Can the Banach algebra structure on $B(E)$ be (almost) retrieved from its Banach space structure?

Question

This is basically just out of curiosity. Also, since my research area is in von Neumann algebras and my knowledge of general Banach algebras as well as general Banach spaces is somewhat limited, I apologize in advance if this question has a well-known answer that I’m simply unaware of.

A result of Kadison from 1951 shows, among other things, that if $H, K$ are Hilbert spaces and $\phi: B(H) \to B(K)$ is an isometric isomorphism as Banach spaces, then $\phi$ is actually a $\ast$-isomorphism or $\ast$-anti-isomorphism (as von Neumann algebras), followed by multiplication on the left by a fixed unitary in $B(K)$. The proof is highly dependent on the $C^\ast$-structure. I’m wondering, however, if something like this is still true without assuming Hilbert spaces. Specifically,

1. Is it true that, given two Banach spaces $E$, $F$ and an isometric isomorphism $\phi: B(E) \to B(F)$ as Banach spaces, $\phi$ is necessarily an isometric isomorphism or isometric anti-isomorphism as Banach algebras, followed by multiplication on the left by a fixed surjective isometry in $B(F)$?
2. If 1 is not true in general, is it at least true that, if $B(E)$ and $B(F)$ are isometrically isomorphic as Banach spaces, then they are (isometrically) isomorphic or (isometrically) anti-isomorphic as Banach algebras?
3. If 2 is not true in general, is there at least some natural class of Banach spaces (more general than just Hilbert spaces, of course) for which a result like 2 is true?
4. Kadison’s result actually also showed that if any two $C^\ast$-algebras are isometrically isomorphic as Banach spaces, then they are isomorphic as Jordan algebras (i.e., there is a linear isomorphism preserving the Jordan product $a \circ b = \frac{ab + ba}{2}$). Does something like this hold for more general classes of Banach algebras? I’m aware that this cannot be true for all Banach algebras — $M^{2 \times 2}(\mathbb{C})$ with the Frobenius norm has two Banach algebra structures — the usual matrix multiplication and the Hadamard (i.e., entrywise) multiplication, whose associated Jordan algebra structures are different. But is there some natural class of Banach algebras, more general than $C^\ast$-algebras, for which this conclusion holds?

Answer 1

This is by no means a complete answer, but a series of bits that would be too long for the comment space.

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By an old result of Eidelheit, if $\phi:B(E)\to B(F)$ is a Banach algebra isomorphism onto $B(F)$ (resp. onto a subalgebra), then there is a Banach space isomorphism $u:E\to F$ (resp. $u:E\to F_0\subset F$) such that $\phi(x)u = ux$. Moreover, if $\phi$ is isometric, then so is $u$ [1]. Thus, if (2) holds, then we necessarily must have that either (in the isomorphic case) $E$ is isomorphic to $F$, or (in the anti-isomorphic case) $E$ is isomorphic to a subspace of $F^*$ & $F$ is isomorphic to a subspace of $E^*$.

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For any surjective isometry $\beta:B(\ell^2)\to B(\ell^2)$, let $\alpha = \beta|_{K(\ell^2)}$ be its restriction, then $\beta = \alpha^{**}$. Particularly, $\beta$ is weak$^*$ continuous. So, for (3), this suggests to consider the class of Banach spaces $X$ for which the compacts $K(X)$ is an M-ideal in its bidual $K(X)^{**}$. $K(\ell^2)$ is also an M-ideal in $B(\ell^2)$.

If $K(X)$ is an M-ideal in $K(X)^{**}$, we have the following: (a) Since $X^*$ is complemented in $K(X)$, then $X^{*}$ is an M-ideal in $X^{***}$, which forces $X$ to be reflexive. Particularly, $B(X) = K(X)^{**}$. (b) If $\beta:B(X)\to B(X)$ is an isometry and $\alpha = \beta|_{K(X)}$ is its restriction, then $\beta = \alpha^{**}$. (c) $K(X)$ and all its closed subspaces have Pelczynski's property (V). Analogously, every $C^*$-algebra has property (V). Classical reference for M-ideals is [2].

Suppose $X$ is reflexive, strictly convex, has the approximation property (AP), and $K(X)$ is an M-ideal in $B(X)$. If $\beta:B(X)\to B(X)$ is a surjective isometry, then there exist isometries $u,v:X\to X$ such that $\beta(x) = uxv$ [3]. In a nutshell, AP and the M-ideal condition together let us reach to the isometries of $B(X)$ from the isometries of the finite dimensional operators on $X$.

Answer 2

It seems that not much is known. I have found a description of isometries on $B(c_0)$ (and some other relevant results) here. A surjective isometry $T:B(c_0)\to B(c_0)$ is of the form $T(a)=UaV$ where $U,V:c_0\to c_0$ are surjective isometries (so we could say that they are of ''standard form''). So we have $T(a)=UV(V^{-1}aV)$ and $a\mapsto V^{-1}a V$ is an isomorphism. Thus in this case (1) is true.

As for (4), I do not know about Banach algebras other than nonselfadjoint operator algebras or some of the form $C(K,E)$ (for certain Banach spaces $E$) where a nice Banach-Stone theorem holds. However, if we allow something that is not an (associative) algebra, there is a certain class of Banach spaces more general C$^*$-algebras (the JB*-triples) endowed with some algebraic structure (a ''triple product'') where the Banach space structure determines the algebraic structure and conversely (a bijective linear map between JB$^*$-triples is an isometry if and only if it preserves the "triple product"). In the case of C$^*$-algebras, every surjective isometry is of the form $uJ(.)$ (where $u$ is a unitary and $J$ a Jordan $^*$-isomorphism). Such a mapping preserves the triple product $\{a,b,c\}=\frac{1}{2}(ab^*c+cb^*a)$.