# 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