Unbranched Riemann Domains over Q-Complete Spaces

Youssef Alaoui

I. INTRODUCTION

Let $\pi: X \to Y$ be a holomorphic map of complex spaces. Then $\pi$ is said to be locally $r$ -complete if there exists for every $x \in Y$ an open neighborhood $U$ in $Y$ such that $\pi^{-1}(U)$ is $r$ -complete.

A Riemann domain over a complex space $Y$ is a pair $(X, \Pi)$ , where $\pi: X \to Y$ is a holomorphic map which is non-degenerate at every point of $X$ , i.e., $\pi^{-1}(\pi(x))$ is a discrete set at each point $x \in X$ . The pair $(X, \pi)$ is called unbranched or unramified if $\pi: X \to Y$ is locally biholomorphic.

Let $X$ and $Y$ be complex spaces and $\pi: X \to Y$ an unbranched Riemann domain such that $Y$ is $q$ -complete and $\pi$ a locally $r$ -complete morphism.

Does it follow that $X$ is $(q + r - 1)$ -complete?

It was shown in [4] that this problem has a positive answer when $q = r = 1$ and $X$ and $Y$ have isolated singularities.

It is known from [9] that if $\pi: X \to \Omega$ is an unbranched Riemann domain between two complex spaces with isolated singularities, $\Omega$ $q$ -complete, and $\pi$ is locally 1-complete, then $X$ is $q$ -complete.

We have shown in [1] that if $\pi: X \to \Omega$ is a locally $q$ -complete unbranched Riemann domain over an $n$ -dimensional Stein complex space $\Omega$ , then $X$ is cohomologically $q$ -complete with respect to the structure $\mathcal{O}_X$ .

As a result, the author has provided a positive answer to the local Steiness problem: he has proved that if $X$ is a Stein space and if $\Omega \subset X$ is a locally Stein open subset of $X$ , then $\Omega$ is Stein. (See [1]).

In this article, we prove that if $\pi: X \to \Omega$ is a locally $r$ -complete unbranched Riemann domain over a $q$ -complete $n$ -dimensional complex space $\Omega$ , then for any coherent analytic sheaf $\mathcal{F}$ on $X$ , the cohomology group $H^l(X, \mathcal{F})$ vanishes for all $l \geq r + q - 1$ , if $q \geq 2$ .

In particular, we obtain the interesting conclusion.

Corollary. If $X$ is a $q$ -complete complex space of dimension $n$ and if $\Omega \subset X$ is a locally $r$ -complete open subset of $X$ , then

(a) $\Omega$ iscohomologically $(q + r - 1)$ -completeif $q\geq 2$
(b) $\Omega$ is cohomologically $r$ -complete with respect to the structure sheaf if $X$ is a Stein space ( $q = 1$ ).

It should be mentioned [13] that if $Y$ is $q$ -complete and if $\pi: X \to Y$ is a locally $r$ -complete morphism, then the space $X$ is cohomologically $(q + r)$ -complete. But in general, $H^{q + r - 1}(X, \mathcal{O}_X)$ does not vanish, even when $\pi: X \to Y$ is locally 1-complete and $q = 1$ [12](See also [6]).

The above question generalizes the following classical problem:

Is a locally $q$ -complete open subset $\Omega$ of a Stein space $X$ necessarily $q$ -complete?

A counter-example to this problem is not known. One can easily verify that $\Omega$ is cohomologically $(q + 1)$ -complete. It is easy to see that a cohomologically $q$ -complete open subset $\Omega \subset \mathbb{C}^n$ is always $q$ -complete with corners. But it is unknown if these two conditions are equivalent.

By the theory of Andreotti and Grauert [3], it is known that if $X$ is a $q$ -complete complex space, then for every coherent analytic sheaf $\mathcal{F}$ on $X$ , the cohomology group $H^{p}(X,\mathcal{F}) = 0$ for all $p \geq q$ . But it is not known if these two conditions are equivalent except when $X$ is a Stein manifold, $\Omega \subset X$ is cohomologically $q$ -complete with respect to the structure sheaf $\mathcal{O}_{\Omega}$ and $\Omega$ has a smooth boundary [7]. In [2], we have shown that there exists for each $n \geq 3$ an open subset $\Omega \subset \mathbb{C}^n$ which is cohomologically $(n - 1)$ -complete, but $\Omega$ is not $(n - 1)$ -complete.

In section 4 of this article, we prove that for each $n \geq 3$ , there exists an integer $q$ with $2 \leq q < n$ such that for any coherent analytic sheaf $\mathcal{F}$ , the cohomology group $H^{p}(\Omega, \mathcal{F})$ vanishes for all $p \geq q$ but $\Omega$ is not $q$ -complete.

II. PRELIMINARIES

We start by recalling some definitions and results concerning $q$ -complete spaces.

Let $\Omega$ be an open set in $\mathbb{C}^n$ with complex coordinates $z_1, \dots, z_n$ . Then it is known that a function $\phi \in C^\infty(\Omega)$ is $q$ -convex if for every point $z \in \Omega$ , the Levi form

Has at most $q - 1$ negative or zero eigenvalues.

A smooth real-valued function $\phi$ on a complex space $X$ is called $q$ -convex if every point $x \in X$ has a local chart $U \to D \subset \mathbb{C}^n$ such that $\phi|_U$ has an extension $\hat{\phi} \in C^\infty(D, \mathbb{R})$ which is $q$ -convex on $D$ .

