(Springer, 2024-06-04) Costa González, Antonio Félix
Let S be a (compact) Riemann surface of genus greather that one. Two automorphism of S are topologically equivalent if they are conjugated by a homeomorphism. The topological classi cation of automorphisms is a classical problem and its study was initiated by J. Nielsen who in the thirties classi ed conformal ones. The case of anticonformal automorphisms is more involved and was solved by K. Yocoyama in the 80s-90s. In order to decide whether two anti-conformal automorphisms are equivalent, it is usually necessary to take into account many invariants, some of which are di cult to compute. In this work we present some situations where the topological equivalence is mainly due to the genus of some quotient surfaces and the algebraic structure
of the automorphism group.
An anticonformal square root of a conformal automorphism f is an anticonformal automorphism g such that g2 = f . Let g1 and g2
be anticonformal square roots of the same conformal automorphism of order m, where m is an even integer. If genus of S= hg1; g2i is even and genus of S= hgii is 6= 2 we prove that hg1i and hg2i are topologically equivalent. If genus of S= hg1; g2i is odd and hg1; g2i is abelian we obtain that hg1i and hg2i are topologically equivalent. We give examples to justify the condition genus of S= hgii 6= 2 and hg1; g2i abelian in each case.
