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 $\alpha$?


1 Answer 1


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 $A \to B$.

  • $\begingroup$ If B is only a discrete associative algebra then the map is instead a map from A into Z(B) (where Z is the center)? $\endgroup$ May 23, 2018 at 0:49
  • $\begingroup$ Also, when would a situation (3,1) occur where HC_*(A) would have non-trivial stuff in negative degrees so that it is not just a map from A? $\endgroup$ May 26, 2018 at 0:07
  • $\begingroup$ This all makes so much more sense now. Coming back to it now. $\endgroup$ Jun 29, 2019 at 11:06

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