L-space Conjecture

L-space Conjecture

Definitions and mathematical statement

A rational homology three-sphere $Y$ has $H_*(Y,{\mathbb{Q}})\cong H_*(S^3,{\mathbb{Q}})$; irreducibility means every embedded two-sphere bounds a ball. An L-space has $\operatorname{rank}\widehat{HF}(Y)=|H_1(Y,{\mathbb{Z}})|$ in the Heegaard Floer convention of the source. A group is left-orderable if it has a strict total order preserved by left multiplication. A codimension-one foliation is coorientable if its normal line bundle is orientable, and taut if every leaf meets a closed transversal. For closed oriented irreducible rational homology three-spheres, the conjecture equates: not being an L-space; having a nontrivial left-orderable fundamental group; and admitting a coorientable taut foliation. The explicit nontriviality clause excludes the trivial-group convention from changing the statement.

Formulation and concept references: [S1].

Short English statement

Are three descriptions of a rational homology three-sphere equivalent: nonminimal Floer homology, an orderable nontrivial fundamental group, and a coorientable taut foliation?

Research attempt

Current frontier.

The three conditions in the original entry must remain separate. The implication from a coorientable taut foliation to nonminimal Heegaard Floer homology and the graph-manifold results are reviewed in [S1]. A 2026 manuscript of Boyer--Gordon--Hu proves circular orderability for closed orientable irreducible toroidal three-manifolds and constructs finite cyclic covers with left-orderable fundamental groups [E1], Theorems 1.1 and 1.4. Neither conclusion is left-orderability of the original group. Their 2025 relative-conjecture manuscript was withdrawn on 18 August 2025 because of an error in the proof of Theorem 5.7; no conclusion from that withdrawn argument is used [E2]. These source distinctions motivate the lifting obstruction examined below.

Attack route.

A Floer-theoretic route needs to turn a nonminimal Floer group into a geometric or dynamical object, rather than merely compute its rank. A foliation route needs enough control of its transverse dynamics to produce an order. A direct group route could attempt to exclude every finite obstruction to an order. We combine the latter two: first make the finite obstruction criterion explicit, then identify the exact integral class that must vanish in a circle-action lifting argument. Rational vanishing or passage to a finite cover is tested rather than presumed sufficient.

Attempt: left-orderability has finite sign obstructions.

Let $G$ be a countable group with identity $e$. The following precise criterion is useful: \begin{equation}\label{wsref-0-65-71-31-38-31-2d-73-69-67-6e-73} G\text{ is left-orderable}\quad\Longleftrightarrow\quad \forall F\subset G\setminus\{e\}\text{ finite},\ \exists (\varepsilon_g)_{g\in F}\in\{\pm1\}^{F}: e\notin\langle g^{\varepsilon_g}:g\in F\rangle_+. \end{equation} The subscript $+$ means nonempty words using only the indicated elements; it is a semigroup, not the subgroup they generate.

Proof. A left order selects the positive one of $g,g^{-1}$, and products of positive elements stay positive, so the right side is necessary. For sufficiency enumerate the nonidentity elements $g_1,g_2,\ldots$. At level $r$ of a binary tree retain the sign choices whose generated positive semigroup does not contain $e$. A retained node has a retained prefix. The hypothesis gives at least one node at every level, and finite branching gives an infinite branch. Let $P$ be the semigroup generated by the selected signed elements along this branch. A word witnessing $e\in P$ would already occur at a finite level, so $e\notin P$. For every $g\ne e$, either $g$ or $g^{-1}$ belongs to $P$; not both, since their product would be $e$. Thus \[ G=P\sqcup\{e\}\sqcup P^{-1},\qquad PP\subseteq P. \] Define $g<h$ when $g^{-1}h\in P$. Trichotomy and transitivity follow from these two properties, and $(ag)^{-1}(ah)=g^{-1}h$ proves left invariance. The finite-group case is obtained from the same finite-prefix argument, or by enumerating with repetitions. $\square$

Consequently, if $G$ is not left-orderable, some finite list of nonidentity elements has the following certificate: for each of its finitely many sign choices there is a positive word equal to $e$. The words may have very different lengths. The proof gives no computable uniform length bound and no efficient procedure to find them. In applying such a certificate one must also justify that the listed elements really are nonidentity in $G$. A torsion element of order $m>1$ is the simplest example: its $m$th power, with either sign, gives the obstruction.

Thus a possible topological route to $(1)\Rightarrow(2)$ would be a theorem excluding all these finite obstructions for fundamental groups of non-L-spaces. Nothing in (eq181-signs) supplies that connection to Floer homology. Merely checking many finite presentations or many sign lists does not establish the universal absence of an obstruction.

Attempt: lift a faithful circle action, with its full obstruction.

Suppose a faithful representation $\rho:G\to\operatorname{Homeo}_+(S^1)$ has been constructed. Choose lifts $\widetilde\rho(g)$ to increasing homeomorphisms of ${\mathbb{R}}$ commuting with $T(x)=x+1$. The failure of these lifts to be a homomorphism is an integer cocycle: \[ \widetilde\rho(g)\widetilde\rho(h) =T^{c(g,h)}\widetilde\rho(gh),\qquad c(g,h)\in{\mathbb{Z}}. \] Associativity gives the cocycle identity. Changing the lifts by integer translations changes $c$ by an integer coboundary. Its class $e(\rho)\in H^2(G;{\mathbb{Z}})$ is the integral Euler class of the action.

Lifting lemma. If $e(\rho)=0$, then $G$ is left-orderable.

