It is proved that if ? : X ? ? is an unbranched Riemann domain and locally r-complete morphism over a q-complete space ?, then X is cohomologically (q + r ? 1)-complete if q ? 2. We have shown in [1] that if ? : X ? ? is an unbranched Riemann domain and locally q-complete morphism over a Stein space ?, then X is cohomologically q-complete for the structure sheaf O X . In section 4 of this article, we prove using a counter-example that there exists for each integer n ? 3 an open subset ? ? C n which is locally (n ? 1)-complete, but ? is not (n ? 1)-complete. The counter-example we give is based on a recent example given by the author [2].

By the theory of Andreotti and Grauert [3] it is known that a q-complete complex space is always cohomologically q-complete. A counter-example to the converse of this theorem was given in [2], where it is shown that there exists for each integer n ? 3 a domain ? ? C n which is cohomologically (n?1)-complete but ? is not (n ? 1)-complete. Since then, the question of whether the joint statements of these theorems are factual has been subject to extensive studies. For example, it was shown that if X is a Stein manifold and if D ? X is an open subset that has a C 2 boundary such that H p (D, O D ) = 0 for all p ? q, then D is q-complete.

In this article, we prove that for any pair of integers (n, q), 2 ? q < n, there exists an open subset ? of C n which is cohomologically (q ? 1)-complete but ? is not (q ? 1)-complete if n = mq + 1, where m = [ n q ] denotes as usual the integral part of n q and q = n ? m + 1.

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

A Riemann domain over a complex space Y is a pair (X, ?), where ? : X ? Y is a holomorphic map which is non-degenerate at every point of X, i.e., ? ?1 (?(x)) is a discrete set at each point x ? X. The pair (X, ?) is called unbranched or unramified if ? : X ? Y is locally biholomorphic. Let X and Y be complex spaces and ? : X ? Y an unbranched Riemann domain such that Y is q-complete and ? 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 ? : X ? ? is an unbranched Riemann domain between two complex spaces with isolated singularities, ? q-complete, and ? is locally 1-complete, then X is q-complete.

We have shown in [1] that if ? : X ? ? is a locally q-complete unbranched Riemann domain over an n-dimensional Stein complex space ?, then X is cohomologically q-complete with respect to the structure 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 ? ? X is a locally Stein open subset of X, then ? is Stein. (See [1]).

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

In particular, we obtain the interesting conclusion.

If X is a q-complete complex space of dimension n and if
? ? X is a locally r-complete open subset of X, then (a) ? is cohomologically (q + r ? 1)-complete if q ? 2.
(b) ? 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 ? : X ? Y is a locally r-complete morphism, then the space X is cohomologically (q + r)-complete. But in general, H q+r?1 (X, O X ) does not vanish, even when ? : X ? 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 ? of a Stein space X necessarily q-complete? A counter-example to this problem is not known. One can easily verify that ? is cohomologically (q + 1)-complete. It is easy to see that a cohomologically q-complete open subset ? ? 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 F on X, the cohomology group H p (X, F) = 0 for all p ? q. But it is not known if these two conditions are equivalent except when X is a Stein manifold, ? ? X is cohomologically q-complete with respect to the structure sheaf O ? and ? has a smooth boundary [7]. In [2], we have shown that there exists for each n ? 3 an open subset ? ? C n which is cohomologically (n ? 1)-complete, but ? is not (n ? 1)-complete.

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

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

Let ? be an open set in C n with complex coordinates z 1 , ? ? ? , z n . Then it is known that a function ? ? C ? (?) is q-convex if for every point z ? ?, the Levi form.
L z (?, ?) = i,j ? 2 ?(z) ?z i ?z j ? i ? j , ? ? C n
Has at most q ? 1 negative or zero eigenvalues.


# II. Preliminaries

Corollary. A smooth real-valued function ? on a complex space X is called q-convex if every point x ? X has a local chart U ? D ? C n such that ?| U has an extension ? ? C ? (D, R) which is q-convex on D.

Two q-convex functions ?, ? on X have the exact positivity directions if, for each point x ? 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 C n , and a complex vector subspace E of C n of dimension ? n ? q + 1 such that the Levi forms of L z (?, ?) and L z (?, ?), z ? U , are positive definite when restricted to E.

