Canonical connection on $CP^n$ I have heard something along the lines of "There is a canonical $U(1)$ connection on $CP^n$" and I am trying to understand what that means.
First I suppose that the sentence refers to a line bundle over $CP^n$, possibly the tautological line bundle.
Second why is there a preferred connection?
I was thinking that $CP^n$ can be realised as the homogenous space $\frac{U(n+1)}{U(1)\times U(n)}$. Hence the Maurer-Cartan form on $U(n+1)$ descends to a unique $U(n+1)$-invariant connection form on $CP^n$ which however takes values in $U(1)\times U(n)$. I was thinking that one way to get a $U(1)$ connection would be to follow the prescription described in
https://math.stackexchange.com/questions/875705/a-construction-on-principal-bundles.
Namely one constructs the bundle $U(n+1) \times_{\rho} U(1)$ where the action $\rho: U(1)\times U(n) \rightarrow U(1)$ is simply projection onto the first factor followed ny $U(1)$ multiplication.
Would such a construction work? Is there a simpler way of thinking of the canonical connection on the tautological (?) line bundle over $CP^n$?
 A: The standard action of $U(n+1)$ on $\mathbb C^{n+1}$ preserves the unit sphere $S^{2n+1}$ and commutes with the action of $U(1)$ (multiplication of vectors by complex unit scalars). The quotient by $U(1)$ is $\mathbb C P^n$ and $S^{2n+1}\to \mathbb C P^n$ is a principal $U(1)$ bundle, on which $U(n+1)$ acts by bundle automorphisms. Now the claim is that there is a unique $U(1)$ connection on $S^{2n+1}\to \mathbb C P^n$  which is $U(n+1)$-invariant. 
Proof: since $U(n+1)$ acts transitivly on $S^{2n+1}$ it is enough to show that at some point  $x\in S^{2n+1}$ there exists a unique complement in $T_xS^{2n+1}$ to the tangent at $x$ of the $U(1)$-orbit through $x$  which is invariant under the stabilizer of $x$ in $U(n+1)$. Let the point be $x=(0,\ldots, 0, 1)$. Then $T_xS^{2n+1}=\mathbb C^n\oplus i\mathbb R$, the 2nd summand is the tangent space at $x$ to the $U(1)$ orbit through $x$, the stabilizer is $U(n)$,  acting on $T_xS^{2n+1}$ by the standard representation on the 1st summand and trivially on the 2nd. Since these are non-isomorphic irreducible $U(n)$-representations, it follows, by Schur lemma, that this is the unique $U(n)$-invariant decomposition of $T_xS^{2n+1}$.  Hence $\mathbb C^n\oplus 0$ is the unique $U(n)$ invariant complement in $T_xS^{2n+1}$ to $0\oplus i\mathbb R$. It follows that there is a unique $U(n+1)$-invariant $U(1)$-connection on the principal $U(1)$-bundle $S^{2n+1}\to \mathbb C P^n$. Its horizontal distribution is  the orthogonal complement of 
the tangents to the $U(1)$-orbits. 
