Complete Hypersurfaces in a Real Space Form
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irk-123456789-1065082016-10-06T00:28:12Z Complete Hypersurfaces in a Real Space Form Shu, Sh. 2008 Article Complete Hypersurfaces in a Real Space Form / Sh. Shu // Журнал математической физики, анализа, геометрии. — 2008. — Т. 4, № 2. — С. 294-304. — Бібліогр.: 12 назв. — англ. 1812-9471 http://dspace.nbuv.gov.ua/handle/123456789/106508 en Журнал математической физики, анализа, геометрии Фізико-технічний інститут низьких температур ім. Б.І. Вєркіна НАН України |
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Shu, Sh. Complete Hypersurfaces in a Real Space Form Журнал математической физики, анализа, геометрии |
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Shu, Sh. |
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Shu, Sh. |
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Complete Hypersurfaces in a Real Space Form |
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Complete Hypersurfaces in a Real Space Form |
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Complete Hypersurfaces in a Real Space Form |
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Complete Hypersurfaces in a Real Space Form |
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Complete Hypersurfaces in a Real Space Form |
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complete hypersurfaces in a real space form |
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Фізико-технічний інститут низьких температур ім. Б.І. Вєркіна НАН України |
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Complete Hypersurfaces in a Real Space Form / Sh. Shu // Журнал математической физики, анализа, геометрии. — 2008. — Т. 4, № 2. — С. 294-304. — Бібліогр.: 12 назв. — англ. |
| series |
Журнал математической физики, анализа, геометрии |
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AT shush completehypersurfacesinarealspaceform |
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2025-07-07T18:35:03Z |
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2025-07-07T18:35:03Z |
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1837014244877402112 |
| fulltext |
Journal of Mathematical Physics, Analysis, Geometry
2008, vol. 4, No. 2, pp. 294�304
Complete Hypersurfaces in a Real Space Form
Shu Shichang
Department of Mathematics, Xianyang Teachers University
Xianyang, 712000, Shaanxi, Peoples Republic of China
E-mail:shushichang@126.com
Received February 24, 2007
Let Mn be an n-dimensional complete hypersurface with the scalar cur-
vature n(n � 1)R and the mean curvature H being linearly related, that
is, n(n � 1)R = k0H(k0 > 0) in a real space form Rn+1(c). Assume that
the mean curvature is positive and obtains its maximum on Mn. We show
that (1) if c = 1; k0 � 2n
p
n(n� 1), for any i,
P
j 6=i
�2
j
> n(n � 1) and
jhj2 � nH2 + (B+
H
)2, then Mn is totally umbilical, or (i) n � 3;Mn is lo-
cally an H(r)-torus with r2 < n�1
n
, (ii) n = 2;Mn is locally an H(r)-torus
with r2 6= n�1
n
; (2) if c = 0 and jhj2 � nH2+( eB+
H
)2, thenMn is isometric to
a standard round sphere, a hyperplane Rn or Sn�1(c1) � R1; (3) if c = �1
and jhj2 � nH2 + ( bB+
H
)2, then Mn is totally umbilical or is isometric to
Sn�1(r) �H1(�1=(r2 + 1)) for some r > 0, where jhj2 denotes the squared
norm of the second fundamental form of Mn; B+
H
; eB+
H
and bB+
H
are denoted
by (1.1), (1.2) and (1.3).
Key words: hypersurface, mean curvature, scalar curvature, real space
form.
Mathematics Subject Classi�cation 2000: 53C40 (primary); 53C20
(secondary).
1. Introduction
Let Rn+1(c) be an (n+ 1)-dimensional connected Riemannian manifold with
constant sectional curvature c. We also call it a real space form. According to
c > 0, c = 0 or c < 0, it is called sphere space, Euclidean space or hyperbolic
space, respectively, and it is denoted by Sn+1(c); Rn+1 and Hn+1(c). As it is well-
known that there are many rigidity results for hypersurfaces with constant mean
curvature or with constant scalar curvature in Sn+1(c); Rn+1 and Hn+1(c), for
example, see [1�3, 5] and [12] etc., but fewer ones are obtained for hypersurfaces
This work is supported in part by the Natural Science Foundation of China and NSF of
Shaanxi, China.
c
Shu Shichang, 2008
Complete Hypersurfaces in a Real Space Form
with the scalar curvature and the mean curvature being linearly related. We know
that an H(r)-torus in a unit sphere Sn+1(1) is the product immersion Sn�1(r)�
S1(
p
1� r2) ,! Rn�R2, where Sn�1(r) � Rn; S1(
p
1� r2) � R2, 0 < r < 1, are
standard immersions. In some orientation, H(r)-torus has principal curvatures
given by �1 = � � � = �n�1 =
p
1�r2
r
and �n = � rp
1�r2
.
