72 Submanifolds and Holonomy
In the remainder of this subsection we will be concerned with the proof of Theo-
rem 2.8.8. Since M is a locally symmetric submanifold, its second fundamental form
α
is parallel. It follows that the nullity distribution E
0
(see Section 1.3) on M has
constant rank. Moreover, we have
α
(
X
Y,Z)=
X
α
(Y,Z)
α
(Y,
X
Z)=0
for all sections Y in E
0
and all vector elds X ,Z on M. This shows that E
0
is a parallel
subbundle of TM.Since
α
(E
0
,E
0
)=0bydenition of E
0
, Moore’s Lemma 1.7.1
now implies that M is locally a submanifold product of R
n
with a Euclidean factor,
a leaf of E
0
, if the rank of E
0
is nonzero. Without loss of generality we can assume
from now on that the nullity distribution on M is trivial.
Since
α
is parallel, the mean curvature vector eld H of M is a parallel normal
vector eld and the shape operator A
H
is a parallel selfadjoint tensor eld on M.
Therefore the principal curvatures
λ
1
,...,
λ
s
of M with respect to H are constant
and h ence, since A
H
is parallel, the corresponding principal curvature spaces form
parallel distributions E
1
,...,E
s
on M.
Since R
(X,Y )H = 0forallX ,Y T
p
M, p M, the Ricci eq uation implies
[A
H
,A
ξ
]=0 for all normal vector elds
ξ
on M. Therefore each eigendistribution E
i
is invariant under all shape operators, that is, A
ξ
E
i
E
i
for all normal vector elds
ξ
on M. This implies in particular that
α
(E
i
,E
j
)=0foralli = j. (2.9)
Since the bundles E
i
are parallel, (2.9) and Moore’s Lemma 1.7.1 imply that M is
locally a submanifold product M = M
1
×...×M
s
,whereM
i
is an integral manifold
of E
i
and a submanifold of a suitable R
¯m
i
R
n
. We denote by
α
i
the second fun-
damental form of M
i
R
¯m
i
and by
π
i
: M M
i
the canonical projection. Then we
have
α
(X,Y )=(
α
1
(d
p
π
1
(X), d
p
π
1
(Y )),...,
α
s
(d
p
π
s
(X), d
p
π
s
(Y )))
for all X,Y T
p
M, p M. This implies that
α
i
is parallel as well. By the theorem
on the reduction of codimension (Theorem 1.5.1) we can reduce the codimension of
each M
i
, since the distribution of the rst normal spaces is parallel (see, for instance,
Exercise 1.8.8). Thus, for each M
i
we get a full immersion M
i
R
m
i
R
¯m
i
.LetH
i
be the m ean curvature vector eld of M
i
.ThenH =(H
1
,...,H
s
) and
α
i
(X,Y ),H
i
=
α
(X,Y ),H =
λ
i
X,Y
for all X ,Y T
p
M
i
, p M
i
.If
λ
i
= 0, this shows that M
i
is a p seudoumbilical subma-
nifold of R
m
i
with parallel mean curvature vector eld and
λ
i
= H
i
. By Proposition
1.6.3 we g et that M
i
is minimal in a hypersphere of R
m
i
. Part (a) of Theorem 2.8.8
then follows from the following lemma and the assumption that the nullity distribu-
tion of M is trivial.
Lemma 2.8.10 Every connected minimal submanifold M with parallel second fun-
damental form in a standard space form
¯
M
n
(
κ
) with
κ
0 is totally geodesic.
Submanifold Geometry of Orbits 73
Proof If the submanifold is complete, then it is a symmetric submanifold by Theo-
rem 2.8.3 and therefore homogeneous, which implies that it is totally geodesic b y the
results in sections 2.5 and 2.6. We will see in Lemma 2.8.13 that every locally sym-
metric submanifold is an open subset of a symmetric submanifold, which implies the
result.
There are also more direct proofs in [128] and [311]. We briey sketch the one
in [311]. We may assume that the dimension of the submanifold M is greater than
one. In this situation some calculations lead to (see also [83])
Δ
α
,
α
= n
κ
α
2
−
α
α
t
2
−R
2
, (2.10)
where Δ is the Laplace oper ator on tensor elds dened by tr(
2
).Since
α
= 0
we have Δ
α
= 0 and therefore
n
κ
α
2
=
α
α
t
2
+ R
2
,
and if
κ
0thisgives
α
= 0.
