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Let $L$ be a finite-dimensional Lie algebra over the real numbers with positive definite Killing form. Why is $L=\{0\}$?

As a hint we should have a look at Gram-Schmidt orthonormalisation.

Everything I tried failed so far, so the usual 'What do I have so far' field is left empty. Sorry for that.

2 Answers2

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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 $L= {0}$.

Dietrich Burde
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  • Thank you, but I can't follow. I have not heard of the term 'reductive' and how this is related to semisimplicity. Do you know how to use the given hint? –  May 26 '17 at 16:05
  • Reductive is very closely related to semisimple, see here. – Dietrich Burde May 26 '17 at 18:18
  • Gram-Schmidt gives the Iwasawa decomposition $G=KAN$ for the Lie group. The Lie algebra of a compact Lie group is reductive. – Dietrich Burde May 27 '17 at 12:24
  • That helped a lot, thank you! –  May 28 '17 at 12:41
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    I have several questions: 1. Why killing form on L/Z(L) is negative definite? 2. Why the induced Killing form on L/Z(L) is positive definite? 3. Why ker(k)=Z(L)? Thanks! – Aolong Li Dec 01 '17 at 16:33
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As Dietrich has said, the Killing form $\kappa$ defines an positive-definite inner product on the Lie algebra $L$, which is also $ad$-invariant. A Lie algebra is a compact Lie algebra iff it admits an $ad$-invariant inner product, which is our case. But a compact Lie algebra's Killing form must be negative-definite. Thus $\kappa$ is both negative- and positive-definite. So we conclude that $L$ must be trivial.

Andre Gomes
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