Two $q$ -convex functions $\phi, \psi$ on $X$ have the exact positivity directions if, for each point $x \in X$ , there exists an open neighborhood $U$ of $x$ that can be identified to a closed analytic subset $B$ of a domain $D$ of some $\mathbb{C}^n$ , and a complex vector subspace $E$ of $\mathbb{C}^n$ of dimension $\geq n - q + 1$ such that the Levi forms of $L_z(\phi, \xi)$ and $L_z(\psi, \xi)$ , $z \in U$ , are positive definite when restricted to $E$ .

We say that $X$ is $q$ -complete if there exists a $q$ -convex function $\phi \in C^{\infty}(X, \mathbb{R})$ which is exhaustive on $X$ , i.e. $\{x \in X; \phi(x) < c\}$ is relatively compact for any $c \in \mathbb{R}$ .

A complex space $X$ is said to be cohomologically $q$ -complete if the cohomology groups $H^{p}(X,\mathcal{F})$ , $\mathcal{F} \in \operatorname{Coh}(X)$ , vanish for all $p \geq q$ .

An open subset $D$ of $\Omega$ is called $q$ -Runge, if for every compact set $K \subset D$ , there is a $q$ -convex exhaustion function $\phi \in C^{\infty}(\Omega)$ such that

This generalizes the classical notion of Runge pairs of Stein spaces.

It is shown in [3] that if $D$ is $q$ -Runge in $\Omega$ , then for every coherent analytic sheaf $\mathcal{F}$ on $\Omega$ , the cohomology groups $H^{p}(D,\mathcal{F})$ vanish for $p \geq q$ and the restriction map

has a dense image for all $p \geq q - 1$ .

A holomorphic map $\pi: X \to \Omega$ of complex spaces is called a $q$ -complete morphism if there exists a $q$ -convex function $\mathbf{\mu}: X \to \mathbb{R}$ such that for every real number $\mu \in \mathbb{R}$ , the restriction of $\Pi$ from $\{x \in X; (x) \leq \mu\}$ to $\Omega$ is proper. The canonical topologies on $H^{p}(X, \mathcal{F})$ are separated for all $p \geq q + 1$ and for every coherent analytic sheaf $\mathcal{F}$ on $X$ .

III. UNBRUNCHED RIEMANN DOMAINS OVER Q-COMPLETE SPACES

Theorem 1. Let $X$ and $Y$ be two $n$ -dimensional complex spaces such that $Y$ is $q$ -complete and $\pi: X \to Y$ is an unbranched Riemann domain and locally $r$ -complete morphism. Then $X$ is cohomologically $(q + r - 1)$ -complete.

Proof. Since $Y$ is $q$ -complete, there exists, according to [14], a smooth $q$ -convex function $\phi: Y \to ]0, +\infty[$ such that for every real number $\lambda$ , $Y(\lambda) = \{y \in Y: \phi(y) < \lambda\}$ is relatively compact in $Y$ and $\{y \in Y: \phi(y) \leq \lambda\} \setminus \partial Y(\lambda)$ contains at most one point. Put $p = q + r - 1$ and let $\mathcal{F}$ be a coherent analytic sheaf on $X$ . We define $X(\lambda) = \Pi^{-1}(Y(\lambda))$ and consider the set $A$ of all real numbers $\lambda$ such that $H^{p}(X(\lambda), \mathcal{F}) = 0$ .

To prove that $H^{p}(X(\lambda),\mathcal{F}) = 0$ for every $\lambda \in \mathbb{R}$ , it will be sufficient to show that

(a) $A\neq \emptyset$ and, if $\lambda \in A$ and $\lambda^{\prime} < \lambda$ , then $\lambda^{\prime}\in A$
(b) if $\lambda_j \to \lambda$ and $\lambda_j \in A$ for all $j$ , then $\lambda \in A$ .
(c) if $\lambda_0\in A$ , there exists $\varepsilon_0 > 0$ such that $\lambda_0 + \varepsilon_0\in A$

We first prove assertion (a). Clearly, $A$ is not empty. Indeed if $\lambda_0 = \min\{\phi(x); x \in Y\}$ , then $[- \infty, \lambda_0] \subset A$ . Also, if $\lambda \in A$ and $\lambda' < \lambda$ , then by theorem 1 of [13], the restriction map

has a dense range. Moreover, $\rho$ is, in addition, injective. In fact, let

be the restriction map, where $\mu$ is any real number with $\mu < Min(\lambda^{\prime},\lambda_{0})$ . Then the composition $\rho^{\prime}o\rho$ is obviously injective. This implies that the restriction $\rho$ is injective, which means that $H^{p}(X(\lambda),\mathcal{F}) = 0$ and $\lambda^{\prime}\in A$ .

To prove $(c)$ , we fix some $\lambda_0 \in A$ and suppose that $\{y \in Y: \phi(y) = \lambda_0\} \setminus \partial Y(\lambda_0) = \{y_0\}$ for some $y_0 \in Y$ .

