Timeline for Hochschild homology of acyclic complex
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| May 11, 2020 at 3:26 | vote | accept | user155668 | ||
| May 9, 2020 at 23:45 | answer | added | Pedro | timeline score: 0 | |
| May 9, 2020 at 23:24 | history | edited | user155668 | CC BY-SA 4.0 |
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| Apr 19, 2020 at 18:58 | comment | added | Denis Nardin | Both derived tensor products and geometric realizations of simplicial objects preserve quasi-isomorphisms, and the bar complex is obtained by composing the two operations. I'm not sure I can say much more than what is in Weibel's Homological algebra. | |
| Apr 19, 2020 at 17:51 | comment | added | user155668 | @DenisNardin Sorry, but while it's clear that a morphism of dg-algebras $A \to B$ induces a morphism $HH_*(A) \to HH_*(B)$, it's not clear why a quasi-isomorphism of dg-algebras induces an isomorphism on Hochschild homology. Maybe I'm using a bad definition for $HH_*(-)$? Could you say a few words about why this is clear? | |
| Apr 19, 2020 at 8:29 | comment | added | Denis Nardin | If you use the derived tensor product (which coincides with the classical one if $A$ is flat over $R$), then the definition of $HH_\ast(A)$ is manifestly invariant under quasi-isomorphism of algebras. In particular the map $A\to 0$ is a quasi-isomorphism... | |
| Apr 19, 2020 at 8:26 | comment | added | user155668 | I think I'd like to define HH_* just using the bar complex (i.e. the same way as I would defined it over a field). However, I'm happy to assume that A is flat over R, so I presume I don't have to derive anything? (In fact, in the case I'm interested in, A is a free associative algebra and R is a polynomial ring in one variable.) | |
| Apr 19, 2020 at 8:05 | comment | added | Denis Nardin | I guess the answer depends on how precisely define the Hochschild homology over a general ring $R$ (i.e., are you using the derived tensor product or not?). | |
| Apr 19, 2020 at 8:02 | history | asked | user155668 | CC BY-SA 4.0 |