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

A complex space X is said to be cohomologically q-complete if the cohomology groups H p (X, F), F ? Coh(X), vanish for all p ? q.

An open subset D of ? is called q-Runge, if for every compact set K ? D, there is a q-convex exhaustion function ? ? C ? (?) such that
K ? {x ? ? : ?(x) < 0} ?? D
This generalizes the classical notion of Runge pairs of Stein spaces.

It is shown in [3] that if D is q-Runge in ?, then for every coherent analytic sheaf F on ?, the cohomology groups H p (D, F) vanish for p ? q and the restriction map
H p (?, F) ?? H p (D, F)
has a dense image for all p ? q ? 1.

A holomorphic map ? : X ? ? of complex spaces is called a q-complete morphism if there exists a q-convex function : X ? R such that for every real number µ ? R, the restriction of ? from {x ? X; (x) ? µ} to ? is proper. The canonical topologies on H p (X, F) are separated for all p ? q + 1 and for every coherent analytic sheaf F on X.

Let X and Y be two n-dimensional complex spaces such that Y is qcomplete and ? : X ? 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 ? : Y ?]0, +?[ such that for every real number ?, Y (?) = {y ? Y : ?(y) < ?} is relatively compact in Y and {y ? Y : ?(y) ? ?} \ ?Y (?) contains at most one point. Put p = q + r ? 1 and let F be a coherent analytic sheaf on X. We
Issue ersion I V V ( F ) define X(?) = ? ?1 (Y (?)) and consider the set A of all real numbers ? such that H p (X(?), F) = 0.
To prove that H p (X(?), F) = 0 for every ? ? R, it will be sufficient to show that (a) A = ? and, if ? ? A and ? < ?, then ? ? A.

(b) if ? j ? ? and ? j ? A for all j, then ? ? A.
(c) if ? 0 ? A, there exists ? 0 > 0 such that ? 0 + ? 0 ? A.
We first prove assertion (a). Clearly, A is not empty. Indeed if  
? 0 = min{?(x); x ? Y }, then ] ? ?, ? 0 ] ? A. Also, if ? ? A and ? < ?,

# 259.

H p (X(?), F)
? ?? H p (X(? ), F)
has a dense range. Moreover, ? is, in addition, injective. In fact, let
H p (X(? ), F) ? ?? H p (X(µ), F)
be the restriction map, where µ is any real number with µ < M in(? , ? 0 ). Then the compostion ? o? is obviously injective. This implies that the restriction ? is injective, which means that H p (X(?), F) = 0 and ? ? A.

