Invariance of Hironaka’s characteristic polyhedron

Abstract

We show that given a face of Hironaka's characteristic polyhedron, it does only depend on the singularity and a flag defined by the linear form determining the face. As a consequence we get that certain numerical data obtained from the characteristic polyhedron are invariants of the singularity. In particular, they do not depend on an embedding.

Characteristic polyhedra of singularities without completion: part II

“…be the coefficients appearing in the expansion of f i of the form above, for i ∈ {1, … , m} . We say (f ) = (f 1 , … , f m ) is strongly normalized (with respect to (y)) if we have B,i ≡ 0 , for every i ∈ {2, … , m} and every 4).…”

“…We have = n (u) (f ) = 5 and D(f ) = (1, 4, 0) . If we use y 1 y 2 g 1 = y 1 y 4 2 + u a y 4 1 y 2 + u b y 1 y 2 to eliminate u e y 1 y 4 2 in f, we obtain…”

“…The reason for this is that h ∶= y 1 y 2 g 1 is not the appropriate element to eliminate u e y 1 y 4 2 in f as there appears u a y 4 1 y 2 and (4, 1, 0) ≺ (1, 4, 0) . We first have to modify h using g 2 (and g 1 ) in order to eliminate u a y 4 1 y 2 (and the bad terms possibly appearing afterwards) as in the proof of Lemma 3.6. We leave the details and the end of the normalization process as an exercise to the reader.…”

“…Since S is a G-ring, the extension ⊂ is separable, so we have 1 , … , r ∈ Ŝ ∩ = Ŝ ∩ Quot(S) = S (the last equality holds by faithful flatness). As 1 , … , r are the one given by Hironaka's procedure, the system (z) = (z 1 , … , z r ) = (y 1 + 1 , … , y r + r ) , which lies in S[y] ⟨u,y⟩ , fulfills (1)(2)(3) (4). ◻…”

“…Let us emphasize that 1 , … , r ∈ R are the elements that we obtain from Hironaka. In theory, as the directrix may be computed (see the appendix of [4]), there is a finite way to compute the elements 1 , … , r given by Hironaka's procedure that may be infinite.…”