ScalingStacks

Proof. We have already explained the two isomorphisms. We now need to understand the essential image of $\boldsymbol{\mathrm{T}}_{2}(*,*)$ under the full embedding into $H^{\bullet}(\operatorname{Ch}_{\mathrm{dg}}(R\mathrm{-mod}^{\mathrm{gr.fr.}}))^{*}$. We claim that the invariant of any $(1,1)$-tangle decomposes into (shifts of) the indecomposable summands $C_{0}$ and $C_{1}$, but never $C_{l}$ for $l\geq 2$. Provided this claim holds, we can compute $\mathrm{HH}_{0}(\boldsymbol{\mathrm{T}}_{2}(*,*))$ as the Hochschild homology of the full additive subcategory of $H^{\bullet}(\operatorname{Ch}_{\mathrm{dg}}(R\mathrm{-mod}^{\mathrm{gr.fr.}}))^{*}$ generated by $C_{0}$ and $C_{1}$, and this again is isomorphic to the Hochschild homology of the full subcategory on the two objects $C_{0}$ and $C_{1}$. Here we use that the zeroth Hochschild homology is preserved under proceeding to the additive and idempotent completion; see Fact 4.12. Following the same arguments as in Proposition 4.14, we see that it is $4$-dimensional, spanned by $[\operatorname{Id}_{C_{l}}]$ and $[RX_{C_{l}}]$ for $l\in\{0,1\}$

The key idea to prove the claim is that all complexes appearing in Khovanov homology come from complexes over $\mathbbm{k}[X]$ by setting $X^{2}=0$ (though certainly not all complexes over $\mathbbm{k}[X]/(X^{2})$ have this property). Indeed, one can use equivariant Khovanov homology, defined over the ring $\mathbbm{k}[X,\alpha]/(X^{2}-\alpha)\cong\mathbbm{k}[X]=:R^{\prime}$ to simplify the complex of a $(1,1)$-tangle into a complex of graded free $\mathbbm{k}[X]$-modules. These decompose, up to homotopy equivalence and shift, into chain complexes of the form

Upon reducing to the ordinary Khovanov theory by tensoring with $\mathbbm{k}[X]/(X^{2})$ over $\mathbbm{k}[X]$, these complexes decompose into (shifts of) copies of $C_{0}$ and $C_{1}$. ∎