# A non-nuclear C \* -algebra with the Weak Expectation Property and the Local Lifting Property

Gilles Pisier

► To cite this version:

Gilles Pisier. A non-nuclear C \* -algebra with the Weak Expectation Property and the Local Lifting Property. Inventiones Mathematicae, 2020, 222 (2), pp.513-544. 10.1007/s00222-020-00977-4. hal-05460809

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# A non-nuclear $C^*$-algebra with the Weak Expectation Property and the Local Lifting Property

by

Gilles Pisier*

Texas A&M University
College Station, TX 77843, U. S. A.

August 4, 2020

## Abstract

We construct the first example of a $C^*$-algebra A with the properties in the title. This gives a new example of non-nuclear A for which there is a unique $C^*$-norm on $A \otimes A^{op}$. This example is of particular interest in connection with the Connes-Kirchberg problem, which is equivalent to the question whether $C^*(\mathbb{F}_2)$, which is known to have the LLP, also has the WEP. Our $C^*$-algebra A has the same collection of finite dimensional operator subspaces as $C^*(\mathbb{F}_2)$ or $C^*(\mathbb{F}_\infty)$. In addition our example can be made to be quasidiagonal and of similarity degree (or length) 3. In the second part of the paper we reformulate our construction in the more general framework of a $C^*$-algebra that can be described as the *limit both inductive and projective* for a sequence of $C^*$-algebras ($C_n$) when each $C_n$ is a *subquotient* of $C_{n+1}$. We use this to show that for certain local properties of injective (non-surjective) $*$-homomorphisms, there are $C^*$-algebras for which the identity map has the same properties as the $*$-homomorphisms.

MSC (2010): 46L06, 46L07, 46L09

The concept of nuclearity has had a major impact on operator algebra theory (see e.g. [2, 23, 24, 7]). It was introduced by Takesaki in [25] but the name itself was coined by Lance [12] in analogy with Grothendieck's nuclear locally convex spaces. A $C^*$-algebra A is called nuclear if there is a unique $C^*$-norm on the algebraic tensor product $A \otimes B$ for any other $C^*$-algebra B. By classical results due to Takesaki and Guichardet there is a minimal and a maximal $C^*$-norm on $A \otimes B$ (see e.g. [26]), so that the nuclearity of A can be written simply as the identity of the respective completions that is: $A \otimes_{\min} B = A \otimes_{\max} B$. In his seminal paper [12], Lance asked whether for the latter to hold for all B it suffices to have it for $B = A^{op}$ (the opposite $C^*$-algebra, i.e. the same one but with product in reversed order). This question was motivated by the case of von Neumann algebras for which the pair $A \otimes A^{op}$ admits a fundamental faithful representation that later came to be called “the standard form” (see [27]). In [10] Kirchberg gave a negative answer (see Remark 1.7) by constructing the first example of a non-nuclear $C^*$-algebra A such that

$$ (0.1) \qquad A \otimes_{\min} A^{op} = A \otimes_{\max} A^{op}. $$

In the first lines of that paper [10], he observed that this could be viewed as the analogue for $C^*$-algebras of the author's result in [16] for Banach space tensor products. It was thus tempting to

<footnote>*ORCID 0000-0002-3091-2049</footnote>
try to adapt the Banach space approach in [16] to the $C^*$-algebra setting to produce new examples satisfying (0.1). In some sense the present paper is the result of this quest but it started to be more than wishful thinking only recently.

Kirchberg ([11] see also [17]) proved that if $A$ has the Weak Expectation Property (WEP in short, defined in §2) and $B$ the Local Lifting Property (LLP in short, defined in §3) then

$$ (0.2) \qquad A \otimes_{\min} B = A \otimes_{\max} B. $$

Thus if a $C^*$-algebra $A$ has both WEP and LLP, then (0.2) holds with $A = B$ and in fact since both LLP and WEP remain valid for $A^{op}$ we have (0.1).

Kirchberg also proved ([11]) that $C^*(\mathbb{F}_\infty)$ has the LLP. This is in some sense the prototypical example of LLP, just like $B(H)$ is for the WEP.

The WEP, originally introduced by Lance [12], has drawn more attention recently because of Kirchberg's work [10] and in particular his proof that the Connes embedding problem is equivalent to the assertion that $C^*(\mathbb{F}_\infty)$ (or $C^*(\mathbb{F}_2)$) satisfies (0.1) or equivalently that it has the WEP. We will refer to this as the Connes-Kirchberg problem.

Note that any $A$ satisfying (0.1) must have the WEP. In fact the WEP of $A$ is equivalent to (0.1) restricted to “positive definite tensors”, see Remark 1.6.

The main result of this paper is the construction of a non-nuclear (and even non exact) separable $C^*$-algebra $A$ with both WEP and LLP. This answers a question, that although it remained implicit in Kirchberg's work (but it is explicitly in [5, p. 383]), was clearly in the back of his mind when he produced the $A$ satisfying (0.1). But since at the time he conjectured the equivalence of WEP and LLP, the question did not seem so natural until the latter equivalence was disproved by Junge and the author in [9]. More specifically, Kirchberg conjectured that $C^*(\mathbb{F}_\infty)$ and $B(H)$ both satisfy (0.1) but the latter was disproved in [9], while the former is the still open Connes-Kirchberg problem.

While we cannot prove (0.1) for $C^*(\mathbb{F}_\infty)$, our algebra $A$ has the same collection of finite dimensional operator subspaces as $C^*(\mathbb{F}_\infty)$. Thus our construction might shed some light, one way or the other, on the Connes-Kirchberg embedding problem.

In the second part, we prove a generalization, which can be viewed as a sort of inductive limit construction for a sequence of $C^*$-algebras ($C_n$) where $C_n$ is for each $n$ a subquotient of $C_{n+1}$. When the $C_n$'s all have the LLP this construction produces a $C^*$-algebra $A$ for which the identity map on $A$ possesses some of the tensor product properties of the linking maps. For instance, if the linking map is the embedding $C^*(\mathbb{F}_\infty) \to B(H)$ (and $B(H)$ is viewed as a quotient of $C^*(\mathbb{F})$ for some free group $\mathbb{F}$), which in some sense has both WEP and LLP, we recover our main example.

**Some abbreviations:** For short we write f.d. for finite dimensional, s.a. for self-adjoint, c.p. for completely positive, c.b. for completely bounded. We denote by $CB(E, F)$ the space of c.b. maps between two o.s. $E, F$ equipped with the c.b. norm. We reserve the notation $E \otimes F$ for the algebraic tensor product of two linear spaces. We denote by $Id_E$ the identity map on $E$. We refer to [7, 19] for background on operator spaces.

# 1 Nuclear pairs

We start by a few general remarks around nuclearity for pairs.

**Definition 1.1.** A pair of $C^*$ algebras $(A, B)$ will be called a nuclear pair if

$$ A \otimes_{\min} B = A \otimes_{\max} B, $$

or equivalently if the min-norm is equal to the max-norm on the algebraic tensor product $A \otimes B$.
**Remark 1.2.** If the min-norm and the max-norm are equivalent on $A \otimes B$, then they automatically are equal.

**Remark 1.3.** Let $A_1 \subset A$ and $B_1 \subset B$ be $C^*$-subalgebras. In general, the nuclearity of the pair $(A, B)$ does not imply that of $(A_1, B_1)$. This “defect” is a major feature of the notion of nuclearity. However, if $(A_1, B_1)$ admit contractive c.p. projections (conditional expectations) $P: A \to A_1$ and $Q: B \to B_1$ then $(A_1, B_1)$ inherits the nuclearity of $(A, B)$. More generally, this holds if we only have approximate versions of $P$ and $Q$. For instance, if $A_1$ and $B_1$ are (closed s.a.) ideals in $A$ and $B$ then the nuclearity of the pair $(A, B)$ does imply that of $(A_1, B_1)$.

Recall that $A$ is called nuclear if $(A, B)$ is nuclear for all $B$.

The basic examples of nuclear $C^*$-algebras include all commutative ones, the algebra $K(H)$ of all compact operators on an arbitrary Hilbert space $H$, $C^*(G)$ for all amenable discrete groups $G$ and the Cuntz algebras.

We wish to single out two fundamental examples

$$ \mathcal{B} = B(\ell_2) \quad \text{and} \quad \mathcal{C} = C^*(\mathbb{F}_\infty). $$

Recall that every separable unital $C^*$-algebra embeds in $\mathcal{B}$ and is a quotient of $\mathcal{C}$. Moreover $\mathcal{B}$ is injective while $\mathcal{C}$ is in some sense projective (see Remark 3.2). Neither $\mathcal{B}$ nor $\mathcal{C}$ is nuclear, nevertheless:

**Theorem 1.4 (Kirchberg [11]).** *The pair $(\mathcal{B}, \mathcal{C})$ is nuclear.*

More generally, for any free group $F$ and any $t \in B(H) \otimes C^*(F)$ or $t \in C^*(F) \otimes B(H)$ we have $\|t\|_{\min} = \|t\|_{\max}$.

A simpler proof appears in [17] (or in [19], or now in [22]).

Since Kirchberg [10] showed that a $C^*$-algebra $A$ has Lance’s WEP if and only if the pair $(A, \mathcal{C})$ is nuclear, we took the latter as our definition of the WEP. Kirchberg [10] also showed that $A$ has a certain local lifting property (LLP) if and only if the pair $(A, \mathcal{B})$ is nuclear. We again take the latter as the definition of the LLP. With this terminology, Theorem 1.4 admits the following generalization:

**Corollary 1.5.** Let $B, C$ be $C^*$-algebras. If $B$ has the WEP and $C$ the LLP then the pair $(B, C)$ is nuclear.

In [9] it was shown that $\mathcal{B}$ failed the LLP, or equivalently that the pair $(\mathcal{B}, \mathcal{B})$ was not nuclear, which gave a negative answer to one of Kirchberg’s questions in [10], namely whether the WEP implies the LLP. However, the following major conjecture (equivalent to the converse implication) remains open:

**Kirchberg’s conjecture :** The pair $(\mathcal{C}, \mathcal{C})$ is nuclear, or equivalently $\mathcal{C}$ has the WEP.

Kirchberg showed at the end of [10] that this conjecture is equivalent to the Connes embedding problem whether any finite von Neumann algebra embeds in an ultraproduct of matrix algebras.

The Kirchberg conjecture asserts that the min and max norms coincide on $\mathcal{C} \otimes \mathcal{C}$. More recently in [15, Th. 29], Ozawa proved that to confirm the Kirchberg conjecture it suffices to show that they coincide on $E_n^1 \otimes E_n^1$ for all $n \ge 1$, where $E_n^1$ is the span of the unit and $n-1$ first free unitary generators of $\mathcal{C}$.

**Remark 1.6.** Recall $A^{op} \simeq \bar{A}$, where $A$ is the complex conjugate of $A$. By an unpublished result of Haagerup (see [22] for complete details) a $C^*$-algebra $A$ has the WEP if and only if the min and max norms coincide on the set of tensors in $\bar{A} \otimes A$ of the form $\sum \bar{a}_j \otimes a_j$ (we call these positive definite tensors in [22]).
Remark 1.7. Kirchberg’s construction in [10, Th. 1.2] of a $C^*$-algebra A satisfying (0.1) is quite difficult to follow. He uses to start with a $C^*$-algebra Q that is a quotient of a WEP $C^*$-algebra (QWEP in short). Then his A is a (quasidiagonal) extension by the compact operators $\mathcal{K}$ of the cone algebra $C(Q)$ of Q. The non-nuclearity (actually non-exactness) of A follows from the fact that for a well chosen Q the short exact sequence $\mathcal{K} \to A \to C(Q)$ does not locally split in his sense. His A must have the WEP (see Remark 1.6), and hence $C(Q)$ and Q are necessarily QWEP but it seems unclear whether a suitable choice of Q might lead to one for which A has LLP. Note however that Q must be chosen to fail the LLP, otherwise (see [10, p. 454]) the exact sequence will locally split. A much clearer presentation (unfortunately without the full details) of his arguments is sketched in Remark 13.4.6 of Brown and Ozawa’s remarkable book [5].

In the next two sections we gather some known facts on the WEP and the LLP, that were probably all known in some form to Kirchberg at the time of [10]. Since we use reformulations best suited for our construction, we include proofs. We refer the reader to Ozawa’s concise survey [14] or to our much longer exposition in [22] for more information.

## 2 The WEP

We define the WEP for a $C^*$-algebra A by the equality $A \otimes_{\min} \mathcal{C} = A \otimes_{\max} \mathcal{C}$, where $\mathcal{C}$ is the full (or maximal) $C^*$-algebra of the free group $F_\infty$. Kirchberg showed that this property is equivalent to a weak form of extension property (analogous to that of $L_\infty$ in Banach space theory), a variant of injectivity that had been considered by Lance [12].

Assume $A \subset B(H)$ as a $C^*$-subalgebra. Then A has the WEP if and only if there is a contractive projection $P : B(H)^{**} \to A^{**}$.

Equivalently, this holds if and only if there is a contractive linear map $T : B(H) \to A^{**}$ such that $T(a) = a$ for any $a \in A$, or in other words such that $T|_A$ coincides with the canonical inclusion $i_A : A \to A^{**}$. Note that when it exists the contractive projection P is automatically completely contractive and completely positive by Tomiyama’s well known theorem. This leads to the following simple (known) criterion which, being almost purely Banach space theoretical, will be particularly well adapted to our needs.

