Erratum to: Spiraling spectra of geodesic lines in negatively curved manifolds

1 Erratum to: Math. Z. (2011) 268:101–142 DOI 10.1007/s00209-010-0662-0

The correct statement of Proposition 1.4 of [2] is the following one.

Proposition 1.4

For the Golden Ratio $\phi =\frac{1+\sqrt{5}}{2}$, we have $K_\phi =3/\sqrt{5}-1\approx 0.34$, and $K_\phi$ is not isolated in $\mathrm{Sp }_{\phi }$.

The proof of Proposition 1.4 follows as in [2] from the corrected version of Proposition4.11 below.

Proposition 4.11

Let $\Gamma _0$ be the cyclic subgroup of $\Gamma =\mathrm{{ PSL}}_{2}({\mathbb Z})$ generated by $\gamma _1=\pm \begin{pmatrix}2 &{}\quad 1\\ 1 &{}\quad 1 \end{pmatrix}$, and let ${\fancyscript{D}}=({{\mathbb H}}^2_{\mathbb R},\Gamma ,\Gamma _0,C_\infty )$. Then $K_{\fancyscript{D}}= 3/\sqrt{5}-1$, and $K_{\fancyscript{D}}$ is not isolated in the approximation spectrum $\mathrm{Sp }({\fancyscript{D}})$.

Proof

The penultimate sentence of the proof of [2, Prop. 4.11] is incorrect. For every $n\in {\mathbb N}$, let $L_1$, $\gamma _n$ be as in the original version of the proof, and let $A_n$ be the geodesic line from $\infty$ to the repelling fixed point $\gamma _n^-$ of $\gamma _n$. In order to compute the (strictly increasing) limit, as $n$ tends to $+\infty$, of the approximation constant $c(\gamma _n^-)$ of $\gamma _n^-$ (using its expression given by Eq. (11) in [2]), we need not only to consider the $\Gamma$-translates of $L_1$ intersecting $A_n$ and to minimise $1-\cos \theta$ where $\theta$ is the intersection angle, but also to consider the $\Gamma$-translates of $L_1$ not intersecting $A_n$ and to minimise $\cosh \ell - 1$ where $\ell$ is the distance to $A_n$.

Consider the common perpendicular arc between the translation axis $L_n$ of $\gamma _n$ and a disjoint $\Gamma$-translate of $L_1$. By the symmetry at $i$ and the computation (done in the original version of the proof) of the translation length of $\gamma _n$, we may restrict to the case when the endpoint on $L_n$ of this common perpendicular arc lies on the subarc of $L_n$ between $i$ and $i+n$. Let $L$ be the translate by $z\mapsto z+1$ of $L_1$, whose points at infinity are $\frac{3\pm \sqrt{5}}{2}$. Clearly (see in particular the picture in the original version of the proof), the common perpendicular arc $\delta _n$ between $L_n$ and $L$ realises the minimum distance between $L_n$ and a $\Gamma$-translate of $L_1$ disjoint from $L_n$ whose closest point on $L_n$ lies between $i$ and $i+n$. As $n\rightarrow \infty$, the segments $\delta _n$ converge (with strictly increasing lengths) to the common perpendicular arc $\delta _\infty$ between the positive imaginary axis and $L$. Since $\delta _\infty$ is contained in the Euclidean unit circle (which is the angle bisector through $i$ of the equilateral geodesic triangle with vertices $i$, $1+i$, $\frac{1+i}{2}$), its hyperbolic length is ${\mathrm{argcosh }}\frac{3}{\sqrt{5}}$ by an easy computation. Since we analysed the contribution of the $\Gamma$-translates of $L_1$ that intersect $L_n$ in the original version of the proof, and since $\frac{3}{\sqrt{5}}-1<1-\frac{1}{\sqrt{5}}$, the (strictly increasing) limit of $c(\gamma _n^-)$ is $\frac{3}{\sqrt{5}}-1$.

To conclude, we also need to improve the last claim of the second paragraph of the proof of [2, Prop. 4.11]. Let $T$ be a triangle as in this second paragraph. The distance from a geodesic line $\gamma$ meeting $T$ to the geodesic line containing the side of $T$ which is not cut by $\gamma$ is maximal when $\gamma$ goes through its opposite vertex and is perpendicular to the angle bisector of $T$ at this vertex. This distance is equal to ${\mathrm{argcosh }}\frac{3}{\sqrt{5}}$ by the above computation. Since we analysed the contribution of the sides of $T$ intersecting $\gamma$ in the original version of the proof, and since $\frac{3}{\sqrt{5}} -1<1-\frac{1}{\sqrt{5}}$, we have $c(\xi )\le \frac{3}{\sqrt{5}}-1$ for every $\xi \in {\mathbb R}-{\mathbb Q}$. The result follows. $\square$

We are grateful to Yann Bugeaud for pointing out the mistake. See [1] for an arithmetic proof of the above result.