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I’m trying to prove that if $G$ is a Lie group, $X$ is a left-invariant vector field on $G$, and $Y$ is a right-invariant vector field on $G$, then $[X,Y] =0$.

When I imagine what it means to be left-invariant, I’m thinking of $X$ as a section of the tangent bundle, and left-invariance says that if I act on $G$ by left multiplication, the vector field is preserved, i.e., $L_{g*}X_e = X_g$, where $X_e \in T_eG$ and $X_g \in T_gG$ are thought of as tangent vectors.

On the other hand, when I think of the Lie bracket, I’m used to thinking of vector fields as derivations of $C^\infty(G)$. That way the $XY$ term in the commutator is simply a composition of derivations.

I’m not sure what it would mean to talk about $XY$ when we think of vector fields as sections of the tangent bundle, and I’m getting confused about what it means for a derivation to be left-invariant.

Any help would be appreciated!

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1 Answer 1

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Let $R_g$ denote right-multiplication by $g$. Then thinking of vector field as operators on functions, for $f \in C^\infty(G)$ we have \begin{align} \def\ddt{\frac{\Bbb d}{\Bbb dt}\Bigg\vert_{t=0}} X_g(Yf) &= ({L_g}_* X_e) (Yf) \\ &= \ddt (Yf)(g\exp(tX_e)) \\ &= \ddt Y_{g\exp(tX_e)} f \\ &= \ddt ({R_{g\exp(tX_e)}}_* Y_e)f \\ &= \ddt {R_{\exp(tX_e)}}_* {R_g}_* Y_e f \\ &= \ddt {R_{\exp(tX_e)}}_* Y_g f \\ &= Y_g \ddt f\circ R_{\exp(tX_e)} \\ &= Y_g(Xf). \end{align} The last line since $$(Xf)(h) = X_h f = {L_h}_* X_e f = X_e (f\circ L_h) = \ddt f \circ L_h(\exp tX) \\ = \ddt f(h\exp tX_e) = \ddt (f\circ R_{\exp tX_e})(h).$$

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  • $\begingroup$ Thanks. Could you clarify why $\frac{d}{dt}\vert_{t=0} {R_{\exp(tX_e)}}_* Y_g f = Y_g\frac{d}{dt}\vert_{t=0} f\circ R_{\exp(tX_e)}$? $\endgroup$
    – Anna
    Feb 5, 2013 at 21:47
  • $\begingroup$ $({R_{\exp(TX_e)}}_* Y_g) f = Y_g (f \circ R_{\exp(tX_e)})$. And then I used the fact that $Y_g$ and $\frac{d}{dt}$ commute since $Y_g$ does not depend on $t$. $\endgroup$ Feb 5, 2013 at 22:38

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