The question is written in section $2)$

1) Introduction

1.1) QCD

For a non-abelian group, the connection term on the lagrangian will be written as

$$\mathcal{A}_{\mu}=A_{\mu}^{a}T_{a}\tag{1}$$

This notation is far from be explicit and this brings me all sorts of confusion. I came from a classical pseudo-Riemannian background so indices are very important to me, I need to see all the indices to be sure and to be confortable with the math.

Therefore, the standard notation for the QCD lagrangian, with quarks $\psi$ and $\bar{\psi}$ interacting with the gluons $\mathcal{A}_{\mu}$ is:

$$\mathcal{L} =-\frac{1}{4}F^{a}_{\mu\nu}F^{\mu\nu}_{a} + \bar{\psi}[i\gamma^{\mu}D_{\mu}-m]\psi= -\frac{1}{4}F^{a}_{\mu\nu}F^{\mu\nu}_{a}+\bar{\psi}[i\gamma^{\mu}(\partial _{\mu}+ig\mathcal{A}_{\mu})-m]\psi \tag{2}$$

But, $(2)$, just like $(1)$, is not the "explicit" form of the lagrangian. Following the reference $[1]$, the explicit form (with colors!) is given by:

$$\mathcal{L} =-\frac{1}{4}\color{teal}{\sum_{a=1}^{8}}F^{\color{teal}{a}}_{\mu\nu}F^{\mu\nu}_{\color{teal}{a}} + \color{brown}{\sum_{i=1}^{3}}\color{brown}{\sum_{j=1}^{3}}\color{magenta}{\sum_{(f)=1}^{6}}\bar{\psi}_{\color{brown}{i}\color{magenta}{(f)}}[i\gamma^{\mu}D_{\mu \color{brown}{j}}^{\color{brown}{i}}-m_{\color{magenta}{(f)}}\delta^{\color{brown}{i}}_{\color{brown}{j}}]\psi^{\color{brown}{j}}_{\color{magenta}{(f)}}$$

$$= -\frac{1}{4}\color{teal}{\sum_{a=1}^{8}}F^{\color{teal}{a}}_{\mu\nu}F^{\mu\nu}_{\color{teal}{a}} + \color{brown}{\sum_{i=1}^{3}}\color{brown}{\sum_{j=1}^{3}}\color{magenta}{\sum_{(f)=1}^{6}}\bar{\psi}_{\color{brown}{i}\color{magenta}{(f)}}[i\gamma^{\mu}\Big(\delta^{\color{brown}{i}}_{\color{brown}{j}}\partial_{\mu}+ig\color{teal}{\sum_{a=1}^{8}}A_{\mu}^{\color{teal}{a}}T_{\color{teal}{a}\color{brown}{j}}^{\color{brown}{i}}\Big)-m_{\color{magenta}{(f)}}\delta^{\color{brown}{i}}_{\color{brown}{j}}]\psi^{\color{brown}{j}}_{\color{magenta}{(f)}} \tag{3}$$

Accordingly to $[1]$, the meaning of the indices are:

$\color{teal}{a = \{1,...,8\}}$ color index in adjoint representation of $SU(3)_{\mathrm{C}}$ (probably here he is actually meaning for the Lie algebra $\mathfrak{su}(3)_{\mathrm{C}}$): indices to take account on how many gluons exist (the famous "Lie algebra indices", maybe "gauge indices"?)

$\color{brown}{i,j = \{1,2,3\} = \{\color{red}{r},\color{green}{g},\color{blue}{b}\}}$ color index in fundamental representation of $SU(3)_{\mathrm{C}}$: indices to take account on how many colors (color charge) a quark can have

$\color{magenta}{(f) = \{1,...,6\} = \{u,d,c,s,t,b\}}$ index to take account on every type of quark that exist

$\mu,\nu = \{0,...,3\}$ our good old spacetime indices.

1.2) QED

Now, using a similar reasoning, I think that explicit indices that appears in QED are:

$\color{teal}{a = \{1\}}$ index in adjoint representation of $U(1)_{\mathrm{em}}$: indices to take account on how many photons exist, i.e., just one photon.

$\color{brown}{i,j = \{1\} = \{\color{brown}{q}\}}$ index in fundamental representation of $U(1)_{\mathrm{em}}$: indices to take account on how many charges a electron can have, i.e., the electric charge.

