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Are there any known examples of groups that virtually split that don't have a codimension-1 subgroup?

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2 Answers 2

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No. For a finitely generated group, the property "have no codimension 1 subgroup" is called Property FW. Property FW passes to finite index subgroups, so excludes virtual splittings.

Actually for an infinitely generated group, the existence of a "codimension 1 subgroup", the transitive version of Property FW, is still discarded by virtual splittings, but there are maybe a few details to check. I'll assume in the absence of context that you're interested by finitely generated groups.

Edit: here are missing references. That Property FW passes to finite index subgroups is Proposition 5.B.1 in this arxiv paper. The non-existence of a "codimension 1 subgroup" is called Property FW' (and equivalent to Property FW for finitely generated groups, but not in general). That is passes to finite index subgroups is obtained by a similar argument. The whole point is that when one induces an action of a finite index subgroup to the whole group (see §4.F in the above reference), this maps a transitive action to an action with finitely many orbits. I don't use CAT(0) cube complexes for this.

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In addition to Yves' answer, here are a few details. So the statement is:

Theorem: Let $G$ be a (not necessarily finitely-generated) group and $H \subset G$ a finite-index subgroup. If $H$ contains a codimension-one subgroup, then so does $G$.

The first step is Niblo and Roller's theorem: A group contains a codimension-one subgroup if and only if its acts fixed-point freely on a connected cube with a single orbit of hyperplanes (Groups acting on cubes and Kazhdan's property (T), 1998). The second step is notice that an action of the entire group on a CAT(0) cube complex can be constructed from such an action of a finite-index subgroup. I think that the idea of the construction goes back to Serre's article Cohomologie des groupes discrets (1971).

Proposition: Let $G$ be a group, $H \subset G$ an finite-index subgroup. Suppose that $H$ acts fixed-point freely on some CAT(0) cube complex $X$. Then $G$ acts fixed-point freely on $X^{[G:H]}$.

Proof. Consider the set $$\mathrm{Hom}_H(G,X) = \{ f : G \to X \mid \forall h \in H, \forall g \in G, \ f(hg)=h \cdot f(g) \}.$$ $G$ acts on $\mathrm{Hom}_H(G,X)$ via: $$g \cdot f : x \mapsto f(g^{-1}x).$$ Fixing some representatives $G=Ha_1 \sqcup \cdots \sqcup Ha_r$, the map $$\varphi(a_1, \ldots,a_r) : \left\{ \begin{array}{ccc} \mathrm{Hom}_H(G,X) & \to & \prod\limits_{i=1}^r X=X^{[G:H]} \\ f & \mapsto & (f(a_1), \ldots, f(a_r)) \end{array} \right.$$ is a bijection. Notice that, if $b_1, \ldots, b_r$ are other representatives, then $$\varphi(b_1, \ldots, b_r) \varphi(a_1, \ldots, a_r)^{-1} : (x_1, \ldots, x_r) \mapsto (h_1x_{\sigma(1)}, \ldots, h_r x_{\sigma(r)}),$$ where $\sigma : \{1, \ldots, r \} \to \{1,\ldots,r\}$ is a permutation and $h_1, \ldots, h_r$ elements of $H$ such that $b_i=h_ia_{\sigma(i)}$ for every $1 \leq i \leq r$. Consequently, if we endow $X^{[G:H]}$ with its natural cubical structure, then then the previous map is an isomorphism.

The conclusion is that, thanks to $\varphi(a_1, \ldots, a_r)$, you can transfer the action $G \curvearrowright \mathrm{Hom}_H(G,X)$ to an action $G \curvearrowright X^{[G:H]}$ by isometries. Finally, notice that this action does not fix a point if $H \curvearrowright X$ is fixed-point free. $\square$

Remark: In the statement of the proposition, it is possible to replace "fixed-point freely" with "properly", but not with "geometrically" (see for instance the Coxeter group I mention here).

Now the theorem can be proved. Suppose that $H$ contains a codimension subgroup. Then $H$ acts fixed-point freely on a connected cube $X$ with a single orbit of hyperplanes, and we deduce from the construction given in the previous proposition that $G$ acts fixed-point freely on $X^{[G:H]}$, which is also a connected cube. It follows from the definition of this action that $X^{[G:H]}$ contains at most $[G:H]$ orbits of hyperplanes. Choose carefully one of them, say $\mathcal{W}$, so that the action of $G$ on the new CAT(0) cube complex $Y$, obtained by cubulating the wallspace $\left( X^{[G:H]}, \mathcal{W} \right)$, is also fixed-point free. The conclusion follows from the observation that $Y$ is again a connected cube, and that by construction it contains a single $G$-orbit of hyperplanes.

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  • $\begingroup$ Niblo-Roller's theorem: I doubt it holds for infinitely generated groups. A counterexample ($\mathrm{SL}_3(\mathbf{Q})$) seems to be given by Theorem 1.17 in arxiv.org/abs/1308.1318 $\endgroup$
    – YCor
    Commented Apr 25, 2018 at 13:40
  • $\begingroup$ If I translate Property FW' in the world of cube complexes, does it mean that I am considering actions with a single orbit of hyperplanes or of halfspaces? My first guess would be the latter case, which may be different from the former one. Also, is it clear that $\mathrm{SL}_3(\mathbb{Q})$ contains a codimension-one subgroup? $\endgroup$
    – AGenevois
    Commented Apr 25, 2018 at 18:35
  • $\begingroup$ Probably this is the restatement? I don't understand the second question, since precisely the reference says that $SL_3(Q)$ has Property FW', i.e., has no "codimension 1 subgroup". (I always use quotation marks since this is a counterintuitive and confusing terminology.) $\endgroup$
    – YCor
    Commented Apr 25, 2018 at 20:01
  • $\begingroup$ The point I do not understand is why, if a group $G$ satisfying Property FW' acts on a connected cube $C$ with a single orbit of hyperplanes, then the action $G \curvearrowright C$ must fix a point. I guess the begining of the proof should be as follows: Fix a vertex $v \in C$ and a halfspace $D$ containing $v$. $G$ acts on $G \cdot D$ by commensurating the subset $V$ of halfspaces containing $v$. Property FW' implies that the action is transfixed. If $V$ is infinite, I agree that $G$ fixes $v$. Otherwise, if $V$ is finite, how do you conclude? $\endgroup$
    – AGenevois
    Commented Apr 28, 2018 at 8:11
  • $\begingroup$ Prop: $G$ acts on a conn. cubing. Let $B_v$ be the set of halfspaces containing $v$. Then $G$ fixes a point iff $B_v$ is transfixed. Indeed, since $|gB_v\Delta B_v|=2d(v,gv)$, $B_v$ is transfixed $\Leftrightarrow$ $g\mapsto |gB_v\Delta B_v|$ is bounded $\Leftrightarrow$ $g\mapsto 2d(v,gv)$ is bounded $\Leftrightarrow$ there's a fixed point. $\endgroup$
    – YCor
    Commented Apr 28, 2018 at 8:30

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