Timeline for Principal bundle over associated bundle
Current License: CC BY-SA 4.0
5 events
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| Apr 29, 2021 at 13:00 | comment | added | user505117 | @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 10:24 | comment | added | user505117 | 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 8:23 | comment | added | Bertram Arnold | 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 28, 2021 at 20:28 | review | First posts | |||
| Apr 29, 2021 at 6:44 | |||||
| Apr 28, 2021 at 20:24 | history | asked | user505117 | CC BY-SA 4.0 |