Let $U$ be a Stein open neighborhood of $y_0$ such that $\Pi^{-1}(U)$ is $r$ -complete and $\overline{U} \cap \overline{Y(\lambda_0)} = \emptyset$ . There exist finitely many Stein open sets $U_i \subset \subset Y$ , $1 \leq i \leq k$ , disjoint from $U$ such that $\partial Y(\lambda_0) \subset \bigcup_{i=1}^{k} U_i$ and $\Pi^{-1}(U_i)$ are $r$ -complete. Let $\theta_i \in C_0^\infty(U_i, \mathbb{R}^+)$ be smooth compactly supported functions such that $\sum_{i=1}^{k} \theta_i(\xi)) > 0$ at every point $\xi \in \partial Y(\lambda_0)$ . We can therefore choose sufficiently small numbers $c_i > 0$ , $0 \leq i \leq k$ , so that the functions $\phi_i: Y \to \mathbb{R}$ , $1 \leq i \leq k$ , defined by

Are $q$ -convex with the same positivity directions. If we set

Moreover, since $\phi$ is exhaustive, there exists $\varepsilon_0 > 0$ such that $Y(\lambda_0 + \varepsilon_0) \subset Y_k \cup U$ . We define for an arbitrary real number $\lambda$ with $\lambda_0 < \lambda < \lambda_0 + \varepsilon_0$ and integer $j = 0, \dots, k$ , the sets $Y_j(\lambda) = Y_j \cap Y(\lambda)$ and $X_j(\lambda) = \Pi^{-1}(Y_j(\lambda))$ .

Since $Y(\lambda) = (Y(\lambda) \cap Y_k) \cup (Y(\lambda) \cap U)$ , then $X(\lambda) = X_k(\lambda) \cup V(\lambda)$ , where $V(\lambda) = \Pi^{-1}(Y(\lambda) \cap U) = \{x \in \Pi^{-1}(U): \phi o\Pi(x) < \lambda\}$ is $p$ -complete, because $\Pi^{-1}(U)$ is $r$ -complete and $\phi o\Pi$ is $q$ -convex. Moreover, $X_k(\lambda) \cap V(\lambda)$ is $p$ -Runge in $V(\lambda)$ . Therefore

To prove $(c)$ , we show inductively on $j$ that $H^{p}(X_{j}(\lambda),\mathcal{F}) = 0$ . For $j = 0$ this is clearly satisfied since $X_0(\lambda) = X(\lambda_0)$ and $\lambda_0\in A$ . Assume now that $j\geq 1$ and that $H^{p}(X_{j - 1}(\lambda),\mathcal{F}) = 0$ . Since $Y_{j} = Y_{j - 1}\cup (Y_{j}\cap U_{j})$ , then $X_{j}(\lambda) = X_{j - 1}(\lambda)\cup V_{j}(\lambda)$ , where

is $p$ -complete since $\Pi^{-1}(U_j)$ is $r$ -complete and $\phi o\Pi$ and $\phi_jo\Pi$ are $q$ -convex with the same positivity directions. Furthermore, as $X_{j - 1}(\lambda)\cap V_j(\lambda)) = X_{j - 1}(\lambda)\cap \Pi^{-1}(U_j) = \{x\in \Pi^{-1}(U_j):\phi_{j - 1}o\Pi (x) < \lambda_0,\phi o\Pi (x) < \lambda \}$ is clearly $p$ -Runge in $\Pi^{-1}(U_j)$ , then the restriction map

has a dense image for all $s \geq p - 1$ . Since $\rho'$ is clearly injective and $p - 1 \geq r$ , then $H^{p - 1}(X_{j - 1}(\lambda) \cap V_j(\lambda), \mathcal{F}) = 0$ . Therefore from the Mayer-Vietoris sequence for cohomology

we deduce that $H^{p}(X_{j}(\lambda),\mathcal{F}) = 0$

To prove statement (b), it is sufficient to show that if $\lambda_j \nearrow \lambda$ and $\lambda_j \in A$ for all $j$ , then

has a dense image.

To complete the proof of theorem 1, it is, therefore, enough, according to (Cf. [3], p. 250), to show the following lemma.

Lemma 1. For every pair of real numbers $\mu < \lambda$ , the restriction map

has a dense range.

Proof. We consider the set $T$ of all real numbers $\lambda$ such that

has a dense range for all $\mu \leq \lambda$ .

To see that $T$ is not empty, we choose $\lambda_0 = \min\{\phi(y); y \in Y\}$ . Then clearly $] - \infty, \lambda_0] \subset T$ .

To prove that $T$ is open in $\rceil -\infty, + \infty [i$ it is, therefore, sufficient to show that if $\lambda \in T$ , there exists $\varepsilon >0$ such that $\lambda +\varepsilon \in T$ . For this, we consider a finite covering $(U_{i})_{1\leq i\leq k}$ of $\{y\in Y:\phi (z) = \lambda \}$ by Stein open sets $U_{i}\subset \subset Y$ and compactly supported functions $\theta_{i}\in C_{o}^{\infty}(U_{i})$ , $\theta_{j}\geq 0$ , $j = 1,\dots,k$ such that $\Pi^{-1}(U_i)$ is $r$ -complete and $\sum_{i = 1}^{k}\theta_{i}(x) > 0$ at any point of $\partial Y(\lambda)$ . Define $Y_{j} = \{z\in Y:\phi_{j}(z) < \lambda \}$ where

$\phi_j(z) = \phi(z) - \sum_{i=1}^{j} c_i \theta_i$ , with $c_i > 0$ sufficiently small so that $\phi_j(z)$ are still $q$ -convex. With the same positivity directions for $1 \leq j \leq k$ .

