Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $<diag(a_i),diag(b_i)>=2n \sum_{i=1}^n a_ib_i$?

share|cite|improve this question
Just look at… and put $tr(A)=tr(B)=0$ condition at the result. –  Adam Baranowski Apr 23 at 16:48

1 Answer 1

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...)

share|cite|improve this answer
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 8 '12 at 3:02
$=\sum_{i=1}^{n-1}a_ib_i+(n-1)a_nb_n+a_nb_n+a_nb_n$, 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 8 '12 at 3:10
@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 8 '12 at 17:55

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.