The horocycle flow is mixing of all orders

— 1. Introduction —

The main purpose of this post is to present a proof, due to Brian Marcus, that the horocycle flow is mixing of all orders. The precise definition of mixing of all orders for ${{\mathbb R}}$-actions is given below in Definition 2; we begin by describing the horocycle flow. Let ${SL(2,{\mathbb R})}$ denote the set of all ${2\times2}$ matrices with real entries and determinant ${1}$, endowed with the usual topology and the (left and right invariant) Haar measure ${\lambda}$. Given a discrete subgroup ${\Gamma\subset SL(2,{\mathbb R})}$, the quotient ${X:=SL(2,{\mathbb R})/\Gamma}$ is given the quotient topology and quotient Haar measure’ ${\mu}$. The latter can be described by

$\displaystyle \int_X\sum_{\gamma\in\Gamma}f(x\gamma)~\mathtt{d}\mu(x)=\int_{SL(2,{\mathbb R})}f(g)~\mathtt{d}\lambda(g)$

for any ${f\in C_c(SL(2,{\mathbb R}))}$. If the total measure ${\mu(X)}$ of ${X}$ is finite (in which case we normalize it so that ${\mu(X)=1}$), we say that ${\Gamma}$ is a lattice. The classical example is when ${\Gamma=SL(2,{\mathbb Z})}$.

The horocycle flow is the (continuous) ${{\mathbb R}}$-action on ${X}$ defined by ${s\cdot(g\Gamma)=(h_sg)\Gamma}$, where ${h_s=\left[\begin{array}{cc} 1 & s \\ 0 & 1 \end{array}\right]}$ for ${s\in{\mathbb R}}$ and ${g\in SL(2,{\mathbb R})}$. We will also need the geodesic flow, corresponding in this case to ${g_t=\left[\begin{array}{cc} e^{t/2} & 0 \\ 0 &e^{-t/2} \end{array}\right]}$.

The theorem we will prove actually deals with a more general situation:

Theorem 1 Let ${X}$ be a (finite dimensional) manifold, let ${\mu}$ a Borel probability measure on ${X}$ and let ${(h_s)_{s\in{\mathbb R}}}$ and ${(g_t)_{t\in{\mathbb R}}}$ be continuous ${\mu}$-preserving ${{\mathbb R}}$-actions. If ${(h_s)_{s\in{\mathbb R}}}$ is ergodic and there exists ${\lambda>1}$ such that

$\displaystyle g_th_sg_t^{-1}=h_{s\lambda^t}, \ \ \ \ \ (1)$

then ${(h_s)_{s\in{\mathbb R}}}$ is mixing of all orders.

The first step of the proof of Theorem 1 is to show that ${(h_s)_{s\in{\mathbb R}}}$ satisfies a weaker property, called property ${P(N)}$ (see Definition 3) which resembles a property equivalent to weakly mixing of order ${N}$ (at least in the case of ${{\mathbb Z}}$-actions). For this step, all we need is that ${(h_s)_{s\in{\mathbb R}}}$ is a continuous probability preserving mixing ${{\mathbb R}}$-action .

— 2. Notions of mixing —

In this section I motivate, state and compare some notions of mixing for probability preserving ${{\mathbb R}}$-actions. A secondary purpose of this section is to familiarize the reader with the notation for ${{\mathbb R}}$-actions.

The most basic notion of mixing is simply ergodicity. According to the mean ergodic theorem, a probability preserving system ${(X,{\mathcal B},\mu,(h_s)_{s\in{\mathbb R}})}$ satisfies this property if and only if

$\displaystyle \forall f\in L^2(X)\qquad\lim_{N-M\rightarrow\infty}\frac1{N-M}\int_M^Nh_sf~\mathtt{d}s=\int_Xf~\mathtt{d}\mu\qquad\text{in }L^2 \ \ \ \ \ (2)$

Here and in the rest of this post we use (as a slight abuse of language) the notation ${h_sf}$ to denote the function ${f\circ h_s}$. The convergence in (2) is in the strong topology of the Hilbert space ${L^2(X)}$, which trivially implies convergence in the weak topology. Not so immediate is the fact that the converse is also true, i.e. the system ${(X,{\mathcal B},\mu,(h_s)_{s\in{\mathbb R}})}$ is ergodic if and only if

