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I was reading a paper the other day that said that all automorphisms of a hyperelliptic curve are liftings of automorphisms of $\mathbb{P}^1$ operating on the set of branch points.

Can someone point me to a good reference for hyperelliptic curves that would explain that (and possibly other important things about them too)?

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According to Weil (projecteuclid.org/…) Ch IV, §9 of Chevalley's book "Introduction to the theory of algebraic functions of one variable", is a "tour de force". –  Felipe Voloch Aug 13 '13 at 22:31
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@FelipeVoloch "a tour de force" is usually not an indication of "a good reference". –  Igor Rivin Aug 14 '13 at 2:07
    
Mumford's Tata lectures on theta has a chapter on hyperelliptic curves. (I remember someone called Chevalley's book "inhuman". I tend to agree with this:-) –  Alexandre Eremenko Aug 14 '13 at 2:24
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@AlexandreEremenko Yes it is Weil who calls it "severely dehumanized", in the above-linked review. –  Francois Ziegler Aug 14 '13 at 2:48

3 Answers 3

Joking aside... Let $x: X \to \mathbb{P}^1$ be of degree $2$ and $\sigma: X \to X$ be an automorphism. Consider $f: X \to \mathbb{P}^1\times\mathbb{P}^1, f=(x,x\circ\sigma)$. If $f$ is injective, then $X$ embeds as a curve of bidegree $(2,2)$ in $\mathbb{P}^1\times\mathbb{P}^1$ and therefore has genus at most $1$. So, if $X$ has genus $>1$ then $f$ is $2-1$ and $f(X)$ has bidegree $(1,1)$ in $\mathbb{P}^1\times\mathbb{P}^1$, so $f(X)$ is the graph of an automorphism $\tau$ of $\mathbb{P}^1$ and $x\circ\sigma = \tau \circ x$. Finally $\tau$ has to send branch points to branch points because $\sigma$ sends ramification points to ramification points, as these points are Weierstrass points, and automorphisms preserve Weierstrass points.

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The previous two answers are really good; another way to see it is the following: Using Riemann-Roch one can see that the $2:1$ morphism $f:X\to\mathbb{P}^1$ is unique modulo automorphisms of $\mathbb{P}^1$. If $\sigma:X\to X$ is an automorphism of $X$ then $f\circ\sigma$ is a $2:1$ morphism from $X$ to $\mathbb{P}^1$ and so there must be an automorphism $\eta:\mathbb{P}^1\to\mathbb{P}^1$ such that $f\circ\sigma=\eta\circ f$. This automorphism clearly sends branch points to branch points.

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More generally: A curve $C$ of genus $g>1$ has a canonical map $\kappa: C \rightarrow {\bf P}^{g-1}$. Since it's canonical, for any automorphism $\alpha: C \rightarrow C$ the composite map $\kappa \alpha: C \rightarrow {\bf P}^{g-1}$ must be the same as $\kappa$ up to some linear automorphism $\lambda$ of ${\bf P}^{g-1}$, so $\kappa\alpha = \lambda\kappa$. For example, the automorphisms of a plane quartic are exactly the linear automorphisms of the plane that preserve the quartic. In our hyperelliptic setting$\ldots$ [cont'd because of 600-character limit] –  Noam D. Elkies Jan 10 at 21:45
    
$\ldots$, $\kappa$ is a $2:1$ map to a rational normal curve, and if $\lambda$ preserves that curve $\kappa(C)$ then it comes from a fractional linear transformation of ${\bf P}^1$, etc. –  Noam D. Elkies Jan 10 at 21:45
    
Dear Noam: yes, this is a great way to see what is happening in general. –  rfauffar Jan 12 at 15:30
    
Thanks. I must have learned this from Joe Harris. –  Noam D. Elkies Jan 12 at 23:56

Felipe's answer is much better, but what about this, just for fun: an automorphism f of X induces an automorphism of effective divisors of degree 2, i.e. of the symmetric square X^(2). This object contains exactly one copy of P^1, corresponding to the divisors of the g(1,2) and thus the automorphism f^(2) of X^(2) induces an automorphism of this P^1, which takes divisors of form 2.P to divisors of form 2.f(P).

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