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Let $P \rightarrow X$ be a principal $G$-bundle, and $P' \rightarrow X'$ be a principal $G'$-bundle. Let $(f',f'')$ be a morphism from $P'$ to $P$, i.e., a pair of maps $f': P' \rightarrow P$ and $f'': G' \rightarrow G$(a Lie group morphism) with $f'(u'a')=f'(u')f''(a'), \forall u'\in P', a'\in G'.$(Cf. Kobayashi & Nomizu, volume 1, p53.) Let $\omega$ be a connection (1-form with values in the Lie algebra of $G$ that is $ad$-invariant) on the bundle $P \rightarrow X$.

I want to know: under what condition the connection $\omega$ on the bundle $P$ can be lifted to a connection $\omega'$ on $P'$, i.e. $f''_*(\omega')=\omega$? Moreover, if this happens, then under what condition $\omega'$ is uniquely determined by $\omega$?

In Kobayashi & Nomizu, volume 1, pp79~83, there are some theorems concerning mappings of connections, but there is no one concern my question.

It seems to be true when both the maps $(f',f'')$ are covering maps, since in the book "Spin Geometry" by Lawson and Michelsohn, page 108, before proposition 4.ll, the authors seem to assume this kind of conclusion. I prefer the "most general" conditions!

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1 Answer 1

up vote 3 down vote accepted

Proposition II.6.2(b) in Kobayashi and Nomizu tells you where Lawson and Michelsohn's spin connection comes from. In this case $X = X'$, $f'$ is the map defining the spin structure, and $f''$ is the $2$-fold covering map $\mathrm{Spin}(n) \longrightarrow \mathrm{SO}(n)$. Kobayashi and Nomizu say that $f''$ should be an isomorphism of Lie groups, but in fact you only need that it induces an isomorphism of Lie algebras. Let me restate this better version of the proposition:

Proposition. Let $(f', f''): P'(M', G') \longrightarrow P(M,G)$ be a map of principal bundles such that $f^{\prime\prime}_\ast: \mathfrak{g}' \longrightarrow \mathfrak{g}$ is an isomorphism. Then for any connection one-form $\omega$ on $P$, there is a unique connection one-form $\omega'$ on $P'$ induced by $\omega$ and furthermore $$\omega'(X') = (f^{\prime\prime}_\ast)^{-1}((f^\ast\omega)(X')).$$

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Thank you very much! I have also this idea anyway, and the condition in you statement is equivalent to that the map $f'': G' \rightarrow G$ is in fact a a Lie group covering (by Lie group theory). –  Lao-tzu Jan 3 at 3:24

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