If we consider the following sets defined in the lemma 2

$Y(\lambda) = \{y\in Y:\phi (y) < \lambda \}$ $X(\lambda) = \Pi^{-1}(Y(\lambda))$ $Y_{i} = \{x\in Y:\phi_{i}(x) < \lambda_{0}\}$ and $Y_{0} = Y(\lambda_{0})$ , then

and $X_{j}(\lambda) = \Pi^{-1}(Y_{j}\cap Y(\lambda)) = X_{j - 1}(\lambda)\cup V_{j}(\lambda)$ , where

V_{j}(\") = \\Pi^{-1}(U_{j} \cap Y_{j}(\lambda)) = \{x \in \Pi^{-1}(U_{j}): \phi o \Pi(x) < \lambda, \phi_{j} o \Pi(x) < \lambda_{0}\}

Now since $X_{j-1}(\lambda) \cap V_j(\lambda)$ is $p$ -Runge in the $p$ -complete set $V_j(\lambda)$ and $H^p(X_j(\lambda), \mathcal{F}) = 0$ , it follows from the long exact sequence of cohomology

that the restriction map

has a dense range.

Moreover, since $\phi$ is exhaustive, there exists $\varepsilon > 0$ such that $Y(\lambda + \varepsilon) \subset Y_k$ . We deduce that the restriction map

has a dense image, which implies that $\lambda +\varepsilon \in T$

Let now $\lambda_j \in T$ , $j \geq 0$ , such that $\lambda_j \nearrow \lambda$ , and let $\mathcal{U} = (U_i)_{i \in I}$ be a countable base of Stein open covering of $X$ . Then the restriction map between spaces of cocycles

has dense image for $j \geq 0$ . Let $\lambda' < \lambda$ and $j \in \mathbb{N}$ such that $\lambda' < \lambda_j$ . By [1, p.246], the restriction map $Z^{n-2}(\mathcal{U}|_{X_\lambda},\mathcal{F}) \to Z^{n-2}(\mathcal{U}|_{X_{\lambda_j}},\mathcal{F})$ has a dense image. Since $\lambda_j \in T$ , then $Z^{n-2}(\mathcal{U}|_{X_{\lambda_j}},\mathcal{F}) \to Z^{n-2}(\mathcal{U}|_{X_{\lambda'}},\mathcal{F})$ has also a dense image, and hence $\lambda \in T$ .

Now since $H^{p}(X(j),\mathcal{F}) = 0$ for all $j \in \mathbb{N}$ and $H^{p - 1}(X(j + 1),\mathcal{F})$ has a dense image in $H^{p - 1}(X(j),\mathcal{F})$ for all $j \geq 0$ , it follows from ([3], p. 250) that

is bijective, which shows that $H^{p}(X,\mathcal{F}) = 0$ .

IV. A COUNTER-EXAMPLE TO THE ANDREOTTI-GRAUERT CONJECTURE

Theorem 2. There exists for each integer $n \geq 3$ a cohomologically $q$ -complete open subset $\Omega \subset \mathbb{C}^n$ , $2 \leq q < n$ , which is not $q$ -complete.

We consider the following example due to Diederich and Forness [4]. Let $(n,q)$ be a pair of integers with $2\leq q < n$ and such that $n = mq + 1$ , where $m = \left[\frac{n}{q}\right]$ is the integral part of $\frac{n}{q}$ . We define the functions.

and

where $\sigma_{j}(z) = Im(z_{j}) + \sum_{i = m + 1}^{n}|z_{i}|^{2} - (m + 1)\sum_{i = m + (j - 1)(q - 1) + 1}^{m + j(q - 1)}|z_{i}|^{2}$ , for $j = 1,\dots,m$

$z = (z_{1},z_{2},\dots,z_{n})$ , and $N > 0$ a positive constant. Then, if $N$ is large enough, the functions $\phi_1,\dots \phi_2$ are $q$ -convex on $\mathbb{C}^n$ and, if $\rho = Max(\phi_1,\dots,\phi_{m + 1})$ , then, for $\varepsilon_o > 0$ small enough, the set $D_{\varepsilon_o} = \{z\in \mathbb{C}^n:\rho (z) < - \varepsilon_o\}$ is relatively compact in the unit ball $B = B(0,1)$ if $N$ is sufficiently large. (See [4]).

We fix some $\varepsilon >\varepsilon_0$ and consider a covering $(U_{i})_{i\in \mathbb{N}}$ of $\partial D_{\varepsilon}$ , by Stein open subsets $U_{i}\subset \subset D_{\varepsilon_{0}}$ and functions $\theta_{i}\in C_{0}^{\infty}(\mathbb{C}^{n},\mathbb{R})$ such that

We can therefore choose sufficiently small positive numbers $c_{1},\dots,c_{k}$ so that the functions $\phi_{i,j} = \phi_i - \sum_{l = 1}^{j}c_l\theta_l$ are $q$ -convex for $i = 1,\dots,m + 1$ and $1\le j\le k$ .

