I have two questions:

Question 1. Suppose that $K$ is a knot in $S^3$. Let $\Sigma(K)$ be the double branched cover of $S^3$ branched along $K$. If $\Sigma(K)=\#_{i=1}^n\Sigma(2,3,5)$, then $K=\#_{i=1}^nT_{3,5}$?

Question 2. There is a conjecture of Ozsvath-Szabo that the only $L$-spaces which are integral homology 3-spheres are connected sums of Poincare homology 3-spheres. If this conjecture and Question 1 is true, then can we (re)prove that Khovanov homology is an unknot-detector using Ozsavth-Szabo's spectral sequence?


1 Answer 1


As Ian Agol points out the Orbifold Theorem will answer question 1 affirmatively. Although the original result is due to Thurston, the common references in the literature are:

Boileau, Michel, Sylvain Maillot, and Joan Porti. Three-dimensional orbifolds and their geometric structures. Vol. 15. Paris: Société mathématique de France, 2003.


Cooper, Daryl, Craig David Hodgson, and Steve Kerckhoff. Three-dimensional orbifolds and cone-manifolds. Vol. 5. Mathematical Society of Japan, 2000.

However, both of these results would still require one to understand the quotients of $\Sigma(2,3,5)$ by its order 2 symmetries. Dunbar provides such a classification, which serves a translation of Threlfall and Seifert's work, see:

Dunbar, William D. "Geometric orbifolds." Revista Matematica de la Universidad Complutense de Madrid 1.1 (1988): 67-100. (available here)


Threlfall, William, and Herbert Seifert. "Topologische Untersuchung der Diskontinuitätsbereiche endlicher Bewegungsgruppen des dreidimensionalen sphärischen Raumes I and II." Mathematische Annalen 104 and 107 (1931 and 1932): 1-70 and 543-586.

More specifically Dunbar gives a proof that if a knot $L$ has the property that $\Sigma(L)=\Sigma(2,3,5)$ then $L=T_{3,5}.$


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