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As of the Wikipedia article on principal bundles connections:

Let $\pi: P \to M$ be a smooth principal bundle, a principal $G$-bundle over a smooth manifold $M$. Then a principal $G$-connection on $P$ is a differential $1$-form on $P$ with values in the Lie algebra $\mathfrak g$ of $G$ which is $G$-equivariant and reproduces the Lie algebra generators of the fundamental vector fields on $P$.

Now for the principal $U_1$-bundle $S^{2N+1} \to CP^N$, I appear to have shown that it does not admit a principal $U_1$-connection. Is this true, or a load of rudbbish?

P.S. My "proof" also implies that there is a unique connection form for the $U_{N-1}$-bundle $SU_{N} \to CP^{N-1}$. Can I assume that this is also incorrect?

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    $\begingroup$ (By the way, I guess you mean $S^{SN+1}$ and not $S^{2N}$.) Connections exist: this is proven in generality in Kobayashi-Nomizu, for example. Hence my guess is that you've made an error in your "proof". $\endgroup$
    – José Figueroa-O'Farrill
    Commented Jan 27, 2011 at 22:56
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    $\begingroup$ For principal circle bundles, one can use Cech cocyles and derive reasonably easily that for a connected and simply connected manifold $X$, given a closed $2$-form $\omega$ such that $\omega/2\pi$ is an integral cohomology class there exists a circle bundle with a connection whose curvature is $\omega$. In your case, this $\omega/2\pi$ is a genarator of $H^2(\mathbb{CP}^N;\mathbb{Z})$. $\endgroup$
    – Somnath Basu
    Commented Jan 27, 2011 at 23:03
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    $\begingroup$ Embed $S^{2n+1} \subset \mathbb{C}^{n+1}$, with the standard hermitian scalar product. Consider the $1$-form $\theta$ that assigns to a tangent vector $v \in T_x S^{2n+1}$ the scalar product $<x,v>$. This is a connection $1$-form. $\endgroup$
    – Johannes Ebert
    Commented Jan 27, 2011 at 23:23
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    $\begingroup$ Connections exist for bundles that trivialise over an open cover with a subordinate partition of unity. Take the trivial connection on each trivialised 'part' of the bundle and patch them together using the partition of unity. If your definition of smooth manifolds assumes paracompactness this is always true for bundles on smooth manifolds, and in particular is true for the case you quote. $\endgroup$
    – David Roberts ♦
    Commented Jan 28, 2011 at 1:17
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    $\begingroup$ As to your edited question: yes, you are incorrect. There are many many connections on a given bundle when they exist. Just take a 1-form on the base space, pull it back to the total space and add it to your connection and you get another connection. $\endgroup$
    – David Roberts ♦
    Commented Jan 28, 2011 at 20:16

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