Direct-sum decomposition formula for S^2_{0,0}(X;α) in terms of cable Khovanov homology
Prove the direct-sum formula asserting that, for a 2-handlebody X obtained by attaching 2-handles along a framed oriented m-component link K ⊂ S^3 with diagram D having no 0-framed unknot components, and for any α ∈ Z^m, the Khovanov skein lasagna module S^2_{0,0}(X;α) is isomorphic to the direct sum over α′ ∈ Z^m with α′ ≡ α (mod 2) and h(α′) = h(α) of KhR_0(K(α′);Q) shifted by Σ_i ((w_i − p_i)(α′_i^2 + 2α′_i)/2 − w(K_i(α′_i))) − |α′|, where P is the framing matrix, W the writhe vector, N the crossing matrix (from D), and h(β) = β^T(P−W+N)β.
References
We expect that in the notation of Theorem 5.5, assuming no component of D is a 0-framed unknot, then (w −p)(α ′2+2α ′) w(K (α )) ′ S 2 (X;α) = ⊕ KhR 0(K (α )){ ∑ i i i i − −∣α ∣}. (30) 0,0 α ∈Zm 2 i 2 ≥′ 2∣′−α h(α )=h(α) … However, we have not been able to work out a proof of (30).