TITLE: Existence of connections on principal bundles
QUESTION [9 upvotes]: Is it always true that a principal $G$-bundle $E$ admits a connection (on the total space, not a local connection on the base manifold $M$)? I know that it must be true, since almost every construction starts off with ...fix a connection on $E$..., I just don't know how to show this rigorously. The only proof I can find is: 
Let $U_i\subset M$ be an open subset of $M$. Then $E$ restricted to $U_i$ is trival and we can 
construct a connection, denoted $\omega_i$, in this case. Now, let $\big( U_\alpha 
\big)$ be an open covering of $M$, and let $\big(f_\alpha\big)$ be a partition of unity subordinate to the cover. Then we can define a connection $\omega = \sum_{\alpha} (f_\alpha \circ \pi) \omega_\alpha$, where $\pi: E\rightarrow M$ is the projection.
However, doesn't the right-hand side of this expression live on $M$? Does this give a connection on $E$?

REPLY [7 votes]: Another point of view can be found in Atiyah's "Complex analytic connections in fibre bundles".  
If $\pi: P \to X$ is principal bundle with fibre a complex (or real) Lie group $G$ on a complex (or differential) manifold $X$, a connection is a $G$-invariant splitting of the following short exact sequence of vector bundles over $P$:
$0 \to T_F P \to TP \to \pi^{-1}TX \to 0$
Here $T_F P$ denotes the bundle of tangent vectors tangent to the fibre. $G$ acts on all these bundles. One can construct an associated sequence of $G$-invariant sections of these bundles to get a sequence of vector bundles on $X$:
$0 \to (T_F P\)^G \to TP^G \to TX \to 0$ 
This is an extension of the vector bundle $TX$ by the vector bundle $T_F P^G$. A connection is now just a splitting of this sequence. By a general result of homological algebra, extensions are classified by 
$H^1(X, Hom(TX, T_F P^G))$      
In the differentiable case, $Hom(TX, T_F P^G)$ is a fine sheaf and the cohomology vanishes. So the sequence above is split and we have connections.