§ Exact sequences for semidirect products; fiber bundles

§ Fiber bundles

In the case of a bundle, we have a sequence of maps FEBF \rightarrow E \rightarrow B where FF is the fiber space (like the tangent space at the identity TeMT_eM). EE is the total space (the bundle TMTM), and BB is the base space (the manifold MM). We require that the inverse of the projection π1:BE\pi^{-1}: B \rightarrow E locally splits as product π1(U)U×F\pi^{-1}(U) \simeq U \times F.

§ Semidirect products

In a semidirect product NKN \ltimes K, we have that NN is normal (because the fish wants to eat the normal subgroup NN / the symbol looks like NGN \triangleleft G which is how we denote normality). Thus, we can only quotient by NN, leaving us with KK. This is captured by the SES:
0NNKπK0 0 \rightarrow N \rightarrow N \ltimes K \xrightarrow{\pi} K \rightarrow 0
  • We imagine this as a bundle, with base space M=KM=K, bundle TM=NKTM=N \ltimes K,and fiber space (like, tangent space at the identity, say) TeM=NT_e M = N.
  • Furthermore, this exact sequence splits; So there is a map s:KNKs: K \rightarrow N \ltimes K(ss for "section/split") such that k,π(s(k))=k\forall k, \pi(s(k)) = k. To see that this is true,define s(k)(e,k)s(k) \equiv (e, k). Since all actions of KK fix the identity eNe \in N, we haves(k)s(k)=(e,k)(e,k)=(e,kk)=s(kk)s(k)s(k') = (e, k) (e, k') = (e, kk') = s(kk') so this is a valid map. To see that π\piis its inverse, just act π\pi; π(s(k))=π(e,k)=k\pi(s(k)) = \pi(e, k) = k.

§ Viewing the semidirect product space as a G-bundle

Consider the space ENKE \equiv N \ltimes K as a bundle over KK given by the projection ENKπKE \equiv N \ltimes K \xrightarrow{\pi} K. We can have NN act on the fibers by a left and a right action. Let's consider both:
  • NN acting on right: (n,k)(n,e)(nnk,ke)(nnk,k)(n, k) \triangleleft (n', e) \equiv (n n'^k, ke) \equiv (n n'^k, k)
  • NN acting on left: (n,e)(n,k)(nne,k)=(nn,k)(n', e) \triangleright (n, k) \equiv (n' n^e, k) = (n' n, k)This is the "easier action" to interpret;it permutes fibers, keeping the base space the same. So this gives a principal bundle action.
  • KK acting on right: (n,k)(e,k)(nek,kk)=(ne,kk)=(n,kk)(n, k) \triangleleft (e, 'k) \equiv (n e^k, kk') = (ne, kk') = (n, kk').This action is easy, it permutes fibers.
  • KK acting on left: (e,k)(n,k)(enk,kk)(nk,kk)(e, k') \triangleright (n, k) \equiv (e n^{k'}, k'k) \equiv (n^{k'}, kk').
So we see that NN acting on the left gives us an action that permutes inside fibers, and KK acting on the right gives us an action that permutes the fibers themselves. So we can write this as NNKKN \triangleright N \ltimes K \triangleleft K to capture the base-space bundle-space relationship, perhaps. Also, see that if we quotient NKN \ltimes K by the action of GNG\equiv N acting on the left, with the quotient map called [][\cdot] for orbit equivalence classes, we get NK[](NK)/N=KN\ltimes K \xrightarrow{[\cdot]} (N \ltimes K)/N = K, which is isomorphic to our starting picture NKπKN \ltimes K \xrightarrow{\pi} K. Hence, it is indeed true that this bundle is a principal GG-bundle.

§ Relationship to gauges

NOTE: this was written before I knew what a G-bundle is. This is perhaps easier to read, but less useful in hindsight. Let XX be the space of all states. Let OO be a group action whose orbits identify equivalent states. So the space of "physical states" or "states that describe the same physical scenario" is the orbit of XX under OO, or X/OX/O. Now, the physical space X/OX/O is acted upon by some group GG. If we want to "undo the quotienting" to have GG act on all of XX, then we need to construct GOG \ltimes O. GG is normal here because OO already knows how to act on the whole space; GG does not, so OO needs to "guide" the action of GG by acting on it. The data needed to construct GOG \ltimes O is a connection. Topologically, we have XX/OX \rightarrow X/O and GX/OG \curvearrowright X/O. We want to extend this to (GO)X(G \ltimes O) \curvearrowright X. We imagine this as:
*1| #1 | @1  X
*2| #2 | @2
*3| #3 | @3
  | |  |
  | v  |
* | #  | @ X/H
where the action of HH permutes amongst the fibers of *, #, @. Next, we have an action of GG on X/HX/H:
*1| #1 | @1  X
*2| #2 | @2
*3| #3 | @3
  | |  |
  | v  |
* | #  | @ [X/H] --G--> # | @ | *
We need to lift this action of H the H-orbits. This is precisely the data a connection gives us (why?) I guess the intuition is that the orbits of XX are like the tangent spaces where XX/OX \rightarrow X/O is the projection from the bundle into the base space, and the GG is a curve that tells us what the "next point" we want to travel to from the current point. The connection allows us to "lift" this to "next tangent vector". That's quite beautiful. We want the final picture to be:
*1| #1 | @1  X          #2| @2|                    
*2| #2 | @2    --G-->   #1|   |   
*3| #3 | @3             #3|   | 
  | |  |                  |   | 
  | v  |                  |   |  
* | #  | @ [X/H] --G--> # | @ | *