I am dealing with the tensor product representation of $SU(3)$ and I have some problems in understanding some decomposition.

1) Let's find the irreducible representation of $3\otimes\bar{3}$

we have that this representation trasforms like

$${T^\prime}^i_j=U^i_k {U^{\dagger}}^l_j T^k_l $$

hence I observe that $$Tr(T)=\delta^j_iT^i_j$$ is an invariant and so


allows me to write $$3\otimes\bar{3}=8\oplus1$$ Here comes my questions: I have heard that this $8$ representation is an "$8_{MA}$" where MA is for "mixed-antisymmetric". The meaning of "mixed-antisymmetric" shold be: "the tensor $\left(T^i_j-\frac{1}{3}\delta^j_iT^i_j\right)$ should be antisymmetric for an exchange of 2 particular indexes but not for a general exchange of 3 indexes". What does this mean? I see only 2 index in that tensor.

2) Consider this representation: $$3\otimes3\otimes3=3\otimes(6\oplus\bar{3})=3\otimes6_S\oplus3\otimes\bar{3}=3\otimes 6_S\oplus8_{MA}\oplus1$$

and now on my notes I have $$3\otimes6_S=10_S\oplus8_{MS}$$

Where "MS" is for "mixed symmetric": symmetric for an exchange of 2 particular indexes but not for a general exchange of 3 indexes.

I could not demonstrate this last decomposition using tensor method. I started noticeing that: $$3\otimes6_S=q^iS^{k,l}$$ where $S^{k,l}$ is a symmetric tensor But then I am not able to proceed in demonstrating the above decomposition (note: I would like to demonstrate this decomposition using only tensor properties, not Young tableaux). I tried to look on Georgi, Hamermesh, Zee and somewhere online but I have not found any good reference which explains well this representatin decomposition...

EDIT: the demonstration should not include the use of Young diagrams...my professor started the demonstration by writing $\epsilon_{\rho,i,k}q^i S^{k,l}=T'^l_\rho=8_{MS}$ and then stopped the demonstration.


2 Answers 2


Since this question looks like homework we will be somewhat brief. OP's notes are apparently describing the symmetry of the corresponding Young diagram for each $SU(3)$ irrep. Each box corresponds to an index. Roughly speaking, indices in same row (column) are symmetric (antisymmetric), respectively.


  1. A single box $[~~]$ corresponds to the fundamental irrep ${\bf 3}$.

  2. Two boxes on top of each other $\begin{array}{c} [~~]\cr [~~] \end{array}$ is the anti-fundamental irrep $\bar{\bf 3}$ if we dualize with the help of the Levi-Civita symbol $\epsilon^{ijk}$. Here we adapt the sign convention $\epsilon^{123}=1=\epsilon_{123}$.

  3. The tensor product ${\bf 3}\otimes{\bf 3}\cong\bar{\bf 3}\oplus{\bf 6}_S$ corresponds to $$ [~~]\quad\otimes\quad[a]\quad\cong\quad\begin{array}{c} [~~]\cr [a] \end{array}\quad\oplus\quad\begin{array}{rl} [~~]&[a] \end{array}$$
    or $T^{ij}=\epsilon^{ijk}A_k+S^{ij}$, where $A_k:=\frac{1}{2}T^{ij}\epsilon_{ijk}$.

  4. The tensor product $\bar{\bf 3}\otimes{\bf 3}\cong{\bf 1}\oplus{\bf 8}_M$ corresponds to $$\begin{array}{c} [~~]\cr [~~] \end{array}\quad\otimes\quad[a]\quad\cong\quad\begin{array}{c} [~~]\cr [~~]\cr [a] \end{array}\quad\oplus\quad\begin{array}{rl} [~~]&[a]\cr [~~] \end{array}$$ or $T^i{}_j=S\delta^i_j+M^i{}_j$, where $S:=\frac{1}{3}T^i{}_i$, and ${\rm Tr}M=0$.

