Compressed subroups of free groups are inert

In this post, I would like to discuss my recent result where I proved that every compressed subgroup of a finitely generated free group is inert. The paper can be accessed through this link. This solves a conjecture of Dicks and Ventura proposed in this monograph.

A finitely generated subgroup $H$ of a free group $F$ is called compressed if for any subgroup $L$ of $F$ containing $H$ , ${\rm rk}(H)\le {\rm rk}(L)$ , and it is called inert if for any finitely generated subgroup $L$ of $F$ the intersection $H\cap L$ is compressed in $L$ . It is clear that an inert subgroup is compressed.

The nature of these two notions seems very different. It follows directly form the definition of inert subgroups that the intersection of two inert subgroups is inert, but it is not clear in the case of compressed subgroups. On the other hand, it is relatively easy to construct an algorithm that decides whether a given finite collection of elements of $F$ generates a compressed subgroup. However, such an algorithm was not known for inert subgroups.

The concept of a subgroup being compressed is quite intuitive; it indicates that the subgroup is not contained in another subgroup of smaller rank. The concept of an inert subgroup was introduced in the work of Dicks and Ventura while they were studying equalizers in free groups.

Given two endomorphisms $\phi$ and $\psi$ of a finitely generated group $F$ , we denote by ${\rm Eq} (\phi, \psi)$ the subgroup of $F$ consisting of $x\in F$ that satisfies $\phi(x)=\psi(x)$ . If we do not put any condition on $\phi$ and $\psi$ , then ${\rm Eq} (\phi, \psi)$ can be infinitely generated. Hovewer, if one of two endomorphisms is injective we arrive to the following conjecture:

The equalizer conjecture: If $\phi$ or $\psi$ is injective, then ${\rm rk} ( {\rm Eq} (\phi, \psi))\le {\rm rk}(F)$ .

This conjecture has its roots in the study of fixed subgroups of endomorphisms ${\rm Fix}(\phi)=\{x\in F\colon \phi(x)=x\}$ (it corresponds to the case where $\psi$ is the identity). In this article Bestvina and Handel showed that ${\rm Fix}(\phi)\le {\rm rk}(F)$ if $\phi$ is an automorphism of $F$ and here Imrich and Turner extended this for arbitrary endomorphisms $\phi$ .

If $H$ is a non-abelian finitely generated subgroup of $F$ and $\alpha$ is an embedding of $F$ into $H$ , then the restrictions of $\alpha\circ \phi$ and $\alpha\circ \psi$ on $H$ , denoted by $\phi_1$ and $\psi_1$ , respectively, are endomorphisms of $H$ . Thus, the equalizer conjecture predicts that $\displaystyle H\cap {\rm Eq} (\phi, \psi)={\rm Eq} (\phi_1, \psi_1)\le {\rm rk}(H)$ , that is, ${\rm Eq} (\phi, \psi)$ is an inert subgroup of $F$ . This was the motivation of Dicks and Ventura to introduce the inertness property.

The inertness of ${\rm Fix}(\phi)$ for an injective endomorphism of $F$ was proved by Dicks and Ventura, and for an arbitrary endomorphism by Antolín and Jaikin in this paper. A key new notion that was introduced in that paper was the notion of $L^2$ -independence. We say that a subgroup $H$ is $L^2$ -independent in $F$ if the map $l^2(G)\otimes _{\mathbb Q[H]} I_{\mathbb Q[H]}\to l^2(G)\otimes _{\mathbb Q[F]} I_{\mathbb Q[F]}$ , induced by the embedding of $H$ into $F$ , is injective. Here, $\mathbb Q[H]$ is the group algebra of $H$ over $\mathbb Q$ , and $I_{\mathbb Q[H]}$ is its augmentation ideal.

The notion of $L^2$ -independent subgroups can be defined in arbitrary groups. In the case of a free group, it can be reformulated in a purely algebraic way, which also allows to speak about its analogues in positive characteristic. Let $K$ be a field and $F$ a free group. Any submodule $M$ of a free left $K[F]$ -module $L$ is again free, and so, we can speak about its rank. If $M$ is finitely generated we say that it is compressed in $L$ if it is not contained in a $K[F]$ -submodule of $L$ of smaller rank.

It turns out that for $K[F]$ -submodules of a free $K[F]$ -module $L$ the property to be compressed can be characterized in homological terms which provides certain flexibility to work with it. The group algebra $K[F]$ can be embedded in the univeral division $K[F]$ -ring of fractions $\mathcal D_{K[F]}$ . There are different ways to construct $\mathcal D_{K[F]}$ , but all of them require some work, so I will omit the construction. We can show that a submodule $M$ of $L$ is compressed if and only if the map $\mathcal D_{K[F]}\otimes _{K[F]} M\to \mathcal D_{K[F]}\otimes _{ K[F]} L$ , induced by the embedding of $M$ into $L$ , is injective, or equivalently, ${\rm Tor}_1^{K[F]}(\mathcal D_{K[F]}, M/L)$ is trivial.

The $K[F]$ -module $I_{K[F]}$ is free. Denote by ${}^F I_{ K[H]}$ the $K[F]$ -submodule of $I_{K[F]}$ generated by $I_{ K[H]}$ . Then we say that $H$ is $\mathcal D_{K[F]}$ -independent in $F$ if ${}^F I_{ K[H]}$ is compressed in $I_{K[F]}$ or, in other words, if the map $\mathcal D_{K[F]}\otimes _{ K[H]} I_{K[H]}\to \mathcal D_{K[F]}\otimes _{ K[F]} I_{K[F]}$ , induced by the embedding of $H$ into $F$ , is injective.

The notions of $L^2$ -independence and $\mathcal D_{\mathbb Q[F]}$ -independence are equivalent. In my paper I also show that for every prime $p$ , $\mathcal D_{\mathbb F_p[F]}$ -independence implies $\mathcal D_{\mathbb Q[F]}$ -independence. Thus, we obtian the followin list of implications for a subgroup of $F$ :

$\mathcal D_{\mathbb F_p[F]}$ -independent $\Rightarrow$ $\mathcal D_{\mathbb Q[F]}$ -independent $\Rightarrow$ $L^2$ -independent $\Rightarrow$ inert $\Rightarrow$ compressed.

I was aware of these implications for a while. A new insight was understanding how to show that a compressed subgroup is also $\mathcal{D}_{\mathbb{F}_2[F]}$ -independent. I strongly believe that, in general, compressed subgroups are $\mathcal{D}_{\mathbb{F}_p[F]}$ -independent for every $p$ , but the argument we currently possess relies essentially on the assumption $p=2$ .