Timeline for Tensor product of linear mappings versus chain complexes
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
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| May 10, 2015 at 15:30 | history | edited | shuhalo |
added tag
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| Apr 12, 2012 at 10:31 | answer | added | Jake | timeline score: 0 | |
| Apr 11, 2012 at 19:54 | comment | added | shuhalo | @Lentner. Thank you very much for your answer. I will read until weekend --- quite busy at the moment. Furthermore, thanks to Qiaochu for your answer and the other commentors. Really great this question allows so many perspectives! | |
| Apr 9, 2012 at 4:46 | answer | added | Qiaochu Yuan | timeline score: 4 | |
| Apr 8, 2012 at 22:48 | answer | added | Simon Lentner | timeline score: 12 | |
| Apr 8, 2012 at 1:38 | comment | added | Tom Goodwillie | This operator $x\otimes y\mapsto dx\otimes y+(-1)^{|x|}x\otimes Dy$ is never called $d\otimes D$; it's called some kind of $d$ again. Try thinking of this as a Leibniz rule (product rule). Boundary maps in chain complexes are a lot like derivations. | |
| Apr 7, 2012 at 21:53 | history | edited | darij grinberg | CC BY-SA 3.0 |
latex backticked
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| Apr 7, 2012 at 21:30 | comment | added | Jack Huizenga | Your "naive" definition of tensor product doesn't yield an object that you would interpret as a chain complex, since the new differentials are changing the grading by 2. Since the even and odd degrees are not related by any maps, this object could be interpreted as a pair of "even" and "odd" complexes, not a single complex. | |
| Apr 7, 2012 at 21:29 | comment | added | Yosemite Sam | @QY: please DO elaborate, I'm very interested in this point of view. | |
| Apr 7, 2012 at 21:22 | comment | added | Qiaochu Yuan | Briefly one should think of the differential as an element of a super-Lie algebra (so for starters think about how a Lie algebra acts on tensor products of its representations). There are MO questions elaborating on this but I can't find them at the moment. | |
| Apr 7, 2012 at 21:19 | history | edited | shuhalo | CC BY-SA 3.0 |
edited body
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| Apr 7, 2012 at 21:16 | comment | added | shuhalo | At least on my computer, the term $(X \otimes Y ){p} := \oplus{k+l=p} X_k \otimes Y_l$ does not seem to render as expected. If others see this as well, I am sorry about that. | |
| Apr 7, 2012 at 21:11 | history | asked | shuhalo | CC BY-SA 3.0 |