Calculating an expression for the trace of generators of two Lie algebra Suppose we have 
$$[Q^a,Q^b]=if^c_{ab}Q^c$$
where Q's are generators of a Lie algebra associated  a SU(N) group. So Q's are traceless. Also we have
$$[P^a,P^b]=0$$
where P's are generators of a Lie algebra associated to an Abelian group. We have the following relation between these generators
$$[Q^a,P^b]=if^c_{ab}P^c$$
I would like to know what we can say about the following trace. Is it equal to zero?
$$tr([Q^a,P^b]Q^c P^d)$$
Cheers!
 A: If $P^{a}$ are finite-dimensional matrices, then I found that your algebra
actually implies that $P^{a}=0$. I think that it is a consequence of the fact
that $SU\left(  N\right)  $ is a simple group, i.e., there is no a normal
subgroup in $SU(N)$. If we assume that $P^{a}$ are hermitian then your
identities:
$$
\left[  P^{a},P^{b}\right]  =0,\qquad\left[  Q^{a},P^{b}\right]
=if^{abc}P^{c},\qquad\qquad(1)
$$
are the algebra of an invariant subgroup.
The formal proof that $P^{a}=0$ is as follows. If $P$ satisfies the algebra (1) then a linear independent subset of $P$ also satisfies Eq.(1), therefore without loss of
generality we can assume that all $P^{a}$ are linear independent. Let me now
split $P$ into the hermitian and anti-hermitian parts:
$$
P=\frac{P+P^{\dagger}}{2}+i\frac{P-P^{\dagger}}{2i}=R+iI,
$$
where $R$ and $I$ are hermitian matrices. Taking into account that all
$f^{abc}$ are real and
$$
\left[  Q^{a},P^{b}\right]  =if^{abc}P^{c}\quad\Longrightarrow\qquad\left[
Q^{a},P^{\dagger b}\right]  =if^{abc}P^{\dagger c},
$$
one can conclude that:
$$
\left[  Q^{a},R^{b}\right]  =if^{abc}R^{c},\qquad\left[  Q^{a},I^{b}\right]
=if^{abc}I^{c}.
$$
Therefore $R^{a}$ and $I^{a}$ are hermitian traceless matrices thus they can be
expressed as linear combinations of $Q$. Thus, we find that matrices
$P^{a}$ are  linear combinations of $Q$:
$$
\qquad\qquad\qquad\qquad\qquad P^{a}=M^{ab}Q^{b},\qquad\qquad\qquad\qquad\qquad (2)
$$
where $M^{ab}$ is some $\left(  N^{2}-1\right)  \times\left(  N^{2}-1\right)
$ compex-valued matrix. Again from Eq.(1) we obtain:
$$
M^{ad}\left[  Q^{d},P^{b}\right]  =\left[  P^{a},P^{b}\right]  =iM^{ad}%
f^{dbc}P^{c}=0.
$$
Since all $P^{e}$ are linear independent then $M^{ad}f^{dbc}=0$, thus
$0=M^{ad^{\prime}}f^{d^{\prime}bc}f^{dbc}=C_{A}M^{ad}=0$, hence $P^{a}=0.$
There is another way to show the same. Let's consider the commutator $\left[  Q^{a},P^{b}\right]$ and use the relation (2):
$$
\left[  Q^{a},P^{b}\right]  =M^{bc}\left[  Q^{a},Q^{c}\right]  =M^{bc}
if^{ace}Q^{e}=if^{abc}P^{c}=if^{abc}M^{ce}Q^{e}
$$
Comparing the coefficient of $Q^{e}$, we find:
$$
M^{bc}f^{ace}=f^{abc}M^{ce}\quad\Longrightarrow\qquad M^{bc}\left(
C^{a}\right)  _{ce}=\left(  C^{a}\right)  _{bc}M^{ce}\quad\Longrightarrow
\qquad\left[  M,C^{a}\right]  =0,
$$
where $\left(  C^{a}\right)  _{ce}=-if^{ace}$ are the generators of
irreducible adjoined representation. Therefore, according to Schur's lemma $M$
is a scalar matrix, i.e., $M^{ab}=\lambda\delta^{ab}$. If one requires that
$P$ should be commutative then $\lambda=0.$
The question is what about the case where $P^{a}$ are not finite-dimensional
matrices. But in this case I don't know how to define the trace.
A: I think the answer is yes if the representation of your abelian group is irreducible, because the generators of an abelian group are numbers if I remember correctly. 
Actually, a system of "operator" forming a group, and which every element commute with all the other element of the group are numbers or proportional to the identity matrix (this is know as the Schuch's lemma if -- once again -- I remember correctly). But if they are proportional to the identity matrix, it means I can easily reduce the representation. At the end I should end up with $U(1)$, which is the only abelian Lie group I know, and this one has a number as generator (an angle say).
So in short, your expression becomes 
$$\text{Tr}\left\{ \left[Q^{a},P^{b}\right]Q^{c}P^{d}\right\} =\mathbf{i}f_{ab}^{e}P^{e}\text{Tr}\left\{ Q^{c}\right\} P^{d}=0$$
if the generator of your abelian group are numbers. I used the commutator relation between $P$ and $Q$, I get out of the trace all the numbers (or what I believe are numbers), and I finally used the fact that the $Q$ are traceless.
I prefer to keep the if style since I may well make a mistake, but you may easily check by examples. If you find a counter example, I would be glad to learn about that. 
