# Differential forms on manifolds, oriented manifolds

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).

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.
• Please fix the $d_1\wedge\dots\wedge d_m$. – Ted Shifrin Feb 25 '18 at 16:59
• 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. – Dog_69 Feb 25 '18 at 19:02