# Nonabelian topological fundamental group of a conjugate variety

Let $$X$$ be a pointed algebraic variety over the field of complex numbers $$\mathbb{C}$$. Let $$\pi_1^{\rm top}(X)$$ and $$\pi_1^{\mathrm{\acute{e}t}}(X)$$ denote the topological and the étale fundamental groups of $$X$$, respectively. Then $$\pi_1^{\mathrm{\acute{e}t}}(X)$$ is the profinite completion of $$\pi_1^{\rm top}(X)$$.

Let $$\sigma$$ be an automorphism of $$\mathbb{C}$$, not necessarily continuous. On applying $$\sigma$$ to the coefficients of the polynomials defining $$X$$, we obtain a new pointed variety $$\sigma X$$. Consider $$\pi_1^{\rm top}(\sigma X)$$ and $$\pi_1^{\mathrm{\acute{e}t}}(\sigma X)$$, then again $$\pi_1^{\mathrm{\acute{e}t}}(\sigma X)$$ is the profinite completion of $$\pi_1^{\rm top}(\sigma X)$$. Furthermore, $$\pi_1^{\mathrm{\acute{e}t}}(X)$$ and $$\pi_1^{\mathrm{\acute{e}t}}(\sigma X)$$ are canonically isomorphic.

Now assume that $$\pi_1^{\rm top}(X$$) is abelian. Then its profinite completion $$\pi_1^{\mathrm{\acute{e}t}}(X)$$ is abelian, and $$\pi_1^{\rm top}(X)$$ embeds into its profinite completion $$\pi_1^{\mathrm{\acute{e}t}}(X)$$ (because $$\pi_1^{\rm top}(X)$$ is a finitely generated abelian group).

Question. Can $$\pi_1^{\rm top}(\sigma X)$$ be nonabelian? (Note that its profinite completion $$\pi_1^{\mathrm{\acute{e}t}}(\sigma X)\cong\pi_1^{\mathrm{\acute{e}t}}(X)$$ must be abelian.)

EDIT (July 18, 2020). I added the tag projective-varieties which became relevant; see the answer of Ian Agol.

• @FrancescoPolizzi: If $\sigma$ is the complex conjugation, then the map on $\mathbb{C}$-points $X\to\sigma X$ is a homeomorphism, hence in this case $X$ and $\sigma X$ have isomorphic topological fundamental groups. In Serre's example $\sigma$ is different from complex conjugation. Apr 5, 2015 at 18:59
• @AndreasThom Other examples: J. S. Milne and J. Suh, Nonhomeomorphic conjugates of connected Shimura varieties, Amer. J. Math. {\bf 132}(3) (2010), 731--750. C. S. Rajan, An example of non-homeomorphic conjugate varieties,} Math. Res. Lett. {\bf 18} (2011), 937--943. I. Bauer, F. Catanese, and F. Grunewald, Faithful actions of the absolute Galois group on connected components of moduli spaces, Invent. Math. {\bf 199} (2015), no. 3, 859--888. Apr 5, 2015 at 19:23
• @MikhailBorovoi: In uam.es/personal_pdi/ciencias/ajaikin/preprints/… we have constructed a counterexample for any $\sigma$ in the absolute Galois group of $\mathbb Q$, different from the identity and the complex conjugation Apr 9, 2015 at 13:43
• Do we even know that if $\pi_1^{\rm top}(X)$ is trivial, then so is $\pi_1^{\rm top}(\sigma X)$? Jul 12, 2020 at 8:52
• @Piotr Achinger I believe this is unknown! (Moreover, I'm not even sure if it is expected...)
– Raju
Jul 12, 2020 at 13:08

The first reduction is to consider varieties with fundamental group $$\mathbb{Z}^{2n}$$ by passing to a finite abelian cover. For $$n=0$$, one has for example projective space $$\mathbb{P}^k$$, and for $$n>0$$, one has abelian varieties. These spaces should have Galois conjugates having fundamental group $$\mathbb{Z}^{2n}$$, e.g. products of elliptic curves.
Next, I think one can reduce to the case of algebraic surfaces. By the Lefschetz hyperplane theorem, the intersection of a smooth projective variety $$X^m \subset \mathbb{P}^k$$, $$m=dim X > 2$$ with a generic (transverse) hyperplane $$\mathbb{P}^{k-1}\subset \mathbb{P}^k$$ will have intersection $$X^m \cap \mathbb{P}^{k-1} = X^{m-1}$$ a smooth subvariety of one less dimension $$m-1$$ and with $$\pi_1(X^{m-1})\cong \pi_1(X^m)$$. Iterating, we see that there is a surface $$X^2\subset X^m$$ such that $$\pi_1(X^2)=\pi_1(X^m)$$ such that $$X^2=X^m\cap \mathbb{P}^{k-2} \subset \mathbb{P}^k$$.
However, there doesn't seem to be much known about surfaces with abelian fundamental group, even simply-connected. I think one case that follows though is the case of surfaces which are complete intersections of smooth hypersurfaces intersecting transversely (see Mandelbaum). In fact, Bott's argument shows that any transverse holomorphic section of a positive line bundle over a variety (smooth ample divisor) satisfies the Lefschetz theorem, and hence will be simply connected. But the property of being a transverse intersection of smooth ample divisors should be a Galois-invariant property, and hence the Galois conjugates $$\sigma X$$ should also be simply-connected.
More generally, one gets smooth projective manifolds with abelian fundamental group by taking a abelian variety $$A$$ and crossing it with projective space $$A \times \mathbb{P}^k$$. Again, this property should be invariant under Galois conjugation, since abelian varieties admit an algebraic group structure which is Galois-invariant. Now take sequences of smooth ample divisors on such spaces to obtain smooth projective surfaces with fundamental group $$\mathbb{Z}^{2n}$$ and whose Galois conjugates should also have abelian fundamental group $$\mathbb{Z}^{2n}$$ by a similar argument.
So the question one can ask: does every smooth projective variety $$X^m$$ with $$\pi_1(X^m)=\mathbb{Z}^{2n}$$ admit a smooth subsurface $$X^2\subset X^m$$ which is a transverse complete intersection of smooth ample divisors in an abelian variety $$A \times \mathbb{P}^k$$? Or maybe $$X$$ itself has this property? This appears to be false for the case of simply-connected smooth projective algebraic surfaces. The thesis of Giancarlo Urzúa mentions that for smooth complete intersections, $$c_1^2(X)\leq 2 c_2(X)$$. However, there are known examples of simply connected smooth projective surfaces with $$c_1^2(X)/c_2(X) > 2$$, going back to Moishezon-Teicher. One could then investigate whether Galois conjugates of these surfaces are simply-connected, and then take smooth hypersurfaces in these times abelian varieties.