How fast does the number of "fixed" points grow compared to the size of the ball in the following group?

Question

I have copied this question from Math.StackExchange, in the hope that some experts here can provide some relevant insight.

Let $M = \oplus_{i\in \mathbb Z} V^{(i)}$ where each $ V^{(i)} \cong \mathbb Z ^5 $. Let $e_1, e_2, e_3, e_4, e_5$ be the standard basis of $\mathbb Z^5$. Let $e_j^{(i)}$ be the element in $M$ whose $i$-th coordinate is $e_j$ and other coordinates are $0$.

Consider the following subgroup $ N = \left\{\begin{array}{c|c} e_1^{(i)}-e_1^{(i+1)} & \\ e_2^{(i)}-e_4^{(i+1)} & i \in \mathbb{Z}\\ e_3^{(i)}-e_5^{(i+1)} & \\ e_3^{(i)}+e_4^{(i)}+e_5^{(i)}+e_1^{(i)}-e_2^{(i+1)} & \end{array}\right\}. $

Let $K = M/N$ and $G = K\rtimes_\phi \mathbb{Z}$ where $\phi(1)$ shifts coordinate one place to the right in $K$. Let $F = \langle e_1^{(i)} \mid i \in \mathbb{Z} \rangle$ be the set of fixed points of $\phi$, this is a normal subgroup of $G$.

Let $T = \{ e_i^{(0)} \mid 1 \leq i \leq 5 \} $ be a finite subset of $K$ and $S = T \cup \{ (0,1) \}$ be the natural generating set of $G$.

Question: Can we show that $ \lim _{r \rightarrow \infty} |S^r \cap F| / |S^r| = 0$? (It is clear that $|S^r|$ grows exponentially)

This is an attempt to answer this question. I find it a bit tricky to determine the growth of $|S^r \cap F|$ because $F$ is not a direct summand of $K$. Any ideas would be much appreciated! Thank you for reading.

Answer 1

The subgroup $F$ (I have not checked this is indeed the set of fixed points of $\phi$) is isomorphic to $\mathbb Z$, the question is just whether it is distorted or not, and if it is how much?

I'll show that $|e_1^r|_S\succeq\sqrt r$, which implies that $|S^r\cap F|\preceq r^2$. Consider $$\pi\colon G \to Q=G/\langle\!\langle e_3^{(0)},e_5^{(0)}\rangle\!\rangle.$$ We have $Q\simeq \mathbb Z^3\rtimes_\phi \mathbb Z$ (torsionfree !), where $\mathbb Z^3=\langle e_1^{(0)},e_2^{(0)},e_4^{(0)}\rangle$ and $\phi$ acts like $$ \begin{pmatrix} 1 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix}.$$ The index-two subgroup $\langle e_1,e_2,e_4,\phi^2\rangle$ is nilpotent of class $2$, hence elements of $Q$ are at most quadratically distorted, in particular $|e_1^r|_S\ge |\pi(e_1)^r|_{\pi(S)}\succeq \sqrt r$.

(I'd guess $|e_1^r|_S\asymp r$, hence $|S^r\cap F|\asymp r$. The point is that, sure you can get quadratically many $e_1$ terms at linear cost using $\phi$, but it also creates $e_3,e_5$ terms, and the clean up will be costly.)

Edit: Here is a proof that $|e_1^r|_S\asymp r$. We consider another quotient, we add the relations $$e_i^{(j)}=e_i^{(j+1)},\; e_3=e_5,\; e_2=e_4 \;\;\text{and}\;\; e_1=-2e_3.$$ The quotient is $Q'=\mathbb Z^3$ where $\mathbb Z^3=\langle e_2,e_3,\phi\rangle$, so $|e_1^r|_S\ge |\pi'(e_1^r)|_{\pi'(S)}\asymp r$.

Answer 2

One can describe $M/N$ by using basic commutative algebra.

Thing of $M$ as a free module of rank 5 over the polynomial ring $\mathbf{Z}[t,t^{-1}]$, with basis $(e_j)_{1\le j\le 5}$. So what you denote by $e_j^{(i)}$ is just $t^ie_j$, and $\phi$ is just multiplication by $t$.

So $N$ is by definition the $\mathbf{Z}[t,t^{-1}]$-submodule generated by the four "relators" $(1-t)e_1$, $e_2-te_4$, $e_3-te_5$, $e_1+e_3+e_4+e_5-te_2$.

When describing $M/N$, we can thus first eliminate $e_2$, $e_3$ and get the quotient of the free module on $e_1,e_4,e_5$ by the relators $(1-t)e_1$, $e_1+(1+t)e_5+(1-t^2)e_4$. In turn, eliminating $e_1$, we see that this is the free module on $e_4,e_5$ modulo the relator $(1-t^2)(e_5+(1-t)e_4)$. Defining $E_5=e_5+(1-t)e_4$, we see that this is the free modulo on $e_4,E_5$ by the relator $(1-t^2)E_5$ --- thus the direct sum of the free module of rank 1 on $e_4$ and a copy of the cyclic module $\mathbf{Z}[t,t^{-1}]/(1-t^2)$.

In terms of these generators, we have $e_1=-(1+t)E_5$.

Now kill $e_4$. You get a quotient module $M/N'$ which is just the cyclic module $\mathbf{Z}[t,t^{-1}]/(1-t^2)$. This is just a copy of $\mathbf{Z}^2$ on which $t$ acts as an involution. In particular, $M/N'$ is torsion-free and virtually $\mathbf{Z}^2$ (it is the $\pi_1$ of the Klein bottle) and in particular its nontrivial elements are undistorted. Since $e_1$ has a nontrivial image in this quotient, we thus see that $e_1$ is undistorted (as already mentioned by Corentin B in his answer).