Posted by: matheuscmss | September 8, 2011

Explicit constants in Ratner’s estimates on rates of mixing of geodesic flows of hyperbolic surfaces

As I promised in this previous post here, today we’re going to compute some explicit constants in Ratner’s paper on the rate of mixing of geodesic flows by simply bookkeeping them while trying to follow the article. In particular, we will borrow the notations from it (and, a fortiori, they will be the same as in this post here).

By the way, let me say that the decision of bookkeeping constants instead of trying more elegant methods is partly motivated by a conversation I had with Livio Flaminio (in the occasion of a summer school at Bedlewo, Poland) where we saw that, even though one disposes of (classical) constructions of the principal, complementary and discrete series, it seems that direct computation with them only give useful information on explicit constants in the case of the principal and discrete series (and in these cases the computation is actually quite elegant). Thus, since our main motivation for this post comes from the complementary series, I decided to stick to the non-elegant straightforward computation presented below the fold.

1. Preliminaries

Let {T} be a non-trivial irreducible unitary {SL(2,\mathbb{R})}-representation in a complex separable Hilbert space {\mathcal{H} = \mathcal{H}(T)}. Denoting by {r(\theta) = \left(\begin{array}{cc}\cos \theta & \sin\theta \\ -\sin\theta & \cos\theta\end{array}\right)\in SO(2,\mathbb{R})}, {\theta\in\mathbb{R}}, we define, for each {n\in\mathbb{Z}},

\displaystyle \mathcal{H}_n(T) = \{v\in\mathcal{H}(T): T(r(\theta))v=e^{in\theta}v \,\forall\,\theta\in\mathbb{R}\}.

Then, one has {\mathcal{H}(T)=\oplus_{n\in\mathbb{Z}}\mathcal{H}_n(T)}. Furthermore, by irreducibility of {T}, we have that {\textrm{dim}(\mathcal{H}_n(T)) = 0} or {1}. In this way, one can construct an orthonormal basis {\{\phi_n\in\mathcal{H}_n(T):n\in\mathbb{Z}\}} of {\mathcal{H}(T)} such that {\phi_n\neq 0} if and only if {\textrm{dim}(\mathcal{H}_n(T)) = 1}.

Denote by {B_{n,m}(t) = \langle \phi_n, T(a(t))\phi_m\rangle}, where {a(t)=\left(\begin{array}{cc}e^t & 0 \\ 0 & e^{-t}\end{array}\right)} is the (positive) diagonal 1-paramter subgroup of {SL(2,\mathbb{R})}. In the sequel, we will be interested in the decay properties of {B_{n,m}(t)} as {t\rightarrow\infty}. To perform this study, we will follow M. Ratner by making a series of preparations.

As it is shown in Lemma 2.1 of Ratner’s paper, {y(t):=B_{n,m}(t)} satisfies the following ODE

\displaystyle y'' + 2y'-4\lambda y = f_1(t)+f_2(t)


\displaystyle f_1(t)=(2e^{2t}\sinh(2t))^{-1}y'(t)


\displaystyle f_2(t)=y(t)\left[\frac{2m(n-me^{-2t})}{\sinh(2t)}-\frac{(n-me^{-2t})^2}{\sinh^2(2t)}\right].

Furthermore, by the discussions after equation (2.12) of Ratner’s paper and by equation (2.13) of this paper, one has {|y(t)|=|B_{n,m}(t)|\leq 1} and {|y'(t)|=|B_{n,m}'(t)|\leq\sqrt{m^2-4\lambda}}. Hence,

\displaystyle |f_1(t)|\leq |(2e^{2t}\sinh(2t))^{-1}|\cdot\sqrt{m^2-4\lambda}


\displaystyle |f_2(t)|\leq \left|\frac{2m(n-me^{-2t})}{\sinh(2t)}-\frac{(n-me^{-2t})^2}{\sinh^2(2t)}\right|.


\displaystyle |(2e^{2t}\sinh(2t))^{-1}|=e^{-4t}|(1-e^{-4t})^{-1}|\leq e^{-4t}\cdot(1-e^{-4})^{-1}

for every {t\geq 1}, we obtain that the constant {C_1} appearing in equation (2.14) of Ratner’s paper is

\displaystyle C_1=(1-e^{-4})^{-1}, \ \ \ \ \ (1)


\displaystyle |f_1(t)|\leq C_1\sqrt{m^2-4\lambda}\cdot e^{-4t} \ \ \ \ \ (2)

with {C_1} as above.

Similarly, {\frac{2m(n-me^{-2t})}{\sinh(2t)}-\frac{(n-me^{-2t})^2}{\sinh^2(2t)}=\frac{1}{\sinh(2t)}[2m(n-me^{-2t})-\frac{(n-me^{-2t})^2}{\sinh(2t)}]}, so that {|f_2(t)|\leq\left|\frac{2m(n-me^{-2t})}{\sinh(2t)}-\frac{(n-me^{-2t})^2}{\sinh^2(2t)}\right|} is bounded by the quantity {\frac{2}{(1-e^{-4})}e^{-2t}[2m(n-me^{-2t})-\frac{(n-me^{-2t})^2}{\sinh(2t)}]} for every {t\geq1}. On the other hand, this last quantity is bounded by

\displaystyle 2C_1e^{-2t}\left[|2mn|+2e^{-2}m^2+2e^{-2}C_1n^2+2e^{-4}C_1|2mn|+2e^{-6}C_1m^2\right].

