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Chose a finite dimensional semi-simple Lie algebra $L$ and a Cartan subalgebra $H$. Why do we have $$\kappa_L (h,k)=\sum_{\alpha\in\Phi (L,H)}\alpha(h)\alpha(k),$$ where $\kappa$ is the Killing form and $\Phi(L,H)$ is the set of roots?

I tried to work something out via the root space decomposition and Lie's theorem but I didn't get that far. Any thoughts?

glS
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user455297
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  • related: https://math.stackexchange.com/q/889551/173147, https://math.stackexchange.com/q/1334606/173147 – glS Jun 08 '21 at 19:42

2 Answers2

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Take $h\in H$. Remember that the action of $\operatorname{ad}(h)$ on $L$ is diagonalisable and that the eigenvalues (other than $0$) are precisely the numbers $\alpha(h)$, with $\alpha\in\Phi(L,H)$. Therefore, with respect to some basis of $L$, the matrix of $\operatorname{ad}(h)$ is a diagonal matrix such that the entries of the main diagonal are the numbers $\alpha(h)$ ($\alpha\in\Phi(L,H)$) and $0$. If $k$ is another element of $H$, the same thing will be true with respect to the same basis. Therefore, the matrix of $\operatorname{ad}(h)\circ\operatorname{ad}(k)$ will be a diagonal matrix whose entries within the main diagonal will be the numbers $\alpha(h)\alpha(k)$ ($\alpha\in\Phi(L,H)$) and $0$. So, its trace is the sum that you have mentioned.

José Carlos Santos
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  • Thank you but why are the eigenvalues of $ad(h)$ PRECISELY the numbers $\alpha (h)$? Every $\alpha(h)$ is an eigenvalue but why is every eigenvalue of this form? – user455297 Jun 15 '17 at 14:33
  • Because, by definition, the roots are those linear forms $\alpha\in H^*$ such that, for each $h\in H$, $\operatorname{ad}(h)$ is diagonalisable and the eigenvalues are the numbers $\alpha(h)$. – José Carlos Santos Jun 15 '17 at 14:44
  • Do you use the root space decomposition at any point in your proof? – user455297 Jun 15 '17 at 14:47
  • Yes, when I write that, with respect to some basis of $L$, the matrix of $\operatorname{ad}(h)$ is a diagonal matrix such that the entries of the main diagonal are the numbers $\alpha(h)$ ($\alpha\in\Phi(L,H)$) and $0$. – José Carlos Santos Jun 15 '17 at 14:49
  • Can you explain this in a few words? My problem is: $ad(h):L\to L$ is diagonalisable w.r.t. some basis by definition of Cartan subalgebra. What does this have to do with the decomposition? – user455297 Jun 15 '17 at 14:55
  • For each $\alpha\in\Phi(L,H)$, pick a basis of $L_\alpha$. Then get the union of all these basis and add to them a basis of $H$. Then you will have a basis of $L$ with the property that I mentioned. – José Carlos Santos Jun 15 '17 at 14:58
  • @user455297 You're welcome. If my answer was useful, please consider upvoting it. – José Carlos Santos Jun 15 '17 at 15:31
  • I have already voted up, but since I have not much reputation it doesn't count... – user455297 Jun 15 '17 at 18:05
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The Cartan decomposition says that $$ L=H\oplus \sum_{\alpha\in \Phi}L_{\alpha}. $$ The matrices of the restrictions of $ad(h)$ to $L_{\alpha}$ for $h\in H$ can be taken simultaneously in lower-triangular form with diagonal $(\alpha(h),\ldots ,\alpha(h))$ by Lie's Theorem. With $\dim L_{\alpha}=1$ and taking the trace we obtain that $$ \kappa(h,k)=tr(ad(h)\cdot ad(k))=\sum_{\alpha\in \Phi}\alpha(h)\alpha(k) $$ for $h,k\in H$.

Dietrich Burde
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  • Thank you very much, but why can we apply Lie's theorem since $L$ is not solvable and why has the diagonal the form you mentioned? – user455297 Jun 15 '17 at 14:46
  • Because of the Cartan subalgebra, which is solvable. Note that we only consider $ad(h)$ for $h\in H$. – Dietrich Burde Jun 15 '17 at 14:53