Foliation Holonomy and the Holonomy Groupoid — Examples

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.

The middle leaf. By [F1] the band is the suspension of $\rho$ over $S^1$ with the displayed action. Let $y=0$. Then $\rho (k)(0)=(-1{)}^{k}0=0$ for every $k$, so $K_0=\mathbb{Z}$ and by [F2] the leaf through $[0,0]$ is $\mathbb{R}\mathrm{/}\mathbb{Z}\cong S^1$ with ${\pi }_1(L_{\mathrm{m}\mathrm{i}\mathrm{d}})\cong \mathbb{Z}$. Its holonomy representation is $k\mapsto {\mathrm{germ}}_0(\rho (k))$. The generator $k=1$ accordingly has holonomy germ the reflection $x\mapsto -x$: the map $\rho (1)$ is not the identity near $0$ (it sends $x$ to $-x$), so its germ is a nonidentity involution, and $k=2$ gives $\rho (2)=\mathrm{i}\mathrm{d}$. Hence the holonomy group is the two-element group generated by this reflection, isomorphic to $\mathbb{Z}\mathrm{/}2$ by [F4].

The other leaves. Let $y\mathrm{\ne }0$. Then $\rho (k)y=(-1{)}^{k}y=y$ holds exactly when $k$ is even, so $K_y=2\mathbb{Z}$ and by [F2] the leaf through $[0,y]$ is $\mathbb{R}\mathrm{/}2\mathbb{Z}\cong S^1$, the leaf wrapping twice around the band. Its holonomy representation sends $k\in 2\mathbb{Z}$ to the identity germ of $\rho (k)=\mathrm{i}\mathrm{d}$, so the holonomy group is trivial.

Conclusion. The middle leaf is a circle whose holonomy group is $\mathbb{Z}\mathrm{/}2$, generated by the reflection germ $x\mapsto -x$, while every other leaf is a circle with trivial holonomy group; the projection to the base circle exhibits the band as an interval bundle over the circle transverse to $F$.

Leaf type. The period set $K_y=\{k\in \mathbb{Z}:f^k(y)=y\}$ is a subgroup of $\mathbb{Z}$. If the orbit of $y$ is periodic of minimal period $k$, then $K_y=k\mathbb{Z}$ with $k\ge 1$ minimal, and $L_y\cong \mathbb{R}\mathrm{/}k\mathbb{Z}\cong S^1$ by [F1] and [F3]; if the orbit is not periodic, $K_y=\{0\}$ and $L_y\cong \mathbb{R}$. This proves claim 1.

The generating loop and its germ. If $y$ has minimal period $k$, the path $a(t)=[(kt,y)]$ is closed because $(k,y)\sim (0,y)$. Under ${\pi }_1(L_y)\cong K_y=k\mathbb{Z}$ it corresponds to the positive generator $k$. Its forward holonomy is the germ of $f^{-k}$ by [F1]. Its representation value is the inverse germ, that of $f^k$; these two germs generate the same cyclic group, proving claim 2.

The identity and finite-order cases. If $f=\mathrm{i}\mathrm{d}$, the quotient is the torus with its product foliation, every $K_y=\mathbb{Z}$, and its leaf-loop holonomy germs are identities. For a rotation by $2\pi p\mathrm{/}q$, with $p\in \mathbb{Z}$ and $q\ge 1$, put $d=q\mathrm{/}\gcd (p,q)$. The rotation has exact order $d$, and $f^k(y)=y$ exactly when $d$ divides $k$, so $K_y=d\mathbb{Z}$ for every $y$. Thus every leaf is a circle and every leaf-loop holonomy germ is the identity, since $f^k=\mathrm{i}\mathrm{d}$ for $k\in d\mathbb{Z}$. A path over one base circuit need not be closed and may have the nonidentity germ of $f^{-1}$; claim 3 concerns loops in the leaves.

Conclusion. Claims 1, 2 and 3 are established in steps 1.1, 1.2 and 1.3: the leaves of $F_f$ are circles exactly for periodic orbits, the holonomy of a periodic leaf is generated by $f^{-k}$, and for the identity or a finite-order rotation all holonomy is trivial even though the leaves are circles.

