POLONICI MATHEMATICI LXXII.2 (1999)
Relative tangent cone of analytic curves
by Danuta Ciesielska (Krak´ow)
Abstract. The purpose of this paper is to give a characterization of the relative tangent cone of two analytic curves in C
mwith an isolated intersection.
1. Introduction. We consider analytic curves X and Y in a neighbour- hood Ω of 0 in C
m(by an analytic curve we mean an analytic set of pure dimension 1) such that X ∩ Y = {0} and study the relative tangent cone to these curves at 0 (see Section 2 of [ATW]). We restrict our attention to analytic curves with irreducible germs at 0. This involves no loss of gener- ality as our considerations are local and the relative tangent cone and the intersection multiplicity of analytic curves are additive.
The main result of this paper, that is, the equality C
0(X, Y ) + C
0(X) = C
0(X, Y ), is proved in Section 3 after preliminary results for analytic curves.
This theorem gives a strong geometric characterization of the relative tan- gent cone of analytic curves.
This research was inspired by the paper [ChKT]. In the last section we present a method which reduces the calculations of the intersection multi- plicity to the calculations of the multiplicity of a holomorphic mapping at a point. We use ideas from [ChKT] and our characterization of the relative tangent cone. The method we apply can simplify the proof of the formula for the intersection multiplicity of analytic curves presented in [ChKT].
2. Preliminary results. We start with the following lemma which will be used in the proof of the main theorem of this paper.
Lemma 2.1. Let d be a positive integer. Suppose that {t
n} is a sequence of complex numbers convergent to 0 and such that {nt
n} is convergent in b C.
Then for each c ∈ C there exists a sequence {h
n} such that :
1991 Mathematics Subject Classification: Primary 32B99; Secondary 14C17.
Key words and phrases: analytic curve, tangent cone, relative tangent cone, intersec- tion multiplicity.
[191]
(1) h
n→ 0,
(2) n
d((t
n+ h
n)
d− t
dn) → c,
(3) for any holomorphic function ϕ : Ω → C defined in an open neigh- bourhood Ω of 0 ∈ C with ord
0ϕ > d we have n
d(ϕ(t
n+ h
n) − ϕ(t
n)) → 0.
P r o o f. We have to consider three cases depending on the behaviour of the sequence {nt
n}. First, in each case we prove assertions (1) and (2).
(i) If nt
n→ 0, then by a simple calculation the sequence h
n= c
1/dn
−1satisfies the assertion of our lemma.
(ii) If nt
n→ a ∈ C\{0}, then take an α ∈ C such that (1+α)
d−1 = ca
−dand define h
n= αt
n. We have h
n→ 0 and
n
d((t
n+ h
n)
d− t
dn) = (nt
n)
d((1 + α)
d− 1) = (nt
n)
dca
−d→ c.
(iii) If nt
n→ ∞, then we take h
n= c(nd)
−1(nt
n)
1−d. We get h
n→ 0 and
n
d((t
n+ h
n)
d− t
dn) =
d
X
i=1
d i
n
dh
int
d−in=
d
X
i=1