ScalingStacks

Proof. We let $s=t=p\;\mathrm{points}$ and again have isomorphisms

$$\mathcal{S}_{0}^{2}(S^{1}\times B^{3};S^{1}\times P_{p},\mathbbm{k})\cong\mathrm{HH}_{0}(\mathcal{S}_{0}^{2}(B^{3};P_{p},\mathbbm{k}))\cong\mathrm{HH}_{0}(\boldsymbol{\mathrm{T}}_{2}(s,t))$$

and we consider the category $\boldsymbol{\mathrm{T}}_{2}(s,t)$ as a full subcategory of the enriched morphism category $H^{\bullet}(\boldsymbol{\mathrm{Foam}}_{2}^{\mathrm{dg}})^{*}(s,t)$.

The $\mathbbm{k}$-linear, additive category $H^{\bullet}(\boldsymbol{\mathrm{Foam}}_{2}^{\mathrm{dg}})^{*}(s,t)$ is Krull-Schmidt and hence idempotent complete; see e.g. the discussion in [30, Sections 4.5, 4.8] based on Bar-Natan’s category, which is equivalent to $\boldsymbol{\mathrm{Foam}}_{2}$ by [6].

Now $\mathrm{Kar}(\boldsymbol{\mathrm{T}}_{2}(s,t))^{\oplus}$ may be considered as an additive, idempotent complete full subcategory of $H^{\bullet}(\boldsymbol{\mathrm{Foam}}_{2}^{\mathrm{dg}})^{*}$; it is thus itself Krull–Schmidt. We have $\mathrm{HH}_{0}(\boldsymbol{\mathrm{T}}_{2}(s,t))\cong\mathrm{HH}_{0}(\mathrm{Kar}(\boldsymbol{\mathrm{T}}_{2}(s,t))^{\oplus})$ by Fact 4.12. Therefore, it suffices to compute its zeroth Hochschild homology of $\mathrm{Kar}(\boldsymbol{\mathrm{T}}_{2}(s,t))^{\oplus}$.

It is straightforward to check that the objects of $\mathrm{Kar}(\boldsymbol{\mathrm{T}}_{2}(s,t))^{\oplus}$ have finite-dimensional endomorphism algebras, and since $\mathbbm{k}$ is perfect, the Chern character

$$h\colon K_{0}(\mathrm{Kar}(\boldsymbol{\mathrm{T}}_{2}(s,t))^{\oplus})\otimes_{\mathbb{Z}}\mathbbm{k}\to\mathrm{HH}_{0}(\mathrm{Kar}(\boldsymbol{\mathrm{T}}_{2}(s,t))^{\oplus})$$

is injective; see Proposition 4.22. To prove that $\mathcal{S}_{0}^{2}(S^{1}\times B^{3};S^{1}\times P_{p},\mathbbm{k})$ is infinite-dimensional in bidegree $(0,0)$, it is thus sufficient to show that $K_{0}(\mathrm{Kar}(\boldsymbol{\mathrm{T}}_{2}(s,t))^{\oplus})\otimes_{\mathbb{Z}}\mathbbm{k}$ is infinite-dimensional.

Moreover, $K_{0}(\mathrm{Kar}(\boldsymbol{\mathrm{T}}_{2}(s,t))^{\oplus})$ is free abelian on the isomorphism classes of its indecomposable objects; cf. Proposition 4.20. Thus, we will be done once we can exhibit infinitely many indecomposable and pairwise non-isomorphic complexes appearing as (direct summands in) tangle complexes.

We will see that such complexes can be constructed as invariants of braids. Clearly, for $p\geq 2$ there are infinitely many braids on $p$ strands. Moreover, the braid complexes are invertible under tensoring with the complex for the respective inverse braid. Since the complex of the trivial braid is indecomposable (its endomorphism algebra $\left(\mathbbm{k}[X]/(X^{2})\right)^{\otimes p}$ is local), so are the complexes for all other braids. It is also known that all braid complexes are pairwise non-isomorphic. This can e.g. be deduced from the faithfulness of the braid group action of Khovanov–Seidel [22]. For us, however, it is enough to consider infinitely many braids that are powers of a single Artin braid generator. For these complexes it is straightforward to check by hand that they are pairwise non-isomorphic. ∎