Yang-Mills Geometry

[Xerxes]

This is a shameless copy/paste of a thread I started on another forum, where I didn't get too much help. See if you guys can do better: Red Alert - I am NOT a physicist

 

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Dashed if I see what's going on here! The mathematics 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 [math]\mathcal{A}[/math] of 1-forms [math]A[/math] 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.

 

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

 

Now for some [math]g \in G[/math], define the [math]g[/math]-orbit of some [math]A \in \mathcal{A}[/math] to be all [math]A',\,\,A''[/math] that can be "[math]g[/math]-reached" from [math]A,\,\,A'[/math], respectively.

 

In other words, the sequence [math]g(A),\,\,g(g(A)),\,\,g(g(g(A)))[/math] is defined. Call this orbit as [math]A^g[/math], and note, from the group law, that any [math]A \in \mathcal{A}[/math] occupies at least one, and at most one, orbit.

 

Thus the partition [math] \mathcal{A}/G[/math] whose elements are simply those [math]A[/math] in the same orbit [math]A^g[/math]. Call this a "gauge equivalence".

 

Now it seems I must consider the orbit bundle [math]\mathcal{A}(G, \mathcal{A}/G)[/math].

 

Here I start to unravel slightly. By the definition of a bundle, I will require that [math]\mathcal{A}[/math] is the total manifold; no sweat, any vector space (within reason) is a manifold. I will also require that [math]\mathcal{A}/G[/math] is the "base manifold". Umm. [math]\mathcal{A},\,\, G[/math] are manifolds (they are - recall that [math]G[/math] is a Lie group), does this imply the quotient is likewise? [math]G[/math] is the structure group for the total manifold, btw.

 

I am now invited to think of the orbit bundle as a principal bundle, meaning the fibres [math]A^g \simeq G[/math], the structure group. Will it suffice to note that this congruence is induced by the fact that each orbit [math]A^g[/math] is uniquely determined by [math]g \in G[/math]

 

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 [math]P(G,M)[/math], where I suppose I am now to assume that the base manifold [math]M[/math] is Minkowski spacetime, and that the structure group is again a Lie group (same one? Dunno)??

 

I'm sorry, but this is confusing me.

 

Now I want to ask if the connection bundle is trivial, i.e. admits of global sections, but this is already far too long a post, so I will leave it. Sufficient to say that a chap called Gribov pops in somewhere around here.

 

Any other take on this would be most welcome - but keep it simple enough for a simpleton!!

 

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