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Let $M$ be a orientable $n$ dimensional manifold. I'm trying to solve the following assertions:

  1. Given a connected system of coordinates $(U,x_1,\cdots, x_n)$, prove that there exists a $n$-form $\omega$ supported in $U$ such that $\int_M \omega > 0$.

  2. Prove that if $M$ is compact $\omega$ is exact. Why do we need the compactness of $M$?

My attempt:

For the first one, we can find a bump function $\lambda: M \to \mathbb R$ that is smooth and is supported in $U$ with $0 \leq \lambda \leq 1$. Define, then, $\omega = \lambda \,dx_1 \wedge \dots dx_n$. Then,

$$\int_M \omega = \int_U \lambda \, dx_1\dots dx_n > 0,$$ since $\lambda > 0$ in some open subset of $U$.

In this argument, I'm not sure where I used that $U$ is connected. And I don't know how to approach the next question; I was thinking of using Stokes theorem, but how?

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  • $\begingroup$ The fact that $\lambda>0$ does not imply necessarily $\int \omega>0$: to prove it you need to use explicitly the orientability of the manifold, as much as its connectedness: for the orientability is equal to the existence of a volume form; and the connectedness is necessary: as a counterexample without this hypotesis, consider the manifold obtained combining two spheres with the same radius but with opposite orientation. Since both the manifolds are orientable and they are disjoint, their union is orientable, but $\int \omega=0$ $\endgroup$
    – user515010
    Dec 5, 2019 at 21:38

1 Answer 1

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  1. Since $M$ is oriented, every chart gives origin to the same orientation on the basis of $T_pM$. Since $U$ is connected, the orientation is the same for every point $p$. We may assume, without loss of generality, that the basis are positively oriented. We may now define the $n$-form $\nu= dx_1\wedge\dots\wedge dx_n$. For every chart in the atlas defined on $U$, thanks to the previous reasoning, we have $\omega(e_1,\dots,e_n)>0$. Multiplying $\omega$ by a suitable bump function $\lambda$, we have $\int_M \lambda\nu>0$

  2. The statement is false in general : given an $n$-form defined on a compact manifold without boundary (i.e. closed), if $\omega=d\zeta$ we would have $$\int_M \omega=\int_{\partial M}\zeta=0$$ From our requirement $\int_M \omega>0$ it follows that our $\omega$ is not exact.

We can, however, take the negation of the question, and ask ourselves if this is true. It actually is if we restrict ourselves to manifolds without boundary.

For this result, the compactness requirement is fundamental: $\mathbb{R}^2$ is not compact, and we have $\xi:=dx\wedge dy$ and $\int_{\mathbb{R}^2}\xi=+\infty>0$.

Similarly, the requirement that the boundary is empty is fundamental: $dx$ on $[0,1]$ is exact, and $\int_{[0,1]}dx=1>0$

Note: we may ask ourselves even if the requirement $\int_M\omega=0$ is a necessary and sufficient condition for $\omega$ to be exact. An answer to this question is given here

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    $\begingroup$ Simpler, the area $2$-form $dx\wedge dy$ is exact on $\Bbb R^2$. But in the first argument, as long as $\bar U$ is compact with smooth boundary, can't you apply Stokes's Theorem to $\partial\bar U$ to get the contradiction, independent of whether $M$ is compact or has boundary? $\endgroup$ Dec 5, 2019 at 22:52
  • $\begingroup$ @TedShifrin You are right, I don't know why I thought about the poincarè plane. I changed my example. As your second point,you are sure right, as long as $\partial \overline{U}=0$, otherwise since it has a boundary the result could be quite different, as $dx$ on $[0,1]\subset \mathbb{R}$ shows. However, i thought of writing the answer as it is not to make it too long. Do you think it should be added or it can be simply stated in the comments? $\endgroup$
    – user515010
    Dec 5, 2019 at 22:59
  • $\begingroup$ Just take $U$ to be a (suitably small) coordinate ball, whose closure is a proper subset of $M-\partial M$? $\endgroup$ Dec 5, 2019 at 23:02
  • $\begingroup$ @TedShifrin Take $[0,1]$ as the manifold, $U=(0,1)$ and take $\omega=\lambda(x)dx$ where $\lambda$ is a generic bump function with support contained in $U$. Then $\int_{[0,1]}\omega=K>0$ and yet $\omega$ is exact $\endgroup$
    – user515010
    Dec 5, 2019 at 23:21
  • $\begingroup$ Oh, what I said is garbage. I was thinking of compactly supported cohomology, in which case the form $\zeta$ would have to have compact support as well. Sorry. $\endgroup$ Dec 5, 2019 at 23:30

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