For a left-invariant $p$-form $w$ and left-invariant vector fields $X_0 , \ldots , X_p$ on a Lie-group $M$, why is $w(X_0 , \ldots , X_p)$ constant?

Question

I saw this in Bredon, on page $305$, but I don't know how to use the left-invariance.

For a vector field, left-invariance means that $(L_{\omega})_{*} \circ X = X \circ L_{\omega}$ for the left-translation map $L_{\omega} : M \to M$ sending $m$ to $\omega m$. For a form $w$, it means $w = L_{\omega}^{*}(w)$ for all $\omega \in M$. By the maps with the stars, I mean the induced maps for vector fields respectively differential forms.

Answer 1

Let $e$ denote the identity element of $M$ and let $m \in M$ be arbitrary.

Let $w_m \in \bigwedge^pT^*_mM$ and $(X_i)_m \in T_mM$ denote the values of $w$ and $X_i$ respectively at $m$. As $L_m(e) = m$, there are induced maps $(L_m)^* : \bigwedge^pT_m^*M \to \bigwedge^pT_e^*M$ and $(L_m)_* : T_eM \to T_mM$. By the left-invariance of $w$ and $X_1, \dots, X_p$, we have $(L_m)^*w_m = w_e$ and $(L_m)_*(X_i)_e = (X_i)_m$. Therefore

\begin{align*} (w(X_1, \dots, X_p))_e &= w_e((X_1)_e, \dots, (X_p)_e)\\ &= ((L_m)^*w_m)((X_1)_e, \dots, (X_p)_e)\\ &= w_m((L_m)_*(X_1)_e, \dots, (L_m)_*(X_p)_e)\\ &= w_m((X_1)_m, \dots, (X_p)_m)\\ &= (w(X_1, \dots, X_p))_m. \end{align*}

So $w(X_1, \dots, X_p)$ is constant.