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Let $F$ be a hyperbolic once-punctured torus, and $G=\pi_1(F)$. Fix a discrete, faithful representation $\rho\colon G\to\mathbb{P}SL(2,\mathbb{R})$ and an element $g\in G$ corresponding to a non-peripheral, simple, closed geodesic $\gamma$ in $F$. I'd like to understand why the following relation holds: $$\frac{1}{1+e^{l(\gamma)}}=\frac{1}{2}\left(1-\sqrt{1-\frac{4}{(\textrm{tr}\rho(g))^2}}\right)$$ where $l(\gamma)$ is the length of the geodesic and tr$\rho(g)$ is assumed to be positive (in particular $>2$ since $\rho(\gamma)$ must be a hyperbolic element). I am trying to write the right-hand side in terms of $l(\gamma)$ by using the relation $$\textrm{tr}\rho(\gamma)=2\cosh(l(\gamma)/2)$$ but I think that at some point I should use other trigonometric relations. I took a look at a list of those but whichever I use I always end up with horrible equations. Could you help me with that? Thank you.

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1 Answer 1

Use $\cosh^2x-1=\sinh^2x$, $\sinh x=\frac{e^x-e^{-x}}{2}$, $\cosh x = \frac{e^x+e^{-x}}{2}$, and some algebra:

\begin{align*} \frac{1}{2}\left( 1 - \sqrt{1 -\frac{4}{4\cosh^2(l/2)}}\right) &= \frac{1}{2}\left( 1 - \sqrt{\frac{4\cosh^2(l/2)-4}{4\cosh^2(l/2)}}\right) \\ &= \frac{1}{2}\left( 1 - \sqrt{\frac{\sinh^2(l/2)}{\cosh^2(l/2)}}\right) \\ &= \frac{1}{2}\left( 1 - \frac{\sinh(l/2)}{\cosh(l/2)}\right) \\ &= \frac{1}{2}\left( 1 - \frac{e^{l/2}-e^{-l/2}}{e^{l/2}+e^{-l/2}}\right) \\ &= \frac{1}{2}\left( \frac{2e^{-l/2}}{e^{l/2}+e^{-l/2}}\right) \\ &= \frac{e^{-l/2}}{e^{l/2}+e^{-l/2}} \\ &= \frac{1}{e^l+1} \end{align*}

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