4
$\begingroup$

On page 204 of Rick Miranda's Algebraic Curves and Riemann Surfaces, he talks about how the canonical map of a hyperelliptic curve is the double cover map composed with the Veronese map $\phi(x) = [1:x:\cdots:x^{g-1}]$. Then, he says:

the double covering map for a hyperelliptic curve of genus $g\ge 2$ is unique since is it the canonical map after all.

I don't understand what is meant by unique and how this follows from the discussion.

$\endgroup$

1 Answer 1

1
$\begingroup$

If $C$ is a hyperelliptic curve (with $g\ne 2$) then there is a map $\phi:C\to{\bf P}^1$ of degree $2$. (It has $2g$ ramification points.) This is the double cover.

How unique is it? If $\psi:C\to{\bf P}^1$ is another such map, then we can have $\psi\ne\phi$, but what we do have is $\psi=\theta\circ\phi$ where $\theta:{\bf P}^1\to{\bf P}^1$ is an automorphism (a fractional linear transformation). So the double cover is unique "up to automorphisms of ${\bf P}^1$",

$\endgroup$
2
  • 3
    $\begingroup$ Why is it unique up to automorphism of $P^1$? $\endgroup$
    – Chris Z
    Commented Aug 18, 2018 at 5:03
  • $\begingroup$ Try to show the following: Suppose $f, g : C \rightarrow \mathbb{P}^n_k$ are two morphisms given by the same invertible sheaf $L$, and global sections $s_0, ... s_n$ and $t_0, ... t_n$. Suppose furthermore that $s_0, ... s_n$ and $t_0, ... t_n$ form a basis of the global sections of $L$. Then, $f$ and $g$ differ by an automorphism of $\mathbb{P}^n_k$. Hint : suppose $g_i= \sum_{j=0}^{n} c_{ij} f_j$. Then, $f^*(\sum_{j=0}^n c_{ij} x_j) = g_i$. Collect the $c$'s into a matrix which is the automorphism of $\mathbb{P}^n_k$ required. $\endgroup$
    – David Lui
    Commented Jul 11, 2021 at 5:23

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .