# growth of a free group automorphism is same for finite index subgroups?

Let $X=\{x_1,\dots,x_N\}$ and $F=F(X)$ be a free group generated by $X$. Let $\phi\colon F\to F$ be an automorphism of $F$. Define a growth function of $\phi$ as: $$\operatorname{gr}_{\phi,X}(n)=\operatorname{max}_{1\le i\le N}\{\|\phi^n(x_i)\|_X\},$$ where $\|.\|_X$ denotes the word length with respect to $X$. We consider these functions up to equivalence $f\simeq g$ defined in the following way. For functions $f,g\colon [0,+\infty)\to[0,+\infty)$ we say that $f \preceq g$ if there exist $C>0$ such that for all $n\in [0,+\infty)$: $$f(n)\le Cg(Cn+C)+C.$$ We say that $f\simeq g$ if $f\preceq g$ and $g\preceq f$. We extend this relation to functions $\mathbb N\to [0,+\infty)$ by assuming them to be constant on each interval $[n,n+1)$.

[EDIT 07/28/17: Since the function $n\mapsto \|\phi^n(x_i)\|_X$ is not monotone (think of a finite order automorphism modeled on some subset of $X$), it is better to use a slightly different definition of $f\preceq g$: we say that $f\preceq g$ if there exist constants $A,B>0$, $C,D\geq0$ such that for all $n\in [0,+\infty)$: $$f(n)\le Ag(Bn+C)+D.]$$

It can be shown that, viewed up to $\simeq$ equivalence, functions $\operatorname{gr}_{\phi,X}(n)$ are independent of generating set $X$ (so can be denoted just $\operatorname{gr}_\phi(n)$). And it looks plausible that:

If $H\le F$ is a subgroup of finite index, invariant under $\phi$, then $\operatorname{gr}_{\phi|_H}(n)\simeq \operatorname{gr}_{\phi}(n)$.

Question: Is the detailed proof of this statement written somewhere in the literature?

(I somehow find it difficult to prove that $\operatorname{gr}_{\phi|_H}(n)\succeq \operatorname{gr}_{\phi}(n)$, for arbitrary automorphism $\phi$.)

• A remark: in general in a f.g. group this does not hold. For instance, in the infinite dihedral group, which has an infinite cyclic subgroup $Z$ of index 2, the inner automorphism with respect to a nontrivial element of $Z$ has linear growth, but its restriction to $Z$ is the identity. – YCor Jul 15 '17 at 13:17

While I cannot point to any place in the literature with this particular statement, I would say that it is an exercise in relative train track theory, which is the very nice normal form theory for $\text{Out}(F_n)$ and $\text{Aut}(F_n)$ found in papers of Bestvina, Feighn, and Handel (in various subgroupings). Usually relative train track theory is applied to $\text{Out}(F_n)$ but you can apply it to $\text{Aut}(F_n)$ with a little trick.

First, you can embed $\text{Aut}(F_n) \hookrightarrow \text{Out}(F_{n+1})$ by letting $F_{n+1}=\langle a_0,a_1,...,a_n\rangle$, and restricting to the subgroup in which $a_0$ is fixed and $a_1,...,a_n$ are mapped to themselves. Every $\phi \in \text{Aut}(F_n)$ is represented by a relative train track map $f : G \to G$ where $G$ has a closed edge labelled $a_0$ with base point $p$ such that $a_0$ is fixed by $f$, and $f$ preserves $G \setminus H$, inducing the automorphism $\phi$ on $\pi_1(G \setminus H,p) \approx F_n=\langle a_1,...,a_n\rangle$. Let's refer to this map $f$ as a "relative train track representative of $\phi \in \text{Aut}(F_n)$" (this is a nonstandard terminology: ordinarily relative train maps are things which represent outer automorphisms, and this $\phi$ does indeed represent an outer automorphism of $F_{n+1}$, but we have used the "fixed loop trick" to make $\phi$ also represent an actual automorphism of $F_n$).

I still have to say what the "normal form" properties of $f$ are, i.e. I have to tell you a little bit about the definition of a relative train track map. After passing to a power, $G$ is written as a nested union $$G_0 \subset G_1 \subset ... \subset G_K=G$$ of $f$-invariant subgraphs, starting with $G_0=a_0$. For each subgraph $H_i = G_i - G_{i-1}$ there are three possibilities: $H_i$ is a "zero stratum" which means that $H_i$ is contractible and its image under $f$ is contained in $G_{i-1}$; or $H_i$ is an "NEG stratum" (meaning "non-exponentially growing") meaning that $H_i$ is a single edge $E$ such that $f(E)=Eu$ with $u$ some path in $G_{i-1}$; or $H_i$ is an "EG stratum" meaning that has $m \ge 2$ edges the $m \times m$ transition matrix of $f$ acting on the edges of $H_i$ is a Perron-Frobenius matrix, and some additional "non cancellation" properties hold which I won't bother to make explicit.

So, the major points are:

1. One can detect the growth type of $f$ from this normal form. For example, $\phi$ has exponential growth if and only if some stratum $H_i$ is an EG stratum. The NEG strata are so named because each crosses itself only once, but the NEG strata can generate polynomial growth by the manner in which they cross over lower stratum edges (think of an example on $F_3 = \langle a_1,a_2,a_3 \rangle$ where $a_1 \mapsto a_1$, $a_2 \mapsto a_2 a_1$, and $a_3 \mapsto a_3 a_2$; in this example, $a_3$ has quadratic growth). One outcome of this is that every automorphism of $F_n$ has either exponential growth or polynomial growth of some integer degree (between $0$ and $n-1$).
2. One can obtain a relative train track map representing $\phi | H \in \text{Aut}(H)$ from a covering space trick as follows. Construct a pointed covering space of $(G \setminus a_0,p)$ representing the subgroup $H$. Then attach a closed edge to each lift of $p_0$, obtaining a covering space $G'$ of $G$, a subgraph $A \subset G'$ which is the union of the lifts of $a_0$, and a base point $p' \in G'$ to which one of the loops $a'_0$ is attached, so that $(G' \setminus A,p')$ is the covering space of $(G \setminus a_0,p)$ representing $H$. The map $f$ lifts to a map $f' : G' \to G'$ fixing $p'$, and preserving $G' \setminus A$, so that the restiction of $f'$ to $(G' \setminus A,p')$ induces the homomorphism $\phi | H$. Thus, we may think of $f' : G' \to G'$ as a train track representative of $\phi | H$ (again a nonstandard terminology). One needs to check that $f'$ satisfies the relative train track axioms, using that $f$ satisfies those axioms, but this is straightforward. We can therefore use $f'$ to detect the growth properties of $\phi | H$.
3. If $\phi$ has exponential growth, then $G$ has an EG stratum $H_i=G_i \setminus G_{i-1}$, implying that the inverse image of $H_i$ in $G'$ is a (union of) EG strata of $f'$, implying that $\phi | H$ has exponential growth.
4. If $\phi$ has polynomial growth, equivalently $\phi$ does not have exponential growth, then every stratum of $G$ is an NEG stratum, implying that every stratum of $G'$ is an NEG stratum, implying that $\phi | H$ does not have exponential growth, implying that $\phi | H$ has polynomial growth.
5. I believe that one can argue similarly regarding degrees of polynomial growth, demonstrating that if $\phi$ has polynomial growth of degree $d$ then $\phi'$ has polynomial growth of degree $d$.
• Dear Professor Mosher, Thank you for such extended answer. What would you recommend for a (possibly gentle) introduction into this train-track machinery? – mathreader Jul 17 '17 at 4:42
• So far the best resources are still the original papers, which are not so gentle. Start with the Bestvina Handel paper "Train tracks and automorphisms of free groups" in which relative train track (RTT) representatives are first defined and their existence is first proved. There are 2nd and 3rd generation of RTT maps: the 2nd generation called improved RTT maps in "The Tits Alternative for Out(F_n) I: Dynamics of exponentially growing outer automorphisms"; the 3rd generation called completely split RTT maps in "The recogniticion theorem for Out(F_n)" by Feighn and Handel. – Lee Mosher Jul 17 '17 at 12:43
• I'll say that there are more gentle, later generation proofs of existence of train track maps under strong hypotheses. But your problem really needs the full power of relative train track maps, and there are no later generation papers about those. – Lee Mosher Jul 17 '17 at 12:43
• Have said all of that, for your problem, at least for the "exponential/nonexponential" dichotomy, you could get by with just the 1st generation relative train track maps. To pin down the exact degree of polynomial growth for an automorphism which is not exponentially growing, you probably need the 2nd generation relative train track maps. – Lee Mosher Jul 17 '17 at 12:46

I have an idea that I believe worked out in detail would give a proof, although I haven't done this:

Without loss of generality we can assume that $H$ is a normal subgroup, because every finite-index subgroup contains a finite-index normal subgroup.

Then if $x$ and $y$ are two elements in the same class modulo $H$, then both $xy^{-1}$ and $y^{-1} x$ are contained in $H$, so we may bound $||\phi^n(xy^{-1})||$ and $||\phi^n(y^{-1}x)||$ using the growth function of $\phi$ restricted to $H$. This is useful, because if we write $x$ and $y$ as a string of symbols, all but $||\phi^n(xy^{-1})||$ symbols on the right side of $\phi^n(x)$ agree with all but $||\phi^n(xy^{-1})||$ symbols on the right side of $\phi^n(y)$, and similarly all but $||\phi^n(y^{-1}x)||$ symbols on the left side of $\phi^n(x)$ agree with all but $||\phi^n(y^{-1}x)||$ symbols on the left side of $\phi^n(y)$.

If the length of $\phi^n(x)$ is much larger than $||\phi^n(xy^{-1})||+||\phi^n(y^{-1}x)||$ then this implies that $\phi^n(x)$ and $\phi^n(y)$ have a peculiar structure. They each have a beginning string, an ending string, and between them a middle string repeated different numbers of times. The periodicity of the middle string comes from applying the equality of the left sides to get between $\phi^n(x)$ and $\phi^n(y)$ and then applying the equality of the right sides to get back, which represents a translation of $||\phi^n(x)|| - ||\phi^n(y) || \leq ||\phi^n(xy^{-1})||$ and preserves every symbol.

So if $\phi^n(x)$ is indeed much larger than the growth function of $\phi$ restricted to $H$, this should imply that for all $y$ sufficiently close to $x$ in the same class modulo $H$, $\phi^n(y) = g_1 g_2^k g_3$ for fixed $g_1,g_2,g_3$ and some $k$ depending on $y$. Hence we have $y = \phi^{-n} ( g_1) \phi^{-n}(g_2)^k \phi^{-n}(g_3)$ for all such $y$, which surely can't ahppen in the free group as long as we take the neighborhood large enough.

• Thank you for your reply, it may turn useful if one would like to avoid dealing with train-tracks. – mathreader Jul 27 '17 at 9:13

An alternate way to address point 5 in Lee Mosher's above answer, using only that $\phi$ and $\phi|_H$ have quasi-isometric mapping tori, is to use a result of Macura (Macura, Nataša, Detour functions and quasi-isometries., Q. J. Math. 53, No. 2, 207-239 (2002). ZBL1036.20033.).

Indeed, let $G=F\rtimes_\phi\mathbb Z$ and let $G'=H\rtimes_\phi\mathbb Z$. Suppose that $\phi$ has polynomial growth of order $r$. Then, as Lee explained above, $\phi|_H$ also has non-exponential, and thus polynomial, growth. But Macura's result (Theorem 1.1 in the above) says that $G$ has detour function a polynomial of degree exactly $r+1$, and another application of the same theorem shows that $G'$ has detour function a polynomial of degree $r'+1$, where $r'$ is the degree of the polynomial growth function of $\phi|_H$. But $G$ is obviously quasi-isometric to $G'$, and Macura shows that the order of the detour function is a quasi-isometry invariant, so $r=r'$.

(The detour function is very similar to the perhaps more familiar divergence function.)

• Mark, thank you for the insight! I was wondering why it is obvious that $G$ is quasi-isometric to $G'$? – mathreader Aug 14 '17 at 7:19
• Hi mathreader. $G'$ is a finite-index subgroup of $G$, and in particular they are qi. One way to see this is to represent $\phi:F\to F$ by a homotopy equivalence $f:\Gamma\to\Gamma$, where $\Gamma$ is a finite graph. Let $X$ be the mapping torus of $f$, so $G\cong\pi_1X$. Let $\Gamma'\to\Gamma$ be the finite cover corresponding to $H$. Since $H$ is $\phi$--invariant, $f$ lifts to $f':\Gamma'\to\Gamma'$ representing $\phi|_H$. The mapping torus $X'$ of $f'$ has $\pi_1X'\cong G'$, and $\Gamma'\to\Gamma$ induces a finite-sheeted covering map $X'\to X$. – Mark Hagen Aug 15 '17 at 15:59
• Thanks! I didn't realize that G' is f.i. in G. It follows from Svarc-Milnor lemma then that they are q.i. – mathreader Aug 15 '17 at 20:02