Action of noncentral mapping classes on curves or arcs on a surface $\DeclareMathOperator\MCG{MCG}$Let $\Sigma$ be a compact oriented surface, with empty or connected boundary. Let $\mathcal{O}$ the space of orbits of nontrivial simple closed curves on $\Sigma$ under $\MCG(\Sigma)$-action. (so, $\mathcal{O}$ has finitely many elements: the sets of nonseparating curves and the sets of separating curves of each possible genus)
If $f\in \MCG(\Sigma)$ is a noncentral element, is it true that for each $o\in \mathcal{O}$, there is a curve $c\in o$ such that the geometric intersection number $i(f(c),c)$ is non zero ?
Also, in the case where $\Sigma$ has one boundary component, I would like to ask the same question but for arcs (for the geometric intersection number, consider homotopy of arcs fixing the boundary and count only intersection points in the interior)
 A: Yes, this is true - there are many ways to prove it, but I'll hit it with a hammer. Let $C(\Sigma)$ denote the curve complex of $\Sigma$. Suppose not, then $f$ would map every curve $[c]$ in $\mathcal{O}\subset C(\Sigma)$ to $[f(c)]$ which has distance $\leq 1$ from $[c]$ since $i(c,f(c))=0$. Since the neighborhood of radius 1 of $\mathcal{O}$ is equal to $C(\Sigma)$, we see that $f$ is quasi-isometric to the identity acting on $C(\Sigma)$. Schleimer and Rafi showed that $Aut(C(\Sigma))\cong QI(C(\Sigma))$, hence $f$ acts trivially on $C(\Sigma)$. Then by Ivanov's theorem $f$ is the identity (or central if $\chi(\Sigma)\geq -2$).
A: If there are infinitely many (isotopy classes of) curves, then yes. Here is sketch of a proof. Let $S$ be the surface and let $\mathcal{C}(S)$ be the curve complex. The diameter of the curve complex is infinite. The mapping class $f$ acts elliptically or hyperbolically on the curve complex, but in either case it has arbitrarily large orbits, so the curve you want exists.
