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Yang-Mills Geometry

  1. Oct 24, 2008 #1
    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 [tex]\mathcal{A}[/tex] of 1-forms [tex]A[/tex] 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 [tex]\text{Aut}(\mathcal{A}): \mathcal{A \to A}[/tex]. It is easy enough to see this is a group under the usual axioms, so set [tex]\text{Aut}\mathcal{A} \equiv G \subseteq GL(\mathcal{A})[/tex] which is evidently a (matrix) Lie group thereby. This is apparently called the gauge (transformation) group.

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

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

    Now it seems I must consider the orbit bundle [tex]\mathcal{A}(G, \mathcal{A}/G)[/tex].
    Here I start to unravel slightly. By the definition of a bundle, I will require that [tex]\mathcal{A}[/tex] is the total manifold; no sweat, any vector space (within reason) is a manifold. I will also require that [tex]\mathcal{A}/G[/tex] is the "base manifold".
    Umm. [tex]\mathcal{A},\,\, G[/tex] are manifolds (they are - recall that [tex]G[/tex] is a Lie group), does this imply the quotient is likewise? I think, not sure.....([tex]G[/tex] 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 [tex]A^g[/tex] and [tex]A^g \cong G[/tex], the structure group. If this is so, will it suffice to note that this congruence is induced by the fact that each orbit [tex]A^g[/tex] is uniquely determined by [tex]g \in G[/tex]?

    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 [tex]P(G,M)[/tex], where I suppose I am now to assume that the base manifold [tex]M[/tex] 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. jcsd
  3. Oct 24, 2008 #2


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    "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.
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