"Color charge" of the adjoint fermion?

Question

What kind of "color charge" does the adjoint fermion carry?

Let us consider the SU(N) gauge theory. The gauge field is in the adjoint representation (rep).

Well-Konwn: If the fermion is in SU(N) fundamental rep, we know the fermions will form a color multiplet of N-component. For $N=3$, we say that there are 3 colors (r,g,b) of a given fermion $\psi$ $$ (\psi_r, \psi_g, \psi_b) $$

1. However, if the fermion is in SU(N) adjoint rep, we know the fermions will form a color multiplet of (N$^2-1)$-component. For N=3, we say that there have 8-component of color multiplet. So,

   What kind of "color charge" does each component of adjoint fermion carry? There should be 8 different choices. $$ (\psi_1, \psi_2, \dots, \psi_{8}) $$ What does the 1,2,3, $\dots$, 8 stand for in terms of color indices?

2. Are the color charges of adjoint fermions organized the same way as the gluons (which are also in adjoint) as in https://en.wikipedia.org/wiki/Gluon#Color_charge_and_superposition? Does both the adjoint fermions carry a color and an anti-color just as a gluon does? Why is that?

3. How can we read this information of color charges from the adjoint Rep of $SU(3)$ Lie algebra?

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p.s. we may say the 8 gluons carry 8 distinct color anti-color pairs: $$ (r\bar{b}+b\bar{r})/\sqrt{2}, -i(r\bar{b}-b\bar{r})/\sqrt{2}, (r\bar{g}+g\bar{r})/\sqrt{2} $$ $$ -i(r\bar{g}-g\bar{r})/\sqrt{2},(b\bar{g}+g\bar{b})/\sqrt{2},-i(b\bar{g}-g\bar{b})/\sqrt{2} $$ $$ (r\bar{r}-b\bar{b})/\sqrt{2},(r\bar{r}+b\bar{b}-2g\bar{g})/\sqrt{6} $$ And how about the 8-multiplet of adjoint fermions? What color charge do they carry?

Answer 1

An adjoint fermion transforms in exactly the same way as an adjoint boson (like the gluon). We can write an adjoint fermion as a matrix valued field $$ \psi_{ab} = \psi^A (T^A)_{ab} $$ where $T^A=\frac{1}{2}\lambda^A$ are the $SU(N)$ generators. The Dirac operator acts as a covariant derivative in the adjoint representation $$ (D_\mu \psi)^A = (\partial_\mu \delta^{AB}+igf^{ACB}A^C_\mu)\psi^B $$
just like it acts on gluons.