## Orderability, and groups of homeomorphisms of the disk

I have struggled for a long time (and I continue to struggle) with the following question:

Question: Is the group of self-homeomorphisms of the unit disk (in the plane) that fix the boundary pointwise a left-orderable group?

Recall that a group $G$ is left-orderable if there is a total order $<$ on the elements satisfying $g if and only if $fg < fh$ for all $f,g,h \in G$. For a countable group, the property of being left orderable is equivalent to the property that the group admits a faithful action on the interval by orientation-preserving homeomorphisms; however, this equivalence is not “natural” in the sense that there is no natural way to extract an ordering from an action, or vice-versa. This formulation of the question suggests that one is trying to embed the group of homeomorphisms of the disk into the group of homeomorphisms of the interval, an unlikely proposition, made even more unlikely by the following famous theorem of Filipkiewicz:

Theorem: (Filipkiewicz) Let $M_1,M_2$ be two compact manifolds, and $r_1,r_2$ two non-negative integers or infinity. Suppose the connected components of the identity of $\text{Diff}^{r_1}(M_1)$ and $\text{Diff}^{r_2}(M_2)$ are isomorphic as abstract groups. Then $r_1=r_2$ and the isomorphism is induced by some diffeomorphism.

The hard(est?) part of the argument is to identify a subgroup stabilizing a point in purely algebraic terms. It is a fundamental and well-studied problem, in some ways a natural outgrowth of Klein’s Erlanger programme, to perceive the geometric structure on a space in terms of algebraic properties of its automorphism group. The book by Banyaga is the best reference I know for this material, in the context of “flexible” geometric structures, with big transformation groups (it is furthermore the only math book I know with a pink cover).

Left orderability is inherited under extensions. I.e. if $K \to G \to H$ is a short exact sequence, and both $K$ and $H$ are left orderable, then so is $G$. Furthermore, it is a simple but useful theorem of Burns and Hale that a group $G$ is left orderable if and only if for every finitely generated subgroup $H$ there is a left orderable group $H'$ and a surjective homomorphism $H \to H'$. The necessity of this condition is obvious: a subgroup of a left orderable group is left orderable (by restricting the order), so one can take $H'$ to be $H$ and the surjection to be the identity. One can exploit this strategy to show that certain transformation groups are left orderable, as follows:

Example: Suppose $G$ is a group of homeomorphisms of some space $X$, with a nonempty fixed point set. If $H$ is a finitely generated subgroup of $G$, then there is a point $y$ in the frontier of $\text{fix}(H)$ so that $H$ has a nontrivial image in the group of germs of homeomorphisms of $X$ at $y$. If this group of germs is left-orderable for all $y$, then so is $G$ by Burns-Hale.

Example: (Rolfsen-Wiest) Let $G$ be the group of PL homeomorphisms of the unit disk (thought of as a PL square in the plane) fixed on the boundary. If $H$ is a finitely generated subgroup, there is a point $p$ in the frontier of $\text{fix}(H)$. Note that $H$ has a nontrivial image in the group of piecewise linear homeomorphisms of the projective space of lines through $p$. Since the fixed point set of a finitely generated subgroup is equal to the intersection of the fixed point sets of a finite generating set, it is itself a polyhedron. Hence $H$ fixes some line through $p$, and therefore has a nontrivial image in the group of homeomorphisms of an interval. By Burns-Hale, $G$ is left orderable.

Example: Let $G$ be the group of diffeomorphisms of the unit disk, fixed on the boundary. If $H$ is a finitely generated subgroup, then at a non-isolated point $p$ in $\text{fix}(H)$ the group $H$ fixes some tangent vector to $p$ (a limit of short straight lines from $p$ to nearby fixed points). Consequently the image of $H$ in $\text{GL}(T_p)$ is reducible, and is conjugate into an affine subgroup, which is left orderable. If the image is nontrivial, we are done by Burns-Hale. If it is trivial, then the linear part of $H$ at $p$ is trivial, and therefore by the Thurston stability theorem, there is a nontrivial homomorphism from $H$ to the (orderable) group of translations of the plane. By Burns-Hale, we conclude that $G$ is left orderable.

