Killing form of a reductive symmetric Lie algebra Suppose $(g; k ,p)$ is a reductive symmetric Lie algebra. i.e. $k$ is a sub-algebra of $g$, $[k,p] \subset p$ , $[p,p] \subset k$ and $g= k \oplus p$. This is actually from Lepowsky and McCollum's paper on _Cartan Subspaces of Symmetric Lie Algebras_ (Transactions of the American Mathematical Society Vol. 216 (Feb., 1976), pp. 217-228), theorem 5.1. "... Since the Killing form $B$ of $g$ is a non-singular trace form on $k$, then $k$ is reductive." My questions: 1. What does _non-singular trace_ form mean? 2. Why is $k$ reductive?

1.) A _nonsingular trace form_ on a Lie algebra $L$ over a field $K$ is a bilinear form $f\colon L\times L\rightarrow K$ given by $f(x,y)=tr(\rho(x)\rho(y))$, which is non-degenerate as a bilinear form; here $\rho\colon L\rightarrow \mathfrak{gl}(V)$ is a Lie algebra representation. For example, with $\rho=ad$ we have the Killing form. It is nonsingular iff the Lie algebra is semisimple, over characteristic zero.

2.) If you read the proof of Theorem 5.1 in the paper, it says "Since the Killing form $B$ of $g$ is a nonsingular trace form on $k$, $k$ is reductive." And then they even give two references for this well-known result: "see either [5, §2.9] or [1, 1.7.3] ". You can also find this in other common books on Lie algebras.