To prove (c), we fix some ? 0 ? A and suppose that {y ? Y :
?(y) = ? 0 }\?Y (? 0 ) = {y 0 } for some y 0 ? Y . Let U be a Stein open neighborhood of y 0 such that ? ?1 (U ) is r-complete and U ? Y (? 0 ) = ?. There exist finitely many Stein open sets U i ?? Y , 1 ? i ? k, disjoint from U such that ?Y (? 0 ) ? k i=1 U i and ? ?1 (U i ) are r-complete. Let ? i ? C ? 0 (U i , R + ) be smooth compactly supported functions such that k i=1 ? i (?)) > 0 at
every point ? ? ?Y (? 0 ). We can therefore choose sufficiently small numbers c i > 0,
0 ? i ? k, so that the functions ? i : Y ? R, 1 ? i ? k, defined by ? 0 = ?, ? i = ? ? i j=1 c j ? j
Are q-convex with the same positivity directions. If we set
Y i = {x ? Y : ? i (x) < ? 0 } and Y 0 = Y (? 0 ), then Y 0 ? Y 1 ? Y 2 ? ? ? ? ? Y k , Y 0 ?? Y k , Y i \ Y i?1 ?? U i for 1 ? i ? k
Moreover, since ? is exhaustive, there exists
? 0 > 0 such that Y (? 0 + ? 0 ) ? Y k ? U .
We define for an arbitrary real number ? with ? 0 < ? < ? 0 + ? 0 and integer
j = 0, ? ? ? , k, the sets Y j (?) = Y j ? Y (?) and X j (?) = ? ?1 (Y j (?)). Since Y (?) = (Y (?) ? Y k ) ? (Y (?) ? U ), then X(?) = X k (?) ? V (?)
, where 
V (?) = ? ?1 (Y (?) ? U ) = {x ? ? ?1 (U ) : ?o?(x) < ?} is p-complete,I V V ( F ) ? ?1 (U ) is r-complete and ?o? is q-convex. Moreover, X k (?) ? V (?) is p-Runge in V (?). Therefore H p (X(?), F) = H p (X k (?), F) ? H p (V (?), F) = H p (X k (?), F)
To prove (c), we show inductively on j that H p (X j (?), F) = 0. For j = 0 this is clearly satisfied since X 0 (?) = X(? 0 ) and ? 0 ? A. Assume now that j ? 1 and that
H p (X j?1 (?), F) = 0. Since Y j = Y j?1 ? (Y j ? U j ), then X j (?) = X j?1 (?) ? V j (?),
where
Notes V j (?) = ? ?1 (U j ? Y j (?)) = {x ? ? ?1 (U j ) : ?o?(x) < ?, ? j o?(x) < ? 0 } is p-complete since ? ?1 (U j
) is r-complete and ?o? and ? j o? are q-convex with the same positivity directions. Furthermore, as
X j?1 (?) ? V j (?)) = X j?1 (?) ? ? ?1 (U j ) = {x ? ? ?1 (U j ) : ? j?1 o?(x) < ? 0 , ?o?(x) < ?} is clearly p-Runge in ? ?1 (U j ), then the restriction map H s (? ?1 (U j ), F) ? ? H s (X j?1 (?) ? V j (?), F)
has a dense image for all s ? p ? 1. Since ? is clearly injective and p ? 1 ? r,
then H p?1 (X j?1 (?) ? V j (?), F) = 0.
Therefore from the Mayer-Vietoris sequence for cohomology
? ? ? ? H p?1 (X j?1 (?) ? V j (?), F) ? H p (X j (?), F) ? H p (X j?1 (?), F) ? ? ? ? , we deduce that H p (X j (?), F) = 0.
To prove statement (b), it is sufficient to show that if ? j ? and ? j ? A for all j, then
H p?1 (X(? j+1 ), F) ?? H p?1 (X(? j ), F)
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.

For every pair of real numbers µ < ?, the restriction map
H p?1 (X(?), F) ? H p?1 (X(µ), F)
has a dense range.

Proof. We consider the set T of all real numbers ? such that
H p?1 (X(?), F) ? H p?1 (X(µ), F)
has a dense range for all µ ? ?.

To see that T is not empty, we choose ? 0 = min{?(y); y ? Y }.  c i ? i , with c i > 0 sufficiently small so that ? j (z) are still q-convex With the same positivity directions for 1 ? j ? k.
ported functions ? i ? C ? o (U i ), ? j ? 0, j = 1, ? ? ? , k such that ? ?1 (U i ) is r-complete
If we consider the following sets defined in the lemma 2
Y (?) = {y ? Y : ?(y) < ?}, X(?) = ? ?1 (Y (?)), Y i = {x ? Y : ? i (x) < ? 0 } and Y 0 = Y (? 0 ), then Y 0 ? Y 1 ? Y 2 ? ? ? ? ? Y k , Y 0 ?? Y k , Y i \ Y i?1 ?? U i for 1 ? i ? k and X j (?) = ? ?1 (Y j ? Y (?)) = X j?1 (?) ? V j (?)
, where
V j (?) = ? ?1 (U j ? Y j (?)) = {x ? ? ?1 (U j ) : ?o?(x) < ?, ? j o?(x) < ? 0 }
Now since X j?1 (?)?V j (?) is p-Runge in the p-complete set V j (?) and H p (X j (?), F) = 0, it follows from the long exact sequence of cohomology
? ? ? ? H p?1 (X j (?), F) ? H p?1 (X j?1 (?), F) ? H p?1 (V j (?), F) ? H p?1 (X j?1 (?) ? V j (?), F) ? H p (X j (?), F) ? ? ? ? that the restriction map H p?1 (X j (?), F) ? H p?1 (X j?1 (?), F)
has a dense range.

Moreover, since ? is exhaustive, there exists ? > 0 such that Y (? + ?) ? Y k . We deduce that the restriction map
H p?1 (X(? + ?), F) ? H p?1 (X(?), F)
has a dense image, which implies that ? + ? ? T .

