## § Exact sequences for semidirect products; fiber bundles

#### § Fiber bundles

In the case of a bundle, we have a sequence of maps $F \rightarrow E \rightarrow B$ where $F$ is the fiber space (like the tangent space at the identity $T_eM$). $E$ is the total space (the bundle $TM$), and $B$ is the base space (the manifold $M$). We require that the inverse of the projection $\pi^{-1}: B \rightarrow E$ locally splits as product $\pi^{-1}(U) \simeq U \times F$.

#### § Semidirect products

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

#### § Viewing the semidirect product space as a G-bundle

Consider the space $E \equiv N \ltimes K$ as a bundle over $K$ given by the projection $E \equiv N \ltimes K \xrightarrow{\pi} K$. We can have $N$ act on the fibers by a left and a right action. Let's consider both:
• $N$ acting on right: $(n, k) \triangleleft (n', e) \equiv (n n'^k, ke) \equiv (n n'^k, k)$
• $N$ acting on left: $(n', e) \triangleright (n, k) \equiv (n' n^e, k) = (n' n, k)$This is the "easier action" to interpret;it permutes fibers, keeping the base space the same. So this gives a principal bundle action.
• $K$ acting on right: $(n, k) \triangleleft (e, 'k) \equiv (n e^k, kk') = (ne, kk') = (n, kk')$.This action is easy, it permutes fibers.
• $K$ acting on left: $(e, k') \triangleright (n, k) \equiv (e n^{k'}, k'k) \equiv (n^{k'}, kk')$.
So we see that $N$ acting on the left gives us an action that permutes inside fibers, and $K$ acting on the right gives us an action that permutes the fibers themselves. So we can write this as $N \triangleright N \ltimes K \triangleleft K$ to capture the base-space bundle-space relationship, perhaps. Also, see that if we quotient $N \ltimes K$ by the action of $G\equiv N$ acting on the left, with the quotient map called $[\cdot]$ for orbit equivalence classes, we get $N\ltimes K \xrightarrow{[\cdot]} (N \ltimes K)/N = K$, which is isomorphic to our starting picture $N \ltimes K \xrightarrow{\pi} K$. Hence, it is indeed true that this bundle is a principal $G$-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 $X$ be the space of all states. Let $O$ 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 $X$ under $O$, or $X/O$. Now, the physical space $X/O$ is acted upon by some group $G$. If we want to "undo the quotienting" to have $G$ act on all of $X$, then we need to construct $G \ltimes O$. $G$ is normal here because $O$ already knows how to act on the whole space; $G$ does not, so $O$ needs to "guide" the action of $G$ by acting on it. The data needed to construct $G \ltimes O$ is a connection. Topologically, we have $X \rightarrow X/O$ and $G \curvearrowright X/O$. We want to extend this to $(G \ltimes O) \curvearrowright X$. We imagine this as:
*1| #1 | @1  X
*2| #2 | @2
*3| #3 | @3
| |  |
| v  |
* | #  | @ X/H

where the action of $H$ permutes amongst the fibers of *, #, @. Next, we have an action of $G$ on $X/H$:
*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 $X$ are like the tangent spaces where $X \rightarrow X/O$ is the projection from the bundle into the base space, and the $G$ 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--> # | @ | *