

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer

Let $E\rightarrow M$ be an orientable vector bundle of rank n equipped with some Riemannian metric, $P:=F_{SO(n)}(E)$ the orthonormal frame bundle. I say that $P^{c}:=F_{SO(n)}(E)\times_{SO(n)} SO(n,\mathbb{C})$.

I am trying to understand why the following statement is true:

If we have a principal connection $\omega^{c}$ on $P^{c}$, we can decompose it uniquely into $(\omega,\phi)$, with $\omega$ a principal connection on $P$ and $\phi\in \Omega^{1}_{M}(iad P)$

Intuitively, this should be possible by writing $so(n,\mathbb{C})=so(n) + iso(n)$ and precomposing $\omega^{c}$ with an appropriate smooth section $\iota\in Hom(TP,TP^{c})$. For $\pi_{j}$ the projection onto $so(n)$ resp. $iso(n)$, we could write $\omega_{p} = \pi_{1}(w^{c}_{p}\iota_{p})$, $\phi_{p} = i(\omega_{p} + i\pi_{2}(\omega^{c}_{p}\iota_{p}))$. But even if this turns out to be correct in its details, I cannot see how this should be unique or in any way "natural".

Thank you!

share|cite|improve this question
I think I have a partial answer. If you have two principal bundles $Q$ contained in $P$ with lie groups $H\subset G$ ($H$ closed) s.t. $Q$ is a reduction of $P$ and $G/H$ is a reductive space, then for each connection $\omega$ on $P$ you can find a unique connection $\omega^{Q}$ that is a reduction of $\omega$. The conditions hold in our case. Then we can view our $\omega^{SO(3)}$ as a connection in $P^c$ as well. We can then simply take the difference $\omega-\omega^{SO(3)}=\phi$ and obtain a $\phi\in\Omega^1(M,adP^c)$. – David Hornshaw Apr 22 '14 at 17:59
I have a fundamental misunderstanding here, I think. For if $\phi\in\Omega^1(M,iadP)$, why isn't $\omega$ already a connection in $P$ itself? Which would make $\phi$ equal 0. – David Hornshaw Apr 22 '14 at 18:01

Your Answer



Browse other questions tagged or ask your own question.