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

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    Notice that $r_al_g=l_gr_a$.2012-05-11
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    thanks, the result follow!2012-05-11
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    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)2012-05-11
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    Simply use the definition!2012-05-11
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    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...2012-05-11
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    you are not using the fact that $r_{ab}^*\omega=f(ab)\omega$2012-05-11
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    I don't know how to use this result...2012-05-11
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    hint: $r_{ab}=r_b \circ r_a$2012-05-11
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    Many people may want to read your question... not only the experts. You could be nicer and define $r_a^*$ and this sort of stuff, so the rest of us can learn from your post as well.2013-06-19

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