A Characterization of Complex Projective Spaces by Sections of Line Bundles ()
1. Introduction
Kobayashi and Ochiai [1] have given Characterizations of the complex projective spaces. Kobayashi-Ochiai Theorem [1] has been applied to obtain many important characterizations of the projective spaces, such as the proof of Frankel conjectures [2] , the proof of Hartshorne conjecture [3] , and many others [4] -[7] . In this note, we want to give a characterization of the complex projective spaces via sections of line bundles.
Results which can be found in [1] [8] and [9] are used freely often without explicit references. Let M be a complex space with a line bundle L.
is the sheaf of germs of sheaf of holomorphic functions,
is the sheaf of germs of holomorphic sections of a line bundle L.
means
.
2. Characterization of the Projective Spaces
In this paper, a characterization of the projective space will be given.
Definition. Let M be a compact complex space with a line bundle L. M is said to be completely intersected with respect to a line bundle L, provided that complex subspace
is irreducible for any linearly independent elements of
of
, where each
is irreducible, and
is the common zeros of
.
From the Lemma 1.1 [1] and the proof of theorem 16.2.1 [8] , we have
Lemma 1. Let V be a compact irreducible complex space. Let F and L be line bundles over V. Let
be an irreducible section of L and put
. The following sequence of sheaf homomorphisms is exact:

where
is the multiplication by
,
is the sheaf defined by
and
is the restriction map.
Lemma 2. Let M be a n-dimensional compact complex space with a line bundle L. Let
be linear independent elements of
, such that each
is irreducible. If M is completely intersected with respect to L, then there is an exact sequence:
![]()
where
is the subspace of
spanned by the sections
and
is the restriction map.
Proof. The proof is by induction on k. The case k = 0 is trivial. Since M is completely intersected with respect to L,
and
are irreducible. Assume the lemma for
, we have the exact sequence:
![]()
If
, then
from which it follows that
are linearly dependent, a contradiction. Thus,
is nontrivial on
; it follows that
defined as the set of zeros of =
on
is an irreducible divisor.
We apply Lemma l to
,
,
. Then,
. The exact sequence in Lemma 1 induces the following exact sequence
![]()
This means that the kernel of the restriction map
is spanned by the restriction of
to
. Combining this with the lemma for
, we obtain the lemma for k.
Now we give the main result of this paper.
Theorem. Let M be a n-dimensional compact irreducible complex space with a line bundle L. If
and M is completely intersected with respect to L, then M is biholomorphic to a complex space
of dimension n.
Proof. Since
, we can choose linearly independent sections
from
with each
irreducible. Put
.
Claim 1. Each
is an irreducible divisor of M.
First of all,
is nonempty. Indeed, if
for some i, then
for all
. Define map
by
. It is clear that h is isomorphic, that is, L is a trivial line bundle over M. It follows that
, which is contradictory to the hypothesis that
. Thus, each
. Since
is irreducible,
is irreducible and
by [Corollary 14, Ch II, 3] and [Theorem 11, Ch III, 3]. Hence,
is an irreducible divisor.
Let
be the common zeros of
, then
. By the hypothesis, M is completely intersected with respect to L; each
is irreducible.
Claim 2.
.
For
,
is an irreducible divisor by Claim 1, thus
. Assume that
. If
, then
by Lemma 2. This induces that
are linearly dependent, a contradiction. Thus,
is nontrivial on the irreducible complex subspace
. It follows by [Theorem 14, Ch. III, 3] that
.
Claim 3.
is base point free.
By Claim 2,
is an irreducible complex subspace of dimension
. In particular,
is one point. If
vanishes at
, then
are linearly dependent by Lemma 2. Since
are defined as linearly independent sections in the beginning of this proof, it is a contradiction. Thus,
does not vanish at
. This shows that
have no common zeros, that is,
is base point free.
Since
, we may let
be the complex projective space of dimension n defined as the set of hyperplanes through the origin in
. For any point
, put
. Since
is base point free,
is a hyperplane through the origin in
and so
. We now obtain a holomorphic mapping
.
Claim 4. The mapping
is bijective.
Giving a point y of
, it is a hyperplane through the origin in
. Let
be a basis for this hyperplane with each
irreducible. Then, y is the complex subspace spanned by
, that is,
. Let
be the set of zeros of
for
. Since M is completely intersected with respect to L,
is irreducible and
by Claim 2, that is,
is a point. It follows that there exists a point x of M, such that
.
Thus,
. By Lemma 2, we have the exact sequence
.
For any
, we have
. Since
and the above sequence is exact, it follows that
and so
. Thus,
is surjective.
On the other hand, let u and v be any two points of M.
is a hyperplane through the origin in
. Let
be a basis for
with each
irreducible. By Claim 2,
is a single point and this induces that
. If
, then
. It follows that
, that is,
. Thus
is injective.
Consequently, we have shown that M is biholomorphic to a complex projective space
of dimension n.
As an application of the theorem above, we give a proof for the famous Kobayashi-Ochiai Theorem [1] .
Corollary ([Theorem 1.1 [1] ). Let M be a n-dimensional complex irreducible complex space with an ample line bundle F. If
and
, then M is biholomorphic to a complex projective space
of dimension n.
Proof. By the theorem in this paper, it suffices to show that M is completely intersected with respect to F. Let
be linearly independent elements of
with each
irreducible. Let
be the common zeros of
on M.
Assume that
is reducible, then write
for some nontrivial
and
. Because
are linearly independent elements of
with each
irreducible, there are no common zeros of
on
. And due to the definition of
, we have
![]()
By the hypothesis,
, and so
.
Since F is ample,
and
are positive integers. Thus,
, a contradiction. It follows that
is irreducible. By the theorem above, M is biholomorphic to
.
NOTES
*Corresponding author.