Totally umbilical submanifolds in irreducible symmetric spaces

Full text

/. Austral.
Math.
Soc.
(Series
A) 32 (1982), 332-338
TOTALLY UMBILICAL SUBMANIFOLDS IN
IRREDUCIBLE SYMMETRIC SPACES
BANG-YEN CHEN and PAUL VERHEYEN
(Received 18 October 1980)
Communicated by G. Virsik
Abstract
A submanifold of
a
Riemannian manifold is called
a
totally umbilical submanifold
if
its first and
second fundamental forms are proportional. In this paper we prove the following best possible result.
THEOREM.
There
is
no totally umbilical submanifold
of
codimension less than rank
of M
in any
irreducible symmetric space M.
1980 Mathematics subject classification (Amer. Math. Soc): 53 B 20, 53 C 35, 53 C 40.
1.
Introduction
Let N be an ^-dimensional submanifold of an w-dimensional Riemannian mani-
fold M(n
>
2) with the first fundamental form g. Let V and V be the covariant
differentiations on N and M, respectively. The second fundamental form h of the
immersion is defined by the equation
(1.1)
h(X,Y)=VxY-vxY,
where
X
and
Y
are vector fields tangent to N. The submanifold
N
is said to be
totally umbilical if
(1.2)
h(X,Y)=g(X,Y)H,
for all vector fields
X,
Y
tangent to N, where H =
l/n
(traceh)
is
the mean
curvature vector of N in M. The length of H is called the mean curvature of N in
M.
A
totally umbilical submanifold with vanishing mean curvature is called
a
totally geodesic submanifold.
»
Copyright Australian Mathematical Society 1982
332

[21 Totally umbilical submanifolds 333
In Chen (1980), the following results were proved.
PROPOSITION
1. Let N be a totally umbilical submanifold in a symmetric space M.
If co-dim N < rank M
—
1, then N has constant mean curvature and N is either
totally geodesic or of constant sectional curvature.
PROPOSITION
2. // N is a totally umbilical submanifold in an irreducible symmet-
ric space M, then co-dim N > rank M — 1.
It is known that the Riemannian product M = R X S" of a real line R and an
w-sphere S" is a rank 2 symmetric space which admits a totally umbilical
hypersurface with nonconstant mean curvature. Thus the estimate of the codi-
mension of N in Proposition
1
is best possible.
It is also known that the real Grassmann manifold M
—
SO(p + q)/SO(p) X
SO(^) (p > q > 1) is an irreducible symmetric space of rank q which admits a
totally umbilical (in fact, totally geodesic) submanifold with codimension equal to
rank M. In this note we shall prove that there is no totally umbilical submanifold
N in an irreducible symmetric space M with codimension equal to rank M
—
1.
By combining this result with Proposition 2, we obtain the following fundamental
result.
MAIN
THEOREM.
Let N be a totally umbilical submanifold in an irreducible
symmetric space M. Then co-dim N > rank M.
From the examples of real Grassmann manifolds, we see that the estimate of
codimension in Main Theorem is best possible.
2.
Basic formulas
Let N be an n-dimensional submanifold of a Riemannian manifold M. For a
vector field £ normal to N we write
(2.1) Vxi =
-A
i
X+D
x
i,
where -A^ X and Dx£ are the tangential and normal components of v x£>
respectively. A normal vector field £ is said to be parallel if D£ = 0 identically.
Let R and R be the curvature tensors associated with V and v, respectively.
For example, S(Ar,lr) =
[v
x
,V
r
]-
V[xYy F°r tne second fundamental form

