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? 
 A: 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. 
A: 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.
A: 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.