The second example does not require infinite differentiability, just $C^1$, the necessary hypothesis to apply the Thurston stability theorem. This is such a beautiful and powerful theorem that it is worth making an aside to discuss it. Thurston’s theorem says that if $H$ is a finitely generated group of germs of diffeomorphisms of a manifold fixing a common point, then a suitable limit of rescaled actions of the group near the fixed point converge to a nontrivial action by translations. One way to think of this is in terms of power series: if $H$ is a group of real analytic diffeomorphisms of the line, fixing the point $0$, then every $h \in H$ can be expanded as a power series: $h(x) = c_1(h)x + c_2(h)x^2 + \cdots$. The function $h \to c_1(h)$ is a multiplicative homomorphism; however, if the logarithm of $c_1$ is identically zero, then if $i$ is the first index for which some $c_i(h)$ is nonzero, then $h \to c_i(h)$ is an additive homomorphism. The choice of coefficient $i$ is a “gauge”, adapted to $H$, that sees the most significant nontrivial part; this leading term is a character (i.e. a homomorphism to an abelian group), since the nonabelian corrections have strictly higher degree. Thurston’s insight was to realize that for a finitely generated group of germs of $C^1$ diffeomorphisms with trivial linear part, one can find some gauge that sees the most significant nontrivial part of the action of the group, and at this gauge, the action looks abelian. There is a subtlety, that one must restrict attention to finitely generated groups of homeomorphisms: on each scale of a sequence of finer and finer scales, one of a finite generating set differs the most from the identity; one must pass to a subsequence of scales for which this one element is constant (this is where the finite generation is used). The necessity of this condition is demonstrated by a theorem of Sergeraert: the group of germs of ($C^\infty$) diffeomorphisms of the unit interval, infinitely tangent to the identity at both endpoints (i.e. with trivial power series at each endpoint) is perfect, and therefore admits no nontrivial homomorphism to an abelian group.

Let us now return to the original question. The examples above suggest that it might be possible to find a left ordering on the group of homeomorphisms of the disk, fixed on the boundary. However, I think this is misleading. The construction of a left ordering in either category (PL or smooth) was ad hoc, and depended on locality in two different ways: the locality of the property of left orderability (i.e. Burns-Hale) and the tameness of groups of PL or smooth homeomorphisms blown up near a common fixed point. Rescaling an arbitrary homeomorphism about a fixed point does not make things any less complicated. Burns-Hale and Filipkiewicz together suggest that one should look for a structural dissimilarity between the group of homeomorphisms of the disk and of the interval that persists in finitely generated subgroups. The simplest way to distinguish between the two spaces algebraically is in terms of their lattices of closed (or equivalently, open) subsets. To a topological space $X$, one can associate the lattice $\Lambda(X)$ of (nonempty, for the sake of argument) closed subsets of $X$, ordered by inclusion. One can reconstruct the space $X$ from this lattice, since points in $X$ correspond to minimal elements. However, any surjective map $X \to Y$ defines an embedding $\Lambda(Y) \to \Lambda(X)$, so there are many structure-preserving morphisms between such lattices. The lattice $\Lambda(X)$ is an $\text{Aut}(X)$-space in an obvious way, and one can study algebraic maps $\Lambda(Y) \to \Lambda(X)$ together with homomorphisms $\rho:\text{Aut}(Y) \to \text{Aut}(X)$ for which the algebraic maps respect the induced $\text{Aut}(Y)$-structures. A weaker “localization” of this condition asks merely that for points (i.e. minimal elements) $p,p' \in \Lambda(Y)$ in the same $\text{Aut}(Y)$-orbit, their images in $\Lambda(X)$ are in the same $\text{Aut}(X)$-orbit. This motivates the following:

Proposition: There is a surjective map from the unit interval to the unit disk so that the preimages of any two points are homeomorphic.

Sketch of Proof: This proposition follows from two simpler propositions. The first is that there is a surjective map from the unit interval to itself so that every point preimage is a Cantor set. The second is that there is a surjective map from the unit interval to the unit disk so that the preimage of any point is finite. A composition of these two maps gives the desired map, since a finite union of Cantor sets is itself a Cantor set.

There are many surjective maps from the unit interval to the unit disk so that the preimage of any point is finite. For example, if $M$ is a hyperbolic three-manifold fibering over the circle with fiber $S$, then the universal cover of a fiber $\widetilde{S}$ is properly embedded in hyperbolic $3$-space, and its ideal boundary (a circle) maps surjectively and finitely-to-one to the sphere at infinity of hyperbolic $3$-space. Restricting to a suitable subinterval gives the desired map.

To obtain the first proposition, one builds a surjective map from the interval to itself inductively; there are many possible ways to do this, and details are left to the reader. qed.

It is not clear how much insight such a construction gives.

