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: 
.
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August 25, 2011 -
Posted by John Armstrong |
Differential Topology, Topology
Do you mean, in your opening paragraph, that V is an n-dimensional vector space, rather than a manifold?
Yeah, sorry. Fixing…
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