# Poles of equivariant meromorphic functions on Riemann surfaces

Let $$p:\Sigma\to \mathbb{P}^1$$ be the cyclic cover of $$\mathbb{P}^1$$ with Galois group $$\Gamma$$. Let $$\Gamma\cdot p$$ be a free $$\Gamma$$-orbit on $$\Sigma$$. Given any character $$\chi$$ of $$\Gamma$$, does there exist a $$(\Gamma, \chi)$$-equivariant meromorphic function $$f$$ on $$\Sigma$$ such that $$f$$ is regular on $$\Sigma\backslash \Gamma\cdot p$$, and has simple poles at every points in $$\Gamma\cdot p$$?

If $$\chi$$ is the trivial character, then it is clearly true.

• No, this is not true, already for $\Gamma =\mathbb{Z}/2$, if $g(\Sigma )\geq 1$. Let $q$ be the image of $p$ in $\mathbb{P}^1$. The pull back map $H^0(\mathcal{O}_{\mathbb{P}^1}(q))\rightarrow H^0(\mathcal{O}_{\Sigma }(\Gamma \cdot p))$ is an isomorphism, hence all elements of the latter space are $\Gamma$-invariant.
– abx
Dec 7, 2018 at 17:55

Such function never exists. Indeed, let $$k$$ be the order of $$\Gamma$$, then the degree of the branch divisor should be $$km$$ for some $$m \ge 1$$. Then $$p_*O_\Sigma \cong O \oplus \psi \otimes O(-m) \oplus \dots \oplus \psi^{k-1} \otimes O(-(k-1)m)$$ (an isomorphism of $$\Gamma$$-equivariant sheaves on $$\mathbb{P}^1$$, where the latter is equipped with the trivial $$\Gamma$$-action), where $$\psi$$ is a primitive character of $$\Gamma$$.
Your question is equivalent to the computation of $$H^0(\Sigma, p^*(\chi \otimes O(x))^\Gamma,$$ where $$x$$ is a point on $$\mathbb{P}^1$$ away from the branch divisor. Since $$O(x) \cong O(1)$$, the projection formula gives you an isomorphism of the above space with $$H^0(\mathbb{P}^1, \chi \otimes O(1) \oplus \chi \otimes \psi \otimes O(1-m) \oplus \dots \oplus \chi \otimes \psi^{k-1} \otimes O(1-(k-1)m))^\Gamma.$$ The first summand $$\chi \otimes O(1)$$ gives $$\chi \oplus \chi$$, and this has $$\Gamma$$-invariants if and only $$\chi = 1$$. The second summand gives $$\chi \otimes \psi$$ (if $$m = 1$$), and this has $$\Gamma$$-invariants only if $$\chi = \psi^{-1}$$. But the corresponding divisor in $$\Sigma$$ is the ramification divisor of $$p$$. The other summands never contribute.
• I don't think this is correct. If $\Sigma =\mathbb{P}^1$, $\Gamma =\mathbb{Z}/2$ acting by $z\mapsto -z$ and $p=1$, the function $z\mapsto \dfrac{z}{1-z^2}$ has simple poles along $\Gamma \cdot p$ and is equivariant w.r.t. the nontrivial character of $\Gamma$.
• @abx But it also has zero at $0$. I was assuming that the poles along the free orbit are the only poles and zeros of $f$. However, if your interpretation is correct, it is easy to modilfy my argument --- you still need to assume that $\chi = \psi^{-1}$, but then you can modify the ramification divisor of $p$ by a pullbsck of a rational function on $\mathbb{P}^1$. Dec 8, 2018 at 8:44
• @abx Or, actually, you can pick up a summand in the above formula with $\psi^i = \chi^{-1}$, and modify the corresponding section (a power of the ramification divisor) by a rational function on $\mathbb{P}^1$. Dec 8, 2018 at 8:55