Since I could use the reputation... letting $\psi: G \to Diff(M)$ \operatorname{Diff}(M)$ be the action then $X' = \psi_* X$, where we identify $Lie(Diff(M))$ \operatorname{Lie}(\operatorname{Diff}(M))$ with vector fields on $M$. This is because they both correspond to the one-parameter subgroup $t\mapsto \psi(\exp tX)$ of $Diff(M)$. \operatorname{Diff}(M)$. Then $[X', Y'] = [X,Y]'$ is immediate.
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Since I could use the reputation... letting $\psi: G \to Diff(M)$ be the action then $X' = \psi_* X$, where we identify $Lie(Diff(M))$ with vector fields on $M$. This is because they both correspond to the one-parameter subgroup $t\mapsto \psi(\exp tX)$ of $Diff(M)$. Then $[X', Y'] = [X,Y]'$ is immediate. |
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Since I could use the reputation... letting $\psi: G \to Diff(M)$ be the action then $X' = \psi_* X$, where we identify $Lie(Diff(M))$ with vector fields on $M$. Then $[X', Y'] = [X,Y]'$ is immediate. |
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