Submanifold Geometry of Orbits 67
Proof We denote by
α
the second fundamental form of M in
¯
M,by
α
the second
fundamental form of M in M
, and by H the mean curvature vector field of M
in
¯
M.
The Gauss formula implies
α
(X,Y )=
α
(X,Y )+X ,YH
for all vector fields X ,Y tangent to M. From Lemma 1.6.1 we know that H is parallel
in the normal bundle of M
.Thisimpliesthat
α
is parallel if and only if
α
is parallel.
The assertion then follows from Theorem 2.8.3.
The following example illustrates how these two methods can be used to con-
struct new symmetric submanifolds of spheres from some given symmetric subman-
ifolds of spheres. Let M
1
be a symmetric submanifold of S
n
1
−1
(r
1
) and M
2
be a
symmetric submanifold of S
n
2
−1
(r
2
).SinceS
n
1
−1
(r
1
) sits totally umbilically inside
R
n
1
, it follows from Lemma 2.8.5 that M
1
is a symmetric submanifold of R
n
1
.Anal-
ogously, M
2
is a symmetric submanifold of R
n
2
. We now apply Lemma 2.8.4 to
see that the extrinsic product M
1
×M
2
is a symmetric submanifold of R
n
,where
n = n
1
+ n
2
. By construction, M
1
×M
2
sits inside S
n
1
−1
(r
1
) ×S
n
2
−1
(r
2
),whichis
a submanifold of S
n−1
(r) with r =
r
2
1
+ r
2
2
.SinceS
n−1
(r) is a totally um bilical
submanifold of R
n
, it follows from Lemma 2.8.5 that M
1
×M
2
is a symmetric sub-
manifold of S
n−1
(r).
2.8.3 Examples of symmetric submanifolds in standard space forms
We will now use Theorem 2.8.3 to present some examples of symmetric subman-
ifolds in standard space forms. In fact, according to this result, it is sufficient to find
connected complete submanifolds with parallel second fundamental form.
Example 2.8.1 (Totally geodesic submanifolds) Every totally geodesic submani-
fold has vanishing second fundamental form and hence also parallel second fun-
damental form. The connected complete totally geodesic submanifolds of standard
space forms were classified in Theorem 1.4.1.
Example 2.8.2 (Extrinsic spheres) The second fundamental form
α
of an extrinsic
sphere is of the form
α
(X,Y )=X,Y H,
where the mean curvature vector field H is parallel in the normal bundle. Therefore,
the second fundamental form of an extrinsic sphere is parallel.
We now discuss the existence problem for extrinsic products of extrinsic spheres
or totally geodesic submanifolds in standard space forms. Let M = M
1
×...×M
s
be an extrinsic product in a standard space form
¯
M
n
(
κ
) and suppose that each leaf
L
i
(p) of the induced totally geodesic foliations L
i
, i = 1,...,s,onM is an extrinsic
sphere or is totally geodesic in
¯
M
n
(
κ
).Wefix a point p ∈ M and denote by H
i
the
mean curvature vector at p of the extrinsic sphere L
i
(p) in
¯
M
n
(
κ
). Hence, the second
fundamental form
α
i
of L
i
(p) at p is given by
α
i
(X,Y )=X,Y H
i
68 Submanifolds and Holonomy
for all X,Y ∈ T
p
L
i
(p).LetX
i
∈T
p
L
i
(p) and Y
j
∈T
p
L
j
(p) be unit vectors. Since M is
an extrinsic product in
¯
M
n
(
κ
), the Gauss equation yields
0 = R(X
i
,Y
j
)Y
j
,X
i
=
κ
+ H
i
,H
j
(i = j).
Therefore, if
κ
= 0, none of the leaves L
i
(p) can be totally geodesic. We define
κ
i
=
κ
+ H
i
,H
i
.
If dimL
i
(p) ≥ 2, then L
i
(p) is a space of constant curvature
κ
i
, which follows easily
from the Gauss equation. If H
1
,...,H
s
are mutually distinct, we get
0 < H
i
−H
j
,H
i
−H
j
= H
i
,H
i
+ H
j
,H
j
−2H
i
,H
j
=
κ
i
+
κ
j
.
Consequently, at most one of the numbers
κ
i
is nonpositive. This implies, for in-
stance, that such an extrinsic product in real hyperbolic space can never contain
two real hyperbolic spaces, or some real hyperbolic space together with a Euclidean
space.
As for the converse, let p ∈
¯
M
n
(
κ
) and V be a linear subspace of T
p
¯
M
n
(
κ
).Let
V = V
1
⊕...⊕V
s
(s ≥ 2) be an orthogonal decomposition of V and H
1
,...,H
s
∈
T
p
¯
M
n
(
κ
) be mutually distinct and perpendicular to V . Does there exist an extrinsic
product M = M
1
×...×M
s
of extrinsic spheres and totally geodesic submanifolds
in
¯
M
n
(
κ
) with p ∈ M, T
p
L
i
(p)=V
i
and such that H
i
is the mean curvature vector
of L
i
(p) at p? Note that at most one factor can be totally geodesic since we assume
the vectors H
i
to be mutually distinct. We have seen above that we necessarily need
H
i
,H
j
= −
κ
for all i = j. In fact, it can be shown that this condition is also sufficient
for the existence of such an extrinsic product. This can be proved easily when
κ
= 0.
In this case the vectors H
1
,...,H
s
are pairwise orthogonal. Let M
i
be the extrinsic
sphere or totally geodesic submanifold in R
n
with p ∈ M
i
, T
p
M
i
= V
i
and for which
the mean curvature vector at p is equal to H
i
. These submanifolds are contained
in mutually perpendicular Euclidean subspaces of R
n
and hence their Riemannian
product yields the extrinsic product we are looking for. The construction in S
n
can be
done by viewing S
n
as a totally umbilical submanifold in R
n+1
and using the previous
construction m ethod. The case of real hyperbolic space is a little more involved. For
further details we refer to [13]. We summarize this in the following.
Theorem 2.8.6 Let p ∈
¯
M
n
(
κ
) and V be a linear subspace of T
p
¯
M
n
(
κ
).LetV = V
1
⊕
...⊕V
s
be an orthogonal decomposition of V with s ≥ 2 and H
1
,...,H
s
∈ T
p
¯
M
n
(
κ
)
be mutually distinct and perpendicular to V . Then there exists a submanifold product
M = M
1
×...×M
s
of extrinsic spheres and totally geodesic submanifolds in
¯
M
n
(
κ
)
with p ∈ M, T
p
L
i
(p)=V
i
, and such that H
i
is the mean curvature vector of L
i
(p) at
p if and only if H
i
,H
j
= −
κ
for all i = j.
This finishes Example 2.8.2.
Submanifold Geometry of Orbits 69
Example 2.8.3 (Standard embeddings of symmetric R-spaces) For details on real
flag manifolds we refer to Section A.4. Let
¯
M be a connected, simply connected,
semisimple Riemannian symmetric space, G = I
o
(
¯
M), o ∈
¯
M and K the isotropy
group of G at o. Note that K is connected since we assume
¯
M to be simply connected.
Each orbit of the isotropy representation of K on T
o
¯
M is a real flag manifold and its
realization as a submanifold of T
o
¯
M is called the standard embedding of the real flag
manifold. A real flag manifold that is also a symmetric space is also called a sym-
metric R-space or, if G is simple, an irreducible symmetric R-space. Let M = K ·X
be the orbit of the action of K through X ∈ T
o
¯
M, X = 0. We will prove below that
each standard embedding of a symmetric R-space is a symmetric submanifold of the
Euclidean space T
o
¯
M.
