Let $(M,g,J)$ be an almost Kähler manifold and let $\rho$ denote its Ricci form $$ \rho(X,Y) = \operatorname{ric}^{\mbox{c}}(JX,Y) $$ where $\operatorname{ric}^{\mbox{c}}$ is the $J$-invariant part of the Ricci tensor of the Riemannian manifold $(M,g)$.

If $\rho(X,JY)=\operatorname{ric}(X,Y)$ (i.e. $(M,g,J)$ has $J$-invariant Ricci tensor), is it true that $\rho$ is a closed $2$-form?

A second question is:

what are well known results from the theory of Kähler manifolds valid in almost Kahler manifolds with $J$-invariant Ricci tensor?


1 Answer 1


For a Kähler manifold $(M^{2n}, g, J)$, the Ricci 2-form $\rho(X, Y):={\rm Ric}(JX, Y)$ is closed, of type $(1, 1)$, and represents (up to a scaling factor $\frac{1}{2\pi}$) the de Rham cohomology of the first Chern class of $(M, J)$.

For generic almost-Kähler manifolds (i.e. the almost complex structure $J$ is not integrable), none of these properties remains true for the tensor $\rho(X, Y)={\rm Ric}'(JX, Y)$, where ${\rm Ric}'$ denotes the $J$-invariant part of the Ricci tensor.

For details on almost Kähler manifolds with $J$-invariant Ricci tensor, see the paper of T. Draghici, Almost Kähler 4-manifolds with $J$-invariant Ricci tensor, Houston J. Math. 25 (1999), 133–145, or papers of V. Apostolov, or K. Sekigawa.

  • $\begingroup$ Dear @314159. Thank you very much for your answer and the references. Could you please give us an elementary example or short analysis to justify your affirmation about $\rho(X, Y)={\rm Ric}'(JX, Y)$ for generic almost-Kähler manifolds. $\endgroup$ Commented Oct 30, 2016 at 9:33
  • $\begingroup$ For an almost-Kähler manifold $(M, g, J)$ the vanishing of the Nijenhueis tensor $N$ is equivalent to saying that the fundamental 2-form $\Omega$ is parallel with respect to the Levi-Civita connection $\nabla$ of $g$. Now, a measure of the difference of the two types of Ricci forms on $(M, g, J)$ is given as follows: $\rho^{*}− \rho= \nabla^{*}\nabla \Omega$, hence when $N\neq 0$ (i.e. $J$ is not integrable) it is also $\nabla\Omega\neq 0$ and we conclude that $\rho$ cannot be closed. $\endgroup$
    – 314159.
    Commented Oct 30, 2016 at 10:30
  • $\begingroup$ You've this for symplectic manifolds because they are almost-Kahler manifolds. $\endgroup$ Commented Oct 30, 2016 at 11:06

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