# Powers in compact coset spaces

Let $G$ be a topological group, let $K$ be a closed cocompact subgroup (i.e. the coset space $G/K$ is compact in the quotient topology) and let $g \in G$. Is there a sequence (edit: or net) of positive powers $g^{i_n}$ of $g$ such that $g^{i_n}K$ converges to $K$ in the coset space $G/K$?

If the answer is `no' in general, what if $G$ is totally disconnected and locally compact? (For the application, I'd be happy if I could at least get powers of $g$ to land in $UKV$ for any pair of identity neighbourhoods $U$ and $V$.)

• Would you take a net? Dec 22, 2014 at 14:45
• Ah yes, I see the issue there if $G$ is not metrisable. Yes, a net is fine. Dec 22, 2014 at 21:05

$$\DeclareMathOperator{\eps}{\varepsilon}$$No. I'll write $$P$$ instead of $$K$$, as $$K$$ often denotes a compact subgroup. Write $$X=G/P$$.

For $$x_0\in X$$ such that $$g\mapsto gx_0$$ induces a homeomorphism $$G/P\to X$$, the question is whether $$(g^nx_0)_{n\ge 1}$$ always accumulates at $$x_0$$. This fails in the most classical (typically non-normal) case: $$G=\mathrm{PGL}_2(\mathbf{K})$$, where $$\mathbf{K}$$ is, say, $$\mathbf{R}$$ or $$\mathbf{Q}_p$$, and $$P$$ is conjugate the upper triangular group. In this case we can identify $$X$$ to $$\mathbb{P}^1(\mathbf{K})$$ and choose to identify $$x_0$$ to $$1$$ (so $$P$$ is a given conjugate of the upper triangular group). Then pick $$g(x)=2x$$. Then $$g^n(x)$$ converges to $$+\infty$$.

Even the "$$g^n\in UPV$$" expectation fails. For this I haven't found a better argument than a brute computation. Here, in the real case, I take $$P$$ the lower triangular group, and $$g=\begin{pmatrix}2 & 1 \\ 0 & 1\end{pmatrix}$$, so $$g^n=\begin{pmatrix}2^n & 2^n-1\\ 0 & 1\end{pmatrix}$$. Then if $$u,v$$ satisfy $$\|u-I\|_\infty,\|v-I\|_\infty\le \eps=1/6$$ then $$ug^nv\notin P$$. For this I just compute the upper-right entry of $$ug^nv$$: writing $$u=\begin{pmatrix} u_1 & e_1\\ * & *\end{pmatrix}$$ and $$v=\begin{pmatrix} * & e_2\\ * & v_2\end{pmatrix}$$ then $$(ug^nv)_{12}=2^nu_1e_2+(2^n-1)u_1v_2+e_1v_2.$$ Hence the condition $$ug^nv\in P$$, that is, $$(ug^nv)_{12}=0$$, yields $$-(2^n-1)u_1v_2=2^nu_1e_2+e_1v_2$$, hence $$|(2^n-1)u_1v_2|\le |2^nu_1e_2+e_1v_2|$$, which in turn yields $$(1-\eps)^2|(2^n-1)|\le \eps(1+\eps)(2^n+1)$$, that is, $$\frac{1-2^n}{1+2^n}\le \frac{\eps(1+\eps)}{(1-\eps)^2}$$. But for $$n\ge 1$$ $$\frac{1-2^n}{1+2^n}\ge 1/3$$ while for $$\eps\le 1/6$$, $$\frac{\eps(1+\eps)}{(1-\eps)^2}\le 7/25<1/3$$.

The $$p$$-adic case should be checked similarly with $$2$$ replaced with $$1/p$$ (and always a bit easier by the ultrametric inequality).

It seems the following.

In general the answer is no, because compactness does not imply sequential compactness. Let $$\Bbb T=\{z\in\Bbb C:|z|=1\}$$ be the unit circle endowed with the standard topology. Put $$G={\Bbb T}^{\Bbb T}$$. By Tychonov Theorem, $$G$$ is a compact space. Let $$K=\{e\}$$ be the trivial subgroup of $$G$$. Select an element $$g=(g_z)_{z\in\Bbb T}\in G$$ such that $$g_z=z$$ for each $$z\in\Bbb T$$. Suppose that there exists an increasing sequence $$\{i_n\}$$ of positive integers such that the sequence $$\{g^{i_n}\}$$ converges to the identity of the group $$G$$. Let $$U_0=\{z\in\Bbb T: \operatorname{Re} z\ge 0\}$$ be a neighborhood of the identity of the group $$\Bbb T$$. For each natural number $$n$$ put $$T_n=\{z\in\Bbb T: i_mz\in U_0\mbox{ for each }m>n\}$$. The continuity of power on the group $$\Bbb T$$ implies that the set $$T_n$$ is closed for each natural number $$n$$. The assumption implies that $$\Bbb T=\bigcup_{n\in\Bbb N} T_n$$. By Baire Theorem, there exists a number $$m$$ such that a set $$T_m$$ has non-empty interior. Therefore there exists an open arc $$U\subset T_m$$ of the circle $$\Bbb T$$. Since the sequence $$\{i_n\}$$ is increasing, there exists a number $$n>m$$ such that $$i_n>1/\mu(U)$$, where $$\mu$$ is the standard measure on $$\Bbb T$$ such that $$\mu(\Bbb T)=1$$. But then $$U_0\supset i_nT_m\supset i_n\overline U=\Bbb T$$, a contradiction.

I hope you will be almost happy with this addendum. Let $$G$$ be a Hausdorff totally disconnected and locally compact topological group, $$K$$ be a cocompact normal subgroup of the group $$G$$ and $$g\in G$$. For each $$n$$ put $$i_n=n!$$. I claim that a sequence $$\{g^{i_n}K\}$$ converges to $$K$$ in the coset space $$G/K$$. Indeed, by [Pon, Theorem 16], the group $$G$$ has a base $$\mathcal B$$ at the identity consisting of its open compact subgroups. Let $$H\in\mathcal B$$ be an arbitrary group. Since the group $$K$$ is normal then $$HK$$ is a group. Since $$\{hHK: h\in G\}$$ is an open cover of the compact space $$G/K$$, there exists a finite subset $$F$$ of the group $$G$$ such that $$G=\bigcup\{hHK: h\in G\}$$. Then the pigeonhole principle implies that there exist natural numbers $$k and an element $$h\in F$$ such that $$g^k,g^l\in hHK$$. Then $$g^{l-k}\in K^{-1}H^{-1}h^{-1}hHK=KHK=HK$$. Since the set $$HK$$ is a group, $$g^{i_n}\in HK$$ for each $$n\ge l-k$$.

[Pon] Lev S. Pontrjagin, Continuous groups, 2nd ed., M., (1954) (in Russian).

• If $G$ is t.d.l.c. and $K$ is normal, it reduces to the profinite case, which is indeed easy. I'm wondering what happens if $K$ is not normal. Dec 22, 2014 at 21:17
• Good point about sequential compactness. It looks like a net would work for the $\mathbb{T}^{\mathbb{T}}$ example. Dec 22, 2014 at 21:23