I'm doing some proofs connected with Lie algebra. And now for the second time I get something like this: let $\mathfrak{g}$ be a Lie algebra. Then $[0, y] = [y, 0] =0$ $\forall y \in \mathfrak{g}$. I wonder why?
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Because the Lie bracket is bilinear. In particular, $[\alpha x,y]=\alpha [x,y]=[x,\alpha y]$ for all $\alpha\in K$.
Dietrich Burde
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Thanks a lot! So $[0, y] = [0\cdot0, y] = [0, 0\cdot y]$? Do we also have $[AB, C] = A[B, C] + [A, C]B; A,B,C \in \mathfrak{g}$? If yes, why? – Hendrra Jan 18 '18 at 10:56
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Yes, this is true for the case that $[A,B]=AB-BA$ for endomorphisms. – Dietrich Burde Jan 18 '18 at 12:04