homework and exercises - $SU(3)$ irreducible representations with tensor method

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

$$T^i_j=\left(T^i_j-\frac{1}{3}\delta^j_iT^i_j\right)+\frac{1}{3}\delta^j_iT^i_j$$

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.

Answer

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.

Examples:

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 ${\rm Tr}M=0$.

   
   

References:

1. 
   

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

   
   


2. 
   

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