# Topology and holonomy of SU(2)xZ_2

• Apr 29th 2011, 01:54 PM
cduston
Topology and holonomy of SU(2)xZ_2
Hey all,
I am considering (nevermind why...) a gauge theory with symmetry group $G=SU(2)\times\mathbb{Z}_2$, and I want a little help getting all the notions sorted out. I am later going to extend $\mathbb{Z}_2$ to actually be an arbitrary finite group, but to make things simple let's stick with this.

First the topology: using the usual $\mathbb{S}^3$ topology on SU(2) (union of two overlapping 3-balls $U_1, U_2$) and the discrete topology on $\mathbb{Z}_2$, we can make G into a compact topological group using the product topology. But is it connected? In the product topology we have two open sets $(U_1\cup U_2,0)$ and $(\emptyset,1)$ which are disjoint and nonempty, so that means this is a disconnected space, right? It seems like it is LOCALLY connected, however.

Now the tricky part: Is it still a smooth manifold, like SU(2)? At first it seems like no, but if I want to make this a gauge group then we must be able to define a derivative (and holonomy) on it. So thinking just something like

$\mathrm{d}(g(x),e)=(\mathrm{d}g(x),e)$,

with the usual derivative. This certainly seems to satisfy Leibnitz so perhaps it is the correct thing. My question is can anyone see an obvious problem with how I define the derivative? I think it just corresponds to saying "G is a 1-parameter group because the $\mathbb{Z}_2$ part does not depend on the parameter". In other words,

$(g,e)(x)=(g(x),e)$.

So I guess the problem is that I'm not sure if I'm broken the group structure or not - to me it seems like no but I need some more intelligent eyes to verify that :-). However, if the above makes sense then I'm pretty ok with going on to define the holonomy and all that which is required for the rest of my analysis.

So, there are three questions up there, with my answers which I need to verify:
1. Is G connected? No.
2. Is G locally connected? Yes.
3. Is G a smooth manifold? Yes, under the construction above.

Thus, G is a disconnected but locally connected smooth manifold. Thoughts?
• Apr 30th 2011, 07:03 PM
Tinyboss
Crossing a manifold with Z2 (under the discrete topology) gives you a disjoint union of two copies of the original manifold. So, (1) it's certainly not connected, because Mx{0} and Mx{1} are open, disjoint, and cover the space. And (2) it's locally connected iff the original manifold is. And (3) it's a manifold in the obvious way, just as any disjoint union of (like-dimensional) manifolds is.
• May 1st 2011, 10:44 AM
cduston
Ok great, sounds like I'm the right track but can someone comment on my definition of the derivative? Basically just restricting the derivative to whichever factor is denoted by the finite element.

I guess I can now call $G$ a Lie group right? It's a group with a well-defined smooth structure?