$\color{magenta}{(f) = \{1\} = \{e\}}$ index to take account on every type of electron that exist, i.e., the electron.

$\mu,\nu = \{0,...,3\}$ our good old spacetime indices.

then, we actually recover the standard notation for Dirac lagrangian of QED:

$$\mathcal{L}= -\frac{1}{4}\color{teal}{\sum_{a=1}^{1}}F^{\color{teal}{a}}_{\mu\nu}F^{\mu\nu}_{\color{teal}{a}} + \color{brown}{\sum_{i=1}^{1}}\color{brown}{\sum_{j=1}^{1}}\color{magenta}{\sum_{(f)=1}^{1}}\bar{\psi}_{\color{brown}{i}\color{magenta}{(f)}}[i\gamma^{\mu}\Big(\delta^{\color{brown}{i}}_{\color{brown}{j}}\partial_{\mu}+ig\color{teal}{\sum_{a=1}^{1}}A_{\mu}^{\color{teal}{a}}T_{\color{teal}{a}\color{brown}{j}}^{\color{brown}{i}}\Big)-m_{\color{magenta}{(f)}}\delta^{\color{brown}{i}}_{\color{brown}{j}}]\psi^{\color{brown}{j}}_{\color{magenta}{(f)}} = $$

$$= -\frac{1}{4}F^{\color{teal}{1}}_{\mu\nu}F^{\mu\nu}_{\color{teal}{1}} + \bar{\psi}_{\color{brown}{1}\color{magenta}{(1)}}[i\gamma^{\mu}\Big(\delta^{\color{brown}{1}}_{\color{brown}{1}}\partial_{\mu}+igA_{\mu}^{\color{teal}{1}}T_{\color{teal}{1}\color{brown}{1}}^{\color{brown}{1}}\Big)-m_{\color{magenta}{(1)}}\delta^{\color{brown}{1}}_{\color{brown}{1}}]\psi^{\color{brown}{1}}_{\color{magenta}{(1)}} \implies$$

$$\mathcal{L} = -\frac{1}{4}F_{\mu\nu}F^{\mu\nu}+\bar{\psi}[i\gamma^{\mu}(\partial _{\mu}+iqA_{\mu})-m_{q}]\psi \tag{4}$$

2) My question

Now, I have two sorts of confusion here. One concerning the "explicit" form of QED lagrangian, and the other concerning the gamma matrices $\gamma^{\mu}$. This seems to be two questions, but they are, in my opinion quite related as one. I'll explain.

The gamma matrices are formalized into the framework of clifford algebras, there the one-index object $\gamma_{\mu}$ or $\gamma^{\mu}$ are just basis vectors. But in the context of field theory, the writting of gamma matrices as $\gamma^{\mu}$ are just a (historical?) notation. These symbols are matrice and matrices have two indices, period. Therefore, they are the matrices given by: $\gamma^{\mu\color{purple}{A}}_{\color{purple}{B}}$ where $\color{purple}{A,B = \{1,2\}}$ in Pauli-Dirac representation and $\color{purple}{A,B = \{1,2,3,4\}}$ in Dirac representation.

Well, in both $(3)$ and $(4)$, I'm not sure what are the spinor indices. If we write the QED lagrangian with $\gamma^{\mu\color{purple}{A}}_{\color{purple}{B}}$, then other sorts of indices appear,

$$\mathcal{L} = -\frac{1}{4}F_{\mu\nu}F^{\mu\nu}+\bar{\psi}_{\color{purple}{A}}[i\gamma^{\mu \color{purple}{A}}_{\color{purple}{B}}(\partial _{\mu}+iqA_{\mu})-m_{q}\delta^{\color{purple}{A}}_{\color{purple}{B}}]\psi^{\color{purple}{B}} \tag{5}$$

It seems to me that the indices $A,B$ are in fact the indices that takes account for spin the up and down. Since in the Pauli-Dirac representation $\color{purple}{A,B = \{1,2\} = \{\uparrow,\downarrow \}}$ and in Dirac representation $\color{purple}{A,B = \{1,2,3,4\} = \{\uparrow_{E_{+}},\downarrow_{E_{+}},\uparrow_{E_{-}},\downarrow_{E_{-}}\}}$, where in the first case are just spin up and down states, and the later case is the spin up and down for positive and negative energy solutions.

