Here's my current understanding of how the Ehrsmann connection works.

We have a $G$-bundle $G \xleftarrow{\triangleleft G} P \xrightarrow{\pi} M$. Let's first think of it as globally trivial so $P \simeq M \times G$. Now at each point $m \in M$, we have the fiber $\{ m \} \times G$ over $m$. We now consider the kernel of the map $\pi^*: T_{m, g} M \times G \rightarrow T_m M$. What are the elements here? We know that $T(M \times G) \simeq TM \oplus TG \simeq TM \oplus \mathfrak g$ where $\mathfrak g$ is the Lie algebra of the lie group $G$. So we know that $\pi^*$ maps $T_p M \oplus \mathfrak g \mapsto T_p M$. Thus the kernel of $\pi^*$ is going to be $\mathfrak g$. This is called as the "vertical subspace" $V_p P \equiv ker(\pi^*) \subseteq T_p P$.

Now, we have a choice in how we pick $H_p P$ for each point $p \in P$ such that $H_p P \oplus V_p P = T_p P$. This choice of $H_p P$ is the connection. We claim that this choice is equally well encoded by a lie-algebra valued one form, $\omega : TP \rightarrow \mathfrak g$. That is, $\omega: TM \times TG \rightarrow \mathfrak g$, which is $\omega: TM \times \mathfrak g \rightarrow \mathfrak g$. Intuitively, this tells us how much of the component along $\mathfrak g$ is not covered by the $H_p P$.

The idea is that since $V_p P \oplus H_p p = T_p P$, given any vector $t_p \in T_p P$, I can compute $t^v_p \equiv t_p - H_p(t_p)$. Then I will have $v^p \in V_p P$ since I've killed the component in $H_p P$. However, I know that $V_p P$ is the same as $\mathfrak g$. Thus, I spit out the value $t^v_p$, treated as an element of $\mathfrak g$. This tells me how much of $V_p$ is not stolen away by $H_p P$ in the decomposition.

So we have the map $\pi_h: T_p P \rightarrow V_p P$ given by $pi_h(t_p) \equiv t_p - H_p(t_p)$. The kernel of this map is $H_p P = ker(\pi_h)$.

§ Accessing the tangent space from the Ehrsmann Connection

• https://mathoverflow.net/a/94139/123769

§ Generalizing to Non-trivial bundles

If we have a non-trivial bundle, then I need some way to link $\mathfrak g$ with $V_p P$ without splitting the bundle as I did here. The idea is that element of $T_p P$ are basically curves $c_p: I \rightarrow P$. We use the curves to build derivations. For each lie algebra element $a \in \mathfrak g$, I can build the curve $c^a_p: I \rightarrow P$ given by $c^a_p(t) \equiv p \triangleleft \exp(at)$. That is, the curve I get by pushing the point $p$ along $a \in \mathfrak g$. Note that all the points in the curve $a^p$ lie on the same point in the base manifold, because the group only moves within fibers. So we have that $\pi(p) = \pi(c^a_p(t))$ for all $t$. This means that when we push forward the curve $c^a_p(t)$, it represents the constant curve, which has zero derivative! Thus, we have that all these curves are in the kernel $c^a_p(t) \in ker(\pi^*)$, and hence $\mathfrak g \subseteq V_p P$. To show the other inclusion, pick some element $v_p \in V_p P \subseteq T_p P$. The group $G$ must be non-trivial, otherwise the bundle will also be trivial. Let $p \in \pi^{-1}(m)$ for some $m$. Note that since the bundle is a principal bundle, we have that the fiber $\pi^{-1}(m)$ is a $G$-torsor. I guess this is ismorphic as a group to $G$. Now, the tangent space $T_p P$ is the tangent space the group $G$, which has the same dimension as the lie algebra $\mathfrak g$. Hence, the function we defined above must be surjective.

NOTE TO SELF: there should be a more direct proof that uses the fact that the fiber is $G$-torsor!