# Duals of Left-invariant Vector Fields closed?

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?

• Just a small typographical suggestion: Using an asterisk for the dual invariant $1$-forms is not an ideal notation, insofar as the upper star is used to pull back differential forms by mappings. In the context of this particular question, you're fine, but in general you'll want to switch notations :) – Ted Shifrin Aug 27 '18 at 16:59
• What would you recommend instead? – YoungMath Aug 27 '18 at 18:33
• The usual custom is to write $\theta^i$ or $\omega^i$ for the invariant $1$-forms. In particular contexts, people sometimes use something else. Notice that then the indices line up better, as in what @AndreasCap wrote. $d\theta^i(e_j,e_k) = -\theta^i ([e_j,e_k]) = -c^i_{jk}$. – Ted Shifrin Aug 27 '18 at 18:38

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 $\bigwedge^2\mathfrak g\to\mathfrak g$.