Proof. By definition, the Laplace transform $\hat \beta _k(\xi _1, \ldots , \xi _k)$ of $\beta _k(t_1, \ldots , t_k)$ is given by

(3.1) $$ \begin{align} &\int_0^\infty \hspace{-1.5pt} \cdots \hspace{-1.5pt} \int_0^\infty \int_U \int_{\max(0,\tau(u)-t_i)}^{\tau(u)}\int_G \varphi_0(u,h,r)\nonumber\\ &\quad\cdot\bigg( \prod_{i=1}^k \varphi_i(u,h,r+t_i) e^{-\xi_i t_i} \bigg) \, d\omega \, dr \, d\nu^u \, dt_1 \, \cdots \, dt_k \end{align} $$

for $\xi _1, \ldots , \xi _k \in \mathbb {C}$ . As the systems of inequalities

are obviously equivalent, we can rewrite (3.1) as

$$ \begin{align*} &\int_U \int_G \int_0^{\tau(u)} \int_{\tau(u)}^\infty \cdots \int_{\tau(u)}^\infty \varphi_0(u, h, r)\\ &\quad\cdot\bigg( \prod_{i=1}^k \varphi_i(u, h, t_i)e^{-\xi_i(t_i-r)} \bigg) \, dt_1 \, \cdots \, dt_k \, dr \, d\omega \, d\nu^u \end{align*} $$

by reparametrizing the domain of integration. Let us focus on the k innermost integrals over $t_1, \ldots , t_k$ , which can be broken up into a sum of integrals

$$ \begin{align*} \sum_{n_1, \ldots, n_k = 1}^\infty \int_{\tau^{(n_1)}(u)}^{\tau^{(n_1 + 1)}(u)} \hspace{-18pt} \cdots \int_{\tau^{(n_k)}(u)}^{\tau^{(n_k+1)}(u)} \varphi_0(u, h, r) \bigg( \prod_{i=1}^k \varphi_i(u, h, t_i)e^{-\xi_i(t_i-r)} \bigg) \, dt_1 \, \cdots \, dt_k \end{align*} $$

over intervals of the form $[\tau ^{(n)}(u), \tau ^{(n+1)}(u)]$ . Of course, we can rewrite this as

$$ \begin{align*} &\sum_{n_1, \ldots, n_k = 1}^\infty \int_{0}^{\tau(\sigma^{n_1}(u))} \hspace{-18pt} \cdots \int_{0}^{\tau(\sigma^{n_k}(u))} \bigg[ \varphi_0(u, h, r)\\ &\qquad\cdot \bigg( \prod_{i=1}^k \varphi_i(u, h, t_i + \tau^{(n_i)}(u))e^{-\xi_i(t_i-r + \tau^{(n_i)}(u))} \bigg) \bigg] \, dt_1 \, \cdots \, dt_k \end{align*} $$

by changing variables, replacing $t_i$ with $t_i + \tau ^{(n_i)}(u)$ . We can rewrite the integrand here as

(3.2) $$ \begin{align} \varphi_0(u, h, r) \bigg( \prod_{i=1}^k\varphi_i( \sigma^{n_i}(u), ( \operatorname{\mathrm{Hol}}^{(n_i)}(u) )^{-1} \circ h, t_i ) e^{-\xi_i(t_i - r + \tau^{(n_i)}(u))} \bigg) \end{align} $$

using the identifications we made in constructing $U_{\tau , \operatorname {\mathrm {Hol}}}$ . Evaluating the integral of (3.2) with respect to $t_1$ through $t_k$ over the intervals $[0,\tau (\sigma ^{n_1}(u))]$ through $[0, \tau (\sigma ^{n_k}(u))]$ then yields

(3.3) $$ \begin{align} \varphi_0(u,h,r) e^{r\xi} \bigg( \prod_{i=1}^k \hat\varphi_{i,\xi_i}( \sigma^{n_i}(u), ( \operatorname{\mathrm{Hol}}^{(n_i)}(u) )^{-1} \circ h ) e^{-\xi_i \tau^{(n_i)}(u)} \bigg), \end{align} $$

where we set $\xi := \displaystyle \smash {\sum _{i=1}^k} \xi _i$ . Similarly, evaluating the integral of (3.3) with respect to r on $[0, \tau (u)]$ yields

$$ \begin{align*} \hat\varphi_{0,-\xi}(u,h) \bigg( \prod_{i=1}^k \hat\varphi_{i,\xi_i}( \sigma^{n_i}(u), ( \operatorname{\mathrm{Hol}}^{(n_i)}(u) )^{-1} \circ h ) e^{-\xi_i \tau^{(n_i)}(u)} \bigg) \end{align*} $$

and so (3.1) becomes

(3.4) $$ \begin{align} \sum_{n_1, \ldots, n_k=1}^\infty \int_U \int_G \hat\varphi_{0,-\xi}(u,h) \bigg( \prod_{i=1}^k \hat\varphi_{i,\xi_i}( \sigma^{n_i}(u), ( \operatorname{\mathrm{Hol}}^{(n_i)}(u) )^{-1} \circ h ) e^{-\xi_i \tau^{(n_i)}(u)} \bigg) \, d\omega \, d\nu^u \end{align} $$

after interchanging the order of integration and summation.

As $\omega $ is bi-invariant, we can replace h with $( \operatorname {\mathrm {Hol}}^{(\max n_j)}(u) ) \circ h$ . Of course, we have the identity $( \operatorname {\mathrm {Hol}}^{(n_i)}(u) )^{-1} \circ \operatorname {\mathrm {Hol}}^{(\max n_j)}(u) = \operatorname {\mathrm {Hol}}^{(n_i - (\max n_j))}(\sigma ^{n_i}(u))$ , and (3.4) becomes

$$ \begin{align*} \sum_{n_1, \ldots, n_k=1}^\infty \int_U \int_G \bigg[ &\hat\varphi_{0,-\xi}(u,\operatorname{\mathrm{Hol}}^{(\max n_j)}(u) \circ h)\\[-3pt] &\cdot \bigg( \prod_{i=1}^k \hat\varphi_{i,\xi_i}( \sigma^{n_i}(u), \operatorname{\mathrm{Hol}}^{(n_i - (\max n_j))}(u) \circ h ) e^{-\xi_i \tau^{(n_i)}(u)} \bigg) \bigg] \, d\omega \, d\nu^u \end{align*} $$

with this change of variables. This is equivalent to

$$ \begin{align*} \sum_{n_1, \ldots, n_k = 1}^\infty \int_G \int_U \bigg[ &\hat\varphi_{0,-\xi}(u,\operatorname{\mathrm{Hol}}^{(\max n_j)}(u) \circ h)\\[-3pt] &\cdot \bigg( \prod_{i=1}^k \hat\varphi_{i,\xi_i}( \sigma^{n_i}(u), \operatorname{\mathrm{Hol}}^{(n_i - (\max n_j))}(u) \circ h ) e^{-\xi_i \tau^{(n_i)}(u)} \bigg)\bigg] \, d\nu^u \, d\omega \end{align*} $$

by simply reversing the order of integration. Applying $\mathcal L_{P(\varsigma ),0}$ to the integrand a total of $\max n_j$ times yields

$$ \begin{align*} \sum_{{\sigma^{\max n_j}(u')=u}} \bigg[ &\hat\varphi_{0,-\xi}(u',\operatorname{\mathrm{Hol}}^{(\max n_j)}(u') \circ h)\\ &\cdot \bigg( \prod_{i=1}^k \hat\varphi_{i,\xi_i}( \sigma^{n_i}(u'), \operatorname{\mathrm{Hol}}^{(n_i - (\max n_j))}(u') \circ h ) e^{-\xi_i \tau^{(n_i)}(u')} \bigg) e^{\alpha_{P(\varsigma)}^{(\max n_j)}(u')}\bigg] \end{align*} $$

for any given values of $n_1, \ldots , n_k \in \mathbb {N}$ . Now, observe that $e^{-\mathrm {Re}(\xi _i) \tau ^{(n_i)}(u)}$ is at most $e^{- \mathrm {Re}(\xi _i) \tau ^{(\max n_j)}(u)}$ so long as each $\mathrm {Re}(\xi _i)$ is negative. Hence, we can bound the magnitude of the integrand from above by

$$ \begin{align*} \sum_{\sigma^{\max n_j}(u')=u} |\hat\varphi_{0,-\xi}(u',\operatorname{\mathrm{Hol}}^{(\max n_j)}(u') \circ h)| \bigg( \prod_{i=1}^k \|\hat\varphi_{i,\xi_i}\|_{C^0} \bigg) e^{-\mathrm{Re}(\xi) \tau^{(\max n_j)}(u') +\alpha_{P(\varsigma)}^{(\max n_j)}(u')} \end{align*} $$

using the triangle inequality. Rearranging this expression, we see that the magnitude of the integrand is bounded above by