Because {|2mn|\leq m^2+n^2} and {2e^{-2}+2C_1e^{-6}=2C_1e^{-2}} (since {C_1=1/(1-e^{-4})}), we see that

\displaystyle \begin{array}{rcl} |f_2(t)|&\leq& \frac{2C_1}{e^{2t}}\left[\left(1+\frac{2}{e^2}+\frac{2C_1}{e^4}+\frac{2C_1}{e^6}\right)m^2+\left(1+\frac{2C_1}{e^2}+\frac{2C_1}{e^4}\right)n^2\right]\\ &=&2C_1e^{-2t}(1+2C_1e^{-2}+2C_1e^{-4})(m^2+n^2). \end{array}

In other words, the constant {C_2} appearing in equation (2.14) of Ratner’s paper is

\displaystyle C_2=\frac{2}{1-e^{-4}}\left(1+\frac{2}{e^2(1-e^{-4})}+\frac{2}{e^4(1-e^{-4})}\right), \ \ \ \ \ (3)


\displaystyle |f_2(t)|\leq C_2(m^2+n^2)e^{-2t} \ \ \ \ \ (4)

with {C_2} as above.

Next, we observe that the constant {C_1} in equation (2.16) of Ratner’s paper is slightly different from the previous {C_1} constant in equation (2.14) of this paper. Indeed, by denoting by {r_1:=r_1(\lambda) := -1+\sqrt{1+4\lambda}} and {r_2 := r_2(\lambda) := -1 - \sqrt{1+4\lambda}} the roots of the characteristic equation {x^2+2x-4\lambda=0} of the ODE satisfied by {y(t):=B_{n,m}(t)}, the fact that {|f_1(t)|\leq C_1e^{-4t}\sqrt{m^2-4\lambda}} implies that

\displaystyle \begin{array}{rcl} \left|\int_t^{\infty}e^{-r_1 s}f_1(s)ds\right|&\leq& C_1\sqrt{m^2-4\lambda}\int_t^{\infty}e^{(-\textrm{Re}(r_1)-4)s} ds \\  &\leq& \frac{C_1}{3}\sqrt{m^2-4\lambda}\cdot e^{-3t}\end{array} \ \ \ \ \ (5)

because {\textrm{Re}(r_1)+2\geq 1}. However, the constant {C_2} in equation (2.16) of Ratner’s paper remains the same of equation (2.14) of this article:

\displaystyle \begin{array}{rcl}\left|\int_t^{\infty}e^{-r_1 s}f_2(s)ds\right|&\leq& C_2(m^2+n^2)\int_t^{\infty}e^{(-\textrm{Re}(r_1)-2)s} ds \\  &\leq& C_2(m^2+n^2)e^{-t}\end{array}\ \ \ \ \ (6)

because {\textrm{Re}(r_1)+2\geq 1}.

Concluding our series of preparations, we recall the definitions of the following two functions

\displaystyle A_1(t):=\int_1^t e^{(r_1-r_2)s}\left(\int_s^{\infty} e^{-r_1 u}f_1(u)\,du\right)\,ds


\displaystyle A_2(t):=\int_1^t e^{(r_1-r_2)s}\left(\int_s^{\infty} e^{-r_1 u}f_2(u)\,du\right)\,ds

introduced after equation (2.18) of Ratner’s paper. These functions appear naturally in our context because the ODE verified by {y(t)=B_{n,m}(t)} can be rewritten as {(D-r_1)(D-r_2)y = f_1(t)+f_2(t):=f(t)} where {D} is the differentiation operator (with respect to {t}). Thus, since {(D-r_1)y=e^{r_1 t}D(e^{-r_1t}y)}, we have {e^{r_1t}D(e^{-r_1t}(D-r_2)y) = f(t)} and, hence,

\displaystyle e^{(r_2-r_1)t}D(e^{-r_2t}y)=-\int_t^{\infty}e^{-r_1 s}f(s)\,ds+P_1

where {P_1} is a constant. In particular, we can write

\displaystyle y(t) = -e^{-t}\int_1^t\left(\int_s^{\infty}e^u\,f(u)\,du\right)\,ds + P_1te^{-t} + P_2 e^{-t} \ \ \ \ \ (7)

if {r_1=r_2}, and

\displaystyle \begin{array}{rcl} y(t) &=& e^{r_2 t}A(t) + e^{r_2 t}\left[P_1\int_1^t e^{(r_1-r_2)s}\,ds + P_2\right] \\  &=& e^{r_2 t}A(t) + \frac{P_1}{2\sqrt{1+4\lambda}}e^{r_1 t} + \left(P_2 - \frac{e^{2\sqrt{1+4\lambda}}P_1}{2\sqrt{1+4\lambda}}\right)e^{r_2 t}\end{array} \ \ \ \ \ (8)

if {r_1\neq r_2}, where {A(t):=A_1(t)+A_2(t)}. Moreover, by using these equations, and the fact that {y(t) = B_{n,m}(t)\rightarrow 0} as {t\rightarrow\infty} (a consequence of the non-triviality of {T}, that is, it has no invariant vectors), we can deduce that

\displaystyle P_1 = \int_1^{\infty}e^{-r_1s} f(s)\,ds - r_2 e^{-r_1}y(1)+e^{-r_1}y'(1) \ \ \ \ \ (9)


\displaystyle P_2 = y(1) e^{-r_2} \ \ \ \ \ (10)

Finally, from the estimates (2), (4) above, and the facts {\textrm{Re}(r_1)-\textrm{Re}(r_2)\geq 0} and {\textrm{Re}(r_1)+2\geq 1}, we can estimate:

