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?

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

1 Answer 1


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

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.