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I’m tackling physics recreationally from a pure math perspective.

Right now I’m looking at just the outline of gauge theory. The Wikipedia article explains that gauge fields correspond to generators of the Lie algebra of the Lie group the Lagrangian is invariant under. And then gauge bosons are the quanta of these fields, so for example there are eight gluons since SU(3) is eight dimensional. Cool! But what I don’t get is the intermediate step from generators of the Lie algebra to gauge fields.

A gauge field is mathematically a connection on a principal bundle, which is a Lie algebra valued 1-form satisfying some conditions. How do these correspond to generators of the Lie algebra? Here are my thoughts:

Let $\pi: P \to M$ be a principal bundle. Let $\omega$ be a principal connection on $P$. Let $\phi: U \times G \to \pi^{-1}(U)$ be a local trivialization of $P$. Then $s(x)=\phi^{-1}(x,e)$ defines a section and $A=s^*\omega$ is a $\mathfrak{g}$-valued 1-form on $U$ Now we can write $A(x)=\sum c_i(x)T^i$ where $c_i$ is a 1-form on $U$ and the $T^i$ form a basis for $\mathfrak{g}$. Are the $c_i$ what we mean by the gauge fields corresponding to the generators of the Lie algebra? Or would it be the whole $c_iT^i$ terms? It seems more likely that it’s the latter.

Writing it out like this I suppose you could do the same thing on $P$. Just take the component 1-forms or project onto those subspaces. If this is the case then the projections would still have to satisfy the axioms of connections on $P$. And maybe that is really obvious to see but I’m too deep in speculation to verify it if it’s the case or not.

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  • $\begingroup$ Every $c_i$ is gauge field. For instance, electromagnetism has the Abelian algebra $u(1)$ and therefore only one gauge field which is associated to the photon. On the other hand, electroweak force has gauge algebra $su(2)$ hence three gauge fields associated to the particles $W^+$, $W^-$ and $Z^0$. $\endgroup$ – Diracology Jul 24 '20 at 23:08
  • $\begingroup$ “Every $c_i$ is a gauge field”. Cool. So there is some discrepancy in terminology in that a gauge field can refer to the component of a Lie algebra valued 1-form (this component being a vanilla non-Lie-algebra valued 1-form) and not just a connection on a principal bundle. $\endgroup$ – Robin Jul 24 '20 at 23:38
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The language is loose. If you press an expert to specify exactly which piece of the principal-bundle picture is called "the gauge field," you might get different answers on different days of the week.

  • One answer is... yes, all of it.

  • Sometimes we might call the $c_i$ the components of the gauge field.$^\dagger$ The idea is that the generators $T^i$ are fixed. They're part of the definition of the type of field (which is part of the theory's definition), whereas the $c_i$ are used to specify the configuration of the field (which is what we're specifying when we specify an initial state, at least in classical field theory).

  • Sometimes we might call the whole Lie-algebra-valued one-form the gauge field, because this is a natural way of packaging the components $c_i$ into a single object. (Mathematically, it's the other way around — the one-form is defined first — but physicists sometimes think in terms of components, viewing the one-form as a nice way of packaging those components.)

  • My own go-to perspective is lattice gauge theory. Replacing continuous space with a lattice reduces most of the fancy math to elementary calculus. (That's for fields having a continuum of possible values at each point in space. For discrete-valued fields, it's elementary arithmetic.) That's hugely empowering for simple-minded non-mathematicians like me. In lattice gauge theory, a gauge field is just an assignment of one element of the gauge group (by "gauge group" I mean $SU(3)$, for example) to each nearest-neighbor pair $x,y$ of lattice sites. We can think of the group element as $\exp(i\, c_k(x,y)T^k)$, or we can just think of it as an element of the group. The continuous-space version of this perspective is to think of the gauge field as the connection on the principal bundle: given a path in the base manifold and a "starting value" from the group at one end of the path, the gauge field (connection) tells us what the "ending value" will be at the other end of the path. On a lattice, this calculation amounts to multiplying the group elements along the path, in the order specified by the path. Even I can do that.

Those are my Friday answers. If you ask me on Saturday, then I might say something different. The important message is that the language is loose. The assertion "gauge fields correspond to generators of the Lie algebra" would be a little too loose if "correspond to" means "are," but if "correspond to" means "are associated with," then I guess it's okay. Whether we count generators or count components, we get the same number: $SU(3)$ has eight of them.

$^\dagger$ The gauge field has two indices: one that specifies which gauge-group generator the component is tied to, and a spacetime index. So an $SU(3)$ gauge field has eight components for each spacetime component. (Ugh, language can be painful.) They're all regarded part of a single "gauge field," at least on Fridays.

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