A geometric proof of the Lelong-Poincaré formula

We propose a geometric proof of the fundamental Lelong-Poincaré formula : dd log |f | = [f = 0] where f is any nonzero holomorphic function defined on a complex analytic manifold V and [f = 0] is the integration current on the divisor of the zeroes of f . Our approach is based, via the local parametrization theorem, on a precise study of the local geometry of the hypersurface given by f . Our proof extends naturally to the meromorphic case.

Since the Lelong-Poincaré formula plays a crucial role in complex analytic geometry, notably in intersection theory (see [4]), it is a natural aim to look for a geometric proof of this fundamental formula.More precisely, we offer a geometric proof of the following Theorem Let V be a connected complex analytic manifold of dimension n and let f : V → C be a holomorphic nonzero function.Then, the meromorphic differential form d 0 f/f defines a current of type (1,0) on V , and furthermore, we have where [f = 0] is the integration current on the divisor of the zeroes of f .
We denote by D(V ) the set of compactly supported differential forms of class C ∞ in V .
Recall that if T is a current of degree s on V , then dT is the current of degree s + 1 acting by the rule : with d = d 0 + d 00 where d 0 and d 00 are holomorphic and antiholomorphic differentiation operator, respectively.So, for any (n − 1, n − 1) -form ϕ in D(V ), we have Recall also that the integration current exists on any analytic set (see [6]), and if ω is a locally integrable (p, q) -form on V , it defines a current [ω] of type (p, q) on V by the formula : where ϕ is any (n − p, n − q) -form in D(V ).Now, here is the outline of our proof.
By an argument of partition of unity, we easily see that our problem is local on V , and then we may assume that V is a domain (open and connected set) of C n containing 0, such that f (0) = 0 and f is nonzero on V .
The proof of the theorem can be divided into five steps : -Existence of the current [d 0 f/f] in the general case.
-Proof of (LP ) for a special class of test forms.
-Proof of (LP ) in the general case.
The two firs steps are quite elementary applications from analysis in one complex variable.We focus now on the three last steps.§1.Existence of the current [d 0 f/f] in the general case.
Then we have and we are going to see that the coefficient of dz n is locally integrable on V .The proof is obviously analogous for the coefficients of dz j when 1 ≤ j ≤ n − 1.
By the Weierstrass preparation theorem, and up to a restriction of the open set V , we may assume that V = Ω × D(0, ε), ε > 0, where Ω is a domain in C n−1 containing 0, and where f can be written, by setting t = (z 1 , . . ., z n−1 ) and z = z n : (1) where I is an analytic function in V with values in C * , and P t is a monic polynomial of degree k (k being the multiplicity of the function ξ 7 → f (0, . . ., 0, ξ) at 0) which depends analytically on t and whose roots z j (t) are in D(0, ε) for any t ∈ Ω.Now consider a compact set K = K 1 × K 2 with K 1 and K 2 are compact sets in Ω and D(0, ε) respectively.Since and as the meromorphic form d 0 I/I has no singularity in the open set Ω × D(0, ε), it is enough to prove that, for each j ∈ {1, . . ., p}, the integral .
For this purpose, take a number r ∈ ]ε/3, 2ε/3[, a point a ∈ D(0, r), and let us prove that the integral J(a) := So, by restricting the compact set K if necessary, we can assume that for any t ∈ K 1 , the roots z j (t) (1 ≤ j ≤ k) belong to D(0, r).By Fubini's theorem, we have This completes the proof of the existence of the current [d 0 f/f] when f is any holomorphic nonzero function on V .§4. Proof of (LP ) for "convenient "differential forms.
In this section, we prove the formula (LP ) for differential forms which are locally given by ϕ = ρ(t, z) dt ∧ dt where ρ is a smooth function with compact support in V, and dt We will say that a such ϕ is "convenient "with respect to the projection First, we consider the case where the function f is such that, for j 6 = j 0 , the roots z j and z j 0 are different at the generic point t of Ω.
By (1), and since the differential form d 0 I/I is holomorphic on the open set Ω × D(0, ε), we have ∂z dz.By the case n = 1 and Fubini's theorem, we get and hence

