# Stable rank one and corners of $C^\ast$-algebras

Thanks to a result of Herman and Vaserstein in [3], Rieffel's notion of stable rank [4] coincides with the Bass stable rank [1] for every $$C^\ast$$-algebra $$A$$: we denote it by $$\mathrm{sr}(A)$$ and we simply call it the stable rank of $$A$$.

The following fact yields a convenient characterization of the stable rank one case.

Lemma (folklore I suppose) A $$C^\ast$$-algebra has stable rank one if and only if the invertible elements are dense in $$\widetilde{A}$$, where $$\widetilde{A}$$ denotes $$A$$ if it is unital, and the unitization $$A\oplus\mathbb{C}1$$ of $$A$$ otherwise. For short, denoting the group of all invertible elements in $$\widetilde{A}$$ by $$G\left(\widetilde{A}\right)$$, we have: $$\mathrm{sr}(A)=1\quad \Leftrightarrow\quad\overline{G\left(\widetilde{A}\right)}=\widetilde{A}.$$

Now here are the facts I'm interested in. They deal with the stable rank of the corners $$pAp$$ of a given $$C^\ast$$-algebra $$A$$.

Proposition (Corollary V.3.1.18 in [2]) Let $$A$$ be a unital $$C^\ast$$-algebra and $$p$$ a full projection in $$A$$ (i.e. the closed ideal of $$A$$ generated by $$p$$ is all of $$A$$).

1. We have $$\mathrm{sr}(pAp)\ge\mathrm{sr}(A)$$.
2. If $$\mathrm{sr}(A)<\infty$$, then $$\mathrm{sr}(pAp)<\infty$$ and $$\mathrm{sr}(A)=1\quad \Leftrightarrow\quad\mathrm{sr}(pAp)=1.$$

What if we remove the assumption that $$p$$ is full?

• Question A: What are the known examples of $$C^\ast$$-algebras $$A$$ and projections $$p\in A\setminus\{0,1\}$$ such that $$\mathrm{sr}(A)=1<\mathrm{sr}(pAp)?$$
• Question B: What are the known examples of unital $$C^\ast$$-algebras $$A$$ and projections $$p\in A\setminus\{0,1\}$$ such that $$\mathrm{sr}(A)=1<\min\{\mathrm{sr}(pAp),\mathrm{sr}((1-p)A(1-p))\}?$$

[1] H. Bass, K-theory and stable algebra, Publications Mathématiques de l’IHÉS 22 (1964), pp. 5–60.

[2] B. Blackadar, Operator Algebras, Encyclopaedia of Mathematical Sciences 122 (2006), Springer, pp. 444-452.

[3] R. Herman and L. N. Vaserstein, The stable range of $$C^\ast$$-algebras, Inventiones Mathematicae 77 (1984) pp. 553-555.

[4] M. A. Rieffel, Dimension and stable rank in the K-theory of $$C^\ast$$-algebras, Proceedings of the London Mathematical Society 46 (1983), pp. 301–333.

• If you are able to view the remarks after the proof of Corollary V.3.18 then they mention explicit counterexamples to parts of that corollary for non-full projections: e.g. taking p to be a finite-rank projection in A=B(H); or taking p to be projection onto the first co-ordinate in ${\mathbb C} \oplus {\mathcal O}_2$. I think these already give counterexamples to your Question A, but maybe I have misread/misunderstood? Jan 19 at 17:03
• @YemonChoi I am asking for examples where A has stable rank one, which implies that A be (stably) finite. Now $B(H)$ and the Cuntz algebras are infinite. Jan 19 at 17:18
• Oh yes, sorry, I wasn't reading carefully. [I have also deleted another comment which arose from not thinking carefully] Jan 19 at 17:51

## 2 Answers

For separable $$C^\ast$$-algebras this phenomenon cannot happen, since any corner of a separable stable rank one $$C^\ast$$-algebra has stable rank one. More generally, this holds for $$C^\ast$$-algebras $$A$$ such that all its two-sided closed ideals are $$\sigma$$-unital.

In fact, suppose $$A$$ is separable and has stable rank one. Then all of its two-sided closed ideals have stable rank one [4, Theorem 4.4]. By Brown's stable isomorphism theorem (L. G. Brown, Stable isomorphism of hereditary subalgebras of C∗-algebras, Pacific J. Math 71 (1977), 335-348.), any corner $$B$$ of $$A$$ is stably isomorphic to the two-sided closed ideal in $$A$$ generated by $$B$$. As stable rank one is preserved by stable isomorphism [4, Theorem 3.6], any corner of $$A$$ has stable rank one.

