Killing Form for $\mathfrak{sl}_{n}(\Bbb{C})$

Question

Show that for $\frak{g} = \frak{sl}_{n}(\Bbb{C})$, the Killing form is given by $K(x,y) = 2n tr(xy)$.

This is problem 5.2 in Kirillov's book on Lie Algebras. Recall that $K(x,y) = tr (\text{ad } x \text{ ad } y)$, where $\text{ad } x \text{ ad } y$ is a composition of two operators acting on $\frak{sl}_{n}(\Bbb{C})$. From my understanding, in order to compute the trace, we need to compute a matrix representation of $\text{ad } x \text{ ad } y$ with respect to some basis of $\frak{sl}_{n}(\Bbb{C})$, and then look at the diagonal entries.

However, this seems like an awful task. I am told that it suffices to compute $K(E_{ij},E_{ji})$. But it still is proving to be an awful task. Am I missing something, or is this just a very calculation heavy problem?

Answer 1

This is a more elementary approach without using the fact that $\mathfrak{sl}_n(\Bbb C)$ is simple that also works over any field.

If $\mathfrak{g}$ is a Lie algebra and $\mathfrak{i} \subset \mathfrak{g}$ is an ideal, then the Killing form of $\mathfrak{g}$ restricts to the killing form of $\mathfrak{i}$. (This is a simple observation, see: https://groupprops.subwiki.org/wiki/Killing_form_on_ideal_equals_restriction_of_Killing_form)

As $\mathfrak{sl}_n(\Bbb C)$ is an ideal in $\mathfrak{gl}_n(\Bbb C)$, we can also compute the Killing form of the latter.

Consider the $\Bbb C$-algebra $M_{n \times n}(\Bbb C)$ and the $\Bbb C$-algebra homomorphism $\phi:M_{n \times n}(\Bbb C) \to \mathrm{End}(M_{n \times n}(\Bbb C))$ given by $A \mapsto \phi_A=(B \mapsto A\cdot B)$

If we take the standard basis $E_{i,j}$ (with a $1$ at $(i,j)$ and zeroes everywhere else) of $M_{n \times n}(\Bbb C)$ to identify $\mathrm{End}(M_{n \times n}(\Bbb C))$ with $M_{n^2\times n^2}(\Bbb C)$, then there's a simple description of this map: If $A=\sum_{i,j} a_{ij} E_{i,j}$, then $AE_{k,l}=\sum_{i,j} a_{ij} E_{i,j}E_{k,l}=\sum_{i}a_{i,k}E_{i,l}$, so for the matrix coefficient $(\phi_A)_{(k,l),(i,j)}$ of $\phi_A$ corresponding to $(k,l),(i,j)$ (indexing our matrix entries by pairs of pairs of indices instead of pairs of indices) we get $(\phi_A)_{(k,l),(i,j)}=a_{i,k}$ if $j=l$ and $0$ else.

Consider analogously the map $\psi:M_{n \times n}(\Bbb C) \to \mathrm{End}(M_{n \times n}(\Bbb C)), A \mapsto \psi_A=(B \mapsto B\cdot A)$

We make an analogous computation: If $A=\sum_{i,j}a_{ij}E_{i,j}$, then $E_{k,l}A=\sum_{i,j}a_{ij} E_{k,l}E_{i,j}=\sum_{j}a_{lj}E_{k,j}$, so $(\psi_A)_{(k,l),(i,j)}=a_{l,j}$ for $i=k$ and $0$ else.

Note that $\psi$ satisfies $\psi_{AB}=\psi_B \circ \psi_A$. and $\phi_A \psi_B=\psi_B \phi_A$ for all $A,B$.

We want to compute the trace of $\phi_A \circ \psi_B=\psi_B \circ \phi_A$:

We know that $(\phi_A)_{(k,l),(i,j)}=a_{i,k} \delta_{j,l}$ and $(\psi_B)_{(u,v),(w,z)}=b_{v,z}\delta_{u,w}$ using the Kronecker delta.

