# Bi-invariant form on compact connected Lie group

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 $(r^*_a \omega)_x(v_1, \ldots,v_k)$.

• Notice that $r_al_g=l_gr_a$. – Mariano Suárez-Álvarez May 11 '12 at 2:57
• thanks, the result follow! – Jr. May 11 '12 at 3:00
• 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) – Jr. May 11 '12 at 3:52
• Simply use the definition! – Mariano Suárez-Álvarez May 11 '12 at 4:00
• 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... – Jr. May 11 '12 at 4:05