Article
Kyungpook Mathematical Journal 2016; 56(2): 541-575
Published online June 1, 2016 https://doi.org/10.5666/KMJ.2016.56.2.541
Copyright © Kyungpook Mathematical Journal.
Jacobi Operators with Respect to the Reeb Vector Fields on Real Hypersurfaces in a Nonat Complex Space Form
U-Hang Ki1, Soo Jin Kim2, Hiroyuki Kurihara3
The National Academy of Siences, Seoul 06579, Korea1
Department of Mathematics Chosun University, Gwangju 61452, Korea2
The College of Education, Ibaraki University, Mito 310-8512, Japan3
Received: August 18, 2015; Accepted: January 19, 2016
Abstract
Let
Keywords: complex space form, real hypersurface, structure Jacobi operator, structure tensor, Ricci tensor.
1. Introduction
A complex
where
The Reeb vector field
Theorem 1.1. (Ki, Kim and Lim [5])
(I)
In cases that M n (c ) =P n ℂwith η (Aξ ) ≠= 0,(
A 1)a geodesic hypersphere of radius r, where 0 <r <π /2and r ≠=π /4; (
A 2)a tube of radius r over a totally geodesic P k ℂfor some k ∈ {1, . . . ,n –2},where 0 <r <π /2and r ≠=π /4.
(II)
In cases M n (c ) =H n ℂ,(
A 0)a horosphere; (
A 1)a geodesic hypersphere or a tube over a complex hyperbolic hyperplane H n –1ℂ; (
A 2)a tube over a totally geodesic H k ℂfor some k ∈ {1, . . . ,n – 2}.
Theorem 1.2. (Ki, Nagai and Takagi [9])
In [6], the authors started the study on real hypersurfaces in a complex space form with
All manifolds in this paper are assumed to be connected and of class
2. Preliminaries
Let
where
We call
A real hypersurface
for any vector fields
for any vector fields
Since we consider that the ambient space is of constant holomorphic sectional curvature 4
for any vector fields
In what follows, to write our formulas in convention forms, we denote by
From the Gauss
for any vector field
Now, we put
where
and hence
which shows that
Making use of (
because
Now, differentiating (
which together with (
Applying (
which connected to (
Using (
for any vector field
From (
Let Ω be the open subset of
At each point of Ω, the Reeb vector field
In what follows we assume that Ω is not an empty set in order to prove our main theorem by reductio ad absurdum, unless otherwise stated, all discussion concerns the set Ω.
3. Real Hypersurfaces Satisfying R ξ φ = φR ξ
Let
Then, using (
Replacing
which implies
because
which together with (
Using (
Since
where a 1-form
Differentiating (
because of (
Interchanging
or, using (
Combining this to (
If we put
where
4. Real Hypersurfaces Satisfying R ξ φ = φR ξ and ∇φ ∇ξ ξ R ξ = 0
We will continue our discussions under the same hypothesis
by virtue of ∇
because
which tells us that
Since
Combining (
If we apply
Using (
for any vector field
Now, if we put
Also, if we take the inner product (
which together with (
If we take the inner product (
Using (
Putting
Combining (
If we apply
which together with (
Now, we can take a orthonormal frame field {
which implies
Taking the inner product with
Putting
Putting
or, using (
Now, putting
which together with (
Because of (
Using (
Using above two
where we have used (
Replacing
Since
which implies that
5. The Exterior Derivative of 1-form u
We will continue our discussions under the hypotheses as those stated in Section 4.
Putting
Because of (
Using this and (
Substituting (
Combining this to (
If we put
or, using (
By the way, since
Using this, above equation is reduced to
If we take the inner product (
Also, taking the inner product (
On the other hand, replacing
or using (
By the way, applying (
Substituting this into (
On the other hand, (
If we apply by
Now, if we put
Using (
or using (
which together with (
Putting
where we have used (
Thus, it follows that
which together with (
We notice here that
Remark 5.1
In fact, because of the hypothesis
which implies that
Now, we prepare the following lemma for later use.
