# On C* -algebras with Local Lifting Property and Weak Expectation Property

Gilles Pisier

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On $C^*$-algebras with Local Lifting Property and Weak Expectation Property

by

Gilles Pisier*

Sorbonne Université

and

Texas A&M University

November 3, 2025

Abstract

We give a new, somewhat simpler, presentation of the author's recent construction of a non-nuclear $C^*$-algebra which has both the local lifting property (LLP) and the weak expectation property (WEP).

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

# 1 Introduction

The main goal of this note is to give a hopefully simpler proof that there exist non-nuclear $C^*$-algebras with both the local lifting property (LLP) and the weak expectation property (WEP). (See [9] for operator spaces with analogous properties). We will obtain this goal in Theorem 4.8. While this paper is mostly expository, the new approach produces some “new” results, e.g. in Remarks 3.13 and 3.14.

We start by an outline of our main results.

We first introduce some basic terminology and notation.

Let $A, B$ be $C^*$-algebras and let $E \subset A$ be an operator subspace. Given a $C^*$-algebra $D$, a linear map $u: E \to B$ will be called $D$-nuclear with constant $c$ if

$$ \|Id_D \otimes u : D \otimes_{\min} E \to D \otimes_{\max} B\| \le c. $$

If this holds with $c = 1$ we say that $u$ is $D$-nuclear.

This terminology is inspired by the well known definition of nuclear $C^*$-algebras: $A$ is nuclear if the identity on $A$ is $D$-nuclear for all $D$.

We first need some basic notation.

For any $C^*$-algebra $C$ we denote

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

<footnote>*ORCID 0000-0002-3091-2049</footnote>
For short we set $L = \ell_\infty(C)$ and $\mathcal{L} = \mathcal{L}(C)$. Let $Q: L \to \mathcal{L}$ denote the quotient map. For any $x = (x_n) \in L$ we have $\|Q(x)\| = \limsup_{n \to \infty} \|x_n\|$. Our basic starting point is an isometric $*$-homomorphism

$$F: C \to \mathcal{L}$$

together with a completely contractive self-adjoint lifting $f = (f_n): C \to L$. Lifting means here as usual that $F = Qf$ and self-adjoint means $f(x^*) = f(x)^*$ for all $x \in C$. The existence of such an $f$ is of course a nontrivial assumption, but it is guaranteed to hold if $C$ has the LP.<fnref n="1" /> Unfortunately we need to impose a more technical condition, as follows:

$$(1.1) \qquad \text{For any f.d. } E \subset C \quad \limsup \|f_n^{-1}|_{f_n(E)}\|_{cb} \le 1.$$

This means that the coordinates of the lifting $f = (f_n)$ are asymptotically locally almost completely isometric.

We will associate to $F$ two $C^*$-algebras $Z \subset L$ and $A \subset \mathcal{L}$. Let $E_n \subset C$ be f.d.s.a. subspaces and let $(m(n))$ be a sequence of integers. We set

$$T_n = f_{m(n)|_{E_{n-1}}}: E_{n-1} \to E_n,$$

assuming $\{E_n\}$ adjusted so that

$$\forall n \ge 0 \qquad T_n(E_{n-1}) + T_n(E_{n-1})T_n(E_{n-1}) \subset E_n.$$

For any $k > n$ let $T_{[k,n]} = T_k T_{k-1} \cdots T_n: E_{n-1} \to E_k$ and hence also $T_{[n,n]} = T_n$.

Let $Z_n \subset L$ be the set formed by all $x = (x_k) \in L$ such that $x_n \in E_n$ and $x_k = T_{[k,n+1]}(x_n)$ for all $k > n$. We set

$$Z = \overline{\cup_n Z_n} \subset L.$$

Clearly $x_k = 0$ for all $k \ge N$ implies $x \in Z_N$ and hence $c_0(C) \subset Z$. Moreover, it is easy to see that $Z_n \subset Z_{n+1}$. If each map $T_{[k,n]}$ was the restriction of a $*$-homomorphism it would follow easily that $Z_n Z_n \subset Z_{n+1}$. But in our setting the $T_n$'s will only be approximately multiplicative and if they are so at a sufficiently fast rate we still find that $Z = \overline{\cup Z_n}$ is a $C^*$-subalgebra of $L$. We then define the $C^*$-algebra $A = Q(Z)$ or equivalently

$$A = Z/c_0(C).$$

**Theorem 1.1.** If $C$ has the lifting property (LP in short), and if (1.1) holds, the spaces $(E_n)$ and the sequence $(m(n))$ can be selected so that $Z$ has the LP and $A$ the LLP. Moreover, if $D$ is a separable $C^*$-algebra for which $F: C \to \mathcal{L}$ is $D$-nuclear, they can be selected so that in addition $A$ is $D$-nuclear.

The proof of this theorem appears as a combination of statements in §3. See Propositions 3.6 (construction of the system), 3.8 ($Z$ has the LP), 3.10 ($A$ has the LLP) and Theorem 3.12 (about $D$-nuclearity).

Thus if we can produce an $F$ that is $\mathcal{C}$-nuclear (here $\mathcal{C} = C^*(\mathbb{F}_\infty)$) on a $C$ with the lifting property (LP in short) then we will obtain a $C^*$-algebra $A$ with both WEP and LLP. Indeed, in Kirchberg's viewpoint $A$ has WEP (resp. the LLP) iff $\mathrm{Id}_A$ is $\mathcal{C}$-nuclear (resp. $\mathcal{B}$-nuclear, where $\mathcal{B} = B(\ell_2)$). In addition if $C$ is non-exact and we can arrange (using the condition (1.1)) for the

<footnote n="1">¹Incidentally, when $C$ is separable, it is open whether $F$ always admits a contractive lifting.</footnote>
finite dimensional obstructions to the exactness of $C$ to be still present in $A$, this will guarantee that $A$ is not exact, and hence not nuclear.

The obvious question is then: for $D = \mathcal{C}$ or $\mathcal{B}$, how to produce nice examples of $D$-nuclear embeddings $F : C \to \mathcal{L}$ satisfying (1.1) ? This is provided by the setting of cone algebras, see §4.

In our initial paper [8] we used only the existence of a local lifting for our $F$, here the use of a global lifting simplifies somewhat the exposition. In particular, we will use freely the following characterization of the LP from [10] (see also [11]).

**Proposition 1.2.** Let $C$ be a $C^*$-algebra with LP. For any f.d. subspace $E \subset C$ there is $t^E$ in the unit ball of $\mathcal{C} \otimes_{\max} E$ (as defined below) such that for any other $t$ in the same unit ball there is a unital $*$-homomorphism $\pi_t : \mathcal{C} \to \mathcal{C}$ such that $[\pi_t \otimes \mathrm{Id}_E](t^E) = t$.

Let $C, B$ be $C^*$-algebras and let $E \subset C$ and $F \subset B$ be operator subspaces. We denote by $E \otimes F$ the algebraic tensor product. We denote (somewhat abusively) by $E \otimes_{\max} F$ the closure of $E \otimes F$ in $C \otimes_{\max} B$. We should emphasize that $E \otimes_{\max} F$ depends strongly on the ambient $C^*$-algebras $C, B$. We denote by $MB(E, F)$ the set of (linear) maps $u : E \to F$ such that $\mathrm{Id}_E \otimes u$ is bounded from $\mathcal{C} \otimes E$ equipped with the norm induced by $\mathcal{C} \otimes_{\max} C$ to $\mathcal{C} \otimes F$ equipped with the norm induced by $\mathcal{C} \otimes_{\max} B$. When this holds, we define

$$ (1.2) \qquad \|u\|_{mb} = \|\mathrm{Id}_E \otimes u : \mathcal{C} \otimes_{\max} E \to \mathcal{C} \otimes_{\max} F\|, $$

and we equip the space $MB(E, F)$ with this norm.