Recall that there is a convenient way to describe the isotropy representation. Let
g = k ⊕p be the Cartan decompo sitio n of the Lie algebra g of G.Thenp is canon-
ically isomorphic to T
o
¯
M and, via this identification, the isotr opy representation is
isomorphic to the adjoint representation Ad : K → SO(p). The orbit M = K ·X is a
symmetric space if and only if the eigenvalues of the transformation ad(X) : g → g
are ±c,0forsomec > 0. Without loss of generality we can assume that X is normal-
ized so that c = 1. We decompose the semisimple Lie algebra g into the direct sum
g = g
1
⊕...⊕g
k
of simple Lie algebras g
i
and put k
i
= k ∩g
i
and p
i
= p ∩g
i
.Then
p = p
1
⊕...⊕p
k
and, by means of this decomposition, we can write X =(X
1
,...,X
k
).
We denote by K
i
the connected Lie subgroup of G
i
with Lie algebra k
i
.Then
M = K ·X is isometric to the Riemannian product
M = K ·X = K
1
·X
1
×...×K
k
·X
k
.
Viewing M as a submanifold of p it is clear that M is the extrinsic product of the
submanifolds K
i
·X
i
of p
i
. In particular, the standard embedding of any symmetric
R-space decomposes as the extrinsic product of the standard embeddings of some
irreducible symmetric R-spaces.
Let M = K ·X be a symmetric R-space regarded as an embedded submanifold
of p. We will now show explicitly that M is a symmetric submanifold of p.Since
K ⊂I
o
(M), the Cartan deco mposition of the Lie algebra of I
o
(M) induces a reductive
decomposition k = k
X
⊕m,wherek
X
is the Lie algebra of the isotropy group of K at
X. General theory about symmetric spaces says that for each U ∈m the curve
γ
: R → M , t → Ad(Exp(tU))X
is the geodesic in M with
γ
(0)=X and
˙
γ
(0)=U, where we identify T
X
M and m in
the usual way. On the other hand, viewing
γ
as a curve in p,wehave
˙
γ
(0)=
d
dt
t=0
Ad(Exp(tU))X =
d
dt
t=0
e
ad(tU )
X = ad(U)X =[U,X].
This implies
T
X
M = {[X,U] : U ∈ m} = ad(X )(m).
Since the inner product on p comes fro m the Killing form of g and since ad(X) is
70 Submanifolds and Holonomy
skewsymmetric with respect to the Killing form, this implies
ν
X
M = {
ξ
∈ p : [X ,
ξ
]=0}.
We denote by g
ν
the eigenspace of ad(X) with respect to
ν
∈{−1, 0, +1}.Then
we have a vector space direct sum decomposition
g = g
−1
⊕g
0
⊕g
1
.
The Jacob i identity im plies that
[g
ν
,g
μ
] ⊂ g
ν
+
μ
,
where we put g
−2
= g
2
= {0}. This just says that the eigenspaces of ad(X ) turn g
into a graded Lie algebra. In order to define an extrinsic symmetry of M at X in p we
consider the transformation ad(X )
2
: g →g. This transformation has two eigenvalues
0and+1 with corresponding eigenspaces g
0
= g
0
and g
1
= g
−1
⊕g
1
, leading to the
vector space direct sum decomposition
g = g
0
⊕g
1
.
Since [p,k] ⊂ p, [p,p] ⊂ k and X ∈ p, it follows that ad(X)
2
k ⊂ k and ad(X )
2
p ⊂ p.In
particular, this implies
k =(k ∩g
0
) ⊕(k ∩g
1
) and p =(p ∩g
0
) ⊕(p ∩g
1
).
The subspaces in these decompositions are precisely
k
X
= k ∩g
0
, m = k ∩g
1
,
ν
X
M = p ∩g
0
, T
X
M = p ∩g
1
.
Next, we define an involution
ρ
: g = g
0
⊕g
1
→ g = g
0
⊕g
1
, Z = Z
0
+ Z
1
→ Z
0
−Z
1
.