(3.5) $$ \begin{align} &\bigg( \prod_{i=1}^k \|\hat\varphi_{i,\xi_i}\|_{C^0} \bigg)\nonumber\\ &\quad\cdot\bigg(\sum_{\sigma^{\max n_j}(u')=u} |\hat\varphi_{0,-\xi}(u',\operatorname{\mathrm{Hol}}^{(\max n_j)}(u') \circ h)|e^{-\mathrm{Re}(\xi) \tau^{(\max n_j)}(u') +\alpha_{P(\varsigma)}^{(\max n_j)}(u')} \bigg), \end{align} $$

which is reminiscent of the expression defining the transfer operator. To make this concrete, recall that we have an $L^2(G)$ -invariant decomposition of the function $\hat \varphi _{0,-\xi } \in C^1(U,L^2(G))$ in terms of its isotypic components

$$ \begin{align*} \hat \varphi_{0,-\xi}( u', \operatorname{\mathrm{Hol}}^{(\max n_j)}(u') \circ h ) = \sum_{\rho} \hat \varphi_{0,-\xi}^\rho( u', \operatorname{\mathrm{Hol}}^{(\max n_j)}(u') \circ h ), \end{align*} $$

where the sum is taken over irreducible representations $\rho $ of G, including the trivial representation. Once again, by the triangle inequality, we can bound (3.5) above by

$$ \begin{align*} &\bigg( \prod_{i=1}^k \|\hat\varphi_{i,\xi_i}\|_{C^0} \bigg)\\ &\quad\cdot\bigg(\sum_\rho \sum_{\sigma^{\max n_j}(u')=u} |\hat\varphi^\rho_{0,-\xi}(u',\operatorname{\mathrm{Hol}}^{(\max n_j)}(u') \circ h)|e^{-\xi \tau^{(\max n_j)}(u') +\alpha_{P(\varsigma)}^{(\max n_j)}(u')} \bigg) \end{align*} $$

where we quickly recognize the transfer operator $\mathcal L_{\rho ,P(\varsigma )+\mathrm {Re}(\xi )}^{\max n_j}$ applied to $|\hat \varphi ^\rho _{0,-\xi } |$ , noting, of course, that

$$ \begin{align*} | \hat \varphi_{0,-\xi}^\rho ( u', \operatorname{\mathrm{Hol}}^{(\max n_j)}(u') \circ h ) | = | \rho( \operatorname{\mathrm{Hol}}^{(\max n_j)}(u') ) \cdot \hat \varphi_{0,-\xi}^\rho ( u', h )| \end{align*} $$

for each $\rho $ . Putting this back together, we see that (3.1) is bounded above by

$$ \begin{align*} \sum_\rho \sum_{n_1, \ldots, n_k=1}^\infty \int_U \int_G (\mathcal L^{\max n_j}_{\rho, P(\varsigma) + \mathrm{Re}(\xi)}|\hat\varphi^\rho_{0,-\xi}|)(u, h) \bigg( \prod_{i=1}^k \|\hat\varphi_{i,\xi_i}\|_{C^0} \bigg) \, d\omega \, d\nu^u \end{align*} $$

in magnitude, as desired.

Proof. We assume that Theorem 3.2 holds, and so we have

(3.6) $$ \begin{align} \|\mathcal L^n_{z,\rho} \varphi \|_{L^2(\nu^u)} \leq C \|\varphi\|_{C^1} r^n \end{align} $$

for all non-trivial irreducible $\rho $ , all $\varphi \in C^1(U, V^\rho )$ and for each $z \in \mathbb {C}$ with $|\mathrm {Re}(z) - P(\varsigma )| < 1$ . Up to choosing larger values of C and r, we can also assume that the same inequality holds when $\rho $ is trivial: this is precisely the main result of either [Reference Dolgopyat11, Theorem  2] or [Reference Butterley and War9, Proposition 3.16], depending on the regularity hypothesis we are using in Theorem A.

We show that the expression bounding $\hat \beta (\xi _1, \ldots , \xi _k)$ in Lemma 3.3 converges whenever the real parts of $\xi _i$ simultaneously lie in the interval $-1/k < \mathrm {Re}(\xi _i) < 0$ . The decay desired follows immediately from applying the inverse Laplace transform with the specific bounds we obtain.

Fix $\xi _1, \ldots , \xi _k$ with $-1/k < \mathrm {Re}(\xi _i) < 0$ . As before, we consider the function

$$ \begin{align*} \hat \varphi_{i,\xi_i}(u,h) = \int_0^{\tau(u)} \varphi_i(u,h,t) e^{-\xi_it_i} \, dt_i \end{align*} $$

and decompose $\hat \varphi _{0,-\xi }$ into its isotypic components

$$ \begin{align*} \hat \varphi_{0,-\xi} = \sum_\rho \hat \varphi_{0,-\xi}^\rho, \end{align*} $$

where $\xi = \xi _1 + \cdots + \xi _k$ as before, noting that the decomposition of $L^2(G)$ into irreducible subspaces commutes with the Laplace transform. By (3.6), whenever $- ( 1/ k) < \mathrm {Re}(\xi _i) < 0$ , we have

$$ \begin{align*} \int_U \int_G ( \mathcal L^{\max n_j}_{\rho,P(\varsigma) + \mathrm{Re}(\xi)} | \hat \varphi^\rho_{0,-\xi} | )(u,h) \, d\omega \, d\nu^u &\leq \| (\mathcal L^{\max n_j}_{\rho, P(\varsigma) + \mathrm{Re}(\xi)} |\hat \varphi^\rho_{0,-\xi} |)\|_{L^2(\nu^u)}\\ &\leq C \|\hat \varphi_{0,-\xi}^\rho\|_{C^1} r^{\max n_j} \end{align*} $$

for each $\rho $ . Combining this with Lemma 3.3, we are reduced to ensuring that

(3.7) $$ \begin{align} &\sum_\rho \sum_{n_1, \ldots, n_k = 1}^\infty C \|\hat \varphi^\rho_{0,-\xi} \|_{C^1} r^{\max n_j} \bigg( \prod_{i=1}^k \| \hat \varphi_{i,\xi_i} \|_{C^0} \bigg)\nonumber\\ &\quad< \sum_\rho \sum_{n=1}^\infty k!n^{k-1}C \|\hat \varphi^\rho_{0,-\xi} \|_{C^1} r^n \bigg( \prod_{i=1}^k \|\hat \varphi_{i,\xi_i} \|_{C^0} \bigg) \end{align} $$

converges. Although this appears promising, note that we can only bound

$$ \begin{align*} \| \hat \varphi_{0,-\xi}^\rho \|_{C^1} \leq D \|\rho\| \big( \sup_{u \in U} \|\hat \varphi_{0,-\xi}^\rho (u, \cdot)\|_{L^2(G)} \big) \end{align*} $$

for some constant $D> 0$ . To salvage this, we assume for the moment that we have chosen $\varphi _0 \in C^{2N'}(U_{\tau ,\operatorname {\mathrm {Hol}}},\mathbb {R})$ for a sufficiently large $N'> 0$ so that we can invoke Theorem 2.11 to bound

$$ \begin{align*} \|\rho\|\|\hat \varphi^\rho_{0,-\xi}(u,\cdot) \|_{L^2(G)} \leq \frac {D' \| \hat \varphi_{0,-\xi} (u,\cdot) \|_{C^{N'}}}{\|\rho\|^N} \end{align*} $$

pointwise with a fixed constant $D'> 0$ , where $N> 0$ is the constant guaranteed by Proposition 2.10. With this, it is clear that the expression on the right-hand side of (3.7) converges absolutely, which says immediately that $\beta _k$ must decay exponentially fast in $t_1, \ldots , t_k$ ; this is almost what we wanted to show, but we need an explicit bound on $\hat \beta _k$ to obtain uniform estimates with the desired constants.

