Trace and adjoint representation of $SU(N)$

Question

In the adjoint representation of $SU(N)$, the generators $t^a_G$ are chosen as

$$ (t^a_G)_{bc}=-if^{abc} $$

The following identity can be found in Taizo Muta's book "Foundations of Quantum Chromodynamics", appendix B Eq. (B.10), page 381:

$$ \mathrm{tr}(t^a_Gt^b_Gt^c_Gt^d_G)=\delta^{ab}\delta^{cd}+\delta^{ad}\delta^{bc}+\frac{N}{4}(d^{abe}d^{cde}-d^{ace}d^{bde}+d^{ade}d^{bce}) $$

where $d^{abc}$ is totally symmetric in $a$, $b$ and $c$ and is defined in the fundamental representation by

$$ \{t^a_N,t^b_N\}=\frac{1}{N}\delta^{ab}+d^{abc}t^c_N,\quad [t^a_N,t^b_N]=if^{abc}t^c_N,\quad \mathrm{tr}(t^a_Nt^b_N)=\frac{1}{2}\delta^{ab} $$

It puzzles me that how does $d^{abc}$ appear in $\mathrm{tr}(t^a_Gt^b_Gt^c_Gt^d_G)$. Could anyone provide a proof for the above identity? Many thanks in advance!

Answer 1

It is known that for an element $U$ of the group, in matrix sence: $$ Ad_Ux=UxU^{-1}.\,\,(1) $$ Now, we note that the target space of the adjoint rep is spanned by $N^2-1$ traceless matrices $t_a$. So, if we add the unity matrix, we get a full basis in $\mathrm{Mat}_N(\mathbb{C})$. We now note that that the adjoint action is trivially extended to this space, so I can write: $$ Tr_{su(N)}(Ad_U)=Tr_{\mathrm{Mat}_N(\mathbb{C})}(Ad_U)-1, $$ where $su(N)$ is the space of traceless hermitian matirices. It is true since the identity matrix is taken to identity for this operator. Now, using (1), we see that $$ Tr_{\mathrm{Mat}_N(\mathbb{C})}(Ad_U)=Tr(U)Tr(U^{-1}), $$ finally arriving at $$ Tr (Ad_U)=Tr(U)Tr(U^{-1})-1. $$ For example, take $U=I$, and then trace on the left is $N^2-1$ the dimension of the adjoint rep, while traces on the right are $N$ the dimension of the fundamental rep. I guess this formula is well-known.

It is of great use here, because it relates the traces of your kind to the traces in fundamental representation, which are easily computed by the argument Joshua provides.

Take $U=\prod_i \exp(t^a\alpha_a^i)$, for $\alpha_a^i$ an arbitrary set of numbers ($i=1..3$, for example). Then $Ad_U=\prod_i \exp(ad(t^a)\alpha_a^i)$. Now I expand our formula in powers of $\alpha$, and I want to examine the $\alpha_{a}^1\alpha_{b}^2\alpha_{c}^3$ term ($\alpha^i=t^a\alpha^i_a)$: $$ Tr(ad(\alpha^1)ad(\alpha^2)ad(\alpha^3))=Tr(\alpha^1\alpha^2\alpha^3)Tr(I)-Tr(I)Tr(\alpha^3\alpha^2\alpha^1). $$ Now, $Tr(t^at^bt^c)$ is trivially (by the formula for the product $t_at_b$) equal to $\frac{1}{4}\left(d^{abc}+if^{abc}\right)$, while $Tr (I)=N$. So finally we get (in your notation): $$ Tr(t^a_G t^b_G t^c_G)=\frac{N}{2}if^{abc}. $$ You can compare it with your book. Yes, I know that it is not what you wanted, but in this way (and following Joshua answer for the traces in fundamental rep) you can get any desired trace. Since you have the homework tag, I leave the case of four generators as an exercise.

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Update:

The fact that $$Tr_{f}(U)Tr_{\bar{f}}(U)=Tr_{ad}(U)+1$$, where $Tr_R$ is the trace in representation $R$, $ad$ is the adjoin rep, $f$ is the fundamental rep and $\bar{f}$ is the fundamental rep where all matrices are taken to transposes of their inverses (dual rep, in this case the same as the conjugate rep), is a direct consequence of the fact: $$ f\otimes\bar{f}\simeq ad\oplus 1 $$, where $1$ is the trivial one-dimensional representation, $Tr_1(U)=1$. This is because trace is additive under $\oplus$ and multiplicative under $\otimes$. For example, for $SU(3)$ this reads $\mathbf{3}\otimes\mathbf{\bar{3}}=\mathbf{8}\oplus\mathbf{1}$. You can get other idenities from, say $\mathbf{3}\otimes\mathbf{3}=\mathbf{6}\oplus\mathbf{\bar{3}}$, for other representations. This is somehow related to the theory of characters.

