0

Maybe this question is rather obvious but I didn't manage to solve it myself.

Assume $M$ is a closed, oriented manifold. take $$ \Omega^k(M)\ni \omega = \begin{cases} d\beta~,~~~ in~ U\\ 0~,~~~ otherwise \end{cases} $$ Where $U\subset M$ is open, and $\beta \in \Omega^{k-1}(U)$. $\omega$ has a compact support in $U$.

Then $\omega$ is an exact form?

GreatGuest
  • 79

2 Answers2

3

COMPLETE EDIT: Now that you have removed cohomology and the question has been clarified in my mind, I believe you are asking the following question.

Suppose $\omega$ is a $k$-form with compact support contained in $U$ and there is a smooth form $\beta$ on U with $d\beta=\omega$ on $U$; is $\omega$ globally exact on $M$?

The answer is NO, not necessarily. Consider $M=S^1$, and let $U=S^1-{(1,0)} \cong \Bbb R$. On $\Bbb R$ take a compactly supported function $f$ with integral $1$, and consider $\omega = f(x)\,dx$. This extends to be a smooth $1$-form on all of $S^1$, since $\omega$ is identically $0$ outside a compact subset of $\Bbb R$. $\int_{S^1}\omega = 1$, so $\omega$ cannot be exact. However, on $U=\Bbb R$, we can set $\beta(x) = \int_0^x f(t)\,dt$. By the Fundamental Theorem of Calculus, $d\beta = \omega$ on $U$.

Ted Shifrin
  • 115,160
  • thank you, I did confused. I don't know that $\beta$ has a compact support, Just $\omega$. – GreatGuest Aug 17 '14 at 12:24
  • @Ted Shifrin Is it possible to see more of your advanced video lectures on Youtube? ;-) – Troy Woo Aug 17 '14 at 14:40
  • @TroyWoo: I'm not sure what you're asking. We will be trying to upload the first half of the course this fall, starting tomorrow. But you seem quite advanced, so I'm not sure what specifically you're looking for. – Ted Shifrin Aug 17 '14 at 15:44
  • @GreatGuest: Since you're pretty new here, perhaps I should let you know that once you're happy with an answer, you should upvote it so that the question no longer appears as "unanswered." But you should feel free to ask for more details until you are satisfied :P – Ted Shifrin Aug 17 '14 at 15:45
  • @TedShifrin In your answer you assumed that $\beta$ is compactly supported in $U$. I just don't understand how do you know that. In my question I only said that $\omega$ is compactly supported in $U$, not $\beta$ – GreatGuest Aug 17 '14 at 17:03
  • @GreatGuest Its the way it works man. Otherwise it doesn't add up. If you are sure it is not the case, please give the reference for his consideration. – Troy Woo Aug 17 '14 at 17:37
  • I'm sorry but I really don't understand why it is impossible that $d\beta$ has a compact support on $U$ and $\beta$'s support is not compact in $U$. I actually believe that it is possible. – GreatGuest Aug 17 '14 at 18:04
  • Your question is very confusingly stated. Are you asking this: Suppose $\omega$ has compact support contained in $U$ and there is a smooth form $\beta$ on $U$ with $d\beta=\omega$ on $U$; is $\omega$ globally exact? I see now that this is your intent. – Ted Shifrin Aug 17 '14 at 20:41
  • OK that is what I was looking for. Thank you very much for your help. – GreatGuest Aug 17 '14 at 23:51
0

If it is $\beta$ that has compact support in $U$, then it is true. Denote support of $\beta$ by $K$, obviously $U^c$ is a closed set that is disjoint from $K$. Then there exists a separating function $f$ on $M$ taking the value in $[0,1]$, with $f(K)\equiv1$ and $f(U^c)\equiv0$. Then, $$ d(f\beta)=df\wedge\beta+fd\beta=\begin{cases} d\beta & \text{on }U \\ 0 & \text{on }U^c \end{cases} $$ and therefore $d(f\beta)=\omega$ on the whole of $M$ and $\omega$ is exact.

Troy Woo
  • 3,579