad=\bigg(\!\!\begin{array}{ll} {<} \xi\vert D\chi(e)(\mathrm{Ad}_{\bar h(\psi^{-1}(w))}\circ D\psi^{-1}(w)(u)){>}\\+{<}\xi\vert D\chi(e)\circ \mathrm{Ad}_{\bar h (\psi^{-1}(w))}\circ[D\bar h({\psi^{-1}(w)})\circ D\psi^{-1}(w)(u),\psi^{-1}(w)]{>}\end{array}\!\!\bigg).\end{align*} $$

Taking $u=(D\psi ^{-1}(w))^{-1}\circ \psi ^{-1}(w)$ , by Lemma 6.8,

$$ \begin{align*}D(S_\xi\circ \psi^{-1})(w)(u)=S_\xi(\psi^{-1}(w))\ne 0.\\[-35pt]\end{align*} $$

Proof. Let $U\subset M$ and $x\in U$ . Let $v : U\to V\subset \mathrm{Lie}(G)$ be a $\mathcal{C}^1$ function and there be subimmersion around x such that $k=\mathrm{rank}(Dv(x))\geq 1$ . Using the local normal form of subimmersions, there exists $V^{\prime }\subset \mathbb {R}^d$ , $U^{\prime }$ open sets and $\mathcal{C}^1$ -diffeomorphisms $\psi : V\to V^{\prime }$ and $\varphi : U\to U^{\prime }$ such that for all $z\in U^{\prime }$ ,

$$ \begin{align*}\psi(v(\varphi^{-1}(z)))=(z_1,\ldots,z_k,0,\ldots,0).\end{align*} $$

Let $a\in U$ . Therefore,

$$ \begin{align*}D(\psi\circ v)(a)(u)=((D\varphi(a)(u))_1,\ldots,(D\varphi(a)(u))_k,0,\ldots,0).\end{align*} $$

Taking $u=(D\varphi (a))^{-1}(\varphi (a))$ , we get

$$ \begin{align*}D(\psi\circ v)(a)(u)=\psi(v(a)).\\[-37pt]\end{align*} $$

Proof. By hypothesis, $v_U$ is a subimmersion with a rank greater than $1\, \mu $ -almost everywhere. Thus, by Lemma 6.10, for $\mu $ -almost all $x\in U$ , there exists a neighborhood W of x, V an open neighborhood of $v_U(x)$ , $\psi : V\to V^{\prime }\subset \mathrm{Lie}(G)$ a $\mathcal{C}^1$ -diffeomorphism such that for all $a\in W$ ,

$$ \begin{align*}\psi(v_U(a))\in \mathrm{Im}(D(\psi\circ v_U)(a)).\end{align*} $$

Let $x\in U$ such that ${<} \xi \vert w_U(x){>} \ne 0$ . Therefore,

$$ \begin{align*}{<}\xi\vert D\chi(e)(\mathrm{Ad}_{\bar h(v_U(x))}(\psi^{-1}(\psi(v_U(x))))){>}\ne 0.\end{align*} $$

We have

$$ \begin{align*}d{<}\xi\vert w_U(x){>}(u)=D(S_\xi\circ \psi^{-1})(\psi\circ v_U(x))(D(\psi\circ v_U)(x)(u)).\end{align*} $$

By Lemma 6.10, we can get u such that

$$ \begin{align*}d{<}\xi\vert w_U(x){>}(u)=D(S_\xi\circ \psi^{-1})(\psi\circ v_U(x))(v_U(x))={<}\xi\vert w_U(x){>}\ne 0.\end{align*} $$

Thus,

$$ \begin{align*}\mu\bigg({\bigcup_{\xi\in \mathbb{Z}^d \setminus \{0\}}} \{x\in U : d{<}\xi\vert w_U(x){>}=0\}\setminus\{x\in U : {<}\xi\vert w_U(x){>}=0\}\bigg)=0.\\[-47pt] \end{align*} $$

Proof of Corollary 6.6.

By Lemma 6.11, we get that $w_{U_j}$ is $\mathcal{C}^1$ and

$$ \begin{align*}\mu\bigg({\bigcup_{\xi\in \mathbb{Z}^d \setminus \{0\}}} \{x\in U_j : d{<}\xi\vert w_{U_j}(x){>}=0\}\setminus\{x\in U_j : {<}\xi\vert w_{U_j}(x){>}=0\}\bigg)=0.\end{align*} $$

Applying Theorem 6.4, we get the result.

Proof. Let $z\in G$ . By Proposition 6.1(ii), we get that $F(z)\ne \emptyset $ . Let $g\in \overline {F(z)}$ . There exists $(g_n)_{n\in \mathbb {N}}\in (F(z))^{\mathbb {N}}$ such that

$$ \begin{align*}\displaystyle g_n\xrightarrow[n\to+\infty]{}g.\end{align*} $$

Thus, for all n, $g_n^{-1}zg_n\in T$ . We know that T is closed and G is a Lie group, thus,

$$ \begin{align*}g^{-1}zg\in T.\\[-35pt]\end{align*} $$

Proof. Let

$$ \begin{align*}P : (g,t)\in G\times T\mapsto gtg^{-1}.\end{align*} $$

Here, P is continuous because G is a Lie group. Let U be an open set of G. We see immediately that

$$ \begin{align*}F^{-1}(U)=P(U\times T).\end{align*} $$

We know that G is a compact connected Lie group; thus, G is metrizable and U is a countable union of compact sets. By compactness of T and continuity of P, $P(U\times T)$ is a countable union of compact set. Then, $F^{-1}(U)$ is a measurable set.