\displaystyle \begin{array}{rcl}&&|e^{r_2t}A_1(t)|\\  &&= |e^{r_2t}\int_1^t e^{(r_1-r_2)s}\int_s^{\infty}e^{-r_1 u}f_1(u)du\,ds| \\  &&\leq C_1\sqrt{m^2-4\lambda} \,\, e^{t\textrm{Re}(r_2)}\int_1^t e^{(\textrm{Re}(r_1)-\textrm{Re}(r_2))s}\int_s^{\infty}e^{(-\textrm{Re}(r_1)-4)u}du \, ds\\  &&\leq\frac{C_1}{3}\sqrt{m^2-4\lambda} \,\, e^{t\textrm{Re}(r_2)} e^{(\textrm{Re}(r_1)-\textrm{Re}(r_2))t}\int_1^t e^{(-\textrm{Re}(r_1)-4)s}ds\\  &&\leq \frac{C_1}{9e^3}\sqrt{m^2-4\lambda} \,\, e^{t\textrm{Re}(r_1)}\end{array} \ \ \ \ \ (11)


\displaystyle \begin{array}{rcl}&&|e^{r_2t}A_2(t)|\\  &&= |e^{r_2t}\int_1^t e^{(r_1-r_2)s}\int_s^{\infty}e^{-r_1 u}f_2(u)du\,ds| \\  &&\leq C_2(m^2+n^2)e^{t\textrm{Re}(r_2)}\int_1^t e^{(\textrm{Re}(r_1)-\textrm{Re}(r_2))s}\int_s^{\infty} e^{(-\textrm{Re}(r_1)-2)u}du\,\,ds\\  &&\leq C_2(m^2+n^2)e^{t\textrm{Re}(r_1)}\int_1^t e^{(-\textrm{Re}(r_1)-2)s}ds\\  &&\leq \frac{C_2}{e}(m^2+n^2)e^{t\textrm{Re}(r_1)}\end{array}\ \ \ \ \ (12)

Thus, we can take

\displaystyle \bar{C}_1=C_1/9e^3 \quad \quad \textrm{and} \quad\quad \bar{C}_2=C_2/e \ \ \ \ \ (13)

in equations (2.19) and (2.20) of Ratner’s paper.

After these preparations, we are ready to pass to the next section, where we will render more explicit the constants appearing in Lemma 2.2 of Ratner’s paper about the speed of decay of the matrix coefficients {B_{n,m}(t)} as {t\rightarrow\infty}.

2. Decay of matrix coefficients of irreducible unitary {SL(2,\mathbb{R})}-representations

By following closely the proof of Lemma 2.2 of Ratner’s paper, we will show the following explicit variant of it:

Lemma 1 For {t\geq1}, {n,m\in\mathbb{Z}},

\displaystyle |B_{n,m}(t)|\leq (\bar{K}_{\lambda}(m^2+n^2)+\tilde{K}_{\lambda})\cdot b_{\lambda}(t),


  • {b_{\lambda}(t)=te^{-t}} if {\lambda\leq-1/4};
  • {b_{\lambda}(t)=te^{r_1t}} if {-1/4<\lambda<0};
  • {b_{\lambda}(t)=te^{-2t}} if {0\leq\lambda};


\displaystyle \bar{K}_\lambda=\left\{\begin{array}{cl} 4C_1/9e^3+2C_2/e+e & \textrm{if }\lambda\leq-1/4\\ 4C_1/9e^3+2C_2/e+e & \textrm{if }-1/4<\lambda<0\\ (C_1+C_2)/2 & \textrm{if } 0\leq\lambda\end{array}\right.,

\displaystyle \tilde{K}_{\lambda}=\left\{\begin{array}{cl} (1+2\sqrt{2})e + (32+\sqrt{2})C_1^2/3e^3 & \textrm{if }\lambda\leq-1/4\\ 3e + e^2 + 4C_1/9e^3 & \textrm{if }-1/4<\lambda<0\\ e^2& \textrm{if } 0\leq\lambda\end{array}\right..

with the constants {C_1} and {C_2} given by (1) and (3) above.

Remark 1 In Ratner’s article, the function {b_{\lambda}(t)} is slightly different from the one above (when {\lambda<0}): indeed, in this paper, {b_{\lambda}(t) = \min\{te^{-t}, e^{-t}(1+\sqrt{1/|1+4\lambda|})\}} if {\lambda\leq -1/4}, and {b_{\lambda}(t)=\min\{te^{r_1t},e^{r_1 t}\sqrt{1/|1+4\lambda|}\}} if {-1/4<\lambda<0}. In particular, this allows to gain over the factor of {t} (in front of the exponential functions {e^{-t}}, {e^{r_1t}}) when {\lambda} is not close to {-1/4} at the cost of permitting larger constants. However, since we had in mind the idea of getting uniform constants regardless of {\lambda} and the factor of {t} does not seem very substantial, we decided to neglect this issue and to stick to the function {b_{\lambda}(t)} as defined in Lemma 1 above.

Proof: We begin with the case {\lambda=-1/4}, i.e., {r_1=r_2=-1}. In this case, from (7), we know that

\displaystyle y(t)=-e^{-t}\int_1^t\left(\int_s^{\infty}e^uf(u)du\right)ds+P_1te^{-t}+P_2e^{-t}.

