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A real Möbius transformation represented by $A\in {\mathrm{SL}}_2(\mathbb{R})$ with $|\mathrm{tr}A|>2$. It has two fixed ideal boundary points and translates along the geodesic joining them; its hyperbolic translation length is $2\mathrm{arcosh}(|\mathrm{tr}A|/2)$. Conjugating by a real Möbius transformation puts it in the form $z\mapsto e^{\ell }z$ with $\ell >0$; its axis is then the imaginary axis.

Use the Gaussian curvature $-1$ normalizationof the Poincare half-plane model. This is a complete conformal metric, invariant under real Möbius transformations. Its quotient by the free effective principal congruence subgroup is again complete: a geodesic lifts to the complete covering plane and extends there for all time. Transporting the quotient metric through the biholomorphism supplies the required complete conformal metric.

A nonconstant closed geodesic corresponds to a hyperbolic Möbius transformation $A$ of $\Gamma (2)$, and its hyperbolic translation length is $2\mathrm{arcosh}(|\mathrm{tr}A|/2)$. Indeed its eigenvalues have magnitudes $\lambda ,{\lambda }^{-1}$ with $\lambda >1$, and its axis quotient has length $2\log \lambda$. For $A=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ in $\Gamma (2)$, $bc\equiv 0(\mathrm{mod}4)$ implies $ad\equiv 1(\mathrm{mod}4)$. The odd numbers $a,d$ therefore have the same residue modulo four, andA hyperbolic Möbius transformation must have $|\mathrm{tr}A|>2$, so its smallest possible absolute trace is six. It is attained byConsequentlyThis element is primitive: a proper power would have a shorter root represented by a hyperbolic Möbius transformation, contradicting the trace bound. No essential simple closed geodesic exists on the three-punctured sphere, since every essential simple loop is peripheral and corresponds to a parabolic Möbius transformation. The minimizing closed geodesic is therefore nonsimple. Finally, the PDF leaves the numerical Gaussian curvature unspecified: the displayed length uses Gaussian curvature $-1$; for Gaussian curvature $-{\kappa }^2$, it is divided by $\kappa$.

In curvature $-1$, its length is $2\mathrm{arcosh}3$. The level-two principal congruence subgroup has traces congruent to two modulo four, so the smallest hyperbolic absolute trace is six, achieved by $\left(\begin{array}{cc}5& 2\\ 2& 1\end{array}\right)$. The hyperbolic translation length formula gives the result; the shortest closed geodesic is nonsimple.