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math.stackexchange.com/questions/295758
The Lie bracket of a left-invariant vector field $X_L$ and a right-invariant vector field $X_R$ on a Lie group $G$ vanishes.
Proof:
The flow $\phi_t^{X_L}$ of $X_L$ is given by right multiplication: $\phi_t^{X_L}(g) = g \exp(tA)$, where $A = X_L(e) \in \mathfrak{g}$.
The flow $\psi_s^{X_R}$ of $X_R$ is given by left multiplication: $\psi_s^{X_R}(g) = \exp(sB) g$, where $B = X_R(e) \in \mathfrak{g}$.
For any $g \in G$, composing the flows in either order yields:
\[\phi_t^{X_L}(\psi_s^{X_R}(g)) = \exp(sB) g \exp(tA), \quad \psi_s^{X_R}(\phi_t^{X_L}(g)) = \exp(sB) g \exp(tA).
\]Since left and right multiplications commute ($L_{\exp(sB)} \circ R_{\exp(tA)} = R_{\exp(tA)} \circ L_{\exp(sB)}$), the flows $\phi_t^{X_L}$ and $\psi_s^{X_R}$ commute.
The Lie bracket $[X_L, X_R]$ measures the failure of flows to commute. Explicitly, for small $t, s$, the commutator of flows satisfies:
\[\phi_{-t}^{X_L} \circ \psi_{-s}^{X_R} \circ \phi_t^{X_L} \circ \psi_s^{X_R}(g) = g + st[X_L, X_R](g) + \mathcal{O}(t^2, s^2).
\]Since the flows commute identically ($\phi_t^{X_L} \circ \psi_s^{X_R} = \psi_s^{X_R} \circ \phi_t^{X_L}$), the coefficient of $st$ vanishes. Hence, $[X_L, X_R] = 0$. |
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