Following the paper by Ayala, Francis, and Tanaka: <https://arxiv.org/pdf/1409.0848.pdf>


If we are talking about knots we are talking about framed 3-manifolds with a framed 1-dimensional sub-manifold and trivialized normal bundle.


Lets assume the monoidal category is dg-vector spaces with $\otimes$.


Corollary 0.3. says there is an equivalence between framed $1\subset3$ disk-algebras and a triple $(A,B,\alpha)$ where $A$ is a 3-disk algebra ($E\_3$-algebra), $B$ is a 1-disk algebra ($E\_1$-algebra or $A\_\infty$) and $$\alpha:HC\_\*(A)\to HC^\*(B)$$ is a map of 2-disk algebras.


My questions are: If $A$ and $B$ are commutative algebras (concentrated in degree 0) over some field $\mathbb{k}$ (smooth, if you wish) then $A$ is in particular also $E\_3$ and $B$ is $E\_1$.


1) What kind of map should $\alpha$ be in this case?


2) If we take $HC\_\*$ to mean the hochschild chain complex of $A$ then $HC\_\*(A)$ is concentrated in (homological) non-negative degree and $HC^\*(B)$ is concentrated in non-positive degree. Should our map have to respect degrees of elements or is there some sort of degree shift going on?


3) It seems this structure should make $B$ an $A$-module (in the usual sense). This should be a ring map from $A$ into $End(B)$ or a series of compatible linear maps one such being $A\otimes A\otimes B\to B$. Is there a way to see this from the map $\alpha$?



Knot Factorization Homology inputs
Let me start with the second question:


2) The map $\alpha$ will indeed respect the degrees. You should think of it as a map between two $E\_2$-algebras in unbounded chain-complexes (whose domain happens to be concentrated in non-negative degrees, while its codomain is concentrated in non-positive degrees).


1) As a result of (2), the data of the map $\alpha$ is actually equivalent to the data of a map of discrete commutative algebras $A \to B$. To see this, observe that the inclusion $\iota\_{\leq 0}: {\rm Ch}\_{\leq 0}(\mathbb{k}) \to {\rm Ch}(\mathbb{k})$ of non-positively graded chain-complexes into unbounded chain-complexes has a left adjoint $\tau\_{\leq 0}: {\rm Ch}(\mathbb{k}) \to {\rm Ch}(\mathbb{k})\_{\leq 0}$, the truncation functor. Both $\iota\_{\leq 0}$ and $\tau\_{\leq 0}$ are symmetric monoidal functors and descend to an adjunction between $E\_2$-algebras in ${\rm Ch}(\mathbb{k})$ and $E\_2$-algebras ${\rm Ch}(\mathbb{k})\_{\leq 0}$. This means that the data of an $E\_2$-algebra map $\alpha: HC\_\*(A) \to HC^\*(B)$ is equivalent to the data of an $E\_2$-algebra map $A = \tau\_{\leq 0}HC\_\*(A) \to HC^\*(B)$. A similar argument using cotruncation instead of truncation shows that the data of such a map is equivalent to the data of an $E\_2$-algebra map $A \to \tau^{\geq 0}HC^\*(B) = B$. Since $A$ and $B$ are discrete this is the same as a map of commutative algebras.


3) Suppose $A$ was an $E\_1$-algebra and $B$ a chain-complex (which we can think of as an $E\_0$-algebra object in ${\rm Ch}(\mathbb{k})$). Then the structure of an $A$-module on $B$ is given by a map of $E\_1$-algebras $A \to {\rm End}(B)$. This is because the monoidal structure on ${\rm Ch}(\mathbb{k})$ is closed, i.e., admits a compatible system of internal mapping objects. If $A$ is now an $E\_2$-algebra and $B$ is an $E\_1$-algebra, then we can think of $A$ as an $E\_1$-algebra in $E\_1$-algebras, and so there is an associated notion of an action of $A$ on $B$ (in $E\_1$-algebras) in which case one usually says that $B$ has the structure of an $A$-algebra. However, the induced tensor product on $E\_1$-algebras is not a closed monoidal structure: given two $E\_1$-algebras $B,B'$, there is no internal mapping object ${\rm Map}(B,B')$ in $E\_1$-algebras. Nonetheless, there is still an object which controls $A$-actions on $B$, and that is the center $HC^\*(B)$ of $B$, considered as an $E\_2$-algebra. In particular, $A$-algebra structures on $B$ are classified by $E\_2$-algebra maps $A \to HC^\*(B)$. Note that for $A$, as an $E\_1$-algebra object ${\rm Alg}\_{E\_1}({\rm Ch}(\mathbb{k}))$, there is also the associated notion of an *$A$-bimodule* structure on a given $B \in {\rm Alg}\_{E\_1}({\rm Ch}(\mathbb{k}))$. This, in turn, can be described by a map of $E\_2$-algebras $A^{\rm op} \otimes A \to HC^\*(B)$. Now let's consider the case at hand where $A$ is an $E\_3$-algebra and $B$ is an $E\_1$-algebra. Then we can think of $A$ as an $E\_2$-algebra in $E\_1$-algebras. Now, just like associative algebras have the associated notion of a bimodule, for $E\_2$-algebras there is the analogous notion of an *$E\_2$-module*. Roughly speaking, an $E\_2$-action of $A$ on $B$ means a "continuous family" of $A$-actions on $B$ parameterized by the circle (with suitable compatibilities with the $E\_2$-structure of $A$). If $A$ is an $E\_2$-algebra in ${\rm Alg}\_{E\_1}({\rm Ch}(\mathbb{k}))$ and $B$ is an object of ${\rm Alg}\_{E\_1}({\rm Ch}(\mathbb{k}))$ then the data of an $E\_2$-action of $A$ on $B$ will be given by an $E\_2$-algebra map of the form $\alpha: HC\_\*(A) \to HC^\*(B)$. When $A$ and $B$ are discrete commutative algebras this is the same as just an $A$-algebra structure on $B$, which, in turn, is given by a map of commutative algebras $A \to B$.


