Left-invariant differential form

A differential form $\alpha$ on a Lie group is left-invariant when $L_g^{\ast }\alpha =\alpha$ for every $g$. It is determined by its value at the identity.

Choose a nonzero element of ${\Lambda }^n{T}_e^{\ast }G$ for an $n$-dimensional Lie group and transport it by inverse left translations. This gives a smooth nowhere-zero left-invariant differential form of top degree, hence an orientation. The identity $L_{(hg{)}^{-1}}{L}_h=L_{g^{-1}}$ proves invariance. The space of invariant top forms is one-dimensional, since evaluation at the identity identifies it with ${\Lambda }^n{T}_e^{\ast }G$.

A differential form on a Lie group is bi-invariant when it is invariant under both left and right translations.

An adjoint-invariant symmetric bilinear form $B$ on a Lie algebra makes the displayed tensor alternating. If $B$ is invariant under the full group Adjoint representation, its left-translated three-form is also right-invariant. On left-invariant fields, the exterior derivative is $d\eta (W,X,Y,Z)=-2B(W,[X,[Y,Z]]+[Y,[Z,X]]+[Z,[X,Y]])=0$ by the Jacobi identity. The Killing form supplies such a $B$ on every semisimple Lie group. An overall sign change in the lowered structure-constant convention negates the form but not its invariance or closedness.

On a connected Lie group, a left-invariant differential 1-form is closed exactly when it is bi-invariant. Connectedness is essential: on $O(2)$ every left-invariant 1-form is closed because its Lie algebra is abelian, while reflection conjugation negates every nonzero one.