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Let $Y$ be a smooth complex projective curve of genus two, $X$ a Galois cover of degree two of $Y$ and $K$ the canonical divisor of $X$. Let $i$ be the involution of $X$ over $Y$. Can one find two points $P$ and $P'$ on $X$ such that

$$ 2P' = 3P + 3i(P) - K \, $$

and $P$ is not conjugate to $i(P)$ (note that $X$ is hyperelliptic)?

(Thanks to abx for his answer to my previous question).

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    $\begingroup$ Why don't you accept the answer of abx to your previous question? $\endgroup$
    – Sasha
    Commented Sep 29, 2022 at 20:02
  • $\begingroup$ I do accept this answer of abx. The question I am asking now is different. $\endgroup$
    – user95246
    Commented Sep 30, 2022 at 2:46
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    $\begingroup$ No again. The divisor class $3P+3i(P)-K$ is invariant under $i$, so one has $2P'\equiv 2i(P')$. Since $i(P')\neq P'$, this implies that $P'$ is a Weierstrass point. $\endgroup$
    – abx
    Commented Sep 30, 2022 at 4:16
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    $\begingroup$ Answer to the edited question: no once more. $X$ is hyperelliptic, let $\pi: X\rightarrow \mathbb{P}^1$ be the corresponding double covering. From my previous comment $P'$ is a Weierstrass point, so $K+2P'\equiv \pi ^*D$ with $\deg D=3$. Riemann-Roch gives $h^0(K+2P')=4$, so the divisors in the linear system $\lvert K+2P' \rvert$ are of the form $\pi ^*(p+q+r)$ with $p,q,r\in\mathbb{P}^1$. The only way such a divisor is of the form $3P+3Q$ is when $P=Q$, so $P=i(P)$, which is again impossible. $\endgroup$
    – abx
    Commented Sep 30, 2022 at 4:55
  • $\begingroup$ Thank you again. $\endgroup$
    – user95246
    Commented Oct 1, 2022 at 7:28

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