Khovanov to Heegaard Floer collapse

Question: Does $\textrm{dim}_\mathbb{F}\widetilde{\mathit{Kh}}(L) =\textrm{dim}_\mathbb{F} \widehat{\mathit{HF}}(\Sigma(L))$ imply that $\textrm{dim}_\mathbb{F}\widetilde{\mathit{Kh}}(L)=\textrm{det}(L)$?

Ozsvath--Szabo prove for any link $L\subset S^3$ that there exists a spectral sequence from the reduced Khovanov homology $\widetilde{\mathit{Kh}}(L)$ to the Heegaard Floer homology $\widehat{\mathit{HF}}(\Sigma(L))$ of the branched double cover of $S^3$ branched along $L$, with coefficients in $\mathbb{F}=\mathbb{Z}/2\mathbb{Z}$. This implies an inequality of dimensions \[\textrm{dim}_\mathbb{F}\widetilde{\mathit{Kh}}(L) \geq \textrm{dim}_\mathbb{F} \widehat{\mathit{HF}}(\Sigma(L)).\] This is an equality---equivalently, the spectral sequence collapses---whenever \[\textrm{dim}_\mathbb{F}\widetilde{\mathit{Kh}}(L)=\textrm{det}(L)\] (which happens for instance whenever $L$ is quasi-alternating or more generally Khovanov-thin), in which case $\Sigma(L)$ is an L-space. The question above asks if this is the only way in which the spectral sequence can collapse.




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