paper

Maximal systole of hyperbolic surface with largest extendable abelian symmetry

arXiv:1911.10474 · doi:10.2140/pjm.2023.325.85

Abstract

We give the formula for the maximal systole of the surface admits the largest -extendable abelian group symmetry. The result we get is . Here \begin{eqnarray*} K &=& \sqrt[3]{\frac{1}{216}L^3 +\frac{1}{8} L^2 + \frac{5}{8} L - \frac{1}{8} + \sqrt{\frac{1}{108}L(L^2+18L+27)} } & & + \sqrt[3]{\frac{1}{216}L^3 +\frac{1}{8} L^2 + \frac{5}{8} L - \frac{1}{8} - \sqrt{\frac{1}{108}L(L^2+18L+27)} } & & + \frac{L+3}{6}. \end{eqnarray*} and .

38 pages, 46 figures

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