Duals of Left-invariant Vector Fields closed?

Question

Suppose we have the Lie group $\text{SU}(2)$. Then its Lie algebra consists of the span $$X_1 = \begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix}, \quad X_2 = \begin{pmatrix} 0 & -\mathrm i\\ \mathrm i & 0 \end{pmatrix},\quad X_3 = \begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix}.$$ Now, $e_i: g \mapsto L_g\vert_e(X_i) $ are left-invariant vector fields. Consider now the dual frame $e_1^*,e_2^*,e_3^*$. Are these 1-forms closed? If so, why? And even then, is it always the case for Lie groups?

Answer 1

The elements of the dual frame are not closed as one-forms. By definition, each of the functions $e_i^*(e_j)$ is constant. Hence looking at the definition of the exterior derivative, you see see that $de_i^*(e_j,e_k)=-e_i([e_j,e_k])$. This is just $-c^i_{jk}$, i.e. the negative of the usual structure constants of the Lie Algebra, which are characterized by $[X_j,X_k]=\sum_ic^i_{jk}X_i$. Otherwise put, the exterior derivative on left invariant forms restricts to a mam $\mathfrak g^*\to\bigwedge^2\mathfrak g^*$ which is the dual of the Lie bracket, viewed as a map $\bigwedge^2\mathfrak g\to\mathfrak g$.