If $C$ has the LP, it is easy to check with Proposition 1.2 that

$$ (1.3) \qquad \|u\|_{mb} = \|[\mathrm{Id}_E \otimes u](t^E)\|_{\max}. $$

See Lemma 3.2 for full details.

Since any $C^*$-algebra $D$ is a quotient of the $C^*$-algebra of some free group, for any such $D$ we have (see [10, Prop. 1.1] for more details)

$$ (1.4) \qquad \|\mathrm{Id}_D \otimes u : D \otimes_{\max} E \to D \otimes_{\max} F\| \le \|u\|_{mb}. $$

**Some abbreviations:** We use c.b. (resp. c.c.) for completely bounded (resp. completely contractive), c.p. (resp. u.c.p.) for completely positive (resp. unital completely positive), and c.i. for completely isometric. Note that a “contractive” operator or a “contraction” is an operator of norm less than or equal to one. Similarly “completely contractive” or “complete contraction” means the cb-norm is less than or equal to one. We also abbreviate self-adjoint by s.a.

**Some background:** A $C^*$-algebra $C$ has the lifting property (LP) if any c.p.c.c. map into a quotient $C^*$-algebra admits a c.p.c.c. lifting. It is known that this holds for $A$ if and only if it does for the unitization of $A$. A unital $C^*$-algebra $C$ has the lifting property (LP) if and only if any u.c.p. map into a quotient $C^*$-algebra admits a u.c.p. lifting. Nuclear $C^*$-algebras have the LP (due to Choi-Effros) but also $\mathcal{C} = C^*(\mathbb{F}_\infty)$ has the LP (due to Kirchberg). See [10] and [3] for more background, precise references and more recent results on the LP.

# 2 Inductive systems

For lack of valuable examples, we cannot use for our lifting $f$ a bona fide $*$-homomorphism, but using instead an approximate $*$-homomorphism (or what we call an $\varepsilon$-morphism) in §4 we will find examples in the framework of cone algebras to which Theorem 1.1 can be applied to produce non-nuclear $C^*$-algebras with the WEP and the LLP. In this section we show by rather elementary arguments how our inductive limits of linear spaces can lead to $C^*$-algebras.
**Definition 2.1.** Let $C, C_1$ be $C^*$-algebras with s.a. subspaces $E \subset C$ and $E_1 \subset C_1$. Let $\varepsilon > 0$
A linear map $\psi : E \to E_1$ (we will restrict in (iii) to the s.a. case for simplicity) will be called an
$\varepsilon$-morphism if

$$
(i) \quad \|\psi\| \le 1 + \varepsilon,
$$

$$
(ii) \text{ for any } x, y \in E \text{ with } xy \in E \text{ we have } \|\psi(xy) - \psi(x)\psi(y)\| \le \varepsilon \|x\| \|y\|,
$$

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

Obviously the restriction to E of any *-homomorphism from C to C₁ is an ε-morphism for any
ε > 0.

Remark 2.2. Let $E_n \subset C_n$ be s.a. subspaces of $C^*$-algebras ($n \ge 0$). Let $\psi_n : E_{n-1} \to E_n$ be an $\varepsilon_n$-morphism. Assuming $\psi_1(E_0)\psi_1(E_0) \subset E_1$, the composition $\psi_2\psi_1 : E_0 \to E_2$ has norm $\le (1+\varepsilon_1)(1+\varepsilon_2)$ and hence satisfies (i) and (ii) with $\varepsilon = \delta_2$ for

$$
(2.1) \qquad \delta_2 = (1 + \varepsilon_2)\varepsilon_1 + \varepsilon_2(1 + \varepsilon_1)^2.
$$

This is immediate by the triangle inequality since

$$
\psi_2 \psi_1(xy) - \psi_2 \psi_1(x) \psi_2 \psi_1(y) = [\psi_2(\psi_1(xy) - \psi_1(x)\psi_1(y))] + [\psi_2(\psi_1(x)\psi_1(y)) - \psi_2\psi_1(x)\psi_2\psi_1(y)].
$$

More generally, assuming $\psi_n(E_{n-1})\psi_n(E_{n-1}) \subset E_n$ for all $n \ge 1$ then $\psi_n \cdots \psi_2\psi_1 : E_0 \to E_n$ has norm $\le (1+\varepsilon_n)\cdots(1+\varepsilon_2)(1+\varepsilon_1)$. Let $\pi_n = (1+\varepsilon_n)\cdots(1+\varepsilon_1)$. Repeating the argument for (2.1) shows that $\psi_n \cdots \psi_2\psi_1$ satisfies (ii) with $\varepsilon = \delta_n$ for $\delta_n$ satisfying for all $n \ge 2$ :

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

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

$$
\delta'_{n} = \delta'_{n-1} + \varepsilon_{n} \pi_{n-1}^{2} / \pi_{n} \le \delta'_{n-1} + \varepsilon_{n} \pi_{n-1}.
$$

Thus if we assume that $\sum \varepsilon_n < \infty$ so that $\sup_n \pi_n = c < \infty$ we find for all $n \ge 2$ :

$$
\delta'_{n} \le \delta'_{1} + c \sum_{2}^{n} \varepsilon_{k}
$$

and hence for any $n \ge 1$ (note $\delta_1 = \varepsilon_1 \ge \delta'_1$)

$$
(2.2) \qquad \delta_n \le \pi_n (\delta'_1 + c \sum_2^n \varepsilon_k) \le c\varepsilon_1 + c^2 \sum_2^n \varepsilon_k.
$$

In conclusion, we have

$$
(2.3) \qquad \delta_n \le e^2 \sum_{1}^{n} \varepsilon_k,
$$

and hence $\psi_n \cdots \psi_2 \psi_1$ satisfies (ii) with $\varepsilon = e^2 \sum_1^n \varepsilon_k$. In addition, we observe that if $\sum \varepsilon_k \le 1$ so that $c \le e$, then $\pi_n \le \exp \sum \varepsilon_k \le 1 + (e-1) \sum \varepsilon_k$, and hence $\psi_n \cdots \psi_2 \psi_1$ satisfies (i) with $\varepsilon = e^2 \sum_1^n \varepsilon_k$.

Thus we have proved the following:

**Lemma 2.3.** Let $\psi_k : E_{k-1} \to E_k$ be $\varepsilon_k$-morphisms with $\sum \varepsilon_k \le 1$. The composition $\psi_n \cdots \psi_2 \psi_1 : E_0 \to E_n$ is an $\varepsilon$-morphism with $\varepsilon = e^2 \sum_1^n \varepsilon_k$.
**Definition 2.4.** Let $C_n$ be $C^*$-algebras. By an admissible inductive system of o.s. we mean a sequence $(E_n, T_n)_{n \ge 0}$ where, for each $n$, $E_n \subset C_n$ is a f.d.s.a. operator subspace, the map $T_n : E_{n-1} \to E_n$ is a $\varepsilon_n$-morphism such that $T_n(E_{n-1})T_n(E_{n-1}) \subset E_n$ and $(\varepsilon_n)$ is such that

$$ \sum \varepsilon_n \le 1. $$

Let $L = \ell_\infty(\{C_n\}), \mathcal{I}_0 = c_0(\{C_n\})$ and $\mathcal{L} = L/\mathcal{I}_0$. Let $Q: L \to \mathcal{L}$ be the quotient map.

Recall that for any $x \in \mathcal{L}$ admitting $y = (y_n) \in L$ as a lift (i.e. $Q(y) = x$) we have

$$ (2.4) \qquad \|x\|_{\mathcal{L}} = \limsup \|y_n\|. $$

Let $\theta_n : E_n \to L$ be the mapping defined for $n \ge 0$ by

$$ (2.5) \qquad \forall x \in E_n \quad \theta_n(x) = (0, \dots, 0, x, T_{n+1}(x), T_{n+2}T_{n+1}(x), T_{n+3}T_{n+2}T_{n+1}(x), \dots) $$

where $x$ stands at the place of index $n$. The summability of $(\varepsilon_n)$ ensures that $\theta_n(x)$ is a bounded sequence.

Note that

$$ (2.6) \qquad \forall x \in E_n \quad \theta_n(x) - \theta_{n+1} T_{n+1}(x) \in c_0(\{C_n\}), $$

and hence $Q\theta_n = Q\theta_{n+1}T_{n+1}$. Let

$$ Z_n = C_0 \oplus \dots \oplus C_{n-1} \oplus \{(x, T_{n+1}(x), T_{n+2}T_{n+1}(x), T_{n+3}T_{n+2}T_{n+1}(x), \dots) \mid x \in E_n\} \subset L $$

or equivalently

$$ (2.7) \qquad Z_n = [C_0 \oplus \dots \oplus C_{n-1} \oplus 0 \oplus 0 \oplus \dots] + \theta_n(E_n) \subset L $$

and

$$ A_n = Q(Z_n). $$

Note that $(Z_n)$ and $(A_n)$ are non decreasing sequences. We set

$$ Z = \overline{\cup Z_n} \subset L \text{ and } A = \overline{\cup A_n} \subset \mathcal{L}. $$

**Proposition 2.5.** For any admissible inductive system as above, $A$ and $Z$ are $C^*$-subalgebras.

*Remark 2.6.* In the unital case, if the spaces $E_n$ and the maps $T_n$ are all unital the resulting algebras $Z$ and $A$ are clearly unital.

*Proof.* Note that the infinite product $\prod_{j \ge 1} (1 + \varepsilon_j)$ converges. We define $\eta_n > 0$ by the equality $1 + \eta_n = \prod_{j \ge n} (1 + \varepsilon_j)$ so that $\eta_n \to 0$.

A priori $Z = \overline{\cup Z_n} \subset L$. is a s.a. subspace. To check that $Z$ is actually a subalgebra of $L$ we will show that for all $n$ and all $a, b$ in $Z_n$ the product $ab$ lies in $Z$.

For $x \in E_n$ let

$$ (2.8) \qquad w_n(x) = (x, T_{n+1}(x), T_{n+2}T_{n+1}(x), \dots) \in \ell_\infty(\{C_n, C_{n+1}, \dots\}) $$

so that

$$ (2.9) \qquad \theta_n(x) = (0, \dots, 0, w_n(x)) $$
By Lemma 2.3, the maps $w_n$ and $\theta_n$ are $\varepsilon$-morphisms for $\varepsilon = e^2 \sum_{k>n} \varepsilon_k$.

Note that a typical element $a$ (resp. $b$) of $Z_n$ is of the form $a = (a_0, \cdots, a_{n-1}, w_n(a_n))$ (resp. $b = (b_0, \cdots, b_{n-1}, w_n(b_n))$). Then we also have $a = (a_0, \cdots, a_{n-1}, a_n, w_{n+1}(T_{n+1}a_n))$ (resp. $b = (b_0, \cdots, b_{n-1}, b_n, w_{n+1}(T_{n+1}b_n))$.

Note $T_{n+1}(a_n)T_{n+1}(b_n) \in E_{n+1}$. Thus

$$d(ab, Z_{n+1}) \le \|w_{n+1}(T_{n+1}(a_n))w_{n+1}(T_{n+1}(b_n)) - w_{n+1}(T_{n+1}(a_n)T_{n+1}(b_n))\|$$

and hence since $\|a_n\| \le \|a\|$ and $\|b_n\| \le \|b\|$

$$d(ab, Z_{n+1}) \le e^2(\sum_{k>n+1} \varepsilon_k) \|T_{n+1}(a_n)\| \|T_{n+1}(b_n)\| \le e^2(\sum_{k>n+1} \varepsilon_k)(1+\varepsilon_{n+1})^2 \|a\| \|b\|.$$

But now since $Z_n \subset Z_{n+m}$ for any $m > 0$, this also implies

$$\forall a, b \in Z_n \quad d(ab, Z_{n+m+1}) \le e^2(\sum_{k>n+m+1} \varepsilon_k)(1+\varepsilon_{n+m+1})^2 \|a\| \|b\| \to 0,$$

when $m \to \infty$ and hence $ab \in \overline{\cup Z_n} = Z$.

Clearly the same conclusion holds for any $a, b \in \cup Z_n$, so that $Z$ (which, as we already noticed, is s.a.) is a $C^*$-subalgebra of $L$. Since $Q(Z)$ is closed, it follows that $A = Q(Z)$ is a $C^*$-subalgebra of $\mathcal{L}$. $\square$

**Definition 2.7.** An o.s. $X$ locally embeds in another one $C$ if for any $\varepsilon > 0$ and any f.d. subspace $E \subset X$ there is a subspace $F \subset C$ and an isomorphism $w : E \to F$ such that $\|w\|_{cb}\|w^{-1}\|_{cb} \le 1+\varepsilon$. Moreover we set

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

where the infimum runs over all possible such $w$.

More generally, we say that $X$ locally embeds in a sequence $\{C_n\}$ of $C^*$-algebras if for any $\varepsilon > 0$ and for any $E$ there is for some $n$ a subspace $F \subset C_n$ such that $d_{cb}(E, F) \le 1 + \varepsilon$.

We will need to discuss some additional properties:

$$(2.10) \qquad \|T_n\|_{cb} \le 1 + \varepsilon_n,$$

and hence

$$(2.11) \qquad \|T_{[k,n]}\|_{cb} \le \prod_{n \le j \le k} (1 + \varepsilon_j) \le 1 + \eta_n.$$

We will also need

$$(2.12) \qquad \|T_n^{-1}|_{T_n(E_{n-1})}\|_{cb} \le 1 + \varepsilon_n.$$

and hence

$$(2.13) \qquad \|T_{[k,n]|_{T_{[k,n]}(E_{n-1})}}^{-1}\|_{cb} \le \prod_{n \le j \le k} (1 + \varepsilon_j) \le 1 + \eta_n.$$

**Proposition 2.8.** Assume that the inductive system satisfies (2.10) and (2.12), then A locally embeds in $\{C_n\}$. In fact for all $n \ge 0$

$$(2.14) \qquad d_{cb}(E_n, A_n) \le (1 + \eta_{n+1})^2.$$
Proof. By perturbation it clearly suffices to show that $A_n$ locally embeds in $\{C_k, k \ge 0\}$ for all $n$. Fix $n$. Recall that $Q: L \to \mathcal{L}$ denotes the quotient map. Consider the map $\theta_n: E_n \to L$ and let $\alpha_n = Q\theta_n: E_n \to A_n$. We have $\|\alpha_n\|_{cb} \le \|\theta_n\|_{cb} \le \max\{1, \sup_{k>n} \|T_{[k,n+1]}\|_{cb}\}$ and hence by (2.10) $\|\alpha_n\|_{cb} \le 1 + \eta_{n+1}$. To estimate $\|\alpha_n^{-1}\|_{cb}$ we will use (2.12). Let $x \in E_n$. We have $\|T_{[k,n+1]}(x)\| \ge (1 + \eta_{n+1})^{-1}\|x\|$ by (2.13) and hence

$$ \|x\| = \limsup \|T_{[k,n+1]}(x)\| \ge (1 + \eta_{n+1})^{-1} \|x\|. $$

The latter remains true for any $x \in M_N(E_n) \simeq M_N \otimes E_n$ and any fixed $N$ when $T_{[k,n+1]}$ is replaced by $Id_{M_N} \otimes T_{[k,n+1]}$. Thus we obtain

$$ (2.15) \qquad \|\alpha_n^{-1}\|_{cb} \le 1 + \eta_{n+1}. $$

This implies $d_{cb}(A_n, E_n) \to 1$ when $n \to \infty$, and since $\cup A_n$ is dense in $A$, this is enough to show that $A$ locally embeds in $\{C_n\}$. $\square$

Remark 2.9. By definition of the LLP it is easy to see that any $C$ with LLP locally embeds in $\mathcal{C}$. In particular, for any nuclear $C^*$-algebra $D$, $D \otimes_{\min} \mathcal{C}$ has LLP and hence locally embeds in $\mathcal{C}$.

