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.  
 A: 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.