In [12], the authors obtained the following:
Theorem 1.1. Let Mn be an n-dimensional complete hypersurface with con-
stant mean curvature H in a unit sphere Sn+1(1). (1) If jhj2 < D0(n;H), then
Mn is totally umbilical. (2) If jhj2 = D0(n;H), then (i) when H = 0;Mn is
locally a Cli�ord torus; (ii) when H 6= 0; n � 3;Mn is locally an H(r)-torus with
r2 < n�1
n
; (iii) when H 6= 0; n = 2;Mn is locally an H(r)-torus with r2 6= n�1
n
,
where
D0(n;H) = n+
n3H2
2(n� 1)
� (n� 2)nH
2(n� 1)
[n2H2 + 4(n� 1)]
1
2 :
In [6], S.Y. Cheng and S.T. Yau obtained the following:
Theorem 1.2. Let Mn be a complete hypersurface with constant mean cur-
vature in Rn+1. If the sectional curvature of Mn is nonnegative, then Mn is
isometric to a standard round sphere, a hyperplane Rn or a Riemannian product
Sk(c1)�Rn�k, 1 � k � n� 1.
In this paper, we study the hypersurfaces in a real space form Rn+1(c) with
scalar curvature n(n � 1)R and the mean curvature H being linearly related.
We obtain the following theorems:
Theorem 1.3. Let Mn be an n-dimensional complete hypersurface with n(n�
1)R = k0H in a unit sphere Sn+1(1), where k0(� 2n
p
n(n� 1)) is a positive
constant. Assume that the mean curvature H is positive and obtains its maximum
on Mn and for any i,
P
j 6=i �
2
j
> n(n� 1), where �j(j = 1; : : : ; i� 1; i+1; : : : ; n)
are the principal curvatures onMn. If the squared norm of the second fundamental
form jhj2 satis�es
jhj2 � nH2 + (B+
H
)2
on Mn, then Mn is totally umbilical, or (i) n � 3;Mn is locally an H(r)-torus
with r2 < n�1
n
; (ii) n = 2;Mn is locally an H(r)-torus with r2 6= n�1
n
, where
B+
H
= �1
2
(n� 2)
r
n
n� 1
H +
s
n3H2
4(n� 1)
+ n: (1:1)
Theorem 1.4. Let Mn be an n-dimensional complete hypersurface with n(n�
1)R = k0H in a Euclidean space Rn+1, where k0 is a positive constant. Assume
Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 2 295
Shu Shichang
that the mean curvature H is positive and obtains its maximum on Mn. If the
squared norm of the second fundamental form jhj2 satis�es
jhj2 � nH2 + ( eB+
H
)2
on Mn, then Mn is isometric to a standard round sphere, a hyperplane Rn or
a Riemannian product Sn�1(c1)�R1, where
eB+
H
=
r
n
n� 1
H: (1:2)
Theorem 1.5. Let Mn be an n-dimensional complete hypersurface with n(n�
1)R = k0H in a hyperbolic space Hn+1(�1), where k0 is a positive constant.
Assume that the mean curvature H is positive and obtains its maximum on Mn.
If the squared norm of the second fundamental form jhj2 satis�es
jhj2 � nH2 + ( bB+
H
)2
on Mn, then Mn is totally umbilical or is isometric to Sn�1(r)�H1(�1=(r2+1))
for some r > 0, where
bB+
H
= �1
2
(n� 2)
r
n
n� 1
H +
s
n3H2
4(n� 1)
� n; (n2H2 � 4(n� 1)): (1:3)
2. Preliminaries
Let Mn be an n-dimensional hypersurface in Rn+1(c). For any p 2 Mn
we choose a local orthonormal frame e1; : : : ; en; en+1 in Rn+1(c) around p such
that e1; : : : ; en are tangential to Mn. Take the corresponding dual co-frame
f!1; : : : ; !n; !n+1g. In this paper we make the following convention on the range
of indices,
1 � A;B;C � � � � n+ 1; 1 � i; j; k; � � � � n:
The structure equations of Rn+1(c) are
d!A =
P
B
!AB ^ !B ; !AB = �!BA;
d!AB =
P
c
!AC ^ !CB � c!A ^ !B:
If we denote by the same letters the restrictions of !A; !AB to Mn, we have
d!i =
X
j
!ij ^ !j; !ij = �!ji; (2:1)
296 Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 2
Complete Hypersurfaces in a Real Space Form
d!ij =
X
k
!ik ^ !kj �
1
2
X
k;l
Rijkl!k ^ !l; (2:2)
where Rijkl is the curvature tensor of the induced metric on Mn.
