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

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.

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 $$|e_1^r|_S\ge |\pi'(e_1^r)|_{\pi'(S)}\asymp r$$.

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).