Since, by definition, {f(t)=f_1(t)+f_2(t)}, we can apply (2), (4) above to obtain

\displaystyle \begin{array}{rcl}|y(t)|&\leq& C_1\sqrt{m^2+1}\cdot e^{-t}\int_1^t\int_s^{\infty}e^{-3u}du\,\,ds \\&+& C_2 (m^2+n^2)e^{-t}\int_1^t\int_s^{\infty}e^{-u}du\,\,ds \\  &+&|P_1|te^{-t}+|P_2|e^{-t}.\end{array}

On the other hand, using that {|y(1)|\leq 1}, {|y'(1)|\leq\sqrt{m^2-4\lambda}}, the equations (5), (6), (9), (10) above, and the fact that {r_1=r_2=-1} in the present case, we get

\displaystyle \begin{array}{rcl}|y(t)| &\leq& \frac{C_1}{9e^3}\sqrt{m^2+1}\cdot e^{-t} + \frac{C_2}{e}(m^2+n^2)e^{-t} \\&+&te^{-t}\left[\frac{C_1}{3e^3}\sqrt{m^2+1}+\frac{C_2}{e}(m^2+n^2)+e+e\sqrt{m^2+1}\right] + e\cdot e^{-t}\end{array}

Since {\sqrt{m^2+1}\leq |m|+1\leq m^2+n^2+1} and {e^{-t}\leq te^{-t}} (because {t\geq 1}), we conclude that

\displaystyle |y(t)|\leq te^{-t}\left[\bar{K}_{[\lambda=-1/4]}(m^2+n^2)+\tilde{K}_{[\lambda=-1/4]}\right] \ \ \ \ \ (14)


\displaystyle \bar{K}_{[\lambda=-1/4]}=\frac{4C_1}{9e^3}+\frac{2C_2}{e}+e


\displaystyle \tilde{K}_{[\lambda=-1/4]}=\frac{4C_1}{9e^3}+3e

Next, we notice that, when {r_1\neq r_2}, by (8), and (11), (12) above,

\displaystyle \begin{array}{rcl}|y(t)|&\leq& \frac{C_1}{9e^3}\sqrt{m^2-4\lambda}\cdot e^{t\textrm{Re}(r_1)}+ \frac{C_2}{e}(m^2+n^2)\cdot e^{t\textrm{Re}(r_1)}\\ &+& |P_1|\cdot te^{t\textrm{Re}(r_1)} + |P_2|\cdot e^{t\textrm{Re}(r_2)}\end{array} \ \ \ \ \ (15)

If {-1/2\leq\lambda<-1/4}, we have that {\textrm{Re}(r_1)=\textrm{Re}(r_2)=-1}, {|r_2|\leq\sqrt{2}} and {\sqrt{m^2-4\lambda}\leq\sqrt{m^2+2}\leq m^2+n^2+\sqrt{2}}, so that (9), (10) and the equations (5), (6) and (15) above imply

\displaystyle |P_1|\leq\frac{C_1}{3e^3}(m^2+n^2+\sqrt{2})+\frac{C_2}{e}(m^2+n^2)+\sqrt{2}\cdot e+e(m^2+n^2+\sqrt{2}),

\displaystyle |P_2|\leq e

and, a fortiori,

\displaystyle \begin{array}{rcl} |y(t)|&\leq&\left(\frac{C_1}{9e^3}+\frac{C_2}{e}\right)(m^2+n^2)te^{-t}+\frac{\sqrt{2}C_1}{9e^3}te^{-t}+(|P_1|+|P_2|)te^{-t}\\  &\leq& t e^{-t} \left(\bar{K}_{[-1/2\leq\lambda<-1/4]}(m^2+n^2)+\tilde{K}_{[-1/2\leq\lambda<-1/4]}\right)\end{array} (16)


\displaystyle \bar{K}_{[-1/2\leq\lambda<-1/4]}=\frac{4C_1}{9e^3}+\frac{2C_2}{e}+e


\displaystyle \tilde{K}_{[-1/2\leq\lambda<-1/4]}=\frac{4\sqrt{2}C_1}{9e^3}+(1+2\sqrt{2})e.

If {-1/4<\lambda<0}, we have that {0<\sqrt{1+4\lambda}<1}, so that {\textrm{Re}(r_1)=r_1=-1+\sqrt{1+4\lambda}\in(-1,0)}, {\textrm{Re}(r_2)=r_2=-1-\sqrt{1+4\lambda}\in (-2,-1)}, {|r_2|=1+\sqrt{1+4\lambda}\in(1,2)} and {\sqrt{m^2-4\lambda}\leq\sqrt{m^2+1}\leq m^2+n^2+1}. Putting this into (9), (10), and the equations (5), (6), and (15) above, we get

\displaystyle |P_1|\leq\frac{C_1}{3e^3}(m^2+n^2+1)+\frac{C_2}{e}(m^2+n^2)+2e+e(m^2+n^2+1),

\displaystyle |P_2|\leq e^2,


\displaystyle \begin{array}{rcl}|y(t)|&\leq& \left(\frac{C_1}{9e^3}+\frac{C_2}{e}\right)(m^2+n^2)te^{tr_1}+\frac{C_1}{9e^3}te^{tr_1}+(|P_1|+|P_2|)te^{tr_1}\\  &\leq&te^{tr_1}(\bar{K}_{[-1/4<\lambda<0]}(m^2+n^2)+\tilde{K}_{[-1/4<\lambda<0]})\end{array}\ \ \ \ \ (17)


\displaystyle \bar{K}_{[-1/4<\lambda<0]}=\frac{4C_1}{9e^3}+\frac{2C_2}{e}+e


\displaystyle \tilde{K}_{[-1/4<\lambda<0]}=\frac{4C_1}{9e^3}+3e+e^2.

Now we pass to the case {\lambda<-1/2}. In this regime, {\sqrt{m^2-4\lambda}} is not bounded, so that we can’t control {e^{r_2t}A_1(t)} by using (11). So, we follow the arguments in page 281 of Ratner’s paper: we recall that

\displaystyle A_1(t)=\int_1^t e^{(r_1-r_2)s}\left(\int_s^{\infty}e^{-r_1u} f_1(u)\,du\right)\,ds


\displaystyle I(s):=\int_s^{\infty} e^{-r_1u}f_1(u) du=2\int_s^{\infty}e^{(-r_1-2)u}\frac{y'(u)}{\sinh(2u)}du.