Integrating the defining expression for $\hat \varphi _{i,\xi _i} (u,h)$ by parts, we have

$$ \begin{align*} \hat \varphi_{i, \xi_i}(u,h) =\frac{\varphi_i(u,h,t_i) e^{-\xi_it_i}}{-\xi_i} \bigg|_0^{\tau(u)} + \frac 1 {\xi_i} \int_0^{\tau(u)} \bigg(\frac{\partial}{\partial t_i} \varphi_i (u,h,t_i)\bigg) e^{-\xi_i t_i} \, dt_i \end{align*} $$

for each u and h. Assuming that we choose $\varphi _i \in C^2(U_{\tau ,\operatorname {\mathrm {Hol}}},\mathbb {R})$ , we can crudely bound

$$ \begin{align*} \| \hat \varphi_{i, \xi_i} \|_{C^1} \leq \frac{D'' \|\varphi_i\|_{C^2}}{1 + |\mathrm{Im}(\xi_i)|} \end{align*} $$

for some constant $D''$ that depends only on $\|\tau \|_{C^1}$ ; note that it is essential here that $\mathrm {Re}(\xi _i)$ is confined to a bounded interval. We can similarly bound

$$ \begin{align*} \| \hat \varphi_{0, -\xi} \|_{C^{N'}} \leq \frac{D'' \|\varphi_0\|_{C^{N'+1}}}{1 + |\mathrm{Im}(\xi)|} \end{align*} $$

by choosing a larger value for $D''$ if necessary. Putting this all together, we have

for all $\xi _i \in \mathbb {C}$ with $-1/k < \mathrm {Re}(\xi _i) < 0$ . Now, we simply take the inverse Laplace transform in the variables $\xi _1$ through $\xi _k$ in succession to obtain

$$ \begin{align*} &|\beta_k(t_1, \ldots, t_k)|\\ &\quad\leq \frac {e^{\mathrm{Re}(\xi_1)t_1+ \cdots + \mathrm{Re}(\xi_k) t_k}}{(2\pi)^k} \hspace{-1pt}\int_{-\infty}^\infty \hspace{-1.5pt}\cdots\hspace{-1.5pt} \int_{-\infty}^\infty \frac {\|\varphi_0\|_{C^{N'+1}}} {1 + |s_1 + \cdots + s_k|} \bigg( \prod_{i=1}^k \frac {\|\varphi_i\|_{C^2}} {1+|s_i|} \bigg)\, ds_1 \, \ldots \, ds_k, \end{align*} $$

where $s_i$ is to be interpreted as the imaginary part of $\xi _i$ . To complete our decay estimate, we simply need to show that the integral

$$ \begin{align*} \int_{-\infty}^\infty \cdots \int_{-\infty}^\infty \frac 1 {1 + |s_1 + \cdots + s_k|} \bigg( \prod_{i=1}^k \frac 1 {1 + |s_i|} \bigg) \, ds_1 \, \ldots \, ds_k \end{align*} $$

converges. Once the convergence of this integral has been established, we will have shown that

$$ \begin{align*} |\beta_k(t_1, \ldots, t_k)| < Cr^{\max t_j} \|\varphi_0\|_{C^{N'+1}} \bigg( \prod_{i=1}^k \|\varphi_i\|_{C^2}\bigg) \end{align*} $$

for all $\varphi _0 \in C^{N'+1}(U_{\tau ,\operatorname {\mathrm {Hol}}}, \mathbb {R})$ and $\varphi _i \in C^2(U_{\tau ,\operatorname {\mathrm {Hol}}},\mathbb {R})$ . An identical argument to that given in Lemma 2.1 extends this to $C^1$ functions, using [Reference Gorodnik and Spatzier13, Lemma 2.4] once again.

Proof. We show, in fact, that the function defined by the integral

(3.8) $$ \begin{align} f(x) := \int_{-\infty}^\infty \frac 1 {(1+|x+y|)^{0.5-\epsilon}} \bigg( \frac 1 {1 + |y|} \bigg) \, dy \end{align} $$

decays at a rate of

$$ \begin{align*} |f(x)| \leq \frac C {(1 + |x|)^{0.5-2\epsilon}} \end{align*} $$

for all sufficiently small $\epsilon> 0$ ; the statement of the lemma follows immediately by direct successive integration. We work in the case when $x> 0$ , and split the domain of integration in (3.8) into regions where $x+y, y < 0$ , where $x + y> 0$ but $y < 0$ and where $x + y, y> 0$ . In the first case, where $x + y$ and y are both negative, we evaluate

$$ \begin{align*} \int_{-\infty}^{-x} \frac 1 {(1 - (x+y))^{0.5-\epsilon}} \bigg( \frac 1 {1 - y} \bigg) \, dy \end{align*} $$

for a fixed $x> 0$ . One can verify that the antiderivative of this expression is given by

$$ \begin{align*} \frac {(({1-(x+y)})/({1-y}))^{0.5-\epsilon} {}_2F_1(0.5-\epsilon,0.5-\epsilon;1.5-\epsilon;{x}/({1-y}))}{(0.5-\epsilon) (1-(x+y))^{0.5-\epsilon}} \end{align*} $$

where ${}_2F_1(\cdot ,\cdot ;\cdot ;\cdot )$ is the principal branch of the analytic continuation of the Gaussian hypergeometric function. Hence, the integral evaluates to

$$ \begin{align*} \int_{-\infty}^{-x} \frac 1 {(1 - (x+y))^{0.5-\epsilon}} \bigg( \frac 1 {1 - y} \bigg) \, dy = \frac {{}_2F_1(0.5-\epsilon,0.5-\epsilon;1.5-\epsilon;{x}/({1+x}))}{(0.5-\epsilon)(1+x)^{0.5-\epsilon}}, \end{align*} $$

which can be bounded above by $C/{(1+x)^{0.5-\epsilon }}$ for an appropriate choice of constant $C> 0$ , because by [Reference Abramowitz and Stegun1, 15.1.1] the defining series for ${}_2F_1(a,b;c;z)$ converges on the unit disc in the complex plane when $c - (a+b)> 0$ . Similarly, on the region where $x+y$ is positive and y is negative, we evaluate

$$ \begin{align*} &\int_{-x}^0 \frac 1 {(1 + (x+y))^{0.5-\epsilon}} \bigg( \frac 1 {1 - y} \bigg) \, dy\\[4pt] &\quad= \bigg( \frac{(1+(x+y))^{0.5+\epsilon} {}_2F_1(0.5+\epsilon,1;1.5+\epsilon;({1+(x+y)})/({2+x})}{(0.5+\epsilon)(2+x)} \bigg)\bigg|_{y=-x}^{y=0}\\ &\quad= \frac{(1+x)^{0.5+\epsilon} {}_2F_1(0.5+\epsilon,1;1.5+\epsilon;({1+x})/({2+x}))}{(0.5+\epsilon)(2+x)}\\[4pt] &\qquad- \frac{{}_2F_1 (0.5+\epsilon,1;1.5+\epsilon; 1 /({2+x}))}{(0.5+\epsilon)(2+x)}, \end{align*} $$

which can once again be bounded above by $ C /{(1+x)^{0.5-\epsilon }}$ for the same reasons, though with a possibly larger choice of $C> 0$ . Finally, when both $x+y$ and y are positive, we evaluate

$$ \begin{align*} &\int_0^\infty \frac 1 {(1+(x+y))^{0.5-\epsilon}} \bigg( \frac 1 {1+y} \bigg) \, dy\\[4pt] &\quad= \bigg( \frac{( ( {1+(x+y)})/({1+y}) )^{0.5-\epsilon} {}_2F_1(0.5-\epsilon,0.5-\epsilon;1.5-\epsilon;{-x}/({1+y}))}{(0.5-\epsilon)(1+(x+y))^{0.5-\epsilon}} \bigg)\bigg|_{y=0}^{y=\infty}\\[4pt] &\quad= -\frac{{}_2F_1(0.5-\epsilon,0.5-\epsilon;1.5-\epsilon;-x)}{0.5-\epsilon} \end{align*} $$

and we simply need to understand the asymptotics of ${}_2F_1(0.5-\epsilon ,0.5-\epsilon ;1.5-\epsilon ;-x)$ as $x \to \infty $ . By [Reference Abramowitz and Stegun1, 15.3.1], we have an integral representation given by

