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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 $E$?

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Did you mean to write "right-hand side"? Since $\omega_\alpha$ is a (local) connection on the bundle, when you sum over the partition of unity you should still be getting an object on the bundle. To be precise, though, you have to think about what exactly $\omega_\alpha$ means. It should be a rule for lifting tangent vectors in $M$ to tangent vectors in $E$. If you write it out carefully, you should see that your idea works. Good sources for connections on principal bundles are Spivak, volume 2 (chapter 9 I think) and Atiyah and Bott's paper on Yang-Mills theory (section 3, I think). – Dan Ramras Mar 15 '11 at 22:50
Yes, I meant right-hand side (it's corrected now). To me the right-hand side looks very weird. First, shouldn't $f_\alpha$ and $\pi$ be the corresponding maps on the diff. forms? Also, if you can define $(f_\alpha \circ \pi)\omega_\alpha$, then wouldn't this object live on some open set of $M$? – Kevin Wray Mar 15 '11 at 22:55
Section 2 of Chapter II in Volume 1 of Kobayashi Nomizu's Foundations of Differential Geometry proves that connections exist on principal fibre bundles over a paracompact base. – José Figueroa-O'Farrill Mar 16 '11 at 0:24
The first step in your proof should be corrected in: Let us choose a atlas of trivializing chart $\{(U_\alpha,\phi_\alpha)\}$ for the fiber bundle. – Giuseppe Mar 17 '11 at 9:34
I think the picture is clear if you view a connection as a projection from the tangent bundle $TE$ to its vertical subbundle (for principal bundles one adds in general an invariance condition under the group). If your $\omega_\alpha$ are the projections given by local trivializations (or the closely related Lie-agebra valued 1-forms), the "right hand side" is still a projection, since the $f_\alpha$ sum to $1$. – BS. Mar 17 '11 at 16:31

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.

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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.

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Not only on the tangent bundle! And notice the OP was asking for principal bundles. – David Roberts Mar 16 '11 at 21:06
I'm confused by the downvotes. I'm still convinced that my answer is correct, maybe it's just a misunderstanding. Did I make a mistake or is it complete nonsense I wrote here? Can someone enlighten me? – Florian Mar 17 '11 at 17:35

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.

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