Parallel orthogonal complex structures on complexified tangent bundle. Let $M$ be a Riemannian manifold.  I'm wondering if there are nice conditions/obstructions or (non-Kahler) examples of the existence of a map $J \in \operatorname{End} (TM \otimes \mathbb C)$ such that


*

*$J^2 = -Id$.

*$g(JX,JY) = g(X,Y)$ for all $X,Y \in TM \otimes\mathbb C$.

*$\nabla J = 0$ where $\nabla$ is the Levi-Civita connection.


Of course, if $J$ preserves $TM$ then $M$ is Kahler which is a pretty strong condition.  But, for example, if $J$ does not preserve $TM$ then the 2-form $\omega(\cdot,\cdot) = g(J\cdot, \cdot)$ is closed and non-degenerate but could now be zero in de Rham cohomology since it's not necessarily real.
 A: In one sense, there are only Kähler examples, but, in another sense, there are non-Kähler examples of such $J$.  Here is what I mean:
Suppose that one has the data $(M,g,J)$ as defined in the question and that they satisfy the Conditions $1$ and $2$ (we'll get to $3$ below).  Let $TM\otimes\mathbb{C} = W_+\oplus W_-$, where $W_+\subset TM\otimes\mathbb{C}$ is the $(+i)$-eigenspace of $J$ and $W_-\subset TM\otimes\mathbb{C}$ is the $(-i)$-eigenspace of $J$.  (This direct sum decomposition exists because of Condition $1$.)
Now, by Condition $2$, $W_+$ and $W_-$ must be $g$-isotropic subbundles of $TM\otimes\mathbb{C}$.  Since they have trivial intersection and sum to $TM\otimes\mathbb{C}$ (on which $g$ is nondegenerate), it follows that they must each be of half the dimension of $TM\otimes\mathbb{C}$, so, in particular, $M$ must have even dimension.  Moreover, each of these spaces has trivial intersection with $TM\otimes 1$ and $TM\otimes i$ (since these latter spaces have no nonzero $g$-null vectors), so it follows that there are maps $K_\pm:TM\to TM$ such that 
$$
W_\pm = \{\ v - i\ K_\pm v\ \bigl|\ v\in TM\ \}.
$$
Now, the condition that $W_\pm$ be $g$-null is equivalent to the condition that each of $K_+$ and $K_-$ be a $g$-orthogonal almost complex structure on $M$.  The condition that these two spaces be disjoint is just that $K_+-K_-:TM\to TM$ have no kernel.  Conversely, if $K_\pm:TM\to TM$ are two $g$-orthogonal almost complex structures on $M$ such that $K_+-K_-$ has no kernel, then defining the complex subbundles $W_\pm\subset TM\otimes\mathbb{C}$ as above and defining $J:TM\otimes\mathbb{C}\to TM\otimes \mathbb{C}$ to be the complex linear map that has $W_+$ as the $(+i)$-eigenspace of $J$ and $W_-$ as the $(-i)$-eigenspace of $J$ will give a triple $(M,g,J)$ that satisfies Conditions $1$ and $2$.  Note that the $J$ that is so-defined comes from the almost complex structure $K_+$ in the usual way if and only if $K_- = - K_+$.  Note that, in general, there could be many pairs $(K_+,K_-)$ that satisfy these conditions without satisfying $K_-=-K_+$.
Now, what about Condition 3 (i.e., $\nabla J=0$)?  This is equivalent to $\nabla K_\pm = 0$, so one can get examples $(M, g, J)$ that satisfy the three Conditions without being derived from a standard Kähler structure only if $g$ has a pair of parallel complex structures $(K_+,K_-)$ that satisfy the above conditions but don't satisfy $K_-=-K_+$.  This does happen; for example, if $g$ has holonomy $\mathrm{Sp}(n)\subset\mathrm{SO}(4n)$, then there is a $2$-sphere of $g$-parallel complex structures on $M$, and choosing any pair $(K_+,K_-)$ from that $2$-sphere that is not antipodal will allow one to generate a $(M,g,J)$ that does not come from a Kähler structure in the usual sense.  However, note that the underlying Riemannian manifold does admit a Kähler structure, in fact, a $2$-sphere of them.  The family of $(M,g,J)$ that one gets for such a manifold has dimension $4$ and contains the 'usual' Kähler structures as an embedded $2$-sphere.  There can be a higher dimensional family of solution pairs if the holonomy of $g$ is a proper subset of $\mathrm{Sp}(n)$, but, in such cases, the metric will be a product (locally), and the problem reduces to working on the irreducible factors.
