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:
.
Do you mean, in your opening paragraph, that V is an n-dimensional vector space, rather than a manifold?
Yeah, sorry. Fixing…
[…] coordinate patch in a manifold is orientable. Indeed, the image is orientable — we can just use to orient — and given a choice of […]
Pingback by Compatible Orientations « The Unapologetic Mathematician | August 29, 2011 |
[…] we orient a manifold by picking an everywhere-nonzero top form , then it induces an orientation on each […]
Pingback by Orientable Atlases « The Unapologetic Mathematician | August 31, 2011 |
[…] we have an oriented manifold , then we know that the underlying manifold has another orientation available; if is a top form […]
Pingback by Switching Orientations « The Unapologetic Mathematician | September 8, 2011 |
[…] take a manifold with boundary and give it an orientation. In particular, for each we can classify any ordered basis as either positive or negative with […]
Pingback by Oriented Manifolds with Boundary « The Unapologetic Mathematician | September 16, 2011 |
[…] say that is an orientable Riemannian manifold. We know that this lets us define a (non-degenerate) inner product on […]
Pingback by The Hodge Star on Differential Forms « The Unapologetic Mathematician | October 6, 2011 |
[…] want to start getting into a nice, simple, concrete example of the Hodge star. We need an oriented, Riemannian manifold to work with, and for this example we take , which we cover with the usual […]
Pingback by A Hodge Star Example « The Unapologetic Mathematician | October 11, 2011 |
[…] continue our example considering the special case of as an oriented, Riemannian manifold, with the coordinate -forms forming an oriented, orthonormal basis at each […]
Pingback by The Curl Operator « The Unapologetic Mathematician | October 12, 2011 |
[…] we take to be a manifold equipped with an orientation given by an orientation form . Then is nowhere zero, and for any positively oriented basis of at any point […]
Pingback by Compact Oriented Manifolds without Boundary have Nontrivial Homology « The Unapologetic Mathematician | November 24, 2011 |
Hope this is not too dumb: must the form be nowhere-zero on the tangent space basis at each point, or nowhere-zero on any tangent vector field? I’m pretty sure the first does not imply the latter. Thanks.
Well, an orientation form doesn’t just take a single tangent vector field, it takes
of them. The requirement is that at each point
, there must be some
-tuple of vectors in
such that when you put them into the
-form at
you get a nonzero value out.
Yes, sorry, I meant a basis (X_1,..,X_n) , or may the more standard (Del/Delx_1, Del/Delx_2,…., Del/Delx_n), but you’re right, we just need _a_ basis (meaning an assignment of an n-ple of V.Fields at each point) where the form is nowhere-zero. Then we need to have a change- -of-basis matrix M between any two points , so that DetM =/ 0.
Sorry, that is the change-of-basis M between any two (X_p1, X_p2,..,X_pn); (X_q1,…,X_qn) with DetM non-zero, for a “coherent orientation”. AFAIK, if w(X_p1,…,X_pn) =/0 and DetM>0 , then w(M(x_p1,…,X_pn))=/0 .