I have posted [this question](https://math.stackexchange.com/questions/761585/how-to-decompose-connections-on-the-complexified-orthonormal-frame-bundle) 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.



Decomposing connections on extensions of the frame bundle
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](http://people.maths.ox.ac.uk/hitchin/hitchinlist/Hitchin%20THE%20SELF-DUALITY%20EQUATIONS%20ON%20A%20RIEMANN%20SURFACE%20%28PLMS%201987%29.pdf)