The bundle structure. Take arcs $U_i$ of $\mathbb{R}\mathrm{/}\mathbb{Z}$ which are the images of $I_1=(-1\mathrm{/}4,3\mathrm{/}4)$ and $I_2=(1\mathrm{/}4,5\mathrm{/}4)$. The quotient map restricts injectively on each interval and is open, since the saturation of an open set is the union of its integer translates. Thus each arc has a unique smooth interval representative $t_i$. Define ${\mathrm{\Phi }}_i:U_i\times {\mathbb{R}}^n\to M_A$ by ${\mathrm{\Phi }}_i([t],v)=[t_i,v]$. Every point over $U_i$ has exactly one such representative, and local quotient charts of the suspension make ${\mathrm{\Phi }}_i$ and its inverse smooth. On each overlap component $t_2-t_1$ is a constant integer (here $0$ or $1$), so the change of fibre coordinate is a constant power of $A$, hence linear and smooth. These trivializations cover $M_A$ and make its projection locally the product submersion. Their locally constant linear transitions supply the asserted flat vector bundle.

The leaf through a fixed vector. If $Av=v$, then $A^{k}v=v$ for every $k$, so $K_v=\mathbb{Z}$ and by [F1] the leaf through $[0,v]$ is $\mathbb{R}\mathrm{/}\mathbb{Z}$, a circle; the leaf loop $a(t)=[(t,v)]$, $t\in [0,1]$, corresponds to the generator of $\mathbb{Z}$, and its holonomy germ is the germ of $A^{-1}$ at $v$.

The leaf through the origin. Let $A$ be diagonal with all eigenvalues different from $1$. Then $A^{k}0=0$ for every $k$, so $K_0=\mathbb{Z}$ and the leaf through $[0,0]$ is a circle. By [F1] its holonomy representation is $k\mapsto {\mathrm{germ}}_0(A^k)$, whose image is the cyclic group generated by the germ of $A^{-1}$ at $0$; by [F2] that image is the holonomy group. If $A\mathrm{\ne }I$, then $A^{-1}\mathrm{\ne }I$; since $A^{-1}$ is linear, it cannot agree with the identity on a neighbourhood of $0$ without being the identity, so the germ of $A^{-1}$ at $0$ is not the identity germ and the holonomy group is nontrivial. If $A=I$ the holonomy is trivial.

Transversality to the fibres. The leaves of $F_A$ are locally the images of the $\mathbb{R}$-directions $t\mapsto (t,v)$ and the fibres of $M_A\to S^1$ are locally the images of the ${\mathbb{R}}^n$-directions $v\mapsto (t,v)$; these two directions are complementary in $T_{[t,v]}{M}_A$, of dimensions $1$ and $n$. Hence $F_A$ is transverse to the fibres at every point (Local transversals to a regular foliation).

Conclusion. Steps 1.1, 1.2, 1.3 and 2.1 establish claims 1, 2 and 3.

In coordinates on ${\mathbb{R}}^2$ and $\mathbb{R}$ the map $f(t)=(0,t^2)$ has Jacobian $(0,2t{)}^{\mathsf{T}}$, so $d{f}_t(v)=(0,2tv)$ for every $v\in T_t\mathbb{R}$, by [F3].

Hence $d{f}_t(v)\in T_{f(t)}F=\mathbb{R}\times \{0\}$ holds exactly when $2tv=0$. Therefore $D_t^{\ast }=\{v:2tv=0\}$ equals $\{0\}$ for $t\mathrm{\ne }0$ and equals $T_0\mathbb{R}$ at $t=0$.