Let $p_{[n,\infty]}$ denote the $*$-homomorphism taking $x \in L$ to $(x_n, x_{n+1}, \dots) \in \ell_\infty(\{C_k, k \in [n, \infty]\})$. Note that $p_{[n,\infty]}(Z)$ is a $C^*$-subalgebra of $\ell_\infty(\{C_k, k \in [n, \infty]\})$. It is convenient to record here the following elementary fact.

**Lemma 2.10.** Let $n \ge 1$. The map $\Phi_n: z \mapsto (z_0, \dots, z_{n-1}) \oplus p_{[n,\infty]}(z)$ is an isomorphism from $Z$ to $C_0 \oplus \dots \oplus C_{n-1} \oplus p_{[n,\infty]}(Z) \subset C_0 \oplus \dots \oplus C_{n-1} \oplus \ell_\infty(\{C_k, k \in [n, \infty]\})$.

**Proof.** It is clear that $\Phi_n$ is an injective $*$-homomorphism with range included in $C_0 \oplus \dots \oplus C_{n-1} \oplus p_{[n,\infty]}(Z)$. Let $y^0 \in C_0 \oplus \dots \oplus C_{n-1}$ and let $y \in p_{[n,\infty]}(\cup Z_m)$. We may assume $y \in p_{[n,\infty]}(Z_m)$ for some $m > n$. Thus $y = (y_n, \dots, y_m, T_{m+1}y_m, T_{m+2}T_{m+1}y_m, \dots)$ with $y_m \in E_m$. Let

$$ z = (y_0^0, \dots, y_{n-1}^0, y_n, \dots, y_m, T_{m+1}y_m, T_{m+2}T_{m+1}y_m, \dots). $$

Then $z \in Z_m$ and $\Phi_n(z) = (y^0, y) \in C_0 \oplus \dots \oplus C_{n-1} \oplus p_{[n,\infty]}(Z)$. This shows that the range of $\Phi_n$ is dense in $C_0 \oplus \dots \oplus C_{n-1} \oplus p_{[n,\infty]}(Z)$ and hence that $\Phi_n$ is surjective. $\square$

# 3 Approximately multiplicative self-adjoint liftings of $F$

Consider a separable $C^*$-algebra $C$ with LP and a $*$-homomorphism

$$ F: C \to \mathcal{L}(C) = \ell_\infty(C)/c_0(C). $$

We know that $F$ admits a c.c. lifting $(f_n): C \to \ell_\infty(C)$ which is (automatically) asymptotically multiplicative and self-adjoint, meaning that

$$ (3.1) \qquad \forall x, y \in C \quad \limsup \|f_n(xy) - f_n(x)f_n(y)\| = 0 \text{ and } f_n(x^*) = f_n(x)^*. $$

Let $(\delta_n)$ be a given positive sequence. We will construct a system of f.d. subspaces $E_n \subset C$ ($n \ge 0$), maps

$$ T_n: E_{n-1} \to E_n, $$

a sequence of integers $(m(n))$ and a sequence $\varepsilon_n > 0$, such that $\sum \varepsilon_n \le 1$, satisfying the following properties:
(i) Each $T_n$ is the restriction of $f_{m(n)}$ to $E_{n-1}$ and $T_n$ is an $\varepsilon_n$-morphism.

(ii) $T_n(E_{n-1}) + T_n(E_{n-1})T_n(E_{n-1}) \subset E_n$.

(iii) For any $n > k$ we have $e^2 \sum_{k+1}^n \varepsilon_j < \varepsilon_k$.

(iv) For any $C^*$-algebra $D$ and any $\varepsilon_n$-morphism $\theta : E_n \to D$ we have

$$ \|[Id_{\mathcal{C}} \otimes \theta T_n](t^{E_{n-1}})\|_{\max} \le 1 + \delta_n. $$

See Proposition 1.2 for the definition of $t^{E_{n-1}}$. By (1.3), (iv) is equivalent to:

(iv)' For any $C^*$-algebra $D$ and any $\varepsilon_n$-morphism $\theta : E_n \to D$ we have

$$ \|\theta T_n\|_{MB(E_{n-1}, D)} \le 1 + \delta_n. $$

More explicitly by (1.4) (iv)' is equivalent to:

(iv)" For any $C^*$-algebra $D$ and any $\varepsilon_n$-morphism $\theta : E_n \to D$ we have for any $C^*$-algebra $B$

$$ \|Id \otimes \theta T_n : B \otimes_{\max} E_{n-1} \to B \otimes_{\max} D\| \le 1 + \delta_n. $$

Since $(f_n)$ is c.c. we have

$$ (3.2) \qquad \|T_n\|_{cb} \le 1. $$

*Remark 3.1.* Note that (iv) applied to the identity map from $E_n$ to $C$ implies $\|[Id \otimes T_n](t^{E_{n-1}})\|_{\max} \le 1 + \delta_n$.

The construction of our system is by induction: once we have $E_n, T_n, \varepsilon_n$ we must produce $E_{n+1}, \varepsilon_{n+1}$, and $T_{n+1} : E_n \to E_{n+1}$. Compared with [8] the novelty is the use of conditions (iii) and (iv).

**Lemma 3.2.** Let $C$ be a $C^*$-algebra with LP. Let $E \subset C$ be a f.d. subspace.

Then for any linear map $\psi : E \to D$ (D any other $C^*$-algebra) we have

$$ \|\psi\|_{mb} \le \|(Id_{\mathcal{C}} \otimes \psi)(t^E)\|_{\mathcal{C} \otimes_{\max} D}. $$

In other words if we set $\hat{\psi} = Id_{\mathcal{C}} \otimes \psi : \mathcal{C} \otimes_{\max} E \to \mathcal{C} \otimes_{\max} D$ we have $\|\hat{\psi}\| = \|\hat{\psi}(t^E)\|$.

*Proof.* Since $C$ has the LP (by Th. 10.5 in [10]) $E$ is “max-controllable” in the sense of [11], i.e. there is $t^E$ in the unit ball of $\mathcal{C} \otimes_{\max} E$ such that for any $t \in B_{\mathcal{C} \otimes_{\max} E}$ there is a unital $*$-homomorphism $\pi : \mathcal{C} \to \mathcal{C}$ such that $t = (\pi \otimes Id_E)(t^E)$. Then $\|(Id_{\mathcal{C}} \otimes \psi)(t)\|_{\max} = \|(\pi \otimes Id)(Id \otimes \psi)(t^E)\|_{\max} \le \|\hat{\psi}(t^E)\|_{\max}$ whence the announced inequality. $\square$

**Lemma 3.3.** Let $C, C_1$ be $C^*$-algebras. Let $E \subset C, F \subset C_1$ be f.d. subspaces. For any $\delta > 0$ there is $\varepsilon > 0$ and a f.d.s.a. superspace $\mathcal{E}$ with $E \subset \mathcal{E} \subset C$ such that for any $\varepsilon$-morphism $\psi : \mathcal{E} \to D$ (D any other $C^*$-algebra) we have

$$ \forall x \in F \otimes E \quad \|(Id_{C_1} \otimes \psi)(x)\|_{C_1 \otimes_{\max} D} \le (1 + \delta)\|x\|_{C_1 \otimes_{\max} C}. $$

*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 $D$, all $\varepsilon$-morphisms $\psi : \mathcal{E} \to D$ and all $x \in F \otimes E$ we have $\|(Id_{C_1} \otimes \psi)(x)\|_{C_1 \otimes_{\max} D} \le (1 + \delta)\|x\|_{C_1 \otimes_{\max} C}$.

