Determinant of Killing form of sl_n
My question is something like a continuation of: Is there a more intelligent way to compute the determinant of the Killing form of $\mathfrak{sl}(3,F)$?
Motivated by that same problem in Humphreys' book, I wrote a quick Sage program to calculate the determinant of the Killing form of $sl_n(\mathbb{Z})$, and for $n\le 17$ I get $\det\kappa=\pm 2^{n^2-1}n^{n^2}$, where $\pm$ is positive if $n=2$ or $3$ mod $4$, negative in the other two cases. So I'm confident the pattern continues, and I'm happy with it since it says that $sl_n(k)$ is semisimple if $\text{char }k>n$.
My question now is: how could this be formally proven? I'm just now reviewing the rep. theory of Lie algebras in general (I'm only fully confident with basic structure theory of them and Lie groups, and related facts), so I'll be keeping this question in mind as I review; but if the answer is immediately apparent to anyone, I'd appreciate a rough sketch.
Solution 1:
Here's a proof of your formula.
- Let's compute things in the standard (not adjoint) representation. For matrices $A,B$, define $s(A,B)=\mathrm{Tr}(AB)$.
First, in the basis of all matrices, the multiplication law on the basis is $E_{ij}E_{kl}=\delta_{jk}E_{kl}$. So $s(E_{ij},E_{kl})=\mathrm{Tr}(E_{ij}E_{kl})=\delta_{jk}\delta_{il}$.
Now we restrict this to traceless matrices and compute the determinant in the integral basis. This has an $s$-orthogonal decomposition between diagonal and zero-diagonal matrices. For the latter, on the basis we just have $s(E_{ij},E_{ji})=1$ and zero elsewhere, so its contribution to the determinant is $(-1)^{n(n-1)/2}$. For the diagonal part, we have the basis $(H_1,\dots,H_n)$, where $H_i=E_{11}-E_{ii}$, and $s(H_i,H_j)$ thus equals 2 for $i=j$ and 1 otherwise. The determinant of the corresponding $(n-1)\times (n-1)$-matrix (2 on the diagonal and 1 outside the diagonal) is $n$, by a simple linear algebra exercise. Thus the determinant of $s$ on $\mathfrak{sl}_n(\mathbf{Z})$ is $(-1)^{n(n-1)/2}n$.
- $s$ being a nonzero invariant symmetric bilinear form on the simple complex Lie algebra $\mathfrak{sl}_n$, the Killing form $\kappa$ is a scalar multiple of it (see 3). So to compute it it is enough to compute a single nonzero value. Namely, let us compute $\kappa(E_{12},E_{21})$.
The operator $\mathfrak{ad}(E_{12})\mathfrak{ad}(E_{21})$ maps: $E_{kl}$ to $0$ when $k,l\ge 3$, maps $E_{21}$ to 0. It maps $E_{12}$ to $2E_{12}$, $E_{1j}$ to $E_{1j}$ and $E_{j2}$ to $E_{j2}$ for all $j\ge 3$. It maps $P_2$ to $2P_2$ and $P_j$ to $P_2$ for $j\ge 3$. The the trace contribution is $+2$ for $E_{12}$, $+2$ for $P_2$, $+(n-2)$ for $E_{1j}$, $+(n-2)$ for $E_{j2}$, and 0 for others ($E_{kl}$, $E_{21}$, $P_j$). The total yields the trace $2n$; thus $\kappa(E_{12},E_{21})=2n$.
Since $s(E_{12},E_{21})=1$, we deduce that $\kappa=2n.s$. Thus the determinant of the Killing form $\kappa$ on the desired basis is $(2n)^{n^2-1}\det(s)$.
Therefore $$\det(\kappa)=\left((-1)^{\frac{n(n-1)}2}\right)n(2n)^{n^2-1}=\left((-1)^{\frac{n(n-1)}2}\right)2^{n^2-1}n^{n^2}.$$
- added after asked in a comment: Let $\mathfrak{g}$ be a simple complex Lie algebra, $\kappa$ its Killing form (which is non-degenerate), and $B$ another invariant bilinear form. Define $L_B(x)(y)=B(x,y)$ and simlarly $L_\kappa$. Then both $L_B$ and $L_\kappa$ are linear homomorphisms $\mathfrak{g}\to\mathfrak{g}^*$, $L_\kappa$ being bijective. So $u=L_BL_\kappa^{-1}$ is a linear endomorphism of $\mathfrak{g}$, and commutes with the adjoint $\mathfrak{g}$-action (check details!). So any eigenspace of $u$ is an ideal of $\mathfrak{g}$. By simplicity (and since $\mathbf{C}$ is algebraically closed), it follows that $u$ is a scalar multiplication, so $B$ is a scalar multiple of $\kappa$.