Orderability of homeomorphism groups

AbbreviationThe closed square#✓fully proved

The closed square \(I^2 = [0, 1] \times [0, 1]\), the product of two copies of the unit interval, used as a model of the closed disc.

abbrev Square : Type := unitInterval × unitInterval
Used by 3

DefinitionBoundary of the square#✓fully proved

The boundary \(\partial I^2\) of the closed square \(I^2 = [0, 1] \times [0, 1]\) is the set of points \((x, y) \in I^2\) with \(x = 0\), \(x = 1\), \(y = 0\) or \(y = 1\).

def squareBoundary : Set Square := {x | x.1 = 0 ∨ x.1 = 1 ∨ x.2 = 0 ∨ x.2 = 1}
Uses 1 · Used by 1

DefinitionBoundary-fixing homeomorphisms of the square#✓fully proved

Let \(I^2 = [0, 1] \times [0, 1]\) be the closed square and \(\partial I^2\) its boundary. \(\mathrm{Homeo}(I^2; \partial I^2)\) is the subgroup of the group of homeomorphisms of \(I^2\) consisting of those homeomorphisms \(h\) with \(h(x) = x\) for every \(x \in \partial I^2\).

def boundaryFixingHomeomorphisms : Subgroup (Square ≃ₜ Square) :=
  fixingSubgroup (Square ≃ₜ Square) squareBoundary
Uses 2 · Used by 1

DefinitionRight orders of a group#✓fully proved

The right orders of G: linear orders (as relations) invariant under right multiplication.

def rightOrders (G : Type*) [Group G] : Set (G → G → Prop) :=
  {r | IsLinearOrder G r ∧ ∀ a b c : G, r a b → r (a * c) (b * c)}
Used by 1

DefinitionRight-orderability of the boundary-fixing homeomorphisms of the disc#✓fully proved

The proposition (Kourovka Problem 17.20, G. M. Bergman, for the closed disc) that the group \(\mathrm{Homeo}(I^2; \partial I^2)\) of homeomorphisms of the closed square \(I^2 = [0, 1] \times [0, 1]\), a model of the closed disc, that fix the boundary \(\partial I^2\) pointwise is right-orderable: it carries a linear order \(\le\) such that \(a \le b\) implies \(ac \le bc\) for all \(a, b, c\).

def discBoundaryFixingHomeomorphismsRightOrderable : Prop :=
  (rightOrders boundaryFixingHomeomorphisms).Nonempty
Uses 3 · Used by 1

TheoremBoundary-fixing homeomorphisms of the disc are not right-orderable#●ready to prove

Let \(I^2 = [0, 1] \times [0, 1]\) be the closed square, a model of the closed disc, and let \(\mathrm{Homeo}(I^2; \partial I^2)\) be the group of homeomorphisms of \(I^2\) that fix every point of the boundary \(\partial I^2\). There is no linear order \(\le\) on \(\mathrm{Homeo}(I^2; \partial I^2)\) such that \(a \le b\) implies \(ac \le bc\) for all \(a, b, c\); that is, this group is not right-orderable.

This answers Kourovka Problem 17.20 negatively: Hyde constructs a finitely generated subgroup of \(\mathrm{Homeo}(I^2; \partial I^2)\) that admits no left order, and left- and right-orderability are equivalent via \(g \mapsto g^{-1}\).

public theorem_wanted discBoundaryFixingHomeomorphismsRightOrderable_refuted :
    ¬ discBoundaryFixingHomeomorphismsRightOrderable
Sources
  • E. I. Khukhro and V. D. Mazurov (eds.), Unsolved Problems in Group Theory: The Kourovka Notebook, 21st issue, arXiv:1401.0300v49 (2026), Archive of Solved Problems, Problem 17.20 (G. M. Bergman); resolution: J. Hyde, The group of boundary fixing homeomorphisms of the disc is not left-orderable, Ann. of Math. (2) 190 (2019), 657-661, abstract (a finitely generated subgroup of Homeo(I²; ∂I²) admits no left order, hence neither does Homeo(I²; ∂I²)); Kourovka Archive, Problem 17.20 (No, not always), https://annals.math.princeton.edu/2019/190-2/p05.
Uses 1
5 fully proved1 ready to prove

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