# Discrete gauge theories

I'm trying to understand a particular case of gauge theories, namely discrete spaces on which a group $$G$$ can act transitively, with a gauge group $$H$$ which is discrete as well.

From what I've already read about continuous and discrete gauge theory (I can provide the necessary references if needed*), it seems that the full gauge transformation group will be the semidirect product of $$G$$ by $$H$$. However, I'm unable to prove it. Can someone point me to the proof of this fact?

*Edit: The bottom of page 2 in this reference describes this kind of thing without giving any proof: https://arxiv.org/abs/1106.2759

• For those who are interested in the question, the OP asked something similar on math.SE the other day: math.stackexchange.com/q/55006 Aug 3, 2011 at 20:13
• Indeed... I first thought it was more mathematic-relevant, but it didn't get much attention. I thought physicists would be more familiar with it... By the way, this semidirect product (even in the case of continuous groups) seem to have been lying all over the place for more than 50 years now, but an explicit proof is still missing (or obviously implicitly implied and in that case I missed it) Aug 3, 2011 at 20:42
• Have you checked out Ref. 1 from that paper? I can't acccess any articles behind a paywall ATM, but it seems like it could be useful. Aug 4, 2011 at 8:35
• @Gerben: I have a lot of papers by the same Kornyak, but in all of them he merely gives the same statement without any proof Aug 4, 2011 at 9:07
• In the case of continuous principal bundles, the bundle automorphism group is an extension of the base space diffeomorphism group by the group of pure gauge transformations. A semidirect product is only a special case of such an extension. Going to the discrete case, one would also expect more general extensions than the semidirect product. Since nontrivial extensions (i.e., non-semidirect product extensions) exist for finite groups, I wonder why these cases are excluded here. Do you have a reference where it is stated that the full gauge group "must" be a semidirect product. Aug 4, 2011 at 11:39

The physical space is a space of orbits under $H$. That is, you start with some huge initial space $X$ but a little later you realize that many states represent the same physical state - any two of these are connected through the action of $H$. So, the physical space is $X / H$. Now, the physical space can be acted on by a symmetry group $G$. Now, to extend this to an action of $G \rtimes H$ on $X$ we need a connection.
Why is that? Well, we know how $H$ acts on all of $X$ and we know how $G$ acts on orbits of $X$ but we have no idea what does $G$ to the points in the orbit (besides the fact that they get all mapped to points in some other orbit). Connection is precisely the tool that gives us this identification of points between various orbits.
In the language of topology (all your sets and groups are finite and so we can give them discrete topology), $X \to X / H$ is a fibre bundle with vertical action of $H$. We are also given an action of $G$ on the base. And we'd like to lift it to a horizontal action of $G$ on all of $X$ but horizontal structure on the bundle is the same thing as connection. Now, once we have actions of both $G$ and $H$ on $X$, we can combine these two in an obvious way by twisting the horizontal action by the vertical action (similarly to how one can twist translations by rotations to get a full group of symmetries of Eucidean space).
• Marek: since you seem to understand what's going on, I think it'd be useful - not only to the OP, but for future reference - if you could explicitly construct this group and show how it acts on a configuration. (Also, most physicists are more comfortable with $x \mapsto x + a$ and $\phi \mapsto U \phi$ than $X \rightarrow X/H$...) Aug 5, 2011 at 0:20
• @Marek can you please explain to me why it is natural to consider the semidirect product $G \rtimes H$, and not some other way of mixing the two groups? Such as say, the direct product? Feb 11, 2020 at 16:42