# Gauge invariance doesn't actually force usual choice of covariant derivative?

As we all know, a gauge invariant theory is of the form $$\mathcal{L} = \bar{\psi} \gamma^\mu \left( i\partial_\mu + A_\mu^a T^a\right) \psi.$$ The multiplet $$\psi$$ and gauge field $$A_\mu = A^a_\mu T^a$$ transform as follows under a gauge transformation: $$\psi \rightarrow G \psi, \quad \quad A_\mu \rightarrow G A_\mu G^{-1} - (\partial_\mu G) G^{-1},$$ where $$G(x)$$ is an element of the gauge group. I know that $$T^a$$ are supposed to be the generators of the gauge group, i.e. they are a basis for the associated Lie Algebra. However, it seems to me that actually this fact is not essential for $$\mathcal{L}$$ to be gauge invariant! I mean, any old matrices $$T^a$$ will do; we just say that under a 'gauge transformation' $$A_\mu$$ transforms as above.

So I am confused. Suppose I was ignorant and all I want to do is construct a gauge invariant theory. It would seem that I should be able to take any matrix (of the right dimension) for $$T^a$$, or am I wrong? If I am right then, from a constructionist point of view, what is the prime reason for choosing $$T^a$$ as we normally do?

• To clarify, are you asking why $A_\mu$ are members of the lie algebra associated to the gauge group? – J. Murray Apr 5 '20 at 21:59
• You seem to be suggesting that “any old matrices” are a basis for some Lie algebra. This is not the case. – G. Smith Apr 5 '20 at 22:03
• @J.Murray Yes please. Why must it be that $A_\mu$ is a general element of the associated Lie Algebra. – Rudyard Apr 5 '20 at 22:06
• @G.Smith No I am asking why does $A_\mu$ have to be from the Lie Algebra at all. My point is it doesn't seem necessary for gauge invariance. – Rudyard Apr 5 '20 at 22:06

## 2 Answers

Suppose that the lagrangian $$\newcommand{\cL}{{\cal L}} \newcommand{\opsi}{{\overline \psi}} \newcommand{\pl}{\partial} \cL=\opsi\gamma^\mu(i\pl_\mu+A_\mu)\psi \hskip2cm A_\mu := \sum_a A^a_\mu T^a \tag{1}$$ is invariant under gauge transformations$$^\dagger$$ $$\begin{gather} \psi\to G\psi \tag{2}\\ (i\pl_\mu+A_\mu)\to G(i\pl_\mu+A_\mu)G^{-1} \tag{3} \end{gather}$$ for all $$G$$ in some matrix group, where the $$T^a$$ are matrices of the same size. In order for this to make sense, the right-hand side must end up with the same matrices $$T^a$$ as the left-hand side, because the fields $$A_\mu^a$$ are the only things being transformed in the last equation. (The components of the matrices are just fixed coefficients in the Lagrangian, like the coefficient $$m$$ in a mass term.) This gives the requirements $$\begin{gather} G\pl_\mu G^{-1} = \text{linear combination of }T^a\text{s} \tag{4}\\ GT^a G^{-1} = \text{linear combination of }T^a\text{s}. \tag{5} \end{gather}$$ By taking $$G$$ to be infinitesimally close to the identity, equation (4) implies that $$G$$ is generated by the $$T^a$$s, and equation (5) implies that the commutator of two $$T^a$$s must be a linear combination of $$T^a$$s.

$$^\dagger$$ The second equation in (3) expresses how $$A_\mu$$ transforms. The partial derivatives on the right-hand side act on both $$G^{-1}$$ and whatever stands to the right of $$G^{-1}$$, just like the partial derivative on the left-hand side acts on whatever stands to the right of the closing parenthesis.

Since a gauge field is spacetime-dependent, a translation in spacetime is necessarily accompanied by some change in gauge. Thus, you can think of the gauge covariant derivative ($$D_\mu$$) as an infinitesimal spacetime translation ($$\partial_\mu$$) along with an infinitesimal transformation in gauge space (any other terms appearing in $$D_\mu$$). Then it should hopefully be more clear that the other terms in $$D_\mu$$ must somehow relate to the gauge group, and cannot be written in terms of completely arbitrary matrices.

To be more concrete, recall/note that an element $${g}$$ of a Lie group $$G$$ is given by $$\begin{equation} {g}=\exp(iA_\mu^aT^a)\,, \end{equation}$$ where $$A_\mu^a$$ are the continuous parameters of $$G$$, and $${T}^a$$ its generators. We can use a Taylor expansion to write this as: $$\begin{equation} {g}={I}+\sum_{n=1}^\infty \frac{1}{n!}\,(iA_\mu^a{T}^a)^n={I}+iA_\mu^a{T}^a+\mathcal{O}\left((A_\mu^a)^2\right), \end{equation}$$ where $$I$$ is the identity element. For the infinitesimal transformation one takes the leading term in the expansion ($$iA_\mu^a{T}^a$$), and this is what you see in your example: $$\begin{equation} \mathcal{L}={\bar\Psi}i\gamma^\mu {D}_\mu {\Psi}={\bar\Psi}i\gamma^\mu (\partial_\mu+iA_\mu^a {T}^a) {\Psi}\,. \end{equation}$$