Non-amenable tight squeezes by Kirchberg algebras

Abstract

We give a framework to produce C$^*$-algebra inclusions with extreme properties. This gives the first constructive nuclear minimal ambient C$^*$-algebras. We further obtain a purely infinite analogue of Dadarlat’s modeling theorem on AF-algebras: Every Kirchberg algebra is rigidly and KK-equivalently sandwiched by non-nuclear C$^*$-algebras without intermediate C$^*$-algebras. Finally we reveal a novel property of Kirchberg algebras: They embed into arbitrarily wild C$^*$-algebras as rigid maximal C$^*$-subalgebras.

Appendix A. Tensor splitting theorem for non-unital simple C\(^*\)-algebras

Here we record a few necessary and useful technical lemmas on non-unital C$^*$-algebras. Although these results would be known for some experts, we do not know an appropriate reference. As a result of these lemmas, we obtain the tensor splitting theorem (cf. [19, 59, 61]) for non-unital simple C$^*$-algebras.

An element of a C$^*$-algebra A is said to be full if it generates A as a closed ideal of A.

Lemma A.1

Let A be a C$^*$-algebra. Let a be a full positive element of A. Then for any finite subset F of A and any $\epsilon >0$, there is a sequence $x_1, \ldots , x_n\in A$ satisfying

$$\begin{aligned} \left\| \sum _{i=1}^n x_i a x_i^*\right\| \le 1 \qquad \mathrm{~and~} \qquad \left\| \sum _{i=1}^n x_i a x_i^*b-b\right\| <\epsilon \quad \mathrm{~for~} b\in F. \end{aligned}$$

Proof

Observe that for any sequence $x_1, \ldots , x_n\in A$ and any $b\in F$, the C$^*$-norm condition implies

$$\begin{aligned} \left\| \sum _{i=1}^n x_i a x_i^*b-b\right\| \le \left\| \sum _{i=1}^n x_i a x_i^*c-c\right\| , \end{aligned}$$

where $c:= \left( \sum _{d\in F}d d^*\right) ^{1/2}$. Therefore we only need to show the statement when F is a singleton in $A_+$. By the fullness of a, we may further assume that the element b in F is of the form $\sum _{i=1}^n y_i a z_i$; $y_1, \ldots , y_n, z_1, \ldots , z_n \in A$. In this case, we have

$$\begin{aligned} b^2 = b^*b =\sum _{i, j=1}^n z_i^*a y_i^*y_j a z_j \le C \sum _{i=1}^n z_i^*a z_i, \end{aligned}$$

where $C:= \Vert a\Vert \Vert (y_i^*y_j)_{1\le i, j \le n}\Vert _{\mathbb {M}_n(A)}$. Put $w:= \left( \sum _{i=1}^n z_i^*a z_i \right) ^{1/2}$. Choose a sequence $(f_k)_{k=1}^\infty$ in $C_0(]0, \infty [)_+$ satisfying $t f_k(t)\le 1$ for all $k\in \mathbb {N}$ and all $t\in [0, \infty [$, and $\lim _{k \rightarrow \infty } tf_k(t)=1$ uniformly on compact subsets of $]0, \infty [$. Then, for each $k\in {\mathbb {N}}$, we have

$$\begin{aligned} \sum _{i=1}^n f_k(w)z_i^*a z_i f_k(w) = f_k(w)^2 w^2\le 1,\\ \left\| \sum _{i=1}^n f_k(w) z_i^*a z_i f_k(w) b-b\right\| \le \sqrt{C} \Vert ( w^2 f_k(w)^2-1)w\Vert . \end{aligned}$$

The last term tends to zero as $k\rightarrow \infty$. Therefore, for a sufficiently large $N\in {\mathbb {N}}$, the sequence $(f_N(w)z_i^*)_{i=1}^n$ satisfies the required conditions. $\square$

For simple C$^*$-algebras, one can strengthen Lemma A.1 as follows.