Now consider
By the hypothesis on the roots z j , we know (see [5]) that R is a closed analytic set with empty interior, hence of Lebesgue measure equal to zero in Ω.
Put {f = 0} = {(t, z) ∈ Ω × D(0, ε) , P t (z) = 0}.The local parametrization theorem for analytic sets exhibits the hypersurface {f = 0} as a branched covering of degree k of Ω via the natural projection and the branching locus is R. Then we have Let S be the singular locus of {f = 0} and put The Lelong theorem (see [6]) gives Let χ ε be the characteristic function of {f = 0} \ S ε in the complex analytic manifold {f = 0} \ π −1 0 (R).We have where the term on the right is independant on ε and integrable since and since the differential form ϕ has continuous coefficients on V (see [4]) and has a compact support.
Then we have which establishes the formula (LP ) when the branching locus does not coincide with Ω.
Now we are ready to prove (LP ) for any holomorphic function on V .
We keep the notations of §1.We know that the vanishing locus of f in Ω × D(0, ε) consists in that of the function (t, z) 7 → P (t, z) which is the branching covering of degree k over the open set Ω.This covering can be seen as an analytic application P : Ω → C k where C k is identified here to the set of monic polynomials of degree k with complex coefficients.
The branched covering P is said to be reduced if the ring O(P ) is reduced, that is, every nilpotent element in O(P ) is zero.P is said to be irreducible if O(P ) is integral.
Since P is reduced if and only if its branching locus R does not coincide with Ω (see [3]), we deduce that the formula (LP ) has been proved when P is reduced.To get this result with any f , it is enough now to use the decomposition theorem (see [3]) which asserts that the branching covering P induced by f can be decomposed in a unique way in the form P = Q j P n j j where P j are reduced (and irreducible) branching coverings, and n j are positive integers.Indeed, if f is any holomorphic function on the open set V = Ω×D(0, ε), then the decomposition theorem allows to write f = where the f j are reduced and irreducible.From the result proved in §4 we deduce The proof of (LP ) is then complete for any (n − 1, n − 1)-form in D(V ) locally given by ϕ(t, z)) = ρ(t, z) dt ∧ dt.
It remains to show that the result above is still true for any compactly supported (n − 1, n − 1) -form of class C ∞ on V .It is the aim of the following section.§3.Proof of (LP ) in the general case.
We want to prove the equality By the local parametrization theorem, we may assume that • {f = 0} is a branching covering of degree k over Ω via the projection π 0 : V → Ω.
• there exists a compact set K in Ω such that the support of ϕ is contained in K × C.
Then, there exists a neighborhood U of 0 in L ³ C n , C n−1 ´such that for every u ∈ U , the projection π u := π 0 + u exhibits Ω as a branching covering of a same neighborhood Ω 0 of K in Ω (more precisely, such that The following lemma shows that it is sufficient to consider "sympathic " forms with respect to the given projection.Proof Following [2], we establish this result by induction on p.

Lemma
For p = 0, the result is obvious.Then we may assume p ≥ 1.
-If v 1,1 6 = 0, then, by induction hypothesis (because a − -Assume now that v 0,0 = v 1,1 = 0 and v 1, 0 6 = 0 (for instance).Let w be a totally decomposed vector in Λ a−1,b (H) * such that hv 1,0 , wi = 1, and put where the w i and the t j are in H * .Since a − 1 ≤ n − 1 < n + p − 1 = dim C H, there exists a nonzero element in Let h * be an element of H * such that hh * , hi = 1.Then We deduce that v 1,0 ∧ h is nonzero in Λ a,b (H).By the induction hypothesis there exists f ∈ L(H, C n ) such that f * (v 1,0 ∧ h) 6 = 0.
Consider now the linear application g : C n+p → H defined by g |H = Id H and g(e) = h.Then, for u = f • g we get u * (v 1,0 ∧ e) 6 = 0.Moreover, if h 0 is close to h in H, and if h 0 is close to f in L(H, C n ), this property will remain true.We deduce f 0 (v 1,0 ) ∧ f 0 (h 0 ) + f 0 (v 0,1 ) ∧ f 0 (h 0 ) = 0 for at least one f 0 which can be assumed to be of rank n and for any h 0 close to h.Since f 0 is of maximum rank, f 0 (h 0 ) will describe a neighborhood of f 0 (h) when h 0 describes a neighborhood of h in H. Thus, we get the desired contradiction.Indeed, if A and B are elements of Λ a−1,b (C n ) and Λ a,b−1 (C n ) respectively, and if A ∧ v + B ∧ v = 0 for any v in an open subset of C n , then A = B = 0.This completes the proof of the proposition.
Remark : In the previous lemma, it is clear that, for a given U, it is sufficient to consider a finite number of u in U .
Let π 0 : C n+p → C n be the canonical projection.For u ∈ L ³ C n+p , C n ´, set π u = π 0 + u.For any couple of integers (a, b) such that a ≤ n et b ≤ n, and for any neighborhood U of 0 in L ³ C n+p , C n ´, we have Λ b (E) * , for a complex vector space E, denotes the space of a-linear and b-antilinear alternating forms on E. By duality and analytic extension with respect to u, this lemma is an immediate consequence of the following result.Proposition Let n ∈ N * and p ∈ N. Let a and b be integers such that a ≤ n and b ≤ n, and let v ∈ Λ a,b ³ C n+p ´ * .If for any u ∈ L(C n+p , C n ) we have u * (v) = 0, then v = 0.