Timeline for Commutator of algebraic subgroups is connected
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
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| Sep 2, 2012 at 15:31 | comment | added | Abhishek Parab | I missed the subtle point that Guntram mentioned. Thanks for pointing it out. | |
| Sep 2, 2012 at 6:04 | comment | added | Todd Trimble | Right, I was only addressing one of the questions of the OP. There is no reason I can think of why the statement for topological groups should imply the statement for algebraic groups. | |
| Sep 2, 2012 at 5:24 | comment | added | Guntram | It is somewhat subtle, but an algebraic group is not a topological group for the Zariski topology. This is because for a topological group one requires that the multiplication map $\mu:G \times G \to G$ be continuous for the product topology, while for algebraic groups one puts on the Zariski topology on $G \times G$, which does not coincide with the product topology in general. So it is not clear to me why the argument for topological groups should imply the statement for algebraic groups. | |
| Sep 2, 2012 at 3:45 | vote | accept | Abhishek Parab | ||
| Sep 2, 2012 at 3:32 | history | answered | Todd Trimble | CC BY-SA 3.0 |