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Let $H$ be an infinite dimensional separable Hilbert space. Let $V$ be a finite dimensional subspace of $H$. Put $$A=\{T\in B(H)\mid T(V)\subseteq V\}.$$

So $A$ is a Banach algebra.

Can we equip $A$ with an involution $*$ such that we get a $C^*$ algebra structure on $A$?

What are K-groups of $A$ as a Banach algebra?

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    $\begingroup$ @MarkSapir Then $A= B(H)$ which is an obvious $C^*$-algebra? $\endgroup$
    – user160032
    Commented Sep 12, 2021 at 22:38
  • $\begingroup$ @MarkSapir I do not care this case since the answer is obvious. $\endgroup$
    – Ali Taghavi
    Commented Sep 12, 2021 at 22:38
  • $\begingroup$ @MarkSapir But if we allow H to be a finite dim space the answer to the first question is negative for ex" dim H=2 and dim V=1 . We get the three dimensional algebra isomorphic to upper triangle matrices. It can not become a C^* algebra. $\endgroup$
    – Ali Taghavi
    Commented Sep 12, 2021 at 22:41
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    $\begingroup$ Whenever nontrivial, your $A$ is not semi-simple and hence cannot be a C*-algebra. $\endgroup$
    – Narutaka OZAWA
    Commented Sep 12, 2021 at 23:08
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    $\begingroup$ @NarutakaOZAWA I was aware of this fact for commutative Banach algebra(May be a corollary ir an exercize in Rudin Book) but what is the statement for noncommutative case? BTW what can be said about k groups of this algebra) $\endgroup$
    – Ali Taghavi
    Commented Sep 13, 2021 at 7:34

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