Let $A$ be a C*-algebra. Suppose that every cyclic representation of $A$ is finite dimensional.
Q. Is $A$ finite dimensional?
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5$\begingroup$ Every separable C*-algebra has a faithful state, hence a faithful cyclic representation. $\endgroup$– RuyJun 13, 2018 at 13:58
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$\begingroup$ The general case then follows: let $B\subseteq A$ be a C$^*$-subalgebra, and let $\mu$ be a state on $B$. A Hahn-Banach extension of $\mu$ is a state on $A$, hence has finite-dimensional GNS space; hence also the GNS space for $\mu$ on $B$ is finite-dimensional. Conclude that all separable subalgebras of $A$ are finite dimensional; hence $A$ is too. $\endgroup$– Matthew DawsJun 13, 2018 at 19:45
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$\begingroup$ I was wondering if there was a Banach algebra variant: e.g. if $A$ has a bai and every cyclic (left) submodule of $A^*$ was finite dimensional, then $A$ is finite dimensional?? $\endgroup$– Matthew DawsJun 13, 2018 at 19:51
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$\begingroup$ Nice argument @MatthewDaws, would you like to put that in the form of an answer or would you rather have me do it? $\endgroup$– RuyJun 14, 2018 at 0:12
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$\begingroup$ Ruy: You should write the formal answer: you had the original idea! $\endgroup$– Matthew DawsJun 14, 2018 at 9:23
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1 Answer
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The answer is yes. To prove it, assume first that $A$ is separable. In this case, by Theorem 7.10 in John B. Conway's book "A course in Operator Theory", we may choose a faithful representation $\pi$ of $A$ on a separable Hilbert space $H$. Choose an orthonormal basis $\{e_n\}_{n\in{\mathbb N}}$ for $H$ and define $$ \varphi(a) = \sum_{n\in{\mathbb N}}{1\over 2^n}\langle\pi(a)e_n,e_n\rangle, \quad \forall a\in A, $$ so that $\varphi$ is a state on $A$. Observe that $\varphi$ is faithful because if $\varphi(a^*a)=0$, then $\pi(a)e_n=0$, for every $n$, which implies that $\pi(a)=0$, and hence also that $a=0$, by virtue of the faithfulness of $\pi$.
Notice that we have just proven that every separable C*-algebra has a faithful state.
It is then an easy exercise to prove that the GNS representation of $A$ given by $\varphi$ is faithfull. It is obviously also cyclic, hence finite dimensional by hypothesis, so $A$ is finite dimensional.
The general case follows from Matthew Daws' reduction to the separable case: Let us prove that every separable closed *-subalgebra $B\subseteq A$ satisfies the condition in the question, that is, every cyclic representation of $B$ is finite dimensional. Indeed, given a cyclic representation $\pi$ of $B$, let $\xi$ be a cyclic vector with unit norm, and consider the associated state $$ \varphi(b) = \langle\pi(b)\xi,\xi\rangle, \quad \forall b\in B. $$ By Corollary 7.6 in Conway's book, $\varphi$ admits an extension to a state $\psi$ on $A$. The GNS representation of $A$ given by $\psi$, which I am going to denote by $\rho$, is then finite dimensional by hypothesis. If $\eta$ is the canonical cyclic vector for $\rho$, then $$ K:=\overline{\hbox{span}}\{\rho(b)\eta:b\in B\} $$ is an invariant subspace for the restriction $\rho|_B$, and if one further restricts $\rho|_B$ to a representation of $B$ acting on $K$, which I am going to cal $\tau$, then $\tau$ is cyclic, with cyclic vector $P(\eta)$, where $P$ is the orthogonal projection from the space of $\rho$ onto $K$. The state on $B$ given by $$ \omega(b) = \langle\tau(b)P(\eta),P(\eta)\rangle, \quad \forall b\in B $$ is then equal to $\varphi$ (to see this one should first prove that $\rho(b)P(\eta)=P\rho(b)\eta = \rho(b)\eta\, $), and then by the uniquenes of the GNS construction (Theorem 7.7 in Conway's book), one has that $\tau$ is equivalent to $\pi$.
Since $K$ is a subspace of the space of $\rho$, it is finite dimensional and hence $\pi$ is also a finite dimensional representation.
This completes the proof that every cyclic representation of every separable closed *-subalgebra of $A$ is finite dimensional.
By the first part of the argument we have that any separable closed *-subalgebra of $A$ is finite dimensional, so it follows easily that $A$ is finite dimensional.