334 Bang-Yen Chen and Paul Verheyen
13 ]
h,
we define the covariant derivative in
(TN)
© {T1- N), to be
(2.2)
(vxh)(Y,
Z) =
Dx(h(Y, Z))
-
A(V
XY,
Z) -
A(y,
V
XZ),
where 77V and
T"1
N
denote the tangent and normal bundles of N, respectively.
We
put
R(X,
Y;Z,W)
=
g(R(X,
Y)Z,W).
For
vector fields
X, Y, Z, W
tangent to
N,
the equations
of
Gauss and Codazzi take the forms:
(2.3)
R(X,Y;Z,W)
=
R(X,Y;Z,W)
+
g(h(X,W),h(Y,Z))
-g(h(X,Z),h(Y,W)),
(2.4)
(R(X,
Y)Z)X=
(vxh)(Y, Z) - {vYh)(X, Z),
where
±
in (2.4) denotes the normal component.
Let
X
and
Y
be
orthonormal vectors tangent
to N.
The sectional curvature
K(
X
A
Y) of
the plane
X
A Y is given by
(2.5)
K(XAY)
=
R(X,Y;Y,X).
If
TV
is
a
totally umbilical submanifold in
a
Riemannian manifold M, (2.4) gives
(2.6)
R(X,
Y,Z, H)
= {{g{Y,
Z)Xa
2
-
g(X,
Z)Ya
2
},
where
a2
=
g{H,
H).
3.
Constancy of mean curvature
An isometry s
of a
Riemannian manifold is said to be involutive
if
its iterate
s2
is the identity map.
A
Riemannian manifold
M
is
a
symmetric space
if, at
each
point p of M, there exists an involutive isometry sp of
M
such that
p
is an isolated
fixed point of
sp.
We denote by G the closure of the group of isometries generated by {sp
\
p G A/}
in the compact-open topology. Then G acts transitively on M; hence the typical
isotropy subgroup
K,
say
at
0, is compact and
M
—
G/K.
Let a0 be the involutive automorphism
of
G given by ao(x)
= s0
•
x
•
s0,
x
G
G.
Then a0 fixes
K
and
it
induces an involutive automorphism
of
the Lie algebra g of
G. The Cartan decomposition of g is given by
(3.1)
fl = ! +
m,
where
f
and m are the eigenspaces of a0 with eigenvalues
1
and
-1,
respectively.
It
is known that
f
is
the Lie algebra
of K
and m can be identified with the tangent
space T0M of
M
at 0. Moreover, we have
[f,f]cf,
[f,m]cm,
[m,m]cf.
The following lemmas
of
E. Cartan are well-known (see Helgason (1978)).

[4] Totally umbilical submanifolds
335
LEMMA 3. The curvature
tensor
RofMatO satisfies
(3.2)
R(X,Y)Z
=
-[[X,Y],Z]
for X,Y,ZG
m.
LEMMA 4.
Let B
be
a
totally
geodesic
submanifold of M
through
0. Then
B
is flat if
and only
if
[IT,
IT]
—
0
where
m
—
T0B
C
T0M
=
m.
We recall
the
following result
of
Chen-Nagano
(see
Chen (1980)).
LEMMA
5.
Every totally
geodesic
submanifold
B
of an
irreducible
symmetric space
M
satisfies
(3.3) co-dim
B >
rank
M.
We give
the
following.
PROPOSITION 6.
Let N
be
a
totally
umbilical
submanifold
in
a
symmetric space
M.
//co-dim
N
<
rank
M
—
\
and M is
Einsteinian, then
the
mean curvature
of N is
constant.
PROOF.
Let
TV
be
a
totally umbilical submanifold
in
a
symmetric space
M.
If
co-dim
N <
rank
M
—
\, N
has
constant mean curvature (Proposition
1).
There-
fore, we only need
to
consider
the
case where co-dim
N =
rank
M
—
1.
For
any
fixed point
0 in N, let B be a
maximal flat totally geodesic submanifold
of
M
through
0
such that
H(0) G T0B. We
have
dim T0B
=
rank
M.
Since
B is
a
flat totally geodesic submanifold
of M,
Lemma
4
implies that [U,
V]
= 0
for all
U,VG
T0B
Cm.
Since
dim N
-
dim
M -
rank
M + 1, dim T0N
n
T0B > 1. Let
Xo
be
a
unit
vector
in T0N
n
T0B.
Then
[Xo, H] = 0.
Thus (2.6)
and
Lemma 3 give
(3.4)
0
=
R(Y,
X
0
;X
0
,H)
=
±Ya2
for
any
vector
Y in
(Z6 T0N\ g(X0, Z)
=
0}.
If
dim T0N D ro5
5*
2,
this
implies that
Wa2 = 0 for all W G T0N.
If dim
r/fl T0B = 1,
then roiV
U T0B
spans
T0M.
Hence,
(3.5)
T
0
N+T
0
B=T
0
M.
For
any
vector
17
G T0M
we
put
(3.6)
1,
= V +
r,B
where
TJ'
E
7
0
iV
and
rf
G ro£
with
g(X0,
TJ')
=
0. We
have
(3-7)
[H,VB]=[x0,-nB]=0.