Define {J(s)=\int_u^{\infty}y'(v)/\sinh(2v)\, dv}. By integration by parts, {J(u)=\frac{y(u)}{\sinh(2u)}+2\int_u^{\infty}y(v)\frac{\cosh(2v)}{\sinh^2(2v)}\,dv},

\displaystyle I(s)=2\left[e^{(-r_1-2)s} J(s) + (r_1+2)\int_s^{\infty} e^{(-r_1-2)u} J(u)\, du\right]

and {A_1(t)=2(F_1(t)+F_2(t))}, where

\displaystyle F_1(t) = \int_1^t e^{(-r_2-2)s} J(s)\,ds


\displaystyle F_2(t) = (r_1+2)\int_1^t e^{(r_1-r_2)s}\left(\int_s^{\infty} e^{(-r_1-2)u}J(u)\,du\right)\,ds.

It follows that

\displaystyle \begin{array}{rcl} |J(u)|&\leq&\frac{2}{(1-e^{-4})}e^{-2u} + 2\frac{(1+e^{-4})}{(1-e^{-4})}\int_u^{\infty}\frac{dv}{\sinh(2v)}\\  &\leq& \frac{2}{(1-e^{-4})}e^{-2u} + 2\frac{(1+e^{-4})}{(1-e^{-4})^2}e^{-2u}\\  &\leq& Q_1\cdot e^{-2u}\end{array}

where {Q_1=4/(1-e^{-4})^2=4C_1^2}. That is, we can take {Q_1=4C_1^2} in the equation (2.22) of Ratner’s paper. Also, since {\textrm{Re}(r_2)=-1}, we get

\displaystyle |e^{r_2t}F_1(t)|=\left|e^{r_2t}\int_1^t e^{-(r_2+2)s}J(s) ds\right|\leq Q_1e^{-t}\int_1^t e^{-3s}ds\leq Q_2e^{-t}

with {Q_2=4C_1^2/3e^3}, that is, this constant {Q_2} works in equation (2.24) of Ratner’s paper. Finally, by integrating by parts,

\displaystyle e^{r_2t}F_2(t) = \frac{e^{r_2 t}(r_1+2)}{r_1-r_2}\left(\left[e^{(r_1-r_2)s}\int_s^{\infty} e^{(-r_1-2)u}J(u)\,du\right]_1^t + F_1(t)\right)

On the other hand, since {\lambda<-1/2}, one has {\left|\frac{r_1+2}{r_1-r_2}\right|=\left|\frac{1+\sqrt{1+4\lambda}}{2\sqrt{1+4\lambda}}\right|\leq 1}. By combining these facts, we see that

\displaystyle |e^{r_2t}F_2(t)|\leq \frac{2Q_1}{3e^3}e^{-t}+Q_2e^{-t}=Q_3e^{-t}

where {Q_3=Q_1/e^3}.

Thus, using these estimates to control {e^{r_2t}A_1(t)} and the estimate (12) above to control {e^{r_2t}A_2(t)}, we obtain

\displaystyle \begin{array}{rcl} |e^{r_2t}A(t)|&\leq& |e^{r_2t}A_1(t)|+|e^{r_2t}A_2(t)|\\  &\leq& 2|e^{r_2t}F_1(t)|+2|e^{r_2t}F_2(t)|+|e^{r_2t}A_2(t)|\\  &\leq& 2(Q_2+Q_3)e^{-t}+\frac{C_2}{e}(m^2+n^2)e^{-t}\\  &=& (\tilde{Q}+\bar{Q}(m^2+n^2))e^{-t}\end{array}

where {\tilde{Q}=2(Q_2+Q_3)=32C_1^2/3e^3} and {\bar{Q}=\bar{C}_2=C_2/e}.

The second step in the analysis of the regime {\lambda<-1/2} is the control of the quantities {|P_1/2\sqrt{1+4\lambda}|} and {|P_2-e^{2\sqrt{1+4\lambda}}P_1/2\sqrt{1+4\lambda}|}. Since {r_1=-1+i\sqrt{|1+4\lambda|}}, {r_2=-1-i\sqrt{|1+4\lambda|}} and {|y(1)|\leq 1}, {|y'(1)|\leq\sqrt{m^2-4\lambda}}, we can estimate the first quantity as follows:

\displaystyle \begin{array}{rcl} \left|\frac{P_1}{2 \sqrt{1+4\lambda}}\right| &\leq& \frac{1}{2 \sqrt{|1+4\lambda|}} \left( \left|\int_1^{\infty} e^{-r_1 s} f_1(s) ds\right| + \left|\int_1^{\infty} e^{-r_1 s} f_2(s) ds\right|\right) \\ &+& \frac{1}{2\sqrt{|1+4\lambda|}} \left(|r_2 e^{-r_1} y(1)|+|e^{-r_1} y'(1)|\right) \\ &\leq& \frac{1}{2\sqrt{|1+4\lambda}|} \left(\frac{C_1}{3e^3}\sqrt{m^2-4\lambda}+\frac{C_2}{e}(m^2+n^2)\right) \\ &+& \frac{1}{2\sqrt{|1+4\lambda}|}\left(\sqrt{1+|1+4\lambda|}\cdot e+\sqrt{m^2-4\lambda}\cdot e\right) \\ &=& \frac{1}{2\sqrt{|1+4\lambda}|} \left(\frac{C_1}{3e^3}\sqrt{m^2+|1+4\lambda|+1} +\frac{C_2}{e}(m^2+n^2)\right) \\ &+& \frac{e}{2\sqrt{|1+4\lambda}|} \left(\sqrt{1+|1+4\lambda|}+\sqrt{m^2+|1+4\lambda|+1}\right) \\ &\leq& \frac{1}{2\sqrt{|1+4\lambda}|}\left(\frac{C_1}{3e^3} + \frac{C_2}{e} +e\right)(m^2+n^2) \\ &+& \frac{\sqrt{1+|1+4\lambda|}}{2\sqrt{|1+4\lambda}|} \left(\frac{C_1}{3e^3}+2e\right) \\ &\leq& \frac{1}{2}\left(\frac{C_1}{3e^3} + \frac{C_2}{e} + e\right)(m^2+n^2)+\frac{1}{\sqrt{2}}\left(\frac{C_1}{3e^3}+2e\right) \end{array}

