Does there exist a knot K such that operatornameKh^i,j(K;Z)cong operatornameKh^i,j(overlineK;Z) for every bidegree (i,j), while the Lipshitz–Sarkar stable-homotopy refinements X_LS(K) and X_LS(overlineK) are not equivalent?

Lipshitz–Sarkar mirror-detection question

Does there exist a knot $K$ such that $\operatorname{Kh}^{i,j}(K;\mathbb{Z})\cong\operatorname{Kh}^{i,j}(\overline{K};\mathbb{Z})$ for every bidegree $(i,j),$ while the Lipshitz–Sarkar stable-homotopy refinements $\mathcal{X}_{\mathrm{LS}}(K)$ and $\mathcal{X}_{\mathrm{LS}}(\overline{K})$ are not equivalent?

References

Primary source

arXiv

Additional references

Lipshitz-Sarkar refines mirrors more than Khovanov — arXiv

Posted 1 month ago

Progress summary

Refreshed 1 month ago

Claimed solved

A September 2026 preprint claims that a refined knot invariant can distinguish a knot from its mirror, but the result has not been independently confirmed.

The question asks whether the Lipshitz–Sarkar stable-homotopy refinement detects mirror reflection in cases where ordinary integral Khovanov homology does not. No proposer or original date is identified in the retrieved material.

September 2026 claimed affirmative example

A preprint claims an explicit knot–mirror pair distinguished by the Lipshitz–Sarkar refinement despite agreement of their ordinary integral Khovanov homologies. It presents this as an affirmative answer, but the construction and computations are not independently confirmed.

Current status (as of September 2026): An affirmative example is claimed in a preprint, but it is unverified, so the question remains mathematically unsettled.

Sources

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