# Principal bundle over associated bundle

Let $$P$$ be a principal $$G$$ bundle. Let $$S$$ be a space with left action of $$G$$, and let $$Q$$ be a principal $$H$$ bundle over $$S$$ with the property that the action of $$G$$ can be lifted to $$Q$$.

Then $$P \times_G Q$$ is a principal $$H$$ bundle over $$P \times_G S$$ with projection map $$\pi([p,q])=[p,\pi_Q(q)]$$, where $$\pi_Q:Q \rightarrow S$$ is the projection of the principal bundle $$Q$$. One checks that $$\pi$$ is well-defined, i.e. $$\pi([pg^{-1},gq])=[pg^{-1},g \pi_Q(q)]$$, which holds because the $$G$$ action on $$Q$$ is lifted from $$S$$.

What ingredients do I need to get an induced connection on the principal $$H$$ bundle $$P \times_G Q \rightarrow P \times_G S$$?

I hope the answer to be similar to the ordinary case of connections on associated bundles, where every principal bundle connection induces a connection on the associated bundle. I played around with a principal bundle connection on $$P$$, and a principal bundle connection that is $$G$$-invariant on $$Q$$, but could not write down anything on the new bundle.

• I might be missing something, but isn't a connection on $P$ and a $G$-invariant connection on $Q$ enough? The connection on $P$ lets you split a tangent vector of $P\times_G S$ into horizontal and vertical parts, which you can lift to tangent vectors of $P\times_G Q$ by the connections on $P$ and $Q$, respectively (for the latter to be well-defined, you need $G$-equivariance). Apr 29, 2021 at 8:23
• How do you define the lifting? We have $T(P \times_G S)=TP \times _{TG} TS$. In order to be able to lift, it seems I would need connection on $Q$, say $\Psi : TQ \rightarrow VQ$, to be $TG$-equivariant. That is, for $\xi \in Lie(G)$ (actually $\xi \in TG$, but that's harder to write later), $V \in T_u Q$ have $\Psi(\xi \cdot V)=\Psi(V+\tilde{\xi}(u))$ be equal to $\xi \cdot \Psi(V)=\Psi(V)+\tilde{\xi}(u)$, where $\tilde{\xi}$ denotes the vector field induced by the $G$-action. But $\tilde{\xi}$ is not vertical, and it seems these things are never equal. Apr 29, 2021 at 10:24
• @BertramArnold I commented to hastily. I think I understand your construction on the level of horizontal subspaces now. If $H^P,H^Q$ are horizontal distributions on $P,Q$, then $H_{[p,q]}:=\{ [ h,0]:h \in H^P_p \} \oplus \{ [0,h]: h \in H^Q_q \} \subset T(P \times _G S)=TP \times_{TG} TS$ is a horizontal distribution on $P \times_G Q$. It should be well defined because of the $G$-invariance of $H^P$ and $H^Q$, even though I want to think more about this last point. Apr 29, 2021 at 13:00