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? $$tr([Q^a,P^b]Q^c P^d)$$


  • $\begingroup$ From the cyclic property of trace we have $$tr[A,B]=0$$ for any matrices. Also $$tr([A,B]C)=0$$ just for symmetric matrices. Maybe these relation help! $\endgroup$ – Vahid Jul 24 '13 at 17:20
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    $\begingroup$ I dont believe this, which rascal has close flagged this? It is a perfectly legitimate technical theoretical questin. Leave open $\endgroup$ – Dilaton Jul 24 '13 at 20:38
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    $\begingroup$ @Dilaton I flagged it for migration to math.SE . It is a (legitimate) math question. $\endgroup$ – Mo_ Jul 24 '13 at 21:46
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    $\begingroup$ the question is definitely of interest to a lot of people here and lie algebras are used extensively in physics. this question is definitely not out of place in this forum. $\endgroup$ – Prathyush Jul 25 '13 at 20:39

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.

  • $\begingroup$ Thanks very much for your answer. I would like to ask more about your proof. How did you say that $$ \left[ Q^{a},R^{b}\right] =if^{abc}R^{c},\qquad\left[ Q^{a},I^{b}\right] =if^{abc}I^{c}. $$ imply that R and I are traceless? Generally it's strange! Really there is no a non-trivial matrix P that satisfy my relations? $\endgroup$ – Vahid Aug 4 '13 at 2:45
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    $\begingroup$ Probably you know that $\left( C^{a}\right) _{bc}=-if^{abc}$ are the generators of adjoined representation. Since for finite matrices the trace of commutator vanishes, thus $\mathrm{tr}\left[ Q^{a},R^{b}\right] =0=if^{abc}\mathrm{tr}R^{c}=-\left( C^{c}\right)_{ab}\mathrm{tr}R^{c}$ implies a linear relation for $\hat{C}^{a}$ which is only possible for all $\mathrm{tr}R^{c}=0$. You can also use the identity $f^{acd}f^{acd^{\prime}}=C_{A}\delta^{dd^{\prime}}$, hence $R^{c}=-iC_{A} ^{-1}f^{abc}\left[ Q^{a},R^{b}\right] $. $\endgroup$ – Grisha Kirilin Aug 5 '13 at 9:44
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    $\begingroup$ It is difficult to comment Jackiw's paper without reading it carefully. You question was about the trace thus my proof is valid only for finite matrices. As I noticed it is a consequence of that fact that $SU(N)$, algebraically, is a simple Lie group, i.e., its Lie algebra is simple. Thus if you require that the set of matrices $R^{a}$ has as many components as $Q^{a}$, then it immediately follows that $R^{a}\equiv Q^{a}$. $\endgroup$ – Grisha Kirilin Aug 7 '13 at 17:07
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    $\begingroup$ But you should keep in mind that it is only about finite matrices. For example, the components of the angular momentum $J^{i}$ are generators for $SU(2)$, and the commutator relation for any "vector" operator $V^{j}$ is $[J^{i},V^{j}]=i\epsilon^{ijk}V^{k}$. Hence if you put $V^{j}=p^{j}$, i.e., the operator of momentum, so that $[p^{i},p^{j}]=\delta^{i,j}$, then you find $[J^{i},p^{j}]=i\epsilon^{ijk}p^{k}$, i.e. the algebra you asked for. But you should remember that translations are not compact (group) thus you cannot define the notion of "trace" for them. $\endgroup$ – Grisha Kirilin Aug 7 '13 at 17:15
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    $\begingroup$ Generally, you can include $SU(N)$ as a subgroup in some noncompact group $G$, and then you can find an Abelian noncompact subgroup of $G$, so that $SU(N)$ would be a normalizer for this subgroup en.wikipedia.org/wiki/Centralizer_and_normalizer Probably Jackiw kept something like this in his mind. $\endgroup$ – Grisha Kirilin Aug 7 '13 at 17:20

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.


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