Restricted to Mn; !n+1 = 0, thus
0 = d!n+1 =
X
i
!n+1i ^ !i; (2:3)
and by Cartan's lemma we can write
!in+1 =
X
j
hij!j; hij = hji: (2:4)
The quadratic form h =
P
i;j
hij!i
!j is the second fundamental form of Mn.
The Gauss equation is
Rijkl = c(ÆikÆjl � ÆilÆjk) + hikhjl � hilhjk; (2:5)
n(n� 1)R = n(n� 1)c+ n2H2 � jhj2; (2:6)
where R is the normalized scalar curvature, H = (1=n)
P
i
hii the mean curvature
and jhj2 =
P
i;j
h2
ij
the squared norm of the second fundamental form of Mn,
respectively.
The Codazzi equation and Ricci identity are
hijk = hikj ; (2:7)
hijkl � hijlk =
X
m
hmjRmikl +
X
m
himRmjkl; (2:8)
where the �rst and the second covariant derivatives of the second fundamental
form are de�ned byX
k
hijk!k = dhij +
X
k
hkj!ki +
X
k
hik!kj; (2:9)
X
l
hijkl!l = dhijk +
X
m
hmjk!mi +
X
m
himk!mj +
X
m
hijm!mk: (2:10)
In order to represent our theorems, we need some notations, for details see
H.B. Lawson [9] and P.J. Ryan [11]. First we give a description of the real hyper-
bolic space Hn+1(c) of constant curvature c(< 0).
For any two vectors x and y in Rn+2, we set
g(x; y) = x1y1 + : : :+ xn+1yn+1 � xn+2yn+2;
Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 2 297
Shu Shichang
(Rn+2; g) is the so-called Minkowski space-time. Denote � =
p
�1=c. We de�ne
Hn+1(c) = fx 2 Rn+2 j g(x; x) = ��2; xn+2 > 0g:
Then Hn+1(c) is a simply-connected hypersurface of Rn+2. Hence, we obtain
a model of a real hyperbolic space.
We de�ne
M1 = fx 2 Hn+1(c) j x1 = 0g;
M2 = fx 2 Hn+1(c) j x1 = r > 0g;
M3 = fx 2 Hn+1(c) j xn+2 = xn+1 + �g;
M4 = fx 2 Hn+1(c) j x21 + : : :+ x2
n+1 = r2 > 0g;
M5 = fx 2 Hn+1(c) j x21 + : : :+ x2
k+1 = r2 > 0;
x2
k+2 + : : :+ x2
n+1 � x2
n+2 = ��2 � r2g:
M1; : : : ;M5 are often called the standard examples of complete hypersurfaces in
Hn+1(c) with at most two distinct constant principal curvatures. It is obvious
that M1; : : : ;M4 are totally umbilical. In the sense of Chen [7], they are called
the hyperspheres of Hn+1(c). M3 is called the horosphere and M4 � the geodesic
distance sphere of Hn+1(c). P.J. Ryan [11] obtained the following:
Lemma 2.1([11]). Let Mn be a complete hypersurface in Hn+1(c). Suppose
that, under a suitable choice of a local orthonormal tangent frame �eld of TMn,
the shape operator over TMn is expressed as a matrix A. If Mn has at most two
distinct constant principal curvatures, then it is congruent to one of the following:
(1) M1. In this case, A = 0, andM1 is totally geodesic. Hence M1 is isometric
to Hn(c).
(2) M2. In this case, A = 1=�2p
1=�2+1=r2
In, where In denotes the identity matrix
of degree n, and M2 is isometric to Hn(�1=(r2 + �2)).
(3) M3. In this case, A = 1
�
In, and M3 is isometric to a Euclidean space Rn.