Therefore my question is:

The spinor indices are the $A,B$?

3) Side notes

1. If the question is answered to be true, then the spinor describing the relativistic electron is: $\psi^{\color{purple}{B}}$ and for a quark is a object with many indices: $\psi^{\color{brown}{j}\color{purple}{B}}_{\color{magenta}{(f)}}$

2. The lagrangian of QCD would be, in my opinion: $$\mathcal{L} = -\frac{1}{4}\color{teal}{\sum_{a=1}^{8}}F^{\color{teal}{a}}_{\mu\nu}F^{\mu\nu}_{\color{teal}{a}} + \color{purple}{\sum_{i=1}^{4}}\color{brown}{\sum_{i=1}^{3}}\color{brown}{\sum_{j=1}^{3}}\color{magenta}{\sum_{(f)=1}^{6}}\bar{\psi}_{\color{purple}{A}\color{brown}{i}\color{magenta}{(f)}}[i\gamma^{\mu \color{purple}{A}}_{\color{purple}{B}}\Big(\delta^{\color{brown}{i}}_{\color{brown}{j}}\partial_{\mu}+ig\color{teal}{\sum_{a=1}^{8}}A_{\mu}^{\color{teal}{a}}T_{\color{teal}{a}\color{brown}{j}}^{\color{brown}{i}}\Big)-m_{\color{magenta}{(f)}}\delta^{\color{purple}{A}}_{\color{purple}{B}}\delta^{\color{brown}{i}}_{\color{brown}{j}}]\psi^{\color{brown}{j}\color{purple}{B}}_{\color{magenta}{(f)}}$$

3. If the question is answered to be true, then the spinor trasformations describing the relativistic electron are: $\psi'^{\color{purple}{B}} = S[\Lambda]^{B}_{C}\psi^{\color{purple}{C}}$, which is ok to say that "a electron is a spinor that transforms with SU(2)". But and for a quark: $\psi'^{\color{brown}{j}\color{purple}{B}}_{\color{magenta}{(f)}}=S[\Lambda]^{B}_{C}\psi^{\color{brown}{j}\color{purple}{C}}_{\color{magenta}{(f)}}$ it seems not ok saying that they transforms via $SU(2)$, I mean, they shouldn't transform with a $SU(3)$ matrix (the matrix associated with the representation of the Lie algebra)?

4. I think the summation notation in $(4)$, would appear to be wrong in a first glance: $$\mathcal{L} = -\frac{1}{4}F^{\color{teal}{1}}_{\mu\nu}F^{\mu\nu}_{\color{teal}{1}} + \bar{\psi}_{\color{brown}{1}\color{magenta}{(1)}}[i\gamma^{\mu}\Big(\delta^{\color{brown}{1}}_{\color{brown}{1}}\partial_{\mu}+igA_{\mu}^{\color{teal}{1}}T_{\color{teal}{1}\color{brown}{1}}^{\color{brown}{1}}\Big)-m_{\color{magenta}{(1)}}\delta^{\color{brown}{1}}_{\color{brown}{1}}]\psi^{\color{brown}{1}}_{\color{magenta}{(1)}} $$

But, if we consider the generators of lie algebra to be $T_{a} = 1$; since, $g_{U(1)_{\mathrm{em}}} = e^{i\theta} = e^{i\theta 1} \implies \theta^{a} = \theta, T_{a}= 1$. Moreover, $(T_{a})^{i}_{j} = (T_{1})^{i}_{j} = (1)^{i}_{j} = \delta^{i}_{j}$. Finally, considering the case that $i,j = \{1\}$, therefore, in fact, seems to be $(T_{1})^{1}_{1} = \delta^{1}_{1} = 1$. So, the identity matrix multipling the Dirac equation is from $A,B$ indices not from $i,j$: $(i\gamma^{\mu}\partial_{\mu}- m\mathbb{I}_{4\times4})\psi$ is actually, $(i\gamma^{\mu \color{purple}{A}}_{\color{purple}{B}} \partial_{\mu}- m\delta^{\color{purple}{A}}_{\color{purple}{B}})\psi^{\color{purple}{B}}$.

$[1]$ https://www2.physics.ox.ac.uk/sites/default/files/2014-03-31/qcdgrad_rojo_oxford_tt14_2_basics_pdf_40958.pdf - Page 7