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Let $M$ be a compact manifold with $\partial M = \varnothing$ and let $\omega$ be the volume form $\sqrt{\det g_{ij}} dx_1 \wedge \dots \wedge dx_n$.

I want to show that $\omega$ is not exact.

My thoughts so far:

Assume that there was an $n-1$-form $\psi$ such that $d\psi = \omega$. Then

$$ \int_M \omega = \int_{ M} d\psi = \int_{\partial M} \psi = 0$$

Intuitively, I get that an integral over something positive can't be zero. My trouble is showing it mathematically rigourusly.

How to prove that $ \int_M \omega \neq 0$?

self-learner
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2 Answers2

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I assume that you are on an orientable compact manifold. To show that $\int_M \omega \neq 0$, recall that $\int_M$ is defined using parition of unity: Let $\{U_i\}$ be an open cover of coordinate charts and $\phi_i$ is a partition of unity subordinate to $\{U_i\}$. Then

$$\int_M \omega = \sum_i \int_M \phi _i \omega = \sum _i \int_{ U_i} \phi_i \omega . $$

Now from the expression of $\omega$ one sees that $\int_{U_i} \phi_i \omega >0$ for all $i$.

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You can prove that $\int_U \omega > 0$ for any oriented coordinate chart $(U, \varphi)$ of $M$ just by definition: We have $\varphi(U) \subseteq \mathbb{R}^n$ and, using the coordinate definition of integration of $n$-forms, $$ \int_U \omega = \int_{\varphi(U)} \sqrt{ \det g_{ij} } \; dx^1 \cdots dx^n, $$ which is positive (since we are now integrating the positive function $\det g_{ij}$ over an open neighborhood in $\mathbb{R}^n$). This forces $\int_M \omega > 0$ as well.

mollyerin
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  • Is the integral $$ \int_{\varphi(U)} \sqrt{ \det g_{ij} } ; dx^1 \cdots dx^n$$ equal to $$\int \dots \int_{\varphi(U)} \sqrt{ \det g_{ij} } ; dx^1 \cdots dx^n,$$ by any change? – self-learner Jan 27 '15 at 07:28
  • Yes, it's the ordinary Lebesgue (or Riemann, or whatever you prefer) integral on $\mathbb{R}^n$. – mollyerin Jan 27 '15 at 07:47