Timeline for Are groups in (Var/k, rational maps) necessarily algebraic groups?
Current License: CC BY-SA 2.5
10 events
| when toggle format | what | by | license | comment | |
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| Apr 22, 2010 at 4:27 | comment | added | BCnrd | Dear unknown (google): unless G = 1, identity never dominant. The defn of "strict" does not involve inverse in any way. Weil had to do real work to create it. The defn of strictness involves the rational "translation" maps $(x,y) \mapsto (\mu(x,y), y)$ and $(x,y) \mapsto (x, \mu(x,y))$ (which do make sense rationally, due to requiring $\mu$ to be dominant). One demands two things: these maps are birational, and the domain of defn of each meets all fibers of both projections $G \times G \rightrightarrows G$ in dense subsets. (Exer: check dense open in actual smooth f.type gp is strict.) | |
| Apr 22, 2010 at 4:26 | history | edited | Qfwfq | CC BY-SA 2.5 |
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| Apr 22, 2010 at 4:17 | comment | added | Qfwfq | @BCnrd: Does the definition of strict birational group law involve the "inverse" in some way? The problem in defining the identity instead, as far as I've understood from the comments above, is that $*\rightarrow G$ is almost never dominant. Right? | |
| Apr 22, 2010 at 4:11 | history | edited | Qfwfq | CC BY-SA 2.5 |
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| Apr 22, 2010 at 4:09 | comment | added | Qfwfq | @BCnrd: Thank you for the clarifications. | |
| Apr 22, 2010 at 4:02 | comment | added | BCnrd | Bjorn, maps are required to be dominant (e.g., $\mu$!), and in particular the definition of a birational group law involves no identity axiom and so no inverses! (The empty scheme is a birational group law.) It's all about translations and assoc. There is a finer notion of "strict birational group law", and thm is that (i) any non-empty bir. gp law on a smooth septd scheme of finite type contains dense open that is strict (for "induced" bir. gp law structure), and (ii) any non-empty strict bir. gp. law is dense open in unique actual smooth group. No control on affineness in the result. | |
| Apr 22, 2010 at 3:37 | comment | added | Bjorn Poonen | Is RVar even a category? You can't always compose rational maps. | |
| Apr 21, 2010 at 23:43 | comment | added | Theo Johnson-Freyd | Just a side question that shows my ignorance. Is the usual product of varieties the categorical product in RVar? It will be if there are not too many more rational maps than polynomials ones. | |
| Apr 21, 2010 at 21:45 | answer | added | Thomas Scanlon | timeline score: 18 | |
| Apr 21, 2010 at 19:56 | history | asked | Qfwfq | CC BY-SA 2.5 |