# On a map between Riemann surfaces of genus $1$

Let $$C$$ be a compact Riemann surface of genus $$1$$, and $$p\in C$$, and $$w$$ be a local holomorphic coordinate on $$C$$ near $$p$$ with $$w=0$$ at $$p$$. As usual, for a divisor $$D$$ denote by $$L(D)$$ the vector space of meromorphic functions $$f$$ (together with $$f\equiv 0$$) such that $$(f) + D$$ is an effective divisor; and $$l(D) = \dim L(D)$$.

One can easily show (by the Riemann-Roch theorem and $$\deg D < 0 \Rightarrow l(D)=0$$) that $$l(mp) = 0$$ for $$m\leq 0$$ and $$m$$ for $$m>0$$ (where $$m\in \mathbb{Z}$$). In particular, we can deduce that $$L(p) = \langle 1 \rangle$$, $$L(2p) = \langle 1,f \rangle$$ and $$L(3p) = \langle 1,f,g \rangle$$ where $$f(w) = w^{-2} + O(w^{-1})$$ and $$g(w) = w^{-3} + O(w^{-2})$$ near $$p$$. Then by considering $$L(6p)$$, which contains the $$\bf{seven}$$ maps $$g^2, f^3, fg, f^2, g, f, 1$$, we deduce that $$g^2 = f^3 + Afg + Bf^2 + Cg + Df + E$$ for unique scalars $$A,\ldots,E$$ (the coefficients at $$g^2$$ and $$f^3$$ must be equal by comparing $$w^{-6}$$ terms on both sides of the equation).

And so we remain with the following two:

a) Hence construct a non-trivial holomorphic map of Riemann surfaces $$\Phi: C \to D$$ where $$D$$ is the cubic curve $$D = \{[x,y,z] \in \mathbb{CP^{2}}: y^2z = x^3 + Axyz + Bx^2z + Cyz^2 + Dxz^2 + Ez^3\}$$

b) Now suppose $$\sigma: C \to C$$ is an isomorphism of Riemann surfaces with $$\sigma^k = id_C$$ for some $$k\geq 2$$ and $$p$$ is an isolated fixed point of $$\sigma$$, and the coordinate $$w$$ is chosen so that $$\sigma: w \to e^{2\pi i/k}w$$ near $$p$$. Then the pullback $$\sigma^{*}$$ maps $$L(mp) \to L(mp)$$ by $$\sigma^{*}: h \to h \circ \sigma$$ for each $$m\in \mathbb{Z}$$ and defines a representation of $$\mathbb{Z}_k = \langle 1, \sigma, \ldots, \sigma^{k-1} \rangle$$ on $$L(mp)$$. We choose $$f,g$$ from above to be eigenvectors of $$\sigma^{*}$$. Show that:

i) If $$k=2$$ then $$A=C=0$$

ii) If $$k=3$$ then $$A=B=D=0$$

iii) If $$k=4$$ then $$A=B=C=E=0$$

iv) If $$k=6$$ then $$A=B=C=D = 0$$

v) If $$k\neq 2,3,4,6$$ then $$A=B=C=D=E = 0$$

By considering $$g/f$$, show that case v) cannot happen.

So in a) I guess it should be something like $$\Phi(p) = [f,g,1]$$ but how to write it down properly and check it is holomorphic?

In b) - absolutely no ideas.

Any help appreciated!