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Let $A$ be a $C^*$-algebra which is not $W^*$ with norm $\|\cdot\|.$ Is it possible that there exists an equivalent norm $\|\cdot\|'$ on $A$ such that $(A,\|\cdot\|')$ has an isometric predual, thus making it a $W^*$-algebra? That is, can a $C^*$-algebra which is not $W^*$ be linearly homeomorphic to a $W^*$-algebra?

We know that every quasi-reflexive Banach space can be renormed to be a dual space, but a $C^*$-algebra which is not $W^*$ is never quasi-reflexive, so the answer is almost certainly no in most cases (likely in all cases). But is there some property of $C^*$'s which forbids this in general?

Miles Gould
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  • Are you asking that $A$ with the new norm remains a C-algebra? In that case no, bc C-norms are unique. I want to say the answer is no even if you don't ask that it remains a C*-algebra bc it seems is $X$ and $Y$ are linerly isomorphic Banach spaces and $X$ admits a predual I should be able to cook up a predual for $Y$ as well, although this confuses me as it seems to crash with the second part of your question. Maybe I'm misunderstanding something. – ham_ham01 Feb 21 '25 at 19:32
  • @ham_ham01 Your argument works. However, two linearly homeomorphic Banach spaces can have different numbers of isometric preduals. – Miles Gould Feb 21 '25 at 20:30
  • Ham_ham01 is right: you say "thus making it a W-algebra". This sounds like you're asking that the new norm is in particular a complete C-norm, which is impossible. – Just dropped in Feb 21 '25 at 22:52
  • If your question is whether a $C^\ast$-algebra can be linearly homeomorphic to a $W^\ast$-algebra but not $W^\ast$ itself, that is possible (of course, as others already noted, the linear homeomorphism in question cannot be a $\ast$-isomorphism). The only examples I know are quite complicated, though. For example, let $X$ be the disjoint union of continuum many copies of $[0,1]$ and $B(X)$ be the algebra of bounded Borel functions on $X$ modulo the ideal of functions vanishing outside a meager set. Then $B(X)$ is linearly homeomorphic to $\ell^\infty(\mathfrak{c})$ but not $W^\ast$ itself. – David Gao Feb 21 '25 at 23:07
  • @DavidGao Is the mapping explicit or only guaranteed to exist? I’m curious how much it damages the multiplication/involution. – Miles Gould Feb 21 '25 at 23:39
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    @MilesGould It’s not explicit, but you can ensure the mapping preserves the involution since the real versions of these spaces are also linearly homeomorphic. The multiplication is a different matter. I don’t think you can preserve anything about that. – David Gao Feb 22 '25 at 00:06
  • @DavidGao I suppose it preserves certain commutation relations, and that’s about it. – Miles Gould Feb 22 '25 at 15:35
  • @MilesGould True, but that’s more an accidental side effect of using commutative $C^\ast$-algebras. $B(X)$ as above is also linearly homeomorphic to $\mathbb{B}(\ell^2(\mathfrak{c}))$, for example, and in that case commutation relations are not preserved either. – David Gao Feb 22 '25 at 20:03
  • @DavidGao What about if we don’t demand the codomain to be $W^*$, instead just a dual Banach space? – Miles Gould Feb 25 '25 at 15:13
  • @MilesGould I’m not sure what you meant? $W^\ast$ algebras are all dual Banach spaces, as you know. Do you mean whether there are simpler examples if we don’t ask the codomain to be $W^\ast$? If that’s the question, I don’t know of any. There aren’t that many pairs of “natural” Banach spaces that are linearly homeomorphic but not isometric. The only general technique for this that I know only applies to certain continuous function spaces, which are already $C^\ast$ and so will be $W^\ast$ as long as they have preduals. – David Gao Feb 25 '25 at 22:33

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