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Recall that Hochschild-Kostant-Rosenberg -type theorems identify certain smoothness conditions guaranteeing an isomorphism between the cotangent complex and (a shift of) the Hochschild homology of an algebra $A$.

The reason one might hope for something like this in the first place is the following. The cotangent complex is

$$L_A = \Sigma^\infty_{CAlg_{A//A}}(A \otimes A) = \varinjlim_n \Omega^n_{CAlg_{A//A}} \Sigma^n_{CAlg_{A//A}} (A \otimes A)$$

whereas the shifted Hochschild homology is

$$\Omega HH(A) = A \times_{A \otimes_{A \otimes A} A} A = \Omega_{CAlg_{A//A}} \Sigma_{CAlg_{A//A}} (A \otimes A)$$

So an HKR theorem says that the colimit $\varinjlim_n \Omega^n \Sigma^n$ stabilizes after one step, under certain smoothness conditions.

Question 1: Does this colimit ever stabilize after a finite number of steps which is greater than 1? If so, does this correspond to some kind of conditions on how singular $A$ is?

Question 2: Is this HKR phenomenon of $\Omega^n \Sigma^n$ stabilizing after one step in any way related to similar "group completion" phenomena in algebraic K-theory? For instance, I believe that the $S_\bullet$ construction generally only needs $\Omega B$ to be applied once before it stabilizes.

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    $\begingroup$ Via cofibrant replacement, applying the HKR theorem levelwise extends it to all simplicial rings, when expressed in terms of the cotangent complex. $\endgroup$
    – Jon Pridham
    Commented Oct 17, 2022 at 18:59
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    $\begingroup$ The premise of the question confuses me - HKR identifies HH_* with the symmetric algebra of the shifted cotangent complex (aka differential forms), not the shifted complex itself.. also no smoothness is necessary but characteristic zero is. $\endgroup$
    – David Ben-Zvi
    Commented Oct 17, 2022 at 23:13

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