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Theorem: If $G$ is a compact semisimple Lie group, then its Killing form is negative definite.

In its proof: Since $G$ is compact there is an Ad-invariant inner product on $\mathfrak g$.

Since each $ad_X$ is skew-symmetric, let $ad_X=(a_{ij})$ relative to an orthonormal basis of $\mathfrak g$. Then $$B(X,X)=tr(ad_X\circ ad_X)=\sum_i\sum_ja_{ij}a_{ji}=-\sum_{i,j}a^2_{ij}\leq0$$ Since $G$ is semisimple, $B$ is nondegenerate, so the above sum is strictly less than zero. $ \diamond $

I have a question here: Where the Ad-invariant inner product is used in the proof above?

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The fact that $B$ is $Ad$-invariant implies that the matrix of $ad_x$ is skew symmetric in a orthogonal basis, the result follows from the fact that if $X$ is skew symmetric, $trace(X^2)\leq 0$ if $X$ is skew-symmetric and $B$ is non degenerated.

N.B: if $X$ is skew-syymetric, $X+X^T=0$ implies that $X(X+X^T)=X^2+XX^T=0$, thus $trace(X^2)=-trace(XX^T)$. We know that $trace(XX^T)\geq 0$.

If $B$ is $Ad$-invariant, $B(Ad(exp(tz))x,Ad(exp(tz))y)=B(x,y)$. If you calculate the differential at $t=0$, you obtain $B([z,x],y)+B(x,[z,y])=0$

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  • $\begingroup$ Thanks Tsemo, but how to prove the fact that: $B$ is $Ad$-invariant implies that $ad_x$ is skew symmetric in an orthonormal basis? $\endgroup$ – Ronald Jun 17 '16 at 0:34
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    $\begingroup$ If $B$ is $Ad$-invariant, $B(Ad(exp(tz))x,Ad(exp(tz))y)=B(x,y)$. If you calculate the differential at $t0$, you obtain $B([z,x],y)+B(x,[z,y])=0$ $\endgroup$ – Tsemo Aristide Jun 17 '16 at 0:36
  • $\begingroup$ Sorry Tsemo, but how $B([z,x],y)+B(x,[z,y])=0$ implies that $ad_x$ is skew symmetric? $\endgroup$ – Ronald Jun 17 '16 at 0:57
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    $\begingroup$ It is equivalent to saying that $ad_z$ is skew-symmetric, $B([z,x],y)+B(x,[z,y])=B(ad_z(x),y)+B(x,ad_z(y))=0$. $\endgroup$ – Tsemo Aristide Jun 17 '16 at 0:59
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    $\begingroup$ @Ronald We are by assumption working with a real Lie algebra. Otherwise as you point out, the entries of the matrix $a_{ij}$ could be complex, and we could not conclude that $- \sum_{i,j} a_{ij}^2$ was negative. $\endgroup$ – Idempotent Jul 14 '16 at 17:37

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