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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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##### References:
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