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Let $C$ be a (projective smooth complex) curve such that $K_C=2(D+p)$, with $D+p$ defining a $g_7^2$; $p$ is a base point and $D$ defines a 2-to-1 map $\varphi:C\rightarrow E\subset\mathbb{P}^2$ onto an elliptic curve. It's easy to see that $C$ is a genus $8$ curve that is not hyperelliptic, so the canonical linear system gives an embedding into $\mathbb{P}^8$. I want to explicitly construct and understand this situation. My approach is to consider the graded ring

$\bigoplus_{n\geq0}H^0(C,nD+np)$

I start with the basis $\{ux_0,ux_1,ux_2\}$ of $H^0(C,D+p)$. Here $u:\mathcal{O}_C\hookrightarrow\mathcal{O}_C(p)$ is the constant section. Passing to degree 2 one should get 8 generators and the second symmetric power of the basis just given has dimension 6, so we need two extra generators to complete a basis for $H^0(C,K_C)$. I have several guesses about how these should look like in terms of the $x_i$'s and with a bit more of work one sees that passing to degree $3$ would require another $2$ new generators in degree $3$. I degree 3 we can see already the equation of the elliptic curve $E$ that we may assume to be

$ac^2=b^3-\alpha ab^2+\beta a^2b$;

Where $a:=ux_0,b:=ux_1,c:=ux_2$ and some constants $\alpha,\beta$. How can I see the double cover branched at 14 points in terms of $a,b,c$ and my new generators in degrees 2 and 3? Name them $y_0,y_1\in H^0(C,K_C)$ and $z_0,z_1\in H^0(C,3D+3p)$. Any useful observation will be greatly appreciated!

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  • $\begingroup$ Since $\phi$ is a double cover, you have $K_C=\phi^*L$ for some $L\in Pic(E)$. It follows that $2P$ is (up to linear equivalence) a pull back from $E$. Since $h^0(C, 2P)=1$, the divisor $2P$ is invariant under the elliptic involution, so $P$ is a ramification point of $\phi$. Maybe this helps. $\endgroup$
    – rita
    Commented Jun 29, 2016 at 8:59

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