I am playing around with coupling a classical $SU(2)$ Yang-Mills theory to Einstein's equations.

Assuming spherical symmetry, the $SU(2)$ connection can be written \begin{equation} A = \omega(r)\tau_1 d\theta + \omega(r)\sin\theta \tau_2 d\theta + \cos\theta \tau_3 d\phi,\tag{1} \end{equation} where the $\tau_i$ are the generators of the $\mathfrak{su}(2)$ algebra.

A static spherically symmetric metric has the form \begin{equation} ds^2 = -T^{-2}(r)dt^2 + B^{-1}(r)dr^2 + r^2d\theta^2 + r^2\sin^2\theta d\phi^2.\tag{2} \end{equation}

The Yang-Mills equations are \begin{equation} D\star F = 0,\tag{3} \end{equation} along with the Bianchi Identity \begin{equation} DF = 0.\tag{4} \end{equation}

Clearly, $D$ contains the usual gauge covariant exterior derivative \begin{equation} DF = dF + [A\wedge F]\tag{5} \end{equation} with respect to the $SU(2)$ connection.

Here is where my question arises: since the Yang-Mills field lives in curved spacetime, shouldn't the gauge covariant exterior derivative include additional terms which describe the usual covariant derivative of $F$ with respect to the Levi-Civita connection on the spacetime manifold?

  • $\begingroup$ A gauge theory is a principal bundle over some smooth manifold, in particular it doesn't depend on the Riemannian structure (connection, metric, etc) on the base manifold, therefore no spacetime curvature doesn't change the connection or covariant derivative in your gauge theory. $\endgroup$
    – zzz
    Aug 3, 2015 at 18:39
  • $\begingroup$ You can also note that in general the covariant derivative is completely fixed by the principal bundle connection. $\endgroup$
    – zzz
    Aug 3, 2015 at 18:41
  • $\begingroup$ Regardless of the above yes a coupling to a background in your lagrangian will introduce a dependence on the Riemannian metric in the dynamical part of the yang mills equations, namely the first yang mills equation will probably not look like that anymore. Still, the covariant derivative won't change. $\endgroup$
    – zzz
    Aug 3, 2015 at 18:46
  • $\begingroup$ @bechira I think the difference will be the gauge covariant derivative will then contain a spin connection term, in addition to the principal bundle connection. $\endgroup$
    – Ryan Unger
    Aug 3, 2015 at 19:23
  • $\begingroup$ @bechira Thanks for your comments. I phrased my question a bit poorly when I spoke of the need to modify the gauge covariant derivative. Obviously the structure of the base manifold will not affect the gauge covariant derivative on the bundle over that manifold. I agree that the Yang-Mills field equations will be modified. I will work on deriving them from the Lagrangian. $\endgroup$
    – Evan Rule
    Aug 3, 2015 at 21:11

2 Answers 2


OP is considering Yang-Mills theory over a curved base space $(M,g)$. If the base space connection is the Levi-Civita connection $\nabla^{LC}=\partial+\Gamma$, then it doesn't matter whether one uses the gauge-covariant derivative $D=\partial+A$ or the full covariant derivative $\nabla=D+\Gamma$ since the Christoffel symbols $\Gamma$ drops out of the Yang-Mills theory and OP's eqs. (3), (4) and (5). This is mainly due to the torsionfreeness of the Levi-Civita connection $\nabla^{LC}$.


Levi-Civita connection makes its appearance only when some quantity is covariant under local Lorentz transformations. But if everything is invariant under local Lorentz transformations, the Levi-Civita connection is irrelevant. The metric and YM fields (which are of concern in Yang-Mills theory in curved space time) are invariant, while Riemann curvature is covariant (which is of concern in general relativity). That is why Christoffel symbol drops out of the Yang-Mills theory and OP's eqs. (3), (4) and (5). This fact has nothing to do with torsionfreeness condition of the Levi-Civita connection as the other answer is suggesting. It's true even under non-zero torsion condition.

A side note: Bianchi Identity is the same regardless of flat or curved space time, since metric does not show up in Bianchi Identity. On the otherhand, the Hodge dual $\star$ in Yang-Mills equation is metric-dependent.


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