Proper transform of the blowing-up

Marek Karaś

IMUJ PREPRINT 1998/03

Abstract

The main goal of this note is the following: for any arbitrary analytic set X ⊂ ℂ^{n} with 0 ∈ X, we have \overline{σ^{−1}(X\\{0\})}∩σ^{−1}(0) = ℙ(S_{0}X), where σ is the blowing-up of the ℂ^{n} with the centre at 0.

1 Introduction

Let n ≥ 2 be any fixed natural number. By (x_{1} : …: x_{n}) we denote the homogeneous coordinates in ℙ^{n−1} and for any cone V ⊂ ℂ^{n} by ℙ(V) we denote the set {(x_{1} : …: x_{n}) | (x_{1}, …, x_{n}) ∈ V } ⊂ ℙ^{n−1}. In particular, for any v = (v_{1}, …, v_{n}) ∈ ℂ^{n}\0 we have ℙ(ℂv) = (v_{1} : …: v_{n}). Consider the variety

Y := {((x_{1}, …, x_{n}), (y_{1} : …: y_{n})) ∈ ℂ^{n} × ℙ^{n−1} |
x_{i}y_{j} = x_{j}y_{i} for 1 ≤ i < j ≤ n}.

Definition 1 The natural projection from ℂ^{n} × ℙ^{n−1} into ℂ^{n} restricted to the variety Y we call the blowing-up of the ℂ^{n} with the centre at 0. We will denote it by σ : Y → ℂ^{n}.

Remark 2 (1) σ^{−1}(0) = 0 × ℙ^{n−1} ≅ ℙ^{n−1},

(2) the mapping σ|_{σ^{−1}(ℂ^{n}\0)} : σ^{−1}(ℂ^{n}\0) → ℂ^{n}\0 is an isomorphism.

Now consider an analytic subset X ⊂ ℂ^{n} with 0 ∈ X. Definition 3 The set

\overline{σ^{−1}(X\0)}

we call the proper transform of the set X under σ.

Because σ|_{σ^{−1}(ℂ^{n}\0)} : σ^{−1}(ℂ^{n}\0) → ℂ^{n}\0 is an isomorphism then to obtain the proper transform of the set X under σ it is sufficient to find the set

\overline{σ^{−1}(X\\{0\})} ∩ σ^{−1}(0).

Consider the following easy example Example 4 Let X = \{(x, y) ∈ ℂ^{2} | xy = 0\}. Now \overline{σ^{−1}(X\\{0\})} ∩ σ^{−1}(0) = {(1 : 0), (0 : 1)}. Definition 5 Let X be any analytic subset of ℂ^{n}, a ∈ X. The set

S_{a}X := {v ∈ ℂ^{n} | ∃a^{k} → a, λ_{k} ∈ ℂ : λ_{k}(a^{k} − a) → v}

we call the tangent cone of the set X at point a.

In the Example 4 we have \overline{σ^{−1}(X\0)} ∩ σ^{−1}(0) = ℙ(X) = ℙ(S_{0}X).

2 The main result

Now consider the mapping

η : ℂ^{n}\0 ∋ (a_{1}, …, a_{n}) ↦ (a_{1} : …: a_{n}) ∈ ℙ^{n−1}

We have the following Lemma 6 (1) The mapping η is continuous, (2) for any λ ∈ ℂ\0 and a = (a_{1}, …, a_{n}) ∈ ℂ^{n}\0 η(λa) = η(a),

(3) if {a^{k}} ⊂ ℂ^{n} is a sequence such that η(a^{k}) → ℙ(ℂv) then there exists a sequence {λ_{k}} ⊂ ℂ such that λ_{k}a^{k} → v.

Proof: (1) and (2) are obvious. To prove (3) we can assume (without loss of generality) that v = (1, 0, …, 0). Because η(a^{k}) → (1 : 0 : …: 0) then there exists k_{0} ∈ ℕ such that for k ≥ k_{0} η(a^{k}) ∈ U_{0} = {(x_{1} : …: x_{n}) ∈ ℙ^{n−1} | x_{1} ≠ 0\}. Convergence η(a^{k}) → (1 : 0 : …: 0) in the map U_{0} means that \frac{a_{i}^{k}}{a_{1}^{k}} → 0 for i = 2, …, n. Now if we take λ_{k} = \frac{1}{a_{1}^{k}} for k ≥ k_{0} then

λ_{k}a^{k} → (1, 0, …, 0)

in ℂ^{n}. □

Now we are in a position to prove the main result.

Theorem 7 Let X ⊂ ℂ^{n} be any analytic set with 0 ∈ X. We have

\overline{σ^{−1}(X\0)} ∩ σ^{−1}(0) = ℙ(S_{0}X)

Proof: Let v ∈ S_{0}X. By Definition 5 there exists a sequence {a^{k}} of points of X\0 and λ_{k} ∈ ℂ such that a^{k} → 0 and λ_{k}a^{k} → v. Now we will show that σ^{−1}(a^{k}) → (0, ℙ(ℂv)). Indeed, we have σ^{−1}(a^{k}) = (a^{k}, η(a^{k})) and a^{k} → 0, so it remains to show that η(a^{k}) → ℙ(ℂv). By (1) and (2) from Lemma 6 we have η(a^{k}) = η(λ_{k}a^{k}) → η(v) = ℙ(ℂv). Thus ℙ(S_{0}X) ⊂ \overline{σ^{−1}(X\0)} ∩ σ^{−1}(0).

To prove the opposite inclusion let {a^{k}} ⊂ σ^{−1}(X\0) be any arbitrary sequence with a^{k} → (0, ℙ(ℂv)) ∈ σ^{−1}(0). Now σ(a^{k}) → 0 in ℂ^{n} and η(σ(a^{k})) → ℙ(ℂv). By lemma 6 (3) there exists a sequence {λ_{k}} ⊂ ℂ such that λ_{k}σ(a^{k}) → v. Thus v ∈ S_{0}X. □

Example 8 Let X ⊂ ℂ^{n} be any analytic cone. By the above theorem we have \overline{σ^{−1}(X\0)} ∩ σ^{−1}(0) = ℙ(S_{0}X) = ℙ(X).

References

[1] I. R. Shavarevich, Osnowy algebraicieskoj geometrii, Nauka, Moskwa 1972.

[2] R. Hartshorne, Algebraic geometry, Springer-Verlag, New York 1983.

Marek Karaś Instytut Matematyki Uniwersytetu Jagiellońskiego ul. Reymonta 4/526 Kraków Poland e-mail: karas@im.uj.edu.pl