We continue with the proof of part (b) of Theorem 2.8.8. Without loss of gen-
erality we can assume that M is a full irreducible symmetric submanifold of R
n+1
.
Recall from part (a) that M is minimal in a hypersphere S
n
. We will assume that this
sphere is centered at the origin and h as radius
2m with m = dimM .
The idea of the proof is the following: First, we associate with M a Lie algebra k,
which can be seen as the Lie algebra o f all isometries of R
n+1
that leave M invariant.
Then we dene a Lie bracket on the vector space g = k R
n+1
in such a way that
(g,k) is a symmetric pair. This proof follows original ideas by Ferus [129] and is
completely algebraic. It will turn out that the d enition of g depends only on the
value of the second fundamental form of M at a single point. Hence, as a by-product,
we also get that a symmetric submanifold is uniquely determined by the value of its
second fundamental form at one point.
We x a point p M and put V = T
p
M and W =
ν
p
M. Since the immersion is
full, we have W = im
α
p
= span{
α
p
(v, v) : v V }, where the latter equality follows
by polarization of
α
.
We now introduce an operator which, although encoding the same piece of infor-
mation as the second fundamental form or the shape operator, turns out to be useful
especially for investigating homogeneous submanifolds. For each x V the innites-
imal transvection
ϕ
x
is the endomorphism
ϕ
x
: V W V W , X →
α
(x,X
T
) A
X
x.
We denote by m the real vector space that is spanned by {
ϕ
x
: x V }.
For x, y V we denote by
¯
R
x,y
the endomorphism on V W given by
¯
R
xy
v =
R(x,y)v for all v V and
¯
R
xy
ξ
= R
(x,y)
ξ
for all
ξ
W ,whereR is the Riemannian
curvature tensor of M and R
is the normal curvature tensor of M. From the equations
by Gauss and Ricci it follows that [
ϕ
x
,
ϕ
y
]=
¯
R
xy
.Wedene a subalgebra h of so(V )
74 Submanifolds and Holonomy
so(W ) so(V W )=so
n+1
by
h = {B so(V ) so(W ) : B ·
α
= 0}
= {B so(V ) so(W ) : [B, A
ξ
]=A
B
ξ
for all
ξ
W } (2.11)
= {B so(V ) so(W ) : [B,
ϕ
x
]=
ϕ
Bx
for all x V }.
Here B ·
α
means that B acts on
α
as a derivation. One should think of the elements
in h as the innitesimal isometries that are generated by one-parameter groups of
isometries of R
n+1
leaving M invariant and xing p.Theseinnitesimal isometries
are just the differentials of F
t
O
n+1
such that
F
t
α
(x,y)=
α
(F
t
x,F
t
y). (2.12)
In other words, if B h,thenF
t
= Exp(tB) satises (2.12). Notice also that
¯
R
xy
h.
The direct sum k = h m is a subalgebra of so
n+1
with bracket relations
[
ϕ
x
,
ϕ
y
]=
¯
R
xy
, [A,
ϕ
x
]=
ϕ
Ax
(x,y V, A h).
Let K be the connected Lie subgroup of SO
n+1
with Lie algebra k.Wenowdene a
Lie algebra structure on the vector space g = k R
n+1
= k V W. For this purpose
we consider the adjoint operator R
⊥∗
: Λ
2
W Λ
2
V of R
, which is characterized by
R
⊥∗
(
ξ
ζ
),x y = R
(x,y)
ξ
,
ζ
= [A
ξ
,A
ζ
]x,y
for x,y V and
ξ
,
ζ
W . Hence we can dene R
⊥∗
(
ξ
ζ
)x =[A
ξ
,A
ζ
]x.For
ξ
,
ζ
W
we dene an endomorphism R(
ξ
ζ
) on V W by R(
ξ
ζ
)x = R
⊥∗
(
ξ
ζ
)x and
R(
ξ
ζ
)
η
equal to the unique element
ρ
W such that
A
ρ
=[R
⊥∗
(
ξ
ζ
),A
η
]=[[A
ξ
,A
ζ
],A
η
]. (2.13)
Note that
ρ
is uniquely determined by (2.13), since M is full in R
n+1
and hence
ρ
→ A
ρ
has trivial kernel. By denition we have [R(
ξ
ζ
),A
η
]=A
R(
ξ
ζ
)
η
,and
thus R(
ξ
ζ
) h.
