6
$\begingroup$

Let us consider the functors from dg algebras to graded spaces given by Hochshcild homology and cyclic homology. Then it is well known that in degree $0$, both functors are given by $A\longmapsto A/[A,A]$. However, if $A$ is a usual associative algebra over a field of characeristic $0$, and if we pick a free model $B=(TV,d)\longrightarrow A$, the homology of $B/[B,B]$ is $\mathrm{HC}_*(A)$. In this sense, HC is the right Taylor expansion for the abelianization functor in characteristic zero. At some point I naïvely thought that this would give HH instead. This motivates the following questions:

Q1. How should I have known that taking the left derived functor of abelianization in the model category $\mathsf{Alg}$ would give HC and not HH, even though at $t=0$ both theories coincide?

Q2. What other 'real life' examples are there of this phenomenon? (If the example is in $\mathsf{Alg}$, better!)

Q3. Can one describe the functors that coincide with HC and HH at $t=0$ in the sense above, but nonetheless give different theories in large degrees? Or at least give examples, as in Q2?

Disclaimer I am aware of the correct definition of $\mathrm{HH}$ as a funtor in $\mathsf{Alg}$ in terms of free models, where one needs to correct the $0$-forms $B$ by $1$-forms on $B$ and build a double complex. Similarly, $\mathrm{HC}$ in arbitrary characteristic needs a double complex built from the previous one, making it $2$-periodic, which incorporates the universal derivation from $0$-forms to $1$-forms. This gives me a "technical" answer, I am looking for something a bit less technical but still convincing.

$\endgroup$
2
  • 7
    $\begingroup$ Since HH of a free algebra is not concentrated in degree 0, trying to compute it by resolving and abelianizing fails in the simplest case. The left derived functor of an operation is the thing that gives exactly that operation on free algebras. That's one reason why it's HC not HH. $\endgroup$ Commented Dec 19, 2018 at 16:59
  • $\begingroup$ @PhilTosteson Oh, of course. $\endgroup$
    – Pedro
    Commented Dec 19, 2018 at 19:06

0

You must log in to answer this question.