(3.9) $$ \begin{align} {}_2F_1 (0.5 - \epsilon, 0.5-\epsilon; 1.5-\epsilon;-x) = \frac{\Gamma(1.5-\epsilon)}{\Gamma(0.5-\epsilon)\Gamma(1)} \int_0^1 \frac 1 {t^{0.5+\epsilon}(1+tx)^{0.5-\epsilon}} \, dt \end{align} $$

for $x> 1$ . Setting $u = tx$ , we can rewrite

$$ \begin{align*} &\int_0^1 \frac 1 {t^{0.5+\epsilon}(1+tx)^{0.5-\epsilon}} \, dt\\[3pt] &\quad= \frac 1 {x^{0.5-\epsilon}} \int_0^x \frac 1 {u^{0.5+\epsilon}(1+u)^{0.5-\epsilon}} \, du\\[3pt] &\quad= \frac 1 {x^{0.5-\epsilon}} \bigg( \int_0^1 \frac 1 {u^{0.5+\epsilon}(1+u)^{0.5-\epsilon}} \, du + \int_1^x \frac 1 {u^{0.5+\epsilon}(1+u)^{0.5-\epsilon}} \, du \bigg), \end{align*} $$

at which point we note that the expression in parentheses can be bounded above by $C (1 + \log x)$ . Hence, (3.9) can be bounded above by $ C/{(1+x)^{0.5 - 2\epsilon }}$ with a possibly larger choice of C, as desired. The proof in the case when $x < 0$ is identical.

Proof. The convergence of the limit is simply a consequence of the fact that $d(g_{t_n}(x),g_{t_n}(y)) \to 0$ as $n \to \infty $ . More precisely, we can rewrite

$$ \begin{align*} \operatorname{\mathrm{Hol}}^{(n+1)}_{I_y,T}(y) ( \operatorname{\mathrm{Hol}}_{I_x,T}^{(n+1)}(x) )^{-1} \end{align*} $$

as

$$ \begin{align*} &\operatorname{\mathrm{Hol}}_{\phi_{n+1,y}}^{\phi_{n,y}}(g_{t_{n+1}}(y),g_{t_n}(y),t_n - t_{n+1}) \\ &\quad\circ\; (\operatorname{\mathrm{Hol}}^{(n)}_{I_y,T}(y) ( \operatorname{\mathrm{Hol}}_{I_x,T}^{(n)}(x) )^{-1}) \\ &\quad\quad\circ (\operatorname{\mathrm{Hol}}_{\phi_{n+1,x}}^{\phi_{n,x}}(g_{t_{n+1}}(x),g_{t_n}(x),t_n-t_{n+1}) )^{-1}, \end{align*} $$

and because $f_t$ is $C^1$ , we see that $\operatorname {\mathrm {Hol}}_{\phi _{n+1,x}}^{\phi _{n,x}}(g_{t_{n+1}}(x),g_{t_n}(x))$ must also be locally $C^1$ in $g_{t_{n+1}}(x)$ . Since $d_N(g_{t_{n+1}}(y),g_{t_{n+1}}(x))$ decay exponentially fast as $n \to \infty $ and the trivializations $I_x$ and $I_y$ eventually agree, we see that

$$ \begin{align*} d_G ( \operatorname{\mathrm{Hol}}_{\phi_{n+1,y}}^{\phi_{n,y}}(g_{t_{n+1}}(y),g_{t_n}(y),t_n-t_{n+1}) , \operatorname{\mathrm{Hol}}_{\phi_{n+1,x}}^{\phi_{n,x}}(g_{t_{n+1}}(x),g_{t_n}(x),t_n-t_{n+1}) ) \end{align*} $$

must also decay exponentially fast. In particular, for any $h \in G$ , this means that

$$ \begin{align*} &d_G ( \operatorname{\mathrm{Hol}}_{\phi_{n+1,y}}^{\phi_{n,y}}(g_{t_{n+1}}(y),g_{t_n}(y),t_n-t_{n+1}) \circ h\\ &\quad\circ (\operatorname{\mathrm{Hol}}_{\phi_{n+1,x}}^{\phi_{n,x}}(g_{t_{n+1}}(x),g_{t_n}(x),t_n-t_{n+1}))^{-1},h ) \end{align*} $$

decays exponentially fast and, hence,

$$ \begin{align*} d_G( \operatorname{\mathrm{Hol}}^{(n+1)}_{I_y, T}(y) ( \operatorname{\mathrm{Hol}}^{(n+1)}_{I_x, T}(x) )^{-1}, \operatorname{\mathrm{Hol}}^{(n)}_{I_y, T}(y) ( \operatorname{\mathrm{Hol}}^{(n)}_{I_x, T}(x) )^{-1} ) \end{align*} $$

decays exponentially fast as $n \to \infty $ . As G is complete, the limit must exist. A similar argument shows that this limit is, in fact, equal to $\Theta ^+_{\phi _{0,x},\phi _{0,y}}(x, y)$ : because the unstable foliation of $f_t$ is invariant under the flow, we can rewrite

$$ \begin{align*} \Theta^+_{\phi_{0,x},\phi_{0,y}}(x, y) \end{align*} $$

as

(4.2) $$ \begin{align} \operatorname{\mathrm{Hol}}^{(n)}_{I_y,T}(y) \circ \Theta^+_{\phi_{n,x},\phi_{n,y}}(g_{t_n}(x), g_{t_n}(y)) \circ ( \operatorname{\mathrm{Hol}}^{(n)}_{I_x, T}(x) )^{-1} \end{align} $$

for any $n> 0$ and any sequences $I_x, I_y$ and T as previously. As $t_n \to -\infty $ , $d_N(g_{t_n}(x), g_{t_n}(y))$ decreases exponentially fast, and so $\Theta ^+_{\phi _{n,x},\phi _{n,y}}(g_{t_n}(x), g_{t_n}(y))$ converges to the identity in G. Of course, this means that, as $n \to \infty $ , (4.2) converges to the limit in (4.1). As the expression in (4.2) is constant at $\Theta ^+_{\phi _{0,x},\phi _{0,y}}(x,y)$ , this proves the proposition.

Proof. This follows immediately from our hypotheses on the regularity of $f_t$ , which imply that the leaves of the strong unstable foliation of $f_t$ are $C^1$ .

Proof. We can relate the unstable holonomies between x and $u \in W^{su}_{g_t}(x)$ with respect to $\phi _1$ and $\phi _2$ by

(4.3) $$ \begin{align} \Theta^+_{\phi_2,\phi_2} ( x,u ) = \operatorname{\mathrm{id}}_{\phi_1}^{\phi_2}(u) \circ \Theta^+_{\phi_1,\phi_1}( x,u ) \circ ( \operatorname{\mathrm{id}}_{\phi_1}^{\phi_2}(x) )^{-1} \end{align} $$

by definition. We now simply take the derivative of each side of (4.3) with respect to u at $u = x$ ; in the notation of Definition 4.3, this becomes

(4.4) $$ \begin{align} X_w^{\phi_2}(x) = \operatorname{\mathrm{Ad}}_{\operatorname{\mathrm{id}}_{\phi_1}^{\phi_2}(x)} ( X_{w\phantom{'}}^{\phi_1}(x) ) + ((d{\kern-1.1pt}R)_{(\operatorname{\mathrm{id}}_{\phi_1}^{\phi_2}(x))^{-1}} \circ d( \operatorname{\mathrm{id}}_{\phi_1}^{\phi_2} )_x )(w) \end{align} $$

for any $w \in T^1_xW^{su}_{g_t}(x)$ , where $d{\kern-1.1pt}R$ denotes the derivative of right multiplication in G. Given any $y \in W^{ss}_{\epsilon }(x)$ and $w'$ corresponding to w as in Definition 4.3, exactly the same calculation yields

