Timeline for Tensor product of linear mappings versus chain complexes
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
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| Apr 9, 2012 at 17:17 | comment | added | Simon Lentner | This answer goes in a very similar direction than my answer above; and I think the sign and the shift do arrise if the algebra and coalgebra are treated together...if you consider the $x\otimes x$ you're actually tliking about the groupring $k[\mathbb{Z}]$, i.e. $x$ has to be invertible. If you want on the other hand $x^2=0$ you MUST have the sign as I explained above - except of course you're in characteristic 2 where $k[x]/(x^2)$ IS the proper universal Lie enveloping :-) | |
| Apr 9, 2012 at 6:44 | history | edited | Qiaochu Yuan | CC BY-SA 3.0 |
added 10 characters in body
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| Apr 9, 2012 at 5:06 | comment | added | Chris Gerig | "Why this is the case" I'm sure is nothing more than because the differential should drop the grading by 1 and not 2. | |
| Apr 9, 2012 at 4:49 | comment | added | Qiaochu Yuan | The above answer doesn't explain the sign; if you're curious about that I think you should ask a separate question, although there are some MO questions that already cover similar background. | |
| Apr 9, 2012 at 4:46 | history | answered | Qiaochu Yuan | CC BY-SA 3.0 |