# Do complex varieties have a dense open subset with residually finite fundamental group?

Let $S$ be a smooth connected variety over the complex numbers. The fundamental group might not be residually finite (i.e., the homomorphism $\pi_1(S(\mathbb C)) \to \pi_1^{\mathrm{et}}(S)$ might not be injective).

Is there a dense Zariski open subset $U\subset S$ such that $\pi_1(U(\mathbb C))$ is residually finite?

• Yes, but I I don't have time to write a long answer right now. If $U$ is a so called Artin neighbourhood (= "bon voisinage" in SGA4, exp XI), then the fundamental group is free by free by free etc. This should be residually finite. Jun 14, 2017 at 15:07
• @DonuArapura Interesting! Many thanks. If you have the time, I would be very happy to see a bit of the details. Jun 14, 2017 at 17:25

In SGA4 exp XI, Artin constructs a nonempty Zariski open $U\subset S$ which admits a sequence $U=U_n \to U_{n-1}\to\ldots$ which are topological fibrations with curves as fibres. Without loss of generality, these fibrations, and the base of the fibrations, can be taken to be non proper. It follows that $\pi_1(U)$ admits a finite filtration by normal subgroups such that the quotients are free and finitely generated (cf. YCor's comment). An easy induction shows that $\pi_1(U)$ is residually finite. Note that the base case of the induction, where $\pi_1(U)$ is free, is discussed here: Why are free groups residually finite?
• If a group has a subnormal series $1=G_0\le G_1\le\dots\le G_n=G$ such that each $G_i$ is normal in $G_{i+1}$ and $G_{i+1}/G_i$ is free and finitely generated, then $G$ is residually finite. (It's probably false if one only assumes that $G$ is finitely generated and that the $G_{i+1}/G_i$ are free.) It's certainly enough here.