Let now ? j ? T , j ? 0, such that ? j ?, and let U = (U i ) i?I be a countable base of Stein open covering of X. Then the restriction map between spaces of cocycles
Z p?1 (U| X ? j+1 , F) ? Z p?1 (U| X ? j , F)
has dense image for j ? 0. Let ? < ? and j ? N such that ? < ? j . By [1, p.246],

the restriction map Z n?2 (U| X ? , F) ? Z n?2 (U| X ? j , F) has a dense image. Since
? j ? T , then Z n?2 (U| X ? j , F) ? Z n?2 (U| X ? , F
) has also a dense image, and hence
? ? T . H p (X, F) ? H p (X(0), F)
is bijective, which shows that H p (X, F) = 0.

There exists for each integer n ? 3 a cohomologically q-complete open subset ? ? C n , 2 ? 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 ? q < n and such that n = mq + 1, where m = [ n q ] is the integral part of n q . We define the functions.
? j (z) = ? j (z) + m i=1 ? i (z) 2 + N ||z|| 4 ? 1 4 ||z|| 2 , j = 1, ? ? ? , m,and? m+1 (z) = ?? 1 (z) ? ? ? ? ? ? m (z) + m i=1 ? i (z) 2 + N ||z|| 4 ? 1 4 ||z|| 2 ,
where
? j (z) = Im(z j )+ n i=m+1 |z i | 2 ?(m+1) m+j(q?1) i=m+(j?1)(q?1)+1 |z i | 2 , for j = 1, ? ? ? , m z = (z 1 , z 2 , ? ? ? , z n ), and N > 0 a positive constant. Then, if N large enough, the functions ? 1 , ? ? ? ? 2 are q-convex on C n and, if ? = M ax(? 1 , ? ? ? , ? m+1
), then, for ? o > 0 small enough, the set D ?o = {z ? C n : ?(z) < ?? o } is relatively compact in the unit ball B = B(0, 1) if N is sufficiently large. (See [4]).

We fix some ? > ? 0 and consider a covering (U i ) i?N of ?D ? , by Stein open subsets U i ?? D ?0 and functions ? i ? C ? 0 (C n , R) such that
? j ? 0, Supp(? j ) ?? U j , k i=1 ? j (x) > 0 at any point x ? ?D ? .
We can therefore choose sufficienly small positive numbers c 1 , ? ? ? , c k so that the
functions ? i,j = ? i ? j l=1 c l ? l are q-convex for i = 1, ? ? ? , m + 1 and 1 ? j ? k. We define ? i,0 = ? i for i = 1, ? ? ? , m + 1, D 0 = D ? and D j = {z ? D ?0 : ? j (z) < ??}, where ? j (z) = ? ? j i=1 c i ? i for j = 1, ? ? ? , k.
Then ? j are q-convex with corners and it is clear that
D 0 ? D 1 ? ? ? ? ? D k , D 0 ?? D k ?? D ?0 and D j \ D j?1 ?? U j for j = 1, ? ? ? = k.
In the situation described above, for any coherent analytic sheaf F on 
D ?0 , the restriction map H p (D j+1 , F) ? H p (D j , F) is surjective for all p ? q ? 1 and all 0 ? j ? k ? 1. In particular, dim C H p (D j , F) < ?, if p ? q ? 1.

