# On the choice of the horizontal space of a principal bundle.

Let $$(P, M, \pi, G)$$ be a (smooth) principal G-bundle with $$M$$ being a smooth manifold and G a Lie group. Given $$p \in P$$, there is a natural vertical space, defined by

$$V_{p}P := \operatorname{ker}(\pi_{*}),$$ where $$\pi_{*}: T_{p}P \rightarrow T_{\pi(p)}M$$ is the pushforward.

I saw in the books that there is no direct choice for horizontal space. Is this reflected in the arbitrary $$\pi$$ or even with the fixed $$\pi$$ projection, is it possible to choose more than one horizontal complement? If it is the latter, could anyone give a concrete example of two choices of horizontal spaces for a vertical space?

Appreciate.

The idea is that given $$p \in P$$, $${\rm Ver}_p(P) = \ker {\rm d}\pi_p$$ is a vector subspace, but it admits infinitely many complements. For instance, assume that we have a trivial principal $$G$$-bundle $$M\times G \to M$$. Then we have $${\rm Ver}_{(x,g)}(M\times G) = \{0\}\times T_gG.$$One choice of horizontal spaces is $${\rm Hor}_{(x,g)}(M\times G) = T_xM\times \{0\}.$$But you can also choose any right-invariant Riemannian metric on $$M\times G$$ and take $${\rm Hor}_{(x,g)}(M\times G)$$ to be the orthogonal complement of $${\rm Ver}_{(x,g)}(M\times G)$$. If the subspaces $$T_xM\times \{0\}$$ and $$\{0\}\times T_gG$$ are not orthogonal, then this gives a different horizontal distribution to the standard one.
In fact, for any principal $$G$$-bundle $$P \to M$$, a horizontal distribution is equivalent to a choice of $$1$$-form $$A \in \Omega^1(P,\mathfrak{g})$$ such that $$A(X^\#)=X$$ for all $$X\in \mathfrak{g}$$ (where $$X^\#\in\mathfrak{X}(M)$$ is the action field of $$X$$) and $$R_g^*A = {\rm Ad}(g^{-1})\circ A$$ for all $$g \in G$$ (such an $$A$$ is called a connection $$1$$-form). The space of connection $$1$$-forms for $$P\to M$$ is an affine space whose translation space consists of all $$B \in \Omega^1(P,\mathfrak{g})$$ such that $$B(X^\#)=0$$ for all $$X\in\mathfrak{g}$$ and $$R_g^*B = {\rm Ad}(g^{-1})\circ B$$ for all $$g \in G$$, so there's a reasonable amount of freedom, except in very particular cases (such as $$G \to \{{\rm pt}\}$$, where the only connection $$1$$-form is the Maurer-Cartan form $$\Theta\in \Omega^1(G,\mathfrak{g})$$).