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$G$: Lie group of dimension $n$.

$\tilde{\Omega}$: Orientation on $G$.

$\Omega=\epsilon^1\wedge \epsilon^2\wedge \cdots \wedge \epsilon^n$ where $\epsilon^1\, \epsilon^2, \cdots ,\epsilon^n$ is positively oriented w.r.t $\tilde{\Omega}$ such that $L_g^*\epsilon^i=\epsilon^i$ $\forall $ $i$ $\forall g,h\in G$ where $L_g(h)=gh$.

I want to show that, for any $f\in C^{\infty}(G)$ and for any $g\in G$, $$ \int_G L_g^* f \Omega=\int_G f\Omega.$$

I think I only need to show that $L_g$ is orientation preserving. Then the problem here is that we don't know the orientation. It seems showing that $L_g^*\Omega=\Omega$ is reasonable but $\Omega$ can be zero at some point so we cannot say that it is an orientation.

Am I thinking wrongly or there is another way to show this?

Thank you in advance!

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  • $\begingroup$ I agree that $L_g^\ast \Omega = \Omega$. I don't think $\Omega$ can be zero at some point, unless it is always zero.... $\endgroup$ – Jason DeVito Dec 12 '18 at 2:59
  • $\begingroup$ @JasonDeVito Honestly, I agree with that, But how is that true? $\endgroup$ – Lev Ban Dec 12 '18 at 3:03
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    $\begingroup$ Sorry - I went to bed after my last comment. Use the fact that $L_g^\ast \epsilon_i = \epsilon_i$ for each $i$, and the fact that pullback commutes with wedge product. $\endgroup$ – Jason DeVito Dec 12 '18 at 13:37

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