Suppose that $A$ is a unitary connection on a Hermittian differentiable vector bundle $E$ over a Kahler manifold $X$, then we have operators $$\bar{\partial}_A: \Omega_{X}^{p,q}(E)\to \Omega_{X}^{p,q+1}(E)$$ $${\partial}_A: \Omega_{X}^{p,q}(E)\to \Omega_{X}^{p+1,q}(E)$$ And we have the Kahler identities: $$\bar{\partial}_{A}^{*}=-i[\partial_A,\Lambda]$$ $${\partial}_{A}^{*}=i[\bar{\partial}_A,\Lambda]$$

Is this result also true for almost Kahler manifolds? Does anyone know a good reference for it? By definition, an almost Kahler manifold is a manifold endowed with $(g,J,\omega)$, where $g$ is a Riemannian metric, $\omega$ is a symplectic form, and $J$ is an almost complex structure, and they satisfy the compatibility condition $$g(X,Y)=\omega(X,JY)$$

  • $\begingroup$ Could you elaborate on your assumptions on $E$ in each case? I assume that at the beginning $E$ is a hermitian holomorphic vector bundle, but in the desired setting, I don't think it makes sense to say $E$ is holomorphic if the base is not a complex manifold. Maybe I'm wrong. $\endgroup$ – Michael Albanese Apr 4 '15 at 18:36
  • $\begingroup$ @MichaelAlbanese Thanks! It's already edited. $\endgroup$ – Boyu Zhang Apr 4 '15 at 19:47
  • $\begingroup$ I think this is true. The essential input which justifies the calculation using standard coordinate is that $\nabla^{LC}\omega=d\omega=0$. $\endgroup$ – user44651 Jan 8 '16 at 5:12

Your Answer

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

Browse other questions tagged or ask your own question.