# How am I to interpret $\text{Tr}(\text{ad}_X\text{ad}_Y)$?

I'm trying to show that the $$(2,0)$$ Killing tensor is invariant under the $$\text{Ad}$$ homomorphism: $$K(\text{Ad}_A(X),\text{Ad}_A(Y))=K(X,Y),$$ with $$X,Y\in \mathfrak{g},\hspace{1mm}A\in G,$$ and $$K(X,Y)\equiv-\text{Tr}(\text{ad}_X\text{ad}_Y)$$. My question isn't about how to do this; rather, it is about how to interpret this trace (once that's accomplished, I'm confident I can carry out the proof). Trying to peer into it, I'd write $$\text{Tr}(\text{ad}_X\text{ad}_Y) = \text{Tr}([X,[Y,\cdot]]),$$ but that just heightens my feeling of this expression needing an argument to make any sense. So, are the $$2$$ inputs of this $$(2,0)$$ tensor to be interpreted as $$X$$ and $$Y$$, and I'm to manipulate this expression as it currently stands? Or am I to gain insight of this Killing form by feeding in two elements of $$\mathfrak{g}$$, and thereby go about this proof? Already I must weigh myself against the latter, as I can't see it making sense for me to feed two inputs into $$\text{Tr}(\text{ad}_X\text{ad}_Y)$$. Boiling it down, my question is how do I interpret this trace?

• Hint: perhaps you should test-drive all this on su(2), where it is virtually impossible to get lost? Jan 21, 2020 at 15:39

The trace is over the Lie algebra $$\mathfrak{g}$$ itself. Note that $${\rm ad}_X\in{\rm End}(\mathfrak{g})~\equiv~{\cal L}(\mathfrak{g},\mathfrak{g})$$ is a linear map from $$\mathfrak{g}$$ to $$\mathfrak{g}$$. If we chose a basis $$(t_j)_{j=1,\ldots,n}$$ for the Lie algebra $$\mathfrak{g}$$, then we can represent the linear map $${\rm ad}_X~\equiv~[X,\cdot]$$ by its corresponding matrix $$({\rm ad}_X)^j{}_k$$, cf. my Phys.SE answer here. The trace is then $$K(X,Y)~:=~{\rm Tr}_{\mathfrak{g}}({\rm ad}_X \circ{\rm ad}_Y) ~=~\sum_{j,k=1}^n({\rm ad}_X)^j{}_k({\rm ad}_Y)^k{}_j.$$ It is of course independent of the choice of basis.

• Excellent, thank you!
– dsm
Jan 21, 2020 at 18:08

A Lie algebra $$\frak g$$ is, among other things, a vector space. For each $$X \in \frak g$$, one can define a linear map $$\mathrm{ad}_X : \frak g \to \frak g$$. Then $$\mathrm{ad}_X \mathrm{ad}_Y$$ is the composition of two linear maps, and therefore a linear map itself. Then the trace of a linear map is just the usual trace!

If you imagine elements of $$\frak g$$ as column vectors then for each $$X$$, $$\mathrm{ad}_X$$ is a matrix and the trace is the sum of its diagonal elements. Based on this intuition and considering a basis for the Lie algebra, it shouldn't be too hard to find an explicit expression for the Killing form.

Perhaps writing out the individual homomorphism will help?

\begin{align} \mathrm{ad}_X: \mathfrak{g} &\to \mathfrak{g} \\ Z &\mapsto [X, Z] \end{align}

Because the bracket is linear in it's second argument, this is a linear map.

Consequently, so is

\begin{align} \mathrm{ad}_X\circ\mathrm{ad}_Y: \mathfrak{g} &\to \mathfrak{g} \\ Z &\mapsto [X, [Y, Z]] \end{align}

The Killing form is invariant under any automorphism $$\varphi$$, not just $$\mathrm{Ad}_A$$. To show this, you could start with $$\mathrm{ad}_{\varphi(X)}\circ\mathrm{ad}_{\varphi(Y)} = \varphi\circ\varphi^ {-1}\circ\mathrm{ad}_{\varphi(X)}\circ\mathrm{ad}_{\varphi(Y)} = \dots$$ The last step will be using the cyclic property of the trace.