The main result of this chapter is Proposition 6.1 from [G-S] (see Chapter 1).
Proposition 6.1.Let${{A}_{1}},\cdots ,{{A}_{n}}$be a sequence in SL2(R)satisfying the conditions\[\underset{1\le j\le n}{\mathop{\min }}\,\left\| {{A}_{j}} \right\|\ge \mu >n\caption {(6.2)}\]\[\underset{1\le j\le n}{\mathop{\max }}\,|\log \parallel {{A}_{j}}\parallel +\log \parallel {{A}_{j+1}}\parallel -\log \parallel {{A}_{j+1}}{{A}_{j}}\parallel |<\frac{1}{2}\log \mu \caption {(6.3)}\]
Then\[|\log \parallel {{A}_{n}}\cdots {{A}_{1}}\parallel +\sum\limits_{j=2}^{n-1}{\log \parallel {{A}_{j}}\parallel -}\sum\limits_{j=1}^{n-1}{\log \parallel {{A}_{j+1}}{{A}_{j}}\parallel |
Some notation used in the proof:
AssumeK∈SL2(R). Denote$u_{K}^{\pm }$the normalized eigenvectors of$\sqrt{{K}^{\ast}{K}}$.
Thus\[Ku_{K}^{+}=\left\| K \right\|\upsilon _{K}^{+}\quad K\upsilon _{K}^{-}={{\left\| K \right\|}^{-1}}\upsilon _{K}^{-}\]where\[\left\| \upsilon _{K}^{+} \right\|=1=\left\| \upsilon _{K}^{-} \right\|\]
GivenK,M∈SL2(R), denote further forε1= ±1,ε2= ±1\[{{b}^{{{\varepsilon }_{1}},{{\varepsilon }_{2}}}}(K,M)=\upsilon _{K}^{{{\varepsilon }_{1}}}\cdot u_{M}^{{{\varepsilon }_{2}}}\](only defined up to the sign).
Proof of Proposition 6.1. First, we observe that\[\begin{align*} \left\| MK \right\|\ge \left\| MKu_{K}^{+} \right\|&=\left\| K \right\|\left\| Mu_{K}^{+} \right\| \\ &\ge\left\| K \right\|({{b}^{+,+}}(K,M).\left\| M \right\|-{{\left\| M \right\|}^{-1}}) \\ \end{align}\]and also\[\left\| MK \right\|\le {{b}^{+,+}}(K,M)\left\| K \right\|.\left\| M \right\|+{{\left\| K \right\|}^{-1}}\left\| M \right\|+\left\| K \right\|{{\left\| M \right\|}^{-1}}\]
In particular\[\frac{\left\| {{A}_{j+1}}{{A}_{j}} \right\|}{\left\| {{A}_{j}} \right\|\left\| {{A}_{j+1}} \right\|}+\frac{1}{{{\left\| {{A}_{j+1}} \right\|}^{2}}}\ge {{b}^{+,+}}({{A}_{j}},{{A}_{j+1}})\ge \frac{\left\| {{A}_{j+1}}{{A}_{j}} \right\|}{\left\| {{A}_{j}} \right\|\left\| {{A}_{j+1}} \right\|}-\frac{1}{{{\left\| {{A}_{j}} \right\|}^{2}}}-\frac{1}{{{\left\| {{A}_{j+1}} \right\|}^{2}}}\caption {(6.5)}\]
Next, one gets for any vectoru\[{{A}_{n}}\cdots {{A}_{1}}u=\sum\limits_{{{\varepsilon }_{1}},\ldots ,{{\varepsilon }_{n}}=\pm 1}{{{\left\| {{A}_{n}} \right\|}^{{{\varepsilon }_{n}}}}\left[ \prod\limits_{j=1}^{n-1}{{{\left\| {{A}_{j}} \right\|}^{{{\varepsilon }_{j}}}}{{b}^{{{\varepsilon }_{j}},{{\varepsilon }_{j+1}}}}({{A}_{j}},{{A}_{j+1}})} \right]}\langle u_{{{A}_{1}}}^{{{\varepsilon }_{1}}},u\rangle \upsilon _{{{A}_{n}}}^{{{\varepsilon }_{n}}}\caption {(6.6)}\]
It follows from (6.5), (6.2),...