2

I'm reading this old question and there are some things I don't understand.

For example, why in the case of $S^1$ can every $1$-form be written in the form $f(\theta)d\theta=c d\theta+dg(\theta)$ where $g$ is differentiable and $c$ is some integral? What is special about $S^1$? The way the question is phrased suggests that on other manifolds there is no such representation of a $1$-form.

Community
  • 1
self-learner
  • 1,217

2 Answers2

4

The general statement has to do with $n$ forms on a compact $n$-dimensional manifold $M$ without boundary. It is a general fact that for such manifolds the $n$th de Rham cohomology group is $H^n_{dR}(M) = \mathbb R$. Now if $\alpha$ is an $n$-form that represents a non-zero class in $H^n_{dR}(M)$ (i.e. $\alpha$ is not exact), then every other $n$th degree cohomology class is represented by a constant multiple of $\alpha$. Now if $\mu$ is any $n$-form it is automatically closed and thus represents a class in $H^n_{dR}(M)$ and so is in the same class as $c \alpha$ for some $c \in \mathbb R$. This means that $\mu$ and $c\alpha$ differ by an exact form, so $\mu = c\alpha + d\beta$ for some $n-1$-form $\beta$.

In your specific example, $\mu = f(\theta)d\theta, \alpha = d\theta$ and $\beta = g(\theta)$.

Eric O. Korman
  • 19,927
  • 1
    This is a wonderful answer. It helped me a lot! Thank you. I have one question to ask if that's ok: you write if $\mu$ is any $n$ form it is automatically closed. I tired to find out if all differential forms on compact manifolds are automatically closed but didn't find anything. So it's probably not true. But why in this case is $\mu$ automatically closed? – self-learner Nov 18 '14 at 23:05
  • 1
    This is because $\mu$ is an $n$-form on an $n$-dimensional manifold. Thus $d\mu$ is an $n+1$-form and so must be zero since $\Lambda^{n+1} T^*M = 0$. – Eric O. Korman Nov 19 '14 at 00:37
  • Just to be sure I understand: $\Lambda^{n+1}$ is the set of all $n+1$-forms. $T^\ast M$ looks like some sort of tangent space... but I have not seen it. I know the notation $T_pM$ for tangent space at point $p$ of a manifold $M$. What does star mean? – self-learner Nov 19 '14 at 03:27
  • The star means the dual space, so $T^*M$ is the space of dual vectors and sections of its exterior product are differential forms. Any $n$-form on an n-dimensional manifold looks like $\mu = f dx_1 \wedge \cdots \wedge dx_n$ in coordinates. Therefore $d\mu = \sum_j \partial_j f dx_j \wedge dx_1 \wedge \cdots \wedge dx_n = 0$. – Eric O. Korman Nov 19 '14 at 18:50
  • But if $\mu$ is closed isn't it equivalent to $0$ since it is in the kernel of the exterior derivative? The de Rham homology groups are the quotients of image divided by kernel of the exterior derivative, no? – self-learner Nov 30 '14 at 03:48
  • @self-learner the cohomology groups are the kernel divided by the image (since $d^2 = 0$, the image is contained in the kernel). – Eric O. Korman Nov 30 '14 at 16:24
0

It means not every form on $S^1$ is necessarily an exact form, meaning it being the differential of a function on $S^1$. In this case, $fd\theta=cd\theta+dg$, being a closed form (you should see that the OP is talking about closed form), is not exact due to the existence of the term $cd\theta$ (note that closed form $\omega$ means $d\omega=0$ and therefore exact forms are automatically closd since $d^2=0$).

Troy Woo
  • 3,647