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Given a Lie group $G$, we say that a vector field $X$ is left-invariant if for each $g\in G$ we have $\mathrm{d}L_g(X)=X$, that is $\mathrm{d}L_g|_h(X_h)=X_{gh}$ for all $h\in G$, where $L_g\colon G\to G$ is the left translation by $g$. We also have left-invariant differential forms, that is forms $\omega$ such that $L_g^*\omega=\omega$. Point by point in the group, however, this translates to $(L_g^*\omega)_h=\omega_h$. Why not $L_g^*(\omega_h)=\omega_{g^{-1}h}$? This would be more in line with the vector field case, that can be rewritten as $\mathrm{d}L_g|_{g^{-1}h}(X_{g^{-1}h})=X_h$ to make it even more similar. What motivates this difference?

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The conditions $(L_g^*\omega)_h=\omega_h$ for all $h$ and $L_g^*(\omega_h)=\omega_{g^{-1}h}$ for all $h$ are exactly the same. Notice that $$(L_g^*\omega)_h=L_g^*(\omega_{gh}),$$ since $L_g(h)=gh$ so $(L_g^*\omega)_h$ is obtained by pulling back the value of $\omega$ at $gh$. Letting $h'=gh$, we thus have $$(L_g^*\omega)_h=L_g^*(\omega_{h'})$$ and $$\omega_h=\omega_{g^{-1}h'}.$$ It follows that $(L_g^*\omega)_h=\omega_h$ is equivalent to $L_g^*(\omega_{h'})=\omega_{g^{-1}h'}$

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