Edit: According to the valuable comment of Yemon Choi I revise the question by replacing "faithful" with "irreducible".

We say that a $C^*$ algebra $A$ satisfies the invariant subspace property if there exist an irreducible representation $\phi: A \to B(H)$, for some Hilbert space $H$, such that $\forall a \in A,\; \phi(a)$ has a nontrivial closed invariant subspace of $H$.

What is an example of a simple $C^*$ algebra with this property not isomorphic to the matrix algebra or the algebra of compact operators?

In particular, does $C^*_{red}(F_2)$ satisfy this property?

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    $\begingroup$ The union of an increasing sequence of finite dimensional matrix algebras? $\endgroup$
    – user6976
    Mar 31, 2018 at 1:17
  • $\begingroup$ So every simple AF algebra works? $\endgroup$ Mar 31, 2018 at 1:26
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    $\begingroup$ I do not know, I am not a specialist in $C^*$ algebras. I just guessed. $\endgroup$
    – user6976
    Mar 31, 2018 at 1:28
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    $\begingroup$ I think youh need to adjust the question, because as it is stated, can't you just do the following? Take any faithful representation $\theta: A\to B(H)$ and then just take $\phi : A\to B(H\oplus H)$, $\phi(a) = {\rm diag}( \theta(a), \theta(a))$. $\endgroup$
    – Yemon Choi
    Mar 31, 2018 at 2:38
  • $\begingroup$ @YemonChoi Thank you. I replaced "faithful" by "Irreducible". $\endgroup$ Mar 31, 2018 at 7:26

1 Answer 1


Any unital C*-algebra has this property. This follows from these two facts:

Fix $a\in A$.

(1) If $a-\lambda\cdot 1$ is not left invertible then there is an irreducible representation $\pi$ such that $\lambda$ is an eigenvalue of $\pi(a)$.

(2) The set of $\lambda$ such that $a-\lambda \cdot 1$ is not left invertible contains the boundary of the spectrum of $a$. In particular, this set is nonempty.

Proof of (1): Assume first that $a$ is not left invertible, so that $\overline{Aa}$ is a proper left ideal in $A$. Choose a maximal left ideal $L$ containing $\overline{Aa}$. Then there exists a pure state $\phi$ such that $$ L=\{x\in A:\phi(x^*x)=0\}. $$ See Theorem 2.9.5 in Dixmier's "C*-algebras". Now do the GNS construction with this state. We get that $\pi_{\phi}(a)\xi_\phi=0$ (because $a\cdot 1\in L$). So $\pi_\phi(a)$ has kernel. Now, in general, if $a-\lambda\cdot 1$ is not left invertible then we apply the same argument to $a-\lambda \cdot 1$.

Notice that, conversely, if $\lambda$ in the spectrum of $a$ can be realized as an eigenvalue in some representation, then $a-\lambda \cdot 1$ cannot be left invertible.

Proof of (2): (This is well known.) Suppose that $a$ is left invertible and in the closure of the set of invertible elements. Say $ba=1$ and $c_n\to a$, with $c_n$ invertible. Then $bc_n\to 1$. So $bc_n$ is invertible for large $n$. Since $c_n$ is invertible, we get that $b$ is invertible, whence $a$ is invertible.

Somewhat digressing into the question of realizing the full spectrum as eigenvalues:

(1) If $A$ is finite (i.e., the unit is a finite projection) then left invertible implies invertible so the full spectrum can be realized as eigenvalues. (Pointed out by Yemon in the comments.)

(2) If $a$ is normal, then again left invertible implies invertible, so the same works. This case is a theorem in Pedersen's book "C*-algebras and their automorhism groups".

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    $\begingroup$ I just realized hat if $a-\lambda$ is left invertible then $\lambda$ can't possibly be realized as an eigenvalue in some representation. So realizing the "left spectrum" as eigenvalues is best possible. $\endgroup$ Apr 26, 2018 at 21:08
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    $\begingroup$ In your last paragraph, you just need $A$ to be (Dedekind) finite, right? $\endgroup$
    – Yemon Choi
    Apr 26, 2018 at 23:39
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    $\begingroup$ Oh yes, finite is what's needed. I'll update the answer with this and a couple more remarks when I get the chance. $\endgroup$ Apr 26, 2018 at 23:55
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    $\begingroup$ @AliTaghavi: I've expanded it. $\endgroup$ Apr 27, 2018 at 14:10
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    $\begingroup$ As far as I understand, this constructs a particular irreducible representation for every fixed element $a\in A$; however, the OP asks for one irreducible representation which would work for all elements simultaneously. Am I missing an obvious trick to achieve that, given the construction? $\endgroup$ Dec 28, 2019 at 8:52

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