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Let $\mathcal{A}$ be an injective $C^*$-algebra irreducibly acting on a Hilbert space $\mathcal{H}$, and let $\phi$ be a completely positive idempotent from $\mathbb{B}(\mathcal{H})$ onto $\mathcal{A}$, and let $u\in\mathbb{B}(\mathcal{H})$ be a unitary.

Question: Can we conclude that $\phi(u)\ne0$? If necessary, we can assume that $u$ normalizes $\mathcal{A}$, that is, $u\mathcal{A}u^*=\mathcal{A}$.

The point is the irreducibility, since without it $\phi(u)$ can easily be $0$. The question appears to be easy, but I have been struggling for months, and I have already asked five experts.

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No. Take any irreducibly represented simple and injective C*-algebra $A\subset B(H)$ and an outer automorphism $\theta$ of period $2$ (which is easy to find when $A$ is the hyperfinite II_1 factor).

Case 1: The representation is covariant, i.e., $\exists u\in{\mathcal U}(H)$ such that $uau^*=\theta(a)$ for $a\in A$. We may assume $u^2=1$ by irreducibility. Since $\theta$ is outer, $A\rtimes_\theta({\mathbb Z}/2{\mathbb Z})$ is simple by Kishimoto's theorem. Hence, $\mathrm{C}^*(A,u)\cong A\rtimes_\theta({\mathbb Z}/2{\mathbb Z})$ and there is a conditional expectation $\phi\colon\mathrm{C}^*(A,u)\ni a+bu\mapsto a\in A$, $a,b\in A$. The map $\phi$ extends on $B(H)$.

Case 2: The representation is not covariant, in which case there is no nonzero $x\in B(H)$ such that $xa=\theta(a)x$ for all $a\in A$, by Schur's lemma. Consider $A_1:=A\rtimes_\theta({\mathbb Z}/2{\mathbb Z})=\mathrm{C}^*(A,v)$ represented on $H\oplus H$, where $a\in A$ acts by $a\oplus\theta(a)$ and the switching unitary operator $v(\xi\oplus\eta)=\eta\oplus\xi$ implements $\theta$. Now, $A_1$ is injective, irreducibly represented on $H\oplus H$, and the dual automorphism $\hat{\theta}\colon a+bv\mapsto a-bv$ is outer and implemented by the unitary operator $u:=1\oplus-1$ on $H\oplus H$. Hence it reduces to Case 1.

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  • $\begingroup$ In Case 2, $A_1$ is ``irreducibly'' represented on $H\oplus H$, but $A$ is not. So how is this case related to my question? $\endgroup$
    – Masayoshi Kaneda
    Commented May 14, 2021 at 15:05
  • $\begingroup$ @Masayoshi Kaneda: It is $A_1$ that serves as a counterexample. $\endgroup$
    – Narutaka OZAWA
    Commented May 17, 2021 at 0:28
  • $\begingroup$ Yeah, you are right. We do not know a priori that the representation is covariant. I was confused because in my context, which I did not present here, an injective $C^*$-algebra $A$ (this is different from the $\mathcal{A}$ in my question) is ``fixed,'' and for my purpose it suffices to have at least one irreducible representation $\mathcal{A}$ (this is same as $\mathcal{A}$ in my question) of $A$ that answers my question positively. $\endgroup$
    – Masayoshi Kaneda
    Commented May 17, 2021 at 10:09

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