Mathematically, what is color charge?
Color charge is the representation of the SU(3) gauge group. The representation theory of SU(3) is described below:
The basic representation is called the "3" or the fundamental, or defining, representation. It is a triplet of complex numbers $V^i$, which transform under a 3 by 3 SU(3) matrix by getting multiplied by the matrix. The value of "i" is sometimes called "red","green","blue", so that a quark which is all in the $V^1$ direction is red, etc. This is reasonable, because every fundamental representation vector is a linear combination with complex coefficient of redgreenblue.
Hermann Weyl proved that every other representation occurs somewhere among the tensors: $V^{i_1,i_2 .. i_n}_{j_1,j_2,...,j_m}$ where the upper indices transform by multiplying the index by the SU(3) matrix, and lower indices by multiplying by the conjugate matrix, which is also the inverse. Addition of representations is just like addition of angular momentum in quantum mechanics, by taking tensor products.
Warmup: Quick Representations SU(2)
For SU(2), the invariant tensors are $\delta^i_j$, $\epsilon_{ij}$, and $\epsilon^{ij}$, which are trace and twodimensional volume. You can raise and lower indices using $\epsilon$ tensors, so every tensor representation is equivalent to one with all the indices down. Any three antisymmetric indices are necessarily zero, and any two antisymmetric indices can be deleted by contracting them with the appropriate epsilon tensor. So there are only clumps of symmetric indices in a representation.
The irreducible representations are exhausted by the fully symmetric tensors with all indices down:
$$ T_{i_1, i_2 , .... i_n}$$
Because when you multiply two of these, you get a tensor
$$ T_{i_1, i_2,....,i_n;j_1, j_2, ... j_n'}$$
with symmetry on permuting the first n indices and the last n' indices. I will write this as (n,n'). By contracting the $\epsilon$ tensor on one of the i's and one of the j's (they all give the same result because they are symmetric), you extract an (n1,n'1) representation from this. The remainder is fully symmetric on n+n' indices, because you have removed the antisymmetric part. The result is the decomposition
$$(n,n') = (n+n') + (n1,n'1)$$
So that, recursively, the tensor product of (n) and (n') decomposes into
$$(n+n'), (n+n'2), (n+n'4), ... (1) or (0)$$
where the last term is 1 if n+n' is odd, or 1 if n+n' is even. If you recognize that the nindex totally symmetric tensor with two possible values for each index has exactly n+1 different components, you realize that the (n) representation is just the spin (n/2) representation, and the decomposition above is the familiar ClebschGordon series for addition of quantum angular momentum.
The tensorial method is never taught for some reason, but it is the quickest way to do ClebschGordon decompositions in real life.
Back to SU(3)
SU(3) transformations preserve inner products, and 3dimensional complex volumes, so there are three basic invariant tensors, $\delta^i_j$, $\epsilon_{ijk}$, and $\epsilon^{ijk}$. The $\epsilon$ tensors allow you to take the antisymmetric part on any two upper indices and turn it into a lower index, or the anyisymmetric part of two lower indices and turn it into an upper index.
The irreducible representations of SU(3) are all tensors
$$ T^{i_1,i_2,....,i_n}_{j_1,j_2,...,j_m}$$
which are fully symmetric on the upper indices, and fully symmetric on the lower indices. To see this, call this representation (n;m), and tensor two such representations to produce
$$(n,n';m,m')$$
Which means n totally symmetric upper indices followed by n' totally symmetric upper indices, and m totally symmetric lower indices followed by m' totally symmetric lower indices.
Then, acting the epsilon tensor between the n and n' clump produces
$$(n,n'; 1,m,m')$$
leaving behind $(n+n';m,m')$, since it takes away the antisymmetric part. The recursive rule is as ollows
$$(n_1,...,n_k; m_1,,...,m_k) \rightarrow $$ $$ (n_1+n_2,n_3,...,n_k;m_1,...,m_k) \oplus (n_11,n_21,n_3,...n_k; 1, m_1, ...,m_k)$$
$$(n_1,...,n_k; m_1,....,m_k) \rightarrow$$ $$ (n1,...,n_k; m_1+m_2,m_3,...,m_k) \oplus (1,n_1,...,n_k;m_11,m_21,...,m_k)$$
These rules correspond to acting with the two epsilon tensors, and they terminate on terms of the form (n;m) in a finite number of steps, because either thing on the right hand side either has a smaller number of clumps, or a smaller sum of indices. Decomposing the traces out of (n;m) gives all the irreducible representations.
Removing the traces
After you reduce the tensors to (n;m), you get rid of all the trace parts, by subtracting $\delta^i_j$ times an (n;m) tensor. This turns every (n,m) of the previous section into a series of tracereduced (nk;mk) parts which go from k=0 to k=min(m,n). These tensors are the true irreducible representations.
The color charge is defined as the representation of SU(3) of the colored object. The representation is indexed by (n,m). The colorcharge of an instance of the actual object is an nlist of rgbrgb..rgb values, where the order doesn't matter, and an mlist of cmycmy...cmy colors, where the order doesn't matter. These are the basic colors, and you superpose them with arbitrary complex numbers, but imposing the trace condition, which is a little hard to state in RGB language it says that the sum of (r,LIST;c,LIST) + (g,LIST;m,LIST') + (b,LIST;y,LIST') is zero for any colors in LIST and LIST'.
To add two color charges, you use the procedure above for tensor products. The sum of two color charges is a complicated mixture of colorcharges, given by decomposing the tensor representation.
These rules are relatively complicated, so be thankful that the only fundamental color representations in the world are quark fundamental triplets, and gluon oneupindex, onedownindex traceless tensors, and that all hadrons are singlets.
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Hugh Allen
Updated on November 26, 2020Comments