The map is transverse to $F$ for $t\mathrm{\ne }0$: there $d{f}_t(T_t\mathbb{R})$ is the vertical line $\{0\}\times \mathbb{R}$, which together with $T_{f(t)}F=\mathbb{R}\times \{0\}$ spans $T_{f(t)}{\mathbb{R}}^2$. At $t=0$ the differential vanishes, so $d{f}_0(T_0\mathbb{R})=\{0\}$ and the sum $d{f}_0(T_0\mathbb{R})+T_{f(0)}F=\mathbb{R}\times \{0\}$ is a proper subspace of $T_{f(0)}{\mathbb{R}}^2$: the map is not transverse at $0$. Thus the failure of the rank conclusion occurs exactly at the point where transversality fails, and the transversality hypothesis of The pullback foliation under a transverse map is essential.

The rank of $D_t^{\ast }$ is $0$ for $t\mathrm{\ne }0$ and $1$ at $t=0$; in particular $D^{\ast }$ is not a smooth distribution of constant rank $\dim N-\mathrm{codim}F=1-1=0$, and it is not a vector subbundle of $T\mathbb{R}$ near $0$. So the first two conclusions of the inference fail.

Finally the horizontal leaf $L_c=\mathbb{R}\times \{c\}$ has preimage $f^{-1}(L_c)=\{t:t^2=c\}$, a finite set or empty; its connected components are points, whose tangent spaces are $\{0\}$, while $D_0^{\ast }=T_0\mathbb{R}$. So the leaf-preimage description is not compatible with the inverse-image spaces at $0$ either, and the inference is false in every one of its clauses. The claim is refuted.

Define the band as $(\mathbb{R}\times (-1,1))\mathrm{/}\mathbb{Z}$ with $k\cdot (t,x)=(t+k,(-1{)}^{k}x)$ and foliation by horizontal lines. The middle leaf is $\mathbb{R}\mathrm{/}\mathbb{Z}$ with fundamental group $\mathbb{Z}$ by $\mathrm{Deg}:{\pi }_1(\mathbb{R}\mathrm{/}\mathbb{Z},[0])\to (\mathbb{Z},+)$ is an isomorphism (the core loop lifts from $0$ to $1$ and has degree one). Transport once around its generator holds $x$ constant on the covering strip and then uses the gluing $(1,x)\sim (0,-x)$; hence its return germ is $x\mapsto -x$, a nonidentity involution. The suspension holonomy formula gives the same germ (the inverse reflection equals itself), and the $k$-fold loop has germ $x\mapsto (-1{)}^{k}x$.

Same germ. By [F1] the class $[\gamma ]\in {\pi }_1(L_{\mathrm{m}\mathrm{i}\mathrm{d}})$ satisfies $\rho ([\gamma ])=r$ with $r^2=\mathrm{i}\mathrm{d}$. By [F3] the double loop $2\gamma$ has holonomy germ $r\circ r=\mathrm{i}\mathrm{d}$, which is also the holonomy germ of the constant loop $e$. Hence $e$ and $2\gamma$ have the same holonomy germ.

Different homotopy classes. By [F4] and [F1], the class of $2\gamma$ in ${\pi }_1(L_{\mathrm{m}\mathrm{i}\mathrm{d}})\cong \mathbb{Z}$ is twice a generator and hence nonzero, whereas the constant loop represents $0$; so $e$ and $2\gamma$ are not leafwise homotopic relative to endpoints.

The kernel is nontrivial. The kernel of $\rho$ contains the class of $2\gamma \mathrm{\ne }0$, so $\ker \rho$ is a nontrivial subgroup of $\mathbb{Z}$, namely $2\mathbb{Z}$. Consequently the holonomy group is $\mathbb{Z}\mathrm{/}2$, and the isotropy of the holonomy groupoid at the base point is $\mathbb{Z}\mathrm{/}2$ while the corresponding isotropy of the monodromy groupoid is $\mathbb{Z}$; the projection $\mathrm{Mon}(F)\to \mathrm{Hol}(F)$ identifies the two loops of steps 1.1 and 1.2 and is therefore not injective.

Conclusion. Steps 1.1 and 1.2 exhibit two leaf loops that are nonhomotopic relative to endpoints yet have the same holonomy germ, refuting the stated inference; step 2.1 shows that the monodromy and holonomy groupoids are genuinely different quotients of the leafwise path groupoid.