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I'm just trying to complete the argument. The aim is to prove that $SO(3)$ does not admit a left-invariant flat Lorentz metric. Well, suppose it does. Then $R(x,y) = 0$ for all $x$ and $y$ in it's Lie algebra, then we have $\nabla_{x}\nabla_{y} - \nabla_{y}\nabla_{x}$ = $\nabla_{[x,y]}$ and the map sending $x$ to $\nabla_{x}$ would be a Lie Algebra homomorphism from $o(3)$ to the Lie Algebra $o(1,2)$ of all skew-symmetric endomorphisms of $o(3)$. Since $o(3)$ is simple and the kernel of this map is an homomorphism, the kernel is an ideal, so it must be ${(0)}$, and we would have an isomorphism of $o(3)$ onto $o(1,2)$ But why does it give a contradiction? Does it have to do with the fact that $o(3)$ is simple? If it does, why isn't $o(1,2)$ simple?

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to finish the argument just notice that $o(3)$ and $o(2,1)$ are not isomorphic, e.g since their Cartan-Killing forms have different signatures –  user8268 Sep 14 '13 at 22:05

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