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](https://mathoverflow.net/questions/192230/is-there-a-manifold-with-fundamental-group-mathbbq#comment479763_192230) or [this answer](https://math.stackexchange.com/a/1362689/39599) 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](https://mathoverflow.net/q/69792/21564). In fact, not every Kähler manifold admits a complete Kähler metric, see [this question](https://mathoverflow.net/q/239677/21564). Despite this, do we still have the Kähler analogue of the two bold statements above?



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> Does every countable group arise as the fundamental group of a complete Kähler manifold?
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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](https://mathoverflow.net/q/270023/21564).



Does every group arise as the fundamental group of a complete Kähler manifold?
Any Stein manifold admits a complete Kähler metric. Start with a connected real analytic manifold with the given fundamental group. A suitable tubular neighbourhood of the complexification will be Stein. This Stein manifold will have the same fundamental group as the original manifold. For details see page 383 of *Narasimhan, R.*, [**On the homology groups of Stein spaces**](http://dx.doi.org/10.1007/BF01428900), Invent. Math. 2, 377-385 (1967). [ZBL0148.32202](https://zbmath.org/?q=an:0148.32202).