Let us first fix $x \in F \otimes E$. For each $\alpha \in I$ with $E \subset \mathcal{E}_\alpha$, there is a $D_\alpha$ and an $\varepsilon_\alpha$-morphism $\psi_\alpha : \mathcal{E}_\alpha \to D_\alpha$, such that

$$ \|(Id_{C_1} \otimes \psi_\alpha)(x)\|_{C_1 \otimes_{\max} D_\alpha} \ge (1 + \delta)^{-1} \sup \|(Id_{C_1} \otimes \psi)(x)\|_{C_1 \otimes_{\max} D} $$
where the last supremum runs over all $D$ and all $\varepsilon_\alpha$-morphism $\psi : \mathcal{E}_\alpha \to D$. Note that this last supremum is finite since the nuclear norm of each $\psi : \mathcal{E}_\alpha \to D$ is at most $(1 + \varepsilon_\alpha) \dim(\mathcal{E}_\alpha)$. Let $\psi'_\alpha : C \to D_\alpha$ 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 $\psi' = (\psi'_\alpha) : C \to \ell_\infty(I; \{D_\alpha\})$ and let $Q : \ell_\infty(I; \{D_\alpha\}) \to \ell_\infty(I; \{D_\alpha\})/c_0(I; \{D_\alpha\})$ be the quotient map. Then $\pi = Q\psi' : C \to \ell_\infty(I; \{D_\alpha\})/c_0(I; \{D_\alpha\})$ is clearly a contractive *-homomorphism. We have a contractive morphism $\mathrm{Id}_{c_1} \otimes \pi :$

$$C_1 \otimes_{\max} C \to C_1 \otimes_{\max} [\ell_\infty(I; \{D_\alpha\})/c_0(I; \{D_\alpha\})] = [C_1 \otimes_{\max} \ell_\infty(I; \{D_\alpha\})]/[C_1 \otimes_{\max} c_0(I; \{D_\alpha\})]$$

where the last = holds by the “exactness” of the max-tensor product (see e.g. [6, p. 285]). Moreover, we have clearly a contractive morphism

$$[C_1 \otimes_{\max} \ell_\infty(I; \{D_\alpha\})]/[C_1 \otimes_{\max} c_0(I; \{D_\alpha\})] \to [\ell_\infty(I; \{C_1 \otimes_{\max} D_\alpha\})]/[c_0(I; \{C_1 \otimes_{\max} D_\alpha\})].$$

Since $\pi(e) = Q((\psi_\alpha(e))_\alpha)$ for all $e \in E$ and since $x \in C_1 \otimes E$, it follows that

$$\limsup_\alpha \|(Id_{C_1} \otimes \psi_\alpha)(x)\|_{C_1 \otimes_{\max} D_\alpha} \le \|x\|_{C_1 \otimes_{\max} C},$$

which proves the desired result for each fixed given $x \in F \otimes E \subset C_1 \otimes E$. But since $F \otimes E$ is a finite dimensional subspace of $C_1 \otimes_{\max} C$ 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 [6, Th. 26.8]. □

**Lemma 3.4.** In the situation of the preceding Lemma 3.2, for any $\delta > 0$ there is $\varepsilon > 0$ and a f.d.s.a. superspace $\mathcal{E}$ with $E \subset \mathcal{E} \subset C$ such that for any other $C^*$-algebra $D$ and any $\varepsilon$-morphism $\psi : \mathcal{E} \to D$ we have

$$\forall x \in \mathcal{C} \otimes E \quad \|(Id_{\mathcal{C}} \otimes \psi)(x)\|_{\mathcal{C} \otimes_{\max} D} \le (1 + \delta)\|x\|_{\mathcal{C} \otimes_{\max} E},$$

or equivalently for any $D$

$$\|\psi|_E\|_{MB(E,D)} \le 1 + \delta.$$

*Proof.* By Lemma 3.3 (actually just by the first part of the proof of it) applied with $x = t^E \in \mathcal{C} \otimes E$ we find $\mathcal{E}$ such that for any $\varepsilon$-morphism $\psi : \mathcal{E} \to D$ we have $\|(Id_{\mathcal{C}} \otimes \psi)(t^E)\|_{\mathcal{C} \otimes_{\max} D} \le 1+\delta$. Therefore by Lemma 3.2 $\|\psi_E\|_{mb} \le 1+\delta$. □

**Lemma 3.5.** Consider a separable $C^*$-algebra $C$ with LP and a $*$-homomorphism

$$F : C \to \mathcal{L}(C) = \ell_\infty(C)/c_0(C)$$

together with a c.c. lifting $(f_n) : C \to \ell_\infty(C)$, which is (automatically) asymptotically $*$-multiplicative. Let $E \subset C$ be a f.d.s.a. subspace. For any $\delta > 0$ there is $\varepsilon' > 0$ such that for any $m$ large enough there is a f.d.s.a. subspace $E_1 \subset C$ containing $f_m(E) + f_m(E)f_m(E)$ so that if we set $T = f_{m|E} : E \to E_1$ then for any other $C^*$-algebra $D$ and any $\varepsilon'$-morphism $\psi : E_1 \to D$ we have

$$(3.3) \qquad \|\psi T\|_{MB(E,D)} \le 1 + \delta,$$

and $T$ itself is an $\varepsilon'$-morphism.
It will be convenient to set for $T: E \to E_1$

$$ (3.4) \qquad \|T\|_{mb,\varepsilon} = \sup \|\psi T\|_{MB(E,D)} $$

where the sup runs over all $C^*$-algebras $D$ and all $\varepsilon$-morphisms $\psi: E_1 \to D$, so that (3.3) could be written as $\|T\|_{mb,\varepsilon'} \le 1 + \delta$. Note that the case when $\psi: E_1 \to C$ acts as the identity yields $\|T\|_{MB(E,C)} \le \|T\|_{mb,\varepsilon'}$ for any $\varepsilon' > 0$.

**Proof.** Let $\mathcal{E}$ and $\varepsilon$ be as in Lemma 3.4. Fix $0 < \varepsilon' < \varepsilon$ to be specified. Since $(f_m)$ satisfies (3.1), for all $m$ large enough $f_{m|\mathcal{E}}: \mathcal{E} \to C$ is an $\varepsilon'$-morphism. A fortiori so is $f_{m|E}: E \to C$ and hence $T$. We set $E_1 = f_m(\mathcal{E}) + f_m(\mathcal{E})f_m(\mathcal{E})$. Note $T(E) + T(E)T(E) \subset E_1$. By Remark 2.2 for any $D$ and any $\varepsilon'$-morphism $\psi: E_1 \to D$, the composition $\psi' = \psi f_{m|\mathcal{E}}$ is an $\varepsilon''$-morphism on $\mathcal{E}$ where $\varepsilon'' = (1 + \varepsilon')\varepsilon' + \varepsilon'(1 + \varepsilon')^2$. Thus we can clearly select and fix $\varepsilon' \approx \varepsilon/2$ so that $\varepsilon'' \le \varepsilon$ or even, say, $\varepsilon'' = \varepsilon$. Then Lemma 3.4 tells us that $\|\psi T\|_{MB(E,D)} = \|\psi'_{|E}\|_{MB(E,D)} \le 1 + \delta$. $\square$

Let $(\varepsilon_n)$ and $(\delta_n)$ be positive sequences such that

$$ (3.5) \qquad \sum \varepsilon_n < \infty \text{ and } \delta_n \to 0. $$

From what precedes we see that to any embedding $C \to \mathcal{L}(C)$ we can associate (at least if $C$ has LP) an inductive system and a resulting $A$:

**Proposition 3.6.** In the situation of Lemma 3.5 for each positive sequence $(\delta_n)$ there is a summable positive sequence $(\varepsilon_n)$ and an admissible inductive system $(E_n, T_n)$ with a sequence $(m(n))$ as before satisfying (i)-(iv)'.

