176 Submanifolds and Holonomy
Proposition 5.1.6 An immersed submanifold f : M → R
n
has curvature normals of
constant length if and only if its adapted third fundamental form has constant eigen-
values.
5.1.3 Higher rank rigidity
We now state a global rigidity result for which the assumption of completeness
is fundamental.
Theorem 5.1.7 (Di Scala, Olmos [108]) Let M be a simply connected complete Rie-
mannian manifold with dim M ≥ 2 and let f : M → R
n
be a full and irreducible
isometric immersion with rank
f
(M) ≥ 1 such that the curvature normals have con-
stant length. Assume that the number of curvature normals is constant on M or that
rank
f
(M)=rank
loc
f
(M).Then f(M) is contained in a sphere.
Moreover, if rank
f
(M) ≥ 2, then M is a submanifold with constant principal cur-
vatures (and hence f (M) is either an isoparametric hypersurface of the sphere or an
orbit of an s-representation).
Corollary 5.1.8 (Olmos [257, 258]) Let M be an extrinsically homogeneous irre-
ducible full submanifold of R
n
with dim M ≥2 and rank(M) ≥1. Then M is contained
in a sphere. Moreover, if rank(M) ≥ 2, then M is an orbit of an s-representation.
By Theorem 2.5.2 there exist no minimal homogeneous submanifolds of R
n
apart
from the totally geodesic submanifolds. Hence we also get the following corollary:
Corollary 5.1.9 (Di Scala, Olmos [105, 257, 258]) Let M be an extrinsically homo-
geneous irreducible full submanifold of R
n
with parallel mean curvature vector field
and dim M ≥2. Then M is either a minimal submanifold of a sphere in R
n
or an orbit
of an s-representation.
The previous corollary cannot be strengthened, since any representation of a com-
pact Lie group has a minimal orbit in the sphere (for example, a principal orbit with
maximal volume, see [154]).
We will now explain the main steps used in the proof of Theorem 5.1.7; details
can be found in [108]. Our aim is to demonstrate that the curvature normals are
parallel in the normal connection and then the result follows from Theorem 4.5.8
and its extension to submanifolds of Euclidean spaces in Section 4.5.3. Our strategy
is to show that if there exists a nonparallel curvature normal, then the submanifold
must split off a curve, which contradicts the assumption of irreducibility.
Simplifying hypothesis: We will make some extra assumptions that spare the
technical details. These assumptions are automatically fulfilled if M satisfies the
hypotheses in Corollary 5.1.8, that is, if M is a homogeneous submanifold with
rank(M) ≥ 1 (simply connectedness is not important, since one can pass to the uni-
versal covering space). Here are our extra assumptions:
Rank Rigidity of Submanifolds and Normal Holonomy of Orbits 177
(ex
1
) The number of curvature normals is constant on M and
rank
f
(M)=rank
loc
f
(M).
(ex
2
) If a curvature normal is parallel in an open and nonempty subset of M,thenit
is globally parallel.
Note that if the number of curvature normals is constant on M, then the curva-
ture normals and their associated distributions are globally defined and smooth (see
Exercise 5.3.5). Also note that the second assumption in (ex
1
) can be replaced by the
assumption that the local rank rank
loc
f
(M)
p
is constant on M (see Exercise 5.3.3).
Suppose that the simplifying hypotheses (ex
1
)and(ex
2
) for Theorem 5.1.7 hold.
Let E
1
,...,E
g
be the (globally defined) distributions and
η
1
,...,
η
g
be the corre-
sponding curvature normals. We can assume that
η
i
≥
η
j
if i < j. If all curva-
ture normals are parallel, then each of them provides a global isoparametric normal
section and we have finished the proof by Theorem 4.5.8. Let us then assume that the
curvature normals are not all parallel. Let k ∈{1,...,g} be the minimal index such
that
η
k
is not parallel. So
η
k
is a nonparallel curvature normal of maximal length and
η
1
,...,
η
k−1
are parallel curvature normals.
For i, j ∈{1,...,g} we d e fine the map h
ij
: M → R by
h
ij
=
η
j
2
−
η
i
,
η
j
=
η
j
−
η
i
,
η
j
.
