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Let $X$ be a reflexive Banach space, and let $B(X)$ denote the set of all bounded linear operators $X\to X$. Does there exist a norm-closed subalgebra $A\subseteq B(X)$ with the following properties?

  1. A is unital.
  2. A is reflexive as a Banach space.
  3. Every closed maximal left ideal has infinite codimension.

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PS: The following are equivalent for $A$

  • Every closed maximal left ideal of $A$ has infinite codimension.
  • Every primitive ideal of $A$ has infinite codimension.
  • Every irreducible representation of $A$ is infinite dimensional.
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    $\begingroup$ Possibly relevant: mathoverflow.net/questions/299311/reflexive-operator-algebra $\endgroup$
    – Tomasz Kania
    Commented May 6, 2021 at 12:14
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    $\begingroup$ I guess the algebra of Hilbert-Schmidt operators on $l^2$ satisfies 2 and 3, but of course it isn't unital. $\endgroup$
    – Nik Weaver
    Commented May 6, 2021 at 15:36
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    $\begingroup$ @NikWeaver $\ell^p$ ($1<p<\infty$) with pointwise operations are reflexive Banach algebras. One may attach a unit without losing reflexivity. $\endgroup$
    – Onur Oktay
    Commented May 6, 2021 at 18:37
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    $\begingroup$ @NikWeaver, when you talk about Hilbert-Schmidt operators, it is not quite an example for 2 and 3 since most likely Onur wants a closed subalgebra of $B(X)$. $\endgroup$
    – Tomasz Kania
    Commented May 6, 2021 at 19:14
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    $\begingroup$ Since this now has a bounty, can you please edit your question to clarify whether A is supposed to be a norm-closed subalgebra of B(X)? This ambiguity seems to have created some confusion in comments. $\endgroup$
    – Yemon Choi
    Commented Jun 6, 2021 at 17:01

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