It leaves k and p invariant, and the above gradation of g shows that [g
0
,g
0
] ⊂ g
0
,
[g
0
,g
1
] ⊂ g
1
and [g
1
,g
1
] ⊂ g
0
, which implies that
ρ
is an automorphism of g. Thus
ρ
is an involutive automorphism of g that commutes with the Cartan involution of g
corresponding to the Cartan decomposition g = k ⊕p. We claim that
σ
X
: p → p , W →
ρ
(W )
is an extrinsic symmetry of M at X. By construction,
σ
X
is an involutive isom etry of
p with
σ
X
(X)=X , d
X
σ
X
(W )=W for all W ∈
ν
X
M = p ∩g
0
and d
X
σ
X
(W )=−W
for all W ∈ T
X
M = p ∩g
1
. Thus, it remains to show that
σ
X
(M)=M. For this we
consider once again the g eodesics
γ
(t)=Ad(Exp(tU))X in M.SinceU ∈ m = k ∩g
1
and
ρ
−1
(X)=
ρ
(X)=X, we obtain
σ
X
(
γ
(t)) =
ρ
(Ad(Exp(tU))X )=
ρ
(Ad(Exp(tU))
ρ
−1
(X))
= Ad(Exp(t
ρ
(U)))X = Ad(Exp(−tU))X =
γ
(−t)
Submanifold Geometry of Orbits 71
for all t ∈ R.Since
γ
is a curve in M this shows that
σ
X
leaves M invariant. At any
other point Ad(k )X ∈ M the isometry k
σ
X
k
−1
of p defines an extrinsic symmetry of
M at Ad(k)X . So we can now conclude that M is a symmetric submanifold of p.
It is clear from the construction that M lies in the sphere S
n−1
(r) ⊂ p with radius
r = X. Since this sphere is totally umbilical in p, it follows from Lemma 2.8.5
that M is also a symmetric submanifold of S
n−1
(r). Note that, by a suitable homoth-
ety, we can realize M also as a symmetric submanifold of the unit sphere S
n−1
.We
summarize the previous discussion in:
Proposition 2.8.7 The standard embedding of any symmetric R-space K ·Xisasym-
metric submanifold both of the Euclidean space p and of the sphere S
n−1
(X).
2.8.4 Classification of symmetric submanifolds of standard space forms
We now classify the symmetric submanifolds of the standard space forms.
Roughly speaking, we will show that the examples given above exhaust all possi-
bilities. Although it is possible to formulate just one classification theorem for all
standard space forms (see [13]), we will investigate the cases of zero, positive and
negative curvature separately for the sake of simplicity.
Case 1: Euclidean spaces.
As a first step, we classify the locally symmetric su bmanifolds of R
n
. This classi-
fication is due to Ferus and can be found in the papers [127–130]. The most concise
proof is in [130] using the algebraic framework of Jordan triple systems. Here we pre-
fer to adopt the more geometric approach of [128] and [129], following also [124]
and [311].
Theorem 2.8.8 (Ferus) Let M be a locally symmetric submanifold of R
n
.Then
(a) M is locally a submanifold product
R
m
0
×M
1
×...×M
s
→ R
m
0
×R
m
1
×...×R
m
s
⊂ R
n
,
where M
i
is a full immersion into R
m
i
that is minimal in a hypersphere of R
m
i
.
(b) Each M
i
as in (a) is locally a standard embedding of an irreducible symmetric
R-space (which is, in particular, an orbit of an s-representation).
Remark 2.8.9 Of course, it may happen that m
0
= 0, that is, M has no Euclidean
factor. Another possible case is that
∑
m
i
= n,thatis,M is full in R
n
. Note also that
the statement in the theorem is a local result. If we assume that M is complete we get
a global result: a complete symmetric submanifold of R
n
is covered by a submanifold
product R
m
0
×M
1
×...×M
s
→ R
m
0
×R
m
1
×...×R
m
s
⊂R
n
,whereM
i
is a standard
embedding of an irreducible symmetric R-space.
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