3
$\begingroup$

I have the following basic question on Čech–Alexander complexes.

Let $R$ be a ring and $A$ be an $R$-algebra. To this datum one can attach a cosimplicial ring which assigns to an object $[n]=\{0,1,\dotsc,n\}$ of the category of totally ordered finite sets $\Delta$ the ring $A^{\otimes [n]}=A\otimes_R \cdots \otimes_R A$ (the Čech nerve of $R\to A$). To this cosimplicial ring, we can consider its Moore Complex, also called in this situation the Čech–Alexander complex: $$(\mathrm{CA})\quad 0\to A\xrightarrow{\partial^1} A\otimes_R A \xrightarrow{\partial^2}A\otimes_R A\otimes_R A\xrightarrow{\partial^3} \cdots$$ with $A^{\otimes [n]}$ in degree $n$ and $$ \partial^n(a_1\otimes\dotsb \otimes a_n)=\sum_{i=0}^n{(-1)^i(a_1\otimes \dotsb \otimes a_i\otimes 1 \otimes a_{i+1}\otimes \dotsb \otimes a_n)}$$

My Question: Is there a canonical multiplicative structure on $(\mathrm{CA})$? If yes, what is the formula? Is this obtained by the maps $A^{\otimes [i]}\otimes A^{\otimes [j]}\to A^{\otimes [i+j]}$, $(a_0\otimes \cdots \otimes a_i)\otimes (b_0\otimes \cdots \otimes b_j)\mapsto a_0\otimes \cdots \otimes (a_i b_0)\otimes \cdots \otimes b_j$?

For instance, I do not see how one can multiply two elements in degree one to obtain one of degree two; that is two elements in $A\otimes_R A$ to get an element in $A\otimes_R A\otimes_R A$. This looks inconveniently shifted….

EDIT: If I am inclined to think that there is a multiplicative structure, this is by transport of structure from the one of cohomology theories on indiscrete/chaotic sites; e.g. as described by Kedlaya in Notes on prismatic cohomology.

$\endgroup$
4
  • $\begingroup$ I know it's not terribly important, but, since your $[n]$ has $n + 1$ elements, I'm confused what $A^{\otimes[n]}$ means. Does it have $n$ or $n + 1$ tensor factors? $\endgroup$
    – LSpice
    Commented Mar 17 at 14:49
  • 1
    $\begingroup$ @LSpice Thanks for your comment! $A^{\otimes [n]}$ indeed has $n+1$ tensor factors (and lives in degree $n$). $\endgroup$
    – Stabilo
    Commented Mar 17 at 14:51
  • $\begingroup$ I think the formula of 20.25.3.2 here should apply? stacks.math.columbia.edu/tag/01FP $\endgroup$ Commented Mar 17 at 16:18
  • $\begingroup$ @NicolasHemelsoet Thanks for your comment! Are you able to unravel what it means in this particular situation? $\endgroup$
    – Stabilo
    Commented Mar 17 at 16:47

0

You must log in to answer this question.