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Does there exist a hyperelliptic curve $X$ of genus $g\geq 2$ over the complex numbers such that $X$ has a hyperelliptic quotient $X\to Y$ (in the sense that $Y$ is hyperelliptic and the morphism $X\to Y$ is finite (not necessarily etale of degree two)) with the property that $\# \mathrm{Aut}(Y) > \# \mathrm{Aut}(X)$.

The answer is yes when asked like this; take $Y$ to be $\mathbb P^1$ and $X\to Y$ the hyperelliptic map. (This is under the pretense that $\mathbb P^1$ is also a hyperelliptic curve.)

But what if we also ask $Y$ to be of genus at least two? That is:

Does there exist a hyperelliptic curve $X$ of genus $g\geq 2$ over the complex numbers such that $X$ has a hyperelliptic quotient $X\to Y$ with the properties that $$g(Y)\geq 2, \ \textrm{and} \quad \# \mathrm{Aut}(Y) > \# \mathrm{Aut}(X).$$

I emphasize that the map $X\to Y$ is only assumed to be finite in this question (it is not necessarily of degree two, or etale)

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Sure: let $Y$ be your favorite hyperelliptic curve $u^2=f(t)$ with many automorphisms, and let $X$ be the curve $u^2=f(t(s))$ for some "random" rational function $f$ of degree at least $2$. For example, let $Y$ be the genus-2 curve $u^2 = t^5-t$ (so $\#({\rm Aut}(Y)) = 2\#(S_4) = 48$); and let $X$ be $u^2 = P(s)^5 Q(s) - P(s) Q(s)^5$ for some random polynomials $P,Q$ of the same degree $d>1$ (corresponding to $t(s) = P(s)/Q(s))$. Then $\#{\rm Aut}(X)$ is almost always $2$ (or $4$ if $d=2$).

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  • $\begingroup$ Even an unramified double cover can work, breaking most of the symmetry of $Y$ without introducing enough new symmetry in $X$. E.g. with $f(t)=t^5-t$ and $t(s)=s^2$ we have $Y$ as before and $X$ isomorphic with $u^2=s^8-s$ which has only $32$ automorphisms. $\endgroup$
    – Noam D. Elkies
    Commented Aug 4, 2014 at 14:27

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