(4) M4. In this case, A =
p
1=r2 + 1=�2In;M4 is isometric to a round sphere
Sn(r) of radius r.
(5) M5. In this case, A = �Ik � �In�k, where � =
p
1=�2 + 1=r2, and
� = 1=�2p
1=r2+1=�2
;M5 is isometric to Sk(r)�Hn�k(�1=(r2 + �2)).
We also need the following algebraic Lemma due to [10] and [1].
Lemma 2.2([10, 1]). Let �i; i = 1; : : : ; n be real numbers, with
P
i
�i = 0
and
P
i
�2
i
= �2 � 0. Then
� n� 2p
n(n� 1)
�3 �
X
i
�3i �
n� 2p
n(n� 1)
�3; (2:11)
298 Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 2
Complete Hypersurfaces in a Real Space Form
and equality holds if and only if either (n � 1) of the numbers �i are equal to
�=
p
n(n� 1) or (n� 1) of the numbers �i are equal to ��=
p
n(n� 1).
3. Proof of Theorems
In order to prove our theorems, we introduce an operator 2 due to S.Y. Cheng
and S.T. Yau [5] by
2f =
X
i;j
(nHÆij � hij)fij; (3:1)
where f is a C2-function on Mn, the gradient and Hessian (fij) are de�ned by
df =
X
i
fi!i;
X
j
fij!j = dfi +
X
j
fj!ji: (3:2)
The Laplacian of f is de�ned by �f =
P
i
fii.
We choose a local frame �eld e1; : : : ; en at each point of Mn, such that hij =
�iÆij . From (3.1) and (2.6), we have
2(nH) = nH�(nH)�
P
i
�i(nH)ii
=
1
2
�(nH)2 �
P
i
(nH)2
i
�
P
i
�i(nH)ii
=
1
2
n(n� 1)�R+
1
2
�jhj2 � n2jrHj2 �
P
i
�i(nH)ii: (3.3)
From (2.7) and (2.8), by a standard and direct calculation, we have
1
2
�jhj2 =
X
i;j;k
h2ijk +
X
i
�i(nH)ii +
1
2
X
i;j
Rijij(�i � �j)
2; (3:4)
where Rijij = c + �i�j(i 6= j) denotes the sectional curvature of the section
spanned by fei; ejg:
From (3.3) and (3.4), we get
2(nH) =
1
2
n(n� 1)�R+ jrhj2 � n2jrHj2 + 1
2
X
i;j
(c+ �i�j)(�i � �j)
2: (3:5)
By making use of the similar method in [4], we can prove the following:
Proposition 3.1. Let Mn be an n-dimensional hypersurface in a real space
form Rn+1(c) with n(n� 1)R = k0H, k0 = constant > 0. Assume that the mean
curvature H > 0. Then we have the operator
L = 2� (k0=2n)�
Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 2 299
Shu Shichang
(1) if c > 0 and for any i,
P
j 6=i �
2
j
> n(n� 1)c; L is elliptic;
(2) if c � 0; L is elliptic.
P r o o f. We choose an orthonormal frame �eld fejg at each point in Mn so
that hij = �iÆij. For any i,
(nH � �i � k0=2n) =
P
j
�j � �i � (1=2)[�
P
j
�2
j
+ n2H2 + n(n� 1)c]=(nH)
= [(
P
j
�j)
2 � �i
P
j
�j � (1=2)
P
l 6=j
�l�j � (1=2)n(n� 1)c](nH)�1
= [
P
j
�2
j
+ (1=2)
P
l 6=j
�l�j � �i
P
j
�j � (1=2)n(n � 1)c](nH)�1
= [
P
i6=j
�2
j
+ (1=2)
P
l6=j
l;j 6=i
�l�j � (1=2)n(n � 1)c](nH)�1
= (1=2)[
P
j 6=i
�2
j
+ (
P
j 6=i
�j)
2 � n(n� 1)c](nH)�1: (3.6)
(1) If c > 0 and for any i,
P
j 6=i
�2
j
> n(n� 1)c; from (3.6), we have
(nH � �i � k0=2n) � (1=2)[
X
j 6=i
�2j � n(n� 1)c](nH)�1 > 0:
Therefore, we know that L is an elliptic operator.
(2) If c � 0, from (3.6) again, we have
(nH � �i � k0=2n) > 0:
Therefore, we also know that L is an elliptic operator. This completes the
proof of Prop. 3.1.