Moreover, if $(f_n)$ satisfies (1.1) then we can find $(E_n)$ and $m(n)$ satisfying (2.12) in addition to (i)-(iv)'.

**Proof.** By induction, suppose we have obtained the system up to $T_n: E_{n-1} \to E_n$ and $\varepsilon_n$ with $T_n = f_{m(n)}$ restricted to $E_{n-1}$. We then apply Lemma 3.5 to $E = E_n$ with $\delta = \delta_{n+1}$ to produce a number $\varepsilon'$ so that choosing $m$ large enough we may set $\varepsilon_{n+1} = \varepsilon'$, $m(n+1) = m$, $E_{n+1} = E_1$ and $T_{n+1} = f_{m(n+1)}|_{E_n}$. Lemma 3.5 then implies

$$ \|T_{n+1}\|_{mb,\varepsilon_{n+1}} \le 1 + \delta_{n+1}. $$

As for condition (iii) since the inequality is strict we can clearly adjust further $\varepsilon_{n+1} < \varepsilon'$ so that the condition remains valid. Thus we obtain (i)-(iv)' for the next level $n+1$, which proves the first part. If in addition $(f_n)$ satisfies (1.1), we can choose $m(n+1)$ large enough so that $\|f_{m(n+1)|f_{m(n+1)}(E_n)}^{-1}\|_{cb} < 1 + \varepsilon_{n+1}$, and this gives us (2.12) at the next induction step. $\square$

**Remark 3.7.** We will use (iii) and (iv)' together as follows. Consider $\theta_n: E_n \to Z$. By Lemma 2.3 we know that $\theta_n$ is a $\delta$-morphism for $\delta = e^2 \sum_{j>n} \varepsilon_j$, and hence by (iii) an $\varepsilon_n$-morphism. By (iv)' we have $\|\theta_n T_n\|_{MB(E_{n-1},Z)} \le 1 + \delta_n$. We claim that this implies

$$ (3.6) \qquad \forall n \ge 1 \quad \|\theta_{n-1}\|_{MB(E_{n-1},Z)} \le 1 + \delta_n $$

and hence also

$$ (3.7) \qquad \forall n \ge 0 \quad \|\theta_n\|_{MB(E_n,Z)} \le 1 + \delta_{n+1}. $$
Indeed, since $\forall x \in E_{n-1}$

$$ \theta_{n-1}(x) = (0, \dots, 0, x, w_n(T_n(x))) = (0, \dots, 0, x, 0, \dots) + \theta_n(T_n(x)) $$

(recall (2.8) and (2.9)) we have

$$ (3.8) \qquad \Phi_n \theta_{n-1}(x) = (0, \dots, 0, x) \oplus p_{[n, \infty]} \theta_n T_n(x) $$

By Lemma 2.10 we have $\|\theta_{n-1}\|_{MB(E_{n-1},Z)} \le \max\{1, \|p_{[n,\infty]}\theta_n T_n\|_{MB(E_{n-1},p_{[n,\infty]}(Z))}\}$, and hence

$$ \|\theta_{n-1}\|_{MB(E_{n-1},Z)} \le \max\{1, \|\theta_n T_n\|_{MB(E_{n-1},Z)}\}. $$

Whence our claim $\|\theta_{n-1}\|_{MB(E_{n-1},Z)} \le 1 + \delta_n$ for all $n \ge 1$.

Following the terminology of [7] we say that the inclusion of a subalgebra $Z$ in a larger one $L$ is max-injective if the associated $*$-homomorphism $D \otimes_{\max} Z \to D \otimes_{\max} L$ is injective (and hence isometric) for all $C^*$-algebras $D$. For that to hold it suffices that it holds for $D = \mathscr{C}$.

**Proposition 3.8.** Consider the situation of Proposition 3.6. Assume $(\delta_n)$ bounded. Then the associated $C^*$-algebra $Z$ has the LP and the inclusion $Z \to \ell_\infty(C)$ is max-injective.

*Proof.* To show that $Z$ has the LP we use the criterion from [10]. Let $(C_i)$ be an arbitrary family of $C^*$-algebras (we could take simply $C_i = \mathscr{C}$ for all $i$). We claim that there is a constant $c$ such that for any $t \in Z \otimes \ell_\infty(\{C_i\})$ the associated family $(t_i)$ with $t_i \in Z \otimes C_i$ satisfies

$$ (3.9) \qquad \|t\|_{Z \otimes_{\max} \ell_\infty(\{C_i\})} \le c \sup_i \|t_i\|_{Z \otimes_{\max} C_i}. $$

Actually, since here we are comparing $C^*$-norms if there is such a $c$ then $c = 1$ works, and we obtain the criterion as in [10]. To verify this claim it suffices to check this for $t \in Z_n \otimes \ell_\infty(\{C_i\})$ with $c$ independent of $n$. The following argument, simpler than the original one, is due to Jean Roydor. The claim is an easy consequence of the fact that there is a linear isomorphism $v_n : E_n \to Z_n$ (defined by $v_n(x) = \theta_n(x)$) such that (recall (1.2))

$$ (3.10) \qquad \|v_n\|_{MB(E_n, Z_n)} \|v_n^{-1}\|_{MB(Z_n, E_n)} \le 1 + \delta_{n+1}. $$

Indeed, this follows by combining (3.7) with the observation that $v_n^{-1} : Z_n \to E_n$ is nothing but the restriction of the $n$-th coordinate $p_n : \ell_\infty(C) \to C$ which is a $*$-homomorphism from $Z$ to $C$, and hence has mb-norm at most 1.

Now, it is just a simple exercise to deduce that (3.9) holds for $t \in Z_n \otimes \ell_\infty(\{C_i\})$ with $c = 1 + \delta_{n+1}$ from the fact that the analogous inequality holds with $c = 1$ for all $t \in E_n \otimes \ell_\infty(\{C_i\})$ (the latter since $C$ has the LP).

Recall the notation $L = \ell_\infty(C)$. To show that $Z \subset L$ is max-injective, let $t \in \mathscr{C} \otimes Z$. We will show that $\|t\|_{\mathscr{C} \otimes_{\max} Z} \le c \|t\|_{\mathscr{C} \otimes_{\max} L}$ for some constant $c$. As in the first part we must have $c = 1$ since both sides of the inequality are $C^*$-norms. By perturbation we may clearly assume that $t \in \mathscr{C} \otimes Z_n$. Then since $x = v_n p_n x$ for any $x \in Z_n$ we have by (3.7) again

$$ \|t\|_{\mathscr{C} \otimes_{\max} Z_n} \le \|Id \otimes v_n\|_{MB(E_n, Z_n)} \|Id \otimes p_n(t)\|_{\mathscr{C} \otimes E_n} \le (1 + \delta_{n+1}) \|t\|_{\mathscr{C} \otimes_{\max} L} $$

where at the last step we used that $\|p_n\|_{MB(L,C)} \le 1$. This yields the announced result with $c = 1 + \sup_n \delta_n$. $\square$
**Remark 3.9.** As pointed out to me by Jean Roydor the preceding argument highlights the following general fact : if a separable $C^*$-algebra $Z$ locally embeds in the mb-sense described by (3.10) into another one with the LP then $Z$ inherits the LP. Thus we may state: a separable $Z$ has the LP if and only if it embeds in $\mathscr{C}$ locally in the mb-sense (meaning the same as in Definition 2.7 but with mb in place of cb). This is similar to the equivalence observed in [12] of the mb-versions of the LLP and the LP. Note that the terms are slightly misleading: it is not clear that a $C^*$-subalgebra locally embeds in the mb-sense, unless the inclusion is max-injective.

