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The trace is simply a (properly normalised) ad-invariant inner product on the Lie algebra; that is, a nondegenerate symmetric bilinear form $\langle-,-\rangle$ which obeys the "associativity" condition $$\langle [x,y],z \rangle = \langle x, [y,z] \rangle$$ for every $x,y,z$ in $\mathfrak{g}$.

Lie algebras admitting such inner products are said to be metric. The normalisation of the inner product is such that $k$ is an integer. This only makes sense for indecomposable metric Lie algebras; that is, those which are not isomorphic to the direct product of perpendicular proper ideals.

The notation "tr" stems from the fact that if $\rho: \mathfrak{g} \to \operatorname{End}(V)$ is a faithful finite-dimensional representation, then $$\langle x, y\rangle := c \operatorname{tr}\rho(x)\rho(y)$$ works for a suitable nonzero $c$. (For a simple Lie algebra, just take $\rho$ to be the adjoint representation.)

For the explicit case of $\mathfrak{g}$ the Lie algebra of SU(2) you can take $\rho$ to be the fundamental representation and $c= -\frac12$, I believe.

Edit

Notice that $\operatorname{tr}(A \wedge dA)$ is really $\langle A \stackrel{\wedge}{,} dA \rangle$, where $\langle -\stackrel{\wedge}{,}-\rangle$ means that we are both taking the wedge product of the forms and the inner product on the Lie algebra. Similarly, $$\operatorname{tr}(A \wedge A \wedge A) = \frac12 \langle [A\stackrel{\wedge}{,}A] \stackrel{\wedge}{,} A \rangle,$$ with a similar notational caveat about $[A\stackrel{\wedge}{,}A]$.

In response to Anirbit's comment, I would say that there is, in general, no trace on vector-valued differential forms; although if the forms take values in endomorphisms, then of course there is: simply compose with the trace of endomorphisms to obtain a map $$\Omega^\bullet(M;\operatorname{End}(V)) \to \Omega^\bullet(M).$$

2 added 84 characters in body

The trace is simply a (properly normalised) ad-invariant inner product on the Lie algebra; that is, a nondegenerate symmetric bilinear form $\langle-,-\rangle$ which obeys the "associativity" condition $$\langle [x,y],z \rangle = \langle x, [y,z] \rangle$$ for every $x,y,z$ in $\mathfrak{g}$.

Lie algebras admitting such inner products are said to be metric. The normalisation of the inner product is such that $k$ is an integer. This only makes sense for indecomposable metric Lie algebras; that is, those which are not isomorphic to the direct product of perpendicular proper ideals.

The notation "tr" stems from the fact that if $\rho: \mathfrak{g} \to \operatorname{End}(V)$ is a faithful finite-dimensional representation, then $$\langle x, y\rangle := c \operatorname{tr}\rho(x)\rho(y)$$ works for a suitable nonzero $c$. (For a simple Lie algebra, just take $\rho$ to be the adjoint representation.)

For the explicit case of $\mathfrak{g}$ the Lie algebra of SU(2) you can take $\rho$ to be the fundamental representation and $c= -\frac12$, I believe.

Edit

Notice that $\operatorname{tr}(A \wedge dA)$ is really $\langle A \stackrel{\wedge}{,} dA \rangle$, where $\langle -\stackrel{\wedge}{,}-\rangle$ means that we are both taking the wedge product of the forms and the inner product on the Lie algebra. Similarly, $$\operatorname{tr}(A \wedge A \wedge A) = \frac12 \langle [A,A] A\stackrel{\wedge}{,}A] \stackrel{\wedge}{,} A \rangle.$$rangle,$$with a similar notational caveat about [A\stackrel{\wedge}{,}A]. 1 The trace is simply a (properly normalised) ad-invariant inner product on the Lie algebra; that is, a nondegenerate symmetric bilinear form \langle-,-\rangle which obeys the "associativity" condition$$\langle [x,y],z \rangle = \langle x, [y,z] \rangle$$for every x,y,z in \mathfrak{g}. Lie algebras admitting such inner products are said to be metric. The normalisation of the inner product is such that k is an integer. This only makes sense for indecomposable metric Lie algebras; that is, those which are not isomorphic to the direct product of perpendicular proper ideals. The notation "tr" stems from the fact that if \rho: \mathfrak{g} \to \operatorname{End}(V) is a faithful finite-dimensional representation, then$$\langle x, y\rangle := c \operatorname{tr}\rho(x)\rho(y)$$works for a suitable nonzero c. (For a simple Lie algebra, just take \rho to be the adjoint representation.) For the explicit case of \mathfrak{g} the Lie algebra of SU(2) you can take \rho to be the fundamental representation and c= -\frac12, I believe. Edit Notice that \operatorname{tr}(A \wedge dA) is really \langle A \stackrel{\wedge}{,} dA \rangle, where \langle -\stackrel{\wedge}{,}-\rangle means that we are both taking the wedge product of the forms and the inner product on the Lie algebra. Similarly,$$\operatorname{tr}(A \wedge A \wedge A) = \frac12 \langle [A,A] \stackrel{\wedge}{,} A \rangle.