Yang-Mills Geometry: An Explainer for Simpletons

[QuarkHead]

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!

 

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

 

[Entropia]

I appreciate your effort in trying to understand Yang-Mills geometry from a mathematical perspective. It is indeed a complex and abstract concept, and it can be difficult to grasp without a strong background in both mathematics and physics. I will try my best to provide a response that may help you in your understanding.

Firstly, I would like to clarify that the use of the term "simpleton" in the title of the article is not meant to be derogatory, but rather to convey the idea that the author will explain the concept in a simplified manner for those who are not familiar with it.

Now, onto the concept of Yang-Mills geometry. You are correct in your understanding that the vector space \mathcal{A} represents the potentials of a physical field. In physics, a field is a physical quantity that has a value at every point in space and time. For example, the electric field is a vector field that has a value at every point in space and time. The existence of a potential for a field implies that the field can be derived from this potential, through a process called differentiation.

Moving on to the gauge group G, it is a group of transformations that can be applied to the potential A without changing the physical field that it represents. In other words, the field remains the same even after the potential has been transformed by an element of G. This is known as a gauge symmetry, and it plays a crucial role in the Yang-Mills theory.

Now, the concept of orbits and gauge equivalence can be a bit confusing, but think of it this way - the gauge group G acts on the potential A, and as a result, A is transformed into a new potential A'. This new potential represents the same physical field, but it is now in a different "gauge". The set of all possible gauges that A can be transformed into by elements of G is called the orbit of A. And the partition \mathcal{A}/G is simply a way of grouping together all the potentials that belong to the same orbit.

The orbit bundle \mathcal{A}(G,\mathcal{A}/G) is then a mathematical object that describes the relationship between the gauge group G, the potential space \mathcal{A}, and the set of gauge-equivalent potentials \mathcal{A}/G. It is a principal bundle, which means that it is a special type of bundle where the fibres (or "pieces")