The question is from Donaldson's paper "scalar curvature and projective embeddings I (MR1916953)".


Let (M, $\omega$) be a compact symplectic manifold, $(L, h)\to (M,\omega)$ be an Hermitian line bundle with curvature $\sqrt{-1}\omega$. Consider the group $\mathcal{G}$ of Hermitian bundle maps from $L$ to $L$ which preserves the connection. Then this group action induces a Lie algebra $C^\infty(M)$ action on the space of sections $\Gamma(L^k)$, which is given by for $s\in \Gamma(L^k)$
$$
R\_f(s)=\nabla\_{\xi\_f}(s)-\sqrt{-1}kfs,
$$
where $\xi\_f$ is the Hamiltonian vector field, such that $i\_{\xi\_f}\omega=df$.


My question is how $R\_f$ comes from? I know it's from the linearization of the group action, but I could not figure out the correct one so that after the linearization, one can obtain $R\_f$. Any help will be appreciated. 



Lie group action in Donaldson's paper
To warm up, note that any $G \in \mathcal{G}$ covers a diffeomorphism $g : M \to M$, which must preserve the curvature of the connection, so is a symplectomorphism. On the other hand, if $\bar G : L \to L$ is an arbitrary bundle lift of a symplectomorphism $g$, then $\bar G^\* \nabla - \nabla$ is a closed (imaginary) 1-form, whose cohomology class is an obstruction to correcting $\bar G$ to an element of $\mathcal{G}$. Hence there is an exact sequence
$$0 \to U(1) \to \mathcal{G} \to \{\textrm{symplectomorphisms}\} \to H^1(M) \to 0. $$
This is at least consistent with the Lie algebra of $\mathcal{G}$ being $C^\infty(M)$, as that fits into
$$ 0 \to \mathbb{R} \to C^\infty(M) \to \{\textrm{Hamiltonian vector fields}\} \to 0. $$


To actually identify the Lie algebra of $\mathcal{G}$, first identify the infinitessimal bundle automorphisms of $L$ with vector fields on the total space of the form $\tilde X - f u$, where $\tilde X$ is the horizontal lift of a vector field $X$ on $M$, $f \in C^\infty(M)$ and $u$ is the vector field generated by the $U(1)$ action. If you persuade yourself that
$$ \mathcal{L}\_{\tilde X + f u}\nabla = i\_X (\sqrt{-1}\omega) - \sqrt{-1} df $$
(perhaps easiest by thinking of the connection as a 1-form on a principal $U(1)$-bundle) then that proves that the infinitessimal automorphisms of $\nabla$ are precisely of the form $\tilde \xi\_f - f u$. The induced action on $\Gamma(L^k)$ is the given $R\_f$.