We can also prove the following:
Proposition 3.2. Let Mn be an n-dimensional hypersurface in a real space
form Rn+1(c) with n(n� 1)R = k0H, k0 = constant > 0. Assume that the mean
curvature H > 0. Then we have:
(1) if c > 0 and k0 � 2n
p
n(n� 1)c, then
jrhj2 � n2jrHj2;
(2) if c � 0, for all k0 > 0, then
jrhj2 � n2jrHj2:
300 Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 2
Complete Hypersurfaces in a Real Space Form
P r o o f. Since H > 0, we have jhj2 6= 0. In fact, if jhj2 =
P
i
�2
i
= 0 at
a point of Mn, then �i = 0, i = 1; 2; : : : ; n, at this point. Therefore H = 0 at this
point. This is impossible.
From (2.6) and n(n� 1)R = k0H, we have
k0riH = 2n2HriH � 2
P
j;k
hkjhkji;
(1
2
k0 � n2H)riH = �
P
j;k
hkjhkji;
(1
2
k0 � n2H)2jrHj2 =
P
i
(
P
j;k
hkjhkji)
2 �
P
i;j
h2
ij
P
i;j;k
h2
ijk
= jhj2jrhj2:
Therefore, we have
jrhj2 � n2jrHj2 � [(
k0
2
� n2H)2 � n2jhj2]jrHj2 1
jhj2
= [
(k0)2
4
� n3(n� 1)c]jrHj2 1
jhj2 : (3.7)
(1)If c > 0 and k0 � 2n
p
n(n� 1)c, from (3.7), we have
jrhj2 � n2jrHj2 � 0:
(2) If c � 0, from (3.7), we also have
jrhj2 � n2jrHj2 � 0:
This completes the proof of Prop. 3.2.
Proposition 3.3. Let Mn be an n-dimensional hypersurface in a real space
form Rn+1(c) with n(n� 1)R = k0H, k0 = constant > 0. Then we have
nLH � (jrhj2 � n2jrHj2) + jgj2fnc+ nH2 � n(n� 2)p
n(n� 1)
jHjjgj � jgj2g;
where jgj2 is a nonnegative C2-function on Mn de�ned by jgj2 = jhj2 � nH2.
P r o o f. From (3.5) we have
nLH = n[2H � (k0=2n)�H]
= 2(nH)� (1=2)�[n(n� 1)R]
= jrhj2 � n2jrHj2 + 1
2
P
i;j
(c+ �i�j)(�i � �j)
2
= jrhj2 � n2jrHj2 + ncjhj2 � n2H2c� jhj4 + nH
P
i
�3
i
: (3.8)
Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 2 301
Shu Shichang
Let jgj2 be a nonnegative C2-function on Mn de�ned by jgj2 = jhj2 � nH2.
Since
P
i
(H � �i) = 0,
P
i
(H � �i)
2 = jhj2 � nH2 = jgj2, by Lem. 2.2 we get
nH
P
i
�3
i
= 3nH2jhj2 � 2n2H4 � nH
P
i
(H � �i)
3
� 3nH2jgj2 + n2H4 � njHj n� 2p
n(n� 1)
jgj3: (3.9)
Therefore, from (3.8) and (3.9), we have
nLH � jrhj2 � n2jrHj2 + jgj2fnc+ nH2 � n(n� 2)p
n(n� 1)
jHjjgj � jgj2g:
This completes the proof of Prop. 3.3.
P r o o f o f T h e o r e m 1.3. From the assumption of Th. 1.3, Prop. 3.2
and Prop. 3.3 for c = 1, we have
nLH � jgj2fn+ nH2 � n(n�2)p
n(n�1)
Hjgj � jgj2g = jgj2PH(jgj); (3:10)
where
PH(jgj) = n+ nH2 � n(n� 2)p
n(n� 1)
Hjgj � jgj2:
PH(jgj) has two real roots B�
H
and B+
H
given by
B�
H
= �1
2
(n� 2)
r
n
n� 1
H �
s
n3H2
4(n� 1)
+ n:
Therefore, we know that
PH(jgj) = (jgj �B�
H
)(�jgj +B+
H
):
Clearly, we know that jgj�B�
H
> 0. From the assumption of Th. 1.3, we infer that
PH(jgj) � 0 onMn. This implies that the right-hand side of (3.10) is nonnegative.
