Timeline for Can a Lie group as an abstract group be given more than one topology making it a Lie group?
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13 events
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| Aug 15, 2013 at 15:22 | comment | added | Mariano Suárez-Álvarez | (It would be nice to have a reference for the last result mentioned in this answer!) | |
| Apr 21, 2011 at 13:26 | comment | added | Selene Routley | Theo, I am intrigued by what you might be thinking in the second half of your question to Algori: "...just using the classification result to assert that ... then there exists a smooth isomorphism". Do you mean that group simplehood is independent of the Lie group structure we can consistently give it, so any isomorphism cannot take the group out of the known classical + exceptional list of possible simple groups. However, you need more to show that the group can't shift its Lie group structure in the process - as in Algori's $\mathbb{R}^n \leftrightarrow \mathbb{R}^m$ example. | |
| Apr 21, 2011 at 10:32 | comment | added | algori | WetSavannaAnimal aka Rod Vance -- I do not think the continuum hypothesis is necessary here: if a set $A$ has cardinality less than continuum, so has $\mathbb{Q}\times A$, the direct product of $A$ copies of $\mathbb{Q}$ (and so the direct sum as well). | |
| Apr 21, 2011 at 3:41 | comment | added | Selene Routley | Algori, in your answer, "they are Q -vector spaces of the same dimension" - does one need to assume the continuum hypothesis to say this: i.e. to show that they have the same cardinality? | |
| Apr 21, 2011 at 2:37 | comment | added | Selene Routley | Theo and algori: I (without the technical background as an optical engineer) am under the impression that the Linus Kramer paper cited above is proving Algori's first statement. I don't claim to understand all the steps in the Kramer paper (far from it), but if it is sound, he is almost interested in exactly the same question as I am groping for in my original, less well formed one. | |
| Apr 20, 2011 at 14:28 | comment | added | algori | Theo -- I am under the impression that the first of these two statements should be true (although I do not remember how to prove it). | |
| Apr 20, 2011 at 14:12 | comment | added | Theo Johnson-Freyd | @algori: Are you claiming that every group isomorphism of semisimple Lie groups is automatically smooth, or just using the classification result to assert that if the groups are isomorphic qua groups, then there exists a smooth isomorphism? | |
| Apr 20, 2011 at 14:10 | comment | added | Theo Johnson-Freyd | Well, there are uniform ways to attach Lie algebras to abstract groups, but it's not quite the way a Lie algebra comes from a Lie group. For example, the lower central series of a discrete group is a filtered pro-nilpotent group, and its associated graded abelian group naturally admits a Lie-ring structure. Of course, this is not what OP is after. | |
| Apr 20, 2011 at 11:27 | comment | added | algori | You are most welcome! Abstract groups do not in general have a Lie algebra attached to them, or they may have several, like a sum of continuously many copies of $\mathbb{Q}$. | |
| Apr 20, 2011 at 11:01 | comment | added | Selene Routley | Many thanks, Algori. So, if I'm reading this correctly, some abstract groups can have many different Lie algebras: $\mathbb{R}^n$ can be a Lie algebra for $\mathbb{R}^m$ as a groupm for any integers n and m? | |
| Apr 20, 2011 at 10:58 | vote | accept | Selene Routley | ||
| Apr 20, 2011 at 8:43 | history | edited | algori | CC BY-SA 3.0 |
added 4 characters in body
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| Apr 20, 2011 at 8:10 | history | answered | algori | CC BY-SA 3.0 |