What is the trace in the Chern-Simons action? Warning: This is a very stupid question regarding a basic misunderstanding that I have. I realize that the question is very elementary, but I guess asking stupid questions is better than remaining ignorant.
To be explicit, consider the $\mathrm{SU}(2)$ Chern-Simons action on some very nice $3$-manifold $M$, i.e. the number
$S(A) = \frac{k}{4\pi}\int_M \mathrm{tr} \left(A\wedge\mathrm{d}A + \frac{2}{3}A\wedge A\wedge A\right)$,
where $A$ is an $\mathfrak{su}(2)$-valued $1$-form on $M$. 
What I simply cannot wrap my head around, and what is obviously a very silly basic question, is: What trace is this?! As I understand it, there is a certain abuse of notation in $\wedge$ on vector bundle-valued forms (namely, with $E$ the bundle, the wedge of two $E$-valued forms is an $E\otimes E$-valued one), but in the case of the Chern-Simons action this answer suggests that the $E\otimes E=\mathfrak{g}\otimes\mathfrak{g}\rightarrow\mathfrak{g}$ is supposed to be the Lie bracket on $\mathfrak{g}$. Anyway, that leads me to think that the trace is the trace on $\mathfrak{g}=\mathfrak{su}(2)$, which of course vanishes everywhere.
What am I misunderstanding here?
 A: I will give the physicist's answer.
I hope you are familiar with this notation.
$A=A_\mu dx^\mu$ and in this notation the action looks like
$S=\frac{k}{4\pi} \int_{\mathcal{M}} d^3x\ \epsilon^{\mu\nu\alpha}\ Tr \left( A_\mu \partial_\nu A_\alpha +\frac{2}{3}  A_\mu A_\nu A_\alpha   \right)$
where $\epsilon^{\mu\nu\alpha}$ is the Levi-Civita tensor in three dimensions. 
$A_\mu=A_\mu^at^a$ where $A_\mu^a$ are real functions, $t^a$ are the matrix representatives of generators of Lie algebra $[t^a,t^b]=if^{abc}t^c$. Here $a,b,c=1,2,...,dim G$ where G is the gauge group and $f^{abc}$ are structure constants.
So, that trace in the action is the trace of products of these $t^a$ matrices.
For example,
for $G=SU(N)$ and in the fundamental representation $Tr(t^at^b)=\frac{1}{2}\delta^{ab}$
$(a,b=1,2,...N^2-1)$
A: 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 $[A\stackrel{\wedge}{,}A]$.

Further addition
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
$$\Omega^\bullet(M;\operatorname{End}(V)) \to \Omega^\bullet(M).$$
