1 5 Se p 19 99 How to calculate A - Hilb C 3 Alastair Craw

Nakamura [N] introduced the G-Hilbert scheme for a finite subgroup G ⊂ SL(3, C), and conjectured that it is a crepant resolution of the quotient C3/G. He proved this for a diagonal Abelian group A by introducing an explicit algorithm that calculates A-HilbC3. This note calculates A-HilbC3 much more simply, in terms of fun with continued fractions plus regular tesselations by equilateral triangles. 1 Statement of the result 1.1 The junior simplex and three Newton polygons Let A ⊂ SL(3,C) be a diagonal subgroup acting on C. Write L ⊃ Z for the overlattice generated by all the elements of A written in the form 1 r (a1, a2, a3). The junior simplex ∆ (compare [IR], [R]) has 3 vertexes e1 = (1, 0, 0), e2 = (0, 1, 0) and e3 = (0, 0, 1). Write R∆ for the affine plane spanned by ∆, and Z 2 ∆ = L ∩ R 2 ∆ for the corresponding affine lattice. Taking each ei in turn as origin, construct the Newton polygons obtained as the convex hull of the lattice points in ∆ (see Figure 1.a): fi,0, fi,1, fi,2, . . . , fi,ki+1, (1.1) where fi,0 is the primitive vector along the side [ei, ei−1], and fi,ki+1 that along [ei, ei+1]. (The indices i, i ± 1 are cyclic. Also, since ei is the origin, the notation fi,j does double duty for a point of ∆ and the corresponding