We denote here by $\ell_1^n$ the operator space dual of $\ell_\infty^n$. One nice realization of $\ell_1^n$ can be given inside $\mathcal{C}$: just let $E_1^n = \operatorname{span}[1, U_1, \dots, U_{n-1}] \subset \mathcal{C}$ where $(U_j)$ are the free unitary generators of $\mathcal{C}$, then $\ell_1^n \simeq E_1^n$ completely isometrically. Equivalently we could take instead the span of $\{U_1, \dots, U_n\}$. Note that $\|v\|_{cb} = \|v\|$ for any $v$ defined on $\ell_1^n$. This is known as the maximal operator space structure of $\ell_1^n$ (see e.g. [19, p. 183]).

**Proposition 2.1.** A $C^*$-algebra $A \subset B(H)$ has the WEP if (and only if) for any $n \ge 1$ and any subspace $S \subset \ell_1^n$ any linear map $u : S \to A$ admits for each $\varepsilon > 0$ an extension $\tilde{u} : \ell_1^n \to A$ with

$$ \|\tilde{u}\|_{cb} \le (1 + \varepsilon)\|u\|_{cb}. $$

*Proof.* This is a well known application of Hahn-Banach. Let B be another $C^*$-algebra. Note the isometric identity

$$ (2.1) \qquad B(B, A^{**}) = (B \otimes_A A^*)^*, $$

where $B \otimes_A A^*$ denotes the normed space $B \otimes A^*$ (algebraic) equipped with the projective norm, denoted by $\|\cdot\|_{B \otimes_A A^*}$. Let $X = B(H) \otimes A^*$ and $Y = A \otimes A^*$ so that $Y \subset X$. Consider the assertion that the inclusion

$$ (2.2) \qquad (Y, \| \cdot \|_{A \otimes_A A^*}) \subset (X, \| \cdot \|_{B(H) \otimes_A A^*}) $$
is isometric. We claim that if this holds then $A$ has the WEP. Indeed, assume that (2.2) is isometric. Consider the linear form $f: Y \to \mathbb{C}$ defined by $f(a\otimes\xi) = \xi(a)$ (which corresponds through (2.1) with $B=A$ to $i_A: A \to A^{**}$). By Hahn-Banach, $f$ extends to a linear form $g: (X, \|\cdot\|_{B(H)\otimes_A A^*}) \to \mathcal{C}$ with $\|g\| \le 1$. Let $T: B(H) \to A^{**}$ be the map associated to $g$ via (2.1). Then $\|T\| \le 1$ and the fact that $g$ extends $f$ is equivalent to $T|_A = i_A$, so that $A$ has the WEP. To complete the proof of the if part it suffices to show that the extension property in Proposition 2.1 implies that (2.2) is isometric. This is easy to show using the factorization of the mappings $v: A \to B(H)$ corresponding to an element $x$ in the open unit ball of $(X, \|\cdot\|_{B(H)\otimes_A A^*})$. Such a $v$ can be written as $v = UV$, where $V: A \to \ell_1^n$ has nuclear norm 1 and $\|U: \ell_1^n \to B(H)\| < 1$. If it so happens that $x \in A \otimes A^*$ then $v(A) = UV(A) \subset A$. Let $S = V(A) \subset \ell_1^n$ and $u = U|_S: S \to A$. Note $\|U\|_{cb} = \|U\|$, and hence $\|u\|_{cb} \le \|U\|_{cb} < 1$. Let $\tilde{u}$ be as in the extension property in Proposition (2.1), then the factorization $v = \tilde{u}V$ now shows that $v$ corresponds to an element $x$ in the open unit ball of $(Y, \|\cdot\|_{A\otimes_A A^*})$. Thus (2.2) is isometric. This proves the if part.

Conversely, if $A$ has the WEP, by the injectivity of $B(H)$ any $u: S \to A$ extends to a map $u^1: \ell_1^n \to A^{**}$ with $\|u^1\|_{cb} = \|u\|_{cb}$ and $u^1|_S = i_A u$. Since $\ell_\infty^n(A^{**}) = \ell_\infty^n(A)^{**}$ isometrically, there is a net of maps $u_i: \ell_1^n \to A$ with $\|u_i\|_{cb} \le 1$ tending pointwise $\sigma(A^{**}, A^*)$ to $u^1$. Then $(u_i - u)|_S$ tends pointwise $\sigma(A, A^*)$ to 0, and by Mazur's theorem passing to convex combinations we obtain a net $u'_i: \ell_1^n \to A$ with $\|u'_i\|_{cb} \le 1$ such that $(u'_i - u)|_S$ tends pointwise to 0 in norm. Thus for $i$ large enough $u'_i$ is "almost" the desired extension of $u$. Then for each $\varepsilon > 0$ a simple perturbation argument (see e.g. [19, p. 69]) gives us a true extension $\tilde{u}$ as in Proposition 2.1. $\square$

**Remark 2.2.** In the preceding situation, assume $A \subset B(H)$. A linear map $u: S \to A$ satisfies $\|u\|_{cb} \le 1$ if and only if it admits an extension $\tilde{u}: \ell_1^n \to B(H)$ with $\|\tilde{u}\| \le 1$. This is immediate by the injectivity of $B(H)$ and the equality $\|\tilde{u}\| = \|\tilde{u}\|_{cb}$.

This allows us to view the extension property in Proposition 2.1 as a Banach space theoretic property of the inclusion $A \subset B(H)$ like this: any $u$ that extends to a contraction into $B(H)$ extends to a map of norm $\le 1 + \varepsilon$ into $A$.

**Remark 2.3.** The interest of Proposition 2.1 is that the apparently weak form of extension property considered there suffices to imply the WEP. But actually, any $A$ with WEP satisfies a stronger extension property, as follows:

Let $C$ be a separable $C^*$-algebra with the LLP and let $A$ be another one with the WEP. Then for any finite dimensional subspace $E \subset C$ and any $\varepsilon > 0$, any $u \in CB(E, A)$ admits an extension $\tilde{u} \in CB(C, A)$ such that $\|\tilde{u}\|_{cb} \le (1 + \varepsilon)\|u\|_{cb}$. See [10] or [17, Th. 20.27] for full details.

<figure><img src="image_2.png" /></figure>

# 3 The LLP

We define the LLP for a $C^*$-algebra $A$ by the equality $A \otimes_{\min} \mathcal{B} = A \otimes_{\max} \mathcal{B}$, where $\mathcal{B} = B(\ell_2)$. Kirchberg showed that this property is equivalent to a certain local lifting property (analogous to that of $L_1$ in Banach space theory), which has several equivalent forms, one of which as follows:

**Proposition 3.1.** A $C^*$-algebra $A$ satisfies $A \otimes_{\min} \mathcal{B} = A \otimes_{\max} \mathcal{B}$ if and only if for any $*$-homomorphism $\pi: A \to C/I$ into a quotient $C^*$-algebra, for any f.d. subspace $E \subset A$ and any $\varepsilon > 0$ the restriction $\pi|_E$ admits a lifting $v: E \to C$ with $\|v\|_{cb} \le (1 + \varepsilon)$.
Note: Actually when A has the LLP the preceding local lifting even holds with $\epsilon = 0$ (see the proof of Proposition 3.1).

**Remark 3.2.** According to Kirchberg [11] a unital $C^*$-algebra A has the lifting property (LP) if any unital c.p. $\pi: A \to C/\mathcal{I}$ as above admits a (global) unital c.p. lifting, and if A is not unital, it is said to have the LP if its unitization does. He proved that $\mathcal{C}$ has the LP (see [14] or [5, p. 376] for a proof). By known results, it follows that $C \otimes_{\min} \mathcal{C}$ has the LP whenever C is nuclear and separable. Roughly this can be checked using the CPAP of C, Kirchberg's theorem that $M_n(\mathcal{C})$ has the LP for any $n$ (see [11, p. 59]) and the fact (due to Arveson [1]) that the set of liftable maps on separable $C^*$-algebras is pointwise closed. Actually, Kirchberg observed in [10] that if his conjecture that $\mathcal{C}$ has the WEP is correct, then the LLP implies the LP in the separable case. Kirchberg's LP (as we just defined it) implies that any completely contractive c.p. map $u: A \to C/\mathcal{I}$ admits a completely contractive lifting from A to C, but the converse does not seem clear, although the analogous converse does hold for the LLP.

**Lemma 3.3.** Assume that an isomorphism $\pi: A \to C/\mathcal{I}$ is locally liftable, meaning that it has the property considered in Proposition 3.1. Let B be any $C^*$-algebra.

Then $(C, B)$ nuclear $\Rightarrow$ $(A, B)$ nuclear.

In particular if C has the LLP then A has the LLP.

**Proof.** Indeed, let $t \in E \otimes B$ with $E \subset A$ f.d. with $\|t\|_{\min} = 1$. Let $q: C \to C/I$ be the quotient map. Then if $(C, B)$ is nuclear

$$ \|(v \otimes Id)(t)\|_{C \otimes_{\max} B} = \|(v \otimes Id)(t)\|_{C \otimes_{\min} B} \le 1 + \epsilon $$

and hence $\|(\pi \otimes Id)(t)\|_{C/I \otimes_{\max} B} = \|(qv \otimes Id)(t)\|_{C/I \otimes_{\max} B} \le 1 + \epsilon$. Since $\pi$ is an isomorphism we obtain $\|t\|_{A \otimes_{\max} B} = 1$, and hence $(A, B)$ is nuclear. $\square$

**Proof of Proposition 3.1.** Let A be unital. Then $A = C^*(\mathbb{F})/I$ for some free group $\mathbb{F}$, and $C^*(\mathbb{F})$ has the LLP. In general let $\tilde{A}$ be the unitization, let $q: C^*(\mathbb{F}) \to \tilde{A}$ be a surjective $*$-homomorphism, and let $C = q^{-1}(A) \subset C^*(\mathbb{F})$. Since C is an ideal in $C^*(\mathbb{F})$, it still has the LLP. By Lemma 3.3 with $\pi: A \to C/\ker(q)$, the if part follows.

Conversely, if A has the LLP, so does its unitization $\tilde{A}$ (see Remark 1.3). If $\pi: A \to C/I$ is as in Proposition 3.1, then $\pi$ extends to a unital $*$-homomorphism $\tilde{\pi}: \tilde{A} \to \tilde{C}/I$ and it is easy to check that if $\tilde{\pi}$ is locally liftable then $\pi$ also is. Thus to prove the only if part we may assume that A, C and $\pi$ are all unital. Again we write $A = C/I$ with $C = C^*(\mathbb{F})$. Let $E \subset A$ be a f.d. subspace. The inclusion map $E \to A$ corresponds to a tensor $t \in E^* \otimes A$ with $\|t\|_{\min} = 1$ when $E^*$ is equipped with its dual operator space structure (see [7, 19]). We may assume $E^* \subset \mathcal{B}$ completely isometrically. Viewing $t \in \mathcal{B} \otimes A$ the LLP assumption implies $\|t\|_{\mathcal{B} \otimes_{\max} A} = 1$. Since $\mathcal{B} \otimes_{\max} A = \mathcal{B} \otimes_{\max} C/\mathcal{B} \otimes_{\max} I$. The tensor t admits a lifting $t \in E^* \otimes C$ with $\|t\|_{\mathcal{B} \otimes_{\max} C} \le 1+\epsilon$. A fortiori, we have $\|t\|_{E^* \otimes_{\min} C} < 1+\epsilon$, and the linear map $\tilde{u}: E \to C$ associated to t gives us the desired local lifting. With more effort (see [19, p. 45]) one can show the same for $\epsilon = 0$. $\square$

Let E, F be operator spaces. Recall

$$ (3.1) \qquad d_{cb}(E, F) = \inf\{\|u\|_{cb} \|u^{-1}\|_{cb}\} $$

where the infimum runs over all complete isomorphisms $u: E \to F$ whenever E, F are completely isomorphic (e.g. if E, F are of the same finite dimension), and $d_{cb}(E, F) = \infty$ if they are not completely isomorphic. Definition 3.4 below is just the obvious analogue of the notion of a Banach space finitely representable (in short f.r.) in another one.
**Definition 3.4.** Let A, C be C*-algebras (or operator spaces). We will say that A “locally embeds” in C if for any ε > 0 the following property holds: for any f.d. Z ⊂ A there is a f.d. Z' ⊂ C such that d<sub>cb</sub>(Z, Z') ≤ 1 + ε.

We say that A and C are “locally equivalent” if each one locally embeds in the other.

Remark 3.5. Concentrating on the case C = C, let us denote (as in [19, p. 343]) for any o.s. A

$$
d_{SC}(A) = \sup\{\inf\{d_{cb}(Z, Z') \mid Z' \subset \mathcal{C}\} \mid Z \subset A, \dim(Z) < \infty\}.
$$

One defines similarly $d_{SC}(A)$ for any $C^*$-algebra $\mathcal{C}$.

Then A locally embeds in C means that d<sub>S C</sub>(A) = 1. The following formula was proved in [9, Th. 4.5] (see also [19, p. 349] or [22, Cor. 20.6]): for any o.s. A ⊂ B(H) we have

$$
(3.2) \qquad d_{S\mathcal{C}}(A) = \sup\{\|t\|_{B(H)\otimes_{\max}\mathcal{B}}/\|t\|_{B(H)\otimes_{\min}\mathcal{B}} \mid t \in A \otimes \mathcal{B}\}.
$$

When A is a C*-algebra, the basic properties of C*-norms imply that if d<sub>S C</sub>(A) ≤ 1 + ε for some
ε > 0 then d<sub>S C</sub>(A) = 1.

