§ Ehrsmann connection

Here's my current understanding of how the Ehrsmann connection works. We have a GG-bundle GGPπMG \xleftarrow{\triangleleft G} P \xrightarrow{\pi} M. Let's first think of it as globally trivial so PM×GP \simeq M \times G. Now at each point mMm \in M, we have the fiber {m}×G\{ m \} \times G over mm. We now consider the kernel of the map π:Tm,gM×GTmM\pi^*: T_{m, g} M \times G \rightarrow T_m M. What are the elements here? We know that T(M×G)TMTGTMgT(M \times G) \simeq TM \oplus TG \simeq TM \oplus \mathfrak g where g\mathfrak g is the Lie algebra of the lie group GG. So we know that π\pi^* maps TpMgTpMT_p M \oplus \mathfrak g \mapsto T_p M. Thus the kernel of π\pi^* is going to be g\mathfrak g. This is called as the "vertical subspace" VpPker(π)TpPV_p P \equiv ker(\pi^*) \subseteq T_p P. Now, we have a choice in how we pick HpPH_p P for each point pPp \in P such that HpPVpP=TpPH_p P \oplus V_p P = T_p P. This choice of HpPH_p P is the connection. We claim that this choice is equally well encoded by a lie-algebra valued one form, ω:TPg\omega : TP \rightarrow \mathfrak g. That is, ω:TM×TGg\omega: TM \times TG \rightarrow \mathfrak g, which is ω:TM×gg\omega: TM \times \mathfrak g \rightarrow \mathfrak g. Intuitively, this tells us how much of the component along g\mathfrak g is not covered by the HpPH_p P. The idea is that since VpPHpp=TpPV_p P \oplus H_p p = T_p P, given any vector tpTpPt_p \in T_p P, I can compute tpvtpHp(tp)t^v_p \equiv t_p - H_p(t_p). Then I will have vpVpPv^p \in V_p P since I've killed the component in HpPH_p P. However, I know that VpPV_p P is the same as g\mathfrak g. Thus, I spit out the value tpvt^v_p, treated as an element of g\mathfrak g. This tells me how much of VpV_p is not stolen away by HpPH_p P in the decomposition. So we have the map πh:TpPVpP\pi_h: T_p P \rightarrow V_p P given by pih(tp)tpHp(tp)pi_h(t_p) \equiv t_p - H_p(t_p). The kernel of this map is HpP=ker(πh)H_p P = ker(\pi_h).

§ Generalizing to Non-trivial bundles

If we have a non-trivial bundle, then I need some way to link g\mathfrak g with VpPV_p P without splitting the bundle as I did here. The idea is that element of TpPT_p P are basically curves cp:IPc_p: I \rightarrow P. We use the curves to build derivations. For each lie algebra element aga \in \mathfrak g, I can build the curve cpa:IPc^a_p: I \rightarrow P given by cpa(t)pexp(at)c^a_p(t) \equiv p \triangleleft \exp(at). That is, the curve I get by pushing the point pp along aga \in \mathfrak g. Note that all the points in the curve apa^p lie on the same point in the base manifold, because the group only moves within fibers. So we have that π(p)=π(cpa(t))\pi(p) = \pi(c^a_p(t)) for all tt. This means that when we push forward the curve cpa(t)c^a_p(t), it represents the constant curve, which has zero derivative! Thus, we have that all these curves are in the kernel cpa(t)ker(π)c^a_p(t) \in ker(\pi^*), and hence gVpP\mathfrak g \subseteq V_p P. To show the other inclusion, pick some element vpVpPTpPv_p \in V_p P \subseteq T_p P. The group GG must be non-trivial, otherwise the bundle will also be trivial. Let pπ1(m)p \in \pi^{-1}(m) for some mm. Note that since the bundle is a principal bundle, we have that the fiber π1(m)\pi^{-1}(m) is a GG-torsor. I guess this is ismorphic as a group to GG. Now, the tangent space TpPT_p P is the tangent space the group GG, which has the same dimension as the lie algebra g\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 GG-torsor!