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I have two $(n-1)$-forms $\omega_{1}$ and $\omega_{2}$ on $\mathbb{R}^n$ and a smooth function $g(x) \colon \mathbb{R}^n \to \mathbb{R}$ ($dg$ doesn't vanish anywhere) such that $dg \wedge \omega_1 = dg \wedge \omega_2$ holds. Let $M = \{x \in \mathbb{R}^n \mid g(x) = 0 \}$. Is it true that $$ \int\limits_{M} \omega_1 = \int\limits_{M} \omega_2 $$

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Did you try to show the result when $g(x)=x_n$? –  Davide Giraudo Feb 4 '12 at 11:02
    
No, but I considered difference $dg \wedge (\omega_1 - \omega_2)$ so it is sufficient to show for $\omega_2 = 0$. I have $dg(X) = 0$ for any $X$ that is tangent to $g(x) = 0$. Then I take $X_1,...,X_n$ such that $X_1,...,X_{n-1}$ are tangent to $g(x) = 0$ and so $dg \wedge \omega_1 (X_1,...,X_n) = \pm dg(X_n) \omega_1(X_1,...,X_n)$, but $dg(X_n) \neq 0$ so $\omega_1$ is zero on tangent bundle of $M$. –  Nimza Feb 4 '12 at 11:12
    
$\omega_1$ and $\omega_2$ are closed? –  Paul Feb 4 '12 at 12:26
    
$\omega_1$ and $\omega_2$ are arbitrary $(n-1)$-forms such that $dg \wedge (\omega_1 - \omega_2) = 0$ –  Nimza Feb 4 '12 at 13:49
    
To the question is reduced to the next: why if the form $\omega$ vanishes on tangent spaces of $M$ then $\int_{M} \omega = 0$? –  Nimza Feb 4 '12 at 17:24

1 Answer 1

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Considering $\omega = \omega_1 - \omega_2$ we have to show that if $dg \wedge \omega = 0$ then $\int_{M} \omega = 0$.

  1. $dg(X_p) = 0$ iff $X \in T_{p}M$ because $dg(X_p) = \langle \nabla g(p), X \rangle$.
  2. Let $X^1,...,X^n$ be such that $X^1,...,X^{n-1}$ are from $T_pM$ and $X^n$ is transversal to $T_pM$. Then $$ 0 = dg \wedge \omega (X^1_p,...,X^n_p) = \pm dg(X^n_p) \omega(X^1_p,...,X^{n-1}_p) $$ but $dg(X^n_p) \neq 0$ then $\omega$ vanishes on tangent spaces to $M$, hence $\int_{M} \omega = 0$.
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