Let $M$ be a real even dimensional endowed with a symplectic form $\omega$ and a flat, torsionless connection $\nabla$ compatible with $\omega$, that is $$\nabla \omega=0.$$

Under this assumptions, taking into account a simplified version of the Markus conjecture, it should follow that $M$ is geodesically complete.

My question is: Under what additional geometric assumptions is the result known to be true?

One concrete example is: If in addition we assume the existence of a parallel almost complex structure $J$ on $M,$ then the result is true for $2$ dimensional and $4$ dimensional manifolds as observed by Misha.

What if instead of the the parallel almost complex structure we assume that there is a nonzero parallel vector field $V$ on $M?$ Does it follow that the manifold is geodesically complete?

Let $M$ be a real, even dimensional, compact manifold endowed with a symplectic form $\omega$ and a flat, torsionless connection $\nabla$ compatible with $\omega$, that is $$\nabla \omega=0.$$

Under this assumptions, taking into account a simplified version of the Markus conjecture, it should follow that $M$ is geodesically complete.

My question is: Under what additional geometric assumptions is the result known to be true?

One concrete example is: If in addition we assume the existence of a parallel almost complex structure $J$ on $M,$ then the result is true for $2$ dimensional and $4$ dimensional manifolds as observed by Misha.

What if instead of the the parallel almost complex structure we assume that there is a nonzero parallel vector field $V$ on $M?$ Does it follow that the manifold is geodesically complete?

Let $M$ be a complex manifold with real tangent bundle $TM$. Let $J$ be its associated almost complex structure ($J\circ J=-\operatorname{id}$) and $\nabla$ a torsion free, flat connection in $TM$ compatible with $J$, that is $$\nabla J=0.$$

Is it true that in this case the manifold is geodesically complete?

If one replaces the complex structure by a symplectic structure, the question becomes a simpler version of the Markus conjecture.

Let $M$ be a real even dimensional endowed with a symplectic form $\omega$ and a flat, torsionless connection $\nabla$ compatible with $\omega$, that is $$\nabla \omega=0.$$

Under this assumptions, taking into account a simplified version of the Markus conjecture, it should follow that $M$ is geodesically complete.

My question is: Under what additional geometric assumptions is the result known to be true?

One concrete example is: If in addition we assume the existence of a parallel almost complex structure $J$ on $M,$ then the result is true for $2$ dimensional and $4$ dimensional manifolds as observed by Misha.

What if instead of the the parallel almost complex structure we assume that there is a nonzero parallel vector field $V$ on $M?$ Does it follow that the manifold is geodesically complete?

Let M be a complex manifold with real tangent bundle TM. Let $J$ be its associated almost complex structure($JoJ=-id$) and $\nabla$ a torsion free, flat connection in $TM$ compatible with $J$, that is $$\nabla J=0.$$ Is it true that in this case the manifold is geodesically complete?

If one replaces the complex structure by a simplectic structure the question becomes a simpler version of Markus conjecture.

Let $M$ be a complex manifold with real tangent bundle $TM$. Let $J$ be its associated almost complex structure  ($J\circ J=-\operatorname{id}$) and $\nabla$ a torsion free, flat connection in $TM$ compatible with $J$, that is $$\nabla J=0.$$

Is it true that in this case the manifold is geodesically complete?

If one replaces the complex structure by a symplectic structure, the question becomes a simpler version of the Markus conjecture.