Since, from Prop. 3.1, the operator L is elliptic, and H obtains its maximum on
Mn, from (3.10) we know that H = const on Mn. Therefore, we know that Mn
is an n-dimensional complete hypersurface with constant mean curvature H(> 0)
in a unit sphere Sn+1(1). By the assumption of Th. 1.3 and the result of Th. 1.1,
we can check directly that jhj2 � nH2 + (B+
H
)2 = n+ n
3
H
2
2(n�1) �
(n�2)nH
2(n�1) [n2H2 +
4(n � 1)]
1
2 = D0(n;H). Therefore we have either Mn is totally umbilical, or (i)
n � 3;Mn is locally an H(r)-torus with r2 < n�1
n
; (ii) n = 2;Mn is locally an
H(r)-torus with r2 6= n�1
n
. This completes the proof of Th. 1.3.
302 Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 2
Complete Hypersurfaces in a Real Space Form
P r o o f o f T h e o r e m 1.4. From the assumption of Th. 1.4, Prop. 3.2
and Prop. 3.3, for c = 0, we have
nLH � jgj2fnH2 � n(n�2)p
n(n�1)
Hjgj � jgj2g = jgj2QH(jgj); (3:11)
where
QH(jgj) = nH2 � n(n� 2)p
n(n� 1)
Hjgj � jgj2:
QH(jgj) has two real roots eB�
H
and eB+
H
given by
eB�
H
= �(n� 1)
r
n
n� 1
H; eB+
H
=
r
n
n� 1
H:
Therefore, we know that
QH(jgj) = (jgj � eB�
H
)(�jgj + eB+
H
):
Clearly, we know that jgj � eB�
H
> 0. From the assumption of Th. 1.4, we infer
that QH(jgj) � 0 on Mn. This implies that the right-hand side of (3.11) is
nonnegative. From Proposition 3.1, we know that L is elliptic, and H obtains its
maximum on Mn. From (3.11), we have H = const on Mn. From (3.11) again,
we get jgj2QH(jgj) = 0. We infer that the equality holds in Lem. 2.2. Therefore,
we know that (n � 1) of the numbers H � �i are equal to jgj=
p
n(n� 1). This
implies that Mn has (n � 1) principal curvatures equal and constant. As H is
constant, the other principal curvature is constant as well. From an inequality of
Chen�Okumura [8], we know that jhj2 � n2H2=(n� 1) implies that the sectional
curvature K of Mn is nonnegative. Therefore, we know that Mn is a complete
hypersurface in Rn+1 with constant mean curvature and nonnegative sectional
curvature. From Theorem 1.2, we have either Mn is isometric to a standard
round sphere, a hyperplane Rn or a Riemannian product Sn�1(c1) � R1. This
completes the proof of Th. 1.4.
P r o o f o f T h e o r e m 1.5. From the assumption of Th. 1.5, Prop. 3.2
and Prop. 3.3, for c = �1, we have
nLH � jgj2f�n+ nH2 � n(n�2)p
n(n�1)
Hjgj � jgj2g = jgj2RH(jgj); (3:12)
where
RH(jgj) = �n+ nH2 � n(n� 2)p
n(n� 1)
Hjgj � jgj2:
RH(jgj) has two real roots bB�
H
and bB+
H
given by
bB�
H
= �1
2
(n� 2)
r
n
n� 1
H �
s
n3H2
4(n� 1)
� n; n2H2 � 4(n� 1):
Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 2 303
Shu Shichang
Therefore, we know that
RH(jgj) = (jgj � bB�
H
)(�jgj + bB+
H
):
Clearly, we know that jgj� bB�
H
> 0. From the assumption of Th. 1.5, we infer that
RH(jgj) � 0 onMn. This implies that the right-hand side of (3.12) is nonnegative.
From Proposition 3.1, we know that L is elliptic. Since H obtains its maximum
on Mn, from (3.12), we have H = const on Mn. From (3.12) again, we get
jgj2RH(jgj) = 0, so jgj2 = 0, and Mn is totally umbilical, or RH(jgj) = 0. In the
latter case, we know that (n�1) of the numbers H��i are equal to jgj=
p
n(n� 1).
This implies thatMn has (n�1) principal curvatures equal and constant. As H is
constant, the other principal curvature is constant as well, soMn is isoparametric.
From the result of Lem. 2.1, Mn is isometric to Sn�1(r) �H1(�1=(r2 + 1)) for
some r > 0. This completes the proof of Th. 1.5.
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