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Call an algebraic variety $\pi_1$-subgroup separable iff, for every $Y\subseteq X$ a closed subvariety and $\hat Y\xrightarrow{i} Y$ a normalisation of $Y$, and subgroup $\Gamma=Im(\pi_1(\hat Y,y)\xrightarrow{i_*} \pi_1(X,x))$, it holds that $\Gamma$ is the intersection of the finite index subgroups contaning $\Gamma$.

This condition is obviously weaker than $\pi_1(X,x)$ is subgroup separable (aka lerf); by definition, a group is subgroup separable iff the condition above holds for each finite generated subgroup $\Gamma$. Thus it holds if $\pi_1(X,x)$ is Abelian or nilpotent.

But is this condition much weaker?
I am interested to to see as many examples of complete projective varieties $X$ such that all $X^n$ have this property; I have a theorem that holds for such varieties.

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  • $\begingroup$ The trouble is that we know very little about such subgroups. For instance, product of surface groups is not LERF but I see no way to prove or disprove that images of fundamental groups of Riemann surfaces in it are separable. $\endgroup$
    – Misha
    Mar 23, 2014 at 23:14
  • $\begingroup$ There are examples of Deligne of lattices which are not residually finite (central extensions of $Sp(2n,\mathbb{Z})$). One might be able to find such an example which is a fundamental group of an algebraic variety, and such that the center is the fundamental group of a subvariety (so it wouldn't be separable). But this is speculation - I don't know how one would carry out such a construction. However, by analogy, I know that $\tilde{Sp(2,\mathbb{Z})}= B_3$ is the fundamental group of an algebraic variety (although this case is subgroup separable). mathoverflow.net/a/79283/1345 $\endgroup$
    – Ian Agol
    Mar 24, 2014 at 4:50
  • $\begingroup$ Thanks, interesting! But I am more interested in positive examples. Are there any large classes of known examples that do have the property (that my theorem applies to), except for varieties with nilpotent fundamental groups? $\endgroup$
    – mmm
    Mar 24, 2014 at 5:41

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