# Why is the Cotangent Space of Complex Projective Space Not $U(1)$-Equivariant?

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 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 ....

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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.

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Great answer, thanks alot. Is Milnor-Stasheff's book a good place start on learning this stuff. It's not material that's central to my interests at the moment, so I'd like something that's not too heavy going - maybe even fun. –  Ago Szekeres Nov 30 '10 at 20:36
Moreover, what is the dimension of the module of vector fields of $S^7$. –  Ago Szekeres Nov 30 '10 at 21:08
One can deal with all the cases as you do for $n=3$. Just observe that $T_{P^n}$ is generated by global sections, since by the Euler sequence there is a surjection $\osum_0^n O_{P^n}(1)\to T_{P^n}$.So every line bundle that is direct summand of $T_{P^n}$ is also generated by global sections, hence it has total Chern class $1+ax$ with $a\ge 0$. Then it is easy to see that the analog of the equations given for $n=3$ have no solution. –  rita Nov 30 '10 at 21:13
Having taught several seminars on that, I can tell you that Milnor-Stasheff is a rather demanding text compared to others (like Hatchers unfinished book "Vector bundles and K-theory"). I also like to advertise the definition of Chern classes that Matthias Kreck gives here: mathi.uni-heidelberg.de/~kreck/stratifolds/DA_22_08.pdf , page 119 –  Johannes Ebert Nov 30 '10 at 21:20
@Ago: $S^7$ is parallelizable, so the dimension is $7$. –  Johannes Ebert Nov 30 '10 at 21:22
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