# Is the space of connections modulo gauge equivalence paracompact?

I find this question interesting, but need to get it out of my system: is the space of connections (modulo gauge) on a compact four-manifold paracompact, in the Sobolev topology?

If so, I believe it would admit partitions of unity, which would surely make life easier in gauge theory. But I haven't seen the experts make use of such a fact. I have also heard that spaces of curves (as used in symplectic geometry) do not always admit partitions of unity.

The question occurred to me while reading about the first of the "five gaps" described by the late Abbas Bahri: http://sites.math.rutgers.edu/~abahri/papers/five%20gaps.pdf

As Bahri himself points out, this so-called "gap" has been filled in several different ways (via the Freed-Uhlenbeck theorem or via holonomy perturbations; that is, if the original approach is indeed flawed). Still, I think it would be useful for the younger generation to understand the core of the difficulty. I had wondered if it can be summed up in a negative answer to this question.

Yes, the space of gauge orbits of connections is paracompact (even when you the use Fréchet topology).

First, the space of all connections is paracompact since it is an affine space modelled on a nuclear Fréchet space (and/or it is metrisable). Narasimhan & Ramadas (Geometry of SU(2) Gauge Fields) showed that the action of the group of gauge transformation is proper (they actually only consider SU(2) gauge theories over $S^3$ but their argument generalizes to arbitrary structure groups and arbitrary compact base manifolds). For proper group actions, the orbit space is Hausdorff and the canonical projection map is closed. Since the image of a paracompact Hausdorff space under a closed continuous map is also paracompact, we conclude that the orbit space is paracompact.

Moreover, the gauge orbit space is stratified by smooth (Fréchet) manifolds (it's the usual orbit type stratification, the top stratum being the space of irreducible connections, see for example The orbit space of the action of gauge transformation group on connections or On the gauge orbit space stratification: a review). Since every stratum is modelled on a nulear Fréchet space, they are even smoothly paracompact (see Convenient Setting of Global Analysis by Kriegl & Michor for further details, especially chapters 14 (Smooth Bump Functions) to 16 (Smooth Partitions of Unity and Smooth Normality)).

• This is a very good answer which will probably convince me after thinking it over. Meanwhile there are still the two subsidiary questions: 1) what is different for spaces of curves, and 2) is there a way to formalize the objection in the above note, from other than a constructivist perspective. – Alex Waldron Aug 22 '17 at 4:25
• I'm not sure what kind of problems you mean concerning the space of curves, but the space of maps $C^\infty(M, N)$ from a compact manifold to a finite-dim. manifold is also paracompact (see the book by Kriegl & Michor, section 42.3). – Tobias Diez Aug 22 '17 at 8:28

I assume that by Sobolev topology you mean the topology induced by the Sobolev norm. Since all normed spaces are metric spaces the affirmative answer to your question follows from the fact that all metric spaces are paracompact. See e.g. A new proof that metric spaces are paracompact by Mary Ellen Rudin (pdf).

edit

I missed that you are asking for connections modulo gauge equivalence. In that case I can refer you to Theorem 2 of arXiv:1012.3180 where it is proved* that the moduli space of all irreducible connections is locally Hausdorff Hilbert manifold and that the space of all irreducible connections forms a $\mathrm{Gau}$-principal bundle over it.

.* In the setting of Lie algebroids and their connections which subsume many classical cases.

• The question was about the quotient of this normed space by the gauge group, – Igor Belegradek Aug 19 '17 at 2:19
• The quotient is a Banach manifold (or Hilbert manifold?), away from the reducible connections (which are a major part of this question), but that does not imply paracompactness. – Alex Waldron Aug 19 '17 at 19:31
• @AlexWaldron It depends on the definition of Hilbert manifold. Sometimes metrizability / paracompactness is part of it (see e.g. map.mpim-bonn.mpg.de/Hilbert_manifold) but I suspect that this author just means manifold modelled on a Hilbert space. I should've included the word irreducible of course. Sorry for the omission. – Vít Tuček Aug 21 '17 at 8:59
• A stupid question: Is the action of the gauge transformations proper? An obvious reason why it wouldn't be is if any of the stabiliser groups are non-compact. – David Roberts Aug 21 '17 at 9:59