Let $G$ be a Lie group and $\omega$ be a left invariant $k$-form, how to prove that $r^*_a \omega$ is left invariant? What I do:

$(l^*_g (r^*_a \omega))_x (v_1, \ldots,v_k)=(r^*_a \omega))_{gx} ({l_g}_*v_1, \ldots,{l_g}_*v_k)= \omega_{gxa} ({r_a}_* {l_g}_*v_1, \ldots,{r_a}_* {l_g}_*v_k)=$

and stuck here...

The right side of the equality should be $(r^*_a \omega)_x(v_1, \ldots,v_k)$.

  • 2
    $\begingroup$ Notice that $r_al_g=l_gr_a$. $\endgroup$ – Mariano Suárez-Álvarez May 11 '12 at 2:57
  • $\begingroup$ thanks, the result follow! $\endgroup$ – Jr. May 11 '12 at 3:00
  • $\begingroup$ taking advantage of the topic, I'd like to ask another question: We have $r^*_a \omega = f(a)\omega$ how to prove that this $f$ defines a homomorphism of groups? ($G$ into the multiplicative reals) $\endgroup$ – Jr. May 11 '12 at 3:52
  • $\begingroup$ Simply use the definition! $\endgroup$ – Mariano Suárez-Álvarez May 11 '12 at 4:00
  • $\begingroup$ I used, we obviously must have $r^*_{ab}\omega=f(a)f(b)\omega$, so I tried to develop the left side, but I stuck... $\endgroup$ – Jr. May 11 '12 at 4:05

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