  5. The tensor product ${\bf 6}_S\otimes{\bf 3}\cong{\bf 8}_M\oplus{\bf 10}_S$ corresponds to $$\begin{array}{rl} [~~]& [~~] \end{array}\quad\otimes\quad[a]\quad\cong\quad\begin{array}{rl} [~~]&[~~]\cr [a] \end{array}\quad\oplus\quad\begin{array}{rcl} [~~]& [~~] & [a] \end{array}$$ or $T^{ij,k}=\left\{M^{i}{}_{\ell}\epsilon^{\ell jk}+(i\leftrightarrow j)\right\} +S^{ijk}$, where $M^i{}_{\ell}:=\frac{1}{3}T^{ij,k}\epsilon_{jk\ell}$, and ${\rm Tr}M=0$.


  1. H. Georgi, Lie Algebras in Particle Physics, 1999, Section 13.2.

  2. J.J. Sakurai, Modern Quantum Mechanics, 1994, Section 6.5.


Since the accepted answer treats the problem through the point of view of Young diagrams and I was having some problems with tensor methods myself, I believe it is worth to add the tensor approach in here.

Mixed Antisymmetric

Concerning the meaning of $MA$, recall that $SU(3)$ representations can be characterized in terms of "upstairs symmetrized indices" and "downstairs symmetrized indices" (see, for example, the beginning of Chapter V.2 of Zee's Group Theory in a Nutshell for Physicists). In other worlds, while $(𝑇^𝑖_π‘—βˆ’\frac{1}{3}𝛿^i_j 𝑇^k_k)$ does have only two indices, we could also raise the lower index with a Levi-Civita symbol, hence obtaining $(𝑇^𝑖_𝑗\epsilon^{jkl} βˆ’\frac{1}{3}\epsilon^{ikl}𝑇^k_k)$, which is now explicitly antisymmetric in two indices. The lesson here is that in $SU(3)$, thanks to the peculiar existence of a Levi-Civita symbol with three indices, we can write every tensor in terms of tensors with symmetrized indices, being these "upstairs" or "downstairs".

Computing $3 \otimes 6_S$

Let us then proceed to use tensor methods to write the tensor product of the representations. We have here a vector $q^i$ transforming in the fundamental representation, $3$, and a symmetric tensor $S^{jk}$ transforming in the $6_S$ representation. As you noticed, an object transforming in the $3 \otimes 6_S$ representation transforms as $q^i S^{jk}$. Since the representations of $SU(3)$ are characterized by symmetric tensors with indices upstairs or downstairs, our goal will be to separate the antissymetric parts in each index combination.

We can't antisymmetrize the indices of $S_{jk}$, since it is a symmetric tensor. We are left to antisymmetrize the index of $q^i$ with each of the indices of $S_{jk}$. Hence, we employ the Levi-Civita symbol to get

$$q^i S^{jk} = \frac{1}{3}\epsilon^{ijl}\epsilon_{lmn}q^mS^{nk} + \frac{1}{3}\epsilon^{ikl}\epsilon_{lmn}q^mS^{jl} + \left(q^i S^{jk} - \frac{1}{3}\epsilon^{ijl}\epsilon_{lmn}q^mS^{nk} - \frac{1}{3}\epsilon^{ikl}\epsilon_{lmn}q^mS^{jl}\right)$$

The coefficients are chosen so that the last term is fully symmetric in the three indices. Another possibility would be to pick the ansatz $S^{ijk} = q^i S^{jk} - a\epsilon^{ijl}\epsilon_{lmn}q^mS^{nk} - b\epsilon^{ikl}\epsilon_{lmn}q^mS^{jl}$ and impose the condition that $S^{ijk}$ is symmetric in all indices to get the correct coefficients.

Either, now the first term is symmetric under $j \leftrightarrow k$ and has no definite symmetry when $i$ is considered. The last term corresponds to a fully symmetric tensor, which has $10$ independent components, meaning it is a $10_S$. Hence, the first term has $8$ independent components and corresponds, as a consequence, to an $8_{MS}$, the $MS$ coming from mixed symmetry.


  1. A. Zee, Group Theory in a Nutshell for Physicists (Princeton University Press: Princeton, 2016). Chapter V.2.

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