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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.

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  • $\begingroup$ Won't $\sum_\chi f_\chi$ have only one simple pole ? $\endgroup$ – reuns Dec 7 '18 at 17:53
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    $\begingroup$ 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. $\endgroup$ – abx Dec 7 '18 at 17:55
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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.

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  • $\begingroup$ 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 $. $\endgroup$ – abx Dec 8 '18 at 8:07
  • $\begingroup$ @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 $\mathbb{P}^1$. $\endgroup$ – Sasha Dec 8 '18 at 8:44
  • $\begingroup$ A non constant meromorphic function has poles and zeros. $\endgroup$ – abx Dec 8 '18 at 8:55
  • $\begingroup$ @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 $\mathbb{P}^1$. $\endgroup$ – Sasha Dec 8 '18 at 8:55

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