I'm looking at the cotangent bundle of $CP^{N}$ at the moment in the context of equivariance. For many reasons, it seems to me that this bundle is not $U(1)$-equivariant, or, in other words, cannot be constructed as from a representation of $U(1)$ in the standard manner (see [here](http://en.wikipedia.org/wiki/Principal_bundle#Associated_vector_bundles_and_frames) for example) where we consider $CP^{N}$ as the base space of the principle bundle $S^{2N}$. However, I cannot find a neat convincing argument for why this should be so. 


**MY ATTEMPT**: I would guess that $U(1)$-equivariance would mean that the bundle could be expressed as a direct sum of line bundles, but I cannot see how to show that this is not the case (even though it seems most probable). Again I would guess that some sort of Chern argument comes in, but I don't know how to caclulate Chern classes without a connection. The only connection here I know is the Grassmannian, and at this point the whole thing just becomes a mess ....



Why is the Cotangent Space of Complex Projective Space Not $U(1)$-Equivariant?
If there were a representation $V$ of $U(1)$ and an isomorphism $TCP^n \cong S^{2n+1} \times\_{U(1)} V$, then the tangent bundle of $S^{2n+1}$ would be the direct sum of the trivial bundle $S^{2n+1} \times V$ plus the trivial real line bundle (the vertical tangent bundle to the $S^1$-bundle. In particular, $S^{2n+1}$ is parallelizable, which implies, by a result by Adams, that $n=0,1$ or $3$. If $n=0$, then the question is void, for $n=1$, $TCP^1$ is obviously a complex line bundle. 


That leaves the case of $CP^3$. If $TCP^3$ were a sum of three complex line bundles, then the total Chern class would be $(1+ax)(1+by)(1+cy)$ for some integers $a,b,c$ ($x$ is an appropriate generator of $H^2(CP^n)$. On the other hand, we know the total Chern class very well; it is $1+4x+6x^2+4 x^3$. This is discussed, without any reference to a connection, in many sources, e.g. Milnor-Stasheffs book. So you get


$$a+b+c=4, ab+bc+ac=6, abc=4.$$


It is easy that these equation do not have an integral solution (reduce modulo $2$ to conclude that $a,b,c$ have to be even). 


Probably it is possible to do the higher dimensional cases in that arithmetical way, instead of nuking it with Adams' result.