We are now able to dene the Lie algebra structure on g. For two elements in
k we use the Lie algebra structure on k.IfB k and X V W ,thenwedene
[B,X]=BX. Finally, f or v,w V and
ξ
,
ζ
W we dene
[v,w]=[
ϕ
v
,
ϕ
w
] , [v,
ξ
]=
ϕ
A
ξ
v
, [
ξ
,
ζ
]=R(
ξ
ζ
).
We have to verify that the Jacob i identity holds. This is obvious if all elements
lie in k. Since the bracket on g is equ ivariant with respect to the action of K on g,it
follows by differentiation tha t the Jacobi identity holds if at least one element lies in
k only. We are thus left to verify the Jacobi identity for elements in V W,thatis,we
have to show that S[X ,[Y, Z]] = 0forallX ,Y,Z V W,whereS denotes the cyclic
sum. This is clear if all the three elements lie in V , since then the Jacobi identity is
just the rst Bianchi identity for R.Ifx V and
ξ
,
ζ
W ,then
S[x,[
ξ
,
ζ
]] = [x, [A
ξ
,A
ζ
]]+[
ξ
,
ϕ
A
ζ
x
][
ζ
,
ϕ
A
ξ
x
]=[A
ξ
,A
ζ
]x+ A
ξ
A
ζ
xA
ζ
A
ξ
x = 0.
Submanifold Geometry of Orbits 75
If x,y V and
ξ
W ,then
S[x,[y,
ξ
]] = [x,
ϕ
A
ξ
y
]+[y,
ϕ
A
ξ
x
] [
ξ
,[
ϕ
x
,
ϕ
y
]]
=
α
(A
ξ
y,x)
α
(A
ξ
x,y)+R
(x,y)
ξ
= 0
by the Ricci equation. Finally, we prove the Jacobi identity for
ξ
,
ζ
,
η
W .SinceM
is full in R
n+1
,itsufces to prove that A
S[
ξ
,[
ζ
,
η
]]
= 0. But this follows from
A
[
ξ
,[
ζ
,
η
]]
= A
[
ζ
,
η
]
ξ
= [[A
ζ
,A
η
],A
ξ
]
by denition of R. Note, in particular, that the Lie algebra structure on g is com-
pletely described by the shape operator of M.
We now de ne
η
= 2mH W ,whereH is the mean curvature vector of M at
p.SinceM is minimal in the sphere with radius
2m,wehavethatA
η
= id
V
.
Lemma 2.8.11 The endomorphism ad(
η
) : g g is semisimple and its eigenvalues
are 0,±1.
Proof For B h we have A
B
η
=[B,A
η
]=[B,id
V
]=0. Since M is full in R
n+1
this
implies ad(
η
)B =[
η
,B]=B
η
= 0. This shows h ker ad(
η
).For
ξ
W we have
R(
η
ξ
)x =[A
η
,A
ξ
]x = 0andR(
η
ξ
)
ζ
=[[A
η
,A
ξ
],A
ζ
]=0
for all x V and
ζ
W ,sinceA
η
= id
V
.Asad(
η
)
ξ
= R(
η
ξ
) this implies
W ker ad(
η
). A simple computation shows that
ad(
η
)|
{x±
ϕ
x
:xV }
= id.
Altogether it now follows that ad(
η
) is semisimple and its eigenvalues ar e 0 and ±1.
Lemma 2.8.12 The Lie algebra g is semisimple and g = k R
n+1
is a Cartan de-
composition. Moreover, the Killing form of g restricted to V W = R
n+1
coincides
with the inner product on R
n+1
.
