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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.

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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 $\Theta\in \Omega^1(G,\mathfrak{g})$).

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    $\begingroup$ Thank you, Ivo. $\endgroup$
    – Joao Vitor
    Oct 25 '20 at 17:27
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    $\begingroup$ My notes might be helpful. $\endgroup$
    – Ivo Terek
    Oct 25 '20 at 17:40

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