Oriented Manifolds
We recall that if is an
-dimensional vector space than the space
of “top forms” on
is one-dimensional. And since we’re working over
this means that the zero form divides this space into two halves. Choosing one — by choosing a nonzero
-form on
or, equivalently, by choosing a basis for
— makes
into an “oriented” vector space.
We can do the same thing, of course, for the tangent space of an
-dimensional manifold; For each point
the stalk
is isomorphic to
, and so we can pick a nonzero value at each point. But of course this isn’t really what we want; we want to be able to choose this orientation “smoothly”. That is, we want a top form
such that
for all
.
If we can find such a form, we say that it “orients” , and — along with the choice of orientation —
is “orientable”. This is not always possible; there are “non-orientable” manifolds for which there are no non-vanishing top forms.
It turns out that is orientable if and only if the bundle
is isomorphic to the product
. That is, if we can find a map
that plays nicely with the projections down to
, and so that the restriction to the stalk
is a linear isomorphism of one-dimensional real vector spaces.
In the one direction, if we have an orientation given by a top form , then at each point we have a nonzero
. Any other point in
is some multiple of
, so we just define the real component of our transformation
to be this multiple, while the
component is the projection of the point from
to
. The smoothness of
guarantees that this map will be smooth.
On the flip side, if we have such a map, then it’s invertible, giving a bundle map . We can take the section of the product bundle sending each
to
and feed it through this inverse map:
.