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I am trying to show that a Lie Group $G$ is orientable. I would like to do this by construccting a nowhere vanishing top form $\omega$ on $G$, which would then imply that $G$ is orientable.

I think that the general idea should be to somehow find a nowhere vanishing form at a point (probably the identity?) and then somehow "shift" the form around $G$ by left multiplication, but I am not sure about how the details would work. Could someone help me this more precise?

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  • $\begingroup$ The Haar measure on the Lie group provides with a n-form $\omega$. $\endgroup$ Nov 9, 2017 at 23:53

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Every Lie group is parallelizable, then orientable. Just consider $\psi:G \times T_1G \rightarrow TG$ given by $\psi(g,v)=(g, d(L_g)_1(v))$.

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  • $\begingroup$ Is there a way to avoid using parallelizability? I saw a proof using this fact, but I'm not sure how to prove that a Lie group is parallelizable! $\endgroup$
    – Tex
    Nov 9, 2017 at 23:41
  • $\begingroup$ @Tex Parallelizable means that the tangent bundle splits into $G \times \mathbb{R}^n$. That's what Victor's map $\psi$ is doing. $\endgroup$ Nov 9, 2017 at 23:47
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    $\begingroup$ The function above is a diffeomorphim with inverse $\psi^{-1}(g,u)=(g,d(L_{g^{-1}})_g(u))$. $\endgroup$ Nov 9, 2017 at 23:48
  • $\begingroup$ Could you expand the answer a bit more? I'm not sure I understand how this works. $\endgroup$
    – Tex
    Nov 9, 2017 at 23:49
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    $\begingroup$ @Tex if you wanted to exhibit a nowhere-vanishing top-degree form on $G$, you could choose a basis $\xi_1, \dots,\xi_n$ of $\mathfrak{g}$, then a dual basis $\mu_1,\dots,\mu_n$ of $\mathfrak{g}^*$, and let $\omega_g = L_{g^{-1}}^* \mu_1 \wedge \dots \wedge\mu_n$. But showing that form doesn't vanish anywhere is tantamount to showing $G$ is parallelizable. $\endgroup$ Nov 9, 2017 at 23:50

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