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In the post about proving that real Lie algebra with positive Killing form is zero: real Lie algebra with positive Killing form is zero:

Let $L$ be a real Lie algebra with positive definite Killing form. Its Killing form $\kappa$ defines an inner product on $L$. Hence $L$ is reductive. Thus the quotient $L/Z(L)$ is semisimple. So, the Killing form is negative definite of $L/Z(L)$. Therefore, this Killing form is both positive definite and negative definite, it follows that $L/Z(L) = {0}$. So we get $L = Z(L)=\ker(\kappa)$. But $\kappa$ is non-degenerate since it’s positive definite. It follows that $L= {0}$.

I am confused with the following gaps:

  1. Why killing form on $L/Z(L)$ is negative definite?
  2. Why the induced Killing form on $L/Z(L)$ is positive definite?

And is there any relation with the fact that $\mathfrak{g}$ is real?

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I admit I don't immediately understand this step either... certainly semisimple Lie algebras do not have negative-definite Killing form in general.

Here is another argument: if $\mathfrak{g}$ has positive-definite Killing form $\kappa$, let $x_1,...,x_n$ be an orthonormal basis of $\mathfrak{g}$ and write $[x_i,x_j] = \sum_k a_{ijk} x_k$ with $a_{ijk} \in \mathbb{R}$. The invariance $\kappa([x,y],z) = \kappa(x,[y,z])$ implies $a_{ijk} + a_{ikj} = 0$. Therefore $$\kappa(x_i,x_i) = \mathrm{tr}( \mathrm{ad}(x_i) \cdot \mathrm{ad}(x_i)) = \sum_{j,k} a_{ijk}a_{ikj} = - \sum_{j,k} a_{ijk}^2 < 0,$$ contradiction.

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  • $\begingroup$ Could you please show how the invariance implies the "skew-symmetry"? $\endgroup$ – Blake Oct 5 '18 at 8:53
  • $\begingroup$ $a_{ijk} = K([x_{i},x_{j}],x_{k}) = K(x_{i}, [x_{j}, x_{k}]) = -K([x_{i}, x_{k}], x_{j}) = -a_{ikj}$ $\endgroup$ – Rob Rockwood Jan 27 at 12:24

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