Let $w_1, \ldots, w_n$ be left invariant $1$-forms on a Lie group $G$. Why is $w = w_1 \wedge \ldots \wedge w_n $ left-invariant?

Question

$\ldots$while the $w_i$ are dual to vector fields $X_1 , \ldots , X_n$ spanning $T_e G$. I cannot apply the definition straight forward because I do not really no how to handle the wedge product in combination with the left invariance.

Answer 1

In general $F^*(\alpha\wedge \beta) = (F^*\alpha)\wedge(F^*\beta)$ so

$$L_g^*(w_1\wedge\dots\wedge w_n) = (L_g^*w_1)\wedge\dots\wedge(L_g^*w_n) = w_1\wedge\dots\wedge w_n.$$

Note, the fact that $\{w_i\}$ are dual to $\{X_i\}$ was not used in the above computation.

Answer 2

If $T : V \to W$ is a linear operator on a vector space $V$, then the induced map $\wedge^k(T) : \wedge^k(V) \to \wedge^k(W)$ acts by $T (v_1 \wedge \ldots \wedge v_n) = (Tv_1) \wedge \ldots \wedge (Tv_n)$.

So if you start with some vectors that are invariant under an endomorphism, then so is their wedge product.

The manifold case is the bundle version of this, but would be good to be clear on the vector space version first.