Another approach to the original question involves trying to construct an explicit (finitely generated) subgroup of the group of homeomorphisms of the disk that is not left orderable. There is a “cheap” method to produce finitely presented groups with no left-orderable quotients. Let $G = \langle x,y \; | \; w_1, w_2 \rangle$ be a group defined by a presentation, where $w_1$ is a word in the letters $x$ and $y$, and $w_2$ is a word in the letters $x$ and $y^{-1}$. In any left-orderable quotient in which both $x$ and $y$ are nontrivial, after reversing the orientation if necessary, we can assume that $x > \text{id}$. If further $y>\text{id}$ then $w_1 >\text{id}$, contrary to the fact that $w_1 = \text{id}$. If $y^{-1} >\text{id}$, then $w_2 >\text{id}$, contrary to the fact that $w_2=\text{id}$. In either case we get a contradiction. One can try to build by hand nontrivial homeomorphisms $x,y$ of the unit disk, fixed on the boundary, that satisfy $w_1,w_2 =\text{id}$. Some evidence that this will be hard to do comes from the fact that the group of smooth and PL homeomorphisms of the disk are in fact left-orderable: any such $x,y$ can be arbitrarily well-approximated by smooth $x',y'$; nevertheless at least one of the words $w_1,w_2$ evaluated on any smooth $x',y'$ will be nontrivial. Other examples of finitely presented groups that are not left orderable include higher Q-rank lattices (e.g. subgroups of finite index in $\text{SL}(n,\mathbb{Z})$ when $n\ge 3$), by a result of Dave Witte-Morris. Suppose such a group admits a faithful action by homeomorphisms on some closed surface of genus at least $1$. Since such groups do not admit homogeneous quasimorphisms, their image in the mapping class group of the surface is finite, so after passing to a subgroup of finite index, one obtains a (lifted) action on the universal cover. If the genus of the surface is at least $2$, this action can be compactified to an action by homeomorphisms on the unit disk (thought of as the universal cover of a hyperbolic surface) fixed pointwise on the boundary. Fortunately or unfortunately, it is already known by Franks-Handel (see also Polterovich) that such groups admit no area-preserving actions on closed oriented surfaces (other than those factoring through a finite group), and it is consistent with the so-called “Zimmer program” that they should admit no actions even without the area-preserving hypothesis when the genus is positive (of course, $\text{SL}(3,\mathbb{R})$ admits a projective action on $S^2$). Actually, higher rank lattices are very fragile, because of Margulis’ normal subgroup theorem. Every normal subgroup of such a lattice is either finite or finite index, so to prove the results of Franks-Handel and Polterovich, it suffices to find a single element in the group of infinite order that acts trivially. Unipotent elements are exponentially distorted in the word metric (i.e. the cyclic subgroups they generate are not quasi-isometrically embedded), so one “just” needs to show that groups of area-preserving diffeomorphisms of closed surfaces (of genus at least $1$) do not contain such distorted elements. Some naturally occurring non-left orderable groups include some (rare) hyperbolic $3$-manifold groups, amenable but not locally indicable groups, and a few others. It is hard to construct actions of such groups on a disk, although certain flows on $3$-manifolds give rise to actions of the fundamental group on a plane.

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### 4 Responses to Orderability, and groups of homeomorphisms of the disk

1. Ian Agol says:

Call a semigroup left-orderable if it satisfies the same definition as for a group. Is the semigroup of continuous maps of an interval which are the identity on the boundary left-orderable? What about continuous maps of a disk which are the identity on the boundary?

2. Danny Calegari says:

Dear Ian – great question! Let me propose something like a “right-order” (I have not carefully checked that it works) for the case of the interval, which has some of the right spirit, but does not quite work.

First fix a point $p$ and define $f if the leftmost preimage of $p$ under $f$ is to the left of the leftmost preimage under $g$. Then $f implies $fh < gh$. One can now pick a countable dense sequence of points $p_i$ and define $f if there is some $n$ so that the leftmost point of $f^{-1}(p_i)$ is equal to the leftmost point of $g^{-1}(p_i)$ for all $i, and the leftmost point of $f^{-1}(p_n)$ is less than the leftmost point of $g^{-1}(p_n)$. This is not quite good enough, since e.g. it does not distinguish between two maps $f,g$ that agree up to the first common preimage of $1$, but disagree thereafter. If we restrict to PL maps, we can resolve the ambiguity by comparing the second preimage (or leftmost preintervals) and so on. Or: we can restrict attention to the subsemigroup of functions which take the interior of the interval to itself, in which case I think the original proposal works. Anyway, I don't see an obvious way to deal with the specific semigroup you propose. For groups, left orderable and right orderable are the same, but I suspect this is not true for semigroups (I can't think of a counterexample off the top of my head) so if you really want left orderable, I'm at a loss.

For the case of a disk, I don't have a feeling either way.

• Andrés Navas says:

Hi Danny,

here is another (nontrivial) nice application of Burns-Hale’s theorem: the group G of GERMS of orientation preserving homeomorphisms of the real line fixing the origin is left-orderable. As usual, I cannot see any explicit left-ordering on G…

3. Andrés Navas says:

Hi again,

I have been thinking on your Question. My guess is that the answer should be negative. Nevertheless, to get a finitely-generated non left-orderable group inside your group, thinking on (finite index subgroups of) SL(3,Z) seems hard to work. Perhaps one should consider “smaller” groups. My favorite one is
the crystallographic group
$$\Gamma = \langle f, g \!: a^2 b a^2 = b, b^2 a b^2 = a \rangle.$$
Notice that, letting $c:= ab$, the subgroup $\langle a^2, b^2 ,c^2 \rangle$ is a torsion-free, rank-3 Abelian, normal subgroup of index 4. In a certain sense, $\Gamma$ is the “smallest” torsion-free non left-orderable group.

Up so far, I have no argument to show that $\Gamma$ dos not embed into the group of orientation-preserving homeomorphisms of the disk fixing the boundary…