Here, we used that {\lambda<-1/2} (so that {|1+4\lambda|>1}) and {\sqrt{(1+x)/x}<\sqrt{2}} whenever {x>1}. Similarly, we can estimate the second quantity as follows:

\displaystyle \left|P_2-\frac{e^{2\sqrt{1+4\lambda}}}{2\sqrt{1+4\lambda}}P_1\right| \leq |P_2|+\left|\frac{P_1}{2\sqrt{1+4\lambda}}\right|\leq \bar{Q}_1(m^2+n^2)+\tilde{Q}_1,

where {\bar{Q}_1=\frac{1}{2}\left(\frac{C_1}{3e^3}+\frac{C_2}{e}+e\right)} and {\tilde{Q}_1=\frac{1}{\sqrt{2}}\left(\frac{C_1}{3e^3}+2e\right)+e}.  Inserting these estimates above into (8), we deduce that

\displaystyle \begin{array}{rcl} |y(t)| &\leq& |e^{r_2 t} A(t)| + \left|\frac{P_1}{2\sqrt{1+4\lambda}}\right|\cdot |e^{r_1 t}| + \left|P_2 - \frac{e^{2 \sqrt{1+4\lambda}}}{2\sqrt{1+4\lambda}}P_1\right|\cdot |e^{r_2 t}| \\ &\leq& \left( (\bar{Q} + 2 \bar{Q}_1)(m^2+n^2) + (\tilde{Q} + 2\tilde{Q}_1 - e) \right) e^{-t} \\ &=& \left( \bar{K}_{[\lambda<-1/2]}(m^2+n^2) + \tilde{K}_{[\lambda<-1/2]}\right) e^{-t}\end{array} \ \ (18)


\displaystyle \bar{K}_{[\lambda<-1/2]}=\frac{C_1}{3e^3}+\frac{2C_2}{e}+e


\displaystyle \tilde{K}_{[\lambda<-1/2]}=\frac{(32+\sqrt{2})C_1^2}{3e^3}+(1+2\sqrt{2})e.

Finally, we consider the case {\lambda\geq0}. We begin by estimating {|e^{r_2t}A_2(t)|} and {e^{r_2t}A_1(t)}: using (2), (4) above and {r_1=-1+\sqrt{1+4\lambda}\geq0}, {r_2=-1-\sqrt{1+4\lambda}\leq-2}, we obtain

\displaystyle \begin{array}{rcl} |e^{r_2 t} A_2(t)| &\leq& C_2 (m^2 + n^2) e^{r_2 t} \int_1^t e^{(r_1-r_2)s} \int_s^{\infty} e^{-r_1 u} e^{-2u} du ds \\ &\leq & \frac{C_2}{2} (m^2+n^2) e^{r_2 t} \int_1^t e^{(-r_2-2)s} ds \\ &\leq& \frac{C_2}{2} (m^2 + n^2) t e^{-t} \end{array}


\displaystyle \begin{array}{rcl} |e^{r_2t}A_1(t)|&\leq& \frac{C_1}{2}\sqrt{m^2-4\lambda}e^{r_2t}\int_1^t e^{-(r_2+2)s}ds\leq\frac{C_1}{2}|m| t e^{-2t}\\ &\leq&\frac{C_1}{2}(m^2+n^2)te^{-2t}. \end{array}


\displaystyle |e^{r_2t}A(t)|\leq |e^{r_2t}A_1(t)|+|e^{r_2t}A_2(t)|\leq\frac{C_1+C_2}{2}(m^2+n^2)te^{-2t},

so that we can take {\bar{C}=(C_1+C_2)/2} and {\tilde{C}=0} in the equation (2.28) of Ratner’s paper.

Next, we observe that {y(t)\rightarrow0} when {t\rightarrow\infty} and {r_1\geq 0} imply {P_1=0} and

\displaystyle y(t)=e^{r_2t}A(t)+y(1)e^{-r_2}e^{r_2t}.

Therefore, from the previous discussion and {r_2+2\leq0}, it follows that

\begin{array}{rcl} |y(t)| &\leq& \frac{C_1+C_2}{2}(m^2+n^2)te^{-2t}+e^{-r_2}e^{(r_2+2)t}e^{-2t} \\  &\leq& (\bar{K}_{[\lambda\geq0]}(m^2+n^2)+\tilde{K}_{[\lambda\geq0]})te^{-2t}.\end{array}


\displaystyle \bar{K}_{[\lambda\geq0]}=\frac{C_1+C_2}{2}


\displaystyle \tilde{K}_{[\lambda\geq0]}=e^2

At this stage, from (14), (16), (17), (18), (19) above, we see that the proof of the desired lemma is complete. \Box

In next (and final) section, we will apply Lemma 1 to derive explicit variants of Theorems 1 and 3 of Ratner’s paper. To do so, we need some notation. We denote by {r(\theta)=\left(\begin{array}{cc}\cos\theta & \sin\theta \\ -\sin\theta & \cos\theta\end{array}\right)\in SO(2,\mathbb{R})}, {\theta\in\mathbb{R}}. Given an unitary {SL(2,\mathbb{R})}-representation {T}, we denote by {K(T,3)} the set of vectors {v\in\mathcal{H}(T)} such that {\theta\mapsto T(r(\theta))v} is {C^3}. Finally, if the map {\theta\mapsto T(r(\theta))v} is {C^1}, we denote by

\displaystyle L_Wv := \lim\limits_{\theta\rightarrow 0}\frac{T(r(\theta))v-v}{\theta}

the Lie derivative of {v} along the direction of {W=\left(\begin{array}{cc}0 & 1 \\ -1 & 0\end{array}\right)} of the infinitesimal generator of the rotation group {SO(2,\mathbb{R})=\{r(\theta):\theta\in\mathbb{R}\}}.

