I am reading the paper "Calculating the fundamental group of an orbit space" by M A Armstrong where he states the following -
Let $\mathbb R$ act on the torus $T\cong S^1\times S^1$ by $$r\cdot(e^{2\pi ix},e^{2\pi iy})=(e^{2\pi i(x+r)},e^{2\pi i(y+r\sqrt{2})})$$
This action lifts to an action of $G=\pi_1(T)\times\mathbb R$ on $\mathbb R^2$ which has the same orbit space, namely $(m,n,r)\in\mathbb{Z\times Z\times R}$ sends $(x,y)$ to $(x+m+r,y+n+r\sqrt{2})$. One easily checks that the closure of this group of homeomorphisms of $\mathbb R^2$ in Homeo$(\mathbb R^2)$ is precisely the group of all translations of $\mathbb R^2$.
My questions are the following -
In what sense is the $G$ action a lift of the $\mathbb R$ action? That is, what do they mean by the action "lifts"?
What is the advantage of the $\sqrt{2}$? If the action was $r\cdot(e^{2\pi ix},e^{2\pi iy})=(e^{2\pi i(x+r)},e^{2\pi i(y+r)})$ what would happen?
Thank you.
$\newcommand{\Reals}{\mathbf{R}}$The covering map $\Pi:\Reals^{2} \to S^{1} \times S^{1}$ is $$ \Pi:(x, y) = (e^{2\pi ix}, e^{2\pi iy}). $$
The action $$ \widetilde{g}_{(m,n,r)}(x, y) = (x + m + r, y + n + \sqrt{2}r) \tag{1} $$ on $\Reals^{2}$ is a lift of the action $$ g_{r}(e^{2\pi ix}, e^{2\pi iy}) = (e^{2\pi i(x+r)}, e^{2\pi i(y+\sqrt{2}r)}) \tag{2} $$ on the torus in the sense that $g_{r} \circ \Pi = \Pi \circ \widetilde{g}_{m,n,r}$. If you prefer, the "downstairs" action is $$ g_{m,n,r}(e^{2\pi ix}, e^{2\pi iy}) = (e^{2\pi i(x+m+r)}, e^{2\pi i(y+n+\sqrt{2}r)}) = (e^{2\pi i(x+r)}, e^{2\pi i(y+\sqrt{2}r)}); $$ the translations by $m$ and $n$ act by the identity on the torus.
Irrationality of $\sqrt{2}$ makes the orbits of (2) dense in the torus. If $\sqrt{2}$ were rational (e.g., $1$), the orbits of the action (2) would be compact curves in the torus, and the lifted action (1) would be a proper closed subgroup of the translation group. One "standard picture" is to start with the integer lattice in $\Reals^{2}$ and to draw the line $\ell$ through the origin with slope $m$. If $m$ is rational, say $m = p/q$ in lowest terms, $\ell$ hits the lattice point $(q, p)$, so the curve $\Pi(\ell)$ in the torus closes up. If $m$ is irrational, no two distinct points of $\ell$ differ by a lattice element, which means $\Pi$ is injective when restricted to $\ell$. Intuitively, $\Pi(\ell)$ winds with constant slope around the torus, never returning to the same location twice.