# Prove that an integration is left invariant.

$$G$$: 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?

• I agree that $L_g^\ast \Omega = \Omega$. I don't think $\Omega$ can be zero at some point, unless it is always zero.... – Jason DeVito Dec 12 '18 at 2:59
• 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. – Jason DeVito Dec 12 '18 at 13:37