# Why is the integral of any orientation form over $\mathbb{S}^1$ non zero?

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 :)

• Ok I understand this now. Thank you both very much for your help!
– CEH
Aug 19, 2014 at 22:39
• 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). Aug 20, 2014 at 0:13
• 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 :)
– CEH
Aug 20, 2014 at 15:54

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.