Remark 3.6. Note that when $A \subset B(H)$, we have obviously $\|t\|_{B(H)\otimes_{\max}\mathcal{B}} \le \|t\|_{A\otimes_{\max}\mathcal{B}}$ for any $t \in A \otimes \mathcal{B}$. Therefore, (3.2) implies that $d_{S\mathcal{C}}(A) = 1$ for any $A$ with the LLP. In other words, any $A$ with the LLP locally embeds in $\mathcal{C}$.

The next result is related to [9]. In the latter, it was shown that WEP $\not\iff$ LLP, and at the same time that there are $A$'s that do not locally embed in $\mathcal{C}$.

**Proposition 3.7.** Let A be a C\*-algebra that locally embeds in C. If A has the WEP then it has the LLP.

*Proof.* We will use several results from [8] (see also [22, chap. 6] for a detailed presentation). Assume that $A$ locally embeds in $\mathcal{C}$ and has the WEP. Let $t \in A \otimes \mathcal{B}$. We will show that $\|t\|_{\max} = \|t\|_{\min}$. Let $Z \subset A$ be a f.d. subspace such that $t \in Z \otimes \mathcal{B}$. For any $\varepsilon > 0$ there is $Z' \subset \mathcal{C}$ and an isomorphism $v: Z' \to Z \subset A$ with $\|v\|_{cb} \|v^{-1}\|_{cb} < 1 + \varepsilon$. Using the factorization of the canonical inclusion $i_A: A \to A^{**}$ through some $B(H)$ (which is one form of the WEP) we find an extension $\tilde{v}: \mathcal{C} \to A^{**}$ of $v$ such that $\|\tilde{v}\|_{dec} = \|v\|_{cb}$ (here we use the dec-norm of [8] and the fact due to Haagerup that the dec-norm and the cb-norm coincide for maps with range $B(H)$). Now we have $(i_A \otimes Id)t = (\tilde{v}v^{-1} \otimes Id)t$ and hence (again using [8])

$$
\|(i_A \otimes \mathrm{Id})t\|_{A^{**} \otimes_{\mathrm{max}} \mathcal{B}} \le \|\tilde{v}\|_{\mathrm{dec}} \|(v^{-1} \otimes \mathrm{Id})t\|_{\mathcal{C} \otimes_{\mathrm{max}} \mathcal{B}} \le \|v\|_{cb} \|(v^{-1} \otimes \mathrm{Id})t\|_{\mathcal{C} \otimes_{\mathrm{max}} \mathcal{B}}.
$$

By Theorem 1.4, $\|(v^{-1} \otimes \mathrm{Id})t\|_{\mathcal{C} \otimes_{\max} \mathcal{B}} = \|(v^{-1} \otimes \mathrm{Id})t\|_{\mathcal{C} \otimes_{\min} \mathcal{B}}$ and hence we find

$$
\|(i_A \otimes \mathrm{Id})t\|_{A^{**} \otimes_{\max} \mathcal{B}} \le \|v\|_{cb} \|(v^{-1} \otimes \mathrm{Id})t\|_{\mathcal{C} \otimes_{\min} \mathcal{B}} \le \|v\|_{cb} \|v^{-1}\|_{cb} \|t\|_{A \otimes_{\min} \mathcal{B}} \le (1+\varepsilon) \|t\|_{A \otimes_{\min} \mathcal{B}}.
$$

But by an easy and well known argument we have $\|(i_A \otimes \mathrm{Id})t\|_{A^{**} \otimes_{\max} \mathcal{B}} = \|t\|_{A \otimes_{\max} \mathcal{B}}$ for any $t \in A \otimes \mathcal{B}$, thus since $\varepsilon > 0$ is arbitrary we conclude $\|t\|_{\max} \le \|t\|_{\min}$. $\square$

4 Outline

We follow the general strategy in [16] (see also [4]). We construct a sequence of operator spaces $E_n \subset E_{n+1}$ such that for any $S \subset \ell_1^n$, any $u: S \to E_n$ admits an extension (or an approximate extension) $\tilde{u} \to E_{n+1}$ but into the larger space $E_{n+1}$, with $\|\tilde{u}\|_{cb} \approx \|u\|_{cb}$, as in the following diagram.
<figure><img src="image_3.png" /></figure>

The rough plan is then to start from a space $E_1$ such that $d_{S\mathscr{C}}(E_1) = 1$ and find successive spaces $E_n$ while maintaining the condition $d_{S\mathscr{C}}(E_n) = 1$ for all $n$. Then the idea is that the union $X = \bigcup E_n$ will satisfy the extension property in Proposition 2.1 (that is equivalent to the WEP for a $C^*$-algebra), while the condition $d_{S\mathscr{C}}(E_n) = 1$ for all $n$ will imply $d_{S\mathscr{C}}(X) = 1$ and hence by Proposition 3.7 the purported WEP of $X$ would imply its LLP. Just starting from a f.d. operator space $E_1$ with exactness constant $> 1$ will ensure that $X$ if it is a $C^*$-algebra is not exact, and hence not nuclear.

Since we can embed $E_n$ into an injective object, namely $B(H)$ (and $\ell_\infty$ in the Banach space case) it is easy to do the first step $n = 1$ and to find some “big” space $E_2$ satisfying the extension but the difficulty is to find $E_2$ still such that $d_{S\mathscr{C}}(E_2) = 1$. The latter expresses that $E_2$ remains relatively “small”. In the Banach space analogue of [16] this is the main problem: there bounded cotype 2 constants are the key tool that replaces $d_{S\mathscr{C}}(E_n) = 1$; note however that the construction there is isomorphic (with uniformly bounded constants), as opposed to the present “almost” isometric one (i.e. with constants asymptotically tending to 1). However, it is actually possible to essentially proceed and maintain the condition $d_{S\mathscr{C}}(E_n) = 1$ for all $n$, using the operator space analogue of the construction in [16], but in the completely isometric setting, with all relevant constants equal to 1. This led to operator space versions of our main result, that we obtained already a few years ago (we gave a talk on this at MSRI in the Fall 2016). We plan to write the details in a separate paper.

But this seemed like a dead end because it gave no clue how to arrange so that the union $\bigcup E_n$ be not only an operator space but a genuine $C^*$-algebra. For this we need to produce embeddings $T_n : E_n \to E_{n+1}$ that are multiplicative, or at least close to multiplicative, in such a way that in the limit we obtain an algebra and multiplicative maps. This is now overcome by a quite different construction of $E_{n+1}$ given $E_n$ based on Lemma 7.1 below. We construct our example as an inductive limit of f.d. operator spaces, with linking maps that are close to multiplicative in some suitable sense. Similar asymptotic morphisms already appear in the $E$-theory of Connes and Higson [6] but our use of them seems unrelated. There are many examples of inductive limits of $C^*$-algebras in the literature, see for example [3]. See also [2, p. 465], [23, ch. 6] or [13] for general background on inductive limits.

# 5 Approximately multiplicative maps

We will need to use liftings that are approximately multiplicative. This will be provided by the use of the cone algebra $C_0(\mathscr{C})$, but it seems worthwhile to first collect some basic general facts about almost multiplicative mappings. The latter facts are known (the ideas go back at least to [6]) but do not seem to have been recorded in the literature.

**Definition 5.1.** Let $B, B_0$ be $C^*$-algebras. Let $\mathcal{E}_0 \subset B$ be a self-adjoint subspace and let $\varepsilon_0 > 0$. A linear map $\psi : \mathcal{E}_0 \to B_0$ (we will restrict in (iii) to the s.a. case for simplicity) will be called an $\varepsilon_0$-morphism if

$$ (i) \quad \|\psi\| \le 1 + \varepsilon_0, $$
(ii) for any $x, y \in \mathcal{E}_0$ with $xy \in \mathcal{E}_0$ we have $\|\psi(xy) - \psi(x)\psi(y)\| \le \varepsilon_0\|x\|\|y\|$,

(iii) $\psi$ is self-adjoint i.e. for any $x \in \mathcal{E}_0$ we have $\psi(x^*) = \psi(x)^*$.

**Remark 5.2.** Let $\psi_0 : \mathcal{E}_0 \to B_0$ be a linear map. Let $\mathcal{E}_1 \subset B_0$ be another self-adjoint subspace such that $\psi_0(\mathcal{E}_0) \subset \mathcal{E}_1$ and $\psi_0(\mathcal{E}_0)\psi_0(\mathcal{E}_0) \subset \mathcal{E}_1$. Let $B_1$ be another $C^*$-algebra. If $\psi_0 : \mathcal{E}_0 \to B_0$ is an $\varepsilon_0$-morphism and $\psi_1 : \mathcal{E}_1 \to B_1$ an $\varepsilon_1$-morphism ($\varepsilon_1 > 0$), then $\psi_1\psi_0 : \mathcal{E}_0 \to B_1$ is an $\varepsilon_0\|\psi_1\| + \varepsilon_1\|\psi_0\|^2$-morphism, and hence a $\delta_1$-morphism with

$$ (5.1) \qquad \delta_1 \le \varepsilon_0(1 + \varepsilon_1) + \varepsilon_1(1 + \varepsilon_0)^2. $$

This is immediate since

$$ \psi_1\psi_0(xy) - \psi_1\psi_0(x)\psi_1\psi_0(y) = \psi_1(\psi_0(xy) - \psi_0(x)\psi_0(y)) + \psi_1(\psi_0(x)\psi_0(y)) - \psi_1\psi_0(x)\psi_1\psi_0(y). $$

More generally if we have a similar composition $\psi_n \cdots \psi_1\psi_0$ with $\psi_j$ being an $\varepsilon_j$-morphism, then $\psi_n \cdots \psi_1\psi_0 : \mathcal{E}_0 \to B_n$ has norm $\le (1 + \varepsilon_n) \cdots (1 + \varepsilon_1)(1 + \varepsilon_0)$. Let $p_n = (1 + \varepsilon_n) \cdots (1 + \varepsilon_0)$. Using (5.1), a simple induction shows that $\psi_n \cdots \psi_1\psi_0$ is a $\delta_n$-morphism with $\delta_n$ satisfying :

$$ \delta_n \le \delta_{n-1}(1 + \varepsilon_n) + \varepsilon_n p_{n-1}^2. $$

Thus the number $\delta'_n = \delta_n p_n^{-1}$ satisfies

$$ \delta'_n \le \delta'_{n-1} + \varepsilon_n p_{n-1}^2 / p_n \le \delta'_{n-1} + \varepsilon_n p_{n-1}. $$

Thus if we assume that $\sum \varepsilon_n < \infty$ so that $\sup_n p_n = c < \infty$ we find

$$ \delta'_n \le \delta'_0 + c \sum_1^n \varepsilon_k $$

and hence for any $n \ge 1$

$$ (5.2) \qquad \delta_n \le p_n(\delta'_0 + c \sum_1^n \varepsilon_k) \le c\varepsilon_0 + c^2 \sum_1^n \varepsilon_k. $$

It will be convenient to work with a general directed set $I$. Given a family $(C_\alpha)$ of $C^*$-algebras, we denote by $\ell_\infty(I, \{C_\alpha\})$ the $C^*$-algebra formed of all the bounded families $(x_\alpha)$ (we view these as “generalized sequences”) with $x_\alpha \in C_\alpha$ for all $\alpha \in I$, equipped with the sup-norm. We denote by $c_0(I, \{C_\alpha\})$ the ideal formed of all $x = (x_\alpha)$ such that $\limsup_\alpha \|x_\alpha\| = 0$. As usual if

$$ Q : \ell_\infty(I, \{C_\alpha\}) \to \ell_\infty(I, \{C_\alpha\}) / c_0(I, \{C_\alpha\}) $$

denotes the quotient map, then for any $x \in \ell_\infty(I, \{C_\alpha\})$ we have

$$ \|Q(x)\| = \limsup_\alpha \|x_\alpha\|. $$

The interest of $\varepsilon$-morphisms is illustrated by the following simple lemma, that we will use only toward the end of this paper.

