$0 \rightarrow N \rightarrow N \ltimes K \xrightarrow{\pi} K \rightarrow 0$

- 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 $\pi$; $\pi(s(k)) = \pi(e, k) = k$.

- $N$ 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: $(e, k') \triangleright (n, k) \equiv (e n^{k'}, k'k) \equiv (n^{k'}, kk')$.

```
*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--> # | @ | *
```