Proof. Write $c$ as a coboundary and change each lift by the corresponding integer translation. The modified lifts form a homomorphism $\widehat\rho:G\to\operatorname{Homeo}_+({\mathbb{R}})$. It is faithful, since an element acting trivially upstairs acts trivially downstairs, where $\rho$ is faithful. To construct an order explicitly, enumerate a dense set $q_1,q_2,\ldots$ in ${\mathbb{R}}$. For distinct $g,h$, continuity and faithfulness ensure that some $q_i$ distinguishes their actions. Compare $\widehat\rho(g)(q_i)$ and $\widehat\rho(h)(q_i)$ at the first such index. This lexicographic comparison is a strict total order, and composition on the left by an increasing homeomorphism preserves the comparison and all preceding equalities. Hence it is left invariant. $\square$

This proof does not require a free action on ${\mathbb{R}}$, or that every group element move every point. It requires a faithful action and the vanishing of an integral cocycle. Similar Euler-class lifting mechanisms occur in the actual three-manifold arguments of [E1], for example its proof of Lemma 4.4; the general hypotheses above are isolated here instead of assumed for all manifolds in the target.

Stress test: rational vanishing is not enough.

A rational homology sphere has $H^2(Y;{\mathbb{Q}})=0$, but its integral $H^2(Y;{\mathbb{Z}})$ can contain torsion. Even identifying this with group cohomology requires an asphericity hypothesis; it is not true for every space solely because its fundamental group is specified. For an aspherical $Y$, integral group cohomology equals that of $Y$, but rational vanishing still leaves exactly the torsion obstruction visible above.

The faithful action of $C_m$ on $S^1$ by rotations through multiples of $1/m$ is a direct test. Its Euler class vanishes rationally, but it cannot lift to a faithful action on ${\mathbb{R}}$: an increasing homeomorphism of finite order is the identity. Indeed, if $h(x)>x$, its successive iterates are strictly increasing, and if $h(x)<x$, they are strictly decreasing; neither is compatible with finite order. This also proves that $C_m$ is not left-orderable. Thus rational Euler-class vanishing cannot replace the hypothesis of the lifting lemma.

Likewise, an orderable finite-index subgroup need not make the whole group orderable. The trivial subgroup has finite index in $C_m$, while the whole group fails the preceding test. This example does not purport to capture all extra geometry of an irreducible toroidal manifold, but it disproves the bare group-theoretic implication that would otherwise incorrectly upgrade the cyclic-cover conclusion of [E1].

Attempt: combine the two diagnostics.

A proposed construction from a taut foliation could produce circle dynamics and then try to lift. The exact sufficient input is now explicit: a faithful circle action with integrally trivial Euler class, or some other faithful real-line action. When such an input is available, the lifting proof yields a positive cone and therefore excludes every finite certificate in (eq181-signs). A circle action with a nontrivial kernel instead orders at most the image by this method; an additional argument would have to handle the kernel compatibly. No such kernel or Euler-class control has been established here for an arbitrary coorientable taut foliation in the catalogue's class.

Conversely, trying to use an arbitrary left order to produce a taut foliation requires a separate realization theorem in the topology of $Y$. The ordered group by itself contains no constructed plane field, integrability condition, or closed transversals. The criterion (eq181-signs) therefore does not silently establish $(2)\Rightarrow(3)$. Nor does the known foliation-to-Floer implication close a cycle of implications when its other arrows remain missing.

The explicit nontriviality convention is also respected: the trivial group admits a formal left order, but condition $2$ of this entry rules it out. Nothing above changes that convention for $S^3$. Infinite fixed sets of an action or a finite cover's foliation are not being substituted for a foliation on the original manifold.

Outcome.

Outcome: Exploratory approach with unresolved gap.

The attempt proves the finite sign-obstruction criterion and a precise Euler-class lifting lemma, then exhibits torsion examples defeating two natural shortcuts. These are group-theoretic tools and diagnostics, not new solved three-manifold cases or claims of priority for these standard constructions.

The missing geometric input is a mechanism tying nonminimal Floer homology or a taut foliation to the required faithful, integrally liftable dynamics, together with the converse realization where needed. The current circular-order and finite-cover results do not supply those assertions for the original manifold. No full equivalence or counterexample is obtained.

Sources

• S1 J. Hanselman, "L-spaces, foliations, and left-orderability," Princeton RTG lecture, notes by J. Van Dyke; conjecture due to Boyer-Gordon-Watson (Math. Ann. 356 $2013$) and Juhasz. https://web.ma.utexas.edu/users/vandyke/notes/princeton_rtg_notes/hanselman/hanselman.pdf. Catalogue formulation source.
• E1 Steven Boyer, Cameron McA. Gordon and Ying Hu, Toroidal 3-manifolds have circularly orderable fundamental groups, arXiv:2607.10529v1, 12 July 2026, Theorems 1.1 and 1.4, and Lemma 4.4. https://arxiv.org/html/2607.10529v1. Consulted 27 September 2026. Recent manuscript; circular order and left-orderable finite covers are distinguished from an order on the original group.
• E2 Steven Boyer, Cameron McA. Gordon and Ying Hu, Slope detection, taut foliations, and the relative L-space conjecture, arXiv:2508.06395v2, 18 August 2025, withdrawn. https://arxiv.org/abs/2508.06395v2. Status only, checked 27 September 2026. The withdrawal notice identifies an error in the proof of Theorem 5.7. No theorem from that manuscript is used.