# Does it really make sense to talk about the color of gluons?

It is my understanding that by enforcing SU(3) gauge invariance on our lagrangian of 3-colored quark fields, we are forced to accept the existence of 8 new massless vector fields, the gluons. The 8 here comes directly from the dimension of SU(3).

That being said I often see discussions about the gluons in terms of linear combinations of $$r\bar r$$, $$b\bar b$$, etc.

This simply cant be the nature of the gluons though can it? Because it seems to imply that the number of colors and the number of gluon fuelds are not independant, while they clearly are.

Certainly gluons are not singlets in color space and so they must have color, but it doesnt make sense to me that this color of the gluons would be some mapping directly from quark color.

Thanks to anyone with the insight and time to share it!

The quarks transform according to the fundamental representation $$\mathbf{3}$$ of SU(3), and the antiquarks according to the conjugate representation $$\mathbf{\overline 3}$$. The gluons transform according to the adjoint representation $$\mathbf{8}$$.

The adjoint representation is contained in the product of the fundamental representation and its conjugate:

$$\mathbf{3}\otimes \mathbf{\overline 3} = \mathbf{8}\oplus \mathbf{1}$$

Therefore gluons are conventionally labeled using color-anticolor combinations, avoiding the color singlet combination $$(r\overline{r}+b\overline{b}+g\overline{g})/\sqrt{3}$$.

• Awesome do you have a any sources/papers/books I could find more about this? – Craig Nov 10 '18 at 18:53
• As well; in a universe where we had say, 4 or 5 colors and 8 gluons, how would this work? – Craig Nov 10 '18 at 18:57
• @Craig, are you familiar with representation theory? A big concept is that the same group can be "represented" with larger or smaller vector spaces. In this case, the quarks have the minimum number of vectors needed to realize SU(3) symmetry, and the gluons have (very loosely) that maximum number of vectors that can have the SU(3) symmetry, one for each dimension. If we had 4 colors, then we would instead want $4\otimes \bar 4 = 15\oplus 1$, and we would have 15 gluon fields for the 4 colors. – Alex Meiburg Nov 10 '18 at 19:01
• The adjoint reprsentation of SU(n) has dimension $n^2-1$. So if you have 4 colors there need to be 15 gluons and if you have 5 colors there need to be 24 gluons. – G. Smith Nov 10 '18 at 19:05
• I’ll let others suggest the best references. But you need to clarify what you main interest is. The mathematics of representation theory? (For example, how do you figure out how an arbitrary product of irreducible representations decomposes into irreducible representations?) The physics of QCD, or the whole Standard Model? The reason why quantum field theory involves group representations? Etc. – G. Smith Nov 10 '18 at 20:25

The gluons are generators of the SU(3) gauge group; whatever notation is used to describe the fundamental representation can be extended to higher representations through their embedding in tensor products of the fundamental (and its dual.) [Also, by "sums" of $$r\bar r$$, $$b\bar b$$, etc., do you really mean products?]

• Oops I mean linear combinations of products of $r\bar r$, $b\bar b$, etc – Craig Nov 10 '18 at 18:54