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What is the automorphism group of the complex Lie algebra of bounded operators on a complex Hilbert space, with the commutator as Lie bracket? What for the real Lie algebra of bounded antihermitian operators? Does their structure depend on whether the space is finite-dimensional, infinite-dimensional separable, or inseparable? Does it depend on continuity assumptions in appropriate topologies?

There are obvious subgroups, namely the group of inner automorphisms given by conjugation by some invertible element of the algebra in the first case, and by conjugation by a unitary or antiunitary operator in the second case. The question is how much more there can be.

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  • $\begingroup$ A natural question is whether automorphisms are automatically continuous for the possible natural topologies on these Lie algebras, and independently to describe bicontinuous automorphisms. $\endgroup$
    – YCor
    Mar 25, 2016 at 12:48
  • $\begingroup$ 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). $\endgroup$
    – YCor
    Mar 25, 2016 at 12:55
  • $\begingroup$ @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. $\endgroup$
    – Arnold Neumaier
    Mar 25, 2016 at 15:30
  • $\begingroup$ 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. $\endgroup$
    – YCor
    Mar 25, 2016 at 16:35
  • $\begingroup$ I rewrote the question to avoid having to assume normality of the inner automorphism group. $\endgroup$
    – Arnold Neumaier
    Mar 25, 2016 at 16:43

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