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A semigroup at a pair of Weierstrass points on a cyclic 4-gonal curve and a bielliptic curve. (English) Zbl 1107.14024
Let \(X\) be a nonsigular complex projective algebraic curve of genus \(g \geq 2\) and let \(\mathbb{C}(X)\) be the field of rational functions of \(X\). The Weierstrass semigroup at \(P \in X\) is the set \(H(P)\) of nonnegative integers \(n\) such that there exists \(f \in \mathbb{C}(X)\) which has \(n P\) as its pole divisor. In the same way one may define the Weierstrass semigroup at \(P, Q \in X\) as being the set \(H(P, Q)\) of pairs of nonnegative integers \((n_1, n_2)\) such that there exists \(f \in \mathbb{C}(X)\) which has \(n_1 P_1 + n_2 P_2\) as its pole divisor.
In the present paper the authors study \(H(P, Q)\) in the case where \(X\) is a cyclic 4-gonal curve. Assume that \(P, Q \in X\) are total ramification points of the fixed cyclic 4-gonal map (then 4 is the least positive element of \(H(P)\) and of \(H(Q)\), moreover \(4 P\) and \(4 Q\) are linearly equivalent divisors), the authors determine all possible semigroups \(H(P, Q)\) and show that every possibility indeed happens. They then proceed to analyse the case of bielliptic curves (i.e. double coverings of elliptic curves); for these curves there are points \(P\) and \(Q\) such that 4 is the least positive element in both \(H(P)\) and \(H(Q)\) but \(4 P\) and \(4 Q\) are not linearly equivalent. Again, they determine all possible semigroups \(H(P, Q)\) and show that every possibility actually happens.

MSC:
14H55 Riemann surfaces; Weierstrass points; gap sequences
14H45 Special algebraic curves and curves of low genus
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