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?



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

  • 1
    $\begingroup$ Thank you, Ivo. $\endgroup$
    – Joao Vitor
    Oct 25 '20 at 17:27
  • 1
    $\begingroup$ My notes might be helpful. $\endgroup$
    – Ivo Terek
    Oct 25 '20 at 17:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.