(4.5) $$ \begin{align} X_{w'}^{\phi_2}(y) = \operatorname{\mathrm{Ad}}_{\operatorname{\mathrm{id}}_{\phi_1}^{\phi_2}(y)} ( X_{w'}^{\phi_1}(y) ) + ((d{\kern-1.1pt}R)_{(\operatorname{\mathrm{id}}_{\phi_1}^{\phi_2}(y))^{-1}} \circ d( \operatorname{\mathrm{id}}_{\phi_1}^{\phi_2} )_y )({w'}), \end{align} $$

assuming, of course, that y is sufficiently close to x that we are able to use the same trivializations $\phi _1, \phi _2$ . As both trivializations are constant along the strong stable foliation of $f_t$ and we chose $y \in W^{ss}_{\epsilon }(x)$ , we clearly have $\operatorname {\mathrm {id}}_{\phi _1}^{\phi _2}(x) = \operatorname {\mathrm {id}}_{\phi _1}^{\phi _2}(y)$ and, hence,

$$ \begin{align*} (d{\kern-1.1pt}R)_{(\operatorname{\mathrm{id}}_{\phi_1}^{\phi_2}(x))^{-1}} = (d{\kern-1.1pt}R)_{(\operatorname{\mathrm{id}}_{\phi_1}^{\phi_2}(y))^{-1}} \end{align*} $$

as functions $T^1G \to T^1G$ . Moreover, because the trivializations are also constant along the flowlines of $f_t$ , we see that $\operatorname {\mathrm {id}}_{\phi _1}^{\phi _2}$ must be constant along the leaves of the centre stable foliation of $g_t$ . Hence, we must have

$$ \begin{align*} (d( \operatorname{\mathrm{id}}_{\phi_1}^{\phi_2} )_x)(w) = ( d( \operatorname{\mathrm{id}}_{\phi_1}^{\phi_2} )_y )(w') \end{align*} $$

for all $w \in T^1_xW^{su}_{g_t}(x)$ . Subtracting (4.4) from (4.5) and using the fact that $\operatorname {\mathrm {id}}_{\phi _1}^{\phi _2}(x) = \operatorname {\mathrm {id}}_{\phi _1}^{\phi _2}(y)$ , we then obtain

$$ \begin{align*} X_{w'}^{\phi_2}(y) - X_{w}^{\phi_2}(x) = \operatorname{\mathrm{Ad}}_{\operatorname{\mathrm{id}}_{\phi_1}^{\phi_2}(y)} ( X_{w'}^{\phi_1}(y) ) - \operatorname{\mathrm{Ad}}_{\operatorname{\mathrm{id}}_{\phi_1}^{\phi_2}(x)} ( X_{w \phantom{'}}^{\phi_1}(x) ) \end{align*} $$

as desired.

Proof. By Proposition 4.1, we have

$$ \begin{align*} \Theta^+_{\phi_2,\phi_2}(g_t(x), g_t(u)) = h(u) \circ \Theta^+_{\phi_1, \phi_1}(x,u) \circ ( h(x) )^{-1} \end{align*} $$

so long as u is sufficiently close to x. Noting the resemblance to (4.3), simply repeating our calculations in Proposition 4.6 yields

$$ \begin{align*} X_{w'}^{\phi_{g_t(x)}}(g_t(y)) - X_{w}^{\phi_{g_t(x)}}(g_t(x)) = \operatorname{\mathrm{Ad}}_{h(x)} ( X_{w'}^{\phi_x}(y) ) - \operatorname{\mathrm{Ad}}_{h(x)} ( X_{w\phantom{'}}^{\phi_x}(x) ) \end{align*} $$

for all $y \in W^{ss}_{\epsilon }(x)$ and all $w \in T^1_xW^{su}_{g_t}(x)$ . This completes the proof; note that we do not obtain equality this time because the strong stable leaves for $g_t$ contract, and there will be $y' \in W^{ss}_{\epsilon }(g_t(x))$ that are not of the form $g_t(y)$ for $y \in W^{ss}_\epsilon (x)$ .

Proof. Let $x \in N$ be a point at which $\mathfrak h_\epsilon ^*(x)$ has maximal dimension. As $\mathfrak h_\epsilon ^*(\cdot )$ is lower semi-continuous, it has maximal dimension on an open neighbourhood W containing x. By Proposition 4.7, $\mathfrak h_\epsilon ^*(\cdot )$ therefore has maximal dimension on an open set containing the forward orbit of $g_t$ . This is a dense set if $g_t$ is topologically transitive.

As $g_t$ is ergodic and $\dim \mathfrak h_\epsilon ^*(\cdot )$ is measurable, it must be constant almost everywhere. The measure $\nu $ is an equilibrium measure with a Hölder potential and therefore has full support; hence, the open and dense set on which $\dim \mathfrak h_\epsilon ^*(\cdot )$ has maximal dimension must also have full measure.

Proof. As $\dim \mathfrak h^*_\epsilon (\cdot )$ is lower semi-continuous, there is an open set $W \subset B_\epsilon (x)$ on which it is maximal. Because x is backwards recurrent, we can find a monotonic sequence of times $T = \{ t_n \}$ with $t_n \to -\infty $ for which $g_{t_n}(x) \in W$ for each $n> 0$ . Moreover, we can suppose that $t_1$ is large enough that we also have $g_{t_n}(y) \in W$ for each $n> 0$ . Now, write

$$ \begin{align*} h^{(n)}(x) := \operatorname{\mathrm{Hol}}_{\phi_x}^{\phi_x}(g_{t_n}(x),g_{t_{n-1}}(x),t_{n-1}-t_n) \circ \cdots \circ \operatorname{\mathrm{Hol}}_{\phi_x}^{\phi_x}( g_{t_1}(x),x,-t_1 ) \end{align*} $$

and

$$ \begin{align*} h^{(n)}(y) := \operatorname{\mathrm{Hol}}_{\phi_x}^{\phi_x}(g_{t_n}(y),g_{t_{n-1}}(y),t_{n-1}-t_n) \circ \cdots \circ \operatorname{\mathrm{Hol}}_{\phi_x}^{\phi_y}( g_{t_1}(y),y, -t_1) \end{align*} $$

for the n-step holonomies at $g_{t_n}(x)$ and $g_{t_n}(y)$ . As we have $g_{t_n}(x), g_{t_n}(y) \in W$ by construction, we have

$$ \begin{align*} \operatorname{\mathrm{Ad}}_{h^{(n)}(x)}( \mathfrak h_\epsilon^{\phi_x}( g_{t_n}(x) ) ) = \mathfrak h_\epsilon^{\phi_x}(x) \end{align*} $$

and

$$ \begin{align*} \operatorname{\mathrm{Ad}}_{h^{(n)}(y)}( \mathfrak h_\epsilon^{\phi_x}( g_{t_n}(y) ) ) = \mathfrak h_\epsilon^{\phi_y}(y) \end{align*} $$

by Proposition 4.7; note that we have implicitly used the fact that $B_{2\epsilon }(x) \subset V_x$ in writing $\mathfrak h_\epsilon ^{\phi _x}( g_{t_n}(x) )$ and $\mathfrak h_\epsilon ^{\phi _x}( g_{t_n}(y) )$ , where $V_x$ is the chart over which $\phi _x$ is defined. By rearranging these equations, we see that

$$ \begin{align*} d_{\operatorname{\mathrm{Gr}}(\dim \mathfrak h^*_\epsilon(x),\mathfrak g)}( \mathfrak h_\epsilon^{\phi_y}(y), \operatorname{\mathrm{Ad}}_{h^{(n)}(y) ( h^{(n)}(x) )^{-1}}( \mathfrak h_\epsilon^{\phi_x}(x) ) ) \end{align*} $$

is equal to

$$ \begin{align*} d_{\operatorname{\mathrm{Gr}}(\dim \mathfrak h^*_\epsilon(x),\mathfrak g)}( \mathfrak h_\epsilon^{\phi_x}( g_{t_n}(y) ), \mathfrak h_\epsilon^{\phi_x}( g_{t_n}(x) ) ), \end{align*} $$