We define $\phi_{i,0} = \phi_i$ for $i = 1,\dots,m + 1$ , $D_0 = D_\varepsilon$ and $D_{j} = \{z\in D_{\varepsilon_{0}}:\rho_{j}(z) < -\varepsilon \}$ , where $\rho_j(z) = \rho -\sum_{i = 1}^{j}c_i\theta_i$ for $j = 1,\dots,k$ . Then $\rho_{j}$ are $q$ -convex with corners and it is clear that

Lemma 2. In the situation described above, for any coherent analytic sheaf $\mathcal{F}$ on $D_{\varepsilon_0}$ , the restriction map $H^p (D_{j + 1},\mathcal{F})\to H^p (D_j,\mathcal{F})$ is surjective for all $p\geq \tilde{q} -1$ and all $0\le j\le k - 1$ . In particular, $\dim_{\mathbb{C}}H^{p}(D_{j},\mathcal{F}) < \infty$ , if $p\geq \tilde{q} -1$ .

Proof. We first prove that the cohomology group $H^{p}(D_{j} \cap U_{l}, \mathcal{F}) = 0$ for all $p \geq \tilde{q} - 1$ , $0 \leq j \leq k$ , and $1 \leq l \leq k$ . In fact, the set $D_{j} \cap U_{l}$ can be written in the form $D_{j} \cap U_{l} = D_{1}' \cap \dots \cap D_{m + 1}'$ , where $D_{i}' = \{z \in U_{l}: \phi_{i,j}(z) < -\varepsilon\}$ are clearly $q$ -complete. Then for every $i_{1}, \dots, i_{m} \in \{1, \dots, m + 1\}$ , $D_{i_{1}}' \cap \dots \cap D_{i_{m}}'$ are $(\tilde{q} - 1)$ -complete. Therefore, by using Proposition 1 of [11], we obtain

if $p\geq \tilde{q} -1$ , which implies that $H^{p}(D_{j}\cap U_{l},\mathcal{F}) = 0$ for all $p\geq \tilde{q} -1$

Now since $D_{j+1} = D_j \cup (D_{j+1} \cap U_{j+1})$ , it follows from the Mayer-Vietoris sequence for cohomology

that the restriction map

is surjective when $p \geq \tilde{q} - 1$ .

Let now $A$ be the set of all real numbers $\varepsilon \geq \varepsilon_0$ such that $H^p(D_{\varepsilon}, \mathcal{F}) = 0$ for all $p \geq \tilde{q} - 1$ .

Lemma 3. - The set $A$ is not empty and, if $\varepsilon \in A$ , $\varepsilon > \varepsilon_0$ , then there exists $\varepsilon' \in A$ such that $\varepsilon_0 \leq \varepsilon' < \varepsilon$ .

Proof. In fact, if $\mu_0 = Min_{z\in \overline{B}}\{\phi_i(z),i = 1,\dots,m + 1\}$ , then one sees easily that $[- \mu_0, + \infty [ \subset A$ .

For the proof of the second assertion, if with the notations of lemma 1 we set $D_0 = D_{\varepsilon}$ , we obtain $D_0 \subset D_1 \subset \dots \subset D_k$ , $D_0 \subset \subset D_k \subset \subset D_{\varepsilon_0}$ and $D_j \setminus D_{j-1} \subset \subset U_j$ for $j = 1, \dots = k$ .

We fix some $1 \leq j \leq k$ and $1 \leq l \leq k$ , and set $D_{j} \cap U_{l} = D_{1}^{\prime} \cap \dots D_{m + 1}^{\prime}$ , where $D_{i}^{\prime} = \{z \in U_{l}: \phi_{i,j}(z) < -\varepsilon\}$ , then $D_{i}^{\prime}$ are $q$ -complete and $q$ -Runge in $U_{l}$ . Therefore because of the proof of lemma 2, one obtains

for $p \geq \tilde{q} - 1$ and, consequently, the restriction map

is surjective for all $p \geq \tilde{q} - 1$ .

We now show inductively on $j$ that $H^{\tilde{q} -1}(D_j,\mathcal{F}) = 0$ . For $j = 0$ , this is clearly satisfied since $D_0 = D_{\varepsilon}$ and $\varepsilon \in A$ . Assume now that this property has already been proved for $j < k$ . Since for every $i_1, \dots, i_m$ , in $\{1, \dots, m + 1\}$ , the open set $D_{i_1}' \cap \dots \cap D_{i_m}'$ is $(\tilde{q} - 1)$ -Runge in $U_l$ , then the restriction map

has a dense range for $p \geq \tilde{q} - 2$ . Since the canonical topologies on $H^{i}(D_{i_{m}}^{\prime} \cap \dots \cap D_{i_{m}}^{\prime}, \mathcal{F})$ are obviously separated for $i \geq 2$ , then $H^{p}(D_{i_{m}}^{\prime} \cap \dots \cap D_{i_{m}}^{\prime}, \mathcal{F}) = 0$ for all $p \geq \tilde{q} - 2$ . We know from Proposition 1 of [11] that $H^{p}(D_{j} \cap U_{l}, \mathcal{F}) \cong H^{p + m}(D_{1}^{\prime} \cup \dots \cup D_{m + 1}^{\prime}, \mathcal{F})$ for $p \geq \tilde{q} - 2 = n - m - 1$ . We can choose the covering $(U_{i})_{1 \leq i \leq k}$ of $\partial D_{\varepsilon}$ such that $U_{l} \setminus D_{1}^{\prime} \cup \dots \cup D_{m + 1}^{\prime}$ has no compact connected components, so it follows from the mean theorem of [5], that the restriction $H^{p}(U_{l}, \mathcal{F}) \longrightarrow H^{p}(D_{1}^{\prime} \cup \dots \cup D_{m + 1}^{\prime}, \mathcal{F})$ has a dense image for $p \geq n - 1$ . This proves that

Now since $H^{\tilde{q} -2}(D_j \cap U_{j+1}, \mathcal{F}) = H^{\tilde{q} -1}(D_{j+1} \cap U_{j+1}, \mathcal{F}) = H^{\tilde{q} -1}(D_j, \mathcal{F}) = 0$ , it follows from the Mayer-Vietoris sequence for cohomology

that $H^{\bar{q} -1}(D_{j + 1},\mathcal{F}) = 0$

On the other hand, since $\rho$ is proper, there exists $\varepsilon' > 0$ such that $\varepsilon - \varepsilon' > \varepsilon_o$ and $D_{\varepsilon - \varepsilon'} = \{z \in D_{\varepsilon_o}: \rho(z) < \varepsilon' - \varepsilon\} \subset \subset D_k$ .