By the Cauchy-Schwarz inequality we have h
ij
> 0ifi > j ≥ k.LetJ be an arbitrary
subset of {1,...,g}{k} (J = /0 is allowed!) and define
Ω
J
= {p ∈ M : h
jk
(p)=0 if and only if j ∈ J}
o
,
where the superscript denotes the interior of the set. Observe that Ω
J
= /0ifJ is not
contained in {1,...,k −1}. Note also that
Ω
/0
= {p ∈ M : h
jk
(p) = 0forall j ∈{1,...,g}{k}}
o
.
In particular, Ω
/0
= M if k = 1 . It is a standard fact that
Ω =
!
J⊂{1,...,k−1}
Ω
J
is an open and dense subset of M.
We will show that the distribution E
k
is autoparallel. It suffices to show that the
restriction E
k
|
Ω
J
is autoparallel for any J ⊂{1,...,k −1}. For this we will follow the
outline in Section 2 of [257]. Let J ⊂{1,...,k −1}. Without loss of generality we
may assume that J = {1,...,s} with s < k.Then
(i)
η
i
,
η
j
is constant if i, j < k (in particular, if i, j ≤s), since
η
i
,
η
j
are parallel.
(ii)
η
i
,
η
i
>
η
i
,
η
k
for all i ≤ s,since
η
i
≥
η
k
.
(iii)
η
k
,
η
k
>
η
l
,
η
k
for all l > k ,since
η
k
≥
η
l
.
178 Submanifolds and Holonomy
(iv)
η
i
,
η
k
is constant on Ω
J
for i ≤ s ,since
η
i
,
η
k
=
η
k
2
and the curvature
normals have constant length.
Let p ∈ Ω
J
and
ξ
p
be the parallel normal vector field on M with
ξ
p
(p)=
η
k
(p).
The shape operator A
ξ
p does not distinguish the distribution E
k
near p unless J = /0.
This is because of the d efinition of Ω
J
, which implies
λ
k
(
ξ
p
(p)) −
λ
i
(
ξ
p
(p)) =
h
ik
(p)=0fori ∈{1,...,s}. Also note that
λ
k
(
ξ
p
(p))−
λ
l
(
ξ
p
(p)) = h
lk
(p) = 0inΩ
J
for all k = l > s. It is clear by (i), (ii), (iii), and (iv) that there exists a parallel normal
vector field
¯
ξ
on M that is a linear combination of
η
1
,...,
η
s
andsuchthat
¯
ξ
+
ξ
p
dis-
tinguishes near p the distribution E
k
(that is,
λ
j
(
¯
ξ
(p)+
ξ
p
(p)) =
λ
k
(
¯
ξ
(p)+
ξ
p
(p))
if j = k). From (iv),
¯
ξ
,
η
k
= c is a constant, so
¯
ξ
+
ξ
p
,
η
k
= c +
ξ
p
,
η
k
.Us-
ing the Cauchy-Schwarz inequality and since
ξ
p
=
η
k
is constant, the function
ξ
p
,
η
k
has a maximum at p and
¯
ξ
+
ξ
p
,
η
k
attains its maximum at p, too. Hence
its differential at p is zero. The selfadjoint tensor field
T
p
= A
¯
ξ
+
ξ
p
−
¯
ξ
+
ξ
p
,
η
k
id
then satisfies the Codazzi equation (only) at p,sinceA
¯
ξ
+
ξ
p
and id satisfy the Codazzi
equation. Namely, (∇
X
T
p
)Y =(∇
Y
T
p
)X,where∇ is the Levi-Civita connection on
M and (∇
X
T
p
)Y = ∇
X
(T
p
Y ) −T
p
∇
X
Y . Equivalently, the tensor field (∇
X
T
p
)Y,Z
is symmetric in all its three entries. Since E
k
= ker(T
p
) near p,wegetforX,Y
tangent to E
k
and Z arbitrary
0 = (∇
Z
T
p
)Y,X
p
= (∇
X
T
p
)Y,Z
p
= −T
p
∇
X
Y,Z
p
and hence (∇
X
Y )
p
∈(ker(T
p
))
p
= E
k
(p).Sincep is arbitrary we conclude that E
k
is
autoparallel on Ω
J
.AsJ is arbitrary, we can now conclude that E
k
is an autoparallel
distribution on Ω and hence on M. Applying Lemma 5.1.3 we obtain rk(E
k
)=1,
since
η
k
is not parallel. Moreover, the distribution E
⊥
k
is integrable, as we will show
in the next lemma.
