Existence of connections on principal bundles

Question

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 $E$?

Answer 1

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.

Answer 2

Use the formula $\omega=\sum_\alpha (f_\alpha \circ \pi) \omega_\alpha$ where $\pi$ is the tangent bundle projection $TM \to M$. Connections are defined on $TM$.

Edit: The last sentence should probably read: Principal bundle connections are mappings defined on $TM$. To clarify, the definition I am using here is the following: On the vector bundle $TE$ consider the action of the group $G$ which is induced by the fibre-wise action of $G$ on $E$. Then $TE/G$ is a vector bundle over $M$ in a natural way, and the principal bundle projection (let's denote this by $\pi_E$) has a derivative $d\pi_E$ which is well-defined on $TE/G$. Then a connection is just a right-inverse of $d\pi_E: TE/G \to TM$ (in the category of vector bundles over $M$). So to construct a connection on $E\to M$, you use connections on the trivial bundles $E_{U_\alpha}\to U_\alpha$, forming a weighted sum of them with weights given by a partition of unity $f_\alpha$. Since the connections are mappings defined on $TU_\alpha$ and $f_\alpha$ is defined on $M$ you have to apply the tangent bundle projection $\pi: TM\to M$ (not $\pi_E$) before applying $f_\alpha$. That's how I interpret the formula.

Answer 3

I do not give you a reference for the existence of connections on any principal fiber bundle, but just of linear connections. Sure your question was about connections over principal bundles and, on page 67 in KN, there is a complete proof, but, on page 68, they just remark the possibility of another proof. This alternative possible proof is the one you do not have understood.

So in order to help you to overcome this difficulty, I refer you to Lee's textbook on Riemannian Manifolds. Even if his Lemma 4.4 and Proposition 4.5 give all the details of the proof only for linear connections, once you understand what you was missing, then the same argument works in general.

I hope to have been useful.