5Riemannian holonomy groups

III Riemannian Geometry

5 Riemannian holonomy groups
Again let
M
be a Riemannian manifold, which is always assumed to be connected.
Let
x M
, and consider a path
γ
Ω(
x, y
),
γ
: [0
,
1]
M
. At the rather
beginning of the course, we saw that
γ
gives us a parallel transport from
T
x
M
to
T
y
M
. Explicitly, given any
X
0
T
x
M
, there exists a unique vector field
X
along γ with
X
dt
= 0, X(0) = X
0
.
Definition
(Holonomy transformation)
.
The holonomy transformation
P
(
γ
)
sends X
0
T
x
M to X(1) T
y
M.
We know that this map is invertible, and preserves the inner product. In
particular, if x = y, then P (γ) O(T
x
M)
=
O(n).
Definition (Holonomy group). The holonomy group of M at x M is
Hol
x
(M) = {P (γ) : γ Ω(x, x)} O(T
x
M).
The group operation is given by composition of linear maps, which corresponds
to composition of paths.
We note that this group doesn’t really depend on the point
x
. Given any
other
y M
, we can pick a path
β
Ω(
x, y
). Writing
P
β
P
(
β
), we
have a map
Hol
x
(M) Hol
y
(M)
P
γ
P
β
P
γ
P
β
1
Hol
y
(M)
.
So we see that Hol
x
(M) and Hol
y
(M) are isomorphic. In fact, after picking an
isomorphism O(
T
x
M
)
=
O(
T
y
M
)
=
O(
N
), these subgroups are conjugate as
subgroups of O(n). We denote this class by Hol(M).
Note that depending of what we want to emphasize, we write
Hol
(
M, g
), or
Really,
Hol
(
M
) is a representation (up to conjugacy) induced by the standard
representation of O(n) on R
n
.
Proposition. If M is simply connected, then Hol
x
(M) is path connected.
Proof. Hol
x
(
M
) is the image of Ω(
x, x
) in O(
n
) under the map
P
, and this map
is continuous from the standard theory of ODE’s. Simply connected means
Ω(x, x) is path connected. So Hol
x
(M) is path connected.
It is convenient to consider the restricted holonomy group.
Definition (Restricted holonomy group). We define
Hol
0
x
(M) = {P (γ) : γ Ω(x, x) nullhomotopic}.
As before, this group is, up to conjugacy, independent of the choice of the
point in the manifold. We write this group as Hol
0
(M).
Of course, Hol
0
(M) Hol(M), and if π
1
(M) = 0, then they are equal.
Corollary. Hol
0
(M) SO(n) .
Proof. Hol
0
(
M
) is connected, and thus lies in the connected component of the
identity in O(n).
Note that there is a natural action of
Hol
x
(
M
) and
Hol
0
x
(
M
) on
T
x
M
,
V
p
T
M for all p, and more generally tensor products of T
x
M.
Fact.
Hol
0
(
M
) is the connected component of
Hol
(
M
) containing the identity
element.
Hol
0
(
M
) is a Lie subgroup of
SO
(
n
), i.e. it is a subgroup and an immersed
submanifold. Thus, the Lie algebra of
Hol
0
(
M
) is a Lie subalgebra of
so(n), which is just the skew-symmetric n × n matrices.
This is a consequence of Yamabe theorem, which says that a path-connected
subgroup of a Lie group is a Lie subgroup.
We will not prove these.
Proposition
(Fundamental principle of Riemannian holonomy)
.
Let (
M, g
) be
a Riemannian manifold, and fix
p, q Z
+
and
x M
. Then the following are
equivalent:
(i) There exists a (p, q)-tensor field α on M such that α = 0.
(ii)
There exists an element
α
0
(
T
x
M
)
p
(
T
x
M
)
q
such that
α
0
is invariant
under the action of Hol
x
(M).
Proof.
To simplify notation, we consider only the case
p
= 0. The general case
works exactly the same way, with worse notation. For α (T
x
M)
q
, we have
(
X
α)(X
1
, ··· , X
q
) = X(α(X
1
, ··· , X
q
))
q
X
i=1
α(X
1
, ··· ,
X
X
i
, ··· , X
q
).
Now consider a loop
γ
: [0
,
1]
M
be a loop at
x
. We choose vector fields
X
i
along γ for i = 1, ··· , q such that
X
i
dt
= 0.
We write
X
i
(γ(0)) = X
0
i
.
Now if α = 0, then this tells us
α
dt
(X
1
, ··· , X
q
) = 0.
By our choice of
X
i
, we know that
α
(
X
1
, ··· , X
q
) is constant along
γ
. So we
know
α(X
0
1
, ··· , X
0
q
) = α(P
γ
(X
0
1
), ··· , P
γ
(X
0
q
)).
So α is invariant under Hol
x
(M). Then we can take α
0
= α
x
.
Conversely, if we have such an
α
0
, then we can use parallel transport to
transfer it to everywhere in the manifold. Given any y M, we define α
y
by
α
y
(X
1
, ··· , X
q
) = α
0
(P
γ
(X
1
), ··· , P
γ
(X
q
)),
where
γ
is any path from
y
to
x
. This does not depend on the choice of
γ
precisely because α
0
is invariant under Hol
x
(M).
It remains to check that
α
is
C
with
α
= 0, which is an easy exercise.
Example.
Let
M
be oriented. Then we have a volume form
ω
g
. Since
ω
g
= 0,
we can take
α
=
ω
g
. Here
p
= 0 and
q
=
n
. Also, its stabilizer is
H
=
SO
(
n
).
So we know Hol(M ) SO(n) if (and only if) M is oriented.
