# 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 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 $$I,J=1,2,\dots,\dim(R)$$.

• Your detailed answer resolves my doubt! Thanks so much. Sep 2 '19 at 5:28