Suppose that $A$ is a separable, simple, non-unital C*-algebra. Let $\varphi$ be an approximately inner automorphism on $A\otimes\cal K$, meaning that there exists a sequence of unitaries $v_n$ in the multiplier algebra $\mathcal{M}(A\otimes\mathcal{K})$ with $v_n x v_n^* \stackrel{n\to\infty}{\longrightarrow} \varphi(x)$ for all $x\in A\otimes\cal K$. We also denote by $\varphi$ the induced automorphism on the multiplier algebra.

Let $e_{11}\in\cal K$ denote the obvious rank one projection.

My question: Is $1\otimes e_{11}$ Murray-von-Neumann equivalent to $\varphi(1\otimes e_{11})$ in the multiplier algebra $\mathcal{M}(A\otimes\mathcal{K})$?

It is obvious that this is true if $A$ is unital, but unclear (to me) otherwise. I don't know whether simplicity of $A$ is a red herring here, but that is the case that I am interested in for now.

  • $\begingroup$ What is $1\otimes e_{11} $ if $A $ is not unital? $\endgroup$
    – vap
    Apr 19, 2017 at 18:11
  • $\begingroup$ It is the multiplier given by $(1\otimes e_{11})(a\otimes e_{kl}) = \delta_{1,k}\cdot a\otimes e_{1l}$ for all $a\in A$ and $k,l\in\mathbb{N}$. (Analogous formula from the right) $\endgroup$ Apr 20, 2017 at 9:16
  • $\begingroup$ Thanks. So, for $n$ large and $x $ a projection, $v_nxv_n^*$ is close to $\varphi (x) $ and therefore unitarily equivalent to it. Then $v_nxv_n^*$ is equivalent to $x $. Is there a problem with this argument? $\endgroup$
    – vap
    Apr 20, 2017 at 14:26
  • $\begingroup$ Yes. The problem is that, as $A$ is non-unital, the element $1\otimes e_{11}$ is a multiplier not in $A\otimes\cal K$, so the limit $\varphi(1\otimes e_{11}) = \lim_{n\to\infty} v_n (1\otimes e_{11}) v_n^*$ is in the strict topology, but a priori not in norm. $\endgroup$ Apr 21, 2017 at 10:36
  • $\begingroup$ This is true if $A$ has stable rank one (no simplicity). Is that interesting already? Also true if $A$ is simple purely infinite (since then it would be stable by Zhang's dichotomy). $\endgroup$ Apr 25, 2017 at 14:46

1 Answer 1


If $A$ has stable rank one then this can be deduced from the following result: If $B$ has stable rank one and $a,b\in B_+$ are Cuntz equivalent then the right ideals $\overline{aB}$ and $\overline{bB}$ are isomorphic as Hilbert modules. This is Theorem 3 of

Coward, Kristofer T.; Elliott, George A.; Ivanescu, Cristian. The Cuntz semigroup as an invariant for $C^*$-algebras. J. Reine Angew. Math. 623 (2008), 161--193.

See also Proposition 1 of

Ciuperca, Alin; Elliott, George A.; Santiago, Luis. On inductive limits of type-I $C^*$-algebras with one-dimensional spectrum. Int. Math. Res. Not. IMRN 2011, no. 11, 2577--2615.

It's a little awkward to translate this into an answer to the question. Let me try.

First, this: Let $p,q\in M(B)$ be multiplier projections such that $pB\cong qB$ as Hilbert modules. Then $p$ is Murray-von Neumann equivalent to $q$. Proof: Let $v\colon pB\to qB$ be an isomorphism. Extend $v$ to $B$ setting it to 0 on $(1-p)B$. Then $v\in M(B)$ (regarded as the adjointable operators on $B$) and $v^*v=p$, $vv^*=q$.

Now let $B$ be separable and of stable rank one. Let $\phi$ be an approximately inner automorphism extended to $M(B)$. Let us show that any projection $p\in M(B)$ is Murray-von Neumann equivalent to $\phi(p)$. It suffices to show that $pB\cong \phi(p)B$. Let $c\in B$ be a strictly positive element of $pBp$ (exists since $B$ is separable). Then $pB=\overline{cB}$. Since $c^{1/n}\uparrow p$ strictly and $\phi$ is strictly continuous, we also have $\phi(p)B=\overline{\phi(c)B}$. But $c$ and $\phi(c)$ are approximately unitarily equivalent, so they are Cuntz equivalent. We can use the theorem recalled above.

To answer the question, we take $B=A\otimes \mathcal K$ and $p=1\otimes e_{11}$. Notice that to answer the question we only need $A$ $\sigma$-unital rather than separable.

Remark: There exist examples of positive elements $a,b$ in an $A$ (of sr=2) that are approximately unitarily equivalent and yet $\overline{aA}$ and $\overline{bA}$ are not isomorphic as Hilbert modules. However, (1) $A$ is non-simple and they way the examples work they can't be tweaked to get this, (2) The unitary conjugations moving $a$ to $b$ don't seem to converge pointwise to an automorphism. So there is a chance that the question has a positive answer without sr1.

  • $\begingroup$ Very nice, thanks! Since this is already very useful to me, I accept this as an answer. $\endgroup$ Apr 27, 2017 at 14:57

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