336 Bang-Yen Chen and Paul Verheyen
[s)
Combining this with Lemma 3 we obtain
(3.8)
R(X0,
VB
;r,',
H) =
R~(X
0
,
V,VB,
») = 0
because
Xo, H,
T)
B
G
TOB
C
m.
Let £,,...
,En be an
orthonormal basis
of
T0N
with
En
—
Xo
and
T/,,.
..
,-qm_n
an orthonormal basis
of
TQ
N
with
ij
m
_
n
parallel
to H.
Equations (2.6) and (3.8)
give
(3.9) /?(£„, T^T,,,//)^^,^)^2, i=\,...,m-n-\,
(3.10)
R(En,EJ,Ej,H)=±Ena2, j^n.
Therefore, the Ricci tensor
S
of M
satisfies
f m-n-\
S{EH,H)={Un-\)+
2 h;
If
M
is Einsteinian, this implies
(3.11) A>2
= 0.
(3.4) and (3.11) gives
Wa2
—
0
for all W
in TQN. Since this
is
true
for
arbitrary
point
in
./V,
a2
is
constant.
4.
Proof of Main Theorem
Let
M
be an irreducible symmetric space. Then
M
is Einsteinian.
If
iV"
is totally
umbilical
in M
with co-dim
N <
rank
M
—
1,
N
has constant mean curvature
a
(Proposition 6). Moreover, the mean curvature
a
is
a
nonzero constant (Lemma 5).
If
the
mean curvature vector
H is
parallel, JV
is an
extrinsic sphere
in M.
Theorem
2 of
Chen (1979) implies that
M
admits
a
totally geodesic submanifold
N
of
constant sectional curvature
of
dimension equal
to
1
+
dim
N.
This
is
impossible (Lemma
5).
Consequently,
the
mean curvature vector
of N is
not
parallel.
If dim
N =
2,
dim
M <
rank
M
+
1.
From
the
list
of
irreducible symmetric
spaces (see Helgason (1978)),
M
is
2-dimensional. This
is a
contradiction. Thus
dim
N >
2.
Case (a).
If M is of
compact type, Theorem
4 of
Chen (1980) shows that
dim
N
<
\
dim
M.
Thus, from the assumption
of
codimension, we get
dim
M
*£
2 rank
M
—
1.
This contradicts the classification
of
irreducible symmetric spaces.

