Let $H$ be a separable Hilbert space and let $A$ be a non-unital $C^*$-subalgebra in $B(H)$ such that the second dual $A^{**} \equiv B(H).$ Does $A$ coincide with the ideal of compact operators $K(H)?$
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3$\begingroup$ What do you mean by coincide? Isomorphic or isometrically isomorphic? The commutative $C^*$-algebras $c$ of all convergent sequences and $c_0$ of all null sequences both have $\ell^\infty$ as the bidual. $\endgroup$– Jochen WengenrothCommented Aug 1 at 6:07
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2$\begingroup$ @JochenWengenroth The corresponding result is false for $K(H)$, since $(K(H) + \mathbb{C})^{\ast\ast}$ has a one-dimensional summand whereas $B(H)$ does not. $\endgroup$– David GaoCommented Aug 1 at 6:33
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3$\begingroup$ @JochenWengenroth Also, the OP is assuming $A$ is a $C^\ast$-algebra, and any isomorphism in the category of $C^\ast$-algebras is automatically isometric. $\endgroup$– David GaoCommented Aug 1 at 6:37
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3$\begingroup$ I don't understand what $A^{**} \equiv B(H)$ means. If $A$ is a separable $C^\ast$-algebra such that $A^{\ast \ast} \cong B(H)$ (isomorphic as $C^*$-algebras), then $A$ must be simple (since $B(H)$ is a factor) and type I (since $B(H)$ is a type I von Neumann algebra). The only simple separable type I $C^\ast$-algebras are $M_n(\mathbb C)$ and $K(H)$, so one must have $A \cong K(H)$. But $K(H)$ embeds (as a $C^*$-algebra) into $B(H)$ in many different ways. $\endgroup$– Jamie GabeCommented Aug 1 at 15:41
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2$\begingroup$ @JamieGabe This is even true if $A^{\ast\ast} \equiv B(H)$ means the two are isometrically isomorphic as Banach spaces, since a result of Kadison implies that $A^{\ast\ast}$ and $B(H)$ are isometrically isomorphic as Banach spaces iff they are isomorphic as von Neumann algebras. Also, separability assumption is not necessary, as $A^{\ast\ast} \cong B(H)$ (just isometrically as Banach spaces) implies $A^\ast \cong S_1(H)$ isometrically. As $S_1(H)$ is separable, $A$ must be separable. $\endgroup$– David GaoCommented Aug 1 at 16:48
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