Lifting Riemannian metrics on principal bundles

Question

Given a principal bundle $\pi:M\rightarrow M/G$, there are natural maps $$\pi_{\mathcal{F}}:\mathcal{F}(M)^G\rightarrow\mathcal{F}(M/G)$$ $$\pi_{\mathfrak{X}}:\mathfrak{X}(M)^G\rightarrow\mathfrak{X}(M/G)$$ $$\pi_{\mathcal{R}}:\mathcal{R}(M)^G\rightarrow\mathcal{R}(M/G)$$ from $G$-invariant functions ($\mathcal{F}$), vector fields ($\mathfrak{X}$) and Riemannian metrics ($\mathcal{R}$) on $M$ to the corresponding objects on the base space $M/G$.

The pull-back map $\pi^*:\mathcal{F}(M/G)\rightarrow\mathcal{F}(M)^G$ and $\pi_{\mathcal{F}}$ are mutual inverses. Also, fixing a section $s:M/G\rightarrow M$ of $\pi$, I know how to construct a section $\pi^*_s:\mathfrak{X}(M/G)\rightarrow\mathfrak{X}(M)^G$ of $\pi_{\mathfrak{X}}$.

My question is:

How can one construct sections of $\pi_{\mathcal{R}}$?

Given a section $s$, any point of $M$ is of the form $\psi_g(s(\pi(x)))$ for some $g\in G$ and $x\in M$ (here $\psi_g$ is the diffeo on $M$ defined by the action of $g$). From $s$ we get a connection on $M$ with horizontal spaces $$H_{\psi_g(s(\pi(x))}=d_{\pi(x)}(\psi_g\circ s)(T_{\pi(x)}(M/G)).$$ This is sufficient to construct the pull-back for vector fields, because a vector field on $M/G$ is lifted to a horizontal vector field on $M$, which is then projected back by $\pi_\mathfrak{X}$ to the original vector field.

For Riemannian metrics it seems that something else is necessary because lifting a metric on $M/G$ tells us how the metric on $M$ looks like when restricted to horizontal spaces, but we need more than that:

How is the metric defined for non-horizontal vectors?

Answer 1

1. You're absolutely correct about $\pi_{\mathcal{F}} : \mathcal{F}(M)^G \to \mathcal{F}(M/G)$ and $\pi^\ast : \mathcal{F}(M/G) \to \mathcal{F}(M)^G$ being inverses.
2. A better way to find a section of $\pi_{\mathfrak{X}} : \mathfrak{X}(M)^G \to \mathfrak{X}(M/G)$ is via a principal connection. Recall that in the Ehresmann picture, a principal connection for $M \to M/G$ is a $G$-equivariant bundle endomorphism $\Pi : TM \to TM$ such that $\Pi^2 = \Pi$ and $$ \operatorname{im}(\Pi) = VM := \ker(d\pi : TM \to T(M/G)); $$ in other words, it's a $G$-equivariant projection onto the vertical subbundle $VM$ of $TM$. Then, given a principal connection $\Pi$, the natural map $\pi_{\mathfrak{X}} : \mathfrak{X}(M)^G \to \mathfrak{X}(M/G)$ actually restricts to an isomorphism $$ \pi_{\mathfrak{X}} : \mathfrak{X}_H(M)^G \to \mathfrak{X}(M/G), \quad \mathfrak{X}_H(M) := \{\xi \in \mathfrak{X}(M) \mid \Pi\xi = 0\}; $$ if you like, you get a bundle isomorphism $$ (HM)/G \to T(M/G), \quad HM := \ker(\Pi : TM \to TM) $$ covering the identity on $M/G$, where $HM$ is called the horizontal subbundle.
3. In light of 2., any Riemannian metric $g_0$ on $M/G$ canonically lifts to a $G$-invariant Euclidean metric $g_h$ on the horizontal subbundle $HM$ of $TM$. To get a $G$-invariant Riemannian metric on $M$, i.e., a $G$-equivariant Euclidean metric on $TM = VM \oplus HM$, just pick a $G$-invariant Euclidean metric $g_v$ on $VM$ and form $g = g_v \oplus g_h$ on $TM = VM \oplus HM$; a typical way to do this is to use the $G$-equivariant isomorphism $VM \cong M \times \mathfrak{g}$ for $\mathfrak{g}$ the Lie algebra of $G$, and then let $g_v$ be the pullback to $VM$ of the Euclidean metric on $M \times \mathfrak{g}$ defined by your favourite bi-invariant Riemannian metric on $G$.

So, to answer your final question, a section of $\pi_{\mathcal{R}} : \mathcal{R}(M)^G \to \mathcal{R}(M/G)$ is induced by a choice of principal connection $\Pi$ on $M \to M/G$ and of $G$-invariant Euclidean metric on $VM := \ker(\Pi)$; in particular, a section of $\pi_{\mathcal{R}}$ is induced by a choice of a principal connection on $M \to M/G$ and a bi-invariant Riemannian metric on $G$.