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.


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 $R_f$.

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    $\begingroup$ The persuasion can be done by Cartan's formula: If $A$ denotes the connection 1-form (satisfying $\iota_u A = i$ and $\mathrm d A = i\pi^*\omega$ and $\pi:L\to M$ the projection, then $\mathcal{L}_{\tilde X + fu} A = \iota_{\tilde X + fu}\mathrm dA + \mathrm d \iota_{\tilde X + fu}A = i(\iota_{\tilde X + fu} \pi^*\omega + \mathrm d \pi^*f) = i\pi^*(\iota_{X}\omega + f)$, so $X$ must be the Hamiltonian vector field of $f$. $\endgroup$ – Bertram Arnold Sep 15 '16 at 11:09

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