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?

  • 5
    $\begingroup$ 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. $\endgroup$ Jun 14, 2017 at 15:07
  • $\begingroup$ @DonuArapura Interesting! Many thanks. If you have the time, I would be very happy to see a bit of the details. $\endgroup$
    – Randy
    Jun 14, 2017 at 17:25

1 Answer 1


(I'm converting my comment to an answer.)

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?

  • 1
    $\begingroup$ 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. $\endgroup$
    – YCor
    Jun 14, 2017 at 21:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.