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It is known that a C*-algebra is finite-dimensional if (and only if) it is reflexive as a Banach space. What is known about the analog of this question for operator algebras? (Here, an operator algebra means a norm-closed subalgebra of the bounded, linear operators on a Hilbert space.) In particular:

Is an operator algebra finite-dimensional if it is reflexive as a Banach space?

Is an operator algebra amenable if it is reflexive as a Banach space?

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    $\begingroup$ For the first question,the answer is no : the "row" (= algebra of operators on $\ell_2$ whose image is contained in $\mathbf C e_1$) is isomorphic to a Hilbert space and hence is reflexive. You can make it unital if you wish. $\endgroup$ Commented May 3, 2018 at 12:13
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    $\begingroup$ For the second question the answer is also no; The 3-dimensional operator algebra of upper triangular matrices in $M_3(\mathbb{C})$ gives an example. The 2001 paper of Runde "Banach space properties forcing a reflexive, amenable Banach algebra to be trivial" and it's bibliography looks like it has lots of related information that you might be looking for. For example the first sentence of his paper recalls an open problem(as of 2001) that is similar to your first question. $\endgroup$ Commented May 3, 2018 at 14:46
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    $\begingroup$ Hi Hannes: following on from Mikael's answer, one source of cheap counterexamples to many questions is to take an arbitrary operator space X, embed it into B(H) in some way, and then consider the algebra $$\left[\begin{matrix} {\bf C} I & X \\ 0 & {\bf C} I \end{matrix} \right] $$ $\endgroup$
    – Yemon Choi
    Commented May 3, 2018 at 16:31
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    $\begingroup$ @CalebEckhardt Indeed the minimal unital counterexample is the 2-dimensional subalgebra of $M_2({\bf C})$ generated by the identity and the matrix unit $E_{12}$. $\endgroup$
    – Yemon Choi
    Commented May 3, 2018 at 16:34

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