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I heard this from Haskell Rosenthal many years ago.

If V is a complex vector space, say the opposite of V is the complex vector space with the same elements, the same operations except switch scalar multiplication to scalar multiplication by the complex conjugate scalar. Of course this definition applies in particular to a complex Banach space. Question: Is every complex Banach space isomorphic to its opposite? (An isomorphism is a complex-linear homeomorphism.)

[ prompted by question Silly question about opposite groups ]

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Does this paper of Kalton do the trick? (disclaimer: I haven't read through the details)

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  • $\begingroup$ It looks like this is the answer. "no". And Kalton says that the first example is due to Bourgain, Proc. Amer. Math. Soc. 96 (1986) 221--226 $\endgroup$
    – Gerald Edgar
    Commented Oct 27, 2009 at 12:00

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