Timeline for What turns $k$-variety into $k$-manifold?
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 14, 2014 at 16:43 | vote | accept | Michael | ||
| Jan 14, 2014 at 9:04 | answer | added | Peter Michor | timeline score: 4 | |
| Jan 14, 2014 at 1:28 | review | Close votes | |||
| Jan 14, 2014 at 10:52 | |||||
| Jan 14, 2014 at 1:12 | comment | added | Qfwfq | I can't make sense of this question (as it stands now). Voting to close. | |
| Jan 14, 2014 at 0:53 | comment | added | Michael | @TomGoodwillie: a manifold with almost quaternionic structure $I,J,K$, as described here. | |
| Jan 14, 2014 at 0:50 | comment | added | Tom Goodwillie | What do you mean by a quaternionic manifold? | |
| Jan 14, 2014 at 0:49 | comment | added | Michael | @QiaochuYuan: there are at least two possible interpretations of quaternionic polynomials: $\Bbb H[x]$ and noncommutative polynomials in $x$ you mentioned. In the former case assigning value to $x$ is not a homomorphism, so varieties won't make sense; however one can still consider zero-set of noncommutative polynomials. The simplest case I had in mind was the zero-set of $x^2+a=0$, which may consist of a 2-sphere. | |
| Jan 14, 2014 at 0:40 | comment | added | Qiaochu Yuan | What do you mean by a quaternionic variety? Over a commutative field you can argue that any polynomial is (in a suitable sense) locally linear and hence its zero set is (in a suitable sense) locally a hyperplane, but over a noncommutative field where you allow noncommutative polynomials like $x \mapsto axb$ this fails. | |
| Jan 14, 2014 at 0:23 | answer | added | Peter Crooks | timeline score: 1 | |
| Jan 14, 2014 at 0:17 | history | edited | Michael | CC BY-SA 3.0 |
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| Jan 14, 2014 at 0:06 | comment | added | Michael | @DanielLoughran: OK, modulo singularities, what properties of $k$ make "regular" $k$-varieties being manifolds? I'm more interested in the properties of $k$ that turn regular points of $k$-varieties into open neighborhoods of $k$-manifold, than in the properties of varieties themselves that make them regular. | |
| Jan 13, 2014 at 23:49 | comment | added | Daniel Loughran | A variety need not be a manifold if it is singular, e.g. $xy=0$ is not a manifold. | |
| Jan 13, 2014 at 23:34 | history | asked | Michael | CC BY-SA 3.0 |