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Let $C$ be a projective irreducible algebraic curve over an algebraically closed field of characteristic $0$ (for simplicity). Assume that there is also a cover $\varphi\!: C \to E$ to an elliptic curve with the zero point $\mathcal{O} = (0 : 1 : 0)$. Given an automorphism $\alpha \in \mathrm{Aut}(C)$, introduce the notation $\varphi_\alpha := \varphi \circ \alpha$ for compactness. Let $f = \sum_{i=0}^{d} f_i \!\cdot\! x^i \in \mathbb{Z}[x]$ be any polynomial such that $f(\alpha) = 0$ in the endomorphism ring $\mathrm{End}(J)$ of the Jacobian $J$ of the curve $C$. In many examples I noted that identically $\sum_{i=0}^d f_i \!\cdot\! \varphi_{\alpha^i} = \mathcal{O}$ in the group $\mathrm{Mor}(C, E)$ of all covers from $C$ to $E$. For example, this turns out to be true if $\alpha$ corresponds to an $n$-th primitive root of unity and $f = \Phi_n$ is the $n$-th cyclotomic polynomial. Is there a proof of this unexpected observation (maybe, under some additional conditions)?

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This is true without additional assumptions if you are working with maps to the elliptic curve modulo constant maps, i.e. you implicitly consider the constant map to be trivial.

The map $\varphi \colon C \to E$ factors through the Abel-Jacobi map $u\colon C \to J$ and a map $\pi \colon J \to E$ by the universal property of the Abel-Jacobi map. So

$$\sum_{i=0}^d f_i \cdot \varphi_{\alpha^i} = \sum_{i=0}^d f_i \cdot \varphi \circ \alpha^i= \sum_{i=0}^d f_i \cdot \pi \circ u \circ \alpha^i =\pi \circ \sum_{i=0}^d f_i \cdot u \circ \alpha^i = \pi \circ \sum_{i=0}^d f_i \cdot \alpha^i \circ u $$ $$= \pi \circ 0 \circ u =0 $$ since $\pi$ is a homomorphism and $u$ is functorial (up to constants) and thus commutes with $\alpha_i$ (modulo constants).

Without working modulo constant maps, the argument still proves that the map is constant, so it suffices to check that any point of $C$ is sent to $\mathcal O$, e.g. if there is an $\alpha$-invariant point sent to $\mathcal O$, or an $\alpha$-invariant point sent to a torsion point of order dividing $\sum_{i=0}^d f_i$, or points invariant under powers $\alpha$ satisfying more complicated conditions.

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  • $\begingroup$ Thank you! I had in mind a similar reasoning. $\endgroup$ Commented Oct 30, 2023 at 12:55

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