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Let $k$ be a base commutative ring, and let $A$ be a (unital but not necessarily commutative) $k$-algebra. The cone on $A$ is the ring $CA$ of infinite matrices $(a_{ij})_{i,j \geq 1}$ that are finitely supported in each row and column. Then, the suspension of $A$ is $CA/\mathcal{M}_\infty(A)$, the quotient by the ideal $\mathrm{colim}_n \mathcal{M}_n(A)$ of finitely-supported matrices.

Here are some facts (lifted from section 1.4 of Loday's Cyclic Homology).

  1. $HH_*(CA) = 0$ and $HH_*(SA) \cong HH_{*-1}(A)$.
  2. There is a map $A[x^\pm] \to SA$ (selecting an infinite string of just-above-the-diagonal 1's) inducing a surjection $HH_*(A[x^\pm]) \twoheadrightarrow HH_*(SA)$.
  3. An external product with an element $\chi = [x] \in HH_1(k[x^\pm])$ induces an inverse $HH_*(SA) \cong HH_{*-1}(A) \xrightarrow{\cdot \chi} HH_*(A[x^\pm])$.

Recall that Hochschild homology can be defined as $HH_*(A) = \pi_*|N^{cyc}(A)|$, the geometric realization of the cyclic nerve. Thus, I would expect that fact 1 is actually a consequence of an equivalence $|N^{cyc}(A)| \simeq \Omega | N^{cyc}(SA)|$. Is this true? More generally, is there a sense in which I can read "$\mathrm{Spec}(SA) \to \mathrm{Spec}(CA) \to \mathrm{Spec}(A)$" as a (co?)fiber sequence with contractible middle term, at least "to the eyes of Hochschild homology"?

The second and third facts, however, I find yet more mysterious: they seem to assert that taking an "algebro-geometric sphere" (read: $\mathbb{G}_m$) over $A$ witnesses this delooping at the level of Hochschild homology. I have no idea why this should be true, but would love to have some intuitive (algebro-)geometric explanation.

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  • $\begingroup$ Well, I'm not sure that's an answer to part I, but the fact that $HH_\ast(CA)=0$ follows from the fact that this is an infinite sum algebra (check arxiv.org/pdf/0903.3983v1.pdf Exercise 2.3.3). Then it shall follow from the homology long exact sequence that $HH_\ast(SA)=HH_{\ast-1}(A)$ and hence (?) $|N^{cyc}(A)|$ is to be a homotopy retract of $\Omega|N^{cyc}(SA)|$. What am I doing wrong? $\endgroup$ – Kolya Ivankov Oct 26 '15 at 19:19
  • $\begingroup$ @KolyaIvankov Thanks for the reference. This provides plenty of algebra background, but I'm also hoping for some sort of "geometric" explanation. On the other hand, perhaps this can be extracted from the definition of an infinite sum ring, but I can't see it. Unfortunately these are necessarily noncommutative, so it's meaningless to ask what's going on with commutative infinite sum rings. $\endgroup$ – Aaron Mazel-Gee Oct 29 '15 at 16:55
  • $\begingroup$ Well, the algebra $CA$, in question, as well as suspension, was noncommutative, therefore the algebraic reference. Do You mean, say, some explanation why $|N^{cyc}(CA)|$ is contractible not using the existing homological argument? I'd be interested in that as well, frankly, if such thing exists. $\endgroup$ – Kolya Ivankov Oct 30 '15 at 21:23

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