# Defining (invariant) $1$-forms on a Lie group

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.

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.

• 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. May 31, 2022 at 11:59
• 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. May 31, 2022 at 18:22
• 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. May 31, 2022 at 18:26