In order to reinforce these ideas, it would be convenient for the reader to re-
produce the above arguments in the important case k = 1 (thus proving that E
1
is
autoparallel on M = Ω
/0
).
Lemma 5.1.10 Under the above assumptions we have:
(i) The distribution E
k
is autoparallel and rk(E
k
)=1.
(ii) The distribution E
⊥
k
is integrable.
Proof Part (i) was proved above. Let
˜
Ω be the subset of M on which
η
k
is not parallel.
Then
˜
Ω is an open and dense subset of M by condition (ex
2
). Using part (i) and
Lemma 3.4.2 we g et for q ∈
˜
Ω:
∇
⊥
Z
q
η
k
= 0 if and only if Z
q
∈ E
⊥
k
(q).
Let X,Y be vector fields on M tangent to E
⊥
k
.Since(
ν
M)
0
is flat, we get
0 = R
⊥
(X,Y )
η
k
= ∇
⊥
X
∇
⊥
Y
η
k
−∇
⊥
Y
∇
⊥
X
η
k
−∇
⊥
[X,Y ]
η
k
= −∇
⊥
[X,Y ]
η
k
.
Rank Rigidity of Submanifolds and Normal Holonomy of Orbits 179
Thus [X ,Y]
q
∈ E
⊥
k
(q) for q ∈
˜
Ω.SoE
⊥
k
is involutive on
˜
Ω and hence on M,which
implies that E
⊥
k
is an integrable distribution.
Since M is simply connected, we have E
k
= RX with some globally defined unit
vector field X on M. The integral curves of X are unit speed geodesics, since E
k
is
autoparallel. Let
φ
t
be the flow of X.Thenwehave
φ
∗
t
E
k
= E
k
and
φ
∗
t
E
⊥
k
= E
⊥
k
(5.2)
for all t ∈ R.Thefirst equ a lity is clear. For the secon d equality, let c(s) be a curve
tangent to E
⊥
k
and define h(s,t)=
φ
t
(c(s)) and the geodesic
γ
(t)=
φ
t
(c(0)).Since
X= 1, the vector field
J(t)=
∂
∂
s
s=0
h(s,t)=d
c(0)
φ
t
(c
(0))
is the Jacobi field along
γ
with initial conditions J(0)=c
(0) ⊥X
c(0)
and
J
(0)=
D
∂
t
t=0
∂
∂
s
s=0
h(s,t)=
D
∂
s
s=0
∂
∂
t
t=0
h(s,t)=
D
ds
s=0
X
c(s)
⊥ X
c(0)
.
Then J(t) is always perpendicular to
γ
(t), which generates E
k
(
γ
(t)). This shows that
φ
∗
t
E
⊥
k
= E
⊥
k
.
Let p ∈ M and L
p
be the maximal integral manifold of E
⊥
k
containing p.Then
there exist an open neighborhood V of p in L
p
and
ε
> 0suchthat
g : [−
ε
,
ε
] ×V → M , (t, q) → g(t, q)=
φ
t
(q)
is a diffeomorphism onto its image. Let ˜c : [0,1] → M be a piecewise differentiable
loop at p that is contained in g ([−
ε
,
ε
] ×V ). If we write ˜c(s)=g(h(s),c(s)) =
φ
h(s)
(c(s)), then both h and c are closed curves starting at 0 and p respectively. Now
define
g
c
: [−
ε
,
ε
] ×[0,1] → M , (t,s)=g(t,c(s)).
From (5.2) we get
R
⊥
∂
g
c
∂
t
,
∂
g
c
∂
s
= 0, (5.3)
where we used the Ricci equation and the fact that E
k
is invariant under all shape
operators. Observe now that c
1
(s)=(0,s) and c
2
(s)=(h(s),s) are both curves in
[−
ε
,
ε
] ×[0,1] from (0,0) to (0, 1). Then, using (5.3) and Exercise 5.3.2, we g et
τ
⊥
g
c
◦c
1
=
τ
⊥
g
c
◦c
2
=
τ
⊥
˜c
.
