Let $Y$ be the Riemann surface defined by the equation $y^d=h(x)$ and $\pi: Y \to \mathbb{C}_\infty$ be the projection map sending $(x,y)$ to $x$. Let $\sigma: Y \to Y$ be the automorphism defined by $(x,y) \mapsto (x,\zeta y)$, where $\zeta$ is a primitive $d^{th}$ rooth of the unity. Let $\mathcal{M}_i$ be the space of those meromorphic functions $f$ on $Y$ such that $f \circ \sigma=\zeta^i f$.

It is easy to prove that the maps $x$ and $y$ (respectively the projections onto the two coordinates) belong to $\mathcal{M}_0$ and $\mathcal{M}_1$. Furthermore, every function $f \in \mathcal{M}_0$ is of the form $r \circ \pi$, where $r$ is a meromorphic function on the Riemann sphere. In fact, we may define $r(x_0):=f(x_0,y_1(x_0))$, where $(x_0,y_1(x_0))$ is one point of the preimage of $\pi^{-1}(x_0)$. Clearly, the invariance of $f$ with respect to the composition with $\sigma$ ensures that the map $r$ is well defined.

Now, how can I prove that every function $f \in \mathcal{M}_i$ is of the form $y^i r \circ \pi$, for some $r$ meromorphic on the Riemann sphere? I tried to define $r$ as before, but the map is not well defined since a change of preimage involves a change in the function $f$.

Finally, how can I deduce that every meromorphic function $f$ on $Y$ is given in a unique way as the sum $\sum\limits_{i=0}^{d-1} f_i$, where $f_i \in \mathcal{M}_i$ for any $i=0,\dots,d-1$?


1 Answer 1


Since $y\circ \sigma = \zeta y$, if $f\in \mathcal{M}_i$ then$f/y^i$ is $\sigma$-invariant, hence the pullback of a meromorphic function on the sphere, say $r$. Now just rewrite $f/y^i = \pi^\ast r$.

For the last part, fix a meromorphic function $f$ and define $$ f_i = \frac{1}{d} \sum\limits_{j=1}^d \zeta^{-ij}(f\circ \sigma^j) \in \mathcal{M}_i$$ (the exponent in $\sigma^j$ is with respect to composition) and check that $f = \sum f_i$. This is a straightforward generalization of what Miranda does for hyperelliptic curves in page 62 of his book.

  • $\begingroup$ The first part is ok. Now, I'm checking that $f=\sum f_i$ but it doesn't work. Can you express the detailed computation when $d=3$, please? In this way, I may compare my computations with yours. $\endgroup$ Jun 23, 2018 at 8:52
  • $\begingroup$ I've done a mistake during computations. I solved it, thank you! $\endgroup$ Jun 23, 2018 at 9:52

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