# Orientation manifold, what is wrong with my argument?

As I learned, a manifold M is oriented if there exists a smooth nowhere-vanishing n-form on M. So, I am very doubting about the following construction of a n-form $\omega$ on any smooth manifold M (M is not necessary oriented):

for each p $\in$ M, choose $\left(U,\left(x_{1},\ldots,x_{n}\right)\right)$} is a arbitrary chart about p, set $\omega\left(p\right)=dx_{1}\wedge\ldots\wedge dx_{n}\in\Lambda^{n}\left(T_{p}^{*}M\right)$. So we have constructed a smooth n-form $\omega$ which is apparent nowhere-vanishing on M.

Where is my wrong view? Thanks.

• Hint: is this really smooth? – Anthony Carapetis Oct 31 '13 at 13:23
• Yes, $\omega$ is a smooth n-form. – user36548 Oct 31 '13 at 13:27
• What happens in the transition regions where two charts overlap? – Cheerful Parsnip Oct 31 '13 at 13:27
• Your recipe for $\omega$ makes sense on each $U$ individually, but will not give you anything remotely well-defined (let alone continuous) if you try to make sense of it on the union of two such charts that overlap. – Ted Shifrin Oct 31 '13 at 13:39
• @user36548: you need to define a smooth form on the whole manifold. If your definition chooses a different chart for each $p$ then your form will not even be continuous. If you define $\omega = dx^1\wedge\cdots\wedge dx^n$ on a whole chart, then it is smooth on that chart; but unless the manifold is (a subset of) $\mathbb R^n$, you will need to use multiple overlapping charts to cover it, and thus you must be careful with how you define $\omega$ on this overlap. – Anthony Carapetis Nov 1 '13 at 2:44