Since $H^{\tilde{q} -1}(D_k,\mathcal{F})\to H^{\tilde{q} -1}(D_{\varepsilon -\varepsilon '},\mathcal{F})$ is surjective, $H^{\tilde{q} -1}(D_k,\mathcal{F}) = 0$ and $dim_{\mathbb{C}}H^{\tilde{q} -1}(D_{\varepsilon -\varepsilon '},\mathcal{F}) < \infty$ , then $H^{\tilde{q} -1}(D_{\varepsilon ' - \varepsilon '},\mathcal{F}) = 0$ , whence $\varepsilon -\varepsilon^{\prime}\in A$ .

Lemma 4. The open set $D_{\varepsilon_0}$ is cohomologically $(\tilde{q} -1)$ -complete.

Proof. For this, we consider the set $A$ of all real numbers $\varepsilon \geq \varepsilon_0$ such that $H^p(D_{\varepsilon}, \mathcal{F}) = 0$ for all $p \geq \tilde{q} - 1$ . Then by lemma 3, $A$ is not empty and open in $[\varepsilon_0, \infty[$ . Moreover, if $\varepsilon = Inf(A)$ , there exists a decreasing sequence of real numbers $\varepsilon_j \in A$ , $j \geq 1$ , such that $\varepsilon_j \searrow \varepsilon$ . Since $H^p(D_{\varepsilon_j}, \mathcal{F}) = 0$ for $p \geq \tilde{q} - 1$ and, by lemma 1, the restriction map $H^p(D_{\varepsilon_{j+1}}, \mathcal{F}) \to H^p(D_{\varepsilon_j}, \mathcal{F})$ is surjective for all $p \geq \tilde{q} - 2$ , then by ([3], p. 250), the restriction map

is an isomorphism for $p \geq \tilde{q} - 1$ , which shows that $\varepsilon \in A$ .

Assume now that $\varepsilon >\varepsilon_0$ . Then there exists, according to lemma 1, $\varepsilon^{\prime}\in A$ such that $\varepsilon_0 < \varepsilon ' < \varepsilon$ , which contradicts the fact that $\varepsilon = Inf(A)$ . We conclude that $\varepsilon = \varepsilon_0\in A$ , and hence $D_{\varepsilon_0}$ is cohomologically $(\tilde{q} -1)$ -complete.

End of the proof of theorem 2

We have shown that $D_{\varepsilon_0}$ is cohomologically $(\tilde{q} - 1)$ -complete. We are now going to prove that for a good choice of the contents $\varepsilon_0$ and $N$ , we can find an $\varepsilon > \varepsilon_0$ such that $D_{\varepsilon}$ is cohomologically $(\tilde{q} - 1)$ -complete but $\Omega$ not $(\tilde{q} - 1)$ -complete.

In fact, it was shown by Diederich-Forness [4] that if $\delta > 0$ is small enough, then the topological sphere of real dimension $n + \tilde{q} - 2$

is not homologous to 0 in $D_{\varepsilon_0}$ . This follows from the fact that the set $E = \{z \in \mathbb{C}^n: x_1 = z_2 = \dots = z_n = 0\}$ does not intersect $D_{\varepsilon_0}$ , since on $E$

and

such that $\rho \geq 0$ on $E$ . So the following real form of degree $n + \tilde{q} - 2$

is well-defined and d-closed on $D_{\varepsilon_0}$ . Since $\omega$ does not depend on $y_1, \dots, y_m$ , then by the standard argument $\int_{S_\delta} \omega \neq 0$ . Therefore $S_\delta$ is not homologous to 0 in $D_{\varepsilon_0}$ .

Let $\mathcal{E}_q$ be the sheaf of germs of $C^\infty$ $q$ -forms on $\mathbb{C}^n$ and $\mathcal{T}_q$ the sheaf of germs of $C^\infty$ d-closed $q$ -forms. Then we have an exact sequence of sheaf homomorphisms

Since by the de Rham theorem for every $p \geq 1$ , the cohomology group $H^{p}(D_{\varepsilon_{0}}, \mathbb{C})$ is isomorphic to

it follows from Stokes formula that $H^{n + \tilde{q} -2}(D_{\varepsilon},\mathbb{C})$ does not vanish.

We are going to show that $H^{r}(D_{\varepsilon_{0}}, \mathcal{O}_{D_{\varepsilon_{0}}}) = 0$ for all $r$ with $1 \leq r \leq \tilde{q} - 3$ .