Hugh Allen almost 3 years
A similar question was asked here, but the answer didn't address the following, at least not in a way that I could understand.
Electric charge is simple  it's just a real scalar quantity. Ignoring units and possible quantization, you could write $q \in \mathbb{R}$. Combination of electric charges is just arithmetic addition: $ q_{net} = q_1 + q_2 $.
Now to color charge. Because there are three "components", I am tempted to conclude that color charges are members of $\mathbb{R}^3$. However, I've read that "red plus green plus blue equals colorless", which seems to rule out this idea. I can only think that either:
 red, green and blue are not orthogonal, or
 "colorless" doesn't mean zero color charge (unlikely), or
 color charges don't combine in a simple way like vector addition
In formulating an answer, please consider that I know some mathematics (vectors, matrices, complex numbers, calculus) but almost nothing about symmetry groups or Lie algebras.

Luboš Motl over 12 yearsDear @Hugh, it's a good opportunity to learn some Lie algebra  color is just a "small" generalization of the electric charge if you learn some group theory. Without nonAbelian groups, you may imagine that red, green, blue are three equally long vectors in a 2dimensional plane separated by the 120 degree angle. So they do add by vector addition and red+green+blue is indeed zero. More properly, red,green,blue are components of a 3dimensional representation, adding colored particles means making tensor product, and colorless means a singlet in the decomposition.

anna v over 12 yearsDear Lubos this is a comment that should become an answer. I know that in answers you go into a thesis mode, but this really answers the question and is enough.

anna v over 12 years@Luboš Motl Maybe this will catch your attention?

Marek over 12 yearsas Luboš said, if you need to understand this (beyond the very superficial coverage I gave in the answer you've linked to), you need to learn some group theory and gauge theory. The 3 in the number of colors has to do with group $SU(3)$ and it's similar to the 2 of $SU(2)$ weak force and $U(1)$ of the electromagnetism. I am mentioning this to you because hopefully you have at least some familiarity with the latter two groups (which have to do with spin and 2D rotations, respectively); in contrast to $SU(3)$ which can appear formidable to newcomers.

Admin over 11 years@Ron: I am new to tensors. Please could you help me a bit. Does your delta symbol with one superscript and one subscript mean the same as the Kronecker delta explained here en.wikipedia.org/wiki/Kronecker_delta.I know the levicivita tensor with three subscripts i,j,k. What does it mean when it has only two subscripts? Lastly you said " any two antisymmetric indices can be deleted by contracting them with the appropriate epsilon tensor". How do you do this. Please could you give me an example. Any help will be much appreciated. Thanks in advance.

Ron Maimon over 11 years@ramanujan: the deltasymbol is the Kronecker delta, but it is only invariant with one up and one down index (the invariant inner product in SU(n) is the vector with its complex conjugate, which transform as upper/lower index respectively). The three component epsilon tensor is for 3 dimensions, 2 components for two dimensions, and SU(2) deals with indices that go over only 2 possibilities, the 2 index epsilon tensor is $i\sigma_y$ (Paulimatrix), or (0,1;1,0). Two antisymmetric indices in 2d SU(2) have only one independent value ($T_{01}=T_{10}$), and this is half $T_{ij}\epsilon^{ij}$.

Riad over 3 yearsRon I am glad to find you again. I miss your good answers in quora. I like to say that charges are connected with forces. But forces are all the same.. the acceleration of a mass. Shouldn't this mean that all charges are related too.