where the distances are measured in the standard metric on the Grassmannian of $(\dim \mathfrak h^*_\epsilon (x))$ -dimensional subspaces of $\mathfrak g$ . As $\mathfrak h_\epsilon ^{\phi _x}(\cdot )$ is lower semi-continuous and has maximal dimension on W, it must be continuous on W. Up to passage to the interior of a compact subset of W, we can assume that $\mathfrak h_\epsilon (\cdot )$ is uniformly continuous on W. Because $d_N(g_{t_n}(y),g_{t_n}(x)) \to 0$ as $n \to \infty $ , we must then have

$$ \begin{align*} d_{\operatorname{\mathrm{Gr}}(\dim \mathfrak h^*_\epsilon(x),\mathfrak g)}( \mathfrak h_\epsilon^{\phi_x}( g_{t_n}(y) ), \mathfrak h_\epsilon^{\phi_x}( g_{t_n}(x) ) ) \end{align*} $$

as $n \to \infty $ . This yields

$$ \begin{align*} \mathfrak h_\epsilon^{\phi_y}(y) = \lim_{n \to \infty} \operatorname{\mathrm{Ad}}_{h^{(n)}(y) ( h^{(n)}(x) )^{-1}}( \mathfrak h_\epsilon^{\phi_x}(x) ), \end{align*} $$

at which point we simply observe that $\operatorname {\mathrm {Ad}}_g(\cdot )$ is continuous in g and that $h^{(n)}(y) (h^{(n)}(x))^{-1}$ converges to $\Theta ^+_{\phi _x,\phi _y}(x,y)$ by Proposition 4.1.

Proof. Suppose that x is bi-recurrent. Fix an isometry $g \in G$ and consider a stable–unstable sequence $x_0, x_1, \ldots , x_k, x_{k+1} = x_0$ in N with $x_0 = x$ and whose total holonomy is g, where $x_{i+1}$ is either on the strong stable or strong unstable leaf through $x_i$ for $g_t$ . As g is the total holonomy, we can write g as the composite

$$ \begin{align*} &\operatorname{\mathrm{id}}_{\phi_{x_k}}^{\phi_{x_0}}(x_0) \circ \Theta^\pm_{\phi_{x_k},\phi_{x_k}}( x_k,x_0 ) \circ \ldots \operatorname{\mathrm{id}}_{\phi_{x_1}}^{\phi_{x_2}}(x_2)\\ &\quad\circ \Theta^\pm_{\phi_{x_1},\phi_{x_1}} ( x_1,x_2 ) \circ \operatorname{\mathrm{id}}_{\phi_{x_0}}^{\phi_{x_1}}(x_1) \circ \Theta^\pm_{\phi_{x_0},\phi_{x_0}}( x_0,x_1 ), \end{align*} $$

where we can freely assume that each consecutive pair $x_i, x_{i+1}$ has a common trivialization $\phi _{x_i}$ for which $B_\epsilon (x_i),B_\epsilon (x_{i+1}) \subset V_{x_i}$ ; this is true up to refining the sequence (Figure 2). Suppose, moreover, that we have chosen $\phi _{x_0} = \phi _x$ .

Figure 2 A refined stable–unstable sequence $x_0, x_1, \ldots , x_k, x_{k+1} = x_0$ . Any stable–unstable sequence can be refined so that for each $0 \leq n \leq k$ , there is a trivialization $\phi _{x_n}$ over a chart $V_{x_n}$ containing both $x_n$ and $x_{n+1}$ . This refinement has the same total holonomy.

We would like to now invoke Proposition 4.6 and Lemmas 4.9 and 4.10 to show that $\mathfrak h^{\phi _x}_\epsilon (x)$ is $\operatorname {\mathrm {Ad}}_g$ -invariant for the g corresponding to the total holonomy along this sequence. The $x_i$ we have chosen, however, may fail to be forwards- or backwards-recurrent as necessary. However, note that $\operatorname {\mathrm {id}}_{\phi _{x_i}}^{\phi _{x_{i+1}}}(\cdot )$ , $\Theta ^+_{\phi _{x_i},\phi _{x_i}}(\cdot ,\cdot )$ and $\Theta ^-_{\phi _{x_i},\phi _{x_i}}(\cdot ,\cdot )$ are all locally continuous in all of their arguments. As bi-recurrent points are dense in N, given any $\delta> 0$ , we can find a sequence of bi-recurrent points $x^{\prime }_0, x^{\prime }_1, \ldots , x^{\prime }_k, x^{\prime }_{k+1} = x^{\prime }_0$ near $x_0, x_1, \ldots , x_k, x_{k+1} = x_0$ with $x^{\prime }_0 = x$ whose total holonomy $g'$ is given by

$$ \begin{align*} &\operatorname{\mathrm{id}}_{\phi_{x_k}}^{\phi_{x_0}}(x^{\prime}_0) \circ \Theta^\pm_{\phi_{x_k},\phi_{x_k}}( x^{\prime}_k,x^{\prime}_0 ) \circ \cdots \circ \operatorname{\mathrm{id}}_{\phi_{x_1}}^{\phi_{x_2}}(x^{\prime}_2)\\ &\quad\circ \Theta^\pm_{\phi_{x_1},\phi_{x_1}} ( x^{\prime}_1,x^{\prime}_2 ) \circ \operatorname{\mathrm{id}}_{\phi_{x_0}}^{\phi_{x_1}}(x^{\prime}_1) \circ \Theta^\pm_{\phi_{x_0},\phi_{x_0}}( x^{\prime}_0,x^{\prime}_1 ) \end{align*} $$

so that we have $d_G(g',g) < \delta $ . As each $x^{\prime }_i$ is bi-recurrent, successive applications of Proposition 4.6 and Lemmas 4.9 and 4.10 show that we have

$$ \begin{align*} \mathfrak h_\epsilon^{\phi_x}(x) = \operatorname{\mathrm{Ad}}_{g'}( \mathfrak h_\epsilon^{\phi_x}(x) ) \end{align*} $$

for $g'$ arbitrarily close to G. As $\operatorname {\mathrm {Ad}}_{g'}$ is continuous in $g'$ , we then obtain

$$ \begin{align*} \mathfrak h_\epsilon^{\phi_x}(x) = \operatorname{\mathrm{Ad}}_{g}( \mathfrak h_\epsilon^{\phi_x}(x) ) \end{align*} $$

as desired.

Proof. Without loss of generality, suppose that we have chosen $\phi _x$ so that $\Theta ^+_{\phi _x,\phi _x}(x,u)$ is trivial for all $u \in (W^{su}_{g_t}(x) \cap B_{2\epsilon }(x) )^\circ $ , and so that $\Theta ^-_{\phi _x,\phi _x}(s_1,s_2)$ is trivial whenever $s_1$ and $s_2$ lie on the same (local) leaf of the strong stable foliation of $g_t$ in $B_{2\epsilon }(x)$ .

We write $\mathfrak h := \mathfrak h_\epsilon ^{\phi _x}(x)$ and suppose for the sake of contradiction that $\mathfrak h \subsetneq \mathfrak g$ is a proper subalgebra. By Proposition 4.9 and our choice of $\phi _x$ , we must have $\mathfrak h_\epsilon ^{\phi _x}(u) = \mathfrak h$ for all $u \in B_\epsilon (x)$ lying on the local leaf through x of the strong unstable foliation of $g_t$ . Hence, for any $u_1 \in B_\epsilon (x)$ , each vector $X_{w'}^{\phi _x}(u_1)$ used in the definition of $\mathfrak h_\epsilon ^{\phi _x}(u)$ must lie in the Lie algebra $\mathfrak h$ . Integrating this, we see that the unstable holonomies $\Theta ^+_{\phi _x,\phi _x}(u_1,u_2)$ are constrained to $\exp (\mathfrak h)$ for all $u_1,u_2 \in B_\epsilon (x)$ that lie on the same local leaf of the strong unstable foliation of $g_t$ .

