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For the Hermitian and traceless generators $T^A$ of the fundamental representation of the $SU(N)$ algebra the anticommutator can be written as $$ \{T^A,T^{B}\} = \frac{1}{d}\delta^{AB}\cdot1\!\!1_{d} + d_{ABC}T^C $$ where $\delta^{AB} = 2\text{Tr}[T^AT^B]$ is the normalization chosen for the generators (note that they are also chosen orthogonal), $d=N$ for the fundamental representation, and $1\!\!1_{d}$ is the $d$-dimensional identity matrix.

For the fundamental representation it seems possible to deduce this expression by arguing that the anticommutator is Hermitian and hence can be written in terms of the $N^2-1$ traceless generators and one matrix with non-vanishing trace.

Does this expression hold for a general representation of the generators? If yes please explain why and/or provide a reference.

The relevance in the above equation appears in trying to express a general product: $$T^AT^B = \frac{1}{2}[T^{A},T^{B}]+\frac{1}{2}\{T^{A},T^{B}\}$$ where the commutator is already known as a consequence of the closure of $SU(N)$.

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Indeed, the anticommutator $$ S^{AB}\equiv \{T^A,T^B\} $$ is not in the Lie algebra, but, rather, in the universal enveloping algebra (formed by sums of products of generators); and, as you appreciated in your answer, only for the fundamental representation is it in the space involving the identity and the Lie algebra — by dint of the completeness of them in spanning $N\times N$ matrices.

In general, for other representations, you spill out of that space.
And, indeed, as you may trivially see for SU(2) spin-1 3×3 matrices, the anticommutators outrange the 4-d space of the identity with the 3 generators.

In your answer, you properly decomposed all anticommutators into projections on the identity, the space of the Lie algebra, and the remainder orthogonal space of the universal Lie Algebra M. In practice, however, M luckily manages to be projected out of all quantities of significance, like the trace of trilinears you represent by $d_{ABC}$, tweaked below.

Nevertheless, these objects do have remarkably simple properties, as you intuited in your answer, although it is not clear you appreciated the systematics of it. The point is these d-coefficients defined by the trace of the trilinear vary with representation (e.g., they vanish for real representations like the adjoint), but they are all proportional to the $d_{ABC}$ of the fundamental representation!

That is to say that the $d_{ABC}$ of the fundamental bequeaths its tensor structure to all other reps, as they are built out of the fundamental (see below). (In fact, it features in the definition of the cubic Casimir invariant of all SU(N)s for N >2. It, of course, vanishes for SU(2).) There are more properties you may find in D B Lichtenberg's 1970 Unitary Symmetry and Elementary Particles, Chapter 6.2.

For a given representation R of the generators $T^A_R$, normalize as is standard in HEP, $$ \mathrm{Tr} (T^A_R T^B_R)= T(R) \delta^{AB}, $$ where the index of the representation T(R), is e.g. for the fundamental and the adjoint of SU(N), $T(F)=1/2,~~ T(A)=N$.

The trace of the trilinear is $$ A(R) d_{ABC}=2\mathrm{Tr} (T^A_R S_R^{BC})=2\mathrm{Tr} (T^A_R \{T^B_R,T^C_R \} ), $$ where the anomaly coefficient A(R) is normalized such that, of course, A(F)=1, as the d-coefficients are defined in the fundamental, as in the statement of your question.

From the trilinearity of the argument of the trace, you may immediately see that $A(R)=-A(\bar{R})$, and so A=0 for any real representation like the adjoint (or, in the case of SU(2), the fundamental as well, since it is pseudoreal!) Moreover, you may see from the properties of the trace that $$ A(R_1\oplus R_2)= A(R_1)+A(R_2). $$ The nice part comes with the Kronecker product, the coproduct of two reps, $$ A(R_1\otimes R_2)= A(R_1)d(R_2)+ d(R_1)A(R_2) , \tag{*} $$ which ensures the nice property mentioned in the trace of the trilinear. d(R) is the dimension of the representation involved.

To be more explicit, the coproduct is (the ring homomorphism) $$ T^A_{R_1\otimes R_2}=T^A_{R_1}\otimes 1\!\!1_{R_2}+1\!\!1_{R_2}\otimes T^A_{R_2}, $$ which satisfies the Lie algebra, alright; even though the anticommutator, in sharp contrast to the commutator, has extra cross-pieces (it is not primitive, in mathematese): $$ S^{AB}_{R_1\otimes R_2}=S^{AB}_{R_1}\otimes 1\!\!1_{R_2} + 1\!\!1_{R_1} \otimes S^{AB}_{R_2} + 2(T^A_{R_1}\otimes T^B_{R_2}+ T^B_{R_1}\otimes T^A_{R_2} ). $$

However, note that, inserted in the trace, these pesky cross terms are projected out, simply by virtue of the fundamental property of the trace, that the trace of a tensor product is the product of the traces of the tensor factors. As a consequence, the cross terms, when multiplied by the generator coproduct will always produce terms that contain a tensor factor of just one power of generator somewhere, either $R_1$ or $R_2$, and so will be projected out by the tracelessness of a single power of the generator! This, then, ensures that the anomaly trace is always proportional to $d_{ABC}$, with the anomaly coefficients combining through the above relation (*). (Mathematicians call this projection a consequence of Friedrich's theorem, but no matter.)

All representations can be reached through tensoring of the fundamental, so their anomaly coefficients may be computed recursively, in principle. (And, of course, some vanish—for the real ones.)