# 289.

Proof. We first prove that the cohomology group H p (D j ?U l , F) = 0 for all p ? q?1, 0 ? j ? k, and 1 ? l ? k. In fact, the set D j ? U l can be written in the form
D j ? U l = D 1 ? ? ? ? ? D m+1 , where D i = {z ? U l : ? i,j (z) < ??} are clearly q-complete. Then for every i 1 , ? ? ? , i m ? {1, ? ? ? , m + 1}, D i1 ? ? ? ? ? D im are
(q ? 1)-complete. Therefore, by using Proposition 1 of [11], we obtain
H p (D j ? U l , F) ? = H p+m (D 1 ? ? ? ? ? D m+1 , F) if p ? q ? 1, which implies that H p (D j ? U l , F) = 0 for all p ? q ? 1. Now since D j+1 = D j ?(D j+1 ?U j+1
), it follows from the Mayer-Vietoris sequence for cohomology
? H p (D j+1 , F) ? H p (D j , F) ? H p (D j+1 ? U j+1 , F) ? H p (D j ? U j+1 , F) ? H p+1 (D j+1 , F) ? that the restriction map H p (D j+1 , F) ?? H p (D j , F) is surjective when p ? q ? 1. Let now A be the set of all real numbers ? ? ? 0 such that H p (D ? , F) = 0 for all p ? q ? 1.
-The set A is not empty and, if ? ? A, ? > ? 0 , then there exits ? ? A such that ? 0 ? ? < ?.
Proof. In fact, if µ 0 = M in z?B {? i (z), i = 1, ? ? ? , m + 1}, then one sees easily that [?µ 0 , +?[? A.
For the proof of the second assertion, if with the notations of lemma 1 we set
D 0 = D ? , we obtain D 0 ? D 1 ? ? ? ? ? D k , D 0 ?? D k ?? D ?0 and D j \ D j?1 ?? U j for j = 1, ? ? ? = k.
We fix some 1 ? j ? k and 1 ? l ? k, and set
D j ? U l = D 1 ? ? ? ? D m+1 , where D i = {z ? U l : ? i,j(
z) < ??}, then D i are q-complete and q-Runge in U l . Therefore because of the proof of lemma 2, one obtains
H p (D j ? U l , F) ? = H p+m (D 1 ? ? ? ? ? D m+1 , F) = 0
for p ? q ? 1 and, consequently, the restriction map
H p (D j+1 , F) ?? H p (D j , F)
is surjective for all p ? q ? 1.

We now show inductively on j that H q?1 (D j , F) = 0. For j = 0, this is clearly satisfied since D 0 = D ? and ? ? A. Assume now that this property has already 


# Ref
been proved for j < k. Since for every i 1 , ? ? ? , i m , in {1, ? ? ? , m + 1}, the open set D i1 ? ? ? ? ? D im is (q ? 1)-Runge in U l , then the restriction map H p (U l , F) ?? H p (D im ? ? ? ? ? D im , F)
has a dense range for p ? q ? 2. Since the canonical topologies on
H i (D im ? ? ? ? ? D im , F) are obviously separated for i ? 2, then H p (D im ? ? ? ? ? D im , F) = 0 for all p ? q ? 2. We know from Proposition 1 of [11] that H p (D j ? U l , F) ? = H p+m (D 1 ? ? ? ? ? D m+1 , F) for p ? q ? 2 = n ? m ? 1. We can choose the covering (U i ) 1?i?k of ?D ? such that U l \ D 1 ? ? ? ? ? D m+1
has no compact connected components, so it follows from the mean theorem of [5], that the restriction
H p (U l , F) ?? H p (D 1 ? ? ? ? ? D m+1 , F) has a dense image for p ? n ? 1. This proves that H p (D j ? U l , F) ? = H p+m (D 1 ? ? ? ? ? D m+1 , F) = 0 for all p ? q ? 2. Now since H q?2 (D j ? U j+1 , F) = H q?1 (D j+1 ? U j+1 , F) = H q?1 (D j , F) = 0, it
follows from the Mayer-Vietoris sequence for cohomolgy
? H q?2 (D j ?U j+1 , F) ? H q?1 (D j+1 , F) ? H q?1 (D j , F)?H q?1 (D j+1 ?U j+1 , F) ? that H q?1 (D j+1 , F) = 0.
On the other hand, since ? is proper, there exists ? > 0 such that ? ? ? > ? o and
D ??? = {z ? D ?o : ?(z) < ? ? ?} ?? D k . Since H q?1 (D k , F) ? H q?1 (D ??? , F) is surjective, H q?1 (D k , F) = 0 and dim C H q?1 (D ??? , F) < ?, then H q?1 (D ? ?? , F) = 0, whence ? ? ? ? A.
The open set D ?0 is cohomologically (q ? 1)-complete.

Proof. For this, we consider the set A of all real numbers ? ? ? 0 such that H p (D ? , F) = 0 for all p ? q ? 1. Then by lemma 3, A is not empty and open in [? 0 , ?[. Moreover, if ? = Inf (A), there exists a decreasing sequence of real numbers ? j ? A, j ? 1, such that ? j ?. Since H p (D ?j , F) = 0 for p ? q ? 1 and, by lemma 1, the restriction map H p (D ?j+1 , F) ? H p (D ?j , F) is surjective for all p ? q ? 2, then by ( [3], p. 250), the restriction map
H p (D ? , F) ?? H p (D ?1 , F)
is an isomorphism for p ? q ? 1, which shows that ? ? A.

