In the definition of a Kähler manifold, is the integrability of the underlying almost complex structure redundant?

In other words, is there a manifold $M$ with almost complex structure $J$, symplectic structure $\omega$ and Riemannian metric $g(X,Y)=\omega(JX,Y)$ but $J$ is not integrable?

  • $\begingroup$ This is a similar question. $\endgroup$ Commented Aug 21, 2017 at 23:24

1 Answer 1


If $\omega$ is a non-degenerate two-form on $M$ (not necessarily closed), then there exists an almost complex structure $J$ such that $g(X, Y) := \omega(X, JY)$ is a Riemannian metric compatible with the almost complex structure (i.e. $g(JX, JY) = g(X, Y)$); see Theorem $1.17$ of these notes. We say that such an almost complex structure is compatible with the symplectic form.

If in addition $\omega$ is closed (i.e. $\omega$ is a symplectic form), then we call $(M, \omega, J, g)$ an almost Kähler manifold. So rephrasing the above, we see that for any symplectic manifold, there is a compatible almost complex structure $J$ such that $(M, \omega, J, g)$ is an almost Kähler manifold. So the only difference between a symplectic manifold and an almost Kähler manifold is a choice of compatible almost complex structure (there are many).

Your question can now be rephrased as follows: is there an almost Kähler manifold $(M, \omega, J, g)$ such that $J$ is not integrable? The answer to this question is yes. In the paper Compact Parallelizable Four Dimensional Symplectic and Complex Manifolds by Fernández, Gotay, and Gray, the authors construct examples of compact four-dimensional manifolds which admit symplectic structures but no complex structures (in particular, any choice of almost complex structure compatible with a fixed symplectic form will not be integrable).

In conclusion, the answer to your initial question is no, the integrability condition on the almost complex structure is not redundant in the definition of Kähler manifolds.

Although not necessary to answer the question, you may be interested in the following.

If $\omega(X, Y) = g(JX, Y)$, then "$\nabla J = d\omega\oplus N_J$" where $\nabla$ is the Levi-Civita connection with respect to $g$, and $N_J$ is the Nijenhuis tensor of $J$. The quotation marks are used to indicate this is not a literal equation, but rather a description of a relationship between the quantities involved. More precisely, $\nabla J = 0$ if and only if $d\omega = 0$ and $N_J = 0$. In particular, on an almost Kähler manifold, $J$ is integrable if and only if it is covariantly constant.

  • $\begingroup$ Thank you very much for your great answer. $\endgroup$ Commented Aug 21, 2017 at 3:34

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .