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A left-invariant vector field on a Lie group $G$ satisfies

$$ (L_g)_*X = X $$

for all $g\in G$. So we have in particular $\omega(X_p)=(L_{g^{-1}})X_p=X_e$ with $\omega$ the Maurer-Cartan form on $G$.

Please have a look at this derivation of the Maurer-Cartan equation. It says that if $X$ and $Y$ are left-invariant vector fields, then they satisfy

$$\tag{1}X(\omega(Y))=Y(\omega(X))=0$$

Why is that true? I only get

$$ X(\omega(Y))=X(Y)=[X,Y] $$

so

$$ X(\omega(Y))-Y(\omega(X))=X(Y)-Y(X)=[X,Y]-[Y,X]=2[X,Y] $$

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You should start from the Cartan Formula which is $$d\omega(X,Y)=X(\omega(Y))-Y(\omega(X))-\omega([X,Y])$$ Then since $X(\omega(Y))$ and $Y(\omega(X))$ are $0$ then you have $$d\omega(X,Y)+\omega([X,Y])=0.$$

In fact $\omega(Y)$ is a constant since $Y$ is left-invariant. If $\omega(Y)$ is a constant then whatever derivativeyou will apply you will obtain $0$. Being more precise if you have a vector field $X$ which at the end of the story is some kind of derivative and therefore you have $$X(\omega(Y))=0$$.

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  • $\begingroup$ Hmm, what you are saying makes sense to me, but what's wrong with the last two equations in my question? If $\omega(X)=X$, then $X(\omega(Y))$ reduces to $X(Y)$ which is per definition equal to $[X,Y]$. $\endgroup$ – Andy Miles Aug 14 '16 at 10:49
  • $\begingroup$ I can't se why $X(\omega(Y))$ should be equal to $[X,Y]$. In general $[X,Y](f)=X(Y(f))-Y(X(f)).$ As far as I can see $X(\omega(Y))=X(Y)$ is not implied by the left invariance of $X$ and $Y$ $\endgroup$ – Dac0 Aug 14 '16 at 11:10
  • $\begingroup$ See here: A left-invariant vector field is a section X of TG such that $(L_g)_*X=X$. This means that $\omega(X)=X$, so $Y(\omega(X))=Y(X)$. And $Y(X)$ is defined as $[Y,X]$. $\endgroup$ – Andy Miles Aug 14 '16 at 11:15
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    $\begingroup$ Why $Y(X)$ should be defined as $[Y,X]$? If $X,Y$ are vector fields (left-invariant or not) you have by the very definition of the commutator that $[Y,X](f)=Y(X(f))-X(Y(f))$. $\endgroup$ – Dac0 Aug 14 '16 at 12:37
  • $\begingroup$ In the place you cited he's referring to a specific 1-form $\omega_g$ and not to a generic 1-form $\omega$ $\endgroup$ – Dac0 Aug 14 '16 at 12:38

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