Proof Since M is full, the action of K cannot x a nonzero vector in R
n+1
.Letz be
the centralizer of R
n+1
in g. Note that z is a commutative ideal and contained in R
n+1
(exercise, cf. [347, page 235]). Since [z,
η
]=0wehavez W .Let
ξ
z.Thenwe
have
0 = [x,
ξ
]y,
η
= −
ϕ
A
ξ
x
y,
η
=
α
(x,y),
ξ
for all x, y V . Thus
ξ
= 0andz = {0}, which implies that g is semisimple (exercise,
cf. [347, page 235]).
The restriction of the Killing form B of g to k is negative denite, and g decom-
poses as a direct sum of simple ideals g = g
0
...g
t
,whereg
i
=[q
i
,q
i
]+q
i
with
q
i
= g
i
R
n+1
and B =
λ
i
·,·on q
i
for some
λ
i
= 0. We decompose
η
=
η
1
+...+
η
t
according to the decomposition of g.Then(ad
g
i
(
η
i
))
2
=(ad(
η
))
2
|
g
i
is semisim-
ple with eigenvalues 0 and 1. Hence
λ
i
η
i
,
η
i
= B(
η
i
,
η
i
)=tr(ad
g
i
η
i
)
2
0. But
η
i
= 0 would imply q
i
W . Then, for any
ξ
q
i
and x V , we would have
76 Submanifolds and Holonomy
A
ξ
x =
ϕ
x
ξ
=[
ϕ
x
,
ξ
] g
i
V = {0}, whence
ξ
= 0andq
i
= g
i
= {0}. Thus,
λ
i
> 0
for all i and the restriction of B to R
n+1
is positive denite. This proves that k R
n+1
is a Cartan decomposition.
Next, we show that the Killing form of g restricted to V W = R
n+1
coincides
with the inner product o n R
n+1
.Forx,y V we have
B(x,y)=2tr(ad(x)ad(y))
(see [201, page 140]). We choose orthonormal bases (x
i
) of V and (
ξ
j
) of W with
respect to the inner product on R
n+1
.Then
1
2
B(x,y)=
i
ad(x)ad(y)x
i
,x
i
+
j
ad(x)ad(y)
ξ
j
,
ξ
j
=
i
[
ϕ
x
i
,
ϕ
y
]x,x
i
+
j
ϕ
A
ξ
j
y
x,
ξ
j
= −R(x
i
,y)x,x
i
+
j
A
ξ
j
x,A
ξ
j
y
=
i
α
(x,x
i
),
α
(y,x
i
)+
i
α
(x
i
,x
i
),
α
(x,y)+
j
A
ξ
j
x,A
ξ
j
y
= −
α
(x,y),
1
2
η
=
1
2
x,y.
On th e other hand, if x, ¯x, y, ¯y V ,weget
B(
α
( ¯x,x),
α
( ¯y,y)) = B(
ϕ
¯x
x,
ϕ
¯y
y)=B([
ϕ
¯x
,x],[
ϕ
¯y
,y]) = B([
ϕ
¯y
,[
ϕ
¯x
,x]],y)
= −[
ϕ
¯y
,[
ϕ
¯x
,x]],y =
ϕ
¯x
x,
ϕ
¯y
y =
α
( ¯x,x),
α
( ¯y,y).
Since W is spanned by
α
(V,V ),wehaveB = ·,· on W . Finally, we have
B(V,W )=B(
ϕ
V
η
,W )=B([
ϕ
V
,
η
],W )=B(
ϕ
V
,[
η
,W ]) = 0 .
This completes the proof.
Note that the above lemma also shows that the isotropy r epresentation can be
identied with the adjoint action of K on R
n+1
.
According to the general construction (as explained in Section 2.8.2), the or-
bit K ·
η
is a symmetric submanifold or, more precisely, a standard embedding of a
symmetric R-space. Since M is, by assumption, full and irreducible, the remarks of
Section 2.8, page 69, imply that K ·
η
is actually a standard embedding of an irre-
ducible symmetric R-space. Moreover, the tangent space of K ·
η
at
η
is easily seen
to coincide with V and, using Corollary 2.7.3, one can see that the second fundamen-
tal form of K ·
η
at
η
coincides with the one of M at p. Thus, the normal space of
K ·
η
at
η
is equal to W .
Lemma 2.8.13 The locally symmetric submanifold M is an open part of K ·
η
.
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