Confusion about trace in the vertex term of Lagrangian I was reading through Mariano Quirós's lecture notes titled "Finite Temperature Field Theory and Phase Transitions". In Sec. 1.2, the author is calculating the one-loop effective potential at $T=0$. On Page 8, he is doing the calculations when gauge bosons are in the loop, and works with a Lagrangian,
$$\mathcal{L}=-\frac{1}{4}\mathrm{Tr}\,F^{\mu\nu}F_{\mu\nu}+\frac{1}{2}\mathrm{Tr}\,(D_\mu\phi_a)^\dagger (D^\mu\phi^a)+\cdots\tag{36}$$
where (as mentioned before in the notes), the $\phi_a$ are $N_s$ complex scalar fields. He then says:

The only vertex which contributes to one-loop is
  $$\mathcal{L}=\frac{1}{2}(M_{gb}^2)_{\alpha\beta}A_\mu^\alpha A^{\mu\beta}+\cdots\tag{39}$$
  where
  $$(M_{gb}^2)_{\alpha\beta}(\phi_c)=g_\alpha g_\beta\mathrm{Tr}\,\left[(T^i_{\alpha\ell}\phi_i)^\dagger T^\ell_{\beta j}\phi^j\right]\tag{40}$$

where $\phi_c$ is the functional derivative of the connected generating functional wrt the source (ref. eqs. (6) and (15)). He then mentions:

... (ii) $T_\alpha$ are the generators of the Lie algebra
  of the gauge group in the representation of the $\phi$-fields and the trace in (40) is over indices of that representation.

My confusion: I do not quite follow the term "... in the representation of the $\phi$-fields..." Does the presence of the term $T^\ell_{\beta j}\phi^j$ imply that the $T_\alpha$ are expressed as $N_s\times N_s$ matrices? I am also unsure about whether $i$ and $\ell$ refer to the components of such a matrix.
Please let me know if I need to provide more specific information about my confusion.
 A: A gauge theory is specified by some (reductive) Lie group $G$, and some (typically reducible) finite-dimensional representation $R$. The latter describes the matter fields.
For example, QCD corresponds to $G=SU(N)$ (with $N=3$) and $R=\square^{\oplus 6}$, where $\square$ denotes the fundamental ($N$-dimensional) representation, and the six copies corresponds to the six flavours. One could consider alternative models, where $R$ is a different representation. Typical examples are $R=\square\!\square$ (the symmetric representation) or $R=\bar\square\!\square$ (the adjoint representation). Incidentally, for three colours, $\square$ and $\begin{matrix}\square\\[-5ex] \square\end{matrix}$ are isomorphic representations, so one need not consider the anti-symmetric (but for large $N$ expansions, it is sometimes useful to replace the fundamental for the anti-symmetric). Also, gluons transform in the adjoint.
With this in mind, we see that matter fields live in a vector space $V$ that realises $R$. For example, $V(\square)\cong \mathbb C^N$ and $V(\square\!\square)\cong\mathbb C^{\frac12N(N+1)}$. That is to say, a fundamental field can be thought of as an $N$-dimensional column vector, and a symmetric field a $\frac12N(N+1)$-dimensional column vector (although it is more common to think of it as a symmetric $N\times N$ matrix, for obvious reasons).
The trace over $R$ means a trace over this vector space. If $M\colon V\to V$ is a matrix in this space, then $\operatorname{tr}_R(M)\sim\sum_IM^IM^I$, where $I=1,2,\dots,\dim(V)$. The normalisation of the trace depends on conventions, but one that is used very often is
$$
\operatorname{tr}_R(M):=\frac{1}{x_R}\operatorname{tr}_V(M)
$$
where $x_R$ is the Dynkin index of $R$, defined such that
$$
\operatorname{tr}_R(t_R^at_R^b)\equiv\delta^{ab}
$$
where $t_R^a$ are the generators of $G$ in the representation $R$, and $a$ is an adjoint index. For example, $x_\square=1$ and $x_{\square\!\square}=N+2$. Other conventions may contain extra factors of $1/2$, or drop $x_R$ entirely.
For example, if $M$ is the mass matrix of fundamental quarks, one has
$$
\frac12 \phi^\dagger_i M^{ij}\phi^j\equiv\frac12\operatorname{tr}_\square(\phi^\dagger M\phi)
$$
where on the l.h.s. the sum is over all colours, $i=1,2,\dots,N$. In the symmetric representation, one can think of $I$ as a pair of colour indices, $I=(ij)$, symmetrised. So
$$
\frac12\phi^\dagger_{(ij)}M^{(ij)(kl)}\phi^{(kl)}\equiv \frac12\phi^\dagger_IM^{IJ}\phi^J\equiv\frac12\operatorname{tr}_{\square\!\square}(\phi^\dagger M\phi)
$$
In the end, it is just a matter of notation. If you feel more comfortable using explicit indices, go ahead. But what you have to keep in mind is that for each $a$, $t_R^a$ is a $\dim(R)\times\dim(R)$ matrix. For fundamental fields, this is $N\times N$, but not for other representations. The matrix $t_R^a$ has indices in $V$, so $(t^a_R)_{IJ}$ is a complex number, for each $a=1,2,\dots,\dim(\mathfrak g)$, and each $I,J=1,2,\dots,\dim(R)$.