**Proposition 3.10.** Consider the situation of Proposition 3.6. Assume $\delta_n$ bounded. Assume in addition that the inductive system satisfies (2.12), then the associated $C^*$-algebra $A$ has the LLP.

*Proof.* We set $\mathscr{B} = B(\ell_2)$. We claim that there is a constant $c$ for which the inclusion $A_n \to A$ satisfies

$$ \|\mathscr{B} \otimes_{\min} A_n \to \mathscr{B} \otimes_{\max} A\| \le c \quad \text{for all } n. $$

This clearly implies that the pair $(\mathscr{B}, A)$ is nuclear, i.e. that $A$ has the LLP. As in the proof of Proposition 2.8 let

$$ \alpha_n = Q\theta_n : E_n \to A_n. $$

On the one hand by (2.15) we have

$$ (3.11) \qquad \|\alpha_n^{-1} : A_n \to E_n\|_{cb} \le 1 + \eta_{n+1}, $$

which means

$$ \|Id \otimes \alpha_n^{-1} : \mathscr{B} \otimes_{\min} A_n \to \mathscr{B} \otimes_{\min} E_n\| \le 1 + \eta_{n+1}. $$

On the other hand by (3.7) we have

$$ (3.12) \qquad \|\alpha_n\|_{mb} \le 1 + \delta_{n+1}, $$

which implies a fortiori

$$ \|Id \otimes \alpha_n : \mathscr{B} \otimes_{\max} E_n \to \mathscr{B} \otimes_{\max} A\| \le 1 + \delta_{n+1}. $$

Recall that here the norm of $\mathscr{B} \otimes_{\max} E_n$ is the one induced by $\mathscr{B} \otimes_{\max} C$. Since $C$ has the LP and a fortiori the LLP we have $\mathscr{B} \otimes_{\min} E_n = \mathscr{B} \otimes_{\max} E_n$ isometrically and hence we conclude that the inclusion $i_{A_n} : A_n \to A$ satisfies

$$ \|Id \otimes i_{A_n} : \mathscr{B} \otimes_{\min} A_n \to \mathscr{B} \otimes_{\max} A\| \le (1 + \eta_{n+1})(1 + \delta_n)(1 + \delta_{n+1}). $$

This proves the claim.

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

**Remark 3.11.** We should emphasize here a subtle point about (3.4) and (iv)' that we already used: a priori $\theta_n$ is an $\varepsilon_n$-morphism from $E_n$ to $L$ but its values being inside $Z$ we may view it as an $\varepsilon_n$-morphism into $Z$, and hence (iv)' implies $\|\theta_n T_n\|_{MB(E_{n-1},Z)} \le 1 + \delta_n$. Similarly (3.12) means

$$ (3.13) \qquad \|\alpha_n\|_{MB(E_n,A)} \le 1 + \delta_{n+1}. $$

In this light the norm in (3.4) appears much stronger than the mb-norm of $T$. Specifically, it dominates the mb-norm of $T$ even when $T$ is viewed as acting into the $C^*$-algebra generated by its range, which a priori may augment the mb-norm.
**Theorem 3.12.** Consider the situation of Proposition 3.10. Let D be a separable $C^*$-algebra. Assume $\delta_n$ bounded and (1.1). If the embedding $F: C \to \mathcal{L}(C) = \ell_\infty(C)/c_0(C)$ is D-nuclear (i.e. $\mathrm{Id}_D \otimes F$ is bounded from $D \otimes_{\min} C$ to $D \otimes_{\max} \mathcal{L}(C)$), the admissible inductive system can be constructed so that the associated $C^*$-algebra A (as in Proposition 2.5) is D-nuclear and locally embeds in C.

*Proof.* The underlying idea is that the inclusion $A_{n-1} \subset A_n$ is “similar” to the map $T_n: E_{n-1} \to E_n$. By what precedes we have a factorization of $A_{n-1} \subset A_n$ as :

$$A_{n-1} \xrightarrow{\alpha_{n-1}^{-1}} E_{n-1} \xrightarrow{T_n} E_n \xrightarrow{\alpha_n} A_n$$

We already saw in (3.11) and (3.12) that $\|\alpha_{n-1}^{-1}\|_{cb} \le 1 + \eta_n$ and $\|\alpha_n\|_{mb} \le 1 + \delta_{n+1}$. We now turn to the assertion about the D-nuclearity of A. Recall that by the projectivity of the max-tensor product we have a canonical contractive $*$-homomorphism $D \otimes_{\max} \mathcal{L}(C) \to \mathcal{L}(D \otimes_{\max} C)$, and hence for any $x \in D \otimes C$ our assumption on F implies

$$(3.14) \quad \limsup_n \|[Id \otimes f_n](x)\|_{D \otimes_{\max} C} \le \|[Id_D \otimes F](x)\|_{D \otimes_{\max} \mathcal{L}(C)} \le \|x\|_{D \otimes_{\min} C}$$

Let $D_n \subset D$ be an increasing sequence of f.d. subspaces with dense union. We indicate how the induction step should be modified to obtain the assertion in the theorem. We ensure at each step that $T_n: E_{n-1} \to E_n$ is such that

$$(3.15) \quad \|Id \otimes T_n : D_{n-1} \otimes_{\min} E_{n-1} \to D_{n-1} \otimes_{\max} E_n\| \le 1 + \delta_n.$$

Suppose we have constructed $E_{n-1}$, $E_n$, $T_n$. Using a suitable fine enough finite net in the (compact) unit ball of $D_n \otimes_{\min} E_n$ and using (3.14) we can find $m$ (that will be our $m(n+1)$) such that, in addition to the properties already imposed in the proof of Proposition 3.6, we have

$$\|Id \otimes f_m : D_n \otimes_{\min} E_n \to D_n \otimes_{\max} C\| \le 1 + \delta_{n+1}$$

Then we argue as before: we let $T_{n+1} = f_m$ and $E_{n+1} = T_{n+1}(E_n) + T_{n+1}(E_n)T_{n+1}(E_n)$, and we obtain (3.15) with $n$ replaced by $n+1$, which is the next step of the induction. We will now show how (3.15) for all $n \ge 1$ implies that A is D-nuclear. Using (3.13) and (3.15) we find

$$\|Id \otimes \alpha_n T_n : D_{n-1} \otimes_{\max} E_{n-1} \to D_{n-1} \otimes_{\max} A_n\| \le (1 + \delta_n)(1 + \delta_{n+1}).$$

Since $\|\alpha_{n-1}^{-1}\|_{cb} \le 1 + \eta_n$, the inclusion map $i_{A_{n-1}}: A_{n-1} \to A$, which is equal to $\alpha_n T_n a_{n-1}^{-1}$, satisfies $\|Id \otimes i_{A_{n-1}}: D_{n-1} \otimes_{\min} A_{n-1} \to D_{n-1} \otimes_{\max} A\| \le (1 + \eta_n)(1 + \delta_n)(1 + \delta_{n+1})$, or equivalently

$$\|Id \otimes i_{A_{n-1}} : D_{n-1} \otimes_{\min} A_{n-1} \to D \otimes_{\max} A\| \le (1 + \eta_n)(1 + \delta_n)(1 + \delta_{n+1}),$$

and since $(1 + \eta_n)(1 + \delta_n)(1 + \delta_{n+1})$ is bounded and $\cup[D_{n-1} \otimes A_{n-1}]$ dense in $D \otimes_{\min} A$ we conclude that A is D-nuclear.