Finally, for the decomposition you posit correctly in form, consistency with the above trace (tracing or multiplying by a T and tracing) dictates, instead, $$ S^{AB}_R= \frac{2T(R)}{d(R)}\delta^{AB} 1\!\!1_{d(R)} + \frac{A(R)}{2T(R)} d_{ABC}T^C_R +M^{AB}_R~. $$


If you wished to investigate the meshing of commutators with anticommutators and the devolution of the d-coefficients to higher representations, you might utilize, beyond the Jacobi identity, $$ [[A,B],C]+ [[B,C],A]+[[C,A],B] =0, $$ a plethora of its mixed analogs, $$ [\{A,B\},C]+ [\{B,C\},A]+[\{C,A\},B] =0,\\ [\{A,B\},C]+ \{[C,B],A\}+\{[C,A],B\} =0, $$ etc.

Many of the fundamental relations are obtained by consideration of $K^A\equiv d_{ABC}T^BT^C=d_{ABC}S^{BC}$, which, even though not primitive, nevertheless transforms simply, $[K^A,T^B]=if_{ABC}K^C$. It can be shown, as above, that this relation holds for all representations, where, however, the d in its definition is still that of the fundamental. Thus, by above, $\mathrm{Tr}(T_R^A K_R^B)=A(R)(N^2/4-1)~\delta^{AB}/N$, and so on.

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  • $\begingroup$ Maybe I'm missing something, but where do you define $d(R)$? $\endgroup$ – Noiralef Oct 10 '17 at 16:55
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    $\begingroup$ I defined it now. It is the dimension of the representation, the generalization of the d in the question. $\endgroup$ – Cosmas Zachos Oct 10 '17 at 17:02
  • $\begingroup$ Out of curiosity, do you have one or two favorite abstract algebra / representation theory applied to physics texts and/or notes? I have a few, but I'm always curious to know better where people learn this stuff and whether there's an amazing text or notes out there that I've missed. $\endgroup$ – joshphysics Oct 11 '17 at 19:20
  • $\begingroup$ No, sorry... I thrash from book to book, or even reviews. For elementary stuff, 't Hooft's notes, Wybourne, Cahn, Gilmore, Wu-Ki Tung, Iachello, Gourdin, Belinfante & Kolman, roughly in that order... $\endgroup$ – Cosmas Zachos Oct 11 '17 at 19:34
  • $\begingroup$ Also, Hall's notes, Willard Miller's two books, Vilenkin, and tasteful appendices of Okun's book, Ramond, ... the remarkable thing is that nobody is complete/perfect... and you have to spend an eternity translating and reconciling.... $\endgroup$ – Cosmas Zachos Oct 11 '17 at 19:40
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I am not sure what you are asking. For every antisymmetric d-dimensional matrix $T$ you can extract the trace part and hence have $$T=\frac{I}{d}\cdot tr{T}+(T-tr{T}).$$ You can check out that the first term is really the trace part and the second term is traceless. So in your equation, it is simply a definition of $d_{ABC}$. Note that $tr{{T^{AB}}}=2C\delta^{AB}$ in your equation. More generally, for any d-dimensional matrix T, you can have $$T=\frac{1}{2}({T+T^{T}})+\frac{I}{d}\cdot tr\left({\frac{1}{2}(T-T^T)}\right)+\left(\frac{1}{2}(T-T^T)-\frac{I}{d}\cdot tr\left({\frac{1}{2}(T-T^T)}\right)\right)$$ with the symmetric, the trace and the traceless antisymmetric parts correspondingly.

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  • $\begingroup$ The question is not whether it is possible to split it up in a traceless and a part with a non-vanishing trace but rather if the part with non-vanishing trace necessarily can be written as $d_{ABC}T^C$ without any additional term $M^{AB}$. From the above then $d_{ABC}=\frac{1}{C}Tr[\{T^{A}T^{B}\},T^{C}]$ so the $M^{AB}$ would have to be Hermitian, symmetric in $A$ & $B$, traceless and furthermore have $Tr[T^{C}M^{AB}] = 0$. I just do not know whether such an object exists or how to show that it does not. $\endgroup$ – AltLHC Dec 2 '15 at 20:08
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For a general representation $t^{A}$ of the generators of $SU(N)$ it is possible to deduce the following form of the anticommutator $$\{t^{A},t^{B}\} = \frac{2N}{d}\delta^{AB}\cdot 1_{d} + d_{ABC}t^{C} + M^{AB}$$ where $$ \mathrm{Tr}[t^{A}t^{B}] = N\delta_{AB}\\ d_{ABC} = \frac{1}{N}\mathrm{Tr}[\{t^{A},t^{B}\}t^{C}] $$ and the object $M^{AB}$ satisfies a number of properties $$ \mathrm{Tr}[M^{AB}] = 0,\quad M^{AA}=0,\quad \mathrm{Tr}[M^{AB}t^{C}] = 0,\quad M^{AB} = M^{BA}, \quad (M^{AB})^{\dagger} = M^{AB} $$ The second last property expresses the orthogonality of $M^{AB}$ to the generators $t^{A}$ showing that it is not contained in the algebra. In the case of the fundamental representation $M^{AB}=0$ as the degrees of freedom have been exhausted (or alternatively; the generators and the identity span the full space of Hermitian matrices).

In the case of the adjoint representations of $SU(2)$ and $SU(3)$ I performed an explicit calculation of $M^{AB}$, verifying the properties above.

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    $\begingroup$ How did you deduce the form of the anti-commutator? Is the $d_{ABC}$ representation-dependent or is it taken to be the value that it has for generators in the fundamental representation? (maybe the second question is easily answered from the first) $\endgroup$ – Jonathan Rayner Nov 5 '16 at 16:29
  • $\begingroup$ Ummm... $N^{AA}\neq 0$, as you may check from the adjoint of SU(2) ... cf. No summation over AA. $\endgroup$ – Cosmas Zachos Sep 4 '19 at 21:44

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