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.


1 Answer 1


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 $R_f$.

  • 1
    $\begingroup$ 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 $f$. $\endgroup$ Sep 15, 2016 at 11:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.