As a consequence of the loop theorem, if $F$ is a closed surface in the boundary of a 3-manifold and if the kernel $N = \ker(\pi_1(F) \to \pi_1(M))$ is nonempty then there is a nontrivial element of $N$ that can be represented by an embedded curve. I read in Hempel's book that there are lots of normal subgroups of $\pi_1(F)$ that contain no nontrivial elements that can be represented by embedded curves. How can I cook up such subgroups and prove that they have this property?

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    $\begingroup$ In fact, there's a finite-sheeted cover with this property. Any embedded non-separating curve is non-trivial on $H_1(F;\mathbb{Z}/2)$, hence will lie in the kernel of the map $\pi_1(F)\to H_1(F;\mathbb{Z}/2)$. Pass to the cover $\tilde{F}$ corresponding to this kernel; one may check that every separating curve lifts to a non-separating curve in this cover. So take the cover of $\tilde{F}$ corresponding to $ker\{ \pi_1(\tilde{F})\to H_1(\tilde{F};\mathbb{Z}/2)\}$ to get a normal subgroup with the desired property. $\endgroup$ – Ian Agol May 4 '18 at 5:00
  • $\begingroup$ @IanAgol it seems what you call "this property" is the negation of what the OP calls "this property". But none of you is explicit, and you're also not explicit on what you claim to prove "a finite-sheeted cover..." of what? $\endgroup$ – YCor May 4 '18 at 8:32
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    $\begingroup$ @YCor : I made a mistake, saying cover when I meant to say subgroup (I internally make this identification between subgroups and covers in my thinking all the time, and many times accidentally express it externally). In any case, the question posed about subgroups is equivalent to asking for a regular connected cover of the surface to which no simple closed loop lifts. Simple loops represent conjugacy classes of the fundamental group, and hence there is no ambiguity in discussing whether they are represented by an element of a normal subgroup. $\endgroup$ – Ian Agol May 4 '18 at 13:50
  • $\begingroup$ @IanAgol I understood that you were talking of finite index subgroups, but couldn't figure out what you claimed. I now guess you mean that for every $N$ there exists a finite index subgroup $F'$ of $F$ such that $F'\cap N$ contains no nontrivial element represented by an embedded curve. $\endgroup$ – YCor May 4 '18 at 16:24
  • $\begingroup$ @YCor : in your notation F' is the requested subgroup N. he's just asking for the existence of such a subgroup. $\endgroup$ – Ian Agol May 4 '18 at 17:26

One source of interesting examples is the lower central series. In my paper

J. Malestein, A. Putman, On the self-intersections of curves deep in the lower central series of a surface group, Geom. Dedicata 149 (2010), no. 1, 73–84.

my coauthor and I show that the minimal number of self-intersections among nontrivial elements of the kth term of the lower central series of a surface group goes to infinity as k goes to infinity.


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