On the topological type of anticonformal square roots of automorphisms of even order of Riemann surfaces

Abstract: Let $S$ be a (compact)\ Riemann surface of genus greater than one. Two automorphism of $S$ are topologically equivalent if they are conjugated by a homeomorphism. The topological classification of automorphisms is a classical problem and its study was initiated by J. Nielsen who in the thirties classified 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 difficult 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 $g^{2}=f$. Let $g_{1}$ and $g_{2}$ be anticonformal square roots of the same conformal automorphism of order $m$, where $m$ is an even integer. If genus of $S/\left\langle g_{1},g_{2} \right\rangle $ is even and genus of $S/\left\langle g_{i}\right\rangle $ is $\neq2$ we prove that $\left\langle g_{1}\right\rangle $ and $\left\langle g_{2}\right\rangle $ are topologically equivalent. If genus of $S/\left\langle g_{1},g_{2}\right\rangle $ is odd and $\left\langle g_{1},g_{2}\right\rangle $ is abelian we obtain that $\left\langle g_{1}\right\rangle $ and $\left\langle g_{2}\right\rangle $ are topologically equivalent. We give examples to justify the condition genus of $S/\left\langle g_{i}\right\rangle $ $\neq2$ and $\left\langle g_{1},g_{2}\right\rangle $ abelian in each case.