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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?

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    $\begingroup$ This is already false for complex 1-dimensional torus, where holonomy of the flat affine structure is cyclic complex affine. $\endgroup$
    – Misha
    Commented Sep 12, 2015 at 22:14
  • $\begingroup$ Thank you Misha. I am trying to understand your swift response. There is a 4 dimensional manifold of flat affine structures, nondiffeomorphically equivalent on the torus. Which flat affine structure are you referring to? I am obviously misreading your response. $\endgroup$
    – Mike Cocos
    Commented Sep 12, 2015 at 23:08
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    $\begingroup$ Take a cyclic group of dilations of complex line fixing the origin. Now quotient the complex line minus origin by this group. $\endgroup$
    – Misha
    Commented Sep 13, 2015 at 3:20
  • $\begingroup$ @Misha Thank you.Clearly the action does not preserve the volume so if I added that constraint(which apparently it is not necessary from the point of view of Markus conjecture) the dialation example would not work.For an affinely flat, complex surface I think I can prove that the existence of a parallel volume form implies the geodesic completeness. $\endgroup$
    – Mike Cocos
    Commented Sep 13, 2015 at 15:26

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Now that you added the volume requirement, you get close to the Marcus conjecture. In complex dimension 1 you then get a locally Euclidean structure and the answer is clearly positive. In complex dimension 2 it is again positive since linear holonomy then has rank 1 and in this situation Marcus conjecture was proven by Yves Carriere in1989.

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  • $\begingroup$ You are not assuming in any way that the symplectic structure is compatible with the almost complex structure of the manifold? Just parallel with respect to the flat connection. Otherwise it would be a metric connection and trivially true. $\endgroup$
    – Mike Cocos
    Commented Sep 14, 2015 at 22:17
  • $\begingroup$ I am not assuming any symplectic information. (Did I even mention a symplectic structure in my answer?) All I am assuming is that the linear holonomy is in $GL(n,C)\cap SL(2n,R)$, where $n$ is the complex dimension of your manifold. The $SL(2n,R)$ comes from your assumption on the parallel volume form; the $GL(n,C)$ assumption is your parallel almost complex structure. By the way, you should revise your question to reflect the question you are now asking. $\endgroup$
    – Misha
    Commented Sep 15, 2015 at 0:33
  • $\begingroup$ Thank you Misha. I will revise the question. I have to find out how first. I am very new to this but I love it already. $\endgroup$
    – Mike Cocos
    Commented Sep 15, 2015 at 0:47
  • $\begingroup$ In complex dimension 2, the linear holonomy has only "complex" discompacity 1. Carrière's proof no longer applies because it uses the fact that a codimension 1 real subspace separates the full affine space in two (no longer true in a complex setting, where a complex hyperplane does not separate). How ever, the classification of affine complex surfaces applies and can be used to show that such surfaces are indeed complete $\endgroup$
    – R. Alexandre
    Commented Mar 9, 2021 at 12:19
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Let me give a standard example of a closed incomplete manifold with flat affine structure, whose 2-dimensional version is essentially the example from the comment of Misha.

Consider $R^n\setminus 0$ and the action of the group $(Z, +)$ generated by the homothety
$$ p\mapsto 2 p.$$ The action preserves the flat affine connection, and also (if the dimension is even and the standard complex structure exists) the standard complex structure.

The quotient $R^n\setminus 0/Z$ is homeomorphic to $S^{n-1}\times S^1$ and is compact. Since the action of $Z$ preserves the flat affine and complex structure they induce a flat affine and complex structure on the quotient. Since we took out the point $0$ from $R^n$, the universal cover is not complete so the affine strcuture is not complete

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  • $\begingroup$ Thank you.Clearly the action does not preserve the volume so if I added that constraint(which apparently it is not necessary from the point of view of Markus conjecture) the dialation example would not work.For an affinely flat, complex surface I think I can prove that the existence of a parallel volume form implies the geodesic completeness. $\endgroup$
    – Mike Cocos
    Commented Sep 13, 2015 at 15:26
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This is an answer to the last question of the revised version of your question which is

"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?"

The answer is the same as to the question without the assumption. Indeed, having an incomplete manifold with $\omega$ parallel with respect to a flat torsionsfree connection, we can directly multiply it by $R^2$ with the standard connection and the standard symplectic structure. The resulting manifold remains incomplete but now it has even two parallel vector fields.

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  • $\begingroup$ Thank you Vladimir. Indeed you are right. Although, since I am optimistic that Markus's conjecture will prove to be true, I was hoping that adding more to the hypothesis will make the proof easier. What i was hoping(for a two dimensional compact manifold) is that the additional hypothesis might imply that the connection is metric. $\endgroup$
    – Mike Cocos
    Commented Sep 17, 2015 at 13:55

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