I look for a reference of the following implication
Let $ X $ be a compact complex manifold,
If : 1) $ \chi (O_X) \neq0 $
2) the Universal covering does not contain compact subvariety
So $ K_X $ is big .
We know that $ K_X $ is big $\implies$ $ K_X $ is nef, when can we have the equivalent?
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5$\begingroup$ $K_X$ big does NOT imply $K_X$ nef, nor in general neither in this case. $\endgroup$– diveriettiCommented Oct 2, 2020 at 12:21
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2$\begingroup$ I suspect the OP claims this only for $X$ satisfying 1) and 2). But the question certainly needs to be clarified. $\endgroup$– abxCommented Oct 2, 2020 at 12:29
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2$\begingroup$ @diverietti: if $X$ is not minimal it contains some rational curve, which lifts to the universal covering. $\endgroup$– abxCommented Oct 2, 2020 at 13:13
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1$\begingroup$ @abx what you say it's ok for projective manifolds. But this is merely compact complex. Even in the compact Kähler case "$K_X$ not nef implies presence of rational curves" is not known! $\endgroup$– diveriettiCommented Oct 2, 2020 at 13:35
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1$\begingroup$ In any case I really don't understand what the OP would like to know... $\endgroup$– diveriettiCommented Oct 2, 2020 at 13:38
1 Answer
At least in the projective setting the following holds true (this is taken from J. Kollár "Shafarevich maps and automorphic forms", Proposition 13.14.2).
Proposition. Let $X$ be a smooth projective variety. If $K_X$ is nef but not big, and $X$ has generically large fundamental group, then $\chi(\mathcal O_X)=0$.
Here, generically large fundamental group roughly means that $\operatorname{Im}\bigl(\pi_1(W)\to\pi_1(X)\bigr)$ is infinite for every positive dimensional subvariety passing through a very general point of $X$. In particular, your hypothesis 2) implies that $X$ has generically large fundamental group, in a strong way.
Thus, if $\chi(\mathcal O_X)\ne 0$, $X$ is projective, and your hypothesis 2) holds true, it follows that $K_X$ must be big and nef.
Observe that (as I wrote in one of the comments) it is nef not because it is big! It is nef because there cannot be rational curves in $X$, as abc remarked in another comment.
I really don't see how Kollár's proof of the proposition above might be transposed to a non projective setting.
Addendum (after the comments of Jason Starr and YangMills). There is a result by J. McKernan which confirms a conjecture by J. Starr (you can find a link to that paper in the comment of YangMills below) roughly stating that the conclusions of the bend-and-break still hold in the setting of $\mathbb Q$-factorial proper algebraic spaces. Over $\mathbb C$, a smooth algebraic space is the same as a Moishezon manifold, i.e. a complex manifold whose space of meromorphic functions has transcendence degree equal to its dimension.
So, if you know in advance that for some reason your $X$ is a little more than merely a compact complex manifold, namely if it is a compact Moishezon manifold (in this compact case, this is equivalent to be bimeromorphic to a projective manifold), then if $K_X$ is not nef you get some rational curve and hence some compact submanifold of the universal cover of $X$. This is not possible by your assumption 2).
Summing up, your assumption 2) implies that $K_X$ must be nef, unless possibly if $X$ is not bimeromorphic to a projective manifold. Of course if $K_X$ is big then $X$ is bimeromorphic to a projective manifold. But still, even with your assumptions 1)+2), I am not able to prove the bigness of $K_X$ if $X$ is only assume to be compact complex, or even compact Moishezon.
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1$\begingroup$ There is a paper of McKernan about the case of smooth, proper algebraic spaces that are not projective. I believe that he claims that each of these contains a rational curve. $\endgroup$ Commented Oct 2, 2020 at 20:38
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2$\begingroup$ Hi Jason! Now that you say this, I have some reminiscence of a conference in Luminy where McKernan gave a talk around this, but I cannot remember the precise statements... I’ll take a look, cheers! $\endgroup$ Commented Oct 3, 2020 at 19:35
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1$\begingroup$ McKernan's paper is here projecteuclid.org/euclid.aspm/1540319494 (paywall) and here math.ucsd.edu/~jmckerna/Papers/space.pdf (free) $\endgroup$ Commented Oct 4, 2020 at 4:00
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1$\begingroup$ @YangMills thanks! I looked at the paper. I'll add something in my answer later or tomorrow! Thanks again to Jason, too! $\endgroup$ Commented Oct 4, 2020 at 9:09