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I am trying to understand the proof of Theorem 17.21 in Lee's Introduction to smooth manifolds; however I am finding myself stuck right at the beginning. The statement I am having trouble with is: "For $n=1$, note first that any orientation form on $\mathbb{S}^1$ has non-zero integral."

Lee defines an orientation form as a non-vanishing $n$-form on an $n$-dimensional manifold.

I can understand why it is not necessarily zero as the circle is not contractible; however I am having trouble seeing why it can't be zero in any case.

My attempts at a solution don't seems to be leading me in the right direction; however, I have considered using stokes theorem and the fact that the circle is the boundary of a closed ball and I having considered saying: if the integral is zero I can break the circle into two pieces, which means the integral along the two pieces have to be equal. Is there some way to use this to show that the $1$-form can be zero nowhere?

Any advice is much appreciated :)

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  • $\begingroup$ Ok I understand this now. Thank you both very much for your help! $\endgroup$
    – CEH
    Aug 19, 2014 at 22:39
  • $\begingroup$ The reason I tossed that statement off without explanation is because I had just proved it in the preceding chapter: see Proposition 16.1(c). $\endgroup$
    – Jack Lee
    Aug 20, 2014 at 0:13
  • $\begingroup$ Ah I had missed that. Thank you for the the input. And while I have this unique opportunity I might as well say I think your book is awesome so thank you :) $\endgroup$
    – CEH
    Aug 20, 2014 at 15:54

2 Answers 2

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Any volume form, on any (connected) manifold of any dimension, if it exists, has a nonzero integral.

The reason for that is in the definition of integration on manifolds, and the role of orientation in this definition. Namely, considering the reduction of a form to a coordinate chart, one can choose a parametrization and then pull the form back to a Euclidean space, where the integral is calculated in the regular manner. Using the Jacobian matrix, one can show that up to sign, the integral does not depend on the choice of parametrization. The orientation comes in for the sign to be well defined.

By definition, the "legal" parametrizations are exactly those under which the volume form is pulled back to a positive function. So given an orientation, the volume forms are divided to two - those who agree with the orientation, and hence have positive integrals, and those who don't agree, and have negative integrals.

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Tangent spaces to the circle are one dimensional, so any $1$-form on it has to be of the form $f(t)dt$ in standard coordinates with $t\in(0,2\pi)$. If this is an orientation form then $f(t)\neq0$ so either $f(t)>0$ or $f(t)<0$. Either way, $\int_0^{2\pi}f(t)\,dt$ is strictly positive or strictly negative.

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