I understand that this is a long shot, especially because it's such a niche question but: has it been mathematically proven that (under sufficient smoothness conditions, etc.) any field configuration can (locally) be brought to the Lorenz gauge? Note that I'm not looking for uniqueness here, but existence.


Let $B^d \subset \mathbb{R}^d$ be a ball and let $E = B \times \mathbb{R}^p \rightarrow B$ have structure group $G \subset SO(p)$, and let $\nabla = d + A$ be a connection on $E$. Does there always exist a gauge transformation $$A \mapsto A' = M^{-1} A M + M^{-1} \nabla M$$ such that $\nabla^a A_a = 0$?

If not in general, under what conditions must be satisfied for this result to hold?


I know that in Gribov's original paper "QUANTIZATION OF NON-ABELIAN GAUGE THEORIES," he makes the statement:

We do not know any examples of situations of the type L”, where one cannot find a field $A$, with a certain divergence, which is gauge-equivalent to a given field $A$'.

The type L” situations referred to are precisely those in which the gauge orbits of a given field configuration do not intersect the Lorenz gauge condition specified above.

There is also a seminal result by Uhlenbeck ("Connections with Lp Bounds on Curvature") in which she proved that, so long as the original connection had a curvature that was sufficiently small (in a norm sense), then this gauge condition could always be satisfied. However, this is too weak, and it seems like the statement should have been strengthened in the last 40 years.

I've been doing a literature review for the past week or so and can't find anything suggesting that this has been mathematically proven along the same lines as Uhlenbeck proved it for the small curvature condition in the 80s.

What would be ideal is the following constraints:

  • $A$ is smooth
  • There exists a neighborhood in $B^d$ where the Coulomb gauge can be reached.

As mentioned above, non-uniqueness isn't a problem. But existence seems to be assumed everywhere, and while it's probably true, I can't find anyone who has proven it.



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