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Consider $sl_n\mathbb{C}$ as aLie-algebra, and choose h the CSA formed by diagonal matrixes. I can i demonstrate that the Cartan-Killing form in $sl_n\mathbb{C}$ is $<diag(a_i),diag(b_i)>=2n \sum_{i=1}^n a_ib_i$?

glS
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balestrav
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  • Just look at http://math.stackexchange.com/questions/135567/computation-of-the-killing-form-of-mathfrakgl-m/1248343#1248343 and put $tr(A)=tr(B)=0$ condition at the result. – Adam Baranowski Apr 23 '15 at 16:48

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Granting that $sl(n)$ is simple, one might first show that, up to scalars, there is a unique $sl(n)$-invariant non-degenerate symmetric bilinear form on it. To see this, note that a non-degenerate bilinear form gives an $sl(n)$-module map $f$ of $sl(n)$ to its dual $sl(n)^*$, by $f(x)(y)=\langle x,y\rangle$. The non-degeneracy is equivalent to trivial kernel, so, by finite-dimensionality, equivalent to $f$ being an isomorphism. Given two such pairings, and corresponding maps $f,g$, the composite $f^{-1}\circ g$ is an $sl(n)$-isomorphism of $sl(n)$ to itself. By Schur's Lemma, this is a scalar. [done]

Thus, it suffices to evaluate Casimir and the "trace pairing" on a single pair of elements of the Cartan subalgebra, to evaluate the constant. For example, $x=y=(1,-1,0,...,0)$ (diagonal) has eigenvalues $+2$ (once), $-2$ (once), $+1$ ($n-2$ times), and $-1$ ($n-2$ times). Applying it twice and adding up gives $2n$.

(To show from scratch that $sl(n)$ is simple is not too hard, if one wants...)

paul garrett
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  • I'm trying to prove it directly, and i have troble in demonstrating this: $\sum_{i \neq j}(a_i-a_j)(b_i-b_j)=2n\sum_{i=1}^na_ib_i$, where in the first sum the indexes vary from 1 to n, and where $a_n=-\sum_{i=1}^{n-1}a_i$ and $b_n=-\sum_{i=1}^{n-1}b_n$. I tried by induction: $\sum_{i \neq j}(a_i-a_j)(b_i-b_j)=2(n-1)\sum_{i=1}^{n-1}a_ib_i + \sum_{i=1}^{n-1}(a_i-a_n)(b_i-b_n)+\sum_{i=1}^{n-1}(a_n-a_i)(b_n-b_i)$. The last two sum are equal so i calculate only one: $\sum_{i=1}^{n-1}(a_i-a_n)(b_i-b_n)=\sum_{i=1}^{n-1}a_ib_i+(n-1)a_nb_n+a_n(-\sum_{i=1}^{n-1}b_i)+b_n(-\sum_{i=1}^{n-1}a_i)$ – balestrav Jan 08 '12 at 03:02
  • $=\sum_{i=1}^{n-1}a_ib_i+(n-1)a_nb_n+a_nb_n+a_nb_n$, so puttin this back i have: $2n\sum_{i=1}^{n-1}a_ib_i-2\sum_{i=1}^{n-1}a_ib_i+2(n-1)a_nb_n+4a_nb_n= 2n\sum_{i=1}^{n}a_ib_i+2a_nb_n$, and the last term is the problem, it shouldn't be there! Where have i done wrong? – balestrav Jan 08 '12 at 03:10
  • @balestrav, ... without looking at the details of your computation... I fear this induction is a little delicate, if we insist on incorporating the trace=0 condition. It might work better to arrange a version for gl(n), instead. – paul garrett Jan 08 '12 at 17:55