**Lemma 5.3.** Let $B, D$ be $C^*$-algebras. Let $E \subset B, F \subset D$ be f.d.s.a. subspaces. For any $\delta > 0$ there is $\varepsilon > 0$ and a f.d.s.a. superspace $\mathcal{E}$ with $E \subset \mathcal{E} \subset B$ such that for any $\varepsilon$-morphism $\psi : \mathcal{E} \to C$ (C any other $C^*$-algebra) we have

$$ \forall x \in E \otimes F \quad \|(\psi \otimes \mathrm{Id}_D)(x)\|_{C \otimes_{\max} D} \le (1 + \delta)\|x\|_{B \otimes_{\max} D}. $$
Proof. Let $I = \{(\mathcal{E}, \varepsilon)\}$ be the directed set of pairs with $E \subset \mathcal{E}$, $\varepsilon > 0$. Fix $\delta > 0$. It suffices to show that there is $(\mathcal{E}, \varepsilon) \in I$ such that for all $C$, all $\varepsilon$-morphisms $\psi : \mathcal{E} \to C$ and all $x \in E \otimes F$ we have $\|(\psi \otimes \mathrm{Id}_D)(x)\|_{C \otimes_{\mathrm{max}D}} \le (1+\delta)\|x\|_{B \otimes_{\mathrm{max}D}}$. For each $\alpha \in I$ with $E \subset \mathcal{E}_\alpha$, there is a $C_\alpha$ and an $\varepsilon_\alpha$-morphism $\psi_\alpha : \mathcal{E}_\alpha \to C_\alpha$, such that

$$ \|(\psi_{\alpha} \otimes \mathrm{Id}_{D})(x)\|_{C_{\alpha} \otimes_{\mathrm{max}D}} \ge (1 + \delta)^{-1} \sup \|(\psi \otimes \mathrm{Id}_{D})(x)\|_{C \otimes_{\mathrm{max}D}} $$

where the last supremum runs over all $C$ and all $\varepsilon_\alpha$-morphism $\psi : \mathcal{E}_\alpha \to C$. Note that this last supremum is finite since the nuclear norm of each $\psi : \mathcal{E}_\alpha \to C$ is at most $(1 + \varepsilon_\alpha) \dim(\mathcal{E}_\alpha)$. Let $\psi'_\alpha : B \to C_\alpha$ be the map that extends $\psi_\alpha$ by 0 outside $\mathcal{E}_\alpha$ (we could use a linear map but this is not needed at this point). Consider $\psi' = (\psi'_\alpha) : B \to \ell_\infty(I; \{C_\alpha\})$ and let $Q : \ell_\infty(I; \{C_\alpha\}) \to \ell_\infty(I; \{C_\alpha\})/c_0(I; \{C_\alpha\})$ be the quotient map. Then $\pi = Q\psi' : B \to \ell_\infty(I; \{C_\alpha\})/c_0(I; \{C_\alpha\})$ is clearly an isometric $*$-homomorphism. We have contractive morphisms

$$ \pi \otimes \mathrm{Id}_D : B \otimes_{\mathrm{max}} D \to [\ell_\infty(I; \{C_\alpha\})/c_0(I; \{C_\alpha\})] \otimes_{\mathrm{max}} D = [\ell_\infty(I; \{C_\alpha\}) \otimes_{\mathrm{max}} D]/[c_0(I; \{C_\alpha\}) \otimes_{\mathrm{max}} D] $$

where the last $=$ holds by the “exactness” of the max-tensor product (see e.g. [19, p. 285]). Moreover, we have clearly a contractive morphism

$$ [\ell_\infty(I; \{C_\alpha\}) \otimes_{\mathrm{max}} D]/[c_0(I; \{C_\alpha\}) \otimes_{\mathrm{max}} D] \to [\ell_\infty(I; \{C_\alpha \otimes_{\mathrm{max}} D\})]/[c_0(I; \{C_\alpha \otimes_{\mathrm{max}} D\})]. $$

Since $\pi(e) = Q((\psi_\alpha(e))_\alpha)$ for all $e \in E$ and since $x \in E \otimes D$, it follows that

$$ \lim \sup_{\alpha} \|(\psi_{\alpha} \otimes \mathrm{Id}_{D})(x)\|_{C_{\alpha} \otimes_{\mathrm{max}D}} \le \|x\|_{B \otimes_{\mathrm{max}D}}, $$

which proves the desired result for each fixed given $x \in E \otimes D$. But since $E \otimes F$ is a finite dimensional subspace of $B \otimes_{\mathrm{max}} D$ we may replace the unit ball by a finite $\delta$-net in it (or invoke Ascoli's theorem). We can deal with the latter case by enlarging $\mathcal{E}$ finitely many times. We obtain the announced result (possibly with $2\delta$ instead of $\delta$).

A different proof can be obtained using the Blecher-Paulsen factorization, as in [19, Th. 26.8]. □

Remark 5.4. Consider a surjective $*$-homomorphism $q : C_1 \to B$ between $C^*$-algebras. It is well known that for any $\varepsilon > 0$ and any f.d. subspace $\mathcal{E} \subset B$ there is $\psi : \mathcal{E} \to C_1$ with $\|\psi\| \le 1 + \varepsilon$ such that $q\psi(x) = x$ for any $x \in \mathcal{E}$ (actually this even holds for $\varepsilon = 0$, see e.g. [19, p. 46]).

Thus, if one replaces the c.b. norm by the usual one, the analogue of the LLP becomes universally valid, just like for local convexity (and the latter fact implies the former).

However, we will need to work with quotient maps that admit a slightly stronger sort of lifting, as follows:

**Definition 5.5.** Let $q : C_1 \to C_1/\mathcal{I}$ be a quotient $*$-homomorphism and let $B$ be a $C^*$-algebra. We say that a $*$-homomorphism $\sigma : B \to C_1/\mathcal{I}$ is “almost multiplicatively locally liftable” if for any $\varepsilon > 0$ and any f.d.s.a. subspace $\mathcal{E} \subset B$ there is an $\varepsilon$-morphism $\psi : \mathcal{E} \to C_1$ such that $\|q\psi(x) - \sigma(x)\| \le \varepsilon\|x\|$ for any $x \in \mathcal{E}$.

We will say that $q$ almost allows liftings if this holds for $B = C_1/\mathcal{I}$ and $\sigma = \mathrm{Id}_B$ (or equivalently whenever $\sigma$ is an isomorphism). In that case, any $*$-homomorphism $\sigma : B \to C_1/\mathcal{I}$ is almost multiplicatively locally liftable.

Let $q : C_1 \to C_1/\mathcal{I}$ be a surjective $*$-homomorphism. Let $I$ be any directed set. We denote by $q^* : \ell_\infty(I; C_1)/c_0(I; C_1) \to \ell_\infty(I; C_1/\mathcal{I})/c_0(I; C_1/\mathcal{I})$ the $*$-homomorphism associated to $q$ acting an each coordinate, and by $\nu : C_1/\mathcal{I} \to \ell_\infty(I; C_1/\mathcal{I})/c_0(I; C_1/\mathcal{I})$ the embedding associated to $\mathrm{Id}_{C_1/\mathcal{I}}$ acting an each coordinate, or equivalently the map that takes $x \in C_1/\mathcal{I}$ to the equivalence class of the function constantly equal to $x$ on $I$.
**Proposition 5.6.** The following properties of a *-homomorphism $\sigma : B \to C_1/\mathcal{I}$ are equivalent.

(i) The map $\sigma$ is almost multiplicatively locally liftable.

(ii) There is a directed set $I$ and a *-homomorphism $\pi : B \to \ell_\infty(I; C_1)/c_0(I; C_1)$ (which will automatically be an embedding) such that $q^\sharp\pi = \nu\sigma$.

if sigma is itself
isometric

<figure><img src="image_4.png" /></figure>

*Proof.* Assume (i). Let $I$ be the directed set formed of all pairs $\alpha = (\mathcal{E}, \varepsilon)$ where $\mathcal{E}$ is a f.d.s.a. subspace of $B$ and $\varepsilon > 0$, equipped with the usual ordering so that $\alpha \to \infty$ in the corresponding net means that $\mathcal{E} \to B$ and $\varepsilon \to 0$. Let $\alpha = (\mathcal{E}, \varepsilon)$ be such a pair. Let $\psi : \mathcal{E} \to C_1$ be an $\varepsilon$-morphism such that $\|q\psi(x) - \sigma x\| \le \varepsilon\|x\|$ for any $x \in \mathcal{E}$. Then we define a linear map $\psi_\alpha : B \to C_1$ by setting $\psi_\alpha(x) = \psi(x)$ if $x \in \mathcal{E}$ and (say) $\psi_\alpha(y) = 0$ whenever $y$ belongs to a complementary (to $\mathcal{E}$) subspace that we can choose arbitrarily. We denote by

$$ (5.3) \qquad \pi : B \to \ell_\infty(I; C_1)/c_0(I; C_1), $$

the mapping that takes $x \in B$ to $(\psi_\alpha(x))$ modulo $c_0(I; C_1)$. It is easy to check that $\pi$ is an isometric
*-homomorphism such that $q^\sharp\pi = \nu\sigma$.

Conversely, assume (ii). Let $Q : \ell_\infty(I; C_1) \to \ell_\infty(I; C_1)/c_0(I; C_1)$ be the quotient map. Let $\mathcal{E} \subset B$
be a f.d.s.a. subspace. We set $\hat{\mathcal{E}} = \mathcal{E} + \text{span}[\mathcal{E}\mathcal{E}]$. By Remark 5.4, there is $\psi : \hat{\mathcal{E}} \to \ell_\infty(I; C_1)$ with
$\|\psi\| \le 1$ such that $Q\psi(x) = \pi(x)$ for any $x \in \hat{\mathcal{E}}$. Replacing $\psi$ by $(\psi + \psi_*)/2$ we may assume $\psi$ s.a.
Let $\psi_\alpha : \hat{\mathcal{E}} \to C_1$ be the coordinates of $\psi$ so that $\psi = (\psi_\alpha)$. Note $q^\sharp Q\psi(x) = q^\sharp\pi(x) = \nu\sigma(x)$ for
any $x \in \hat{\mathcal{E}}$. Equivalently, if we set $\nu_\alpha(x) = x$ for any $x \in C_1/\mathcal{I}$, we have $(q\psi_\alpha(x))_\alpha - (\nu_\alpha\sigma(x))_\alpha \in$
$c_0(I; C_1/\mathcal{I})$, which means $\limsup_\alpha \|q\psi_\alpha(x) - \sigma(x)\| = 0$ for any $x \in \hat{\mathcal{E}}$. A fortiori this holds for any
$x \in \mathcal{E}$. Now, since $\mathcal{E}\mathcal{E} \subset \hat{\mathcal{E}}$, if $x, y \in \mathcal{E}$ we have $Q(\psi(xy) - \psi(x)\psi(y)) = \pi(xy) - \pi(x)\pi(y) = 0$,
which means $\limsup_\alpha \|\psi_\alpha(xy) - \psi_\alpha(x)\| = 0$. Thus choosing $\alpha$ large enough (and invoking
the compactness of the unit ball of $\hat{\mathcal{E}}$) we find $\psi = \psi_\alpha$ satisfying the properties required to check
(i). $\square$

*Remark 5.7.* It is known that a general $q$ does not almost allow liftings. Indeed, if $q$ is the sur-
jection from $\mathcal{B}$ to the Calkin algebra and if $S \in \mathcal{B}$ is the shift then the subspace $\mathcal{E}$ spanned by
$\{q(1), q(S), q(S^*)\}$ does not satisfy the local lifting described in Definition 5.5. This can be checked
using the Fredholm index. According to D. Voiculescu who kindly explained it to me, this kind of
example was known around the time of the Brown-Douglas-Filmore theorem.

However, it is known that the associated surjective *-homomorphism $\hat{q} : C_0(C_1) \to C_0(C_1/\mathcal{I})$
(extended to the cone algebras) almost allows liftings. This is the content of Lemma 6.1 below.
For a general $q$, Remark 5.4 gives us only a linear isometric embedding $\pi$.

# 6 The cone algebra

Let $C_0 = C((0,1])$ and $C = C([0,1])$. For any $C^*$-algebra $A$, we denote by $C_0(A) = C_0 \otimes_{\min} A$ the
so-called cone algebra of $A$. When dealing with a mapping $u : A \to B$ between $C^*$-algebras (or
operator spaces) we will denote by $u_0 : C_0(A) \to C_0(B)$ the map extending $\text{Id}_{C_0} \otimes u$. We denote
by $\ell_\infty(A)$ the $C^*$-algebra formed of all bounded sequences of elements of $A$, and by $c_0(A) \subset \ell_\infty(A)$ the ideal formed by the sequences that tend to 0.

Let $q: C \to B$ be a surjective $*$-homomorphism and let $\mathcal{I} = \ker(q)$. Let $(\sigma_n)$ be a quasicentral approximate unit in $\mathcal{I}$. Our construction would be much simpler of we could find $(\sigma_n)$ (and hence $(1 - \sigma_n)$) formed of projections. Then the mappings $x \mapsto (1 - \sigma_n)x$ would be approximatively multiplicative. The next lemma somehow produces a way to go around that difficulty by passing to the cone algebras. It is closely related to Kirchberg's [10, §5], but we learnt it from [5, Lemma 13.4.4]. A similar idea already appears in [6, Lemma 10] for the suspension algebra in the context of approximatively multiplicative families indexed by a continuous parameter in $(0, \infty)$.

**Lemma 6.1.** Let $q: C \to B$ be a surjective $*$-homomorphism between $C^*$-algebras. Let $q_0: C_0(C) \to C_0(B)$ be the associated one on $C_0(C) = C_0 \otimes_{\min} C$. Then $q_0$ almost allows liftings.

