Ok so I think understand lie algebra valued one forms, and the maurer cartan form, but what about just regular one forms on a Lie group? I feel like if the Lie algebra has a basis $\frac{\partial}{\partial g^i}|_e$ then the vector fields: $$\frac{\partial}{\partial g^i}|_h=L_h^*\left(\frac{\partial}{\partial g_i}|_e\right)=g\left(\frac{\partial}{\partial g_i}|_e\right)$$ Are a global frame for $TG$, so shouldn't there be some dual frame $\omega^i_h$ such that: $$\omega^i_h\left(\frac{\partial}{\partial g^j}|_h\right)=\delta^i_j$$ I think that if we define a dual basis in the Lie algebra, we can just define the global frame of one forms as the pullback by $g^{-1}$, i.e. if we have a frame $\omega^i_e\in\mathfrak{g}^*$, and $v\in T_hG$ then: $$\omega^i_h(v)=L_{g^{-1}}^*\omega_e^i(v)=\omega_e^i(g^{-1}v)$$ Does that make sense though? Something feels weird about having linear functionals $\mathfrak{g}\rightarrow \mathbb{R}$, but I'm kinda just assuming we do it the same way we do on manifolds without a group structure.


1 Answer 1


Up to some minor notational issues your summary is basically correct.

Given a Lie group $G$ and a vector $X \in \mathfrak{g} := \operatorname{Lie(G)} \cong T_e G$, the (smooth) vector field $\widetilde X \in \Gamma(TG)$ defined by $$\widetilde X\!_g := T_e L_g \cdot X$$ is left-invariant, i.e., $\widetilde{X}$ is $L_g$-related to itself for all $g \in G$. As usual, $L_g$ denotes the map $G \to G$ multiplying on the left by $g$, i.e., $L_g(h) := gh$.

Likewise, given a $1$-form $\alpha \in \mathfrak{g}^* \cong T_e^* G$, the (smooth) $1$-form $\widetilde\alpha \in \Gamma(T^*G)$ defined by $$\widetilde \alpha_g := L_{g^{-1}}^* \alpha$$ is left-invariant, i.e., $L_g^* \widetilde \alpha = \widetilde \alpha$ for all $g \in G$.

Applying these constructions to a basis $(E_a)$ of $\mathfrak{g}$ and its dual basis $(\omega^b)$ gives a left-invariant frame $(\widetilde E_a)$ and a left-invariant coframe $(\widetilde \omega^b)$. Unwinding definitions shows that they are dual, i.e., that $\widetilde \omega^b(\widetilde E_a) = \delta^b{}_a$ for all indices $a, b$.

Replacing left multiplication $L_g$ with right multiplication $R_g$ in the above constructions gives analogous right-invariant objects. In general a left-invariant object need not be right-invariant, and vice versa.

  • $\begingroup$ Can you point out the notational issues? Are they actually issues or just not the usual notation? Also, thank you for the explanation that made a lot of sense. $\endgroup$
    – Chris
    May 31, 2022 at 11:59
  • $\begingroup$ In the first line of the question, the last entry should have $h$ acting on the basis element, not $g$. Also, in a few places the basis elements are denoted by $\partial/\partial g_i \vert_e$, but the index $i$ should be an upper index. $\endgroup$ May 31, 2022 at 18:22
  • $\begingroup$ It's partly a matter of taste, but I would avoid using the notation $\partial / \partial g^i$ for the left-invariant vector fields determined by a basis of $\mathfrak{g} \cong T_e G$: The notation might be taken to mean that they are coordinate vector fields associated to some coordinate chart, but such vector fields always commute, and the vector fields of a left-invariant frame all commute with one another if and only if the Lie algebra is trivial, i.e., if the connected component of the identity of $G$ is abelian. $\endgroup$ May 31, 2022 at 18:26

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .