Do the “$SU(3)$ colors” live in a 3-dimensional vector space? Previously I asked a question about the visualized colors:
Do the "colors" live in a 3-dimensional vector space?
(My earlier question is unfortunately closed)
Now I like to ask the “$SU(3)$ colors” of strong nuclear force.
Do the “$SU(3)$ colors” live in a 3-dimensional vector space?
Should we label the (red, green, blue) in terms of the 3-component vector
$$(r,g,b) \in \mathbb{C}\times \mathbb{C}\times\mathbb{C}= \mathbb{C}^3,$$
living in a 3-dimensional complex vector space?
 A: Maybe this link  will help:

Color charge is a property of quarks and gluons that is related to the particles' strong interactions in the theory of quantum chromodynamics (QCD).


The "color charge" of quarks and gluons is completely unrelated to the everyday meaning of color. The term color and the labels red, green, and blue became popular simply because of the loose analogy to the primary colors. Richard Feynman referred to his colleagues as "idiot physicists" for choosing the confusing name.

It is color charge  and assigned to particles analogously  to the way the electric charge is assigned to particles, even though electric charge  obeys SU(2) symmetries at given situations it does not mean that in the interactions it behaves as a vector quantity. It is a fixed number to the particles and can change only with interactions. The same is true for the three color charges. A particle can change its color only through interaction:


Gluon interactions are often represented by a Feynman diagram. Note that the gluon generates a color change for the quarks. The gluons are in fact considered to be bi-colored, carrying a unit of color and a unit of anti-color as suggested in the diagram at right. The gluon exchange picture there converts a blue quark to a green one and vice versa.

It s not two vectors in a vector space on the gluon. It is two charges.
A: Yes, the color charge of a quark can be represnted by a vector in $\mathbb{C}^3.$ What's more, if you were to "hold" the quark (which you cannot actually do, but let's put that aside) and move it around a background gluon field, that $\mathbb{C}^3$ vector would be be "rotated" by an $SU(3)$ matrix. This is a lot like how if you parallel transport a tangent vector in Riemannian geometry is ends up getting rotated by an orthogonal matrix. The general principle is the same. More mathematically, if  you have a spacetime manifold $M = \mathbb{R}^4$, the gluon field is a connection on the bundle $SU(3) \times M$. (Well, this is all in classical physics. Quantum mechanically, the gluon field is a wavefunctional of connections, with additional subtleties involved.) Now, if there is some "field strength," i.e. you might say there are 'gluons' present, then when you drag your quark in a closed loop it'll come back rotated by an overall $SU(3)$ because there is some curvature in the bundle. Now, a gauge transformation applies a spacetime dependent function $g : M \to SU(3)$  that changes one gluon field to a physically equivalent one. Importantly, the holonomy of dragging the quark in a closed loop will remain unaffected by this gauge transformation, giving us a gauge invariant set objects to characterize the gluon field state with. Note that, while the quark has a $\mathbb{C}^3$ vector, the gluons are best understood as enacting group actions on the quark. This means that their states live in the Lie algebra of $\mathfrak{su}(3)$. Hence the labelling of gluons as, say, "red-blue," which rotates infinitesimally between the red and blue colored quark.
