Timeline for Automorphism group of Lie algebra of bounded operators
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| Mar 25, 2016 at 16:43 | comment | added | Arnold Neumaier | I rewrote the question to avoid having to assume normality of the inner automorphism group. | |
| Mar 25, 2016 at 16:40 | history | edited | Arnold Neumaier | CC BY-SA 3.0 |
added detail from discussion
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| Mar 25, 2016 at 16:35 | comment | added | YCor | OK. So, in the antihermitian case, it's unclear to me that "inner automorphism" form a normal subgroup (thus "outer automorphism group" possibly doesn't make sense). Of course, the question of describing the automorphism group remains. | |
| Mar 25, 2016 at 15:33 | history | edited | Arnold Neumaier | CC BY-SA 3.0 |
added question inspired by Ycor's comment
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| Mar 25, 2016 at 15:30 | comment | added | Arnold Neumaier | @YCor: If the topology makes a difference to the answer, it would be interesting to know the dependence on the topology. Yes to your second question in the first case; in the second case, conjugation by a unitary or antiunitary operator. | |
| Mar 25, 2016 at 12:55 | comment | added | YCor | I guess you call "inner automorphism" an automorphism given by conjugation by some invertible element of the algebra in the first case, and by some unitary operator in antihermitian case? (In the latter case, it's not immediate to me that it is a normal subgroup, as automorphisms are not defined on conjugating elements). | |
| Mar 25, 2016 at 12:48 | comment | added | YCor | A natural question is whether automorphisms are automatically continuous for the possible natural topologies on these Lie algebras, and independently to describe bicontinuous automorphisms. | |
| Mar 25, 2016 at 11:29 | history | asked | Arnold Neumaier | CC BY-SA 3.0 |