$\displaystyle \forall f,g\in L^2(X)\quad\lim_{N-M\rightarrow\infty}\frac1{N-M}\int_M^N\int_Xh_sf\cdot g~\mathtt{d}\mu~\mathtt{d}s=\int_Xf~\mathtt{d}\mu\int_Xg~\mathtt{d}\mu \ \ \ \ \ (3)$

One can arrive at stronger notions of mixing by replacing the (uniform) Cesàro convergence with stronger modes of convergence. For instance, using strong Cesàro convergence (discussed in a previous post) we obtain the notion of weak mixing:

$\displaystyle \forall f,g\in L^2(X)\quad\lim_{N-M\rightarrow\infty}\frac1{N-M}\int_M^N\left|\int_Xh_sf\cdot g\mathtt{d}\mu\mathtt{d}s-\int_Xf\mathtt{d}\mu\int_Xg\mathtt{d}\mu\right|=0$

Using regular limits, we obtain the notion of strong mixing (or just mixing):

$\displaystyle \forall f,g\in L^2(X)\qquad\lim_{s\rightarrow\infty}\int_Xh_sf\cdot g~\mathtt{d}\mu=\int_Xf~\mathtt{d}\mu\int_Xg~\mathtt{d}\mu \ \ \ \ \ (4)$

In words, equation (4) states that for any function ${f\in L^2}$, the orbit ${h_sf}$ converges to the constant function ${\int fd\mu}$ in the weak topology. The analogue statement with convergence in the strong topology would be rather trivial (only the one point system would satisfy it!), because ${\|h_sf\|_{L^2}=\|f\|_{L^2}}$ for all ${s\in{\mathbb R}}$, and hence any non-zero function ${f\in L^2(X)}$ with ${\int fd\mu=0}$ would contradict such a strong form of mixing’.

To get stronger forms of mixing than (4), one can consider higher order correlations, for instance of the form ${\int_Xh_sf_1\cdot h_uf_2\cdot f_3~\mathtt{d}\mu}$, when ${s,u,s-u}$ all go to infinity. In general we have the notion of strong mixing of order ${N}$.

Definition 2 (Strong mixing of order ${N}$) Let ${(X,{\mathcal B},\mu,(h_s)_{s\in{\mathbb R}})}$ be a probability preserving system and let ${N\in{\mathbb N}}$. We say that the system is strongly mixing of order ${N}$ if for any ${f_0,\dots,f_N\in L^\infty(X)}$ and any sequences ${s_0,\dots,s_N:{\mathbb N}\rightarrow{\mathbb R}}$ satisfying ${s_i(\ell)-s_{i-1}(\ell)\rightarrow\infty}$ as ${\ell\rightarrow\infty}$ for any ${i=1,\dots,N}$ we have

$\displaystyle \lim_{\ell\rightarrow\infty}\int_Xh_{s_0(\ell)}f_0\cdots h_{s_N(\ell)}f_N~\mathtt{d}\mu=\int_Xf_0~\mathtt{d}\mu\cdots\int_Xf_N~\mathtt{d}\mu\ \ \ \ \ (5)$

We say that the system is strongly mixing of all orders if it is strong mixing of order ${N}$ for every ${N\in{\mathbb N}}$.

For each ${N\in{\mathbb N}}$, it is currently unknown whether strong mixing of order ${N}$ implies strong mixing of order ${N+1}$. Observe that when ${N=1}$, (5) reduces to (4). In the same way that (4) is a strengthening of (3), which is then equivalent to (2), we can find a weakening of (5) using a Cesàro limit but the strong topology.

The most straightforward way to do this would arguably be to replace (5) with

$\displaystyle \lim_{N-M\rightarrow\infty}\frac1{N-M}\int_M^Nh_{s_1(u)}f_1\cdots h_{s_N(u)}f_N~du=\int_Xf_1~d\mu\cdots\int_Xf_N~d\mu\quad\text{in }L^2$

One issue with this definition is that the average is over ${u}$, but the orbits are at times ${s_i(u)}$. Thus one is comparing an interval of size ${s_1(N)-s_1(M)}$ in the orbit of ${f_1}$ with an interval of size ${s_2(N)-s_2(M)}$ in the orbit of ${f_2}$.

