Can Khovanov homology have arbitrarily large torsion?

That is, given $N\gg 0,$ does there exist $k>N$, a knot (diagram) $D$ and $i,j \in \mathbb{Z}$ such that $\operatorname{Kh}^{i,j}(D) = \mathbb{Z}/k\mathbb{Z}$?

MathOverflow is a question and answer site for professional mathematicians. It only takes a minute to sign up.

Sign up to join this community
Anybody can ask a question

Anybody can answer

The best answers are voted up and rise to the top

$\begingroup$
$\endgroup$

Can Khovanov homology have arbitrarily large torsion?

That is, given $N\gg 0,$ does there exist $k>N$, a knot (diagram) $D$ and $i,j \in \mathbb{Z}$ such that $\operatorname{Kh}^{i,j}(D) = \mathbb{Z}/k\mathbb{Z}$?

$\begingroup$
$\endgroup$

This paper from earlier this year (Jan 18, to be precise) proves the existence of $\mathbb{Z}/n\mathbb{Z}$-torsion for $n\le 8$ and $\mathbb{Z}/2^s\mathbb{Z}$-torsion for $s\le23$. It also states at the beginning of Section 3.4:

Until now, no knot or link with torsion larger than $\mathbb{Z}/8\mathbb{Z}$ was known.

I believe this is the state of art.

Computations suggest that $T(p^k,p^k+1)$ should have $\mathbb{Z}/p^k\mathbb{Z}$-torsion for each $p$ prime and $k\ge 1$.