In particular, in the case of an irreducible unitary {SL(2,\mathbb{R})}-representation {T}, since {T(r(\theta))\phi_n=e^{in\theta}\phi_n} when {\phi_n\in\mathcal{H}_n(T)}, we have that

\displaystyle L_W\phi_n = i n\phi_n

for every {n\in\mathbb{Z}}.

3. Two applications of Lemma 1

Theorem 2 Let {T} be a non-trivial irreducible unitary representation of {SL(2,\mathbb{R})} in {\mathcal{H}(T)} and let {\lambda=\lambda(T)}. Let {v,w\in K(T,3)} and {B(t)=\langle v,w\circ a(t)\rangle}. Then, for all {t\geq 1},

\displaystyle \begin{array}{rcl} |B(t)|&\leq&\sqrt{2\zeta(2)}\cdot\bar{K}_{\lambda}\cdot\|L_W^3v\|\cdot(\|w\|+\sqrt{2\zeta(6)}\|L_W^3w\|)\cdot b_{\lambda}(t) \\ &+&\sqrt{2\zeta(2)}\cdot\bar{K}_{\lambda}\cdot(\|v\|+\sqrt{2\zeta(6)}\|L_W^3v\|)\cdot\|L_W^3w\|\cdot b_{\lambda}(t)\\ &+&\tilde{K}_{\lambda}\cdot(\|v\|+\sqrt{2\zeta(6)}\|L_W^3v\|)\cdot(\|w\|+\sqrt{2\zeta(6)}\|L_W^3w\|)\cdot b_{\lambda}(t) \end{array}

where {\bar{K}_{\lambda}}, {\tilde{K}_{\lambda}} and {b_{\lambda}(t)} are as in Lemma 1.

Proof: Following the proof of Theorem 1 of Ratner’s paper (at page 283), we write

\displaystyle v=\sum\limits_{n\in\mathbb{Z}}c_n\phi_n, \quad w=\sum\limits_{n\in\mathbb{Z}}d_n\phi_n

with {c_n=\langle v,\phi_n\rangle}, {d_n=\langle w,\phi_n\rangle} (and {\phi_n\in\mathcal{H}_n(T)}, {n\in\mathbb{Z}}) as in page 276 of this paper. We have

\displaystyle B(t)=\sum\limits_{n,m\in\mathbb{Z}} c_n d_m B_{n,m}(t)

so that

\displaystyle |B(t)|\leq b_{\lambda}(t)\sum\limits_{n,m\in\mathbb{Z}}|c_n|\cdot|d_m|\cdot(\bar{K}_{\lambda}(m^2+n^2)+\tilde{K}_{\lambda})

by Lemma 1.

On the other hand, since {L_W^3\phi_n = -i n^3\phi_n} for all {n\in\mathbb{Z}}, we know that

\displaystyle \begin{array}{rcl} \sum\limits_{n\in\mathbb{Z}}|c_n|&\leq& |c_0|+\left(\sum\limits_{n\in\mathbb{Z}-\{0\}}n^6|c_n|^2\right)^{\frac{1}{2}}\left(\sum\limits_{n\in\mathbb{Z}-\{0\}}\frac{1}{n^6}\right)^{\frac{1}{2}}\\ &\leq&\|v\| + \sqrt{2\zeta(6)}\cdot\|L_W^3v\|, \end{array}

\displaystyle \begin{array}{rcl} \sum\limits_{m\in\mathbb{Z}}|d_m|&\leq& |d_0|+\left(\sum\limits_{m\in\mathbb{Z}-\{0\}}m^6|d_m|^2\right)^{\frac{1}{2}}\left(\sum\limits_{m\in\mathbb{Z}-\{0\}}\frac{1}{m^6}\right)^{\frac{1}{2}}\\ &\leq&\|w\| + \sqrt{2\zeta(6)}\cdot\|L_W^3w\|, \end{array}

\displaystyle \begin{array}{rcl} \sum\limits_{n\in\mathbb{Z}}|c_n|\cdot n^2&\leq& \left(\sum\limits_{n\in\mathbb{Z}-\{0\}}n^6|c_n|^2\right)^{\frac{1}{2}}\left(\sum\limits_{n\in\mathbb{Z}-\{0\}}\frac{1}{n^2}\right)^{\frac{1}{2}}\\ &\leq&\sqrt{2\zeta(2)}\cdot\|L_W^3v\|, \end{array}


\displaystyle \begin{array}{rcl} \sum\limits_{m\in\mathbb{Z}}|d_m|\cdot m^2&\leq& \left(\sum\limits_{m\in\mathbb{Z}-\{0\}}m^6|d_m|^2\right)^{\frac{1}{2}}\left(\sum\limits_{m\in\mathbb{Z}-\{0\}}\frac{1}{m^2}\right)^{\frac{1}{2}}\\ &\leq&\sqrt{2\zeta(2)}\cdot\|L_W^3w\|, \end{array}

The desired result follows. \Box

In the sequel, {\Omega_T} denotes the Casimir operator associated to {T} (see this post).

Theorem 3 Let {T} be an unitary representation of {SL(2,\mathbb{R})} having no non-zero invariant vectors in {\mathcal{H}(T)}. Write {\Lambda=\Lambda(\Omega_T)} the spectrum of the Casimir operator and

\displaystyle A(T)=\Lambda\cap(-1/4,0).