To motivate the next definition, we first look at an equivalent way to express strong mixing of order ${N}$. If we let ${K_i=s_i/s_N}$ it’s not hard to see that a system is strongly mixing of order ${N}$ if and only if for any ${f_0,\dots,f_N\in L^\infty(X)}$, any sequences ${K_0,K_1,\dots,K_N:{\mathbb N}\rightarrow{\mathbb R}}$ satisfying ${0\equiv K_0\leq K_1\leq\cdots and ${\big(K_i(\ell)-K_{i-1}(\ell)\big)\ell\rightarrow\infty}$ as ${\ell\rightarrow\infty}$ for ${i=1,\dots,N}$ we have

$\displaystyle \lim_{\ell\rightarrow\infty}\int_Xf_0h_{K_1(\ell)\ell}f_1\cdots h_{K_{N-1}(\ell)\ell}f_{N-1} h_\ell f_Nd\mu=\int_Xf_0d\mu\cdots\int_Xf_Nd\mu\ \ \ \ \ (6)$

One can now formulate an analogue of (6) in the strong topology, but using a Cesàro limit.

Definition 3 (Property ${P(N)}$) Let ${(X,{\mathcal B},\mu,(h_s)_{s\in{\mathbb R}})}$ be a probability preserving system and let ${N\in{\mathbb N}}$. We say that the system satisfies the property ${P(N)}$ if for any ${f_1,\dots,f_N\in C_c(X)}$ and any sequences ${m,n,K_0,\dots,K_N:{\mathbb N}\rightarrow{\mathbb R}}$ satisfying

• ${n(\ell)-m(\ell)\rightarrow\infty}$ as ${\ell\rightarrow\infty}$,
• For all ${\ell\in{\mathbb N}}$, ${0=K_0(\ell),
• For ${i=1,\dots,N}$ we have ${\big(K_i(\ell)-K_{i-1}(\ell)\big)n(\ell)\rightarrow\infty}$ as ${\ell\rightarrow\infty}$,

we have

$\displaystyle \lim_{\ell\rightarrow\infty}\frac1{n(\ell)-m(\ell)}\int_{m(\ell)}^{n(\ell)}\prod_{i=1}^Nh_{K_i(\ell)u}f_i~\mathtt{d}u= \prod_{i=1}^N\int_Xf_i~\mathtt{d}\mu\qquad\text{in }L^2 \ \ \ \ \ (7)$

Observe that when ${N=1}$, (7) reduces to (2). Hence ${P(1)}$ is equivalent to ergodicity.

— 3. Mixing implies ${P(N)}$ for all ${N\in{\mathbb N}}$. —

In this section we prove that the horocycle flow (and indeed any mixing continuous probability preserving ${{\mathbb R}}$-action) satisfies the property ${P(N)}$ for every ${N\in{\mathbb N}}$. The proof proceeds by induction on ${N}$ . The main idea is to adapt a trick of Furstenberg, showing that a weak mixing system is weak mixing of all orders, in a certain sense. I have presented a modern version of this trick in a previous post, using the so-called van der Corput trick.

In this post I will adapt the van der Corput trick to this situation.

Lemma 4 [Robust van der Corput trick] Let ${H}$ be a Hilbert space, let ${(\phi_{\ell,u})_{\ell\in{\mathbb N},u\in{\mathbb R}}\subset H}$ be a bounded family of vectors in ${H}$, depending continuously on ${u}$, and let ${n,m:{\mathbb N}\rightarrow{\mathbb R}}$ be sequences such that ${n(\ell)-m(\ell)\rightarrow\infty}$ as ${\ell\rightarrow\infty}$. If

$\displaystyle \lim_{D\rightarrow\infty}\frac1D\int_0^D\limsup_{\ell\rightarrow\infty}\left|\frac1{n(\ell)-m(\ell)} \int_{m(\ell)}^{n(\ell)}\langle\phi_{\ell,u},\phi_{\ell,u+d}\rangle~\mathtt{d}u\right|\mathtt{d}d=0,\ \ \ \ \ (8)$

then

$\displaystyle \lim_{\ell\rightarrow\infty}\left\|\frac1{n(\ell)-m(\ell)}\int_{m(\ell)}^{n(\ell)}\phi_{\ell,u}~\mathtt{d}u\right\|=0$

where the integral is in the Bochner sense.

