Rank Rigidity of Submanifolds and Normal Holonomy of Orbits 181
The following lemma r elates the curvature n ormals of the isometric immersion
f : M → R
n
to the curvature normals of the integral manifolds of E
⊥
k
.
Lemma 5.1.13 Under the above assumptions we have:
(i) The eigenspaces of the simultaneous diagonalization of the shape operators of
(
ν
p
L
p
)
0
coincide with E
1
(p),...,E
k−1
(p),E
k+1
(p),...,E
g
(p).
(ii) Let
˜
η
1
(p),...,
˜
η
k−1
(p),
˜
η
k+1
(p),...,
˜
η
g
(p) be the curvature normals of L
p
at p (L
p
regarded as a submanifold of R
n
) associated with decomposition
T
p
L
p
= E
1
(p) ⊕...⊕E
k−1
(p) ⊕E
k+1
(p) ⊕...⊕E
g
(p).Then
η
i
(
γ
p
(t)) =
˜
η
i
(p)
1 −
˜
η
i
(p),
˜
γ
p
(t)
−
"
˜
η
i
(p)
1 −
˜
η
i
(p),
˜
γ
p
(t)
,
˜
γ
p
(t)
#
˜
γ
p
(t),
where i = k,
γ
p
is the integral curve of X with
γ
p
(0)=p and
˜
γ
p
= f ◦
γ
p
− f (p).
Proof Let
˜
A denote the shape operator o f the integral manifolds of E
⊥
k
,regardedas
submanifolds of R
n
,andletX be the unit tangent vector field on M generating E
k
.
By (5.4), it suffices to show that
˜
A
X
leaves E
i
invariant and that it is a multiple of the
identity (i = k). Let X
i
,Y
j
be vector fields on M tangent to E
i
and E
j
respectively. Let
ξ
be a (locally defined) parallel normal vector field on M distinguishing all the dif-
ferent eigenvalues
λ
1
=
η
1
,·,...,
λ
g
=
η
g
,· (i, j = k). From the Codazzi equation
(∇
Y
j
A
ξ
)X, X
i
= (∇
X
i
A
ξ
)X,Y
j
we easily obtain
ξ
,
η
k
−
η
i
∇
Y
j
X,X
i
=
ξ
,
η
k
−
η
j
∇
X
i
X,Y
j
. (5.7)
Since ∇
X
i
X,Y
j
= −
˜
A
X
X
i
,Y
j
= ∇
Y
j
X,X
i
, equation (5.7) implies
˜
A
X
X
i
,Y
j
= 0if
i = j. Thus, E
i
is invariant under the shape operators of L
p
.Nowleti = j and assume
that X
i
is perpendicular to Y
i
. A direct computation shows that (∇
X
A
ξ
)X
i
,Y
i
= 0.
By the Codazzi equation we have 0 = (∇
X
i
A
ξ
)X,Y
i
=
ξ
,
η
k
−
η
i
∇
X
i
X,Y
i
,which
implies
˜
A
X
X
i
,Y
i
= 0. Thus
˜
A
X
|
E
i
must be a multiple of the identity. This implies
statement (i).
From the “tube formula”, which relates the shape operators of parallel manifolds,
the curvature normals of L
γ
p
(t )
are (1 −
˜
η
i
(p),
˜
γ
p
(t))
−1
˜
η
i
(p) for all i = k.If
π
t
denotes the orthogonal projection onto (R
˜
γ
p
(t))
⊥
, it is not hard to see that
η
i
(p)=
π
t
((1 −
˜
η
i
(p),
˜
γ
p
(t))
−1
˜
η
i
(p)), wh ich implies statement (ii) .
Remark 5.1.14 Assuming that X
i
= 1and∇
⊥
ξ
= 0, the Codazzi equation implies
d
ξ
,
η
i
(X)=
ξ
,
η
i
−
η
k
˜
A
X
X
i
,X
i
.
Proof of Theorem 5.1.7 (under the assumptions (ex
1
)and(ex
2
))
By Lemma 5.1.13(ii), the curvature normals
η
i
, i = k, satisfy the equation
η
i
(
γ
p
(t))
2
=
1
(1 −
˜
η
i
(p),
˜
γ
p
(t))
2
˜
η
i
(p)
2
−
1
1 −
˜
η
i
(p),
˜
γ
p
(t)
˜
η
i
(p),
˜
γ
p
(t)
2
.