Now, consider $H := \exp (\mathfrak h)$ and consider an element $g \in G \setminus H$ that lies in the complement. By our construction of $\phi _x$ , all unstable holonomies are constrained to H and all stable holonomies are trivial; hence, no local sequence of stable and unstable holonomies along a sequence of points $x, x_1, x_2, \ldots x_k, x$ lying in $B_\epsilon (x)$ can result in a total holonomy of g. Moreover, because $\mathfrak h$ is an ideal by Theorem 4.11, H is a normal subgroup of G; hence, we cannot obtain a total holonomy of g for any choice of trivialization. As $f_t$ is locally G-accessible at x, this is a contradiction.

Proof. The existence of such a t satisfying $|t| < \tau (y), \tau ( \mathcal P^{-1}(y) )$ follows immediately from the construction of the Markov partition $(R, \mathcal P)$ . We then clearly have

$$ \begin{align*} \Theta^+_{\mathrm{symb}}(x,y) = \operatorname{\mathrm{Hol}}_{\phi_y}^{\phi_y}(g_t(y),y,-t) \circ \Theta^+_{\phi_x,\phi_y}(x, g_t(y)) \end{align*} $$

by Proposition 4.1 and Definition 4.4. By our choice of trivialization and the fact that t does not exceed the return time of y, $\operatorname {\mathrm {Hol}}_{\phi _y}^{\phi _y}(g_t(y),y,-t)$ is the identity in G, as desired.

Proof. By Proposition 4.13, the unstable holonomies used in Definitions 4.3 and 4.5 are the same. Hence, the regular and symbolic transitivity groups agree, and so by Theorem 4.11, we have $\mathfrak h_{\mathrm {symb}}(x) = \mathfrak g$ .

Proof. By definition, we have

for all $u \in B_\delta (x)$ . The derivatives of the terms in the sequence converge exponentially fast in n, so we can interchange the limit and the derivative to obtain

for any consistent past $v_i$ . As the $X_{w(u),v_i}^{\mathrm {symb}} - X_{w(u),v_j}^{\mathrm {symb}}$ taken over pasts $v_i, v_j$ and vectors w form a basis of $\mathfrak g$ , there is a finite $n_0$ so that the approximations

can also be used to form a basis $X_{w(u),v_i^{(n_0)}}^{\mathrm {symb}}(x) - X_{w(u),v_j^{(n_0)}}^{\mathrm {symb}}(x)$ , taken again over pasts $v_i, v_j$ and vectors w. Hence, we have a Casimir

$$ \begin{align*} \Omega = \sum ( g_{ij} )^{-1} ( X_{w,v_i^{(n_0)}}^{\mathrm{symb}}(x) - X_{w,v_j^{(n_0)}}^{\mathrm{symb}}(x) ) ( X_{w,v_i^{(n_0)}}^{\mathrm{symb}}(x) - X_{w,v_j^{(n_0)}}^{\mathrm{symb}}(x) ) \end{align*} $$

for some finite collection of pasts $v_i, v_j$ and a given vector w. This acts on $V^\rho $ by scalar multiplication by $\|\rho \|^2$ , and so we obtain

$$ \begin{align*} d\rho\bigg(\sum (g_{ij})^{-1} ( X_{w,v_i^{(n_0)}}^{\mathrm{symb}}(x) - X_{w,v_j^{(n_0)}}^{\mathrm{symb}}(x) )^2\bigg)( \varphi(x) ) = \|\rho\|^2 \varphi(x) \end{align*} $$

for this collection of pasts, and the same vector w. Now, there is a uniform constant $\epsilon> 0$ , depending only on the $g_{ij}$ and $\mathfrak g$ , so that

$$ \begin{align*} \| d\rho( X_{w,v_i^{(n_0)}}^{\mathrm{symb}}(x) - X_{w,v_j^{(n_0)}}^{\mathrm{symb}}(x) ) ( \varphi(x) ) \|_{L^2(G)} \geq \epsilon \|\rho\| \varphi(x) \end{align*} $$

for some choice of $v_i, v_j$ and w. As $X_{w,v}^{\mathrm {symb}}(u)$ varies continuously in w and u, there is a neighbourhood $B_\delta (x)$ of x in U and a $C^1$ vector field $w \colon B_\delta (x) \to T^1B_\delta (x)$ for which

$$ \begin{align*} \| d\rho( X_{w(u),v_i^{(n_0)}}^{\mathrm{symb}}(u) - X_{w(u),v_j^{(n_0)}}^{\mathrm{symb}}(u) ) ( \varphi(u) ) \|_{L^2(G)} \geq \frac \epsilon 2 \|\rho\| \varphi(u) \end{align*} $$

holds for all $u \in B_\delta (x)$ , as desired.

Proof. As we chose $\Phi \in \mathcal K_C$ , $\log \Phi $ is C-Lipschitz and we therefore have

$$ \begin{align*} | \log ( \Phi(v^{(n)}(y)) ) - \log ( \Phi( v^{(n)}(x) ) ) | &\leq \frac C {f\kappa^n} d(x,y)\\ &\leq \frac C f d(x,y) \end{align*} $$

for any $x,y \in U$ . Exponentiating both sides, this means

(5.1) $$ \begin{align} \Phi( v^{(n)}(y) ) \leq e^{ (C/{f}) d(x,y)} \Phi( v^{(n)}(x) ) \end{align} $$

for any $x,y \in U$ . Now, suppose that $d(x,y) \leq 2\delta $ for some $\delta> 0$ , and fix a unit speed path $\gamma \colon [0,2\delta ] \to U_0$ with $\gamma (0) = y$ and $\gamma (2\delta ) = x$ . We then have

$$ \begin{align*} | \|\varphi(v^{(n)}(x)) |_{L^2(G)} - \|\varphi( v^{(n)}(y) ) \|_{L^2(G)} | &\leq \int_0^{2\delta} |\langle \operatorname{\mathrm{grad}}( \varphi \circ v^{(n)} )(\gamma(t)), \gamma'(t) \rangle_{T^1N} | \, dt\\ &\leq \frac C {f\kappa^n} \int_0^{2\delta} ( \Phi \circ v^{(n)} )(\gamma(t)) \, dt\\ &\leq \frac C {f \kappa^n} {2\delta} e^{( C /{f\kappa^n}) d(x,y)} \Phi( v^{(n)}(y) )\\ &\leq \frac C {f} \delta e^{(C/{f}) 2\delta} \Phi( v^{(n)}(y) ) \end{align*} $$

by the fundamental theorem of calculus and our bounds on $\varphi $ . Fix $\delta $ small enough to ensure

$$ \begin{align*} \frac C {f} 2\delta e^{ (C/f) 2\delta} \leq \frac 1 8 \end{align*} $$

and

$$ \begin{align*} e^{ (C/f) 2\delta} \leq 2 \end{align*} $$

hold simultaneously; note that we really only require that $C \delta $ is sufficiently small, and so $\delta $ can be chosen inversely proportional to C. We therefore have

(5.2) $$ \begin{align} \Phi( v^{(n)}(y) ) \leq 2 \Phi( v^{(n)}(x) ) \end{align} $$

and

(5.3) $$ \begin{align} \| \varphi( v^{(n)}(x) ) \|_{L^2(G)} \leq \| \varphi( v^{(n)}(y) ) \|_{L^2(G)} + \tfrac 1 8 \Phi( v^{(n)}(y) ) \end{align} $$

whenever $d(x,y) < 2\delta $ . To conclude, suppose that we had

$$ \begin{align*} {\| \varphi(v^{(n)}(y)) \|_{L^2(G)}} \leq \tfrac 1 4{\Phi(v^{(n)}(y))} \end{align*} $$

at some $y \in B_{\delta }(u_0)$ , for a given $u_0 \in U$ and n. Then by (5.3) and (5.2), we must have

$$ \begin{align*} \| \varphi( v^{(n)}(x) ) \|_{L^2(G)} &\leq \tfrac 1 4 \Phi( v^{(n)}(y) ) + \tfrac 1 8 \Phi( v^{(n)}(y) )\\[3pt] &\leq \tfrac 2 4 \Phi( v^{(n)}(x) ) + \tfrac 2 8 \Phi( v^{(n)}(x) )\\[3pt] &\leq \tfrac 3 4 \Phi( v^{(n)}(x) ) \end{align*} $$

for any $x \in B_{\delta }(u_0)$ , as desired.