Assume now that ? > ? 0 . Then there exists, according to lemma 1, ? ? A such that ? 0 < ? < ?, which contradicts the fact that ? = Inf (A). We conclude that ? = ? 0 ? A, and hence D ?0 is cohomologically (q ? 1)-complete.

Unbranched Riemann Domains over Q-Complete Spaces We have shown that D ?0 is cohomologically (q ? 1)-complete. We are now going to prove that for a good choice of the contants ? 0 and N , we can find an ? > ? 0 such that D ? is cohomologically (q ? 1)-complete but ? not (q ? 1)-complete. In fact, it was shown by Diederich-Forness [4] that if ? > 0 is small enough, then the topological sphere of real dimension n + q ? 2
S ? = {z ? C n : x 2 1 + ? ? ? + x 2 m + |z m+1 | 2 + ? ? ? + |z n | 2 = ?, y j = ? n i=m+1 |z i | 2 + (m + 1) m+j(q?1) i=m+(j?1)(q?1)+1 |z i | 2 for j = 1, ? ? ? , m}
is not homologous to 0 in D ?0 . This follows from the fact that the set E = {z ? C n :
x 1 = z 2 = ? ? ? = z n = 0} does not intersect D ?0 , since on E ? j = y j + 3 4 m i=1 y 2 i + N ( m i=1 y 2 i ) 2 for j = 1, ? ? ? , mand? m+1 = ?y 1 ? ? ? ? ? y m + 3 4 m i=1 y 2 i + N ( m i=1 y 2 i ) 2
such that ? ? 0 on E. So the following real form of degree n + q ? 2 ? = (
n i=1 x 2 i + n i=m+1 y 2 i ) ?2n+m ( n i=1 (?1) i x i dx 1 ? ? ? ? dx i ? ? ? ? ? dx n ? dy m+1 ? ? ? ? ? dy n + n?m i=1 (?1) n+i y m+i dx 1 ? ? ? ? ? dx n ? dy m+1 ? ? ? ? ? dy m+i ? ? ? ? ? dy n )
is well-defined and d-closed on D ?0 . Since ? does not depend on y 1 , ? ? ? , y m , then by the standard argument S ? ? = 0. Therefore S ? is not homologous to 0 in D ?0 .

Let E q be the sheaf of germs of C ? q-forms on C n and T q the sheaf of germs of C ? d-closed q-forms. Then we have an exact sequence of sheaf homomorphisms 0 ? T q ? E q d ? T q+1 ? 0 Since by the de Rham theorem for every p ? 1, the cohomology group
H p (D ?0 , C) is isomorphic to {? ? Î?"(D ?0 , E p ) : df = 0} {d? : ? ? Î?"(D ?0 , E p?1 )} , it follows from Stokes formula that H n+q?2 (D ? 0 , C) does not vanish.
We are going to show that H r (D ?0 , O D? 0 ) = 0 for all r with 1 ? r ? q ? 3.