Proof: Without loss of generality, we will assume that ${\|\phi_{\ell,u}\|\leq1}$ for all ${\ell,u}$. Then for each ${d>0}$ and ${n>m}$ we have

$\displaystyle \left\|\int_m^n\phi_{\ell,u}~\mathtt{d}u- \int_m^n\phi_{\ell,u+d}~\mathtt{d}u\right\|<2d$

and hence for ${D>0}$

$\displaystyle \left\|\frac1{n-m}\int_m^n\phi_{\ell,u}~\mathtt{d}u- \frac1D\int_0^D\frac1{n-m}\int_m^n\phi_{\ell,u+d}~\mathtt{d}u~\mathtt{d}d\right\|<\frac{2D}{n-m} \ \ \ \ \ (9)$

Fix ${\epsilon>0}$ arbitrary and take a large ${D}$ and ${\ell}$, depending on ${\epsilon}$. More precisely, take ${D_0}$ so that for any ${D>D_0}$

$\displaystyle \frac1D\int_0^D\limsup_{\ell\rightarrow\infty}\left|\frac1{n(\ell)-m(\ell)} \int_{m(\ell)}^{n(\ell)}\langle\phi_{\ell,u},\phi_{\ell,u+d}\rangle~\mathtt{d}u\right|\mathtt{d}d<\epsilon\ \ \ \ \ (10)$

Next let ${D=D_0/\epsilon}$ and define the sequence of functions ${(f_\ell)_{\ell\in{\mathbb N}}}$ on ${[D_0,D]}$ by

$\displaystyle f_\ell(D')=\frac1{D'}\int_0^{D'}\left|\frac1{n(\ell)-m(\ell)} \int_{m(\ell)}^{n(\ell)}\langle\phi_{\ell,u},\phi_{\ell,u+d}\rangle~\mathtt{d}u\right|~\mathtt{d}d$

It follows from Fatou’s lemma and (10) that ${\limsup_{\ell\rightarrow\infty}f_\ell(D')<\epsilon}$ for each ${D'\in[D_0,D]}$. Moreover, it is not hard to check that ${|f_\ell(D_1)-f_\ell(D_2)|\leq|D_1-D_2|}$, so we can choose ${\ell}$ so that ${n(\ell)-m(\ell)>D/\epsilon}$ and

$\displaystyle \forall D'\in[D_0,D]\qquad\frac1{D'}\int_0^{D'}\left|\frac1{n(\ell)-m(\ell)}\int_{m(\ell)}^{n(\ell)}\langle \phi_{\ell,u},\phi_{\ell,u+d}\rangle~\mathtt{d}u\right|~\mathtt{d}d<\epsilon\ \ \ \ \ (11)$

In view of (9), it suffices to show that

$\displaystyle \left\|\frac1D\int_0^D\frac1{n(\ell)-m(\ell)}\int_{m(\ell)}^{n(\ell)}\phi_{\ell,u+d}~\mathtt{d}u~\mathtt{d}d\right\| <8\epsilon$

From now on until the end of the proof ${\ell}$ is fixed, so we will omit it (i.e., we will write ${m,n,\phi_u}$ instead of ${m(\ell),n(\ell),\phi_{\ell,u}}$). Finally, using Jensen’s inequality we compute

