Mathematically, what is color charge?

Question

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.

Answer 1

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 red-green-blue.

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.

Warm-up: Quick Representations SU(2)

For SU(2), the invariant tensors are $\delta^i_j$, $\epsilon_{ij}$, and $\epsilon^{ij}$, which are trace and two-dimensional 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 (n-1,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

$$(n,n') = (n+n') + (n-1,n'-1)$$

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 n-index 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 Clebsch-Gordon series for addition of quantum angular momentum.

The tensorial method is never taught for some reason, but it is the quickest way to do Clebsch-Gordon decompositions in real life.

Back to SU(3)

SU(3) transformations preserve inner products, and 3-dimensional 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

$$(n_1,...,n_k; m_1,,...,m_k) \rightarrow $$ $$ (n_1+n_2,n_3,...,n_k;m_1,...,m_k) \oplus (n_1-1,n_2-1,n_3,...n_k; 1, m_1, ...,m_k)$$

$$(n_1,...,n_k; m_1,....,m_k) \rightarrow$$ $$ (n1,...,n_k; m_1+m_2,m_3,...,m_k) \oplus (1,n_1,...,n_k;m_1-1,m_2-1,...,m_k)$$

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 trace-reduced (n-k;m-k) 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 color-charge of an instance of the actual object is an n-list of rgb-rgb-..rgb values, where the order doesn't matter, and an m-list of cmy-cmy...-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 color-charges, 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 one-up-index, one-down-index traceless tensors, and that all hadrons are singlets.

Answer 2

I'm not sure about color charges, but in real colors, red, green, and blue aren't orthogonal, but they are linearly independent, so they can be used to span $\mathbb{R}^3$ (i.e. every color can be written as a linear combination of $R, G, B$).

Perhaps a diversion to $L^*a^*b^*$ color space will help. In this color space, the color difference between two colors is $\Delta E = \sqrt{\Delta L^2 + \Delta a^2 + \Delta b^2}$. One $\Delta E$ is supposed to be the minimum (human) perceptible color change, anywhere in color space. Actually they found later that this wasn't quite right, and you can move about $4 \Delta E$ for yellow or about $0.5 \Delta E$ for gray before you see a color change. So they came up with the 2000 formula for $\Delta E$, which linearizes the color difference in $L^*a^*b^*$ space. But let's see what red, green, and blue look like in this color space:

$$Red: (L^*,a^*,b^*) = (47, 66, 50)$$

$$Green: (L^*,a^*,b^*) = (49, -66, 33)$$

$$Blue: (L^*,a^*,b^*) = (20, 25, -48)$$

(This is for a print example. The actual $(L^*,a^*,b^*)$ value for red, green, and blue depends on the media or device.)

But you can see that the colors are not orthogonal. For example, the dot product of red and green is:

$$Red \cdot Green = -403$$

In a perceptual sense, all colors have some brightness ($L^*$), so when you add them together, the $L^*$ increases, making the resulting color brighter. This is what happens on your computer monitor. To get a neutral color, you want $a^* = b^* = 0$.

Note that $L^*$ is always positive, so while mathematically, you can span $\mathbb{R}^3$, true colors (whether in $L^*a^*b^*$ or RGB space) only span the "upper half" of $\mathbb{R}^3$. The only physical way to get a "zero" color is with $(L^*,a^*,b^*) = (0, 0, 0)$ or $(R, G, B) = (0, 0, 0)$.

So in terms of real colors, colorless means that $R \approx G \approx B$ (or $a^* = b^* = 0$). Color charges may be combined using $R, G, B$ as basis vectors, but those basis vectors are not orthogonal.

From the other question, I get the sense that the connection to real color is somewhat of a stretch.