Let \[ X=\widehat{\mathbb C}\setminus\{0,1,\infty\}. \] Let \(\gamma\) be the closed hyperbolic geodesic on \(X\) that makes a figure eight around \(0\) and \(1\). Show that its length is \[ L(\gamma)=\log(17+12\sqrt2). \]
Hint: identify the corresponding element of \[ \pi_1(X)\cong\Gamma(2) \] with a matrix of trace \(6\).
Solution
The figure-eight class is represented in \(\Gamma(2)\) by a hyperbolic matrix of trace \(6\). Its translation length is therefore \[ \boxed{L(\gamma)=\log(17+12\sqrt2).} \]
1. Choose primitive parabolic elements about the punctures \(0\) and \(1\).
Under the uniformization \[ X\cong\mathbb H/\Gamma(2), \] the punctures correspond to the three cusp orbits. A primitive parabolic fixing \(0\) is \[ P_0= \begin{pmatrix} 1&0\\ 2&1 \end{pmatrix} \in\Gamma(2). \] Let \[ T= \begin{pmatrix} 1&1\\ 0&1 \end{pmatrix}. \] Conjugating by \(T\), which sends \(0\) to \(1\), gives a primitive parabolic fixing \(1\): \[ P_1=TP_0T^{-1} = \begin{pmatrix} 3&-2\\ 2&-1 \end{pmatrix} \in\Gamma(2). \] Both have trace \(2\), as expected for primitive parabolic cusp elements.
2. The figure-eight free homotopy class is represented by \(P_0P_1^{-1}\) up to conjugacy and inversion.
Choose the self-intersection point of the figure eight as basepoint. Its two lobes are simple peripheral loops, one enclosing \(0\) and one enclosing \(1\). Along a single oriented figure-eight curve, these two lobes are traversed with opposite peripheral orientations. Thus, if the positively oriented peripheral classes are represented by \(P_0\) and \(P_1\), the figure-eight word is \[ P_0P_1^{-1} \] up to changing the basepoint, reversing the whole curve, or interchanging the two lobes. Those operations only conjugate or invert the group element and hence do not change its trace in absolute value or its translation length.
3. This element has trace \(6\).
Since \[ P_1^{-1} = \begin{pmatrix} -1&2\\ -2&3 \end{pmatrix}, \] we obtain \[ P_0P_1^{-1} = \begin{pmatrix} 1&0\\ 2&1 \end{pmatrix} \begin{pmatrix} -1&2\\ -2&3 \end{pmatrix} = \begin{pmatrix} -1&2\\ -4&7 \end{pmatrix}. \] Therefore \[ \operatorname{tr}(P_0P_1^{-1})=6. \] Since \(6>2\), this element is hyperbolic.
4. A hyperbolic element of trace \(6\) has the required translation length.
If \(M\in\operatorname{PSL}_2(\mathbb R)\) is hyperbolic with translation length \(L\), then \[ |\operatorname{tr}M|=2\cosh(L/2). \] Applying this to the matrix in 3 gives \[ 2\cosh(L/2)=6, \qquad \cosh(L/2)=3. \] Let \[ u=e^{L/2}>1. \] Then \[ \frac{u+u^{-1}}2=3, \] so \[ u^2-6u+1=0. \] The root greater than \(1\) is \[ u=3+2\sqrt2. \] Hence \[ \begin{aligned} L &=2\log(3+2\sqrt2)\\ &=\log\bigl((3+2\sqrt2)^2\bigr)\\ &=\boxed{\log(17+12\sqrt2)}. \end{aligned} \] The closed geodesic in the corresponding free homotopy class realizes exactly this translation length.