It is well-known that there are some manifolds which are not oriented (at least for the one I know is Mobius).

Moreover, we have a criterion which asserts that: any nonvanishing m-form $w$ on $M$ determines a unique orientation of $M$ for which $w$ is positively oriented at each point. (M is smooth m-dimensional manifold).

From that, if we define the form as follows :$w=1 dx_{1}\wedge dx_{2}\ldots\wedge dx_{m} $ for m-dimensional manifold smooth M, then the manifold will be oriented. (cause the form is not vanish).

So, please could you let me know that what is wrong here?.

I think that I have some problems with differential forms, I am still not able to distinguish the differences between the differential forms (the same for vector fields) of those-different manifolds. Because in locally, when we are dealing with computations, it is look like the same (formulations).


1 Answer 1


Your definition of $\omega$ only makes sense locally. So when you write $dx_1 \wedge ... \wedge dx_m$ you are working in local coordinates on your manifold. However, usually your manifold won't have a single chart that covers all of it. Now it is not clear (and not always true..), that you can "glue" those differential forms you've got locally together to form a global differential form.

  • $\begingroup$ Please fix the $d_1\wedge\dots\wedge d_m$. $\endgroup$ Commented Feb 25, 2018 at 16:59
  • $\begingroup$ Exactly. Your $dx^i$ are constructed from $\partial/\partial x^i$. But not always the set $\{\partial_i\}$ is a basis for all $T_pM$. When that happens, we say M is a parallelizable. Since Parallelizable $\Longrightarrow$ Orientable, you get a volume 1-form. But just because you are imposing a stronger condition. $\endgroup$
    – Dog_69
    Commented Feb 25, 2018 at 19:02

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .