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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?

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

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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.

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  • $\begingroup$ Excellent, thank you! $\endgroup$
    – dsm
    Jan 21, 2020 at 18:08
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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.

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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.

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