1
$\begingroup$

Let $L$ be a link in $\mathbb{R}^3$, with $D, D'$ be diagrams in $\mathbb{R}^2$ representing $L$. Khovanov constructed two graded chain complexes $$C_{D} = (Ch_{D}, d_{D}) \quad C_{D'}=(Ch_{D'}, d_{D'})$$ and proved that their homologies $Kh(D), Kh(D')$ are isomorphic. This defines an invariant for the link $L$, namely the Khovanov homology $Kh(L)$.

Mantra in algebraic topology suggests that we should not be taking homology too quickly (to avoid forgetting useful information).

I thus wonder if there are some weaker equivalences between $C_{D}$ and $C_{D'}$. Indeed there is: the Khovanov homology is known to be a degeneration of Khovanov homotopy type (i.e. a refined invariant $\tilde{Kh}(L))$ which is a homotopy type up to homotopy equivalence). How about a version in terms of equivalence of $A_\infty$ algebras? If such Khovanov $A_\infty$ algebra indeed exists, is it stricly weaker than the Khovanov homotopy type?

$\endgroup$
2
  • 2
    $\begingroup$ I think this is very speculative because apparently you don’t even know whether such an A-infinity algebra structure exists. $\endgroup$ Nov 26, 2022 at 21:01
  • 2
    $\begingroup$ Khovanov homology is not a ring. If you know a ring structure, write a paper and wow a lot of people. The refinement is to a stable homotopy type (finite CW-spectrum), and spectra do not have cohomology rings: stabilization destroys the cup product. (You should write a proof of that as an exercise if you don't already know it.) What you do get is an action of the Steenrod algebra on Khovanov homology and people certainly do investigate this at the chain level, eg arXiv:1204.5776 $\endgroup$
    – mme
    Nov 27, 2022 at 23:05

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.