182 Submanifolds and Holonomy
Put c
i
=
η
i
2
and ˜c
i
=
˜
η
i
(p)
2
. The function f
i
(t)=1 −
˜
η
i
(p),
˜
γ
p
(t) satisfies
f
i
(t)
2
c
i
= ˜c
i
− f
i
(t)
2
where c
i
and ˜c
i
are constants. By taking derivatives it is not hard to con-
clude that f
i
(t)=0or f
i
(t)c
i
+ f
i
(t)=0. Then either f
i
(t)=1or f
i
(t)=
sin(
√
c
i
(t + t
0
))/sin(
√
c
i
t
0
),wheret
0
satisfies cot
2
(
√
c
i
t
0
)=(˜c
i
−c
i
)/c
i
(observe that
( f
i
)
2
(0)=(˜c
i
−c
i
f
i
(0)
2
)/c
i
and that f
i
(0)=1). The last case cannot occur because
it would imply that we cannot pass to a parallel leaf when
√
c
i
(t + t
0
) is a root of
sin(x)=0 (recall that M is complete). So
˜
η
i
(p),
˜
γ
p
(t) vanishes. Differentiating
twice it follows that
η
i
,
η
k
= 0onM,since
η
k
(
˜
γ
p
(t)) =
˜
γ
p
(t) by Remark 5.1.12.
We will now prove that f : M → R
n
splits. Note that E
k
is invariant by the shape
operators of M.SinceM is simply connected, it suffices to show that E
⊥
k
is au-
toparallel (recall that if the orthogonal complement of an autoparallel distribution is
autoparallel then both distributions must be parallel). Since
π
1
(M)=0, M must split
intrinsically and we can apply Moore’s Lemma 1.7.1 to split the immersion.
Let us show that E
⊥
k
is an autoparallel distribution on M.Let
˜
A be the shape
operator of the integral manifolds of E
⊥
k
, regarded as submanifolds of R
n
.Observe
that
˜
A
X
coincides with the shape operator of the integral manifolds of E
⊥
k
regarded
as hypersurfaces of M. We claim that
˜
A
X
= 0. In fact, let q ∈ M be fixed and let
ξ
q
be the parallel normal vector field on M with
ξ
q
(q)=
η
k
(q). Then the left hand side
of the equation in Remark 5.1.14 vanishes at q , because the function
ξ
q
,
η
k
has a
maximum at q (using Cauchy-Schwarz inequality, since
ξ
q
and
η
k
have both constant
length). The other side of the equality of Remark 5.1.14 implies that
˜
A
X
X
i
,X
i
q
= 0,
for
ξ
q
,
η
i
(q) = 0. Since q is arbitrary, we obtain that
˜
A
X
= 0. In summary, we have
shown that if some curvature normal is not ∇
⊥
-parallel, we can glob ally split the
immersion f : M → R
n
. This completes the proof of Theorem 5.1.7.
Remark 5.1.15 Let f : M → R
n
be an immersed submanifold with flat normal bun-
dle. The inverse of the length o f any nonzero curvature normal
η
(p) coincides with
the distance in
ν
p
M to the focal hyperplane given by the equation
η
(p),· = 1.
Therefore, M has curvature normals of constant length if and only if the distances to
the focal hyperplanes are constant on M (this is always the case if M has, in addition,
algebraically constant second fundamental form). Theorem 5.1.7 then allows us to
give a global (equivalent) definition of an isoparam etric submanifold: an immersed
complete irreducible submanifold f : M →R
n
with dim M ≥2andflat normal bundle
is isoparametric if the distances to their focal hyperplanes are constant on M.
Remark 5.1.16 Let f : M → R
n
be an immersed submanifold with flat normal bun-
dle. Assume that the curvature normals have all the same length
−1
. This is equiv-
alent to saying that the Gauss map is homothetic, that is, the metric induced by the
Gauss map is a constant multiple of the Riemannian metric on M. In this situation
N¨olker [250] proved that M is a product of spheres of radius and curves with cur-
vature . This is also true locally. Roughly speaking, the proof goes like this: any
curvature distribution on M is autoparallel since it is associated to a curvature nor-
mal of maximal length. If M is not isoparametric, there exists a nonparallel curvature
Rank Rigidity of Submanifolds and Normal Holonomy of Orbits 183
normal, so the immersion must split off a curve. If M is an irreducible isoparametric
submanifold d ifferent from a sphere, then the curvature normals cannot all have the
same length. In fact, must equal the distance to any focal hyperplane.
5.1.4 Local counterexamples
Theorem 5.1.7 fails without the completeness assumption o n M. There exist non-
isoparametric (non-complete) submanifolds of R
n
with flat normal bundle and al-
gebraically constant second fundamental form [108]. We outline the construction of
such examples: one begins with a non-full unit sphere S
n
in R
n+k
.Intheaffine nor-
mal space p +
ν
p
S
n
at a fixed point p ∈ S
n
,acurvec
p
is constructed starting at p and
satisfying certain requirements (in particular, the curvature of c
p
has to be constant).
