Unicity of a vector field on $S^1$-bundle

Question

Let M be a complex smooth manifold,and let $\zeta $ be a vector filed on $M$, why always there exists a unique vector field $\hat{\zeta }$ on $L^{\times}$ which project down to $\zeta $ and $\alpha( \hat{\zeta })=0 $, where $\alpha$ here is connection form and $L^{\times}$ is $S^1$-bundle obtained from line bundle $L$, by removing the zero section.

Answer 1

Let $(M,\omega)$ be a symplectic manifold. The first idea behind geometric quantization is that there exists a $S^1$-principal fiber bundle $\pi : Y \to M$, equipped with a connection $\lambda$ with curvature $\omega$, that is, $d\lambda = \pi^*(\omega)$. Let $\xi$ be the vector field generating the $S^1$ action, then, for all $y \in Y$, $$ T_yY = \ker(\lambda_y) \oplus {\bf R} \xi(y). $$ Proof. From $d\lambda = \pi^*(\omega)$ you get immediately that ${\bf R} \xi \subset \ker(d\lambda)$. On the other hand, $d\lambda(u,v) = \omega(\pi_*(u),\pi_*(v))$, then $d\lambda(u,v) = 0$ for all $v$ and $\pi_*$ surjective implies $\omega(\pi_*(u),w)= 0$ for all $w$. Since $\omega$ is symplectic, $\pi_*(u)=0$. Thus $\ker(d\lambda) = \ker(\pi_*) = {\bf R}\xi$. Now, if $u \in \ker(\lambda) \cap \ker(d\lambda)$ then $u = a\xi$ and $\lambda(a\xi) =0$, but $\lambda (a\xi)= a \lambda(\xi) = a$, because $\lambda(\xi)=1$, therefore $a=0$, that is $u=0$ and the two subspaces $\ker(\lambda)$ and $\ker(d\lambda)$ are complementary. $\square$

In conclusion, every vector $\delta x \in T_xM$ has a "horizontal" lift $\delta y \in T_yY$, with $\pi(y) = x$, that is, $\delta y \in \ker(\lambda)$.

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