Gabai's property R theorem is:

If the 0-surgery manifold of a knot $K$ is homeomorphic to $S^1\times S^2$, then $K$ is the unknot.

Recently, 3-manifold topology has been developed rapidly by Agol, Wise and many other mathematicians.

Is there an another simple proof for property R conjecture?

  • 2
    $\begingroup$ Where by "Gabber" I assume you mean "Gabai". $\endgroup$
    – Igor Rivin
    Jun 9 '15 at 3:29
  • 3
    $\begingroup$ You can prove this (indeed, a more general theorem that $r$-surgery on a knot $K \subset S^3$ is diffeomorphic to $r$-surgery on the unknot iff $K$ is the unknot, $r \in \mathbb Q$) using monopole Floer homology or Heegaard Floer homology; see "Monopoles and Lens spaces surgeries" or "Holomorphic disks and genus bounds" respectively. As far as I know, though, this is in a much different direction than Agol and Wise's work. $\endgroup$
    – mme
    Jun 9 '15 at 4:54
  • 1
    $\begingroup$ @MikeMiller: true, but these results actually depend on Property R. $\endgroup$
    – Ian Agol
    Jun 9 '15 at 8:45
  • $\begingroup$ @IanAgol: Thanks for the correction. I guess the interdependence comes from the use of the foliations constructed in "Foliations and the topology of 3-manifolds"? $\endgroup$
    – mme
    Jun 9 '15 at 12:02
  • $\begingroup$ @MikeMiller: yes, since 0-framed surgery is irreducible, the Seiberg-Witten Floer homology is non-trivial. $\endgroup$
    – Ian Agol
    Jun 9 '15 at 12:15

Property R was reproved by Gordon and Luecke in the course of solving the knot complement problem - see Corollary 3.2. They prove the stronger result (as did Gabai) that zero-frame surgery on a knot is irreducible (hence cannot be $S^1\times S^2$).

Gabai actually proved something slightly stronger, which might account for the increased difficulty of the proof, namely that there is a taut finite-depth foliation which intersects the boundary of the knot complement transversely in a foliation of the boundary torus by longitudes. Capping off this foliation with a foliation by disks in the solid torus gives a taut foliation of the zero-framed surgery, hence irreducibility.

Gordon and Luecke use many of the techniques of Gabai (thin position, Scharlemann cycles and generalizations), but omit the foliations and sutured manifold hierarchies. A technical simplification to the argument was subsequently made by Walter Parry.


A new proof of Property R has appeared by Jamie Conway and Bülent Tosun. A slight shortcut of their argument may be summarized as:

  • If $K \subset S^3$ is a knot, and 0-framed surgery $S^3_0(K)\cong S^1\times S^2$, then (essentially) using the Floer exact triangle, one may show that $S^3_1(K)$ is an L-space. An L-space is a rational homology 3-sphere $Y$ for which $HF^{red}(Y)=0$. $S^3$ and the Poincaré homology sphere are the only known irreducible L-spaces. This follows from the argument in Proposition 1.2 of Akbulut-Karakurt.

  • Theorem 1.2 of

Ozsváth, Peter; Szabó, Zoltán, On knot Floer homology and lens space surgeries, Topology 44, No. 6, 1281-1300 (2005). ZBL1077.57012.

implies that knot Floer homology of $K$ in each grading is rank 1.

  • Now by the main result of

Ni, Yi, Knot Floer homology detects fibred knots, Invent. Math. 170, No. 3, 577-608 (2007); erratum ibid. 177, No. 1, 235-238 (2009). ZBL1138.57031.

$K$ is a fibered knot.

  • Hence 0-framed surgery is fibered. But $S^1\times S^2$ fibers in only one way, and hence the fibering must be genus 0, which implies that $K$ is the unknot.

This proof still makes use of much of Gabai's theory, including his construction of taut foliations, which Ni's proof relies on. However, it eliminates the combinatorial topology technology of Scharlemann cycles which previous proofs relied heavily on.

A simplified version of Yi Ni's theorem was proved by Andras Juhasz using sutured Floer homology.

Juhász, András, Floer homology and surface decompositions, Geom. Topol. 12, No. 1, 299-350 (2008). ZBL1167.57005.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.