On −k for Normal Surface Singularities

In this paper we show the lower bound of the set of non-zero−K for normal surface singularities establishing that this set has no accumulation points from above. We also prove that every accumulation point from below is a rational number and every positive integer is an accumulation point. Every rational number can be an accumulation point modulo Z. We determine all accumulation points in [0, 1]. If we fix the value −K, then the values of pg, pa, mult, embdim and the numerical indices are bounded, while the numbers of the exceptional curves are not bounded. 0. Introduction For a normal surface singularity (X, x) over C, we have two kinds of plurigenera γm(X, x) and δm(X, x) which are defined by Knöller [7] and Watanabe [13] respectively. Both plurigenera grow in order at most 2 and the coefficients of the term of degree 2 are rational numbers. The leading coefficient of γm(X, x) is −K /2, where K is the numerical canonical divisor on the minimal resolution (cf. [9]), and that of δm(X, x) is −P /2 (cf. [12]). It is well known that −K = 0 if and only if the singularity is a rational double point. In this paper, we study the set of the values of −K. The set {−P } is studied by Ganter in [5]. Her results are: if one fixes the numerical index m of singularities, then non-zero −P 2 has the lower bound 1/42m and the set of −P 2 has no accumulation points from above, which is equivalent to the descending chain condition (D.C.C. for short) because of the lower bound. Here one should note that there are accumulation points of −P 2 from above, if one does not fix the numerical index (see 3.10). Our results on −K are simpler. The discussions go well without fixing the numerical index. We prove that non-zero −K has the lower bound 1/3 and the set of −K has D.C.C.. Then we show that all accumulation points from below are rational numbers and every positive integer is an accumulation point. There are many accumulation points so that every rational number can be an accumulation point modulo Z. The accumulation points in [0, 1] turn out to be {1, m/(m + 1)|m ∈ N}. 1