Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It only takes a minute to sign up.
Sign up to join this community
Suppose $M$ is a connected orientable manifold, then if $\omega,\omega'$ are two exterior differential-forms giving orientations to $M$, then there is a nonvanishing continuous function $f$ such that $$\omega'=f\omega$$
Why is this true? I read this statement in the proof of "there exists exactly two orientations on a connected orientable manifold."
Thank you:))
The orientation on a manifold is given by a nowhere vanishing top form (form of top dimension).
Now locally a top form looks like $fdx_{i_{k_{1}}} \wedge ...dx_{{i_{k2}}}$. Since the orientation form is nowhere vanishing any two such forms can differ by a function that is throughout positive or negative.
This dichotomy implies that there can be only two orientations.
Now try to see that globally also this makes sense if your manifold is connected.
A differential form that gives rise to an orientation is of maximal degree and there aren't that many diffenrent differential forms of maximal degree. Consider a single point of the manifold and its tangent space. Let's assume it is of dimension $3$. Then there are several different one-forms, say $dx$ or $dy$. There are also several different two-forms, say $dx\wedge dy$ or $dy\wedge dz$. But there is only one three-form, namely $dx\wedge dy \wedge dz$, up to a real constant.
The same thing happens on the manifold. There is only one differential form of maximal degree at each point, up to a constant. But you can choose a different constant at each point. Hence if you have two different differential forms of maximum degree on a manifold they differ by a function.