*Proof.* Let $\mathcal{I} = \ker(q)$. Let $(\sigma_\alpha)$ be a net forming a quasicentral approximate unit of $\mathcal{I}$, indexed by a directed set $I$. This means $\sigma_\alpha \ge 0$, $\|\sigma_\alpha\| \le 1$, $\|\sigma_\alpha x - x\| \to 0$ for any $x \in \mathcal{I}$ and $\|\sigma_\alpha c - c\sigma_\alpha\| \to 0$ for any $c \in C$. We identify $C_0(C) = C_0 \otimes_{\min} C$ with the set of $C$ valued functions $f: [0, 1] \to C$ such that $f(0) = 0$. Note that

$$C_0(C)/C_0(\mathcal{I}) = C_0(C/\mathcal{I}) = C_0(B).$$

The set of polynomials $\mathcal{P}_0 = \operatorname{span}[t^n | n > 0]$ is dense in $C_0$. Let $\rho_\alpha: C_0 \otimes C \to C_0(C)$ be the map taking $t \mapsto f(t)c$ ($f \in C_0, c \in C$) to $t \mapsto f(t(1 - \sigma_\alpha))c$. For instance (monomials) $\rho_\alpha$ takes $t \mapsto t^n c$ to $t \mapsto t^n(1 - \sigma_\alpha)^n c$. Note $\|1 - \sigma_\alpha\| \le 1$ therefore for any $f \in C_0$ the function $t \mapsto f(t(1 - \sigma_\alpha))$ is in $C_0(C)$ with norm $\|f\|_{C_0}$. This shows that $\sup_\alpha \|\rho_\alpha(y)\| < \infty$ for any $y \in C_0 \otimes C$, so that $(\rho_\alpha)$ defines a map $\rho: C_0 \otimes C \to \ell_\infty(C_0(C))$. Let

$$\mathcal{L}(C_0(C)) = \ell_\infty(C_0(C))/c_0(C_0(C)),$$

and similarly for $C_0(B)$. Let $Q_C: \ell_\infty(C_0(C)) \to \mathcal{L}(C_0(C))$ be the quotient map and similarly for $Q_B$. Since $\sigma_\alpha \in \mathcal{I}$ we have for all $y \in \mathcal{P}_0 \otimes C$ (and hence all $y \in C_0 \otimes C$)

$$(6.1) \qquad \limsup_\alpha \|q_0\rho_\alpha(y) - q_0(y)\| = 0.$$

Since $\sigma_\alpha$ is quasicentral we have

$$\forall x, y \in C_0 \otimes C \quad \limsup_\alpha \|\rho_\alpha(xy) - \rho_\alpha(x)\rho_\alpha(y)\| = 0,$$

indeed this reduces to the case of monomials which is obvious, and also

$$\limsup_\alpha \|\rho_\alpha(x)^* - \rho_\alpha(x^*)\| = 0.$$

It follows that after composing $\rho$ by the quotient map $Q_C: \ell_\infty(C_0(C)) \to \mathcal{L}(C_0(C))$ we obtain a map $\hat{\pi}: C_0 \otimes C \to \mathcal{L}(C_0(C))$ which is a $*$-homomorphism, such that $\hat{\pi} = Q_C\rho$ on $C_0 \otimes C$. Since $C_0$ is nuclear, this extends to a $*$-homomorphism (still denoted by $\hat{\pi}$) defined on the whole of $C_0(C)$, whence $\hat{\pi}: C_0(C) \to \mathcal{L}(C_0(C))$. We have

$$(6.2) \qquad \forall f \in C_0 \otimes \mathcal{I} \quad \limsup_\alpha \|\rho_\alpha(f)\| = 0,$$

and hence $\hat{\pi}(C_0 \otimes \mathcal{I}) = Q_C\rho(C_0 \otimes \mathcal{I}) = 0$. Therefore, after passing to the quotient by $C_0 \otimes \mathcal{I} \subset \ker(\hat{\pi})$, we derive from $\hat{\pi}$ a $*$-homomorphism

$$\pi: C_0(B) \to \mathcal{L}(C_0(C)),$$
such that $\hat{\pi} = \pi q_0$. Let $q^\sharp : \mathcal{L}(C_0(C)) \to \mathcal{L}(C_0(B))$ denote again the quotient map associated to $q_0$. We claim that for any $b = q_0(y) \in C_0 \otimes B$ (with $y \in C_0 \otimes C$)) we have $q^\sharp\pi(b) = \nu(b)$ where $\nu(b)$ is as earlier the element of $\mathcal{L}(C_0(B))$ that is the equivalent class of the family $(b_\alpha)$ defined by $b_\alpha = b$ for all $\alpha$. Indeed, denoting simply by $\mathrm{Id} \otimes q_0 : \ell_\infty(C_0(C)) \to \ell_\infty(C_0(B))$ the coordinate wise extension of $q_0$, we have $q^\sharp Q_C = Q_B(\mathrm{Id} \otimes q_0)\rho(y) = Q_B([q_0(y)])$ where $[q_0(y)]$ denotes the constant family equal to $q_0(y)$, so that $Q_B([q_0(y)]) = \nu(q_0(y))$. Thus $q^\sharp\pi(b) = q^\sharp\pi q_0(y) = q^\sharp\hat{\pi}(y) = q^\sharp Q_C \rho(y) = Q_B(\mathrm{Id} \otimes q_0)\rho(y) = \nu(q_0(y)) = \nu(b)$. This completes the proof by Proposition 5.6. $\square$

**Lemma 6.2.** Let $C, C_1, B$ be $C^*$-algebras. We assume given an injective (and hence isometric) $*$-homomorphism $i : C \to B$ and a surjective one $q : C_1 \to B$. Thus we have $B = C_1/\mathcal{I}$ with $\mathcal{I} = \ker(q)$. Assume that $C$ has the LLP and that $q$ almost allows liftings. Then for any pair of f.d.s.a. spaces $E \subset C$ and $\mathcal{E} \subset B$ such that $i(E) \subset \mathcal{E}$ and for any $\varepsilon > 0$ there is an $\varepsilon$-morphism $\psi : \mathcal{E} \to C_1$ such that $\|\psi i_{|E}\|_{cb} \le 1 + \varepsilon$ that approximately lifts $q$ on $\mathcal{E}$ in the sense that $\|q\psi(x) - x\| \le \varepsilon\|x\|$ for any $x \in \mathcal{E}$.

<figure><img src="image_5.png" /></figure>

*Proof.* Let $(\psi_\alpha)$ and $\pi$ be the maps in (5.3) for the case $\sigma = \mathrm{Id}_B$. By the definition of $\pi$, we have $\|q\psi_\alpha(x) - x\| \le \varepsilon_\alpha\|x\|$ for any $x \in \mathcal{E}_\alpha$ (here $\alpha = (\mathcal{E}_\alpha, \varepsilon_\alpha)$). Let $\chi : \ell_\infty(I; C_1) \to \ell_\infty(I; C_1)/c_0(I; C_1)$ denote the quotient map. By the LLP of $C$ we can lift $\pi i_{|E} : I$ this gives us a map $u = (u_\alpha) : E \to \ell_\infty(C_1)$ with $\|u\|_{cb} = \sup_\alpha \|u_\alpha\|_{cb} \le 1$ such that $\chi u(x) = \pi i(x)$ for all $x \in E$. This means that

$$ \forall x \in E \quad \limsup_\alpha \|u_\alpha(x) - \psi_\alpha i(x)\| = 0. $$

Since $\dim(E) < \infty$, it follows that $\limsup_\alpha \|u_\alpha - \psi_\alpha i_{|E}\|_{cb} = 0$ and hence $\limsup_\alpha \|\psi_\alpha i_{|E}\|_{cb} \le 1$. Thus the lemma follows by taking $\psi = \psi_\alpha$ for $\alpha$ sufficiently “large” in the net. $\square$

# 7 Main construction

Recall $\mathcal{C} = C^*(\mathbb{F}_\infty)$ and $\mathcal{B} = B(\ell_2)$. By the universality of $\mathcal{B}$, there is an embedding $\mathcal{C} \subset \mathcal{B}$. It is known (see [10, Lemma 2.4]) that any separable $C^*$-algebra embeds in a separable one with the WEP. Thus there is a separable $C^*$-algebra $B$ with the WEP such that $\mathcal{C} \subset B$. Let $i : \mathcal{C} \to B$ be an embedding (i.e. a faithful $*$-homomorphism). Let $q : \mathcal{C} \to B$ be a surjective $*$-homomorphism. The relevant diagrams are as follows:

<figure><img src="image_6.png" /></figure>

We will now tackle the lifting problem expressed by these diagrams: we will locally lift $\mathrm{Id}_{C_0(B)}$ by an approximatively multiplicative map using Lemma 6.2. To shorten the notation we set

$$ \mathcal{L} = \ell_\infty(C_0(\mathcal{C}))/c_0(C_0(\mathcal{C})). $$
We denote by $Q: \ell_{\infty}(C_0(\mathscr{C})) \to \mathcal{L}$ the quotient map so that, as before, we have

$$ \forall x = (x_n) \in \ell_{\infty}(C_0(\mathscr{C})) \quad \|Q((x_n))\|_{\mathcal{L}} = \limsup \|x_n\|_{C_0(\mathscr{C})}. $$

The next lemma (for a single $u: S \to E$) is the basic step.

**Lemma 7.1.** Let $F, E \subset C_0(\mathscr{C})$ be f.d.s.a. subspaces such that $F.F \subset E$. Fix $n$ and $\varepsilon > 0$. There is a f.d.s.a. subspace $E_1 \subset C_0(\mathscr{C})$ and a s.a. map $T: E \to E_1$ such that

(i) For any subspace $S \subset \ell_1^n$ and any $u$ in $CB(S, E)$ there is $\tilde{u}: \ell_1^n \to E_1$ such that

$$ \tilde{u}|_S = Tu \text{ and } \|\tilde{u}\|_{cb} \le (1+\varepsilon)\|u\|_{cb}. $$

(ii) $\|T\|_{cb} \le 1 + \varepsilon$ and $\|T|_{T(E)}^{-1}\|_{cb} \le 1 + \varepsilon$.

(iii) $T(F)T(F) \subset E_1$ and for all $x, y \in F$ we have $\|T(xy) - T(x)T(y)\| \le \varepsilon\|x\|\|y\|$.

*Proof.* We first show that to check (i) for a fixed $\varepsilon > 0$, it suffices to check it with $\varepsilon$ replaced by (say) $\varepsilon/4$ for a suitably chosen finite set of subspaces $S \subset \ell_1^n$ and a suitable finite set of $u$'s in the unit ball of $CB(S, E)$. Since $n$ is fixed the set of $k$-dimensional subspaces $S \subset \ell_1^n$ can be viewed as being compact for the Hausdorff distance, so that it admits a finite $\delta$-net for any $\delta > 0$. In other words, by perturbation, to obtain (i) for a given $\varepsilon > 0$, it suffices to check (i) with $\varepsilon$ replaced by $\varepsilon/2$ for a suitably chosen finite set of subspaces $S \subset \ell_1^n$. Then $S$ and $\varepsilon > 0$ being fixed, since the unit ball of $CB(S, E)$ is also compact, to show (i) with $\varepsilon$ replaced by $\varepsilon/2$ for all $u$ in the unit ball of $CB(S, E)$ it suffices to show (i) with $\varepsilon$ replaced by $\varepsilon/4$ for only a suitably chosen finite set of $u$'s in it.

Being now left with a finite set $(u_i)_{1 \le i \le N}$ of $u$'s to extend, we may observe that we may handle these simply one after the other: having handled the first $u_1$ by producing $\tilde{u}_1$ and $T_1: E \to E_1$ we replace $F$ by $T_1(F)$, $u_2: S \to E$ by $T_1u_2: S \to E_1$ and we apply the same procedure to it, and so on. Since the number $N$ of $u$'s is fixed, and $N\varepsilon$ can be chosen arbitrary small, after $N$ steps, we will have $\|T_1 \cdots T_N\| \le (1+\varepsilon)^N$ and the same for the inverses, and taking (5.2) into account to tackle (iii), we obtain the announced result.

Thus it suffices to prove Lemma 7.1 for a fixed $S$ and a single $u: S \to E$. By homogeneity we may assume $\|u\|_{cb} = 1$. Since $B$ and hence $C_0(B)$ has the WEP, the map $i_0u: S \to C_0(B)$ admits an extension $v: \ell_1^n \to C_0(B)$ with $\|v\|_{cb} \le (1+\varepsilon)\|i_0u\|_{cb} = (1+\varepsilon)\|u\|_{cb} = 1 + \varepsilon$. We have $v|_S = i_0u$. Let us choose a f.d.s.a. subspace $\mathcal{E} \subset C_0(B)$ large enough so that $i_0(E) + v(\ell_1^n) \subset \mathcal{E}$. By Lemma 6.2, for any $0 < \varepsilon' < \varepsilon$ there is an $\varepsilon'$-morphism $\psi: \mathcal{E} \to C_0(\mathscr{C})$ with $\|\psi i_{0|E}\|_{cb} \le 1 + \varepsilon'$ such that $\|q_0\psi(x) - x\| \le \varepsilon'\|x\|$ for any $x \in \mathcal{E}$. We then set

$$ E_1 = \psi(\mathcal{E}) + \psi i_0(F)\psi i_0(F) \text{ and } T = \psi i_{0|E}: E \to E_1. $$

Note $T(F)T(F) \subset E_1$. Moreover, we define $\tilde{u}: \ell_1^n \to E_1$ by $\tilde{u}(z) = \psi v(z)$ for all $z \in \ell_1^n$. Then $\tilde{u}|_S = Tu$, $\|\tilde{u}\|_{cb} \le (1+\varepsilon')(1+\varepsilon)$ and $\|T\|_{cb} \le 1+\varepsilon'$. Since $\varepsilon'$ can be chosen arbitrarily small, we may assume $\varepsilon' < 1/\dim(\mathcal{E})$. Let $\kappa: \mathcal{E} \to C_0(B)$ denote the inclusion map. Since $\|q_0\psi - \kappa\| \le \varepsilon'$ we have (see e.g. [19, p. 75]) $\|q_0\psi - \kappa\|_{cb} \le \varepsilon'\dim(\mathcal{E}) < 1$. Similarly, $\|q_0T - \kappa i_{0|E}\|_{cb} \le \dim(E)\|q_0T - \kappa i_{0|E}\| \le \varepsilon'\dim(E) < 1$, and hence $\|T(e)\| \ge \|q_0T(e)\| \ge \|\kappa i_0(e)\| - \varepsilon'\dim(E)\|e\| = (1 - \varepsilon'\dim(E))\|e\|$ for any $e \in E$. This shows $\|T|_{T(E)}^{-1}\| \le (1 - \dim(E)\varepsilon')^{-1}$. A similar reasoning shows that $\|T|_{T(E)}^{-1}\|_{cb} \le (1 - \dim(E)\varepsilon')^{-1}$ (a variant of this reasoning actually shows $\|\psi|_{\psi(\mathcal{E})}^{-1}\|_{cb} \le (1 - \dim(\mathcal{E})\varepsilon')^{-1}$).

