 Solution

Represent the color charges by $r=(1,0,0{)}^T$, $g=(0,1,0{)}^T$ and $b=(0,0,1{)}^T$. The Gell-Mann matrices satisfyThe physical quark-gluon Feynman vertex contains $g_s{T}^a$ with $T^a={\lambda }^a/2$; the off-diagonal combination therefore allows $r\to b$. Color charge conservation assigns the emitted gluon the colour-flow label $r\bar{b}$.

The claimed impossibility of $r\to r$ is false. A direct counterexample from the printed matrices isThus both diagonal gluons have nonzero $r\to r$ vertex elements, $(T^3{)}_{rr}=1/2$ and $(T^8{)}_{rr}=1/(2\sqrt{3})$. They correspond to the traceless combinationsA Gell-Mann matrix is a generator rather than a finite rotation, but exponentiating a diagonal generator also leaves the red basis direction fixed up to a phase. The actual excluded object is the identity generator, proportional to the colour singlet $(r\bar{r}+g\bar{g}+b\bar{b})/\sqrt{3}$. Adding it would extend SU(3) to unitary group and introduce a ninth singlet vector boson. A colour-preserving vertex among the existing eight gluons does not require that extra boson and does not make the ordinary diagonal gluons colour singlets.