Timeline for Name for algebra and its tensor products
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 13, 2020 at 19:38 | history | edited | Vladimir Dotsenko | CC BY-SA 4.0 |
added 6 characters in body
|
| Dec 22, 2015 at 23:49 | comment | added | Vladimir Dotsenko | I do not understand what you mean by "something constant" in your last sentence. Divisibility by 5 matters simply because for all $n$ divisible by 5 the algebra is automatically infinite-dimensional. (Abelianization for such $n$ is still geometrically 2-dimensional). | |
| Dec 22, 2015 at 22:45 | comment | added | Turbo | It might turn out interesting if $2^k+1$ are indeed special that way (I am not sure if this has anything to do with divisibility by $5$). I am also thinking may be for even $n$ something constant should suffice. | |
| Dec 22, 2015 at 22:42 | comment | added | Vladimir Dotsenko | I gave all numerical information that I have at the moment. I think that your questions about $2^t+1$ etc. are a bit far fetched. Looking at the commutative case, I would expect qualitative difference depending on whether or not $n$ is divisible by 5. Powers of 2 are a red herring. | |
| Dec 22, 2015 at 22:39 | comment | added | Turbo | One more query what is the dimensions if $n=2t$ form where $t\in\Bbb N$ ? | |
| Dec 22, 2015 at 22:14 | comment | added | Turbo | At $n=5$ you say 'so the corresponding abelianisation corresponds to something 2-dimensional geometrically' could you say whether at $n=2^k+1$ where $k\in\Bbb N_{>0}$ this is something $k$-dimensional geometrically and for other $n=2t+1$ where $t\in\Bbb N_{>0}$ and not of form $n=2^k+1$ where $k\in\Bbb N_{>0}$ can we tell anything at all? | |
| Dec 22, 2015 at 18:21 | history | answered | Vladimir Dotsenko | CC BY-SA 3.0 |