We first assert that we can choose N , ? 0 , and ? > ? 0 such that, if, with the notations of Proposition 1, we set
? j (z) = ? j (z) + m i=1 ? i (z) 2 + N ||z|| 4 ? 1 4 ||z|| 2 , j = 1, ? ? ? , m,and? m+1 (z) = ?(z) + m i=1 ? i (z) 2 + N ||z|| 4 ? 1 4 ||z|| 2 , where ?(z) = ? m i=1 ? i (z), ? j (z) = Im(z j ) + n i=m+1 |z i | 2 ? (m + 1) m+j(q?1) i=m+(j?1)(q?1)+1 |z i | 2 , for j = 1, ? ? ? , m, (z) = N ||z|| 4 ? 1 4 ||z|| 2 +? 0 and ?(z) = M ax(? 1 (z), ? ? ? , ? m+1 )+ m i=1 ? i (z) 2 + (z)?? 0 ,
then we obtain
D ? = {z ? D ? : ?(z) < ? 0 ? ?} where m = M in z?D? 0 (z),and?(z) = ?(z) + m i=1 ? i (z) 2 + m
In fact, we can choose ? > ? 0 sufficiently big and ? > 0 small enough so that
? 0 ? ? < m < (1 + ?).M in z?D? (z) and ?? ? (1 + ?)? 0 > 0.
On the other hand, if ? = M in z?D? 0 ||z|| 2 , then we have
0 < ? ? ||z|| 2 < 1 4N ? ? 0 N for every z ? D ?0
Therefore by suitable choice of ? 0 , ? and N we can also achieve that
(N ||z|| 4 ? 1 4 ||z|| 2 ) ? M in z?D? 0 (N ||z|| 4 ? 1 4 ||z|| 2 ) < M in( ? ? ? 0 2 , ?? ? (1 + ?)? 0 ),andM ax 0 (N ||z|| 4 ? 1 4 ||z|| 2 ) ? |(N ||z|| 4 ? 1 4 ||z|| 2 ) < M in( ? ? ? 0 2 , ?? ? (1 + ?)? 0 ), for every z ? D ? . Because (z) < ? 0 ? ? on D ? , then clearly we obtain ?(z) = ?(z)+ m i=1 ? i (z) 2 +m < ?(z)+ m i=1 ? i (z) 2 +(1+?).?(z)) < (1+?)(? 0 ??), if z ? D ? , which shows that D ? = {z ? D ? : ?(z) < ? 0 ? ?}
We are now going to show that for every none-positive real number ? with ? < ? 0 ? ?, the open sets
B ? = {z ? D ? : ?(z) < ?} are relatively compact in D ? .
To see this, we consider a sequence (z j ) j?0 ? B ? , which converges to a point z ? D ? . Then one has for every sufficiently large integer j
?(z j ) = M ax(? 1 (z j ), ? ? ? , ? m (z j ), ?(z j )) + m i=1 ? i (z j ) 2 + N ||z j || 4 ? 1 4 ||z j || 2 < ?? Since ?(z j ) < ? 0 ? ? + ??(z j ) < (1 + ?)(? 0 ? ?) Issue ersion I V V ( F )
We deduce that the map
H q?2 (D ? , ? n ) ? ? H n+q?2 (D ? , C)
is surjective. The map ? is defined as follows : If a differential form ? ? C ? n,q?2 (D ? ) satisfies the equation ?? = 0, then ? is also d-closed and therefore defines a cohomology class in H n+q?2 (D ? , C). 
1![Y. Alaoui, On the cohomology groups of unbranched Riemann domains over Stein spaces. To appear in Rendiconti del Seminario Matematico. Politecnico di Torino](image-2.png "Ref 1 .")
![then by theorem 1 of[13], the restriction map III. Unbrunched Riemann Domains Over Q-Complete Spaces Theorem 1.](image-3.png "")
3![A. Andreotti and H. Grauert, Théorèmes de finitude de la cohomologie des espaces complexes. Bull. Soc. Math. France 90 (1962;) 193 -](image-4.png "Ref 3 .")
![Then clearly ] ? ?, ? 0 ] ? T . Unbranched Riemann Domains over Q-Complete Spaces To prove that T is open in ] ? ?, +?[ it is, therefore, sufficient to show that if ? ? T , there exists ? > 0 such that ? + ? ? T . For this, we consider a finite covering (U i ) 1?i?k of {y ? Y : ?(z) = ?} by Stein open sets U i ?? Y and compactly sup-](image-5.png "")
i=1![i (x) > 0 at any point of ?Y (?). Define Y j = {z ? Y : ? j (z) < ?} where Lemma 1.Ref 3. A. Andreotti and H. Grauert, Théorèmes de finitude de la cohomologie des espaces complexes. Bull. Soc. Math. France 90 (1962;) 193 -259. ? j (z) = ?(z) ? j 1](image-6.png "and k i=1 ?")
4![End of the proof of theorem 2Ref 11. K. Matsumoto. On the cohomological completeness of qcomplete domains with corners. Nagoya Math. J. Vol. 165 (2002), 105-112.](image-7.png "Lemma 4 .")
42122![Moreover, since, by theorem 1 in[8], every d-closed differential form ? ? C ? n,q?2 (D ? ) Ref 7. H. Diederich, J. E. Fornaess, Smoothing q-convex functions and vanishing theorems. Invent. Math. 82.291-305 (1985).where? 1 (z) = Im(z 1 ) + n i=3 |z i | 2 ? |z 2 | 2 , z = (z 1 , z 2 , ? ? ? , z n ),and N > 0 a positive constant. Then, if N is large enough, the functions ? 1 and ? 2 are (n ? 1)-convex onC n and, if ? = M ax(? 1 , ? 2 ), then, for ? o > 0 small enough, the set D ?o = {z ? C n : ?(z) < ?? 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 ? 0 > 0 such that if ? = M in z?D? 0 ||z|| 2 , then we have 0 < ? ? ||z|| 2 < 1 4N ? ? 0 N for every z ? D ?0and that by a suitable choice of ? > ? 0 ,D ? = {z ? C n : ?(z) < ??}is cohomologically (n ? 1)-complete but not (n ? 1)-complete.Now if we suppose that at a boundary point z0 ? ?D ? , we have ? 1 (z 0 ) = ? 2 (z 0 ), then ? 1 (z 0 ) = 0 and, hence N |z 0 | 4 ? |z0| 2 ?. This |z 0 | 2 = 1 8N (1 + ? 1 + 64N ? < 1 4N . Therefore 1 64N ? < 1 2, which is a contradiction. This implies that ? 1 (z) = ? 2 (z) at every boundary point z ? ?D ? . We conclude that with such a choice of ? 0 , N and ?, D ? is obviously locally (n ? 1)-complete inC n .Unbranched Riemann Domains over Q-Complete SpacesThere exists for each integer n ? 3 a cohomologically (n?1)-completeopen subset ? of C n which is locally (n ? 1)-complete in C n but ? is not (n ? 1)complete.Proof. We consider for n ? 3 the functions ? 1 , ? 2 : C n ? R defined by? 1 (z) = ? 1 (z) + ? 1 (z) 2 + N ||z|| 4 (z) = ?? 1 (z) + ? 1 (z) 2 + N ||z|| 4 ? 1 4 ||z|| 2 ,](image-8.png "4 = 2 ? 1 +? 1 4 ||z|| 2 , ? 2")
Unbranched Riemann Domains over Q-Complete SpacesYear 20221 16ersionIssueVolume XXIIFrontier Researchof ScienceGlobal Journal© 2022 Global Journals
		
