The Kronecker foliation of the torus has dense leaves and trivial leaf holonomy

The multiples of $\alpha$ are dense. For $N\ge 1$ consider the $N+1$ numbers $\{k\alpha \}$, $k=0,\dots ,N$, where $\{z\}$ denotes the fractional part, and assign $k$ to the interval $[j\mathrm{/}N,(j+1)\mathrm{/}N)$ containing $\{k\alpha \}$. By [F2] two of them, say $k<l$, lie in the same interval, so $0<\mathrm{\mid }\{l\alpha \}-\{k\alpha \}\mathrm{\mid }<1\mathrm{/}N$. Hence there is an integer $m$ with $1\le m\le N$ and a real $\beta \in (0,1\mathrm{/}N)$ with $\beta \equiv \pm m\alpha (\mathrm{m}\mathrm{o}\mathrm{d}1)$: the difference $m\alpha$ lies within $1\mathrm{/}N$ of an integer, and $\beta$ is its positive distance to the nearest integer, nonzero because $\alpha$ is irrational.

Every leaf is diffeomorphic to $\mathbb{R}$. For a point $p\in {\mathbb{R}}^2$ consider $\phi :\mathbb{R}\to {\mathbb{T}}^2$, $\phi (t):=[p+t(1,\alpha )]$. It is continuous by [F1], its image is the leaf through $[p]$, and it is injective: if $\phi (t)=\phi (s)$ with $t\mathrm{\ne }s$ then $(t-s)(1,\alpha )\in {\mathbb{Z}}^2$, so $t-s\in \mathbb{Z}$ and $(t-s)\alpha \in \mathbb{Z}$, forcing $\alpha =(t-s{)}^{-1}(t-s)\alpha \in \mathbb{Q}$, contrary to hypothesis. Near any $t_0$, the image of $\phi$ is, in a small box of the torus adapted to the constant vector field $(1,\alpha )$, the graph of a straight line over the first coordinate; hence $\phi$ is a local homeomorphism onto the leaf with the leaf topology and an immersion. By the uniqueness clause of [F3] the smooth structure on the leaf making the inclusion an immersion is unique, so $\phi$ is a diffeomorphism $\mathbb{R}\to L$ onto the leaf with its intrinsic structure.

The orbit of $0$ under the flow is dense. For any $w\in [0,1)$ and $N$ as in step 1.1, choose the natural number $j$ with $j\beta \le w<j\beta +\beta$; then $\mid w-j\beta \mid <\beta <1\mathrm{/}N$. Since $j\beta \equiv \pm jm\alpha (\mathrm{m}\mathrm{o}\mathrm{d}1)$, there is an integer $k$ with $\mid w-k\alpha \mid <1\mathrm{/}N$ in $\mathbb{R}\mathrm{/}\mathbb{Z}$. As $N$ is arbitrary, the set $\{k\alpha :k\in \mathbb{Z}\}$ is dense in $\mathbb{R}\mathrm{/}\mathbb{Z}$.

The holonomy group of every leaf is trivial. Let $L$ be a leaf and $x\in L$. By step 1.2 and [F4], ${\pi }_1(L,x)\cong {\pi }_1(\mathbb{R},0)=0$. Hence the holonomy representation ${\rho }_x$ is a homomorphism from the trivial group, so its image $\mathrm{Hol}(L,x)$ is the trivial subgroup of ${\mathrm{Diff}}_x(T)$; equivalently, the only leaf loop class is the trivial one and its germ is the identity germ, and every leaf loop — being null-homotopic — has identity holonomy germ by [F5].

Every leaf is dense. The leaf through $[0]$ contains the points $[k(1,\alpha )]=[(k,k\alpha )]$, $k\in \mathbb{Z}$, so for a nonempty open box $(u-\delta ,u+\delta )\times (v-\epsilon ,v+\epsilon )$ choose by step 2.1 an integer $k$ with $\mid (v-u\alpha )-k\alpha \mid <\epsilon$ in $\mathbb{R}\mathrm{/}\mathbb{Z}$; then $t:=u+k$ gives $[t(1,\alpha )]=[(u,u\alpha +k\alpha )]$, a point of the leaf through $[0]$ lying in the box, because $t<mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>1=u<mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>1$ and $u\alpha +k\alpha$ is within $\epsilon$ of $v$ modulo $1$. Hence the leaf through $[0]$ is dense, and since translation by the class of $p$ is a homeomorphism of ${\mathbb{T}}^2$ carrying the leaf through $[0]$ onto the leaf through $[p]$ by [F1], every leaf is dense.

Conclusion. Every leaf of $F_{\alpha }$ is dense by step 3.1 and diffeomorphic to $\mathbb{R}$ by step 1.2, and every leaf holonomy group is trivial by step 2.2. Thus dense leaves coexist with completely trivial holonomy.