$\begin{array}{rl} &\displaystyle \left\|\frac1D\int_0^D\frac1{n-m}\int_m^n \phi_{u+d}~\mathtt{d}u~\mathtt{d}d\right\|^2 \\ \text{(Jensen's inequality)}\leq~&~\displaystyle \frac1{n-m}\int_m^n\left\|\frac1D\int_0^D \phi_{u+d}~\mathtt{d}d\right\|^2~\mathtt{d}u \\=~&~\displaystyle \frac1{n-m}\int_m^n\frac1D\int_0^D\frac1D\int_0^D \langle\phi_{u+d_1},\phi_{u+d_2}\rangle~\mathtt{d}d_2~\mathtt{d}d_1~\mathtt{d}u \\=~&~\displaystyle \frac1{n-m}\int_m^n\frac1D\int_0^D\frac1D\int_{d_1}^D 2\,\mathrm{Re}\langle\phi_{u+d_1},\phi_{u+d_2}\rangle~\mathtt{d}d_2~\mathtt{d}d_1~\mathtt{d}u \\=~&~\displaystyle 2\,\mathrm{Re}\frac1D\int_0^D\frac1D\int_{d_1}^D\frac1{n-m}\int_{m+d_1}^{n+d_1} \langle\phi_u,\phi_{u+d_2-d_1}\rangle~\mathtt{d}u~\mathtt{d}d_2~\mathtt{d}d_1 \\=~&~\displaystyle 2\,\mathrm{Re}\frac1D\int_0^D\frac1D\int_0^{D-d_1}\frac1{n-m}\int_{m+d_1}^{n+d_1} \langle\phi_u,\phi_{u+d_3}\rangle~\mathtt{d}u~\mathtt{d}d_3~\mathtt{d}d_1 \\ \leq~&~\displaystyle \frac{4D}{n-m}+2\,\mathrm{Re}\frac1D\int_0^D\frac1D\int_0^{D-d_1}\left|\frac1{n-m}\int_m^n \langle\phi_u,\phi_{u+d_3}\rangle~\mathtt{d}u\right|~\mathtt{d}d_3~\mathtt{d}d_1 \\ \leq~&~\displaystyle 4\epsilon+\frac{2D_0}D+2\,\mathrm{Re}\frac1D\int_0^{D-D_0}\frac1D\int_0^{D-d_1}\left|\frac1{n-m}\int_m^n \langle\phi_u,\phi_{u+d_3}\rangle~\mathtt{d}u\right|~\mathtt{d}d_3~\mathtt{d}d_1 \\ \leq~&~\displaystyle 4\epsilon+2\epsilon+2\epsilon~\leq~8\epsilon \end{array}$

where in the last step we used (11).

$\Box$

In order to prove the Property P using the van der Corput trick, we need to bootstrap by induction a stronger property which we call robust property P. The difference is that we now allow the functions ${f_i}$ to also depend on ${\ell}$, as long as they ultimately converge.

Definition 5 (Robust property ${P}$) Let ${(X,{\mathcal B},\mu,(h_s)_{s\in{\mathbb R}})}$ be a probability preserving system and let ${N\in{\mathbb N}}$. We say that the system satisfies the property ${RP(N)}$ if for any ${f_{1,\ell},\dots,f_{N,\ell}}$ and any sequences ${m,n,K_0,\dots,K_N:{\mathbb N}\rightarrow[0,1]}$ satisfying

• For each ${i=1,\dots,N}$ and ${\ell\in{\mathbb N}}$ we have ${f_{i,\ell}\in C_c(X)}$.
• For each ${i=1,\dots,N}$, there exists some ${f_i\in C_c(X)}$ such that ${f_{i,\ell}\rightarrow f_i}$ as ${\ell\rightarrow\infty}$ in ${C_c(X)}$.
• ${n(\ell)-m(\ell)\rightarrow\infty}$ as ${\ell\rightarrow\infty}$,
• For all ${\ell\in{\mathbb N}}$, ${0=K_0(\ell),
• For ${i=1,\dots,N}$ we have ${\big(K_i(\ell)-K_{i-1}(\ell)\big)n(\ell)\rightarrow\infty}$ as ${\ell\rightarrow\infty}$,

we have

$\displaystyle \lim_{\ell\rightarrow\infty}\frac1{n(\ell)-m(\ell)}\int_{m(\ell)}^{n(\ell)}\prod_{i=1}^Nh_{K_i(\ell)u}f_{i,\ell}~\mathtt{d}u =\prod_{i=1}^N\int_Xf_i~\mathtt{d}\mu\qquad\text{in }L^2 \ \ \ \ \ (12)$

Since Proposition 6 holds for a general continuous mixing flow, I use ${T}$ to denote the flow. In this post we only need this proposition for ${T=h}$.

