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Let $X$ be a closed Riemann surfaces of genus $g$ and let $p_1:Y_1 \rightarrow X$ and $p_2:Y_2 \rightarrow X$ be two $K$-sheeted, connected, unramified coverings of the Riemann surface. By the theorem of Riemann-Hurwitz the genus of $Y_1$ and $Y_2$ is $K(g-1)+1$. So there is an orientation preserving diffeomorphism between $Y_1$ and $Y_2$.

My question is if there is always a mapping class which maps $Y_1$ to $Y_2$.

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Perhaps this is not what you meant to ask? A "mapping class" means a map from a manifold to itself. $Y_1$, $Y_2$ are different surfaces, ergo no mapping class. –  Lee Mosher May 5 '12 at 21:14
    
... or, to be more precise, an isotopy class of a homeomorphism from a manifold to itself. –  Lee Mosher May 5 '12 at 21:18
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If what you meant to ask is whether there exists a homeomorphism $f : X \to X$ that lifts to a homeomorphism $F : Y_1 \to Y_2$, by the lifting lemma that happens if and only if the subgroups $p_{1*}(\pi_1 Y_1)$ and $p_{2 *}(\pi_1 Y_2)$ are conjugate in $\pi_1 X$. –  Lee Mosher May 5 '12 at 21:27
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Perhaps. Then the answer is clearly negative if images of the homomorphisms are not the same (two finite groups of the same order need not be isomorphic, let alone, equal as subgroups of the symmetric group). However, if both are epimorphisms, the answer could be positive: Greg Kuperberg mentioned such result to me few years ago, but I forgot the exact statement. It is related to problems about actions of mapping class groups on character varieties (of surface group representations to compact groups). In this case, the compact group is the symmetric group. –  Misha May 5 '12 at 21:54
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About 10 years ago I checked that two homomorphisms from a closed oriented surface group to SU(2) which have finite image, are equivalent under the action of the automorphism group of the surface if and only if the images are the same. This was in the context of understanding orbit closures of the action of mapping class group on SU(2) character variety. However, from conversation with Greg few years ago it became clear that the result should be much more general. Maybe Greg remembers the actual result. –  Misha May 5 '12 at 22:03

2 Answers 2

This is a simple exercise and the answer is no (cf. comments of Lee Mosher and Misha). Consider the two degree-4 coverings of the torus to itself with monodromies $\omega_1, \omega_2 : \pi_1(T^2) \to \Sigma_4$ given by $\omega_1(e_1) = (1\ 2\ 3\ 4)$, $\omega_1(e_2) = 1$ and $\omega_2(e_1) = (1\ 2) (3\ 4)$, $\omega_2(e_2) = (2\ 3) (1\ 4)$ (with respect to a basis $\{e_1, e_2\}$ for $\pi_1(T^2)$). These are in different orbits with respect to the action of $\text{Map}(T^2) \cong SL(2, \Bbb Z)$. As a further exercise prove that for coverings of the torus of degree $\leq 3$ the action is transitive.

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Let $F$ be a finite group, let $\pi_g$ be the fundamental group of the closed oriented genus $g$ surface $S_g$ and let $Mod_g$ denote the mapping class group of $S_g$. Consider the action of $Mod_g$ on the "character variety'' $R(\pi_g,F)=Hom(\pi_g, F)/Aut(F)$ by precompositions.

The question is "how transitive is the action of $Mod_g$ on $R(\pi_g,F)$?" It is clear (see e.g. Daniele's answer) that $Mod_g$ cannot send an epimorphism to a non-epimorphism. Thus, consider the subset $E(\pi_g,F)\subset R(\pi_g, F)$ consisting of equivalence classes of epimorphisms. There is one more invariant that $Mod_g$ has to preserve, namely, every homomorphism $f: \pi_g\to F$ represents an element $c_f$ of $H_2(F, {\mathbb Z})$, which is the image of the fundamental class of $S_g$ under the induced map $$H_2(f): H_2(S_g)\to H_2(K(F,1)).$$ Thus, we get a map $c: R(\pi_g,F)\to Q_F:=H_2(F, {\mathbb Z})/Aut(F), f\mapsto c_f$. The classes $c_f$ (modulo $Aut(F)$) are, clearly, preserved by the action of $Mod_g$, so $Mod_g$ preserves each fiber of the map $c$.

Amazingly, it turns out that for every simple nonabelian group $Q$, "stably" (i.e., for fixed $F$ and all sufficiently large genera $g$), the map $c$ completely classifies the $Mod_g$-orbits on $E(\pi_g,F)$:

For every simple nonabelian $Q$, if $g$ is sufficiently large, two epimorphisms $f_1, f_2$ belong to the same orbit $\iff$ $c_{f_1}=c_{f_2}$. This is proven in Theorem 1.3 of the paper by N.Dunfield and W.Thurston "Finite covers of random 3-manifolds". Furthermore, for all large $g$, the action of $Mod_g$ on every orbit $O$ is via the full alternating group of $O$. What happens for small $g$'s is very interesting but unclear.

A similar result, based on the previous work of Convey and Parker, in the context of the action of the braid group, was proven in: M. D. Fried and H. Volklein, "The inverse Galois problem and rational points on moduli spaces," Math. Ann. 290 (1991), 771–800, see footnote on page 5 of the paper by Dunfield and Thurston.

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