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The formal treatment of the gauge covariant derivative in most reference texts for students is too informal and too ad hoc, so that some general issues remain unclear. For example, the gauge covariant derivative applies to scalar or spinor fields, rarely to vector fields over the manifold $\mathbb{R}^4$. Under these conditions, it is not clear to me whether in fact the gauge covariant derivative is a particular case of covariant derivative or a construction that shares somo analogical properties, but is not really a covariant derivative.

In particular, the fact that even if $\mathbb{R}^4$ is a real differentiable manifold, the covariant derivative on a Cartesian coordinat chart is defined as:

$$D_\mu (\cdot) = \partial_\mu(\cdot) + ie A_\mu(x)(\cdot)$$

however, when this operator is applied to a real or complex function, it does not result in a real tangent vector. My questions are:

  1. Is the gauge covariant derivative a covariant derivative, in the sense of the differential geometry of ordinary differential manifolds?
  2. Does the use of complex numbers imply that in reality the underlying variety is not real but must be understood as a type of complexification $\mathbb{C}\otimes\mathbb{R}^4$ of the Minkowskian space $(\mathbb{R}^4,\eta)$?
Qmechanic
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Davius
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2 Answers2

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  1. The gauge covariant derivative is a genuine covariant derivative in the ordinary sense of differential geometry, but in the general sense of a (principal) connection form $A$ inducing covariant derivatives on its associated bundles, not in the narrow sense of being a covariant derivative on the tangent bundle like the covariant derivative associated with the Levi-Civita connection. See also this answer of mine for a discussion how to formalize gauge theory in terms of principal bundles.

  2. No, the $\mathrm{i}$ is merely an artifact of the physics convention that the generators of Lie groups should be Hermitian, not skew-Hermitian: If you write a $X\in\mathrm{SU}(N)$ as $X = \mathrm{e}^{T}$, then $T$ is skew-Hermitian, but Hermitian if you choose $X = \mathrm{e}^{\mathrm{i}T}$. This has nothing to do with the spacetime manifold: The fields $\phi$ that transform non-trivially under the gauge group take values in some associated bundle whose fibers are a representation $(V,\rho)$ (possibly tensored with a spinor or tangent bundle if they are additionally non-scalars under spacetime transformations, but that transformation behaviour is entirely irrelevant for the gauge theory) of the gauge group and a mathematician would probably prefer to write $$ D_\mu \phi = \partial_\mu \phi + e\rho(A_\mu)\cdot\phi,$$ where I misused $\rho$ to also denote the induced representation of the Lie algebra. Whether $V$ is a real or complex vector space is immaterial at that point.

ACuriousMind
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Whether space-time symmetries (as in GR) or internal symmetries (as in Yang-Mills) are gauged, covariant derivatives are defined such that $D_\mu U(g) X = U(g) D_\mu X$.

To get a feel for this, I would recommend looking at non-abelian Yang-Mills theory. Maxwell theory is deceptive because the adjoint of $U(1)$ is trivial which mean vector fields happen to be gauge singlets and covariant derivatives happen to act on them as $\partial_\mu$. The nonabelian literature is also where you might see more pleasing reality properties. The presence of $i$ reflects a choice to use Hermitian instead of anti-Hermitian generators. One goes between them by multiplying all generators by $i$ (not a subset which would change the real form of the gauge group). The presence of $g$ is conventional for the same reason so I'm pretty sure mathematicians don't use it. When they write a Yang-Mills kinetic term, it is always $\frac{1}{2g^2} \mathrm{tr} F_{\mu\nu} F^{\mu\nu}$, not the $-\frac{1}{4} F_{\mu\nu} F^{\mu\nu}$ which electrodynamics textbooks always write.

Connor Behan
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