# The Unapologetic Mathematician

## Oriented Manifolds

We recall that if $V$ is an $n$-dimensional vector space than the space $\Lambda^n(V)$ of “top forms” on $V$ is one-dimensional. And since we’re working over $\mathbb{R}$ this means that the zero form divides this space into two halves. Choosing one — by choosing a nonzero $n$-form on $V$ or, equivalently, by choosing a basis for $V$ — makes $V$ into an “oriented” vector space.

We can do the same thing, of course, for the tangent space $\mathcal{T}_pM$ of an $n$-dimensional manifold; For each point $p\in M$ the stalk $\Lambda^*_k(M)_p$ is isomorphic to $\mathbb{R}$, 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 $\omega\in\Omega^n(M)$ such that $\omega(p)\neq0$ for all $p\in M$.

If we can find such a form, we say that it “orients” $M$, and — along with the choice of orientation — $M$ 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 $M$ is orientable if and only if the bundle $\Lambda^*_k(M)$ is isomorphic to the product $M\times\mathbb{R}$. That is, if we can find a map $f:\Lambda^*_k(M)\to M\times\mathbb{R}$ that plays nicely with the projections down to $M$, and so that the restriction to the stalk $f_p:\Lambda^*_k(M)_p\to\{p\}\times\mathbb{R}$ is a linear isomorphism of one-dimensional real vector spaces.

In the one direction, if we have an orientation given by a top form $\omega$, then at each point we have a nonzero $\omega(p)\in\Lambda^*_k(M)_p$. Any other point in $\Lambda^*_k(M)_p$ is some multiple of $\omega(p)$, so we just define the real component of our transformation $f$ to be this multiple, while the $M$ component is the projection of the point from $\Lambda^*_k(M)_p$ to $M$. The smoothness of $\omega$ 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 $f^{-1}:M\times\mathbb{R}\to\Lambda^*_k(M)$. We can take the section of the product bundle sending each $p\in M$ to $1\in\mathbb{R}$ and feed it through this inverse map: $\omega(p)=f^{-1}_p(1)\in\Lambda^*_k(M)_p$.

August 25, 2011 - Posted by | Differential Topology, Topology

1. Do you mean, in your opening paragraph, that V is an n-dimensional vector space, rather than a manifold?

Comment by David Roberts | August 26, 2011 | Reply

2. Yeah, sorry. Fixing…

Comment by John Armstrong | August 26, 2011 | Reply

3. […] coordinate patch in a manifold is orientable. Indeed, the image is orientable — we can just use to orient — and given a choice of […]

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4. […] we orient a manifold by picking an everywhere-nonzero top form , then it induces an orientation on each […]

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5. […] we have an oriented manifold , then we know that the underlying manifold has another orientation available; if is a top form […]

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6. […] 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 […]

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7. […] say that is an orientable Riemannian manifold. We know that this lets us define a (non-degenerate) inner product on […]

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8. […] 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 […]

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9. […] continue our example considering the special case of as an oriented, Riemannian manifold, with the coordinate -forms forming an oriented, orthonormal basis at each […]

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10. […] 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 […]

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11. 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.

Comment by Jerry P. | January 18, 2014 | Reply

• Well, an orientation form doesn’t just take a single tangent vector field, it takes $n$ of them. The requirement is that at each point $p$, there must be some $n$-tuple of vectors in $\mathcal{T}_pM$ such that when you put them into the $n$-form at $p$ you get a nonzero value out.

Comment by John Armstrong | January 19, 2014 | Reply

12. 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.

Comment by Jerry P. | January 20, 2014 | Reply

13. 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 .

Comment by Jerry P. | January 20, 2014 | Reply