We first assert that we can choose $N$ , $\varepsilon_0$ , and $\varepsilon > \varepsilon_0$ such that, if, with the notations of Proposition 1, we set

then we obtain

where $m^{\prime} = Min_{z\in \overline{D}_{\varepsilon_0}}(z)$ , and

In fact, we can choose $\varepsilon >\varepsilon_0$ sufficiently big and $\lambda >0$ small enough so that $\varepsilon_0 - \varepsilon < m' < (1 + \lambda).Min_{z\in \overline{D}_{\varepsilon}}(z)$ and $\lambda \varepsilon -(1 + \lambda)\varepsilon_0 > 0$

On the other hand, if $\delta = Min_{z\in \overline{D}_{\varepsilon_0}}||z||^2$ , then we have

Therefore by suitable choice of $\varepsilon_0$ , $\varepsilon$ and $N$ we can also achieve that

and

for every $z\in \overline{D}_{\varepsilon}$

Because $(z) < \varepsilon_0 - \varepsilon$ on $\overline{D}_{\varepsilon}$ , then clearly we obtain

which shows that

We are now going to show that for every none-positive real number $\alpha$ with $\alpha < \varepsilon_0 - \varepsilon$ , the open sets

are relatively compact in $D_{\varepsilon}$

To see this, we consider a sequence $(z_{j})_{j\geq 0}\subset B_{\alpha}$ , which converges to a point $z\in \overline{D}_{\varepsilon}$ . Then one has for every sufficiently large integer $j$

Since

then

A passage to the limit shows that

because $(z) < \varepsilon_0 - \varepsilon$ , which implies that $z \in D_{\varepsilon}$ . We conclude that with such a choice of $\varepsilon_0, N$ , and $\varepsilon$ the limit $z \in D_{\varepsilon}$ , and hence the open set

is relatively compact in $D_{\varepsilon}$ for all real numbers $\alpha$ , with $\alpha < \varepsilon_0 - \varepsilon$ .

Now since $\phi$ is in addition $(m + 2)$ -convex, then a similar proof of theorem 15 of [3] shows that, if $\Omega^i$ is the sheaf of germs of holomorphic $i$ -forms on $\mathbb{C}^n$ , $i \geq 0$ , $(\Omega^0 = \mathcal{O}_{\mathbb{C}^n})$ , and $B_c = \{z \in D_{\varepsilon}: \phi(z) < c\}$ for $c \leq \varepsilon_0 - \varepsilon$ , then the map

is injective for every $r < n - m - 1$ and $c < \varepsilon_0 - \varepsilon$ . Then obviously $H^{r}(D_{\varepsilon},\Omega^{i}) = 0$ for $1 \leq r \leq n - m - 2$ and $i \geq 0$ . In fact, let $c_0 = \text{Max}_{z \in \overline{D}_{\varepsilon}} \phi(z)$ . Then there exists $z_1 \in \partial D_{\varepsilon}$ such that $\phi(z_1) = c_0$ . Since $c_0 = \phi(z_1) = \sigma(z_1) + \sum_{i=1}^{m} \sigma_i(z_1)^2 + m' < \rho(z_1) + \varepsilon_0 \leq \varepsilon_0 - \varepsilon$ , then $B_{c_0} = D_{\varepsilon}$ , and hence $H^{r}(D_{\varepsilon},\Omega^{i}) = 0$ for $1 \leq r \leq n - m - 2$ .

Now if we suppose that $D_{\varepsilon}$ is $(\tilde{q} - 1)$ -complete, then there exists a $C^\infty$ strictly $(\tilde{q} - 1)$ -convex function $: D_{\varepsilon} \to \mathbb{R}$ such that $D_{\varepsilon, c} = \{z \in D_{\varepsilon}: (z) < c\}$ is relatively compact in $D_{\varepsilon}$ for every $c \in \mathbb{R}$ .

We now consider the resolution of the constant sheaf $\mathbb{C}$ on $D_{\varepsilon}$

If we set $Z^{j} = \operatorname{Im}\left(\Omega^{j-1} \xrightarrow{d} \Omega^{j}\right)$ for $1 \leq j \leq n-1$ , then we get short exact sequences

Since, by Proposition 1, $D_{\varepsilon}$ is cohomologically $(\tilde{q} - 1)$ -complete, then $H^{r}(D_{\varepsilon}, \Omega^{i}) = 0$ for all $r \geq \tilde{q} - 1$ and $i \geq 0$ . So we obtain the isomorphisms

and the exact sequence

We deduce that the map

is surjective. The map $\varphi$ is defined as follows: If a differential form $\omega \in C_{n,\tilde{q} -2}^{\infty}(D_{\varepsilon})$ satisfies the equation $\overline{\partial}\omega = 0$ , then $\omega$ is also $d$ -closed and therefore defines a cohomology class in $H^{n + \tilde{q} -2}(D_{\varepsilon},\mathbb{C})$ .

Moreover, since, by theorem 1 in [8], every $d$ -closed differential form $\omega \in C_{n,\tilde{q} -2}^{\infty}(D_{\varepsilon})$ is cohomologous to a $\overline{\partial}$ -closed $(n,\tilde{q} -2)$ differential form $\omega^{\prime}\in C_{n,\tilde{q} -2}^{\infty}(D_{\varepsilon})$ , it follows that the map

is bijective.

