# Does annular Khovanov homology detect the unknot (in annulus)?

Recently Kronheimer and Mrowka showed that Khovanov homology detects the unknot. It's still not known if the Jones polynomial detects the unknot.

Does annular Khovanov homology detect the unknot in an an annulus? More generally, does annular Jones polynomial detect the unknot in an annulus (perhaps there's a counterexample?)

• It's possible that you already realize the following fact, but I'll state it anyway. There is a spectral sequence from annular Khovanov homology to standard Khovanov homology. So the rank of the annular Khovanov homology of a diagram in the annulus is at least the rank of the same diagram in the plane. Combining with Kronheimer and Mrowka says that if a nontrivial knot in the thickened annulus has trivial annular Khovanov homology, then the knot must be the unknot when the thickened annulus is embedded into $S^3$ in the standard way. – Adam Lowrance Apr 4 '18 at 18:55
• What's a reference for the spectral from annular Khovanov homology to standard Khovanov homology? Based on your argument - is the problem now solved? – Vinoth Apr 5 '18 at 12:46
• Computing annular Khovanov homology has some discussion of the spectral sequence. I don't think this argument solves the stated problem. The issue is there could be a knot in the thickened annulus that does not bound a disk in the thickened annulus, but does bound a disk when the thickened annulus is embedded into $S^3$ in the standard way. It's possible that this knot could have annular Khovanov homology equal to the standard Khovanov homology of the unknot. If such an example were to exist, all differentials in the spectral sequence would be zero maps. – Adam Lowrance Apr 6 '18 at 0:57