Thm - Amitsur Cochain as E_1 cotangent complex

thm - Koszul Duality exchanges invertibility and categorification

Background: Amitsur Filtration

Setup: Adams Completion via Descent

Let

• $V^{\otimes }$ be a stable presentably symmetric monoidal $\infty$-category.
• $A\in Al{g}^{(1)}(V)$ be a map of $E_1$ algebra in $V$.

Descent along the unit map $1\stackrel{u}{\to }A$ gives a functor

$V\stackrel{\delta }{\to }coLMo{d}_{A\otimes A}(LMo{d}_A(V))$ lifting $A{\otimes }_1(-):LMo{d}_1(V)\to LMo{d}_A(A)$. $\delta$ admits a right adjoint $\phi$, which reconstruct an object from descent data. The unit $I{d}_V\to \phi \circ \delta :=\widehat{(-{)}_A}$ is said to implement $A$-completion.

Amitsur Filtration The comonadic-reconstruction functor $\phi$ in general is given by totalization of a simplicial object. In particular, we can write

$$\widehat{X_A}\simeq Tot(C{B}^{\bullet }(X))$$

for some cosimplicial object $C{B}^{\bullet }(X)$, we defer to the paper of MNN for details. Partial totalization then give rise to a downward filtration on $\widehat{X_A}$, which we’ll call the Amitsur filtration, denoted by

$A{m}_n(X):=To{t}_{\le n}(C{B}^{\bullet }(X))$ Explicit form of the filtration For the remainder we’ll on the case $X=1$. MNN shows the following

$A{m}_n(X)\simeq Fib(I^{\otimes n+1}\to 1)$ Where

• $I:=Fib(1\stackrel{u}{\to }A)\to 1$
• We get maps $I^{\otimes n+1}\to I^{\otimes n}$ as $I^{\otimes n}\otimes (I\to 1)$. Iterating them gives $I^{\otimes n+1}\to I^{\otimes n+1}\to \cdots I^{\otimes 1}\to 1$ this is the map appearing in the equation above.

Amitsur Cochains and Cotangent complex

TODO: add diagrams from our earlier notes on the subject

A simple diagram chase yields

Lemma $g{r}^n(A{m}_{\bullet }(1))\simeq \Sigma A\otimes I^{\otimes n+1}$ Now we also have the following

Lemma $Fib(A\otimes A\to A)\simeq \Sigma A\otimes I$ Proof: Split the diagram using $1\otimes A\stackrel{(u,Id)}{\to }A\otimes A$, compare fibers.

Now we recognize $Fib(A\otimes A\to A)$ as the $E_1$ cotangent complex ${\mathbb{L}}_A$. Putting the two above lemmas together gives an identification

$g{r}^n(A{m}_{\bullet }(1))\simeq {\mathbb{L}}_A^{{\otimes }_{A}n+1}$

Action: Account for the degrees here carefully, does the degree shift indicate maybe we can recover this from THH?