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Let $\mathfrak{g}$ be a real semisimple Lie algebra. Then, we have an obvious $\mathrm{Ad}$-invariant inner product (I don't care about positive definiteness) on $\mathfrak{g}$, namely the Killing form. However, in the concrete case of matrix Lie algebras, when we need such an inner product, we don't typically use the Killing form, but rather the trace: $$ \left< X,Y\right> :=-\mathrm{tr}(XY). $$

In the case of an abstract Lie algebra, when we aren't also handed a canonical finite-dimensional faithful representation, is there a precise way to 'pick out' an $\mathrm{Ad}$-invariant inner product that agrees with the trace in the case that the Lie algebra is a matrix Lie algebra?

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The Killing form is defined by $$\newcommand{\ad}{\operatorname{ad}} K(u,v) = \operatorname{tr} (\ad_u \circ \ad_v).$$ Here tr is the trace in the sense of linear algebra, which is canonically defined, and $\ad$ is the adjoint representation of the Lie algebra on itself. In the case of a matrix Lie algebra, since the adjoint representation is given by $$\ad_u (w) = [u, w]$$ and the bracket is the matrix commutator, you can work out that the Killing form actually agrees with the trace formula, at least up to a scale factor. This is to be expected because the Killing form is with conditions, the, up to scale, unique $\operatorname{Ad}$-invariant non-degenerate quadratic form, and the trace formula has this invariance too.

The condition for this uniqueness is that $G$ should be compact and simple. If $G$ is semi-simple, there is one choice of scale per factor, $a\kappa_1 \oplus b\kappa_2$ is a $\operatorname{Ad}$-invariant on $\mathfrak{g}_1 \oplus \mathfrak{g}_2$ if $\kappa_1, \kappa_2$ are on $\mathfrak{g}_1, \mathfrak{g}_2$. Also for the case of $G = U(1)$, there is a unique, up to scale, $\kappa$: linearly map $\mathfrak{u}(1) \to i\mathbb{R}$ in any way (i.e. choose any basis vector) and define $\kappa$ through multiplication of real numbers. It is a theorem that for a direct sum of Lie algebras where each term is either compact and simple or $\mathfrak{u}(1)$ the symmetric, $\operatorname{Ad}$-invariant bilinear exists and is positive definite. (I learned about this and read the proof in Weinberg's The Quantum Theory of Fields, vol II, but I'm sure it's covered in a Lie algebra textbook.)

On Wikipedia you find listed the Killing form for some common Lie algebras. In particular in every case there except $\mathfrak{gl}(n)$ the listed Killing form is proportional to the trace. $GL(n)$ is of course not compact, but it is covered by the previous paragraph. You can see that when $X,Y$ are traceless, and belong to $\mathfrak{sl}(n)\subset\mathfrak{gl}(n)$, we recover the Killing form for the former.

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    $\begingroup$ A minor correction: It is unique up to scale in the simple case, but not in the semisimple case: Each factor contributes one degree of freedom. $\endgroup$ – Moishe Kohan May 1 '14 at 16:40
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    $\begingroup$ Thank you for the correction. I will edit the answer. $\endgroup$ – Robin Ekman May 1 '14 at 22:43
  • $\begingroup$ The Killing form is sometimes defined more generally with respect to a representation of a Lie algebra. Let $\mathfrak{g}$ be a Lie algebra, $V$ a vector space, and $\rho: \mathfrak{g} \to \mathfrak{gl}(V)$ a representation of the Lie algebra. Then the Killing form with respect to $(\rho , V)$ is $\mbox{Tr} \, (\rho(X) \rho(Y))$. The usual Killing form is that with respect to the adjoint representation $\mbox{ad} : \mathfrak{g} \to \mbox{End} ( \mathfrak{g} )$, while for a matrix Lie algebra, the trace form $\mbox{Tr} \, (XY)$ is the Killing form w.r.t. the defining representation. $\endgroup$ – Idempotent Jul 12 '16 at 19:20

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