Proposition 6 If a flow is mixing (i.e., satisfies (4)), then ${RP(N-1)\Rightarrow RP(N)}$. Therefore a mixing system satisfies ${RP(N)}$ for every ${N}$.

In fact it suffices to have a weakening of (4) with a strong Cesàro convergence instead of regular convergence (i.e. weak mixing).

Proof: Assume everything is given as in Definition 5 and that ${RP(N-1)}$ holds. We first pass to a subsequence so that the limit ${\tilde K_i=\lim K_i(\ell)}$ exists for each ${i}$. When ${f_N}$ is a constant, it is not hard to see that (12) follows directly from ${RP(N-1)}$. Therefore we can assume that ${\int_Xf_Nd\mu=0}$ (and hence the right hand side of (12) is ${0}$). In this case, letting

$\displaystyle \phi_{\ell,u}=\prod_{i=1}^NT_{K_i(\ell)u}f_{i,\ell},$

we see that our goal (12) becomes the conclusion of Lemma 4. Therefore we only need to show that (8) holds. Fixing ${d,u}$ and ${\ell}$, we compute

$\displaystyle \begin{array}{rcl} \displaystyle \langle \phi_{\ell,u+d},\phi_{\ell,u}\rangle&=&\displaystyle \int_X\prod_{i=1}^NT_{K_i(\ell)(u+d)}f_{i,\ell}\prod_{i=1}^NT_{K_i(\ell)u}f_{i,\ell}~\mathtt{d}\mu\\&=&\displaystyle \int_X\prod_{i=1}^NT_{K_i(\ell)u}\big(f_{i,\ell}\cdot T_{K_i(\ell)d}f_{i,\ell}\big)~\mathtt{d}\mu\\&=&\displaystyle \int_Xf_{1,\ell}\cdot T_df_{1,\ell}\cdot\prod_{i=2}^NT_{\big(K_i(\ell)-K_1(\ell)\big)u}\big(f_{i,\ell}\cdot T_{K_i(\ell)d}f_{i,\ell}\big)~\mathtt{d}\mu\\ \end{array}$

Now let ${\psi_{i,\ell}=f_{i,\ell}\cdot T_{K_i(\ell)d}f_{i,\ell}}$. Since, for each ${i=1,\dots,N}$, the sequence ${K_i(\ell)\rightarrow\tilde K_i}$ and ${f_{i,\ell}\rightarrow f_i}$ as ${\ell\rightarrow\infty}$ in ${C_c(X)}$, we deduce that ${\psi_{i,\ell}\rightarrow\psi_i:=f_i\cdot T_{\tilde K_id}f_i}$ as ${\ell\rightarrow\infty}$ in ${C_c(X)}$ as well. Thus, using ${RP(N-1)}$ we deduce that

$\displaystyle \lim_{\ell\rightarrow\infty}\frac1{n(\ell)-m(\ell)}\int_{m(\ell)}^{n(\ell)}\langle \phi_{\ell,u+d},\phi_{\ell,u}\rangle\mathtt{d}u=\prod_{i=1}^N\int_Xf_i\cdot T_{\tilde K_id}f_i~\mathtt{d}\mu \ \ \ \ \ (13)$

Recall that ${\int_Xf_N~\mathtt{d}\mu=0}$ and ${\tilde K_N=1}$. Thus it follows from (4) that the ${N}$-th term in the product converges to ${0}$ as ${d\rightarrow\infty}$. All the other terms in the product are bounded; therefore the (Cesàro) limit of the product in the right hand side is also ${0}$. This shows that (8) holds and this finishes the proof. $\Box$

— 4. ${P(N)~~+}$ (1) imply mixing of order ${N}$

In this section we prove the following result.

Proposition 7 Under the assumptions of Theorem 1, if ${(h_s)_{s\in{\mathbb R}}}$ satisfies ${P(N)}$, then it is mixing of order ${N}$.

Before we prove Proposition 7, we see how, together with Proposition 6, it implies Theorem 1.