It only remains to adjust $\varepsilon$ and $\varepsilon'$ to match (i) and (ii) as stated above. Lastly, (iii) follows since $\psi$ is an $\varepsilon'$-morphism on $\mathcal{E}$ and $i_0$ an isometric $*$-homomorphism. $\square$
**Theorem 7.2.** Let $(Z_n)$ be a sequence of finite dimensional self-adjoint (f.d.s.a.) (operator) subspaces of $C_0(\mathscr{C})$. There is a non-nuclear separable $C^*$-algebra $A$ with both the WEP and the LLP. Moreover, for any $n$ and any $\varepsilon > 0$ there is a subspace $Z \subset A$ such that $d_{cb}(Z_n, Z) < 1 + \varepsilon$.

**Remark 7.3.** Since the WEP and the LLP pass to the unitization the unitization of $A$ has the properties in Theorem 7.2. Moreover, by a result due to Voiculescu (see e.g. [5, p. 251]) for any separable $C^*$-algebra $A$, the cone $C_0(A)$ is quasidiagonal. Thus replacing $A$ by the unitization of $C_0(A)$ we obtain a unital quasidiagonal example as in Theorem 7.2.

**Remark 7.4.** It is easy to deduce from (3.2) that for any nuclear $C^*$-algebra $C$, the algebra $\mathscr{C}$ is locally equivalent to $C \otimes_{\min} \mathscr{C}$. In particular $\mathscr{C}$ is locally equivalent to $C_0(\mathscr{C})$.

**Remark 7.5.** Since the set of f.d. subspaces of $C_0(\mathscr{C})$ is $d_{cb}$-separable if we choose for $(Z_n)$ a dense sequence, then for any f.d. subspace $Z'$ in $C_0(\mathscr{C})$ (or in $\mathscr{C}$) and any $\varepsilon > 0$ there is a subspace $Z \subset A$ such that $d_{cb}(Z', Z) < 1 + \varepsilon$. In the converse direction, by Remark 3.6 the LLP of $A$ implies that $d_{S\mathscr{C}}(A) = 1$. Thus, with this choice of $(Z_n)$, $A$ and $\mathscr{C}$ are locally equivalent.

**Lemma 7.6.** Let $(Z_n)$ be as in Theorem 7.2. Let $\varepsilon_n > 0$ be such that $\sum \varepsilon_n < \infty$. There is a sequence of f.d.s.a. subspaces $E_n \subset C_0(\mathscr{C})$ and s.a. maps $T_n : E_n \to E_{n+1}$ such that we have for any $n \ge 1$

(i) $\forall S \subset \ell_1^n, \forall u : S \to E_n \exists \tilde{u} : \ell_1^n \to E_{n+1}$ such that

$$ \tilde{u}_{|S} = T_n u \text{ and } \|\tilde{u}\| \le (1 + \varepsilon_n) \|u\|_{cb}. $$

(ii) $\|T_n\|_{cb} \le 1 + \varepsilon_n$ and $\|T_n^{-1}|_{T_n(E_n)}\|_{cb} \le 1 + \varepsilon_n$.

(iii) For any $n \ge 2$ we have $T_{n-1}(E_{n-1})T_{n-1}(E_{n-1}) \subset E_n$ and

$$ (7.1) \qquad \forall x, y \in T_{n-1}(E_{n-1}) \quad \|T_n(x)T_n(y) - T_n(xy)\| \le \varepsilon_n \|x\| \|y\|. $$

(iv) For any $n \ge 2$ we have $Z_n \subset E_n$.

*Proof.* We construct $E_n, T_n$ by induction on $n$ starting from an arbitrary $E_1 = Z_1$. At the initial step $n = 1$ (i) is trivial and (iii) is void so that we simply may set $E_2 = E_1 + Z_2 + \text{span}[E_1, E_2]$ and let $T_1 : E_1 \to E_2$ be the natural inclusion. We have the required properties with $\varepsilon_1 = 0$. Assume that $(E_k)_{k \le n}$ and $(T_k)_{k < n}$ have been constructed satisfying (i) (ii) (iii) (iv). For the induction step we must produce $E_{n+1}$ and $T_n$. We find $T_n : E_n \to E_{n+1}$ using Lemma 7.1 applied to $E = E_n$ with $\varepsilon = \varepsilon_n$, and taking $F = T_{n-1}(E_{n-1})$. This gives us $E_{n+1}$ and $T_n : E_n \to E_{n+1}$ (equal to the $T$ given by Lemma 7.1) satisfying (i) (ii) and (7.1). But since (i) (ii) remain unchanged if we enlarge $E_{n+1}$, we may replace our subspace by $E_{n+1} + Z_{n+1} + \text{span}[T_n(E_n)T_n(E_n)]$ to ensure that $T_n(E_n)T_n(E_n) \subset E_{n+1}$ and $Z_{n+1} \subset E_{n+1}$, so that (iv) also holds at the next step. This completes the proof by induction. $\square$

*Proof of Theorem 7.2.* Let $\varepsilon_n > 0$ be such that $\sum \varepsilon_j < \infty$. Let $\delta_n = \sum_{j>n} \varepsilon_j$. Note that the infinite product $\prod_{j \ge 1} (1 + \varepsilon_j)$ converges. We define $\eta_n > 0$ by the equality

$$ (7.2) \qquad 1 + \eta_n = \prod_{j \ge n} (1 + \varepsilon_j), $$

so that $\eta_n \to 0$.

Let $(E_n)$ be as in Lemma 7.6. We will work in the ambient $C^*$-algebra $\mathcal{L} = \ell_\infty(C_0(\mathscr{C}))/c_0(C_0(\mathscr{C}))$, with quotient map $Q : \ell_\infty(C_0(\mathscr{C})) \to \mathcal{L}$. We denote by $L \subset \ell_\infty(C_0(\mathscr{C}))$ the subspace formed of the sequences $(x_n)$ such that $x_n \in E_n$ for all $n$, so that $L \simeq \ell_\infty(\{E_n\})$.
We introduce a mapping $\theta_n : E_n \to \ell_\infty(C_0(\mathscr{C}))$, with values in $L$, defined by

$$ \forall x \in E_n \quad \theta_n(x) = (0, \dots, 0, x, T_n(x), T_{n+1}T_n(x), T_{n+2}T_{n+1}T_n(x), \dots) $$

where $x$ stands at the $n$-th place.

By (ii) in Lemma 7.6, we have

$$ \forall n \ge 1 \quad \|\theta_n\|_{cb} \le 1 + \eta_n. $$

We define the subspace $Y_n \subset \mathcal{L}$ by setting $Y_n = Q\theta_n(E_n)$. Then $Y_n$ is a f.d.s.a. subspace of $\mathcal{L}$ such that $Y_n \subset Y_{n+1}$ for all $n \ge 1$. Indeed, we have

$$ \forall x \in E_n \quad Q\theta_n(x) = Q\theta_{n+1}(T_n(x)), $$

because $\theta_n(x) - \theta_{n+1}(T_n(x))$ is a finitely supported element of $\ell_\infty(C_0(\mathscr{C}))$. Let

$$ A = \overline{\cup Y_n} \subset \mathcal{L}. $$

A priori this is a s.a. subspace. We will see below that it is actually a $C^*$-subalgebra.

We will first show that $d_{cb}(Y_n, E_n) \le \prod_{j \ge n} (1 + \varepsilon_j) = 1 + \eta_n$. Note first of all that the map $w_n = Q\theta_n : E_n \to Y_n$ satisfies $\|w_n\|_{cb} \le \|\theta_n\|_{cb} \le 1 + \eta_n$. By definition of $Y_n$ we know $Y_n = w_n(E_n)$. We claim that

$$ (7.3) \qquad \|w_n^{-1}|_{Y_n}\|_{cb} \le \prod_{j \ge n} (1 + \varepsilon_j). $$

Indeed, for any $x \in E_n$ we have $\|w_n(x)\| = \limsup_k \|T_{n+k} \cdots T_{n+1}T_n(x)\|$ and hence using (ii) in Lemma 7.6 and one more telescoping argument we find $\|w_n(x)\| \ge \prod_{j \ge n} (1 + \varepsilon_j)^{-1}\|x\|$. This shows $\|w_n^{-1}|_{Y_n}\| \le \prod_{j \ge n} (1 + \varepsilon_j)$. A simple modification gives us the same for the cb-norm, whence the claim.

Now we have a commuting diagram factorizing the inclusion $Y_n \subset Y_{n+1}$, as follows:

<figure><img src="image_7.png" /></figure>

Recall $\delta_n = \sum_{j>n} \varepsilon_j$. Let $c = \prod_{1}^{\infty} (1 + \varepsilon_j)$. To check that $A$ is a subalgebra of $\mathcal{L}$ we will show that

$$ (7.4) \qquad \forall a, b \in Y_n \quad d(ab, Y_{n+1}) \le (2c^3 + c^2)\delta_n(1 + \eta_n)^2 \|a\| \|b\|. $$

Assume $a = w_n(x) = w_{n+1}(T_n(x))$ and $b = w_n(y) = w_{n+1}(T_n(y))$. Then

$$ ab = Q\theta_{n+1}(T_n(x))Q\theta_{n+1}(T_n(y)) = Q(\theta_{n+1}(T_n(x))\theta_{n+1}(T_n(y)) $$

$$ = Q((0, \dots, 0, T_n(x)T_n(y), T_{n+1}T_n(x)T_{n+1}T_n(y), T_{n+2}T_{n+1}T_n(x)T_{n+2}T_{n+1}T_n(y), \dots)), $$

where $T_n(x)T_n(y) \in E_{n+1}$ stands at the $n+1$-place. We will compare $ab$ with $Q\theta_{n+1}(T_n(x)T_n(y))$. To lighten the notation, we set

$$ T_{n+k,n+1} = T_{n+k} \cdots T_{n+1}. $$

We have

$$ \|ab - Q\theta_{n+1}(T_n(x)T_n(y))\| = \overline{\lim_k} \|T_{n+k,n+1}T_n(x)T_{n+k,n+1}T_n(y) - T_{n+k,n+1}[T_n(x)T_n(y)]\|. $$
We claim that

$$ (7.5) \quad \|T_{n+k,n+1}T_n(x)T_{n+k,n+1}T_n(y) - T_{n+k,n+1}[T_n(x)T_n(y)]\| \le (2c^3+c^2)(\sum_{n<j \le n+k} \varepsilon_j)\|x\|\|y\|. $$

By homogeneity we may assume $\|x\| = \|y\| = 1$. For convenience observe $\|T_{n+k,n+1}T_n\| \le c$ and $\|T_n\| \le c$, whence the following trivial a priori bound for the left hand side of (7.5)

$$ (7.6) \quad \|T_{n+k,n+1}T_n(x)T_{n+k,n+1}T_n(y) - T_{n+k,n+1}[T_n(x)T_n(y)]\| \le 2c^3. $$

Let us check (7.5) by induction on $k$. The case $k=1$ is clear by (7.1). Assume (7.5) proved for a given $k$ and let us deduce the same for $k+1$. We will use an easy telescoping sum argument. Since $\|T_{n+k+1}\| \le 1 + \varepsilon_{n+k+1}$, taking (7.6) into account, we have by (7.5)

$$ \begin{aligned} & \|T_{n+k+1}[T_{n+k,n+1}T_n(x)T_{n+k,n+1}T_n(y)] - T_{n+k+1}T_{n+k,n+1}[T_n(x)T_n(y)]\| \\ & \le (2c^3+c^2)(\sum_{n<j \le n+k} \varepsilon_j) + 2c^3\varepsilon_{n+k+1}, \end{aligned} $$

while replacing $n$ by $n+k+1$ in (7.1) yields

$$ \|T_{n+k+1}T_{n+k,n+1}T_n(x)T_{n+k+1}T_{n+k,n+1}T_n(y) - T_{n+k+1}[T_{n+k,n+1}T_n(x)T_{n+k,n+1}T_n(y)]\| \le \varepsilon_{n+k+1}c^2 $$

and hence adding these last two inequalities we find

$$ \begin{aligned} & \|T_{n+k+1,n+1}T_n(x)T_{n+k+1,n+1}T_n(y) - T_{n+k+1,n+1}[T_n(x)T_n(y)]\| \\ & \le (2c^3+c^2)(\sum_{n<j \le n+k} \varepsilon_j + \varepsilon_{n+k+1}), \end{aligned} $$

which completes the induction, whence proving of the claim. From the claim we deduce

$$ \|ab - Q\theta_{n+1}(T_n(x)T_n(y))\| \le (2c^3+c^2)\delta_n\|x\|\|y\| $$

and (7.4) follows by (7.3) and (7.2). But now since $Y_n \subset Y_{n+k}$, (7.4) also implies

$$ (7.7) \quad \forall a, b \in Y_n \quad d(ab, Y_{n+k+1}) \le (2c^3+c^2)\delta_{n+k}(1+\eta_{n+k})^2\|a\|\|b\| \to 0, $$

and hence $ab \in \overline{\cup Y_n} = A$. Clearly the same conclusion holds for any $a, b \in \cup Y_n$, so that $A$ (which, as we already noticed, is s.a.) is a $C^*$-subalgebra of $\mathcal{L}$.

We will now show that $A$ has the WEP. By Proposition 2.1, this reduces to the following:

*Assertion 1.* Fix $n$ and let $u : S \to A$ with $S \subset \ell_1^n$ and $\|u\|_{cb} \le 1$. For any $\varepsilon > 0$ there is an extension of $u$ denoted by $\tilde{u} : \ell_1^n \to A$ such that $\tilde{u}|_S = u$ and $\|\tilde{u}\| \le 1 + \varepsilon$.

To check *Assertion 1* we may obviously assume by density that $u(S) \subset \cup Y_m$, or equivalently that $u(S) \subset Y_m$ for some $m \ge n$ that can be chosen as large as we wish. Note that we have a natural embedding $\ell_1^n \subset \ell_1^m$, with which any $S \subset \ell_1^n$ can be viewed without loss of generality as sitting in $\ell_1^m$, and for the map $v = w_m^{-1}u : S \to E_m$ we have $\|v\|_{cb} = \|w_m^{-1}u\|_{cb} \le 1 + \eta_m$. Taking this last remark into account, by (i) in Lemma 7.6 applied to $E_m$, after restricting the resulting map to $\ell_1^n$, we find a map $\tilde{v} : \ell_1^n \to E_{m+1}$ such that $\tilde{v}|_S = T_m v$ and

$$ \|\tilde{v}\| \le (1 + \varepsilon_m)\|v\|_{cb} \le (1 + \varepsilon_m)\|w_m^{-1}\|_{cb}\|u\|_{cb} \le (1 + \varepsilon_m)(1 + \eta_m). $$

Let $\tilde{u} = w_{m+1}\tilde{v} : \ell_1^n \to Y_{m+1}$. Then $\tilde{u}|_S = w_{m+1}T_m w_m^{-1}u = u$ and

$$ \|\tilde{u}\| \le \|w_{m+1}\|(1 + \varepsilon_m)\|w_m^{-1}\|_{cb}\|u\|_{cb} \le (1 + \eta_{m+1})(1 + \varepsilon_m)(1 + \eta_m)\|u\|_{cb} $$
Since $m$ can be chosen arbitrarily large and both $\varepsilon_m \to 0$ and $\eta_m \to 0$, we obtain Assertion 1. By Proposition 2.1, $A$ has the WEP.

By Remark 3.6, for any f.d. subspace $E \subset C$ of a $C^*$-algebra $C$ with LLP we have $d_{S\mathscr{C}}(E) = 1$. This holds in particular for $C = C_0(\mathscr{C})$. Since $d_{cb}(Y_n, E_n) \le \|w_n\|_{cb}\|w_n^{-1}\|_{cb} \le (1 + \eta_n)^2$ and $E_n \subset C_0(\mathscr{C})$ we have $d_{S\mathscr{C}}(Y_n) \le (1 + \eta_n)^2$ for all $n$. Since $A = \bigcup Y_n$ with $(Y_n)$ increasing, we have $d_{S\mathscr{C}}(Y_k) \le d_{S\mathscr{C}}(Y_n)$ for all $k \le n$ and hence $d_{S\mathscr{C}}(Y_k) = 1$ for all $k$. By perturbation this implies

$$ \forall E \subset A \quad d_{S\mathscr{C}}(E) = 1. $$

Since $A$ has the WEP, Proposition 3.7 implies that $A$ also has the LLP.

Lastly, since we have $Z_n \subset E_n$ for all $n$ there is $Z'_n \subset Y_n$ such that $d_{cb}(Z'_n, Z_n) \le d_{cb}(Y_n, E_n) \le (1 + \eta_n)^2$. This is not quite what is stated in Theorem 7.2. But if we arrange the sequence $(Z_n)$ so that each space in it is repeated infinitely many times, then for any given space $Z$ in the sequence $\{Z_n\}$ there will be $Z'_n \subset Y_n$ satisfying $d_{cb}(Z'_n, Z) \le (1 + \eta_n)^2$ for infinitely many $n$'s. Choosing $n$ large enough so that $(1 + \eta_n)^2 < 1 + \varepsilon$, we obtain the second part of Theorem 7.2. $\square$

# 8 Possible variants

1. We can avoid the use of the separable $C^*$-subalgebra $B \subset B(H)$ in our construction: we use $B(H)$ instead, a quotient map $C^*(\mathbb{F}) \to B(H)$ (for some large enough free group $\mathbb{F}$) and the fact that any separable $C^*$-subalgebra of $C^*(\mathbb{F})$ lies in a copy of $\mathscr{C}$ embedded in $C^*(\mathbb{F})$.

2. Using perturbation arguments, one could work with subspaces $E$ such that $E \subset P_0 \otimes \mathscr{C}$ where $P_0 = \operatorname{span}[t^m | m \ge 1] \subset C_0$ is the space of polynomials.

3. We can actually work with non s.a. subspaces, and impose an additional condition that $\{T_n(x)^* | x \in E_n\} \subset E_{n+1}$ together with $\|T_n(x)^* - T_n(x)^*\| \le \varepsilon_n \|x\|$ for any $x \in T_{n-1}(E_{n-1})$. We then will be able to conclude just the same that $A$ is s.a.

4. The construction works just as well if we use all subspaces of $\mathscr{C}$ instead of $\{\ell_1^n\}$, in the style of Remark 2.3 (with $C = \mathscr{C}$). More precisely, let $X_n \subset \mathscr{C}$ be an increasing family of f.d. subspaces with dense union. We may replace $S \subset \ell_1^n$ by $S \subset X_n$, and again study the extension problem of $u : S \to E$ by $\tilde{u} : X_n \to E_1$. This shows that, while using $\ell_1^n$ seems simpler, there is nothing special about it, except for the duality used in the proof of Proposition 2.1 and the fact that $\|v\| = \|v\|_{cb}$ for any $v$ defined on it.

5. By the main result described in [21] the following property of a $C^*$-algebra $A$ is sufficient (and necessary) for the WEP:

For any $n \ge 1$, any map $u : \ell_\infty^n \to A$ with $\|u\|_{cb} \le 1$ and any $\varepsilon > 0$, there are $a_j, b_j \in A$ such that $\sup_{1 \le j \le n} \|u(e_j) - a_j b_j\| \le \varepsilon$ and $\|\sum a_j a_j^*\|^{1/2} \|\sum b_j^* b_j\|^{1/2} \le 1 + \varepsilon$.

Indeed, this implies that $\|u\|_{dec} \le 1$ for any such $u$. One can use this criterion instead of the one in Proposition 2.1 to construct our main example.

6. In fact we can avoid the use of the preceding result using all the algebras $M_N(A)$ in place of $A$. Then we may restrict to $n = 3$. Indeed, by the criterion in [17], the pair $(A, \mathscr{C})$ is nuclear (i.e. $A$ has the WEP) if and only if for any $N \ge 1$ the algebra $M_N(A)$ satisfies the factorization in the preceding point 5, restricted to $n = 3$.

# 9 A more general viewpoint

Perhaps the most general way to describe the applicability of the preceding construction is as follows. We assume given an isometric $*$-homomorphism $i : \mathscr{C} \to B$ and a quotient $*$-homomorphism
q : C → B where C, B are separable C*-algebras and C is assumed to have the LLP. We assume that q almost allows liftings as defined above in Definition 5.5 (this automatically holds when we pass to the cone algebras). Suppose we are given a “suitable” (as defined next) property P of *-homomorphisms between C*-algebras. Then if the inclusion i : C → B satisfies P, we can construct a separable C*-algebra A with d<sub>SC</sub>(A) = 1 (and hence d<sub>S<sup>′</sup>C</sub>(A) = 1 by Remark 3.6) so that the identity of A satisfies that same property. Our goal in this section is to prove this in an even more general setting that we spell out in Theorem 9.2.

Let P be a property of *-homomorphisms σ : C → B between C*-algebras. We say that P is suitable if it is inherited by any σ₁ : C₁ → B₁ satisfying for some constant c the following local factorization through σ: for any f.d.s.a. subspaces Y ⊂ C₁, E₀ ⊂ B and ε > 0 there are f.d.s.a. subspaces E ⊂ C, E ⊂ B such that σ(E) ⊂ E with E ⊃ E₀ together with a map β : Y → E with ‖β‖<sub>cb</sub> ≤ c and an ε-morphism (in the sense of Definition 5.1) γ : E → B₁ such that σ₁|<sub>Y</sub> : Y → B₁ admits a factorization of the form

$$Y \xrightarrow{\beta} E \xrightarrow{\sigma|_E} E \xrightarrow{\gamma} B_1.$$

For instance if D is another C*-algebra, we may consider the property that Id<sub>D</sub> ⊗ σ extends to a contraction from D ⊗<sub>min</sub> C to D ⊗<sub>max</sub> B. This is an example of suitable property. The case D = C corresponds to the WEP. We give details on this in Corollary 9.5.

Another example of suitable property appears in the context of the similarity length in the sense of [19, p. 401]; let ‖·‖<sub>(d)</sub> be the norm on M<sub>n</sub>(B) appearing in [19, p. 401] when B is an arbitrary C*-algebra. We say that σ : C → B has P<sub>d</sub> if there is a constant K such that, for any n, any x ∈ M<sub>n</sub>(B) satisfies ‖(Id<sub>M<sub>n</sub></sub> ⊗ σ)(x)‖<sub>(d)</sub> ≤ K‖x‖<sub>M<sub>n</sub>(C)</sub>. A C*-algebra B is called of length d if its identity map satisfies P<sub>d</sub>. It is known (see [18, Cor. 6]) that this holds for B(H) with d = 3 (with K = 1). From this it is easy to check that there is a separable C*-algebra B containing C such that Id<sub>B</sub> satisfies P<sub>3</sub>. Using the latter we can find a C*-algebra A satisfying the properties in Theorem 7.2 and additionally of length 3.

**Remark 9.1.** [A general setup] To achieve the greatest generality we are led to consider the following situation. Let C<sub>n</sub>, B<sub>n</sub> be C*-algebras (n ≥ 0). Assume given, for each n ≥ 0, an isometric *-homomorphism i<sub>n</sub> : C<sub>n</sub> → B<sub>n</sub> and a surjective *-homomorphism q<sub>n</sub> : C<sub>n+1</sub> → B<sub>n</sub> that almost allows liftings. Let L = l<sub>∞</sub>({C<sub>n</sub>}), I<sub>0</sub> = c<sub>0</sub>({C<sub>n</sub>}) and L = L/I<sub>0</sub>. We assume given, for each n ≥ 1, a certain correspondence E ↦ (E[n, E], ε[n, E]) associating to a f.d.s.a. subspace E of C<sub>n</sub> a f.d.s.a. subspace E[n, E] of B<sub>n</sub> and a positive number ε[n, E] > 0. We also give ourselves a sequence of f.d.s.a. subspaces E<sup>0</sup><sub>n</sub> ⊂ C<sub>n</sub>. The condition E<sub>n</sub> ⊃ E<sup>0</sup><sub>n</sub> in the next statement should be interpreted as expressing that E<sub>n</sub> is “arbitrarily” large.

<figure><img src="image_8.png" /></figure>

In short we are considering a sequence of C*-algebras (C<sub>n</sub>) such that C<sub>n</sub> is a subquotient of C<sub>n+1</sub> for each n, and we assume that the quotient maps almost allow liftings.

The goal of the next theorem is to show that there exists a C*-algebra A that has the same asymptotic “local” properties as the sequence of maps i<sub>n</sub> : C<sub>n</sub> → B<sub>n</sub>. As before in §4, we construct our A as an inductive limit of operator spaces.

**Theorem 9.2.** In the situation described in Remark 9.1, given δ<sub>n</sub> > 0 with δ<sub>n</sub> → 0, if all the C*-algebras C<sub>n</sub> have the LLP, there is a C*-subalgebra A ⊂ L and an increasing sequence of f.d.s.a.
subspaces $Y_n \subset A$ with $\overline{\cup_n Y_n} = A$ such that for each $n \ge 0$ there are f.d.s.a. subspaces $E_n \subset C_n$ and $\mathcal{E}_n \subset B_n$ such that

$$i_n(E_n) \subset \mathcal{E}_n \text{ and also } E_n \supset E_n^0 \text{ and } \mathcal{E}_n \supset \mathcal{E}[n, E_n],$$

for which the inclusion $Y_n \to Y_{n+1}$ admits a factorization of the following form

$$Y_n \xrightarrow{\beta_n} E_n \to \mathcal{E}_n \xrightarrow{\gamma_n} Y_{n+1}$$

where $\beta_n$ is a linear isomorphism satisfying

$$\max\{\|\beta_n\|_{cb}, \|\beta_n^{-1}\|_{cb}\} \le 1 + \delta_n,$$

and where $\gamma_n$ and $\beta_n^{-1}$ are $\delta_n$-morphisms into $\mathcal{L}$ while $E_n \to \mathcal{E}_n$ is the restriction of the embedding $i_n : C_n \to B_n$.

Moreover, we can ensure that, for each $n$, $\gamma_n$ is an $\varepsilon[n, E_n]$-morphism.