(6] Totally umbilical submanifolds
337
Case
(b).
If M
is
of
non-compact type,
M
is
nonpositively curved, that
is,
the
sectional curvature
K of
M
satisfies
(4.1)
K^O.
Since
TV
is
totally umbilical submanifold
of
constant mean curvature
a
¥=
0
in
a
symmetric space,
TV
is
one
of
the
following spaces
(see the
proof
of
Theorem
4
in
Chen (1980)):
(1)
a
space
of
constant sectional curvature
c,
(2)
a
local product
of
two
spaces TV,(c)
and
N2(-c)
of
constant sectional
curvatures
c and
-c,(c
¥=
0)
respectively,
or
(3)
a
local product
of a
curve
and
a
space
N2(c) of
constant sectional curvature
c^O.
Since
TV
has constant mean curvature, (2.6) implies
(4.2)
R~(X,Y;Z,H)=0
for
any
vector fields
X,
Y
and
Z
tangent
to
TV.
Taking
the
derivative
of
(4.2)
with
respect
to
a
tangent vector U
in
TN,
we
have
a2R(X, Y;Z,U) =
g(U,
X)R(H, Y;Z, H) -
g(U,
Y)R(H, X;Z, H)
+ g(Y,
Z)g(DxH, DuH) - g(X, Z)g(DYH, D^).
Let
X =
U,
Y
—
Z
be
orthonormal vectors tangent
to
N.
This implies
(4.3)
a2K(H
A Y) =
a2K(X
A Y) - \DXH\2.
For
an
arbitrary fixed point
0 in N
let
B be a
maximal flat totally geodesic
submanifold
of M
through
0
such that
H(0)
G T0B.
Because co-dim
N <
rank
M
—
1, we
have
dimro7V (1TOB^I.
Let Yo
be
a
unit vector
in
T0N D
T0B. We
have
K(Y0 A
H) -
0.
Thus,
(4.3)
gives
(4.4)
a2/?(
X
A
y0)
=
|
D^T/1
2
.
Comparing (4.1)
and (4.4) we
obtain
(4.5)
DxH
=
0
iorXG{ZET
0
N\g(Z,Y
0
)=0}.
Case (b.l).
If N is of
constant sectional curvature
c, (4.3)
implies that
| DXH
|
is
independent
of
the
choice
of
the
unit vector
X
in
T0N.
Thus,
(4.5)
gives
DXH
=
0
for
all
X £
T0N.
Since this argument applies
to an
arbitrary
in
TV,
H
is
parallel.
This gives
a
contradiction.
Case (b.2).
If
TV
is
the
local product
of
two
spaces
N^c)
and
N2(-c)
of
constant
sectional curvatures
c and -c, c
¥=
0,
respectively, equation
(4.3)
proves that
\DXH\
is
the
same
for all
unit vectors
X
in
roTV,
and
\DZH\ is
the
same
for all

338 Bang-Yen Chen
and
Paul Verheyen
[71
unit vectors
Z in T0N2.
Since both
N] and N2 are of
dimensions s*
2,
(4.5) shows
that
DVH = 0 for all
U in
T0N.
Because this
is
true
for
arbitrary point
in N, H is
parallel. This gives
a
contradiction.
Case (b.3).
If N is the
local product
of a
curve
and a
space
N2(c) of
constant
sectional curvature
c.
Then (4.3)
and
(4.5) imply that
| DXH
|
is
independent
of
the choice
of
unit vector
X
in
T0N2
and
in
fact
(4.1)
DXH = 0 for X G
T0N2.
Since
N2 =
N2(c) *s totally geodesic
in N and N is
totally umbilical
in M, N2 is
totally umbilical
in M.
Moreover,
the
mean curvature vector
of
N2
in M
is
in
fact
the restriction
of H on N2.
Moreover,
it can be
easily proved that
the
normal
connection
D2 of N2 in M
satisfies,
DXH = DXH = 0.
Since this
is
true
for
arbitrary point
in N, N2 is an
extrinsic sphere
in M.
Theorem
2 of
Chen (1979)
then implies that
M
admits
a
totally geodesic submanifold
of
dimension equal
to
1
+
dim
N2.
This contradicts Lemma
5 and our
assumption.
References
B.
Y.
Chen (1979), 'Extrinsic spheres
in
Riemannian manifolds', Houston
J.
Math.
5,
319-324.
B.
Y.
Chen (1980), 'Classification
of
totally umbilical submanifolds
in
symmetric spaces',
J.
Austral.
Math.Soc.Ser.
A
30, 129-136.
S. Helgason (1978), Differential geometry.
Lie
groups
and
symmetric spaces (Academic Press,
New
York).
Department
of
Mathematics Department
of
Mathematics
Michigan State University Katholieke Universiteit Leuven
East Lansing, Michigan 48824 Celestijnenlaan 200B
U.S.A. B-3030 Leuven
Belgium