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I have posted this question on math.stackexchange, without success. I'll make it brief:

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

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 $\phi\in \Omega^{1}_{M}(iad P)$.

What I understand is how to decompose the connection so that $\phi\in\Omega^1_M(iadP^c)$. It should be possible in general, as long as we have a setting where $W\subset Q$ is a reduction of principal bundles s.t. the quotient of their respective lie groups form a reductive space.

I do not understand how we can reduce $\phi$ even further. Is it possible that $w^c$ needs to be flat?

Thank you very much.

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  • $\begingroup$ I don't understand your question. You want to know why you can decompose the connection on $P^c$. You know how to decompose it. Do you not know how to prove uniqueness of the decomposition? Or do you want to further decompose $\phi$ (which is not part of the statement you want to prove)? The adjoint representation of $SO(n)$ doesn't decompose at all, unless $n=4$. $\endgroup$ – Ben McKay Apr 23 '14 at 19:59
  • $\begingroup$ In my (false) previous understanding, $\phi$ took values in the wrong bundle. Andy Sanders answer cleared things up for me! $\endgroup$ – David Hornshaw Apr 24 '14 at 10:01
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The bundle $P$ is obtained by reducing the structure group of the $SO(n,\mathbb{C})$ bundle $P^c$ to the maximal compact subgroup $SO(n).$ Let's call the inclusion \begin{align} i: P\rightarrow P^c. \end{align} Pulling back the connection form gives $i^{*}(\omega^c)\in \Omega^1(P,\mathfrak{so}_n(\mathbb{C})).$ The Lie algebra of $SO(n,\mathbb{C})$ admits an invariant (with respect to the Adjoint action of $SO(n)$) splitting, \begin{align} \mathfrak{so}_n(\mathbb{C})=\mathfrak{so}(n)\oplus i\mathfrak{so}(n). \end{align} Decomposing the pull-back into these pieces gives, \begin{align} i^{*}\omega^c=\omega + \phi. \end{align} Then, $\omega$ is a connection on $P$ and $\phi$ descends to the base taking values in the adjoint bundle, \begin{align} P\times_{Ad(SO(n))} i\mathfrak{so}(n)=i\ \text{ad}(P). \end{align} That is to say, $\phi\in \Omega^1(M,i\ \text{ad}(P)).$

If the original connection is flat, then the pair $(\omega,\phi)$ satisfy some equations: \begin{align} F(\omega)+\frac{1}{2}[\phi,\phi]=0 \\ d^{\omega}\phi=0. \end{align} Here, $F(\omega)$ is the curvature of $\omega,$ the bracket operation combines the wedge product of forms and the Lie bracket, hence is symmetric so that the term $[\phi,\phi]$ does not (necessarily) vanish. The operator $d^{\omega}$ is the exterior covariant derivative operator associated to the connection $\omega.$ Notice that the first equation is of two forms with values in $ad(P)$ and the latter is of two forms with values in $i\ ad(P).$ The first equation is one of a pair of equations known as Hitchin's self-duality equations. You can read all about them (in the case of $SO(3,\mathbb{C})$) in the original beautiful paper of Hitchin: http://people.maths.ox.ac.uk/hitchin/hitchinlist/Hitchin%20THE%20SELF-DUALITY%20EQUATIONS%20ON%20A%20RIEMANN%20SURFACE%20(PLMS%201987).pdf

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  • $\begingroup$ I understand it now, thank you! Sadly, I can't upvote you yet (too little rep). :( $\endgroup$ – David Hornshaw Apr 24 '14 at 10:03

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