Thus we get from the matrix multiplication $(\phi_A \psi_B)_{(k,l),(w,z)}=\sum_{(i,j)}(\phi_A)_{(k,l),(i,j)}(\psi_B)_{(i,j),(w,z)}=\sum_{(i,j)}a_{i,k}b_{j,z} \delta_{j,l}\delta_{i,w}=a_{w,k}b_{l,z}$

Using this, we get $\mathrm{Tr}(\phi_A \psi_B)=\sum_{(i,j)} (\phi_A \psi_B)_{(i,j),(i,j)}=\sum_{(i,j)}a_{i,i}b_{j,j}=(\sum_i a_{i,i})(\sum_j b_{j,j})=\mathrm{Tr}(A) \mathrm{Tr}(B)$.
As a special case, using that $\phi_{\mathrm{Id}_n}=\psi_{\mathrm{Id}_n}=\mathrm{Id}_{n^2}$, we get that $\mathrm{Tr}(\phi_A)=\mathrm{Tr}(\psi_A)=n\mathrm{Tr}(A)$.

Now identifying $\mathfrak{gl}_n(\Bbb C)$ with $M_{n\times n}(\Bbb C)$, we have $\mathrm{ad}(A)=\phi_A-\psi_A$ We get $\mathrm{ad}(A)\mathrm{ad}(B)=(\phi_A-\psi_A)(\phi_B-\psi_B)=(\phi_A\phi_B)-\phi_A \psi_B-\psi_A \phi_B + \psi_A \psi_B)=\phi_{AB}-\phi_A \psi_B -\phi_B \psi_A + \psi_{BA}$.

Thus we can apply our computations to see that

$$\mathrm{Tr}(\mathrm{ad}(A)\mathrm{ad}(B))=\mathrm{Tr}(\phi_A\phi_B)-\mathrm{Tr}(\phi_A\psi_B)-\mathrm{Tr}(\phi_B \psi_A)+\mathrm{Tr}(\psi_{BA})$$ $$=n\mathrm{Tr}(AB)-\mathrm{Tr}(A)\mathrm{Tr}(B)-\mathrm{Tr}(B)\mathrm{Tr}(A)+n\mathrm{Tr}(BA)=2n\mathrm{Tr}(AB)-2\mathrm{Tr}(A)\mathrm{Tr}(B)$$

Restricting this Killing form to $\mathfrak{sl}_n(\Bbb C)$, the second term vanishes and we get $K(x,y)=2n\mathrm{Tr}(xy)$

Answer 2

If we grant outselves that $\mathfrak g=\mathfrak s\mathfrak l_n$ is simple, one way to simplify the computation is as follows.

By a very easy version of Schur's lemma, the simple-ness imples that there is a unique $\mathfrak g$-equivariant map $\mathfrak g\to \mathfrak g$, up to scalars, and, indeed, that all such maps are scalar mappings themselves. Use $$ {\mathrm {Hom}}(\mathfrak g,\mathfrak g) \approx {\mathrm{Hom}}(\mathfrak g\otimes \mathfrak g^*,\mathbb C) $$ and identify $\mathfrak g^*\approx \mathfrak g$ via any non-zero $\mathfrak g$-equivariant pairing on $\mathfrak g$. We see that the space $\mathrm{Hom}(\mathfrak g\otimes\mathfrak g,\mathbb C)$ of ($\mathfrak g$-equivariant) bilinear maps on $\mathfrak g$ is one-dimensional.

So $\mathrm{tr}(\mathrm{ad} x\circ \mathrm{ad} y)$ is a constant multiple of $\mathrm{tr}(xy)$. To determine the constant, evaluate both of these for any choice of $x,y$ producing a non-zero value for one of them. E.g., $x=y=E_{11}-E_{22}$?