In what sense is this action of $\mathbb R$ on $T$ lifted to an action of $\pi_1(T)\times\mathbb R$ on $\mathbb R^2$? 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.
 A: $\newcommand{\Reals}{\mathbf{R}}$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.