By means of parallel transport using the normal connection of S
n
one constructs a
curve c
q
in the affine normal space at any q. The union of the images of such curves
gives the desired submanifold. This submanifold h as only two curvature distribu-
tions. One of them, say E
1
, is autoparallel with integral curves c
q
. If one starts with a
circle in R
4
, the simplest nontrivial example produced is a surface in R
4
.Allexam-
ples constructed in such a way are intrinsically (n + 1)-spheres. Observe that the two
curvature normals must satisfy
η
1
,
η
2
= ||
η
2
||
2
,otherwiseE
2
would b e autoparallel
and M would split.
5.2 Normal holonomy of orbits
To simplify the exposition, we will assume in this section that the submanifolds
are embedded, but everything holds also for immersed submanifolds.
5.2.1 Transvections
Let M be a submanifold of R
n
. We will be concerned here with the group of
transvections Tr(M,∇
⊥
) of the normal connection ∇
⊥
that we already encountered
in Section 4.4. Recall that Tr(M,∇
⊥
) is the group of isometries of R
n
leaving M
invariant and preserving any normal holonomy subbundle. More explicitly, an isom-
etry g ∈ I(R
n
) with g (M)=M is in Tr(M, ∇
⊥
) if for every p ∈ M there exists a
piecewise differentiable curve c : [0,1] → M with c(0)=p and c(1)=g(p) such
that d
p
g|
ν
p
M
=
τ
⊥
c
,where
τ
⊥
c
is the parallel transport along c with respect to ∇
⊥
.
In a similar way, we define Tr
0
(M,∇
⊥
) and Tr
s
(M,∇
⊥
) by replacing the last con-
dition with d
p
g|
(
ν
p
M)
0
=
τ
⊥
c
|
(
ν
p
M)
0
and d
p
g|
(
ν
p
M)
s
=
τ
⊥
c
|
(
ν
p
M)
s
respectively. Recall
that (
ν
M)
s
=((
ν
M)
0
)
⊥
is the subbundle of the normal bundle on which the normal
holonomy group acts as an s-representation.
If M is a full submanifold of R
n
with constant principal curvatures, then the asso-
ciated curvature normals
η
1
,...,
η
g
(with respect to (
ν
M)
0
) are parallel and generate
184 Submanifolds and Holonomy
(
ν
M)
0
, since the first normal space coincides with the normal space. If g ∈I(R
n
) with
g(M)=M, then its differential dg maps curvature normals into curvature normals.
More precisely, if
η
is a curvature normal, then dg(
η
) is a curvature normal, where
dg(
η
)(gp)=d
p
g(
η
p
).Ifg can be continuously deformed to the identity through
extrinsic isometries of M,thendg(
η
)=
η
(since there are finitely many curvature
normals). This observation, together with Corollary 5.1.8, implies the following re-
mark:
Remark 5.2.1 Let G be a connected Lie subgroup of I(R
n
) and M = G · p beafull
irreducible homogeneous submanifold of R
n
with dim M ≥2. Then G ⊂Tr
0
(M,∇
⊥
).
We will see in next theorem that the inclusion G ⊂ Tr
s
(M,∇
⊥
) is a general fact.
It depends on the following well-known result, for which we include a proof using
standard theory about Riemannian holonomy.
Lemma 5.2.2 Let G be a connected Lie subgroup of SO
n
and assume that G acts
on R
n
as an s-representation. Then G = N
SO
n
(G)
o
,whereN
SO
n
(G)
o
is the identity
component of the normalizer of G in S O
n
.
Proof For an irreducible Riemannian symmetric space M the holonomy and isotropy
representations coincide. Moreover, local and global holonomy coincide as well.
Then the proof follows from the next proposition because M cannot be Ricci-flat.
Proposition 5.2.3 (cf. [92]) Let M be a Riemannian manifold and assume that M
is irreducible at p ∈ M. Let g be the Lie algebra of the local (Riemannian) holon-
omy group Hol
loc
p
at p and n be the normalizer of g in so(T
p
M).Thenn contains g
properly if and only if M is K
¨
ahler and Ricci-flat near p.
