6
$\begingroup$

Gauging a global symmetry $G$ introduces several free parameters to the theory. For example,

  • In $d=4$, gauging a simple and simply-connected Lie group introduces a coupling constant and a theta term, $(g,\theta)\in\mathbb R\times U(1)$.

  • In $d=3$, gauging a simple and simply-connected Lie group introduces a coupling constant and a Chern-Simons term, $(g,k)\in\mathbb R\times\mathbb Z$.

Gauging semi-simple Lie groups just adds more copies of these terms. Non-simply connected groups usually have extra theta terms (which may be integral instead of continuous). Similar story for discrete symmetries.

What is the mathematical object that controls these parameters? I would expect some map $\mu\colon\mathbf{Diff}\times \mathbf{Group}\to\mathbf{Ring}$ or something like that (with e.g., $\mu(M_4,SU)=\mathbb R\times U(1)$, as above). Is this map well-defined at all? If so, does it have a name/has it been studied?

$\endgroup$
  • $\begingroup$ @DanYand Good point. Yes indeed, I'm thinking of renormalizable terms only. Or relevant. Or whatever makes the problem well-defined. I'm sure there must be something -- I know people have been thinking about this kind of problems. Thank you for pointing it out and for the interest! $\endgroup$ – AccidentalFourierTransform Mar 1 at 1:59
  • $\begingroup$ I’m not quite what you mean here. It looks like you are confusing the notion of classifying principal bundles, the geometric language of gauge theory, with the physical notion of gauging a global symmetry group as in Yang-Mills theories. $\endgroup$ – Mozibur Ullah Mar 1 at 17:48
  • $\begingroup$ The examples that you gave include minimal couplings and topological terms. Is there any other object or term that you consider as gauging? A second remark: gauge symmetries appear in effective low energy limits which are not renormalizable in general. I think that this is the physically more interesting case. $\endgroup$ – David Bar Moshe Mar 3 at 11:21
0
$\begingroup$
  1. We do not really "gauge global symmetries". It is a common turn of phrase, but it does not capture what gauge theories are nor how they arise, see e.g. this answer of mine. However, this is irrelevant for the rest of your question.

  2. There is no dependence of the coupling constant on the group. The coupling constant really just arises because when you introduce the minimal coupling by replacing the ordinary derivative $\partial_\mu$ by the gauge covariant derivative $\partial_\mu + \lambda A_\mu$ (modulo factors of $\mathrm{i}$), you can put any $\lambda\in\mathbb{R}$ in there. This doesn't depend on the gauge group, it's simply how minimal coupling works. If you couple your gauge field to your matter sector non-minimally, you may incur more or less coupling constants. The constants are a function of the coupling prescription, not of the group.

  3. The $\theta$ and Chern-Simons terms are topological terms. The Chern-Simons terms occur in odd dimensions and are so-called secondary characteristic classes with values in the gauge group, the $\theta$-terms in even dimensions are the integrals of the top-dimensional Chern classes of the principal $G$-bundle of the theory. That the Chern class can be expressed as an invariant polynomial form in the curvature is the content of the Chern-Weil homomorphism.

    The level $k$ of a Chern-Simons term is usually discrete and its allowed values indeed depend on the gauge group. The physicist determines the allowed values of $k$ by computing for which values the Chern-Simons action is actually gauge invariant, see e.g. this answer for an explicit such computation. For some mathematicians, the level is a certain element in the integral cohomology of the classifying space of the gauge group.

    The $\theta$ in front of the Chern class, however, is just another coupling constant, unless you promote it to a dynamical field like the axion in Peccei-Quinn theory.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.