Simple, Z-stable, projectionless C*-algebras

Von | September 22, 2020

Do simple, \mathcal{Z}-stable, stably projectionless C*-algebras have stable rank one?

Definitions: A C*-algebra A is said to be \mathcal{Z}-stable if it tensorially absorbs the Jiang-Su algebra \mathcal{Z}, that is, A\cong\mathcal{Z}\otimes A. Further, a simple C*-algebra is projectionless if it contains no nonzero projections, and it is stably projectionless if A\otimes\mathcal{K} is projectionless. (In the nonsimple case, one should require that no quotient of A contains a nonzero projection – see ​[1]​.) A unital C*-algebra A is said to have stable rank one if the invertible elements in A are dense. A nonunital C*-algebra A has stable rank one if its minimal unitization \widetilde{A} does.

Background/Motivation: Rørdam showed in Theorem 6.7 in ​[2]​ that every unital, simple, finite \mathcal{Z}-stable C*-algebra has stable rank one. Using this, one can show that every simple, finite, \mathcal{Z}-stable C*-algebra A that is not stably projectionless has stable rank one. Indeed, one may first reduce to the separable case (to show that a given element x_0+\lambda 1\in\widetilde{A} with x_0\in A is approximated by invertible elements, consider any separable, simple, \mathcal{Z}-stable sub-C*-algebra of A that is not stably projectionless and that contains x.) Let p\in A\otimes\mathcal{K} be a nonzero projection. By Brown’s stabilization theorem, we have A\otimes\mathcal{K}\cong (p(A\otimes\mathcal{K})p)\otimes\mathcal{K}. Then the hereditary sub-C*-algebra p(A\otimes\mathcal{K})p is separable, unital, simple, finite and \mathcal{Z}-stable (by Corollary 3.2 in ​[3]​) and therefore has stable rank one. Since stable rank one is invariant under stable isomorphism, it follows that A has stable rank one.

Thus, a simple, \mathcal{Z}-stable C*-algebra that is not stably projectionless has stable rank one if and only if it is (stably) finite. A simple, stably projectionless C*-algebra is automatically stably finite, and it is therefore natural to expect that every simple, \mathcal{Z}-stable, stably projectionless C*-algebra has stable rank one.

Let A be a simple, \mathcal{Z}-stable, stably projectionless C*-algebra. By Corollary 3.2 in ​[1]​, A almost has stable rank one, that is, every hereditary sub-C*-algebra B\subseteq A satisfies B\subseteq\overline{\mathrm{Gl}(\widetilde{B})}. In particular, every element in A can be approximated by invertible elements in \widetilde{A}. To show that A has stable rank one, one would need to show that every element in \widetilde{A} is approximated by invertibles. In Theorem 6.13 in ​[4]​ it is shown that A has stable rank at most two, that is, the tuples (x,y) in (\widetilde{A})^2 such that x^*x+y^*y is invertible are dense in (\widetilde{A})^2. One can also show:

Simple, \mathcal{Z}-stable, stably projectionless C*-algebras have general stable rank one.

Here, the general stable rank of a unital C*-algebra A, denoted \mathrm{gsr}(A), is the least integer n\geq 1 such that \mathrm{Gl}_m(A) acts transitively on \mathrm{Lg}_m(A) for all m\geq n. (We use \mathrm{Gl}_m(A) to denote the set of invertible elements in the matrix algebra M_m(A). Further, \mathrm{Lg}_m(A) denotes the set of m-tuples (a_1,\ldots,a_m)\in A^m that generate A as a left ideal, that is, such that a_1^*a_1+\ldots+a_m^*a_m is invertible.) For a nonunital C*-algebra \mathrm{gsr}(A):=\mathrm{gsr}(\widetilde{A}). In general, one has \mathrm{gsr}(A)\leq\mathrm{sr}(A)+1, as noted in Theorem 3.3. of the overview article ​[5]​. If A is unital, then the action of \mathrm{Gl}_m^0(A), the connected component of the unit in \mathrm{Gl}_m(A), on \mathrm{Lg}_m(A) has open orbits. It follows that the orbits of the action of \mathrm{Gl}_m(A) on \mathrm{Lg}_m(A) are also open (and hence also closed).

Let A be a simple, \mathcal{Z}-stable, stably projectionless C*-algebra. Let us prove that \mathrm{gsr}(A)=1. By Theorem 6.13 in ​​[4]​, we have \mathrm{sr}(A)\leq 2, and so \mathrm{gsr}(A)\leq 3. To verify that \mathrm{gsr}(A)\leq 2, we need to show that \mathrm{Gl}_2(\widetilde{A}) acts transitively on \mathrm{Lg}_2(\widetilde{A}). So let x=(\lambda_1+a_1,\lambda_2+a_2)\in\mathrm{Lg}_2(\widetilde{A}) with \lambda_1,\lambda_2\in\mathbb{C} and a_1,a_2\in A. First, we find u\in\mathrm{Gl}_2(\mathbb{C})\subseteq\mathrm{Gl}_2(\widetilde{A}) such that ux=(b_1,\kappa_2+b_2) with \kappa_2\in\mathbb{C} and b_1,b_2\in A. (If \lambda_1=0, we use the identity matrix; if \lambda_2=0, we use the flip matrix; and if \lambda_1\neq 0 and \lambda_2\neq 0 we use the upper-triangular matrix with diagonal entries 1 and upper right entry -\lambda_1/\lambda_2.) Given \varepsilon>0, we use that A\subseteq\overline{\mathrm{Gl}(\widetilde{A})} to approximate b_1 by some invertible \kappa_1+c\in\widetilde{A} such that \| x - u^{-1}(\kappa_1+c,\kappa_2+b_2) \|<\varepsilon. We note that (\kappa_1+c,\kappa_2+b_2) is in the orbit of (1,0), that is, there exists v\in\mathrm{Gl}_2(\widetilde{A}) such that (\kappa_1+c,\kappa_2+b_2)=v(1,0). Then \| x - u^{-1}v(1,0) \|<\varepsilon. Since the orbits of the action of \mathrm{Gl}_2(\widetilde{A}) on \mathrm{Lg}_2(\widetilde{A}) are closed, it follows that x=w(1,0) for some w\in\mathrm{Gl}_2(\widetilde{A}). Finally, since \widetilde{A} is finite, it follows that \mathrm{gsr}(A)=1.

We remark that \mathrm{sr}(A)=1 implies \mathrm{gsr}(A)=1, but not conversely in general.

  1. [1]
    L. Robert, Remarks on Z-stable projectionless C*-algebras, Glasgow Math. J. (2015) 273–277.
  2. [2]
    M. Rørdam, The stable and the real rank of  {\mathcal Z}-absorbing C*-algebras, Int. J. Math. (2004) 1065–1084.
  3. [3]
    A.S. Toms, W. Winter, Strongly self-absorbing C^{*}-algebras, Trans. Amer. Math. Soc. (2007) 3999–4029.
  4. [4]
    L. Robert, L. Santiago, A revised augmented Cuntz semigroup, ArXiv:1904.03690. (2019).
  5. [5]
    B. Nica, Homotopical stable ranks for Banach algebras, Journal of Functional Analysis. (2011) 803–830.

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