Now if we suppose that $D_{\varepsilon}$ is $(\tilde{q} - 1)$ -complete, then there exists a $C^\infty$ strictly $(\tilde{q} - 1)$ -convex function $: D_{\varepsilon} \to \mathbb{R}$ such that $D_{\varepsilon, c} = \{z \in D_{\varepsilon}: (z) < c\}$ is relatively compact in $D_{\varepsilon}$ for every $c \in \mathbb{R}$ .

Notice that for the given $\varepsilon$ , if $\delta > 0$ is small enough, the topological sphere

Since is exhaustive on $D_{\varepsilon}$ , there exists $c' > 0$ such that $S_{\delta}$ is not homologous to 0 in $D_{\varepsilon, c'}$ . Let $c > c'$ . Then $D_{\varepsilon, c}$ and $D_{\varepsilon, c'}$ are $(\tilde{q} - 1)$ -complete and, similarly $H^{p}(D_{\varepsilon, c}, \Omega^{i}) = H^{p}(D_{\varepsilon, c'}, \Omega^{i}) = 0$ for $1 \leq p \leq n - m - 2$ and $i \geq 0$ . Also the maps $H^{\tilde{q} - 2}(D_{\varepsilon, c}, \Omega^{n}) \to H^{n + \tilde{q} - 2}(D_{\varepsilon, c}, \mathbb{C})$ and $H^{\tilde{q} - 2}(D_{\varepsilon, c'}, \Omega^{n}) \to H^{n + \tilde{q} - 2}(D_{\varepsilon, c'}, \mathbb{C})$ are bijective. Moreover, since the Levi form of has at least $m + 1$ strictly positive eigenvalues, then by using Morse theory (See for instance [7]) we find that

It follows from the commutative diagram of continuous maps

that the restriction homomorphism

is bijective. Since in addition $D_{\varepsilon, c'}$ is relatively compact in $D_{\varepsilon, c}$ , the function being exhaustive on $D_{\varepsilon}$ , then, according to theorem 11 of [1], one obtains

Since the sheaf $\Omega^n$ is isomorphic to $\mathcal{O}_{D_{\varepsilon}}$ , then we have also $\dim_{\mathbb{C}}H^{\tilde{q} -2}(D_{\varepsilon,c'},\mathcal{O}_{D_{\varepsilon}}) < \infty$ . Furthermore, since $D_{\varepsilon,c'}$ is cohomologically ( $\tilde{q} -1$ -complete and $H^{r}(D_{\varepsilon},\mathcal{O}_{D_{\varepsilon}}) = 0$ for $1\leq r\leq n - m - 2$ ), it follows from theorem 1 of [6] that $D_{\varepsilon,c'}$ is Stein, which is in contradiction with the fact that $H^{n + \tilde{q} -2}(D_{\varepsilon,c'},\mathbb{C})\neq 0$ , since $S_{\delta}\subset D_{\varepsilon,c'}$ is not homologous to 0 in $D_{\varepsilon,c'}$ . We conclude that $D_{\varepsilon}$ is cohomologically ( $\tilde{q} -1$ -complete but not ( $\tilde{q} -1$ -complete.

Theorem 3. There exists for each integer $n \geq 3$ a cohomologically $(n - 1)$ -complete open subset $\Omega$ of $\mathbb{C}^n$ which is locally $(n - 1)$ -complete in $\mathbb{C}^n$ but $\Omega$ is not $(n - 1)$ -complete.

Proof. We consider for $n \geq 3$ the functions $\phi_1, \phi_2: \mathbb{C}^n \to \mathbb{R}$ defined by

where $\sigma_{1}(z) = Im(z_{1}) + \sum_{i = 3}^{n}|z_{i}|^{2} - |z_{2}|^{2}, z = (z_{1},z_{2},\dots,z_{n})$ , and $N > 0$ a positive constant. Then, if $N$ is large enough, the functions $\phi_1$ and $\phi_{2}$ are $(n - 1)$ -convex on $\mathbb{C}^n$ and, if $\rho = Max(\phi_1,\phi_2)$ , then, for $\varepsilon_o > 0$ small enough, the set $D_{\varepsilon_o} = \{z\in \mathbb{C}^n:\rho (z) < -\varepsilon_o\}$ is relatively compact in the unit ball $B = B(0,1)$ , if $N$ is sufficiently large.

According to ([2], p. 20), we can choose $\varepsilon_0 > 0$ such that if $\delta = Min_{z\in \overline{D}_{\varepsilon_0}}||z||^2$ , then we have

and that by a suitable choice of $\varepsilon >\varepsilon_0$

is cohomologically $(n - 1)$ -complete but not $(n - 1)$ -complete.

Now if we suppose that at a boundary point $z_0 \in \partial D_{\varepsilon}$ , we have $\phi_1(z_0) = \phi_2(z_0)$ , then $\sigma_1(z_0) = 0$ and, hence $N|z_0|^4 - \frac{|z_0|^2}{4} = \varepsilon$ . This implies $|z_0|^2 = \frac{1}{8N}(1 + \sqrt{1 + 64N\varepsilon} < \frac{1}{4N}$ . Therefore $\frac{1}{2}\sqrt{1 + 64N\varepsilon} < \frac{1}{2}$ , which is a contradiction. This implies that $\phi_1(z) \neq \phi_2(z)$ at every boundary point $z \in \partial D_{\varepsilon}$ . We conclude that with such a choice of $\varepsilon_0$ , $N$ and $\varepsilon$ , $D_{\varepsilon}$ is obviously locally $(n-1)$ -complete in $\mathbb{C}^n$ .