A Rado Theorem for Complex Spaces

A Rado Theorem for Complex Spaces

A Rad´otheorem for complex spaces Viorel Vˆıjˆıitu Abstract. We generalize Rad´o’s extension theorem from the complex plane to reduced complex spaces. 2010 Mathematics Subject Classification: 32D15, 32D20, 32C15 Key words: Rad´o’s theorem, complex space, c-holomorphic function 1 Introduction A theorem due to Rad´oasserts that a continuous complex valued function on an open subset of the complex plane is holomorphic provided that it is holomorphic off its zero set. Essentially this theorem was proved in [10]. Since then many other proofs have been proposed, e.g. [2], [3], [6], and [7]. The articles [1], [11] and [14] give some generalizations. Rad´o’s statement remains true for complex manifolds (or, more generally, for normal complex spaces) as well as in the complex plane. In this short note we investigate a natural extension of Rad´o’s theorem when the ambient space has (non normal) singularities. Complex spaces, unless explicitely stated, are assumed to be reduced and countable at infinity. Let N = {1, 2,...}. arXiv:2101.03596v1 [math.CV] 10 Jan 2021 Here we state our main results. Proposition 1 There is an irreducible Stein curve X and a continuous func- tion f : X −→ C that is holomorphic off its zero set but no power f n, n ∈ N, is globally holomorphic. Theorem 1 Let X be a complex space and Ω ⊂ X be a relatively compact open set. Then, there is nΩ ∈ N such that, for every continuous function f : X −→ C that is holomorphic off its zero set, for every integer n ≥ nΩ, the power f n is holomorphic on Ω. 1 2 Vˆıjˆıitu Recall the following definition. Let X be a complex space. A function f : X −→ C is c-holomorphic if it is continuous and the restriction of f to the subset Reg(X) of manifold points of X is holomorphic [15]. 2 Proof of Proposition 1 The singularities that we implant at the points 2, 3,..., of C are generalized cusp singularities. Let p and q be integers ≥ 2 that are coprime. Consider the cusp like irreducible and locally irreducible complex curve C2 p q C2 Γ= {(z1, z2) ∈ : z1 = z2} ⊂ . Its normalization is C and π : C −→ Γ, t 7→ (tq, tp), is the normalization map. Note that π is a homeomorphism. A continuous function h : Γ −→ C that is holomorphic off its zero set, but fails to be globally holomorphic is produced as follows. Select natural numbers m and n with mq−np = 1, and define h :Γ −→ C by setting for (z1, z2) ∈ Γ, m n z1 /z2 if z2 =06 , h(z1, z2) := 0 if z2 =0. It is easily seen that h is continuous (as π is a homeomorphism, the continuity of h follows from that of h ◦ π, that equals id of C), h is holomorphic off its zero set (incidentally, here, this is the set of smooth points of Γ), and h is not holomorphic about (0, 0) (use a Taylor series expansion about (0, 0) ∈ C2 of a presumably holomorphic extension). Furthermore, hk is globally holomorphic provided that k ≥ (p − 1)(q − 1). (Because every integer ≥ (p − 1)(q − 1) can be written in the form αp + βq with α, β ∈ {0, 1, 2,...}, and since hp and hq are holomorphic being the restrictions of z2 and z1 to Γ respectively.) a b Also, z1 z2h is holomorphic on Γ provided that q⌊(m + a)/p⌋ + b ≥ n, where ⌊·⌋ is the floor function. It is perhaps interesting to note that the stalk of germs of weakly holo- morphic functions O0 at 0 is generated as an O0-module by the germs at 0 of 1,h,...,hr, where r = −1 + min{p, q}.. e Rad´otheorem for complex spaces 3 C2 k k+1 Now, for each integer k ≥ 2, let Γk := {(z1, z2) ∈ ; z1 = z2 }. As previously noted, Γk is an irreducible curve whose normalization map is k+1 k πk : C −→ Γk, t 7→ (t , t ), and the function hk : Γk −→ C defined for (z1, z2) ∈ Γk by z1/z2 if z2 =06 , h(z1, z2) := 0 if z2 =0. has the following properties: ak) The function hk is weakly holomorphic. k−1 bk) The power hk is not holomorphic. k−1 k ck) The function z1 hk is holomorphic because it is the restriction of z2 to Γk. Here, with these examples of singularities at hand, we change the standard complex structure of C at the discrete analytic set {2, 3,...} by complex surgery in order to obtain an irreducible Stein complex curve X and a discrete subset Λ = {xk : k = 2, 3,...