Let $G$ be a Lie group. A vector field $X\in\mathfrak{X}(G)$ is left-invariant if the diagram below is commutative:

enter image description here

for every $g\in G$ where $L_g$ stands for the left translation by $g$. Now a differential $1$-form $\omega\in \Omega^1(G)$ is left invariant if $$L^*_g(\omega)=\omega,$$ for every $g\in G$ where $L_g^*(\omega)$ is the pullback form. Is there a way to define a left-invariant form using a commutative diagram as above (maybe using $T^*G$)?

  • $\begingroup$ What is the problem with drawing the same diagram, replacing $X$, $TG$ and $dL_g$ by $\omega$, $T^*G$ and $L_g^*$ respectively? $\endgroup$ – Silvia Ghinassi Dec 5 '15 at 19:43
  • $\begingroup$ The problem is because $L_g^*$ is defined on $\Gamma(T^*G)=\Omega^1(G)$ and not on $T^*G$. $\endgroup$ – PtF Dec 5 '15 at 19:50

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