Proof: } Observe that ${P(1)}$ is equivalent to ergodicity, hence ${(h_s)_{s\in{\mathbb R}}}$ satisfies ${P(1)}$. Now Proposition 7 implies that ${(h_s)_{s\in{\mathbb R}}}$ is strong mixing. It follows from Proposition 6 that ${(h_s)_{s\in{\mathbb R}}}$ satisfies ${P(N)}$ for all ${N\in{\mathbb N}}$. Finally, appealing once more to Proposition 7 we conclude that ${(h_s)_{s\in{\mathbb R}}}$ is mixing of all orders. $\Box$

All that remains to show now is Proposition 7. We need the following notion.

Definition 8 A local section for the flow ${(g_t)_{t\in{\mathbb R}}}$ is a compact subset ${\Sigma\subset X}$ for which there exists ${a>0}$ such that the map ${\phi:[0,a]\times\Sigma\rightarrow X}$ defined by ${\phi(t,y)=g_ty}$ is a homeomorphism onto its image. The image of ${\phi}$, denoted by ${A(a,\Sigma)}$, is called a flowbox.

The following two technical lemmas explain why flowboxes are useful.

Lemma 9 For each local section ${\Sigma}$ there exists a unique Borel measure ${\mu_\Sigma}$ on ${\Sigma}$ such that, for any ${a}$ for which ${A(a,\Sigma)}$ is a flowbox, we have

$\displaystyle \phi_*\big(l|_{[0,a]}\times\mu_\Sigma\big)=\mu|_{A(a,\Sigma)}$

where ${l}$ is the (non-normalized) Lebesgue measure, ${\phi}$ is the map from Definition 8 and ${\phi_*}$ denotes the push forward of a measure under ${\phi}$.

The measure ${\mu_\Sigma}$ can be defined by ${\mu_\Sigma(B)=\mu\Big(\phi\big([0,a]\times B\big)\Big)/a}$ for every measurable ${B\subset\Sigma}$. It is not difficult to show that this does not depend on ${a}$ (as long as it is small enough so that ${\phi}$ is injective).

Lemma 10 seems to be the only place in the proof where we use the fact that the flow runs on a manifold ${X}$, and not an arbitrary space.

Lemma 10 Let ${a_0>0}$. The family of all flowboxes ${A(a,\Sigma)}$ for ${(g_t)_{t\in{\mathbb R}}}$ with ${\mu_\Sigma(\Sigma)\leq1}$ and ${a generates the Borel ${\sigma}$-algebra on ${X}$.

We can now prove Proposition 7.

Proof: In order to prove (5), and in view of Lemma 10, it suffices to assume that ${f_0}$ is the indicator function of a flowbox ${A(a,\Sigma)}$ with ${\mu_\Sigma(\Sigma)\leq1}$ and ${a}$ arbitrarily small (${a}$ can depend on ${f_1,\dots,f_n}$, and ${\lambda}$ from (1), but nothing else, including the ${s_i}$). We will also assume that each ${f_i}$ is in ${C_c(X)}$ and is bounded in absolute value by ${1}$. Finally, replacing each ${s_i}$ with ${s_i-s_0}$ if necessary (and using the fact that ${(h_s)_{s\in{\mathbb R}}}$ preserves ${\mu}$) we can assume that ${s_0=0}$.

Using Lemma 9, the space average in the left hand side of (5) can be rewritten as

$\displaystyle \int_X\prod_{i=0}^Nh_{s_i}f_i~\mathtt{d}\mu= \int_0^a\int_\Sigma\prod_{i=1}^Nf_i\big(h_{s_i}g_ty\big)~\mathtt{d}\mu_\Sigma(y)~\mathtt{d}t$

Using the relation between ${g_t}$ and ${h_s}$ given by (1) this expression becomes

$\displaystyle \int_0^a\int_\Sigma\prod_{i=1}^Nf_i\big(g_th_{\lambda^{-t}s_i}y\big)~\mathtt{d}\mu_\Sigma(y)~\mathtt{d}t$

Since each ${f_i\in C_c(X)}$ and ${(g_t)_{t\in{\mathbb R}}}$ is a continuous action, for small enough ${a}$, the effect of ${g_t}$ in the above integral will be very small. In other words:

$\displaystyle \int_X\prod_{i=0}^Nh_{s_i}f_i~\mathtt{d}\mu=\int_0^a\int_\Sigma\prod_{i=1}^Nf_i\big(h_{\lambda^{-t}s_i}y\big)~\mathtt{d}\mu_\Sigma(y)~\mathtt{d}t+o_{a\rightarrow0}(1)$

Next we make the change of variable ${u=\lambda^{-t}s_N}$ and obtain

$\displaystyle \begin{array}{rcl} &&\displaystyle \int_0^a\int_\Sigma\prod_{i=1}^Nf_i(h_{\lambda^ts_i}y)~\mathtt{d}\mu_\Sigma(y)~\mathtt{d}t\\ &=&\displaystyle \int^{s_N}_{\lambda^{-a}s_N}\frac1{u\log\lambda}\int_\Sigma\prod_{i=1}^Nf_i(h_{us_i/s_N}y) ~\mathtt{d}\mu_\Sigma(y)~\mathtt{d}u\\&=&\displaystyle \frac{1-\lambda^{-a}}{a\log\lambda}\cdot\frac{s_N}{u_0}\cdot\frac a{s_N-\lambda^{-a}s_N} \int^{s_N}_{\lambda^{-a}s_N}\int_\Sigma\prod_{i=1}^Nf_i(h_{us_i/s_N}y) ~\mathtt{d}\mu_\Sigma(y)~\mathtt{d}u \end{array}$

for some ${u_0\in[\lambda^{-a}s_N,s_N]}$. Note that the first two factors in the last expression are very close to ${1}$ when ${a}$ is small (independently on any ${s_i}$). Thus all we need to show is that

$\displaystyle \frac a{s_N(\ell)-\lambda^{-a}s_N(\ell)} \int^{s_N(\ell)}_{\lambda^{-a}s_N(\ell)}\int_\Sigma\prod_{i=1}^Nf_i(h_{us_i(\ell)/s_N(\ell)}y) ~\mathtt{d}\mu_\Sigma(y)~\mathtt{d}u\xrightarrow[\ell\rightarrow\infty]{ }\prod_{i=0}^N\int_Xf_i\mathtt{d}\mu\quad (14)$

To ease the notation, let ${n(\ell)=s_N(\ell)}$, ${m(\ell)=\lambda^{-a}s_N(\ell)}$ and ${K_i(\ell)=s_i(\ell)/s_N(\ell)}$ for each ${i=1,\dots,N}$. Finally let

$\displaystyle \Phi_\ell(t)=\frac1{n(\ell)-m(\ell)}\int_{m(\ell)}^{n(\ell)}\int_\Sigma\prod_{i=1}^Nf_i(h_{K_i(\ell)u}g_ty)~\mathtt{d}\mu_\Sigma(y)~\mathtt{d}u$

Then (14) becomes

$\displaystyle \lim_{\ell\rightarrow\infty}\Phi_\ell(0)=\mu_\Sigma(\Sigma)\cdot\prod_{i=1}^N\int_Xf_i~\mathtt{d}\mu \ \ \ \ \ (15)$

It is easy to see from (1) (and a change of variable) that ${(\Phi_\ell)_{\ell\in{\mathbb N}}}$ is an equicontinuous family at ${t=0}$. Therefore (15) follows from

$\displaystyle \forall b>0\qquad\lim_{\ell\rightarrow\infty}\frac1b\int_0^b\Phi_\ell(t)~\mathtt{d}t\rightarrow\mu_\Sigma(\Sigma)\cdot\prod_{i=1}^N\int_Xf_i~\mathtt{d}\mu \ \ \ \ \ (16)$

Finally, since

$\displaystyle \int_0^b\Phi_\ell(t)~\mathtt{d}t=\int_X1_{A(b,\Sigma)}\frac1{n(\ell)-m(\ell)}\int_{m(\ell)}^{n(\ell)}\prod_{i=1}^Nh_{K_i(\ell)u}f_i~\mathtt{d}u~\mathtt{d}\mu,$

the limit in (16) follows directly from the property ${P(N)}$. This finishes the proof. $\Box$

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