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The covariant derivative for a fermion with a symmetry group $SU(N)$ is given by $$ D_\nu \psi = \partial_\nu \psi -i g A^A_\nu t_A \, \psi, \tag{1} $$ where $A^A_\nu$ is a gauge field, $g$ is a coupling constant, $t_A$ is a generator of the Lie algebra $\mathfrak{su}(n)$, and the indices $A,B,\cdots=1,2,\cdots,N^2-1$.

For fermions in the fundamental representation, eq. (1) can be written as $$ D_\nu \psi^a = \partial_\nu \psi^a -i g A^A_\nu (t_A)^a{}_b \, \psi^b, \tag{2} $$ where the indices $a,b=1,2,\cdots,N$. So if $N=2$, the fermion $\psi$ can represented by $$\psi =\left( \begin{matrix} \psi_1 \\ \psi_2 %\\\psi_3 \end{matrix} \right) $$

But for fermions in the adjoint representation, the generators $(t_A){}^B{}_C = i f_A{}^B{}_C$, which are $3\times 3$ matrices in the case of $SU(2)$. The covariant derivative can be written as $$ D_\nu \psi^A = \partial_\nu \psi^A -i g A^A_\nu (t_A)^A{}_B, \psi^B.\tag{3} $$ Therefore the fermion field must be represented by
$$\psi=\left( \begin{matrix} \psi_1 \\ \psi_2 \\\psi_3 \end{matrix} \right). $$

But I am not sure that what I have expressed above is correct.

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Yes, everything you have written is correct.

By definition, a representation is a vector space $V$ of a particular dimension upon which the group acts by linear transformations $\rho: G\to\mathrm{GL}(V)$. When we say that a certain object such as $\psi$ "transforms in the representation $V$", then we mean that $\psi$ is an element of $V$. So, since the fundamental representation of $\mathrm{SU}(N)$ is the space $\mathbb{C}^N$ with $\mathrm{SU}(N)$ acting as the special unitary matrices, $\psi$ in the fundamental may be written as its components $(\psi_1,\dots,\psi_N)$ after a choice of basis. Likewise, the adjoint representation is given by the Lie algebra with the action as the adjoint action - for $\mathrm{SU}(N)$, it is $N^2-1$ dimensional, so after a choice of basis you may write $\psi$ as its components $(\psi_1,\dots,\psi_{N^2-1})$.

Note, however, that the $\psi_i$ themselves may themselves not be mere numbers, but elements of representations of other groups. A fermion $\psi$ also transforms in a spin representation of the Lorentz group, so each of the $\psi_i$ also has Lorentz components $\psi_i^\mu$. And possibly even more indices if there are further groups under which it transforms non-trivially. Formally, if we have groups $G_1,\dots,G_k$ under which $\psi$ transforms in the representations $V_1,\dots,V_k$, it is altogether an element of the tensor product $V_1\otimes\dots\otimes V_k$.

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  • $\begingroup$ When you say that the fermion field $\psi $ has Lorentz components labelled by $\mu $, might it be better to say spinor components labelled by e.g. $a $? The former makes it sound like a four vector. $\endgroup$
    – innisfree
    Feb 27 '17 at 11:41
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You are correct and these different representations are crucial to particle physics. In order to "fit" these groups and algebras with physical theories one has to adopt a particular representation and this actually defines the states that transform among themselves according to that group.

For example, the $SU(3)$ has irreducible representations of dimension 1, 3, 6, 8, etc. Let us say that this groups is associated to the local color symmetry. Then if we choose the fermion field to be in the triplet (three-dimensional) it means that they show up in three states, blue, green and red which are mixed by gauge transformations. This representation is suitable to describe quarks. On the other hand if we choose the fermion field to be a singlet of $SU(3)$ color (one-dimensional) it will not transform under this group which physically means it does not interact through strong interactions. Thus this may be a lepton, such as the electron.

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