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Showing posts from April, 2024

Bordered HF and Fox-Milnor for slice fibered knots

Problem: Categorify the Fox-Milnor criterion for slice fibered knots. If $K\subset S^3$ is a slice knot (either topologically or smoothly) then its Alexander polynomial satisfies $$\Delta_K(t) = f(t)f(t^{-1})$$ for some $f\in \mathbb{Z}[t,t^{-1}]$. This is called the Fox-Milnor criterion . Knot Floer homology is said to categorify  the Alexander polynomial; more precisely, its graded Euler characteristic satisfies $$\chi(\widehat{\mathit{HFK}}(K)) = \bigoplus_{m,a}(-1)^m t^a \dim \widehat{\mathit{HFK}}(K)  = \Delta_K(t).$$ It is therefore natural to ask (the admittedly vague question of) whether there are extra features of the knot Floer homology of slice knots which recover the Fox-Milnor criterion for the Alexander polynomial. One would hope moreover that these features provide a stronger obstruction to slicing a knot than the classical Fox-Milnor criterion. This is one thing that could be meant by the charge to categorify  Fox-Milnor.  It's not at all clear however what these e

Log-concavity of annular Khovanov homology

Annular Khovanov homology assigns to a link $L$ in a solid torus (aka a thickened annulus) a tri-graded abelian group $$\mathit{AKh}(L) = \bigoplus_{i,j,k} \mathit{AKh}^{i,j}(L,k).$$ Grigsby-Licata-Wehrli proved that there's an $\mathfrak{sl}_2(\mathbb{C})$ action on annular Khovanov homology with complex coefficients in which the weight space grading corresponds to the $k$-grading on $\mathit{AKh}.$ It follows that annular Khovanov homology is trapezoidal with respect to the $k$-grading. More precisely, if we let $a_k = \dim\mathit{AKh}(L,k)$, then $$a_k \geq a_{k'}$$ when $k\equiv k'\,\mathrm{mod}\, 2$ and $|k|<|k'|$. In particular,  the sequence $$\{a_k\}_{k\equiv i\, \mathrm{mod}\, 2}$$ is unimodal for any fixed $i$. The question below asks whether something stronger is true: Question: For a fixed $i$, is the sequence $\{a_k\}_{k\equiv i\, \mathrm{mod}\, 2}$ in fact log-concave ? This is asking whether $$a_{k}^2 \geq a_{k-2}a_{k+2}$$ for all $k$. If so, then th