The “only if” part is not difficult, because we can use parallel transport to
transfer an orientation at a particular point to everywhere.
Example. Let x M, and suppose
Hol
x
(M) U(n) = {g SO(2n) : gJ
0
g
1
= J
0
},
where
J
0
=
0 I
I 0
.
By looking at
α
0
=
J
0
, we obtain
α
=
J
Γ(
End T M
) with
J
= 0 and
J
2
=
1. This is a well-known standard object in complex geometry, and such
a J is an instance of an almost complex structure on M.
Example. Recall (from the theorem on applications of Bochner–Weitzenock)
that a Riemannian manifold (
M, g
) is flat (i.e.
R
(
g
)
1) iff around each point
x M
, there is a parallel basis of parallel vector fields. So we find that (
M, g
)
is flat iff Hol
0
(M, g) = {id}.
It is essential that we use
Hol
0
(
M, g
) rather than the full
Hol
(
M, g
). For
example, we can take the Klein bottle
γ
with the flat metric. Then parallel transport along the closed loop γ has
P
γ
=
1 0
0 1
.
In fact, we can check that Hol(K) = Z
2
. Note that here K is non-orientable.
Since we know that
Hol
(
M
) is a Lie group, we can talk about its Lie algebra.
Definition
(Holonomy algebra)
.
The holonomy algebra
hol
(
M
) is the Lie algebra
of Hol(M).
Thus hol(M ) so(n) up to conjugation.
Now consider some open coordinate neighbourhood
U M
with coordinates
x
1
, ··· , x
n
. As before, we write
i
=
x
i
,
i
=
i
.
The curvature may also be written in terms of coordinates
R
=
R
i
j,k`
, and we
also have
R(
k
,
`
) = [
k
,
`
].
Thus, hol(M ) contains
d
dt
t=0
P (γ
t
),
where γ
t
is the square
x
k
x
`
t
t
By a direct computation, we find
P (γ
t
) = I + λtR(
k
,
`
) + o(t).
Here
λ R
is some non-zero absolute constant that doesn’t depend on anything
(except convention).
Differentiating this with respect to
t
and taking the limit
t
0, we deduce
that at for p U, we have
R
p
= (R
i
j,k`
)
p
V
2
T
p
M hol
p
(M),
where we think
hol
p
(
M
)
End T
p
M
. Recall we also had the
R
ij,k`
version, and
because of the nice symmetric properties of R, we know
(R
ij,k`
)
p
S
2
hol
p
(M)
V
2
T
p
M
V
2
T
p
M.
Strictly speaking, we should write
(R
i
j
k
`
)
p
S
2
hol
p
(M),
but we can use the metric to go between T
p
M and T
p
M.
So far, what we have been doing is rather tautological. But it turns out this
allows us to decompose the de Rham cohomology groups.
In general, consider an arbitrary Lie subgroup
G GL
n
(
R
). There is a
standard representation
ρ
of
GL
n
(
R
) on
R
n
, which restricts to a representation
(ρ, R
n
) of G. This induces a representation (ρ
k
,
V
k
(R
)) of G.
This representation is in general not irreducible. We decompose it into
irreducible components (ρ
k
i
, W
k
i
), so that
V
k
(R
) =
M
i
W
k
i
.
We can do this for bundles instead of just vector spaces. Consider a manifold
M
with a
G
-structure, i.e. there is an atlas of coordinate neighbourhoods where the
transition maps satisfy
x
α
x
0
β
!
p
G
for all
p
. Then we can use this to split our bundle of
k
-forms into well-defined
vector sub-bundles with typical fibers W
k
i
:
V
k
T
M =
M
Λ
k
i
.
We can furthermore say that every
G
-equivariant linear map
ϕ
:
W
k
i
W
`
j
induces a morphism of vector bundles φ : Λ
k
i
Λ
`
j
.
Now suppose further that
Hol
(
M
)
G
O(
n
). Then we can show that
parallel transport preserves this decomposition into sub-bundles. So
restricts
to a well-defined connection on each Λ
k
i
.
Thus, if
ξ
Γ(Λ
k
i
), then
ξ
Γ(
T
M
Λ
k
i
), and then we have
ξ
Γ(Λ
k
i
).
But we know the covariant Laplacian is related to Laplace–Beltrami via the
curvature. We only do this for 1-forms for convenience of notation. Then if
ξ
1
(M), then we have
ξ =
ξ + Ric(ξ).
We can check that
Ric
also preserves these subbundles. Then it follows that
∆ : Γ(Λ
1
j
) Γ(Λ
1
j
) is well-defined.
Thus, we deduce
Theorem.
Let (
M, g
) be a connected and oriented Riemannian manifold, and
consider the decomposition of the bundle of
k
-forms into irreducible representa-
tions of the holonomy group,
V
k
T
M =
M
i
Λ
k
i
.
In other words, each fiber
k
i
)
x
V
k
T
x
M
is an irreducible representation of
Hol
x
(g). Then
(i) For all α
k
i
(M) Γ(Λ
l
i
), we have α
k
i
(M).
(ii) If M is compact, then we have a decomposition
H
k
dR
(M) =
M
H
k
i,dR
(M),
where
H
k
i,dR
(M) = {[α] : α
k
i
(M), α = 0}.
The dimensions of these groups are known as the refined Betti numbers.
We have only proved this for
k
= 1, but the same proof technique can be
used to do it for arbitrary k.
Our treatment is rather abstract so far. But for example, if we are dealing
with complex manifolds, then we know that
Hol
(
M
)
U(
n
). So this allows us
to have a canonical refinement of the de Rham cohomology, and this is known
as the Lefschetz decomposition.