Lemma 5.2
Since (
From (
Now, suppose that
From this and (
on Ω. So we have
From (
Differentiating (
Taking the inner product with
Using (
On the other hand, (
Taking the inner product with
Hence, it follows from (
which shows that for any vector fields
Differentiating this covariantly and using (
Taking the skew-symmtric part of this, we find
Replacing
Substituting this into (
Putting
On the other hand, putting
Combining this to (
If we put
which, together with
Thus, (
which implies
Using (
Now, suppose that
which implies
Accordingly we see that
Taking the inner product this to
The skew-symmetric part of this is given by
which implies that ∇
If
Lemma 5.3
Putting
or using (
On the other hand, combining (
Substituting this and (
From (
By the way, using (
Thus, we have
Differentiating (
Using this and (
where we have put
This completes the proof of Lemma 5.3.
6. Lemmas
We will continue our discussions under the same hypotheses as those in Section 4. Further we assume that Tr
where
So we have
which tells us that
Combining the last two equations, it follows that
Applying this by
Taking the inner product with
First of all, we prove the following:
Lemma 6.1
If not, then we have from (
on this subset. Because of Remark 5.1 and Lemma 5.2, it is verified that
which together with
for any vector field
where we have used the Codazzi
where we have used (
On the other hand, we can write (
where we have used (
Combining the last three equations, it is seen that
which shows that (2
If we combine (
If we apply this by
From (
by virtue of Lemma 6.1, which together with (
From (
Because of Lemma 6.1, (
with the aid of (
which together with (
In the next place, we will prove that
Lemma 6.2
Differentiating (
Replacing
By the way, using (
Substituting this and (
On the other hand, we have
which implies
where we have used (
By the way, putting
which implies
because of (
because of
Now, suppose that
Because of Lemma 6.2, we can write (
which tells us that
where the function
We also have from (
where we have put
Remark 6.3
If not, then we have
because of Lemma 5.2. Combining these two equations, we obtain
Suppose that 2
Lemma 6.4
(2
From (
Using the same method as that used to derive (
where, we have used (
If (2
We discuss our arguments on such a place. Using (
Applying this by
Combining this to (
Because of Remark 5.1 and Lemma 5.2, we conclude that
which together with (
Lemma 6.5
Putting
which enables us to obtain
for some
which shows that
for some
for some
By the way, applying (
Because of (
Combining the last three equations, we obtain
for some
Remark 6.6
In fact, we have
which together with (
However, if we take the inner product with
7. Proof of the Main Theorem
We will continue our discussions under the same assumptions as those in Section 6. Taking the inner product
Taking the skew-symmetric part of this and using (
where we have used (
or using (
If we put
By the way, putting
for some
It follows from this and (
If we take account of (
On the other hand, from (
where we have used (
Combining this to (
which together with (
that is,
According to Lemma 5.2, we see that
From this fact and Lemma 6.4, we see that 2
with the aid of Remark 6.3, where we have put
Differentiating this covariantly and taking the inner product with
As in the same method as that used from (
which together with (
On the other hand, applying (
which together with (
From Lemma 6.1 we have
By the way, applying (
which together with (
Owing to Lemma 5.2, we see that 2
by virtue of 2
According to Remark 5.1, it follows that
Therefore we verify that Ω = ∅︀, that is,
Let
Consequently we conclude that
Theorem 7.1
(I)
In cases that M n (c ) =P n ℂwith η (Aξ ) ≠= 0,(
A 1)a geodesic hypersphere of radius r, where 0 <r <π /2and r ≠=π /4; (
A 2)a tube of radius r over a totally geodesic P k ℂfor some k ∈ {1, . . . ,n –2},where 0 <r <π /2and r ≠=π /4.
(II)
In cases M n (c ) =H n ℂ,(
A 0)a horosphere; (
A 1)a geodesic hypersphere or a tube over a complex hyperbolic hyperplane H n –1ℂ; (
A 2)a tube over a totally geodesic H k ℂfor some k ∈ {1, . . . ,n – 2}.
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