quantum field theory - Why is the chiral symmetry only $SU(3) times SU(3)$ and not $SU(6)$?

In the limit where the masses vanish, low energy QCD has a well known chiral symmetry (see http://arxiv.org/abs/hep-ph/0505265 for a very extensive review, and pg 19 for the section relevant for my question). The Lagrangian of interest is (I use 2 component notation),

\begin{equation} {\cal L} = \sum _{ i \in u, d , s}q _{ R , i} ^\dagger i \bar{\sigma} ^\mu D _\mu q _{ R , i} + q ^\dagger _{ L , i} i \bar{\sigma} ^\mu D _\mu q _{L , i} - \frac{1}{4} G _{ \mu \nu } G ^{ \mu \nu } \end{equation} Well below the weak scale we can rewrite the following as, \begin{equation} {\cal L} = \sum _{ \ell \in u _L , d _L , s _L , u _R , d _R , s _R }q ^\dagger _{ \ell } i \bar{\sigma} ^\mu D _\mu q _{ \ell } - \frac{1}{4} G _{ \mu \nu } G ^{ \mu \nu } \end{equation} Thus it seems that if we aren't worried about electroweak effects (and hence the covariant derivatives in the right handed fields are the same as in the left handed ones) then we in fact have an $ SU(6) \times U(1) _V $ symmetry, \begin{equation} \left( \begin{array}{c} u _L \\ d _L \\ s _L \\ u _R \\ d _R \\ s _R \end{array} \right) \rightarrow e ^{ i T _A \alpha _A }\left( \begin{array}{c} u _L \\ d _L \\ s _L \\ u _R \\ d _R \\ s _R \end{array} \right) \end{equation} instead of the more commonly discussed $ SU(3) \times SU(3) \times U(1) _V $. However, the fact that I've never seen this discuss makes me skeptical of its validity. Is the above a legitimate way to discuss chiral symmetry?

Answer

Okay, I think I have an answer but I would interested to hear some feedback on this. I initially thought that by writing the left and right handed fields in 2 component notation ($q_L $ and $q_R^c$), they were 'equivalent' objects and can freely rotated between one another. However, I realized now this is not the case.

The left and right handed fields are in the fundamental and anti-fundamental representation of $SU(3)_C$ (due to the often omitted conjugate symbol on the right handed fields) and so can't be simply rotated between one another. This restricts left fields to rotate into left fields and right fields and right fields and hence the $SU(3) _L\times SU(3)_R $ instead of a rotation between all the fields ($SU(6)$).

Note that for the above discussion I was using common beyond Standard Model notation where the fundamental fields are all left-chiral fields and hence transform the same way (for details on this see hep-ph/9709356, bottom of pg 8). An alternative to this viewpoint is to define the fundamental fields as left and right chiral fields and then as @Thomas mentioned they transform the same way under $SU(3)_C$ but you can't rotate between them since they have different representations under the Lorentz group.