$\mathrm{SU(3)}$ decomposition of $\mathbf{3} \otimes \mathbf{\bar{3}} = \mathbf{8} \oplus \mathbf{1}$? I have a question about the tensor decomposition of $\mathrm{SU(3)}$. According to Georgi (page 142 and 143), a tensor $T^i{}_j$ decomposes as:
\begin{equation}
\mathbf{3} \otimes \mathbf{\bar{3}} = \mathbf{8} \oplus \mathbf{1}
\end{equation}
where the $\mathbf{1}$ represents the trace. However, I do not understand why we cannot further decompose the traceless part into a symmetric and an antisymmetric part.
In order to understand my logic: A general tensor $\varphi^i$ transforms as:
\begin{equation}
\varphi^i \rightarrow U^i{}_j \varphi^j
\end{equation}
whereas $\varphi_i$ transforms as:
\begin{equation}
\varphi_i \rightarrow (U^*)_i{}^j \varphi_j
\end{equation}
where $U \in \mathrm{SU(3)}$ is a $3 \times 3$ matrix. Now, I will let $S^i{}_j$ denote the traceless part of $T^i{}_j$ (i.e. $S^i{}_j$ has dimensions $\mathbf{8}$) and we can decompose this in the "symmetric" and "antisymmetric" part as usual:
\begin{equation}
S^i{}_j = \frac{1}{2}(S^i{}_j + S_j{}^i) + \frac{1}{2}(S^i{}_j - S_j{}^i)
\end{equation}
Then under an $\mathrm{SU(3)}$ transformation:
\begin{equation}
S^i{}_j + S_j{}^i \rightarrow U^i{}_k (U^*)_j{}^l S^k{}_l + U^i{}_k (U^*)_j{}^l S^k{}_l = U^i{}_k (U^*)_j{}^l (S^i{}_j + S_j{}^i)
\end{equation}
and:
\begin{equation}
S^i{}_j - S_j{}^i \rightarrow U^i{}_k (U^*)_j{}^l S^k{}_l - U^i{}_k (U^*)_j{}^l S^k{}_l = U^i{}_k (U^*)_j{}^l (S^i{}_j - S_j{}^i)
\end{equation}
Therefore, the symmetric part keeps its symmetry and the antisymmetric part keeps its antisymmetry. Thus two invariant subspaces are created and the representation is reducible? To sum up, I would think we decompose $T^i{}_j$ as:
\begin{equation}
\mathbf{3} \otimes \mathbf{\bar{3}} = \mathbf{3} \oplus \mathbf{5} \oplus \mathbf{1}
\end{equation}
where $\mathbf{3}$ denotes the dimensions of the antisymmetric part and $\mathbf{5}$ denotes the dimensions of the symmetric part. Where am I going wrong?
Edit: I got my convention from "Invariances in Physics and Group Theory" by Jean-Bernard Zuber:

 A: Ok, I think there is a mistake here:

A general tensor $\varphi^i$ transforms as: $$\varphi^i\rightarrow
U^i_{\phantom{1}j}\varphi^j$$ whereas $\varphi_i$ transforms as:
  $$\varphi_i\rightarrow
 (U^\boldsymbol{\ast})_i^{\phantom{1}j}\varphi_j$$

