# Does every group arise as the fundamental group of a complete Kähler manifold?

The fundamental group of a manifold is countable, and every countable group $$G$$ arises as the fundamental group of a (smooth) manifold; see this comment or this answer for a construction of an open subset $$U \subset \mathbb{R}^5$$ with $$\pi_1(U) \cong G$$.

Note that every smooth manifold admits a complete Riemannian metric. In fact, every conformal class contains a complete metric, see The Existence of Complete Riemannian metrics by Nomizu and Ozeki. Therefore, every countable group arises as the fundamental group of a complete Riemannian manifold.

As the hermitian property is preserved under conformal change, every conformal class of a hermitian metric on a complex manifold contains a complete hermitian metric. Replacing $$U$$ with $$V := U\times\mathbb{R} \subset \mathbb{R}^6 = \mathbb{C}^3$$, we see that every countable group arises as the fundamental group of a complete hermitian manifold.

Note that $$V$$ also admits a Kähler metric. However, unlike the hermitian case, the Kähler property is not preserved under non-constant conformal change, see this question. In fact, not every Kähler manifold admits a complete Kähler metric, see this question. Despite this, do we still have the Kähler analogue of the two bold statements above?

Does every countable group arise as the fundamental group of a complete Kähler manifold?

It's worth pointing out that the question of which groups arise as the fundamental groups of compact Kähler manifolds (which are necessarily complete) is an active area of research. Such groups are known as Kähler groups and much is known about them, see this question.

• Isn't it true that every fundamental group of a Kaeler manifold has even first Betti number? Mar 4, 2021 at 0:26
• Only true for compact Kaehler manifolds Mar 4, 2021 at 0:34
• Sorry, missed the absence of "compact". Mar 4, 2021 at 0:34
• Finite volume + complete is probably the right condition in this context. Mar 4, 2021 at 14:42
• Compactness is essential for evenness of odd betti numbers. The Riemann sphere minus two point has a complete Kahler metric of finite volume. Mar 4, 2021 at 14:53

• Thanks for your answer. Just for my own edification, let me record the following. The lemma you reference shows that if $U \subset \mathbb{R}^n$ is open, then there is a Runge domain $D \subset \mathbb{C}^n$ which deformation retracts onto $U$. A Runge domain is an open set which is holomorphically convex, so any such set in $\mathbb{C}^n$ is Stein (it is holomorphically separable due to the coordinate functions $z_i$). Mar 4, 2021 at 13:58
• @MichaelAlbanese Because they all admit proper holomorphic embeddings in $C^n$ you can pull back the metric. Mar 4, 2021 at 14:29