What is the trace in the definition of a Killing Form?

Question

Let $\underline G$ be an arbitrary Lie algebra. For arbitrary $X,Y\in\underline G$, we can define the adjoint of $X$ by $\text{ad}_X(Y)=[X,Y]$. Thus, $\text{ad}_X$ is a linear map on $\underline G$.

Given the adjoint, one defines the Killing Form $g(X,Y)=\text{Tr}(\text{ad}_X\circ \text{ad}_Y)$.

I don't understand what $\text{Tr}$ means in this context. I understand that $\underline G$ is a vector space, so $\text{ad}_X\circ \text{ad}_Y$ is a linear map on the vector space, thus something that we should be able to take a trace of. But as far as I can tell, there is no canonical basis of $\underline G$, so I don't know what basis to take the trace in. There's no inner product defined on $\underline G$, so I can't just pick an orthonomal basis.

What does this trace mean? It would be helpful to see a simple example, the specific context I'm interested in is when $\underline G$ is the Lie algebra of $SU(n)$.

Answer 1

The trace can be done in any basis. The trace is invariant under an arbitrary change of basis. If $M$ is an arbitrary matrix, and $A\in \mathbb{GL}(n,\mathbb{C})$, then $$ \text{Tr}(A^{-1}MA)=\text{Tr}(M) $$ by the cyclic property of the trace.

So for $SU(N)$, you just pick any basis of the Lie algebra of skew-symmetric, traceless matrices, and you figure out the diagonal elements.

Answer 2

It really means the trace of the resultant matrix. Let's take $\mathfrak{sl}\left(2\right)$ and consider the following base $$e=\left(\begin{array}{cc} 0 & 1\\ 0 & 0 \end{array}\right),$$$$h=\left(\begin{array}{cc} 1 & 0\\ 0 & -1 \end{array}\right),$$$$f=\left(\begin{array}{cc} 0 & 0\\ 1 & 0 \end{array}\right)$$ You have the following relations $$\left[h,\,e\right]=2e,\,\,\,\,\left[h,\,f\right]=-2f,\,\,\,\left[e,f\right]=h.$$ Then considering as order of the base $\left\{ e,\,h,\,f\right\}$ you get $$\text{ad}\left(e\right) =\left(\begin{array}{ccc} 0 & -2 & 0\\ 0 & 0 & 1\\ 0 & 0 & 0 \end{array}\right),$$ $$\text{ad}\left(h\right) =\left(\begin{array}{ccc} 2 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & -2 \end{array}\right),$$ $$\text{ad}\left(f\right) =\left(\begin{array}{ccc} 0 & 0 & 0\\ -1 & 0 & 0\\ 0 & 2 & 0 \end{array}\right).$$ Then do the multiplication of the matrices (ex $\text{ad}\left(e\right)\text{ad}\left(e\right)$) take the Trace and you will get the following coefficient for the killing form: $$\left(B\right)_{ij}=\left(\begin{array}{ccc} 0 & 0 & 4\\ 0 & 8 & 0\\ 4 & 0 & 0 \end{array}\right).$$ It's really just a straightforward calculation. You might find a lot of examples of this kind in Carter