Intuition for the Killing form Given a real Lie algebra $\mathfrak{g}$, the Killing form $K:\mathfrak{g} \times \mathfrak{g} \to \mathbb{R}$ is defined as $K(X,Y)=tr(ad(X)\circ ad(Y))$. I want a way to visualize this.
Suppose $\mathfrak{g}$ is the Lie algebra of a Lie group $G$. Then $Ad(x):\mathfrak{g} \to \mathfrak{g}$ is the action induced on $\mathfrak{g}$ from conjugation by $x \in G$. $ad(X)(Y)$ is then the differential of $Ad(e^{tX})$ at $t = 0$, so it can be thought of as the direction that conjugation by $e^{tX}$ directs $Y$ toward, for small $t$. These are things I can sort of visualize.
Given $X$,$Y$ tangent vectors to $G$ at the identity, how can I visualize $K(X,Y)$?
 A: The important point about the Killing form are its invariance properties. To see those, you can follow @Dietrich_Burde's comment, and start thinking about matrices. As you may recall, viewing $M_n(\mathbb R)$ as $\mathbb R^{n^2}$, the standard inner product can be written as $\langle X,Y\rangle=tr(X^tY)$. Now if you look at $(X,Y):=tr(XY)$ instead, you loose positive definiteness, but still get a non-degenerate bilinear form. This has the advantage however, that it has much nicer invariance-properties since for any invertible matrix $A$, you get $(AXA^{-1},AYA^{-1})=(X,Y)$. (The corresponding property for $\langle X,Y\rangle$ holds only for unitary matrices $A$.) Now this invariance property tells asyou, that the bilinear form $(\ ,\ )$ is actually not only well defined on matrices but on linear maps, since different matrix representations are related by conjugation.
Now if you move to an abstract Lie algebra $\mathfrak g$, this does not consist of matrices. But the adjoint representation, which is a canonical object maps $\mathfrak g$ to a Lie algebra consisting of linear maps, naemly into $L(\mathfrak g,\mathfrak g)$. If $\mathfrak g$ has trivial center, then the adjoint representation is injective, and thus identifies $\mathfrak g$ with a Lie algebra of linear maps. And then you just pull back the trace form to $\mathfrak g$ via this isomorphism. By construction, the result must have nice invariance properties and if $\mathfrak g$ is semisimple, it turns out to be non-degenerate.
