4
$\begingroup$

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)

$\endgroup$
5
$\begingroup$

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 $d=2$).

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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 Aug 4 '14 at 14:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.