• Hi Jamie: I had a very quick look at Brown's paper and the main results seem to be stated for full corners (stronger condition) but get stable isomorphism to A itself (stronger conclusion). Could you maybe add some clarification about how one gets from those result to the one stated in your answer? Jan 20 at 16:08
• Hi Yemon, I extended it a tiny bit, but could go into more details if you think that would help? Jan 20 at 18:48
• Oh that makes sense now. I think I see what's going on now, although of course if the OP wants more details they can let us know. Jan 20 at 19:32
• @JamieGabe Thank you for the reference. So you're saying that if $A$ is a separable $C^\ast$-algebra with stable rank one, then $pAp$ has stable rank one for every nonzero projection $p$ (without assuming that $p$ is full), right? And this leaves Questions A and B open, right? Jan 22 at 14:08
• Separability is not important here. We can use Proposition 4.1 of Blackadar's "The stable rank of full corners in C*-algebras" to get that if $sr(A)=1$ then $sr(pAp)=1$ for any projection $p$ in $A$. Mar 8 at 22:45

Here's a community wiki answer based on Leonel Robert's comment to Jamie Gabe's answer, to answer the question in general for arbitrary Banach algebras.

Given the lemma established below, the answer to questions A and B is the same: there can't exist such an algebra, as spotted by Jamie Gabe in the separable case.

Immitating the argument given by Bruce Blackadar to prove Proposition 4.1 in [1] (thanks to Leonel Robert for the reference), which itself mimicks Marc Rieffel's proof of Lemma 3.4 in [2], we obtain the following property in the more general setting of Banach algebras, without assuming that $$p$$ be full.

Lemma: Let $$A$$ be a Banach algebra with identity element, and let $$p$$ be an idempotent in $$A$$.

If the invertible elements are dense in $$A$$, then the invertible elements are dense in $$pAp$$.

Proof: Let $$x\in pAp$$ and let $$0<\eta<\frac{1}{\|p^\perp\|^2}.$$ We are going to work with the 2x2 Peirce decomposition of $$A=pAp\oplus pAp^\perp\oplus p^\perp Ap\oplus p^\perp Ap^\perp$$ with respect to $$p$$ and $$p^\perp:=1-p$$.

By assumption, we can find an invertible element $$y=\begin{bmatrix}pyp&pyp^\perp\\p^\perp yp&p^\perp yp^\perp\end{bmatrix}\in A$$ which approximates $$x+p^\perp=\begin{bmatrix}x&0\\0&p^\perp\end{bmatrix}$$ within $$\eta$$, i.e. $$\|z\|\le\eta$$ with $$z:=x+p^\perp-y$$, so that

1. $$\|x-pyp\|=\|pzp\|\le\eta\|p\|^2$$;
2. $$\|pyp^\perp\|=\|pzp^\perp\|\le\eta\|p\|\|p^\perp\|$$;
3. $$\|p^\perp yp\|=\|p^\perp zp\|\le\eta\|p\|\|p^\perp\|$$;
4. $$\|p^\perp yp^\perp-p^\perp\|=\|p^\perp zp^\perp\|\le\eta\|p^\perp\|^2$$.

By 4, we see that $$p^\perp yp^\perp$$ is invertible in $$p^\perp Ap^\perp$$ with inverse $$\delta$$ such that $$\|\delta\|=\left\|\sum_{n\ge 0}\left(p^\perp-p^\perp yp^\perp\right)^n\right\|\le\frac{1}{1-\eta\|p^\perp\|^2}.$$ Using the fact that $$\delta\cdot p^\perp yp^\perp=p^\perp yp^\perp\cdot\delta=p^\perp$$ and a routine 2x2 matrix computation, we obtain the following formula: $$\begin{bmatrix}p&-pyp^\perp\delta\\0&p^\perp\end{bmatrix}y \begin{bmatrix}p&0\\-\delta p^\perp yp&p^\perp\end{bmatrix} =\begin{bmatrix}pyp-(pyp^\perp)\delta(p^\perp yp)&0\\0&*\end{bmatrix}.$$ Since $$pyp^\perp\delta$$ and $$\delta p^\perp yp$$ are nilpotent, the three elements on the left are invertible in $$A$$, and it follows that $$\alpha:=pyp-(pyp^\perp)\delta(p^\perp yp)$$ is invertible in $$pAp$$. It only remains to check that the above estimates yield \begin{align*} \|x-\alpha\|&\le\|x-pyp\|+(pyp^\perp)\delta(p^\perp yp)\\ &\le\eta\|p\|^2+\frac{\eta^2\|p\|^2\|p^\perp\|^2}{1-\eta\|p^\perp\|^2}\\ &=\frac{\eta\|p\|^2}{1-\eta\|p^\perp\|^2}. \end{align*} Since this upper bound tends to $$0$$ when $$\eta\rightarrow 0$$, the result follows.

QED

[1] Bruce Blackadar, The stable rank of full corners in $$C^\ast$$-algebras, Proceedings of the American Mathematical Society 132 (2004), pp. 2945–2950.

[2] Marc A. Rieffel, Dimension and stable rank in the K-theory of $$C^\ast$$-algebras, Proceedings of the London Mathematical Society 46 (1983), pp. 301–333.