10Vector bundles
III Algebraic Topology
10.2 Vector bundle orientations
We are now going to do something that actually involves algebraic topology.
Unfortunately, we have to stop working with arbitrary bundles, and focus on
orientable bundles instead. So the first thing to do is to define orientability.
What we are going to do is to come up with a rather refined notion of
orientability. For each commutative ring
R
, we will have the notion of
R
-
orientability. The strength of this condition will depend on what
R
is — any
vector bundle is
F
2
orientable, while
Z
-orientability is the strongest — if a vector
bundle is Z-orientable, then it is R-orientable for all R.
While there isn’t really a good geometric way to think about general
R
-
orientability, the reason for this more refined notion is that whenever we want
things to be true for (co)homology with coefficients in
R
, then we need the
bundle to be R-orientable.
Let’s begin. Let π : E → X be a vector bundle of dimension d. We write
E
#
= E \ s
0
(X).
We now look at the relative homology groups
H
i
(E
x
, E
#
x
; R),
where E
#
x
= E
x
\ {0}.
We know
E
x
is a
d
-dimensional vector space. So we can choose an isomorphism
E
x
→ R
d
. So after making this choice, we know that
H
i
(E
x
, E
#
x
; R)
∼
=
H
i
(R
d
, R
d
\ {0}; R) =
(
R i = d
0 otherwise
However, there is no canonical generator of
H
d
(
E
x
, E
#
x
;
R
) as an
R
-module, as
we had to pick an isomorphism E
x
∼
=
R
d
.
Definition
(
R
-orientation)
.
A local
R
-orientation of
E
at
x ∈ X
is a choice of
R-module generator ε
x
∈ H
d
(E
x
, E
#
x
; R).
An
R
-orientation is a choice of local
R
-orientation
{ε
x
}
x∈X
which are com-
patible in the following way: if
U ⊆ X
is open on which
E
is trivial, and
x, y ∈ U
,
then under the homeomorphisms (and in fact linear isomorphisms):
E
x
E|
U
U × R
d
R
d
E
y
h
x
ϕ
α
∼
=
π
2
h
y
the map
h
∗
y
◦ (h
−1
x
)
∗
: H
d
(E
x
, E
#
x
; R) → H
d
(E
y
, E
#
y
; R)
sends
ε
x
to
ε
y
. Note that this definition does not depend on the choice of
ϕ
U
,
because we used it twice, and they cancel out.
It seems pretty horrific to construct an orientation. However, it isn’t really
that bad. For example, we have
Lemma. Every vector bundle is F
2
-orientable.
Proof. There is only one possible choice of generator.
In the interesting cases, we are usually going to use the following result to
construct orientations:
Lemma.
If
{U
α
}
α∈I
is a family of covers such that for each
α, β ∈ I
, the
homeomorphism
(U
α
∩ U
β
) × R
d
E|
U
α
∩U
β
(U
α
∩ U
β
) × R
d
∼
=
ϕ
α
∼
=
ϕ
β
gives an orientation preserving map from (
U
α
∩ U
β
)
× R
d
to itself, i.e. has a
positive determinant on each fiber, then E is orientable for any R.
Note that we don’t have to check the determinant at each point on
U
α
∩ U
β
.
By continuity, we only have to check it for one point.
Proof.
Choose a generator
u ∈ H
d
(
R
d
, R
d
\ {
0
}
;
R
). Then for
x ∈ U
α
, we define
ε
x
by pulling back u along
E
x
E|
U
α
U
α
× R
d
R
d
ϕ
α
π
2
. (†
α
)
If
x ∈ U
β
as well, then the analogous linear isomorphism
†
α
differs from
†
β
by
post-composition with a linear map
L
:
R
d
→ R
d
of positive determinant. We
now use the fact that any linear map of positive determinant is homotopic to the
identity. Indeed, both
L
and
id
lies in
GL
+
d
(
R
), a connected group, and a path
between them will give a homotopy between the maps they represent. So we
know (
†
α
) is homotopic to (
†
β
). So they induce the same maps on cohomology
classes.
Now if we don’t know that the maps have positive determinant, then (
†
α
)
and (
†
β
) might differ by a sign. So in any ring
R
where 2 = 0, we know every
vector bundle is
R
-orientable. This is a generalization of the previous result for
F
2
we had.