# Codimension-1 subgroups of 3-manifold groups

Let $G$ be a finitely generated group and let $H$ be a subgroup of $G$. $H$ is a codimension-1 subgroup of $G$ if $C_{G}/H$ has more than one end, where $C_{G}$ is the Cayley graph of $G$.

Do all codimension-1 subgroups of a $3$-manifold group correspond to the fundamental group of an immersed surface in the $3$-manifold?

• There's not a bijective correspondence: any $\pi_1$-injective 3-manifold $N$ with boundary that embeds into $M$, for instance, will also be codimension-one. – HJRW Sep 19 '18 at 7:53
• "codimension 1" is a quite misleading terminology as it conflicts with the intuition of codimension; let me call this "coforked". Let $\Gamma$ be a surface group, so $\Gamma\times\mathbf{Z}$ is a 3-manifold group. Let $F$ be an infinite free quotient of $\Gamma$, with kernel $N$. So $N$ is not finitely generated, and $N\times\mathbf{Z}$ is coforked in $\Gamma\times\mathbf{Z}$. – YCor Sep 19 '18 at 7:58
• If $M$ is a 3-manifold with fundamental group $G$, then $G$ is a codimension-one subgroup of $\pi_1(M \# M \# M)$, as $\pi_1(M \# M \#M) \simeq G \ast G \ast G \simeq (G \ast G ) \underset{G}{\ast} (G \ast G)$ splits over $G$. – AGenevois Sep 19 '18 at 8:03

Let me make some remarks on this. As far as I know, the terminology "codimension-1 subgroup" originated from a paper of Micah Sageev

Sageev, Michah, Ends of group pairs and non-positively curved cube complexes, Proc. Lond. Math. Soc., III. Ser. 71, No. 3, 585-617 (1995). ZBL0861.20041. MR1347406.

A subgroup $$H< G$$ of a finitely generated group $$G$$ is codimension-1 if the number of ends $$e(G,H) > 1$$. If $$\Gamma$$ is a Cayley graph for $$G$$, then $$e(G,H)$$ is the number of ends of the graph $$\Gamma/H$$. In this case, Sageev calls the group "semi-splittable". Then the main theorem of the paper is

Theorem Suppose $$G$$ is a finitely generated group. Then $$G$$ acts essentially on a cubing if and only if $$G$$ is semi-splittable.

A cubing is a complete non-positively curved (or CAT(0)) cube complex (where "cube" refers to what is more commonly known as a "hypercube"). If it is finite-dimensional, then "essential" is equivalent to the existence of an unbounded orbit. In a CAT(0) cube complex, one has "hyperplanes" which intersect each cube in a codimension-1 cube (setting one coordinate to = 0, if a cube is parameterized as $$[-1,1]^n$$) or the empty set. Now the stabilizer of a hyperplane in the theorem gives rise to a codimension-1 subgroup of $$G$$, and conversely, a codimension-1 subgroup will stabilize a hyperplane in Sageev's construction. Note that the construction of the cube complex from the subgroup is not canonical, but relies on some choices (these choices have been codified in the notion of "wall spaces"). Hopefully this helps explain the choice of terminology.

Now, assume $$G$$ is the fundamental group of a connected closed irreducible 3-manifold, and assume that $$G$$ has a codimension-1 subgroup (in particular, $$G$$ is infinite). Then $$G=\pi_1(M)$$, where $$M$$ is an aspherical 3-manifold by the sphere theorem. Corresponding to a codimension-1 subgroup $$H, there is a connected covering space $$N\to M$$ such that $$\pi_1(N)=H$$. Since $$e(G,H)>1$$, one has that $$N$$ has more than one end. Hence $$H_2(N;\mathbb{Z}) \neq 0$$. Choose a non-zero homology class $$z\in H_2(N)$$, then we may find an embedded closed surface $$\Sigma \subset N$$ such that $$[\Sigma]=z$$ (meaning that the map induced by the inclusion of the fundamental class of $$\Sigma$$ into $$H_2(N)$$ is $$z$$). If $$\Sigma$$ is not $$\pi_1-$$injective, then we may use the loop theorem to compress $$\Sigma$$ to get a surface of larger Euler characteristic. So by induction, we may assume that (each component of) $$\Sigma$$ is incompressible and hence is $$\pi_1-$$injective. If $$\Sigma$$ is not connected, we may take a component which is still homologically non-trivial. This surface cannot be a sphere because $$N$$ is also irreducible (again by the sphere theorem). Thus, $$\Sigma$$ is a $$\pi_1-$$injective surface. Corollary 5.3 in Sageev's paper states a slightly weaker version of this result. So we may associate to $$H$$ a $$\pi_1$$-injective surface, but in a highly non-canonical way. This argument is due to Swarup. Conversely, any surface subgroup of an aspherical 3-manifold group is codimension-1 (and actually may be realized by an embedding in the corresponding cover by a theorem of Freedman-Hass-Scott).

Evidently the answer to your question is "no". However if $$M$$ is a 3-manifold with a word hyperbolic fundamental group $$G$$ and $$H\leq G$$ is a quasiconvex embedding, then there is always an immersed surface $$\Sigma$$ that is "related" to the codimension-1 subgroup $$H$$.

First note that there is a standard construction due to Stallings's called a resolution which does the following: if $$G$$ acts on a tree $$T$$, then there is an embedded surface $$\Sigma \subset M$$ whose lifts cut the universal cover $$\tilde{M}$$, and give rise to a resolving tree $$\hat T$$, such that there is a $$G$$ equivariant surjection: $$\hat{T} \twoheadrightarrow T.$$ That is to say if $$G$$ splits over a subgroup $$H$$, then it also splits over a subgroup $$H' \leq H$$ that is the fundamental group of an embeded 2-sided 2-manifold. This result comes from basic 3-manifold topology and doesn't depend on the hyperbolicity of $$G$$, nor that $$H\leq G$$ is quasiconvex.

Now if $$G$$ is word hyperbolic then there is a result of Sageev Michah Sageev, Codimension-1 Subgroups and Splittings of Groups, that says that $$G$$ will split over a subgroup of quasiconvex subgroup $$H$$. By the previous remark you'll get a related surface subgroup.

Another way this could go would be to use the Sageev cube complex given by the quasiconvex subgroup $$H$$, and use some form of subgroup separability (which works in the special à la Haglund-Wise case) to show that $$G$$ virtually splits over $$H$$, i.e. by "untangling" the subgroup $$H$$. In this case you'll have an immersed surface related to $$H$$.

The general 3-manifold case, is probably interesting to consider. Hopefully, this can give you some ideas on where to take your question.

• I don't think you need to invoke subgroup separability (which is of course a very big machine) to get a surface. Any codimension-one subgroup $H$ is necessarily of infinite index, so by the Scott core theorem $H$ is represented by an immersed 3-manifold with boundary $N\to M$. The boundary of $N$ is then an immersed surface "related to" $H$. – HJRW Sep 21 '18 at 13:21
• Whoa! That's a good point! Actually the boundary of $N$ will again be an immersed codimension 1 surface! I think this is the true answer to this question. – NWMT Sep 21 '18 at 13:26