How flexible is the infinite-dimensional torus?

Question

Let $\mathbb T=\mathbb R/\mathbb Z$ be the circle group and $\mathbb T^\omega$ be the infinite-dimensional torus, considered as an abelian compact topological group.

Problem 1. Is it true that for any finite set $F\subset\mathbb T^\omega$ and any neighborhood $U\subseteq \mathbb T^\omega$ of zero there exists an automorphism $\alpha$ of $\mathbb T^\omega$ such that $\alpha(F)\subset U$?

This problem can be reformulated in the language of the special linear groups $SL(n,\mathbb Z)$.

Problem 2. Is it true that for any $n\in\mathbb N$, neighborhood of zero $U$ in $\mathbb R^n$ and vectors $x_1,\dots,x_n$ in $\mathbb R^\omega$ there exists $m>n$ and a matrix $A\in SL(m,\mathbb Z)$ such that $\mathrm{pr}_n\circ A\circ \mathrm{pr}_m(x_i)\in U$ for all $i\in\{1,\dots,n\}$?

Here $\mathrm{pr_k}:\mathbb R^\omega\to\mathbb R^k$, $\mathrm{pr}_k:x\mapsto x{\restriction}k$, is the projection onto the first $k$ coordinates.

Remark 1. For any field $\mathbb F$ and vectors $x_1,\dots,x_n\in\mathbb F^{2n}$ there exists a linear transformation $A\in SL(2n,\mathbb F)$ of $\mathbb F^{2n}$ such that $A(\{x_1,\dots,x_n\})\subset\{0\}^n\times\mathbb F^n$.

Remark 2. For any vector $(x,y)\in\mathbb R^2$ and any $\varepsilon>0$ there exists a matrix $A\in SL(2,\mathbb Z)$ such that $(x,y)\cdot A\in (-\varepsilon,\varepsilon)\times\mathbb R$. Such matrix $A$ can be constructed by finding relatively prime integer numbers $p,q$ such that $|xp+yq|<\varepsilon$ and then finding integer numbers $a,b$ such that $pb-qa=1$ (using the extended Euclidean algorithm). Then the matrix $A=\left(\begin{array}&p&a\\q&b\end{array}\right)\in SL(2,\mathbb Z)$ has the required property.

Taking into account Remarks 1 and 2, I would expect that the following stronger form of Problem 2 has an affirmative answer.

Problem 3. Is it true that for any $n\in\mathbb N$ and vectors $x_1,\dots,x_n\in\mathbb R^{2n}$ there exists a linear transformation $A\in SL(2n,\mathbb Z)$ such that $A(\{x_1,\dots,x_n\})\subset(-\varepsilon,\varepsilon)^n\times\mathbb R^n$?

Answer 1

This is a draft proof of an affirmative answer to Problem 3.

Proposition. For any $n\in\mathbb N$, $\varepsilon>0$ and vectors $x_1,\dots,x_n\in\mathbb R^{2n}$ there exists a linear transformation $A\in SL(2n,\mathbb Z)$ such that $A(\{x_1,\dots,x_n\})\subset[-\varepsilon,\varepsilon]^n\times\mathbb R^n$.

Proof. Let $m=2n$, $x_i=(x_{i1},\dots,x_{im})$ for each $i$ and $X=\|x_{ij}\|$. We shall call a column of a matrix small, provided all its entries have absolute value at most than $\varepsilon$, and big, otherwise. Let $k$ be the maximal number of small columns in a matrix $XB$, where $B\in SL(m,\mathbb Z)$ and $C$ be an arbitrary matrix in $SL(m,\mathbb Z)$ such that $XC$ has $k$ small columns. It suffices to show that if $k<n$ then there exists a matrix $D\in SL(m,\mathbb Z)$ such that a matrix $XCD$ has $k+1$ small columns. Without loss of generality we can suppose that the big columns $y_1,\dots, y_{l}$ of the matrix $XC$ are from the first to the $l$-th. Pick a positive integer $M$ such that the maximal absolute value of an entry of these columns is at most $M\varepsilon$. Pick an arbitrary positive integer $K>(2lM)^n$ and define a map $f$ from the subset $Q^l$ of points of the set $[-K, K]^l$ with all integer coordinates to $\mathbb R^n$ by putting $f(d)=d_1y_1+\dots + d_ly_l$ for each $d=(d_1,\dots,d_l)\in Q^l$. Since all $|d_i|\le K$ and all entries of columns $y_i$ have absolute value at most $M\varepsilon$, each coordinate of a vector $f(d)$ is at most $lKM\varepsilon$. Therefore the image $f(Q^l)$ can be covered by $(2lKM)^n$ axis-parallel cubes with side $\varepsilon$. Since $|Q^l|=(2K+1)^l>(2K)^{n+1}>(2lKM)^n$, there exist two distinct elements $d’$ and $d’$ of $Q^l$ such that each coordinate of a vector $y=f(d’)-f(d’’)$ has an absolute value at most $\varepsilon$. Put $d=d’-d’’$. Dividing entries of $d$ by their greatest common divisor, if needed, we can assume that the greatest common divisor of entries of $d$ is $1$. It is well-known (see, for instance, this MSE thread]) that there exists a matrix $D’\in SL(l,\mathbb Z)$ whose first column is $d$. Put $D=\begin{pmatrix} D’ & 0\\ 0 & I\end{pmatrix}$, where $I$ is the $(m-l)\times (m-l)$ identity matrix. Then the first column of the matrix $XCD$ is small, whereas its last $k$ columns are the same as in the matrix $XC$. $\square$