In Riemannian geometry the non-negativity of the Ricci curvature $R$ of a manifold $X$ has strong implications on the size of the fundamental group $\pi_1(X)$:

  1. If $R>0$, then $\pi_1(X)$ is finite.
  2. If $R=0$, it is known that $\pi_1(X)$ is almost abelian, i.e., it contains an abelian subgroup of finite index. Also, $\pi_1(X)$ has polynomial growth.

In the case $X$ is a smooth complex projective variety, the positivity of Ricci curvature is related to ampleness properties of of $-K_X$, so it would be interesting to see whether analogous results of the above hold in algebraic geometry, with Ricci curvature replaced by the Kodaira dimension $\kappa(X)=\sup_n\dim\phi_{nK}(X)$. So my question is:

What implications do non-positive Kodaira dimension have for the fundamental group of $X$? In particular, does some versions of the above results hold with Ricci curvature replaced by Kodaira dimension?

For example, if $X$ is a smooth projective variety with $\kappa(X)=0$, is $\pi_1(X)$ almost abelian?

One could also ask for refined versions of the above statements. For example, when $X$ is Fano it is well-known that $\pi_1(X)=0$. Does the same condition hold for all $X$ with big $-K_X$?

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    $\begingroup$ Hi John, regarding your last question: there are ruled, non-rational surfaces with big anticanonical divisor. In fact, the projectivization of a sufficiently unstable rank two vector bundle on a curve is an example of such a surface. $\endgroup$
    – damiano
    Mar 7, 2011 at 12:05
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    $\begingroup$ The following article ams.org/mathscinet/search/… of Campana gives a condition for the fundamental group to be almost abelian, mimicking what is known for the rationally connected picture. $\endgroup$
    – Frank
    Mar 7, 2011 at 14:11
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    $\begingroup$ Maybe I should expand a little on my comment. For every curve $C$, there are ruled surfaces $S \to C$ with big anticanonical divisor. In particular, the fundamental group of a surface with big anticanonical divisor can be the fundamental group of any curve. Thus, for surfaces, the answer to your last question is 'no'. $\endgroup$
    – damiano
    Mar 9, 2011 at 10:30

1 Answer 1


I realize this is an old question, but I just noticed it. Testa is correct that without stronger hypotheses, the fundamental group can be arbitrary. However, there are very strong results of Chenyang Xu that prove (following a conjecture of Kollár) that the local étale fundamental group of every KLT singularity is finite, and hence the étale fundamental group of the smooth (orbifold / stacky) locus of every log Fano variety is also finite. This uses the fantastic boundedness theorem of Hacon-McKernan-Xu. So if there is a birational model of your projective manifold that is a log Fano variety, the étale fundamental group is finite.

Also, recently Fujino has proved that for a quasi-log canonical variety that has anti-ample dualizing sheaf (i.e., is Fano), the variety has trivial étale fundamental group. That is simultaneously stronger than Xu's theorem -- because the group is trivial, not just finite -- yet also weaker, since this is not (obviously) a birational invariant: there is no reason that an étale cover of a birational model extends to an étale cover of the singular Fano variety, although it obviously extends to an étale cover of the smooth locus (by purity, etc.).


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