**Proof.** We will construct the sequence $(E_n, \mathcal{E}_n)$ by induction, starting from $E_0 = E_0^0, \mathcal{E}_0 = \mathcal{E}[0, E_0]$. The induction reasoning will additionally produce, for each $n$, a number $\varepsilon'_n > 0$ and an $\varepsilon'_n$-morphism $\psi_n : \mathcal{E}_n \to C_{n+1}$ such that $\psi_n(\mathcal{E}_n) \subset E_{n+1}$ (so that we may view $\psi_n$ as taking values in $E_{n+1}$), and $T_n = \psi_n i_{n|E_n} : E_n \to E_{n+1}$ will be such that $\max\{\|T_n\|_{cb}, \|T_n^{-1}|_{T_n(E_n)}\|_{cb}\} \le 1 + \varepsilon'_n$. Moreover, we will ensure that $\psi_n(\mathcal{E}_n)\psi_n(\mathcal{E}_n) + E_{n+1}^0 \subset E_{n+1}$ (so that a fortiori $T_n(E_n)T_n(E_n) + E_{n+1}^0 \subset E_{n+1}$). The number $\varepsilon'_n$ can be defined as

$$\varepsilon'_n = c2^{-n-1} \min\{1, \varepsilon[j, E_j], \delta_j \mid j \le n\}$$

where $c > 0$ is a numerical constant to be adjusted in the end. Note that

$$(9.1) \qquad \varepsilon'_n \le c2^{-n-1}, \quad \sum_n^\infty \varepsilon'_k \le c2^{-n}\varepsilon[n, E_n] \quad \text{and} \quad \sum_n^\infty \varepsilon'_k \le c2^{-n}\delta_n.$$

Assume $(E_n, \mathcal{E}_n)$ has been constructed as well as $(T_k, \psi_k, \varepsilon'_k)$ for $k < n$. Let $\psi_n : \mathcal{E}_n \to C_{n+1}$ be the $\varepsilon$-morphism (for an $\varepsilon$ to be specified) given by Lemma 6.2 and let

$$(9.2) \qquad E_{n+1} = \psi_n(\mathcal{E}_n) + \psi_n(\mathcal{E}_n)\psi_n(\mathcal{E}_n) + E_{n+1}^0,$$

and

$$\mathcal{E}_{n+1} = \mathcal{E}[n+1, E_{n+1}] + i_{n+1}(E_{n+1}) + i_{n+1}(E_{n+1})i_{n+1}(E_{n+1}).$$

Note $\psi_n(\mathcal{E}_n) \subset E_{n+1}$. We define $T_n : E_n \to E_{n+1}$ by $T_n = \psi_n i_{n|E_n}$. By Lemma 6.2 we have $\|T_n\|_{cb} \le 1 + \varepsilon$. Since $\|q_n\psi_n - Id\| \le \varepsilon$ we have also $\|q_nT_n - i_{n|E_n}\| \le \varepsilon$ and hence (see e.g. [19, p. 75]) $\|q_nT_n - i_{n|E_n} : E_n \to B_n\|_{cb} \le \varepsilon \dim(E_n)$. Since $i_n$ is isometric this gives us $\|T_n(x)\| \ge \|q_nT_n(x)\| \ge (1-\varepsilon)\|x\|$, and hence $\|T_n^{-1}|_{T_n(E_n)}\| \le (1-\varepsilon)^{-1}$ and similarly (assuming $\varepsilon \dim(E_n) < 1$) $\|T_n^{-1}|_{T_n(E_n)}\|_{cb} \le (1-\varepsilon \dim(E_n))^{-1}$. Thus, choosing $0 < \varepsilon < \varepsilon'_n$ small enough we can obtain $E_{n+1}, \mathcal{E}_{n+1}, \psi_n, T_n$ with all the required properties, to complete the induction reasoning.

Note that $T_n = \psi_n i_{n|E_n}$ is, just like $\psi_n$, an $\varepsilon'_n$-morphism into $C_{n+1}$.

Let $\theta_n : E_n \to \ell_\infty(\{C_n\})$ be defined by

$$(9.3) \qquad \theta_n(x) = (0, \dots, 0, x, T_n x, T_{n+1} T_n x, \dots)$$

with $x$ standing at the $n$-th place, and let $w_n : E_n \to \mathcal{L}$ be such that

$$(9.4) \qquad w_n = Q\theta_n,$$
where $Q: \ell_{\infty}(\{C_n\}) \to \mathcal{L}$ is the quotient map.

Let $\eta_n$ be such that $1 + \eta_n = \prod_n^\infty (1 + \varepsilon'_k)$. Note $\eta_n \to 0$. Since $\|T_k\|_{cb} \le 1 + \varepsilon'_k$ we have clearly

$$ \|w_n\|_{cb} \le \|\theta_n\|_{cb} \le 1 + \eta_n. $$

Moreover, since $\|T_k^{-1}|_{T_k(E_k)}\|_{cb} \le 1 + \varepsilon'_k$, we have for any $x \in E_n$

$$ \|w_n(x)\| = \limsup_k \|(T_{n+k} \cdots T_{n+1}T_n)(x)\| \ge (1 + \eta_n)^{-1}\|x\|, $$

and hence $\|w_n^{-1}|_{w_n(E_n)}\| \le 1 + \eta_n$. A similar reasoning applied to $x \in M_N(E_n)$ with $N$ arbitrary yields $\|w_n^{-1}|_{w_n(E_n)}\|_{cb} \le 1 + \eta_n$. We set $Y_n = w_n(E_n) \subset \mathcal{L}$. Note

$$ (9.5) \qquad \forall x \in Y_n \quad w_n(x) = w_{n+1}T_n(x). $$

Therefore $Y_n \subset Y_{n+1}$. Moreover, $Y_n$ is a f.d.s.a. subspace of $\mathcal{L}$. The proof that $A = \overline{\cup Y_n}$ is a $C^*$-subalgebra is entirely analogous to that of Theorem 7.2 so we skip it.

By (5.2) with (9.3) and (9.4), $\theta_n$, and hence also $w_n$, is a $\delta'_n$-morphism for some $\delta'_n \to 0$ with $\delta'_n \approx \sum_n^\infty \varepsilon'_k$. We set $\beta_n = w_n^{-1}: Y_n \to E_n$ and $\gamma_n = w_{n+1}\psi_n: \mathcal{E}_n \to Y_{n+1}$. Then by (5.2) again $\gamma_n$ is a $\delta''_n$-morphism for some $\delta''_n \to 0$ with $\delta''_n \approx \sum_n^\infty \varepsilon'_k$. We have $\gamma_n i_n \beta_n = w_{n+1}T_n w_n^{-1}$ and hence by (9.5) $\gamma_n i_n \beta_n(x) = x$ for any $x \in Y_n$. Lastly $\delta'_n$ and $\delta''_n$ being dominated (up to constant) by $\sum_n^\infty \varepsilon'_k$ as in (5.2), it is clear that our initial choice of the constant $c$ in the definition of $(\varepsilon'_n)$ can be adjusted small enough (see (9.1)) in order to have $\max\{\delta'_n, \delta''_n, \varepsilon'_n, \eta_n\} \le \delta_n$ and also $\delta''_n \le \varepsilon[n, E_n]$. The latter shows that $\gamma_n$ is a fortiori a $\varepsilon[n, E_n]$-morphism. This completes the proof. $\square$

**Remark 9.3.** If we drop the LLP assumption, we only obtain $\|w_n\| = \|\beta_n^{-1}\| \le 1 + \delta_n$.

The following diagram summarizes the preceding proof.

<figure><img src="image_9.png" /></figure>

**Remark 9.4.** In the general situation described in Remark 9.1, let $\mathcal{L}(\{C_n\}) = \ell_{\infty}(\{C_n\})/c_0(\{C_n\})$ and $\mathcal{L}(\{B_n\}) = \ell_{\infty}(\{B_n\})/c_0(\{B_n\})$. With the notation of the preceding proof, we will say (in the style of [3]) that $A \subset \mathcal{L}(\{C_n\})$ is the inductive limit of the system $\{E_n, T_n\}$, and we denote $A = A(\{E_n, T_n\})$. We define $T_n^b: \mathcal{E}_n \to \mathcal{E}_{n+1}$ by $T_n^b(x) = i_{n+1}\psi_n(x) (x \in \mathcal{E}_n)$. Then we have a $C^*$-subalgebra $A(\{\mathcal{E}_n, T_n^b\}) \subset \mathcal{L}(\{B_n\})$ that is similarly the inductive limit of the system $\{\mathcal{E}_n, T_n^b\}$. Let $Q^b: \ell_{\infty}(\{B_n\}) \to \mathcal{L}(\{B_n\})$, and $w_n^b: \mathcal{E}_n \to \ell_{\infty}(\{B_n\})$ be the analogues of $Q$ and $w_n$ for the system $\{\mathcal{E}_n, T_n^b\}$ Let $q^b: \mathcal{L}(\{C_n\}) \to \mathcal{L}(\{B_n\})$ and $i^b: \mathcal{L}(\{C_n\}) \to \mathcal{L}(\{B_n\})$ be the morphisms associated respectively to $(q_n)$ and $(i_n)$. Let $\sigma: A \to \mathcal{L}(\{C_n\})$ be the (inclusion) embedding. With this notation, we have $i^b\sigma = q^b\sigma$ and moreover $i^b(A) = q^b(A) = A(\{\mathcal{E}_n, T_n^b\})$. The details are easy to check using $\|q_{n+k}T_{n+k} - i_{n+k}|_{E_{n+k}}\| \le \varepsilon_{n+k}$ which shows that $i^b\sigma = q^b\sigma$ on $Y_n$, together with $i_{n+1}T_n - T_n^b i_n = 0$ on $E_n$ and $T_{n+k} \cdots T_{n+1}T_n^b - T_{n+k}^b \cdots T_{n+1}^b T_n^b = 0$ on $\mathcal{E}_n$.

The $C^*$-algebra

$$ \mathcal{A} = \{x \in \mathcal{L}(\{C_n\}) \mid q^b(x) = i^b(x)\} = \{Q((x_n)) \mid (x_n) \in \ell_{\infty}(\{C_n\}), q_n(x_{n+1}) = i_n(x_n) \forall n\}, $$
which contains our A as a $C^*$-subalgebra, should probably be viewed as the joint inductive/projective limit of the system of subquotients ($C_n$).

<figure><img src="image_10.png" /></figure>

**Corollary 9.5.** In the preceding situation, assume $(C_n, B_n, i_n, q_n) = (C, B, i, q)$ for all $n \in \mathbb{N}$. Assume C separable. Then for any separable $C^*$-algebra D such that $i \otimes Id_D : C \otimes_{\min} D \to B \otimes_{\max} D$ is continuous (and hence isometric) we can obtain a separable $C^*$-algebra A such that $(A, D)$ is a nuclear pair, and moreover such that A and C are locally equivalent (see Definition 3.4).

*Proof.* This follows from the theorem by Lemma 5.3. We may assume $D = \bigcup F_n$ for an increasing sequence of f.d.s.a. subspaces $F_n \subset D$, and similarly $C = \bigcup \mathcal{F}_n$ for an increasing sequence of f.d.s.a. subspaces $\mathcal{F}_n \subset C$. By Lemma 5.3 applied to $i(E) \otimes F_n$, we can select a number $\varepsilon[n, E] > 0$ and a f.d.s.a. subspace $\mathcal{E}[n, E] \subset B$ containing $i(E)$ associated to $\delta = 1/n$ according to Lemma 5.3.

Let $x \in Y_n \otimes D_n$. We then apply the theorem with $\{E_n^0\} = \{\mathcal{F}_n\}$ where we make sure that each $\mathcal{F}_n$ is repeated infinitely many times in the sequence $\{E_n^0\}$. Since $i: C \to B$ transforms min-norms to max-norms, we have

$$ \|(i\beta_n \otimes Id_D)(x)\|_{B \otimes_{\max} D} \le \|(\beta_n \otimes Id_D)(x)\|_{C \otimes_{\min} D} \le (1+\delta_n)\|x\|_{A \otimes_{\min} D}. $$

Since $\gamma_n$ is an $\varepsilon[n, E_n]$-morphism and $(\gamma_n \otimes Id_D)(i\beta_n \otimes Id_D)(x) = x$ we then obtain

$$ \forall x \in Y_n \otimes D_n \quad \|x\|_{A \otimes_{\max} D} \le (1+1/n) \|(i\beta_n \otimes Id_D)(x)\|_{B \otimes_{\max} D} \le (1+1/n)(1+\delta_n) \|x\|_{A \otimes_{\min} D}. $$

Since $n$ can be chosen arbitrarily large this implies that $(A, D)$ is a nuclear pair. To show that $C$ locally embeds in $A$ it suffices to show by perturbation that each $\mathcal{F}_n$ locally embeds in $A$. Then since the inclusions $\mathcal{F}_n = E_m^0 \subset E_m \subset C$ are valid for infinitely many $m$'s and $E_m$ is completely $(1 + \delta_m)^2$-isomorphic to $Y_m$, we conclude that $C$ locally embeds in $A$. Since $Y_n$ is completely $(1 + \delta_n)$-isomorphic to $E_n \subset C$, we know that $A$ is locally embeds in $C$. $\square$

*Alternate proof of Theorem 7.2.* Let $B$ be as before a separable $C^*$-algebra with WEP containing $C$ as a $C^*$-subalgebra. Then the inclusion $i: C \to B$ satisfies

$$ \|i \otimes Id_C : C \otimes_{\min} C \to B \otimes_{\max} C\| = 1 $$

simply because the identity of $B$ satisfies this by definition of the WEP. We apply Corollary 9.5 with $C_n = C_0(C)$ and $B_n = C_0(B)$ for all $n \ge 1$ with $D = C$. Lemma 6.1 shows that the corresponding $q_n$'s almost allow liftings. By Corollary 9.5 the resulting $A$ has the WEP and locally embeds in $C_0(C)$ (which has the LLP) and hence in $C$. It follows that $A$ has the LLP by Lemma 3.3. $\square$

*Remark 9.6.* For $C^*$-algebras $A, C$ the property that $A$ locally embeds in $C$ implies that $A$ embeds in an ultrapower of $C$. In general the converse does not hold. However it does if $A$ has the LLP. Thus it does not significantly weaken Theorem 7.2 if we just say that each of $A$ and $C$ embeds in an ultrapower of the other.

*Acknowledgement.* I thank the referee for his careful reading.
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