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Let $\mathfrak{g}$ be the lie algebra of a compact lie group, so that the Killing form $(,)$ is negative definite. Let the root space decomposition $\mathfrak g =\mathfrak h \oplus \oplus_{\alpha \neq 0} \mathfrak g_\alpha$ be given.

If $e \in \mathfrak{g_\alpha}$ for $\alpha \neq 0$, then $\alpha(h)(e,e)=([h,e],e)=([e,h],e)=-\alpha(h)(e,e)$ and $(e,e)=0$. This contradicts the form being negative definite.

Where is this contradiction coming from?

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    In general invariance of the inner product gets you that $([x,y],z)=([z,x],y)$. To be sure we have the signs correct we can check in the case of $sl_n$: $([x,y],z)=tr((xy-yx)z)=tr(xyz)-tr(yxz)=tr(zxy)-tr(xzy)=tr([z,x]y)=([z,x],y)$ Now let $x=h$, $y=e$ and $z=e$. – 2017-02-26
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    (the above comment was a response to another comment, since deleted, asking for clarification on the step $([h,e],e)=([e,h],e)$) – 2017-02-26

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One does not have a root space decomposition when $\mathfrak g$ is the lie algebra of a compact lie group. This is because any complex lie group is necessarily not compact.

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    The last statement is not true, see [here](https://en.wikipedia.org/wiki/Complex_Lie_group). There are compact complex Lie groups. – 2017-02-27
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    Thanks for pointing this out. I forgot about tori. But other than these and discrete groups I think I am right. – 2017-02-28