It is known that there is a surjective homomorphism from the genus $g$ mapping class group $M_g$ to $Sp(2g,\mathbb{Z})$. Is there any non-trivial extension of $Sp(2g,\mathbb{Z})$ by $\mathbb{Z}_2$ , which we call $G$, such that the homomorphism $M_g\rightarrow Sp(2g,\mathbb{Z})$ factors through $G$?
1 Answer
$\begingroup$
$\endgroup$
For $g \geq 3$, the map $$H^*(Sp(2g,\mathbb{Z});\mathbb{Z}) \longrightarrow H^*(M_g;\mathbb{Z})$$ is known to be an isomorphism for $* \leq 2$. In particular, by universal coefficients the map $$H^2(Sp(2g,\mathbb{Z});\mathbb{Z}/2) \longrightarrow H^2(M_g;\mathbb{Z}/2)$$ is an isomorphism, and both sides are $\mathbb{Z}/2$. Thus there is a unique nontrivial extension of $Sp(2g,\mathbb{Z})$ by $\mathbb{Z}/2$, but because the extension class remains nonzero in the cohomology of $M_g$, the map from $M_g$ cannot factor through it.