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Suppose we have real Lie algebras. What I want to show is $[x_i, x_j]=a_{ijk}x_k=-a_{ikj}x_j$. I have seen some hints from other questions that in order to prove it one should use the Killing form identity with $ad(X)$ but I cannot explicitly proove it. Can anyone show? Thank you!

About some details about the proof that real Lie algebra with positive Killing form is zero This is link to the other question. I am sorry for pasting it like that but am having some troubles making the hyperlink.

Blake
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    Where does the free suffix k in the last $a_{ikj}x_j$ come from? – user10354138 Oct 05 '18 at 18:47
  • $[x_{i},x_{k}]=a_{ikj}x_{j}$ It is meant to be like that. – Blake Oct 05 '18 at 18:50
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    Does not compute. What are the elements $x_i$ here? Basis elements of some Lie algebra? Why would there be only a single basis element $x_k$ in the formula for the Lie bracket $[x_i,x_j]$? That is certainly not true in general. Are you assuming some summation convention? Are the coefficients $a_{ijk}$ defined via the relation $[x_i,x_j]=\sum_k a_{ijk} x_k$, and you want to prove that $a_{ijk}=a_{ikj}$ for all $i,j,k$? Where's the skew-symmetry? We would expect to have a minus sign somewhere :-) – Jyrki Lahtonen Oct 08 '18 at 04:40
  • You are right about the minus sign. Will edit it in the question. I want to prove it only for basis element. If it is true for basis it will be true in general because of the linearity. – Blake Oct 08 '18 at 17:28

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