ScalingStacks

Proof. We abbreviate $\mathcal{C}^{\prime}:=H^{\bullet}(\operatorname{Ch}_{\mathrm{dg}}(R\mathrm{-mod}^{\mathrm{gr.fr.}}))^{*}$. Let $\mathcal{C}$ denote the full subcategory generated by the indecomposable objects $C_{k}$. By Fact 4.11 and the discussion of the beginning of the section, it suffices to compute the bigraded zeroth Hochschild homology of $\mathcal{C}$. To this end, we study closed homogeneous endomorphisms of the objects $C_{k}$ and trace relations between them.

We note that the components of a chain map between shifts of such objects can have quantum degree zero or two (a scalar multiple of $\operatorname{Id}_{R}$ or $X_{R}$). Since the differential in every complex is of quantum degree two, this means that closed morphisms with components of quantum degree zero are homotopic if and only if they are equal.

First we investigate the chain maps between shifts of objects $C_{l}$ with components of quantum degree zero. For positive homological shifts (right shift) there are simply no closed morphisms, i.e. no chain maps. In shift zero we have the identity on every $C_{l}$ (which does not factor through any $C_{m}$ with $m\neq l$) and for negative homological shifts we have closed maps that factor into a composite of closed maps through a shift of a $C_{m}$ with $m<l$ (by induction, one can show that their trace classes actually vanish). Thus in bidegree $(0,0)$ we have a basis of trace classes $[\operatorname{Id}_{C_{l}}]$ for $l\geq 0$.

Second we are interested in chain maps between shifts of objects $C_{l}$ with components of quantum degree two. In negative homological shifts (left shift) all such maps are nullhomotopic. In non-negative homological shift, every such map is homotopic to a scalar multiple of $RX^{(l)}_{C_{k}}$. However, one easily checks that the trace class of $RX^{(l)}_{C_{k}}$ equals the trace class of $\pm RX^{(l)}_{C_{l}}$. Since these have bidegree $(l,2l+2)$ in the enriched $\operatorname{End}$ of $C_{l}$, we see that they are linearly independent. ∎

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 $C_{1}$. ∎