= Solution
Suppose first that the <left-invariant differential form>[left-invariant 1-form] $\alpha$ is <closed differential form>[closed]. The scalar function $\alpha(l_\xi)$ is constant for every $\xi$, because both the form and vector field are left-invariant. <Cartan's magic formula> therefore gives
$$
\mathcal L_{l_\xi}\alpha=d(\alpha(l_\xi))+\iota_{l_\xi}d\alpha=0.
$$
Part b says that $g\mapsto R_g^*\alpha$ is locally constant. Since $G$ is <connected space>[connected], it is constant, and its value at the identity is $\alpha$. Hence $R_g^*\alpha=\alpha$ for every $g$, so $\alpha$ is <bi-invariant differential form>[bi-invariant].
Conversely, if $\alpha$ is bi-invariant, then all these <Lie derivative of a differential form>[Lie derivatives] vanish. Cartan's formula and the constancy of $\alpha(l_\xi)$ give $\iota_{l_\xi}d\alpha=0$. The left-invariant vector fields span every tangent space, so $d\alpha=0$.
Connectedness cannot be omitted. The <orthogonal group> $O(2)$ has an <Abelian Lie algebra>, so every left-invariant 1-form is closed by the <Maurer-Cartan equation in a Lie-algebra basis>. Conjugation by a reflection acts as $-1$ on its one-dimensional Lie algebra, so a nonzero left-invariant 1-form is not right-invariant. This is the standard obstruction recorded by the <closed left-invariant 1-form> criterion.
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