		
			

then ?(z j ) = ?(z j ) + N (||z j || 4 ? 1 4 ||z j || 2 ) ? m < ? 0 ? ? + ??(z j ) + ?? ? (1 + ?)? 0

A passage to the limit shows that

because (z) < ? 0 ? ?, which implies that z ? D ? . We conclude that with such a choice of ? 0 , N , and ? the limit z ? D ? , and hence the open set

is relatively compact in D ? for all real numbers ?, with ? < ? 0 ? ?. Now since ? is in addition (m + 2)-convex, then a similar proof of theorem 15 of [3] shows that, if ? i is the sheaf of germs of holomorphic i-forms on

In fact, let c 0 = M ax z?D? ?(z). Then there exists

Now if we suppose that D ? is (q ? 1)-complete, then there exists a C ? strictly (q ? 1)-convex function :

We now consider the resolution of the constant sheaf 

Since, by Proposition 1, D ? is cohomologically (q ? 1)-complete, then H r (D ? , ? i ) = 0 for all r ? q ? 1 and i ? 0. So we obtain the isomorphisms

and the exact sequence is cohomologous to a ?-closed (n, q ? 2) differntial form ? ? C ? n,q?2 (D ? ), it follows that the map

Notice that for the given ?, if ? > 0 is small enough, the topological sphere

Since is exhaustive on D ? , there exists c > 0 such that S ? is not homologuous to 0 in D ?,c . Let c > c . Then D ?,c and D ?,c are (q ? 1)-complete and, similarly

Moreover, since the levi form of has at least m + 1 strictly positive eingenvalues, then by using Morse theory (See for instance [7] ) we find that

It follows from the commutative diagram of continuous maps

, the function being exhaustive on D ? , then, according to theorem 11 of [1], one obtains

Furthermore, since D ?,c is cohomologically (q?1)-complete and H r (D ? , O D? ) = 0 for 1 ? r ? n ? m ? 2, it follows from theorem 1 of [6] that D ?,c is Stein, which is in contradiction with the fact that H n+q?2 (D ?,c , C) = 0, since S ? ? D ?,c is not homologous to 0 in D ?,c . We conclude that D ? is cohomologically (q ? 1)-complete but not (q ? 1)-complete.
			
			

				


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