Where did you find these equations? The unitary matrix element in the second line should not be a complex conjugate. I don't remember Giorgi's conventions but the customary notation I'm used to is this one:
$$U_i^{\phantom{i}j}=U_{ij},\quad  \varphi_i\rightarrow
 U_i^{\phantom{1}j}\varphi_j\\
U^i_{\phantom{i}j}=U^\ast_{ij},\quad  \varphi^\ast_i\equiv \varphi^i\rightarrow
 U^i_{\phantom{1}j}\varphi^j\equiv(U_i^{\phantom{i}j}\varphi_j)^\ast.$$
Hence, in your equations I'd understand:
$$(U^\ast)_i^{\phantom{i}j}\equiv U^\ast_{ij}=U^i_{\phantom{i}j}$$
and it doesn't provide the right transformation law for $\varphi_i$.
EDIT: well, provided the previous comments, let me clarify some issues with the notation, that may led to confuse the meaning of these transformation laws. Let us choose the convention to denote $SU(N)$ transformations, that is $N\times N$ unitary matrices with unit determinant, with uppercase letters, like $U$, and base states (scalars, vectors and tensors) with lowercase Greek letters, $\psi\in \mathbb{C}^N$. For example vector states transform as:
$$\psi\to U\psi,\quad \psi_i\to U_{ij}\psi_j\equiv U_i^{\ j}\psi_j$$
Note that here I followed the convention of writing base states of the fundamental or vector representation with lower indices, as Georgi does and as you can find here. This is the convention I'm used to, but nothing stops you to do the contrary, choosing upper indices! Note also that $U\psi$ represents the ordinary product of an $N\times N$ matrix by a vector $\psi=(\psi_1,\ldots,\psi_N)^T$, and produce a vector of the same type. In the notation $U_{ij}$ the index $i$ represents the rows whereas the second index $j$ represents the columns. It's customary to write it like $U_i^{\ j}$ to distinguish rows and columns. $\psi_i$ is a column vector and $i$ counts its rows.
You can define the conjugate representation by means of the conjugate vectors $\psi_i^\ast$, whose transformation law is
$$\psi^\ast\to  (U\psi)^\ast=\psi^\ast U^\ast,\quad \psi_i^\ast\to (U^\ast)_{ij}\psi_j^\ast=\psi_j^\ast(U^\dagger)_{ji}$$
Since these conjugate vectors transform in a different way with respect to $\psi_i$, it's useful to introduce upper indices to distinguish them:
$$\psi^i\equiv \psi_i^\ast \to U^\ast_{ij}\psi_j^\ast\equiv U^i_{\ \ j} \psi^j.$$
As you can see, now indices are "summed on the bottom-right". The extension to any arbitrary  $(p,q)$-tensor is trivial, their transformation law are those of the direct (diagonal) product of $p$ type $\psi^i$ vectors and $q$ type $\psi_i$ vectors: 
$$\psi^{i_1\ldots i_p}_{j_1\ldots j_q}\to \big(U_{j_1}^{\ \ j'_1}\cdot\ldots\cdot U_{j_q}^{\ \ j'_q}\big)\big(U^{i_1}_{\ \ i'_1}\cdot\ldots\cdot U^{i_p}_{\ \ i'_p}\big)\psi^{i'_1\ldots i'_p}_{j'_1\ldots j'_q}.$$
Since upper and lower indices represents different objects it has no sense mixing them.
A: First, if you take the fundamental representation (representation $N$) of $SU(N)$ made of $N$ objects $\varphi^i$, the transformation law is : 
$\varphi^i \to U^i{}_j \varphi^j$. 
By taking the complex conjugate, you get : $\varphi^{*i} \to (U^*)^i{}_j \varphi^{*j}= (U^\dagger)^j{}_i \varphi^{*j}$.
Now, looking at the last expression with $U^\dagger$, one sees that it is more practical to define objects $\varphi_i$, wich transform like $\varphi^{*i}$  : 
$\varphi_{i} \to  (U^\dagger)^j{}_i \varphi_{j}$, 
This is the representation $\bar N$
Now clearly, when you make the product of the two representations $N$ and $\bar N$, you have a representation $T^i_j$ which transforms as $\varphi^i\varphi_j$ : 
$T^i_j \to (U)^i_k (U^\dagger)^l_j T^k_l$
Secondly, you cannot symmetrize or anti-symmetrize the representation $N \otimes \bar N$, that is  $T^i_j$,  because the indices $i$ and $j$ have a different nature, and correspond to different representations.
Now, if you consider the representation $N \otimes N$, that is some representation $S^{ij}$, then here you may separe in a symmetric and anti-symmetric part, for instance, you have : 
$3 \otimes 3 = 6 \oplus \bar 3 $ 
The $6$ is the symmetric part, while the $\bar 3$ is dual (equivalent) to the anti-symmetric part, thanks to the Levi-Civita tensor : $\varphi_i = \epsilon_{ijk}  \varphi^{jk}$
[EDIT]
Due to OP comments, some precisions : 
You have $U^\dagger = (U^*)^T$, where $T$ means transposed operation. Transposition means exchange of the row and columns of the matrix, that is exchange of the $i$ and $j$ indices. If you put the row indice as an upper indice and the column indice as a lower indice, then the exchange necessarily will put the row indice as a lower indice, and the column indice as a upper indice. Your notation  $(U^*)^i{}_j = (U^\dagger)_j{}^i$ is a not-too-good equivalent notation, I say not-too-good, because you loose the orginal meaning that I describe above   .About the representations, this is a different thing (these are not the same $i$ and $j$...), the upper indice transforms as a $N$ representation, and the lower indice transforms as a $\bar N$ representation, so it is like apples and bananas, you can only symmetrize or anti-symmetrize equivalent quantities which transform in the same manner ($2$ apples or $2$ bananas), but not $1$ banana + $1$ apple
A: The problem is: you are not allowed to (anti)symmetrize upper indices with lower indices. In other words
$$
S^{i}_{\;j} = S^{\;i}_{j}\,.
$$
Therefore the decomposition into symmetric + antisymmetric is just the decomposition whole vector space + zero vector space, which is trivial.
Why aren't we allowed to mix upper and lower indices?


*

*Because strictly speaking they act on different spaces, so there is no natural action of the permutation group over it. A tensor of type $(p,q)$ is a map
$$
T(p,q) : V^{\otimes p} \otimes \tilde{V}{}^{\otimes q} \to \mathbb{R}\,,
$$
where $V$ is a vector space and $\tilde{V}$ its dual (space of functionals $f:V\to\mathbb{R}$).

*Because you can actually get rid of the upper indices and work only with lower ones by doing 
$$
\epsilon_{ijk}\,v^{i}_{\ldots} = v_{jk\ldots}\,.
$$
then there is not meaning into permuting a single index with a pair of antisymmetrized indices.

*The irreducible representation dimensions that one can get are given by the Weyl dimension formula
$$
\mathrm{dim}[a,b] = \tfrac12 [(a+1)(b+1)(a+b+2)]\,, \quad a \geq 0, \,\,b\geq 0\,,\, a,b\in \mathbb{N}
$$
and you can check that there is no way to get $5$ out of it.

