Some quantum mechanics

\(\newcommand{\mb}{\mathbb}\)Okay, not actual quantum mechanics. Rather, some aspects of particle physics that can be related to what has already been posted here.

Consider the basic multi-quark objects that we know of so far: hadrons, which are things like protons and neutrons that are made of three quarks, and things like mesons, which are made of a quark and an antiquark.
The basic foundations of quantum mechanics say that to each particle involved we should assign a complex vector space for the possible states of the particle, and for multiple particles we should take the tensor product of those vector spaces for those particles to get a vector space of states for the set of particles.
So for each quark we have a vector space \(V\), and then a set of three quarks has the vector space \(V^{\otimes 3}\). An antiquark should then have the vector space \(V^\vee\), and so a meson should have \(V\otimes V^\vee\).

To each force we assign a group called a gauge group, and we say that if a vector is invariant under the action of that group, then the corresponding object or system of objects is stable with respect to that group. We say that quarks are bound to each other via the strong force, and now ask what is the gauge group of the strong force.

So what could the gauge group for the strong force be? What are our invariants? We have 3-quark systems that are stable, and quark-antiquark systems that are stable. And everything else that is stable is built from these two types of systems, so there aren't any other invariants; nothing in \(V\otimes V\), or in \(V\otimes V\otimes V^\vee\), etc.
So we have a corresponding invariant tensor in \(V^{\otimes 3}\) and an invariant tensor in \(V\otimes V^\vee\). We can guess that the tensor in \(V\otimes V^\vee\) is the identity tensor \(I_j^i e_i \otimes e^j\). What about the one in \(V^{\otimes 3}\)?
If we consider only groups that have an invariant with 3 lower indices, its dual with 3 upper indices, and the identity, and nothing else that can't be written in terms of those invariants, we end up with two possibilities: \(SL(3, \mb C)\) and \(E_6\). For \(SL(3, \mb C)\), \(V\) would be \(\mb C^3\) and the invariant with 3 lower indices is the Levi-Civita tensor. For \(E_6\) we have \(V = \mb C^{27}\) and the tensor is quite a bit more complicated, relating to the multiplication of a Jordan algebra.
So which one is it?
One more bit of information is that quarks, being fermions, prefer antisymmetric tensors. The Levi-Civita tensor is antisymmetric; swap any two indices and you pick up a minus sign. The tensor for \(E_6\) is symmetric. So the gauge group for quarks is \(SL(3, \mb C)\).

Only not quite \(SL(3, \mb C)\). The last bit of quantum mechanics is that we want everything to be unitary. One interpretation of these vectors is that \(v\cdot v^*\), where \(v^*\) is the complex conjugate of \(v\) and \(\cdot\) is the dot product, corresponds to a probability. So we want this quantity to not be affected by the symmetries of the situation.
Hence we want not \(SL(3, \mb C)\) but \(SL(3, \mb C)\cap U(3) = SU(3)\), which is the gauge group for the strong force.

We call the 3 basis directions for \(V\) "red", "green" and "blue", because 3 always means primary colors, except for when it doesn't. So a quark whose state vector is pointing in the red direction is said to have red color charge and so on. Antiquarks come in "antired", "antigreen" and "antiblue", dual to "red", "green" and "blue" respectively.
A quantum mechanical force is carried by particles whose state vectors occur in the Lie algebra of the gauge group for that force. So the force carriers for color, called gluons, occur in \(su(3)\), which is \(8\)-dimensional. The force carriers take a vector in \(V\) to a vector in \(V\), and so are in \(V\otimes V^\vee\), and thus come with a color and an anticolor. So we can talk about "red-antigreen" gluons or "blue-antired" gluons. Note that we don't talk about "red-antired" gluons, because that would require a matrix with nonzero trace, but everything in \(su(3)\) is traceless. Instead we can talk about, say, "red-antired - blue-antiblue" or "red-antired - green-antigreen" gluons, or "red-antired - (1/2)green-antigreen - (1/2)blue-antiblue" gluons.
Oftentimes physicists will say that there are 8 gluons, by which they mean the space of gluons has 8 basis elements.