} such that, at the level of germs (X, xk) is biholomorphic to (Γk, 0). The surgery, that we recall for the commodity of the reader because in some monographs like [8] the subsequent condition (⋆) is missing, goes as follows, Let Y and U ′ be complex spaces together with analytic subsets A and A′ of Y and U ′ respectively, such that there is an open neighborhood U of A in Y and ϕ : U \ A −→ U ′ \ A′ that is biholomophic. Then define ′ ′ X := (Y \ A) ⊔ϕ U := (Y \ A) ⊔ U /∼ by means of the equivalence relation U \ A ∋ y ∼ ϕ(y) ∈ U ′ \ A′. Then there exists exactly one complex structure on X such that U ′ and Y \ A can be viewed as open subsets of X in a canonical way provided that the following condition is satisfied: (⋆) For every y ∈ ∂U and a′ ∈ A′ there are open neighborhoods D of y in Y , D ∩ A = ∅, and B of a′ in U ′ such that ϕ(D ∩ U) ∩ B ⊆ A′. Thus X is formed from Y by ”replacing“ A with A′. In practice, the condition (⋆) is fulfilled if ϕ−1 : U ′ \ A′ −→ U \ A extends to a continuous function ψ : U ′ −→ U such that ψ(A′)= A. In this case, if D 4 Vˆıjˆıitu and V are disjoint open neighborhoods of ∂U and A in Y respectively, then B = A′ ∪ ϕ(V \ A) is open in U ′ because it equals ψ−1(V ) and (⋆) follows immediately. (This process is employed, for instance, in the construction of the blow-up of a point in a complex manifold!) Coming back to our setting, consider Y = C, A = {2, 3,...} and for each k = 2, 3,..., let ∆(k, 1/3) be the disk in C centered at k of radius 1/3 that is mapped holomorphically onto an open neighborhood Uk of (0, 0) ∈ Γk through the holomorphic map t 7→ πk(z − k). Applying surgery, we get an irreducible Stein curve X and the discrete subset Λ with the aforementioned properties. It remains to produce the function f as stated. For this we let I ⊂ OX mk−1 be the coherent ideal sheaf with support Λ and such that Ixk = xk for m k = 2, 3 ..., where xk is the maximal ideal of the analytic algebra of the stalk of OX at xk. From the exact sequence 0 −→ I −→ O −→ O/I −→ 0, where O stands for the sheaf of germse ofe weakly holomorphic functions in X, we obtain a weakly holomorphic function f on X such that, for each e k =2, 3,..., at germs level f equals fk (mod Ixk ). By properties ak), bk) and ak) from above, it follows that no n ∈ N exists n k−1 for which f becomes holomorphic on X. (For instance, if f = fk + gk , for m k−1 certain gk ∈ xk , then f is not holomorphic about xk.) 3 Proof of Theorem 1 Before starting the proof, we collect a few definitions and facts. Let (A, a) be a germ of a local analytic subset in Cn. The third Whitney n cone Ca(A) consists of all vectors v ∈ C such that there is a sequence (aj)j of points in A converging to a and a sequence (nj)j of N such that nj(aj − a) 7→ v as j 7→ ∞. If A is pure k-dimensional, then Ca(A) is also a pure k-dimensional analytic set. This cone Ca(A) lies in the tangent space Ta(A) to A at a (Ta(A) is the sixth Withney cone; see [4] for more details). Ck Cm−k Assume that the analytic germ (A, 0) ⊂ z × w , with x = (z,w), is pure k-dimensional such that the projection π(z,w)= z is proper on it, and Rad´otheorem for complex spaces 5 −1 π (0) ∩ C0(A) = {0}. Then π is a branched covering on A with covering number d := deg0A and critical set Σ. Now, whenever x 6∈ Σ, we may define, for a non-constant c-holomorphic germ f :(A, 0) −→ (C, 0), the polynomial ω(x, t)= (t − f(x′)). π(x′Y)=π(x) Since f is holomorphic on the regular part Reg(A) of A and f is continuous on A, we obtain a distinguished Weierstrass polynomial of degree d with d d−1 holomorphic coefficients, W (z, t) = t + a1(z)t + ··· + ad(z), such that W (x, f(x))= 0 for x ∈ A. Note that W does not depend on w. Besides, if W (t, z) = 0, then |t| = O(kzk1/d) as (z, t) → 0 meaning that there are positive constants M and ǫ such that, if W (z, t)=0 and max{|t|, kzk} < ǫ, then |t| ≤ Mkzk1/d. Therefore, if f : (A, 0) −→ (C, 0) is a non-constant c-holomorphic germ on a pure k-dimensional analytic germ at 0 ∈ Cm, then |f(ζ)| = O(kζk1/d) as ζ → 0, where d equals the degree of the covering of the projection π satisfying −1 π (0) ∩ C0(A)= {0}.

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