Yang-Mills Geometry

Discussion in 'Physics & Math' started by QuarkHead, Sep 25, 2008.

1. QuarkHeadRemedial Math StudentValued Senior Member

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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 $\mathcal{A}$ 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.

Anyhoo, 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 $\text{Aut}\mathcal{A} \equiv G \subset GL(\mathcal{A})$ 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 sequence $g(A),\,\,g(g(A)),\,\,g(g(g(A)))$ 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.

Thus 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. If $\mathcal{A},\,\, G$ are manifolds (they are - recall that $G$ is a Lie group), does this imply the quotient is likewise? $G$ 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 $A^g \simeq G$, the structure group. 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. 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!!

3. temurman of no wordsRegistered Senior Member

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This is what I remember:

The usual definition of an orbit is that for a given group K, and a given element A, the K-orbit of A is all B such that B=kA for some k in K. Your definition corresponds to the case K is the group generated by the single element g. I think you have to consider G-orbits.

And it is not always $G=\mathrm{Aut}\,\mathcal{A}$, but G is a subgroup of the whole automorphism group.

Note that this $\mathcal{A}$ is not the space of potentials. Potential is an element of the Lie algebra of G.

Up to now you are still on a single fiber. Since G is the symmetry group, A/G is in some sense your "real degrees of freedom". This freedom is at each point of the spacetime manifold. So you need a base space, which would be the model of the physical spacetime (e.g. Minkowski space, but remember there is no metric at this point).

5. QuarkHeadRemedial Math StudentValued Senior Member

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temur: You are confusing me
Did I not make the definition $G \equiv \mathrm{Aut}(\mathcal{A}) \subset GL(\mathcal{A})$? How is what you wrote different to what I wrote?

I seriously doubt this is true - would you like to explain why you think this is so?

7. prometheusviva voce!Registered Senior Member

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I'm not a mathematician so if what I say makes things less clear it's probably because I have misunderstood your post, but I can confirm that the manifold upon which Yang Mills is defined is Minkowski. It is possible to define the theory on other, more complicated manifolds but it is very hard to extract anything out of it, and you're pretty much limited to a U(1) gauge group.

I approach it from the other way to you guys but the potential A, or as we crazy physicists like to call it the gauge field $A_\mu$ is indeed an element of the lie algebra of the gauge group. This is why there are as many gauge fields as there are generators of the gauge group, so for the case of QCD which is SU(3) YM + fermions there are 8 gauge fields, in other words there are 8 gauge bosons.

8. temurman of no wordsRegistered Senior Member

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What is the difference between $\mathrm{Aut}(\mathcal{A})$ and $GL(\mathcal{A})$ ?
I would like to, but it is going to take some time. Somebody might explain it here before I do.

9. QuarkHeadRemedial Math StudentValued Senior Member

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OK, temur and prometheus, it seems you are right after all - potential 1-forms = connection 1-forms are Lie algebra-valued, according to my texts here.

However, I now seem to have a major conceptual problem that you can help me out with.

I had always assumed that, given some Lie group $G$, with identity $e$, and the vector space $T_e(G)$ over the identity, then the Lie algebra $\mathfrak{G}$ was the set of all endomorphisms $T_e(G) \to T_e(G)$.

Specifically, the algebra on $T_e(G)$ is given by the commutators $[\quad, \quad]$ such that, for $X,\,Y,\, Z\in T_e(G)$ that $[X,Y] = \alpha Z$ (say) - this, as far as I know, defines the algebra $\mathfrak{G}$.

In other words, the elements in $T_e(G)$ are the objects, vectors, that the algebra $\mathfrak{G}$ acts upon, they are not (an essential) part of the algebra itself - they are place-holders. But we seem to have that the 1-form $A$ is a vector, not an algebraic operation like, say, $[\quad,\quad]$.

Let's look at this: define $[\quad,\quad]: \mathfrak{G \to G}$ by $[\quad,\quad]: [\quad,\quad] \to [\quad,\quad]$, which looks utterly insane, right?

So it does, that is, until we realize that $[\quad,Z]: [X,Y] \to [[X,Y],Z],\,\,\, [\quad,X]: [Y,Z] \to [[Y,Z],X],\,\,\, [\quad,Y]: [[Z,X],Y]$ and that, by the definition of the skew-symmetric commutator $[X,Y] = -[Y,X]$, etc, it follows (moderately) straightforwardly, that we have the essential element in our algebra $\mathfrak{G}$ called the Jacobi identity: $[[X,Y],Z] + [[Y,Z],X]+ [[Z,X],Y]=0$

So, if any of the above it true (is it?) , how can it be that the 1-form (vector, or co-vector, if you prefer) $A \in \mathfrak{G}$?

Did I fall off my bar-stool at some point?

10. temurman of no wordsRegistered Senior Member

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1,330
$T_e(G)$ is not just placeholders, it is (isomorphic to) the Lie algebra itself. Specifically, a vector space g is called a Lie algebra if we have a bilinear, antisymmetric operation [.,.] : g x g -> g satisfying the Jacobi identity.

An example of Lie algebra is the space of vector fields on a manifold with the bracket defined by [X,Y]f = X(Yf) - Y(Xf) for vector fields X, Y and a scalar field f.

If G is a Lie group, and a is its element, then one can define the left action of a on G by L(a) b = ab for any element b of G. A vector field on G is called left invariant if it is invariant under the push-forward action of L(a) for any a in G. The space of left invariant vector fields is isomorphic to the tangent space at any c in G and called the associated Lie algebra.