Answer 2

With the help of Peter Kravchuk and joshphysics, I have completed a proof of the trace identity. I will post it here as a reference. By the method of Kravchuk, we find

$$\begin{split} \mathrm{tr}\big(t^a_Gt^b_Gt^c_Gt^d_G\big)&=2\big[ \mathrm{tr}\big(t^a_Nt^b_N\big)\mathrm{tr}\big(t^d_Nt^c_N\big)+ \mathrm{tr}\big(t^a_Nt^c_N\big)\mathrm{tr}\big(t^d_Nt^b_N\big)+ \mathrm{tr}\big(t^a_Nt^d_N\big)\mathrm{tr}\big(t^c_Nt^b_N\big)\big] \\ &\qquad +N\big[\mathrm{tr}\big(t^a_Nt^b_Nt^c_Nt^d_N\big)+ \mathrm{tr}\big(t^d_Nt^c_Nt^b_Nt^a_N\big)\big] \end{split} $$

where the trace $\mathrm{tr}\big(t^a_Nt^b_Nt^c_Nt^d_N\big)$ can be calculated as

$$\mathrm{tr}\big(t^a_Nt^b_Nt^c_Nt^d_N\big)=\frac{1}{4N}\delta^{ab} \delta^{cd}+\frac{1}{8}(d^{abe}+if^{abe})(d^{cde}+if^{cde}) $$

Using the antisymmetry of $f^{abc}$ and symmetry of $d^{abc}$, we obtain that

$$\mathrm{tr}\big(t^a_Nt^b_Nt^c_Nt^d_N\big)+\mathrm{tr} \big(t^d_Nt^c_Nt^b_Nt^a_N\big)=\frac{1}{2N}\delta^{ab}\delta^{cd}+ \frac{1}{4}(d^{abe}d^{cde}-f^{abe}f^{cde}) $$

With the identity $\big[t^a_N,\big[t^b_N,t^c_N\big]\big]+\big\{t^b_N,\big\{t^c_N,t^a_N\big\}\big\}-\big\{t^c_N,\big\{t^a_N,t^b_N\big\}\big\}=0$, we can rewrite $f^{abc}f^{cde}$ as

$$ f^{abe}f^{cde}=\frac{2}{N}(\delta^{ac}\delta^{bd}- \delta^{ad}\delta^{bc})+d^{ace}d^{bde}-d^{ade}d^{bce} $$

Now, it is strightforward to finish the proof.

Answer 3

Here is my recommendation on how to proceed. Notice that you are given the trace of the product of any two generators. It would therefore be useful to convert the product of four generators inside of the trace you're trying to compute into a sum of products of two generators. This can be done by noting the following commutator-anticommutator identity: $$ t^at^b = \frac{1}{2}\Big([t^a,t^b]+\{t^a,t^b\}\Big) $$ If you use this on each of the pairs $t^at^b$ and $t^ct^d$ in the trace you want to compute, then it will reduce to a sum of traces of pairs of generators and multiples of the identity which you can then easily evaluate.

Edit. As user Peter Kravchuk points out, this method depends on being able to either compute anticommutators in the adjoint rep, or relate fundamental rep traces to adjoint rep traces.

Answer 4

OP has already answered his own question with help from other answers, especially Peter Kravchuk's answer. Here we make some comments on how the fusion rule mentioned by Peter Kravchuk should be concretely realized.

The first point is that the adjoint representation $Ad_{SU(N)}$ of $SU(N)$ is the real vector space of Hermitian traceless $N\times N$ matrices, while the fundamental representation $F_{SU(N)}$ (and its complex conjugate representation $\bar{F}_{SU(N)}$) of $SU(N)$ are complex representations.

Thus the fusion rule can only be realized in a complexified setting. The complexification of $SU(N)$ is $SL(N,\mathbb{C})$.

Obviously all of the Lie algebra relations and trace identities mentioned by OP remain unchanged if we think of the generators $t_a$'s as a complex basis for $sl(N,\mathbb{C})$ rather than a real basis for $su(N)$.

Now the fusion rule follows in two steps

$$\tag{1} F_{SL(N,\mathbb{C})}\otimes_{\mathbb{C}} \bar{F}_{SL(N,\mathbb{C})} ~\cong~ Ad_{GL(N,\mathbb{C})}~\cong~1\oplus Ad_{SL(N,\mathbb{C})}.$$

Here the Lie group representation $Ad_{GL(N,\mathbb{C})}$ is the vector space of all complex $N\times N$ matrices, while the Lie group representation $Ad_{SL(N,\mathbb{C})}$ is the vector space of complex traceless $N\times N$ matrices. The one-dimensional trivial representation $1$ is spanned by the $N\times N$ identity matrix ${\bf 1_{N\times N}}$.

The representation $\bar{F}_{SL(N,\mathbb{C})}$ is the dual/contragredient representation, which should not be confused with the complex conjugate representation. However, for unitary Lie groups, like $SU(N)$, the dual/contragredient representation and the complex conjugate representation are the same.

The fusion rule (1) is concretely proven by choosing bases for the various vector spaces involved, and check that the bases transform covariantly under the $SL(N,\mathbb{C})$ group action.