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: .