Lemma A.2

Let A be a simple C$^*$-algebra. Let $a\in A_+ {\setminus } \{0\}$. Then for any $b\in A_+$ and any $\epsilon >0$, there is a sequence $x_1, \ldots , x_n\in A$ satisfying

$$\begin{aligned} \left\| \sum _{i=1}^n x_i x_i^*\right\| \le \Vert a\Vert ^{-1}\Vert b\Vert ,\qquad \sum _{i=1}^n x_i a x_i^*\approx _\epsilon b. \end{aligned}$$

Proof

We may assume $\Vert a\Vert =\Vert b\Vert =1$. Take $f \in C([0, 1])_+$ satisfying $f(1)\ne 0$ and ${{\,\mathrm{supp}\,}}(f)\subset [1- \epsilon /2, 1]$. Then

$$\begin{aligned} f(a)\ne 0,\qquad \left( 1- \frac{\epsilon }{2}\right) f(a)^2 \le f(a) a f(a) \le f(a)^2. \end{aligned}$$

Applying Lemma A.1 to $f(a)^2$ and $F=\{b^{1/2}\}$, we obtain a sequence $y_1, \ldots , y_n \in A$ satisfying

$$\begin{aligned} \left\| \sum _{i=1}^n y_i f(a)^2 y_i^*\right\| \le 1,\qquad \left\| \sum _{i=1}^n y_i f(a)^2 y_i^*b^{\frac{1}{2}}-b^{\frac{1}{2}}\right\| < \frac{\epsilon }{2}. \end{aligned}$$

Set $x_i:=b^{1/2}y_i f(a)$ for $i=1, \ldots , n$. Then

$$\begin{aligned} \sum _{i=1}^n x_ix_i^*= \sum _{i=1}^n b^{\frac{1}{2}} y_i f(a)^2 y_i^*b^{\frac{1}{2}} \le 1. \end{aligned}$$

Straightforward estimations show that

$$\begin{aligned} \sum _{i=1}^n x_i a x_i^*\approx _{ \epsilon /2} \sum _{i=1}^n b^{\frac{1}{2}} y_i f(a)^2 y_i^*b^{\frac{1}{2}} \approx _{ \epsilon /2} b. \end{aligned}$$

Therefore $x_1, \ldots , x_n$ form the desired sequence. $\square$

As an application of Lemma A.2, one can remove the unital condition from the tensor splitting theorem [59, 61] (cf. [19]). For a C$^*$-subalgebra C of $A\otimes B$, we define the subset $\mathcal {S}_A(C)$ of B to be

$$\begin{aligned} \mathcal {S}_A(C):=\left\{ (\varphi \otimes \text{ id}_B)(c): \varphi \in A^*, c\in C\right\} . \end{aligned}$$

Theorem A.3

Let A be a simple C$^*$-algebra and B be a C$^*$-algebra. Let C be a C$^*$-subalgebra of $A\otimes B$ closed under multiplications by A. Then $\mathcal {S}_A(C)$ forms a C$^*$-subalgebra of C and satisfies $A \otimes \mathcal {S}_A(C) \subset C$. Thus, when A satisfies the strong operator approximation property [20] or when the inclusion $\mathcal {S}_A(C) \subset B$ admits a completely bounded projection, we have $C= A\otimes \mathcal {S}_A(C)$.

Proof

To show the first statement, it suffices to show the following claim. For any pure state $\varphi$ on A and any $a\in A_+$, $c\in C$, with $b := (\varphi \otimes \text{ id}_B)(c)$, we have $a\otimes b \in C$. Indeed the claim implies that, since the set of pure states on A spans a weak-$*$ dense subspace of $A^*$ and $A_+$ spans A, for any $a\in A$ and any $\psi \in A^*$ with $\psi (a)\ne 0$, the subspace $X:=(\psi \otimes \text{ id}_B)(C)$ of B satisfies $a \otimes X =(a\otimes B) \cap C$. This implies $X=\mathcal {S}_A(C)$, and proves the first statement. To show the claim, for any $\epsilon >0$, by the Akemann–Anderson–Pedersen excision theorem [1] (Theorem 1.4.10 in [9]), one can take $e\in A_+$ with $\Vert e\Vert =1$, $e c e \approx _{\epsilon } e^2\otimes b$. By Lemma A.2, there is a sequence $x_1, \ldots , x_n \in A$ satisfying $\sum _{i=1}^n x_ie c ex_i^*\approx _{\epsilon } a\otimes b$. The left term is contained in C by assumption. Thus $a\otimes b \in C$.

For the last statement, when A satisfies the strong operator approximation property, the claim follows from Theorem 12.4.4 in [9]. When we have a completely bounded projection $P:B \rightarrow \mathcal {S}_A(C)$, it is not hard to see that for any $\varphi \in A^*$, $(\varphi \otimes \text{ id}_B)((\text{ id}_A \otimes P)(c)-c)=0$ for all $c\in C$. This proves $(\text{ id}_A \otimes P)|_C=\text{ id}_C$ and thus $C\subset A \otimes \mathcal {S}_A(C)$. $\square$