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Let $\mathfrak g$ be a lie algebra with a root space decomposition $\mathfrak g =\mathfrak h \oplus \oplus_{\alpha \neq 0} \mathfrak g_\alpha$ and let it have an invariant billinear form $(,)$.

Then if $e \in \mathfrak g_\alpha,\alpha \neq 0 $ and $ h \in \mathfrak h$, then $([h,e],e)=(\alpha(h)e,e)=\alpha(h)$. On the other hand, $([h,e_j],e_j)=([e_j,e_j],h)=(0,h)=0$.

Where is this contradiction coming from?

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    Why did you say $(\alpha(h)e,e)=\alpha(h)$? – 2017-02-26
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    Wait this is weird: By this same reasoning, if $x,y$ are in the weight space $\mathfrak g_\alpha$, then for every $h \in \mathfrak h$, $2 \alpha(h)(x,y)=0$ so $(x,y)=0$. In particular $(\alpha(h) e,e)=0$. – 2017-02-26
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    Wait I thought the killing form was supposed to be negative definite? if this is the case then it can't be that $(e,e)=0$. – 2017-02-26
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    @LiLi good point. This raises another question http://math.stackexchange.com/questions/2162645/contradiction-in-the-negative-definiteness-of-killing-form – 2017-02-26
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    Why is it negative definite?In this case, you can show that $(g_\alpha,g_\beta)=0$ as long as $\alpha+\beta\neq0$. – 2017-02-27
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    AHA! Here I am taking $\mathfrak g$ to be a complex lie algebra(to get the root space decomposition) which is why I am not getting the negative definitiveness(which happens when $\g$ is the lie algebra of a (necessarily real ) compact lie group). – 2017-02-27

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