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I'm struggling with the proof of the following, well-known result for the Levi-Civita connection of a Lie Group with biinvariant metric, i.e. satisfying \begin{equation} g(D_bL_a X_b, D_b L_a Y_b)= g(X_a, Y_a) \end{equation} with $L_a$ the left (or right) translation by $a$.

The claim is \begin{equation} \nabla_X Y= \frac{1}{2}[X,Y] \end{equation} which one should (supposedly) be able to prove through Koszul's formula. My first question, though, is more basic: why is this a connection at all? I should have $\nabla_{\phi X}Y=\phi\nabla_XY$ and $\nabla_X\phi Y = \phi\nabla_X Y + X(\phi)Y$: but the commutator is antisymmetric in $X$, $Y$: how can they follow different rules? Having $X(\phi)=0$ for every function and left-invariant vector field would seem to be the only way out, but I can't find any reason for this to hold. As for Koszul, I just keep getting something different - but I have the feeling there's something I really failed to understand. What are the terms one can simplify in the formula? One has $g(X,Y)=const.$ from the biinvariance - is there more?

nelv
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    The formula $\nabla_X Y = \tfrac 1 2 [X,Y]$ holds only when $X$ and $Y$ are both left-invariant. Usually neither $\phi X$ nor $\phi Y$ will be left-invariant. – Jack Lee Feb 07 '16 at 18:28
  • @JackLee yike, that's the point! thanks. I also realized that $\phi X$ is again left invariant iff $\phi=constant$ - so that indeed $X(\phi)=0$. Now I just need to look a bit more into Koszul. – nelv Feb 07 '16 at 19:24
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    Hint: In addition to Koszul's formula, you may want to use (and prove) that the adjoint representation $\mathrm{ad}_X$ of $\mathfrak g$ is antisymmetric in the bi-invariant metric, i.e. $\langle \mathrm{ad}_X Y, Z \rangle = -\langle Y, \mathrm{ad}_X Z \rangle$. – D Ford Jul 03 '19 at 22:04

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