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S May 26, 2023 at 5:59 vote accept Александр Худяков
May 26, 2023 at 3:14 history became hot network question
S May 26, 2023 at 3:09 vote accept Александр Худяков
S May 26, 2023 at 5:59
S May 26, 2023 at 3:09 vote accept Александр Худяков
S May 26, 2023 at 3:09
May 26, 2023 at 1:11 history edited LSpice CC BY-SA 4.0
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May 26, 2023 at 0:18 answer added Moishe Kohan timeline score: 16
May 25, 2023 at 21:58 answer added Geoff Robinson timeline score: 7
May 25, 2023 at 21:31 vote accept Александр Худяков
S May 26, 2023 at 3:09
May 25, 2023 at 21:25 comment added YCor The conclusion is negative as [Dave Benson's answer] shows. However, I think it can be shown that one can block-triangulate so that diagonal blocks are unitary. In particular, the conclusion is positive in the irreducible case.
May 25, 2023 at 21:21 history edited YCor CC BY-SA 4.0
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May 25, 2023 at 20:25 vote accept Александр Худяков
May 25, 2023 at 20:27
May 25, 2023 at 20:18 comment added Александр Худяков Fedor Petrov, thank you for noting this! The hard part is to prove that the operators are bounded.
May 25, 2023 at 20:14 history edited Александр Худяков CC BY-SA 4.0
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May 25, 2023 at 20:12 comment added Fedor Petrov If all operators in $G$ are uniformly bounded, then $G$ preserves the norm $\|x\|:=\sup_{g\in G} |gx|$ (where $|x|$ is some fixed norm). Thus it preserves the John ellipsoid of the unit ball of $\|\cdot\|$, that yields affirmative answer.
May 25, 2023 at 20:08 answer added Dave Benson timeline score: 14
May 25, 2023 at 19:26 history edited Александр Худяков CC BY-SA 4.0
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S May 25, 2023 at 19:07 review First questions
May 25, 2023 at 19:16
S May 25, 2023 at 19:07 history asked Александр Худяков CC BY-SA 4.0