Proof We endow so(T
p
M) with the usual inner product A, B = −tr(A.B). Assume
that n = g. If we decompose n = g ⊕k orthogonally, then g and k are ideals of n and
so [g, k]=0. Now choose 0 = J
p
∈ k.ThenJ
2
p
is a self-adjoint endomorphism and
commutes with g. Thus, J
2
p
commutes with Hol
loc
p
and each eigenspace of J
2
p
defines
a parallel distribution o n M near p.SinceM is irreducible at p we conclude, by the de
Rham Decomposition Theorem, that J
2
p
= −c
2
id. By rescaling J
p
we can assume that
J
2
p
= −id. Extending J
p
by parallelism we obtain a parallel almost complex structure
J on M. Thus, M is K¨ahler near p. The Ricci tensor ric of a K¨ahler manifold M
satisfies (see Appendix A.1)
2ric(X ,JY )=R(X,Y ),J.
If
γ
is any curve in a small neighborhood of p joining p to some nearby point q and
τ
γ
is the parallel transport along
γ
,then
R(X
q
,Y
q
),J
q
=
τ
−1
γ
R(X
q
,Y
q
)
τ
γ
,J
p
= 0
since J
p
is perpendicular to g. Thus M is Ricci-flat near p.
The previous two equations, together with the Ambrose-Singer Holonomy The-
orem, show that the converse is true.
Rank Rigidity of Submanifolds and Normal Holonomy of Orbits 185
Theorem 5.2.4 Let G be a connected Lie subgroup of I(R
n
) and M = G · pbea
homogeneous submanifold of R
n
. Then:
(i) G ⊂ Tr
s
(M,∇
⊥
).
(ii) G ⊂ Tr(M, ∇
⊥
) if M is an irreducible full submanifold of R
n
and dimM ≥ 2.
Proof Let g ∈G and ˜g : [0,1] → G be a differentiable curve in G with ˜g(0)=id and
˜g(1)=g.Letq ∈ M and
γ
(t)= ˜g(t)q. The restricted normal holonomy groups at q
and
γ
(t) are conjugate via the differential of ˜g,thatis,
Φ
∗
γ
(t )
= d ˜g(t)Φ
∗
q
(d ˜g(t))
−1
.
The normal holonomy groups are also conjugate under parallel transport
Φ
∗
q
=(
τ
⊥
γ
t
)
−1
Φ
∗
γ
(t )
τ
⊥
γ
t
= h
t
Φ
∗
q
h
−1
t
,
where
τ
⊥
γ
t
is the ∇
⊥
-parallel transport along
γ
t
=
γ
|
[0,t]
and h
t
=(
τ
⊥
γ
t
)
−1
d ˜g(t). Hence
(
τ
⊥
γ
)
−1
dg ∈ N
SO
n
(Φ
∗
q
)
o
. By the Normal Holonomy Theorem the restricted normal
holonomy group Φ
∗
q
acts on (
ν
q
M)
s
as an s-representation. Then, by Lemma 5.2.2,
there exists
τ
⊥
c
∈ Φ
∗
q
,wherec is a null-homotopic loop at q, such that d
q
g|
(
ν
q
M)
s
=
τ
⊥
γ
τ
⊥
c
|
(
ν
q
M)
s
. Hence d
q
g|
(
ν
q
M)
s
=
τ
⊥
c∗
γ
|
(
ν
q
M)
s
, which proves (i).
From Remark 5.2.1 we have
τ
⊥
γ
|
(
ν
q
M)
0
= d
q
g|
(
ν
q
M)
0
. Moreover,
τ
⊥
c
|
(
ν
q
M)
0
= id
since (
ν
M)
0
is flat and c is null-homotopic. Then d
q
g|
(
ν
q
M)
0
=
τ
⊥
c∗
γ
|
(
ν
q
M)
0
and so
dg =
τ
⊥
c∗
γ
, which proves (ii).
Part (ii) of Theorem 5 .2.4 can be restated as follows:
Theorem 5.2.5 Let G be a connected Lie subgroup of I(R
n
) and M = G · pbean
irreducible full homogeneous submanifold of R
n
with dimM ≥ 2. For all g ∈ G and
q ∈ M there exists a piecewise differentiable curve c : [0, 1] → M with c(0)=q and
c(1)=g(q) such that
d
q
g|
ν
q
M
=
τ
⊥
c
,
where
τ
⊥
c
is the ∇
⊥
-parallel transport along c.
The following corollary of Theorem 5.2.4 has an analogue in the theory of Rie-
mannian holonomy: Let M be a homogeneous Riemannian manifold without flat de
Rham factor. Then the isotropy group of M is contained in the holonomy group of M
(see [134, 4.5, page 110]).
Corollary 5.2.6 Let G be a connected Lie subgroup of I(R
n
) and M = G ·p be a full
homogeneous submanifold of R
n
. Then the image under the slice representation of
the isotropy group G
p
is contained in the normal holonomy group of M at p.
In this corollary we do not have to assume that M is irreducible or dimM ≥ 2.
By Remark 2.1.4 the connected component of the extrinsic group of isometries is
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