Proof. Differentiating the fraction, we see that we need to bound the supremum of

(5.4)

for any $u \in U$ and $w \in T^1_u B_\delta (u_0)$ . Note that we always have

$$ \begin{align*} ( d( \| \varphi \circ v^{(n)} \|_{L^2(G)} ) )_u(w) \leq \| ( d( \varphi \circ v^{(n)} ) )_u(w) \|_{L^2(G)} \end{align*} $$

and so (5.4) is at most

(5.5)

by the triangle inequality. Cancelling terms, we can reduce (5.5) to

which is at most

(5.6)

by the chain rule. As $f \kappa ^n$ is the slowest expansion rate of any vector in $T^1U$ under $d\sigma ^n$ , we can bound

$$ \begin{align*} {\sup_{w \in T^1_uU}} \| ( dv^{(n)} )_u(w) \|_{T^1N} \leq \frac 1 { f \kappa^n} \end{align*} $$

for all $u \in U$ . By hypothesis, we also have

$$ \begin{align*} \sup_{w' \in T^1_{v^{(n)}(u)}U} \| ( d\varphi )_{v^{(n)}(u)}(w')\|_{L^2(G)} \leq C \Phi( v^{(n)}(u) ) \end{align*} $$

for all $u \in U$ , and so (5.6) can be bounded above by

which is, in turn, at most

$$ \begin{align*} \frac {8 C \Phi( v^{(n)}(u) )}{f\kappa^n \Phi( v^{(n)}(u) )} \end{align*} $$

by hypothesis. Cancelling $\Phi $ , we obtain the result desired.

Proof. We deal with a fixed $n> 0$ throughout the proof, and specify how large n needs to be as we proceed; it is important to note that we cannot deal with arbitrarily large n without foregoing the uniformity of the bounds we wish to obtain. We proceed in cases depending on which alternative of Lemma 5.1 holds. For any pasts $v_1^{(n)}$ and $v_2^{(n)}$ , if we have

$$ \begin{align*} \|\varphi(v_{1}^{(n)}(u))\|_{L^2(G)} \leq \tfrac 3 4 \Phi( v_{1}^{(n)}(u) ) \end{align*} $$

for all $u \in B_\delta (x)$ , then because $\rho $ is unitary we can clearly bound

above by

$$ \begin{align*} \tfrac 3 4 (e^{\alpha_{\mathrm{Re}(z)}^{(n)}(v_1^{(n)}(u))} \Phi( v_1^{(n)}(u) ) + e^{\alpha_{\mathrm{Re}(z)}^{(n)}(v_2^{(n)}(u))} \Phi( v_2^{(n)}(u) ) ) \end{align*} $$

for all $u \in B_\delta (x)$ . In this case, we are done by simply setting $y := x$ . Similarly, if we had

$$ \begin{align*} \|\varphi(v_{2}^{(n)}(u))\|_{L^2(G)} \leq \tfrac 3 4 \Phi( v_{2}^{(n)}(u) ) \end{align*} $$

for all $u \in B_\delta (x)$ , then we are again done by setting $y := x$ , up to interchanging our choice of $v_1$ and $v_2$ . So we may as well assume that the second alternative of Lemma 5.1 holds for both $v_1^{(n)}$ and $v_2^{(n)}$ , and that we therefore have

(5.7) $$ \begin{align} \|\varphi(v_{\ell}^{(n)}(u))\|_{L^2(G)} \geq \tfrac 1 4 \Phi( v_{\ell}^{(n)}(u) ) \end{align} $$

for all $u \in B_\delta (x)$ . We temporarily abbreviate

and

for the sake of clarity. Note that $\hat \varphi _\ell $ and $\hat \psi _\ell $ are well-defined on $B_\delta (x)$ as an immediate consequence of (5.7). Now, by reversing the triangle inequality,

(5.8)

is at least

(5.9)

for any $y \in B_\delta (x)$ . As group elements act by isometries with respect to the $L^2(G)$ norm, (5.9) is equal to

which we can bound below by

(5.10)

using the reverse triangle inequality once again. We can rewrite (5.10) as

(5.11)

by replacing $\hat \psi _1(y)$ with $g_1^{-1}(y)g_1(y)\hat \psi _1(y)$ in the first term of the first line, and multiplying both terms on the first line by $g_2(x)g_2^{-1}(y)$ . Hence, (5.8) is bounded below by (5.11), and we see that

is at least

(5.12)

after rearranging. By Theorem 4.16, there is an $N> 0$ , $\epsilon> 0$ , a choice of pasts $v_1^{(n)}$ and $v_2^{(n)}$ , and a y with $d(x,y) = \delta /2$ for which we have

(5.13) $$ \begin{align} &\| \rho( g_{1}(x)g_1^{-1}(y)g_1(y) ) \hat\psi_{1}( y ) - \rho( g_2(x)g_2^{-1}(y)g_1(y) )\hat\psi_{1}( y )\|_{L^2(G)}\nonumber\\ &\quad\geq \epsilon (1+|\mathrm{Im}(z)|) \|\rho\| \frac \delta 2 \end{align} $$

for all $n \geq N$ ; note that we have applied the theorem to

which is certainly a smooth function in $C^1(U, V^\rho )$ . On the other hand, by Lemma 5.2, $\hat \varphi _\ell (u)$ is at worst ${8C}/{f\kappa ^n}$ -Lipschitz in u on $B_\delta (x)$ , and we can estimate

for all $u \in U$ , $w \in T^1_uB_\delta (x)$ and $w' \in T^1_{v_\ell (u)}B_\delta (x)$ . Hence, we can bound

$$ \begin{align*} \| \hat \psi_\ell(x) - \hat \psi_\ell(y) \| \leq \bigg(\frac {8C}{f\kappa^n} + |\mathrm{Im}(z)| \frac{\|\tau\|_{C^1}}{f(\kappa - 1)}\bigg)\,d(x,y) \end{align*} $$

because we chose $y \in B_\delta (x)$ . Suppose that n and $\|\rho \|$ are large enough so that we have

$$ \begin{align*} \bigg( \frac {8 C}{f \kappa^n} + |\mathrm{Im}(z)| \frac{\|\tau\|_{C^1}}{f(\kappa -1)} \bigg) < \frac {\epsilon} 8 (1+|\mathrm{Im}(z)|) \|\rho\| \end{align*} $$

for some constant $K> 0$ . Note that we can make this choice of n so that it depends only on $ C /{\|\rho \|}$ and not $\|\rho \|$ directly; moreover, the requirement that $\|\rho \|$ be sufficiently large can be made absolute, and in particular is independent of z. This ensures

(5.14) $$ \begin{align} \| \hat \psi_\ell(x) - \hat \psi_\ell(y)\|_{L^2(G)} \leq \frac \epsilon 8 ( 1 + |\mathrm{Im}(z)| )\| \rho \| \delta \end{align} $$

because $d(x,y) \leq \delta $ . Note that our choice of n here depends only on $ C/ {\|\rho \|}$ . Plugging (5.13) and (5.14) into (5.12), we conclude that

$$ \begin{align*} &\|\rho(g_1(x))\hat \psi_1(x) - \rho( g_2(x) )\hat \psi_2(x) \|_{L^2(G)} + \|\rho(g_1(y)) \hat\psi_1(y) - \rho( g_2(y) )\hat \psi_2(y)\|_{L^2(G)}\\ &\quad\geq \frac \epsilon 4 \delta (1+|\mathrm{Im}(z)|) \|\rho\| \end{align*} $$

and so

(5.15) $$ \begin{align} \|\rho(g_1(x))\hat \psi_1(x) - \rho( g_2(x) )\hat \psi_2(x) \|_{L^2(G)}> \frac \epsilon 8 \delta (1+|\mathrm{Im}(z)|) \|\rho\| \end{align} $$

or

(5.16) $$ \begin{align} \|\rho(g_1(y))\hat \psi_1(y) - \rho( g_2(y) )\hat \psi_2(y) \|_{L^2(G)}> \frac \epsilon 8 \delta (1+|\mathrm{Im}(z)|) \|\rho\| \end{align} $$

must hold. Without loss of generality, suppose (5.16) holds. Using the Lipschitz estimate on $\hat \varphi _\ell $ , we can bound