If {A(T)\neq\emptyset}, let {\beta(T)=\sup A(T)} and {\sigma(T)=-1+\sqrt{1+4\beta(T)}}. Assume that {\beta(T)<0} when {A(T)\neq\emptyset}. Let {B(t)=\langle v,w\circ a(t)\rangle} with {v,w\in K(T,3)}. Then, for all {t\geq 1},

\displaystyle \begin{array}{rcl} |B(t)|&\leq&\sqrt{2\zeta(2)}\cdot\bar{K}\cdot\|L_W^3v\|\cdot(\|w\|+\sqrt{2\zeta(6)}\|L_W^3w\|)\cdot b_T(t) \\ &+&\sqrt{2\zeta(2)}\cdot\bar{K}\cdot(\|v\|+\sqrt{2\zeta(6)}\|L_W^3v\|)\cdot\|L_W^3w\|\cdot b_T(t)\\ &+&\tilde{K}_T\cdot(\|v\|+\sqrt{2\zeta(6)}\|L_W^3v\|)\cdot(\|w\|+\sqrt{2\zeta(6)}\|L_W^3w\|)\cdot b_T(t) \end{array}

where {\bar{K}=\frac{4C_1}{9e^3}+\frac{2C_2}{e}+e},

\displaystyle \tilde{K}_T=\left\{\begin{array}{cl}\frac{(32+\sqrt{2})C_1^2}{3e^3}+(1+2\sqrt{2})e & \textrm{ if } A(T)=\emptyset, \\ \frac{4C_1}{9e^3}+3e+e^2 & \textrm{ if } A(T)\neq\emptyset\end{array}\right.,

\displaystyle b_T(t)=\left\{\begin{array}{cl}t\cdot e^{-t} & \textrm{ if } A(T)=\emptyset, \\ t\cdot e^{\sigma(T)t} & \textrm{ if } A(T)\neq\emptyset\end{array}\right.

Proof: This is an immediate consequence of Theorem 2 and the arguments from pages 285–286 of Ratner’s paper. \Box

Closing this post, we note that the arguments in Ratner’s paper and the previous variants of Ratner’s theorems have nice consequences to the study of rates of mixing of the geodesic flow on hyperbolic surfaces. More precisely, we consider the regular representation of {SL(2,\mathbb{R})} on {L^2(S)} where {S=SO(2,\mathbb{R})\backslash SL(2,\mathbb{R})/\Gamma=\mathbb{H}/\Gamma} is a hyperbolic surface of finite area (i.e., {\Gamma} is a lattice of {SL(2,\mathbb{R})}). Then, by noticing that the lift to the unit tangent bundle {T^1S=SL(2,\mathbb{R})/\Gamma} of {S} of a function {L^2(S)} is constant along the orbits of {SO(2,\mathbb{R})}, one has that the Lie derivative {L_W} of such lifts vanish. Therefore, by direct application of Theorem 3, one gets:

Corollary 4 Let {\Gamma} be a lattice of {SL(2,\mathbb{R})} and let {T=T_{\Gamma}} be the regular representation of {SL(2,\mathbb{R})} on {L^2(S)} where {S=SO(2,\mathbb{R})\backslash SL(2,\mathbb{R})/\Gamma=\mathbb{H}/\Gamma}. Given {v,w\in L^2(S)} with {\int_S vd\mu=\int_S wd\mu=0}, it holds

\displaystyle |\langle v,T(a(t))w\rangle|:=|\langle v,w\circ a(t)\rangle|\leq\tilde{K}_{\Gamma}\cdot\|v\|_{L^2(S)}\cdot\|w\|_{L^2(S)}\cdot b_{\Gamma}(t)


\displaystyle \tilde{K}_{\Gamma}=\left\{\begin{array}{cl}\frac{(32+\sqrt{2})C_1^2}{3e^3}+(1+2\sqrt{2})e & \textrm{ if } \lambda_1(\Delta_{\Gamma})\leq-1/4, \\ \frac{4C_1}{9e^3}+3e+e^2 & \textrm{ if } -1/4<\lambda_1(\Delta_{\Gamma})<0\end{array}\right.,

\displaystyle b_{\Gamma}(t)=\left\{\begin{array}{cl}t\cdot e^{-t} & \textrm{ if } \lambda_1(\Delta_{\Gamma})\leq-1/4, \\ t\cdot e^{\sigma(\Gamma)t} & \textrm{ if } -1/4<\lambda_1(\Delta_{\Gamma})<0\end{array}\right.,

{\Delta_{\Gamma}} is the hyperbolic Laplacian on {S=\mathbb{H}/\Gamma}, {\lambda_1(\Delta_{\Gamma})} is its first eigenvalue, {\sigma(\Gamma)=-1+\sqrt{1+4\lambda_1(\Delta_{\Gamma})}} is the size of the spectral gap (if {\lambda_1(\Delta_{\Gamma})\in(-\frac{1}{4},0)}), and the constants {C_1,C_2>0} are given by the equations (1), (3).

Remark 2 In Appendix A of this paper here, it was observed that the this corollary can be used to show that certain cyclic covers of the hyperbolic surface {\mathbb{H}/\Gamma(2)} of degree {N\geq 168} have complementary series. Here, of course, the lower bound {N\geq 168} depends on the fact that the constant {\tilde{K}_{\Gamma}} is very explicit (for instance, since C_1=(1-e^{-4})^{-1}, we have \tilde{K}_{\Gamma}\leq \frac{(32+\sqrt{2})}{3e^3(1-e^{-4})^2}+(1+2\sqrt{2})e < 10.9822).

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