# Yang-Mills Geometry

1. Oct 24, 2008

Since it appears (so far) I am infringing no rule, here is another shameless copy/paste of a thread I started on another forum, where I didn't get too much help - rather, folk tried, but confused me even further! See if you guys can do better. (Note:I am not a physicist)

The mathematics here is not especially exotic, but I cannot get the full picture. As I am working from a mathematics, not a physics, text, I will lay it out roughly as I find it.

So. We start, it seems, with a vector space $$\mathcal{A}$$ of 1-forms $$A$$ called "potentials". Is it not the case that the existence of a potential implies the existence of a physical field? (I say "physical field" as I am having some trouble relating this to the abstract math definition - a commutative ring with multiplicative inverse, say).

Anyway, I am invited to consider the set of all linear automorphisms $$\text{Aut}(\mathcal{A}): \mathcal{A \to A}$$. It is easy enough to see this is a group under the usual axioms, so set $$\text{Aut}\mathcal{A} \equiv G \subseteq GL(\mathcal{A})$$ which is evidently a (matrix) Lie group thereby. This is apparently called the gauge (transformation) group.

Now for some $$g \in G$$, define the $$g$$-orbit of some $$A \in \mathcal{A}$$ to be all $$A',\,\,A''$$ that can be "$$g$$-reached" from $$A,\,\,A'$$, respectively. In other words, the (finite?) sequence $$g(A),\,\,g(g(A)),\,\,g(g(g(A))),....,g^n(A)$$ is defined. Call this orbit as $$A^g$$, and note, from the group law, that any $$A \in \mathcal{A}$$ occupies at least one, and at most one, orbit.

This induces the partition $$\mathcal{A}/G$$, whose elements are simply those $$A$$ in the same orbit $$A^g$$. Call this a "gauge equivalence".

Now it seems I must consider the orbit bundle $$\mathcal{A}(G, \mathcal{A}/G)$$.
Here I start to unravel slightly. By the definition of a bundle, I will require that $$\mathcal{A}$$ is the total manifold; no sweat, any vector space (within reason) is a manifold. I will also require that $$\mathcal{A}/G$$ is the "base manifold".
Umm. $$\mathcal{A},\,\, G$$ are manifolds (they are - recall that $$G$$ is a Lie group), does this imply the quotient is likewise? I think, not sure.....($$G$$ is the structure group for the total manifold, btw.)

But surely, this bundle can only be an "orbit bundle" if it is a principal bundle, i.e. the fibres are the orbits $$A^g$$ and $$A^g \cong G$$, the structure group. If this is so, will it suffice to note that this congruence is induced by the fact that each orbit $$A^g$$ is uniquely determined by $$g \in G$$?

Anyway, it seems that, under this circumstance, I may call the (principal?) orbit bundle the bundle of Yang-Mills connection 1-forms on the principal bundle $$P(G,M)$$, where I suppose I am now to assume that the base manifold $$M$$ is Minkowski spacetime, and that the structure group is again a Lie group (same one? Dunno)??

I'm sorry, but this is confusing me. Any other take on this would be most welcome - but keep it simple enough for a simpleton!!

2. Oct 24, 2008

### Haelfix

"does this imply the quotient is likewise?"

Im pretty sure you need some additional structure, namely the above is true so long as G contains all of its accumulation points. In general A/G can be made a manifold so long as G is a closed subgroup and we will have for the base space dimension (dim A- dim G) (I would call it a coset space).

So, A is the principle bundle with structure group G, and you will have a projection p : A --> A/G from the bundle space onto the base.

An example.. S^2 is the coset space SO(3)/SO(2) with dimension 3-1.

Btw theres something a little muddled with the last paragraph I think, I'll get back to it when I have more time.