This we have proved the following result:
Lemma 5.1.11 For every p ∈M there exists an open neighborhood U of p in M such
that for every loop c at p contained in U there exists a loop ¯c a t p contained in the
integral manifold L
p
of E
⊥
k
such that
τ
⊥
c
=
τ
⊥
¯c
.
180 Submanifolds and Holonomy
For every p ∈ M we have
ν
p
L
p
=
ν
p
M ⊕RX
p
when considering L
p
as a sub-
manifold of R
n
. Moreover, the restriction of X to L
p
defines a parallel normal vector
field on L
p
. In fact, let Z be a vector field on M tangent to E
⊥
k
and let
¯
∇ be the Levi-
Civita connection on R
n
.Then
¯
∇
Z
X,X = 0, since X = 1, and the projection of
¯
∇
Z
X to the normal space of M is
α
(X, Z)=0, where
α
is the second fundamental
form of M. This observation, together with Lemma 5.1.11, implies that
(
ν
p
M)
0
⊕RX
p
=(
ν
p
L
p
)
0
(5.4)
for all p ∈ M.Letp ∈ M and
γ
p
: R → M be the integral curve of X with
γ
p
(0)=p
(observe that
γ
p
(t)=
φ
t
(p) is the geodesic with initial condition X
p
). Then we get
d
p
f (E
⊥
k
(p)) = d
p
f (E
⊥
k
(
γ
(t))) for all t ∈R, or equivalently,
d
p
(T
p
L
p
)=d
p
f (T
φ
t
(p)
L
φ
t
(p)
) for all t ∈ R, (5.5)
where f : M → R
n
is the given immersion. In fact, let ¯v ∈ E
⊥
k
(p) and let v(t) be the
parallel transport in M of ¯v along the geodesic
γ
p
. Identifying w ∈ TM with df(w),
we ge t v
(t)=0 (using the fact that
γ
p
is parallel along
γ
p
and that
α
(
γ
p
,v)=0).
Thus v(t) is constant in R
n
, which implies the above equality.
We now fix t ∈ R andanintegralmanifoldL
p
of E
⊥
k
and define the map
ξ
t
: L
p
→ R
n
, q → f (
φ
t
(q)).
Equalities (5. 2) and (5.5) imply that
ξ
t
can be regarded as a parallel normal vector
field on L
p
. Now, (5.2) gives
φ
t
(L
p
)=L
φ
t
(p)
, hence L
p
and L
φ
t
(p)
are parallel
submanifolds of R
n
. More precisely,
(L
p
)
ξ
t
= L
φ
t
(p)
. (5.6)
Remark 5.1.12 Let
γ
p
: R → M be the integral curve of X with
γ
p
(0)=p and let
ξ
be a parallel normal vector field on M. Identifying w ∈ TM with df(w),wehave
0 =
d
dt
t=0
ξ
γ
p
(t )
,
γ
p
(t) = −A
ξ
p
X
p
,X
p
+
ξ
p
,
α
(X
p
,X
p
),
and so
ξ
p
,
η
k
(p) =
ξ
p
,
α
(X
p
,X
p
)
for all
ξ
p
∈(
ν
p
(M))
0
. Thus, the curvature normal
η
k
(p) coincides with the orthogo-
nal projection of
α
(X
p
,X
p
) to (
ν
p
(M))
0
. Moreover, we have the equality
η
k
=
α
(X, X ).
In fact, since L
p
→R
n
, q →
ξ
t
(q)= f (
φ
t
(q))−q is a parallel normal vector field on
L
p
for all t ∈ R, we obtain that
d
2
dt
2
|
t=0
ξ
t
=
α
(X, X ) must also be a parallel normal
vector field on L
p
(X is restricted to L
p
). In order to prove this, observe that the
parallel transport
τ
⊥
p,q
from p to q in (
ν
L
p
)
0
is an isometry mapping the curve t →
ξ
t
(p) into t →
ξ
t
(q) and so
τ
⊥
p,q
(
d
2
dt
2
|
t=0
ξ
t
(p)) =
d
2
dt
2
|
t=0
ξ
t
(q) defines a global parallel
normal vector field on L
p
.But
d
2
dt
2
|
t=0
ξ
t
(p)=
γ
p
(0), and this is perpendicular to
γ
p
(0)=X
p
since ||X||= 1. By (5.4),
γ
p
(0) ∈(
ν
p
M)
0
. This gives the desired equality.
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