By Proposition 2.8 we already know that A locally embeds in C.

<figure><img src="image_3.png" /></figure>

**Remark 3.13.** In the situation of Theorem 3.12 with $\delta_n \to 0$, the conclusion can be summarized as follows:

The inclusion $A_{n-1} \to A_n$ can be factorized as

$$A_{n-1} \xrightarrow{\alpha_{n-1}^{-1}} E_{n-1} \xrightarrow{T_n} E_n \xrightarrow{\alpha_n} A_n$$

where $\|\alpha_{n-1}^{-1}\|_{cb} \to 1$, $\|\alpha_n\|_{mb} \to 1$ and $\|Id \otimes T_n : D_{n-1} \otimes_{\min} E_{n-1} \to D_{n-1} \otimes_{\max} E_n\| \to 1$ (where for the latter we use the norm induced by $D \otimes_{\max} C$).
Remark 3.14. More generally given a sequence $(D(i))_{i \ge 1}$ of separable $C^*$-algebras such that $F: C \to \mathcal{L}(C)$ is $D(i)$-nuclear for each $i$, we claim that our system can be adjusted so that $A$ is $D(i)$-nuclear for each $i$.

Indeed, for each $i$, let $\{D(i)_N | N \ge 1\}$ be an increasing sequence of f.d. subspaces with union dense in $D(i)$. Let $\{\Delta_n | n \ge 0\}$ be an enumeration of the collection $\{D(i)_N | i \ge 1, N \ge 1\}$ where each element is repeated countably infinitely many times. Fix a subspace $\Delta \in \{\Delta_n | n \ge 0\}$. Then $\Delta = \Delta_{n-1}$ for some $n \ge 1$ that we can choose as large as we wish. Clearly the inductive argument of the preceding proof can be modified to ensure that

$$ \forall n \ge 1 \quad \|Id \otimes i_{A_{n-1}} : \Delta_{n-1} \otimes_{\min} A_{n-1} \to \Delta_{n-1} \otimes_{\max} A\| \le (1 + \eta_n)(1 + \delta_n)^2. $$

Thus $\Delta = \Delta_{n-1}$ implies

$$ (3.16) \qquad \|Id \otimes i_{A_{n-1}} : \Delta \otimes_{\min} A_{n-1} \to \Delta \otimes_{\max} A\| \le (1 + \eta_n)(1 + \delta_n)^2. $$

Let $c = \sup(1 + \eta_n)(1 + \delta_n)^2$. Since (3.16) holds for infinitely many $n$'s the min and max norms are $c$-equivalent on $\Delta \otimes [\cup A_n]$, and since the latter holds for any $\Delta$, for each $i$, the same $c$-equivalence holds on $[\cup_N D(i)_N] \otimes [\cup_{n-1} A_{n-1}]$. Since the latter is dense in $D(i) \otimes_{\min} A$ this proves our claim.

Unfortunately we do not see how to remove the assumption that the family $(D(i))_{i \ge 1}$ is countable.

Remark 3.15. In our original approach in [8] it suffices to assume that $C$ has the LLP. We then use for each $n \ge 1$ the existence of a c.c. lifting for $F|_{E_{n-1}}$ and obtain a non-nuclear WEP $C^*$-algebra $A$ that locally embeds in $\mathcal{C}$. Then the LLP of $A$ follows from the general fact that WEP $\Rightarrow$ LLP for algebras that locally embed in $\mathcal{C}$. A priori the non-separability of $\mathcal{B}$ prevents us from taking $D = \mathcal{B}$ to prove that $A$ is $\mathcal{B}$-nuclear i.e. has the LLP, but once we know that $A$ locally embeds in $\mathcal{C}$, it suffices for the LLP of $A$ to know that $A$ is $j$-nuclear where $j: \mathcal{C} \to \mathcal{B}$ is any embedding and where by $A$ is $j$-nuclear we mean that $j \otimes Id_A$ is bounded from $\mathcal{C} \otimes_{\min} A \to \mathcal{B} \otimes_{\max} A$ (see the appendix for a proof). With this in mind the separability of $\mathcal{C}$ allows us to use the same argument as for separable $D$'s to check the LLP of $A$.

Remark 3.16. As already mentioned in [8], we can perform our construction of inductive systems in a more general setting that we will just briefly sketch here. We give ourselves a sequence of isometric $*$-homomorphisms $F(N) : C_N \to \mathcal{L}(C_{N+1})$ together with c.c.s.a. liftings $f(N) : C_N \to L(C_{N+1})$ so that if $Q(N+1) : L(C_{N+1}) \to \mathcal{L}(C_{N+1})$ denotes the quotient map we have $Q(N+1)f(N) = F(N)$. Then if we assume all the $C_N$'s have the LP and that the coordinates of each $f(N)$ are asymptotically locally almost completely isometric (i.e. they satisfy the condition (1.1)) then we can produce an inductive system with $E_{n-1} \subset C_{n-1}$ and with $T_n = f(n-1)_{m(n)}$ on $E_{n-1}$, and we obtain analogous properties.

# 4 Cone algebras

Proposition 2.5 tells us how to associate a $C^*$-algebra $A$ to a $*$-embedding $F: C \to \mathcal{L}(C)$, equipped with a completely contractive lifting $f: C \to L(C)$. By Theorem 3.12 if $f$ satisfies (1.1), if $C$ has the LP and if $F$ is $\mathcal{C}$-nuclear, the algebra $A$ has WEP and LLP. To valuably apply this result, we need to exhibit $F$'s with the required properties. Their existence is not so immediate but cone algebras provide us with that crucial missing ingredient.

Let $C_0 = C((0, 1])$ and $C_1 = C([0, 1])$. For any $C^*$-algebra $A$, we denote by $C_0(A) = C_0 \otimes_{\min} A$ the so-called cone algebra of $A$ and by $C(A)$ its unitization. (Warning: $C(A) \ne C_1 \otimes_{\min} A$ !)
When dealing with a bounded 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 bounded map extending $Id_{C_0} \otimes u$.

The algebra $C$ considered throughout §3 will now be replaced by the cone algebra $C_0(C)$ or $C(C)$. We use the bold letter $C$ for a $C^*$-algebra to avoid a notational confusion.

Let $q : C \to B$ be a surjective $*$-homomorphism and let $\mathcal{I} = \ker(q)$. We will use freely the natural identifications

$$B \simeq C/\mathcal{I} \quad \text{and} \quad C_0(B) \simeq C_0(C/\mathcal{I}) \simeq C_0(C)/C_0(\mathcal{I}), \quad C(B) \simeq C(C/\mathcal{I}) \simeq C(C)/C_0(\mathcal{I}).$$

Let $(\sigma_n)$ be a quasicentral approximate unit in $\mathcal{I}$. Our construction is simpler if there is such a $(\sigma_n)$ (and hence $(1 - \sigma_n)$) formed of projections. Then the mappings $x \mapsto (1 - \sigma_n)x$ are approximatively multiplicative. In general this does not exist. However, if we replace $C, B, q$ by $C_0(C), C_0(B), q_0$ then it does exist and the construction of the preceding section can be applied.

This trick of passing to cone algebras is closely related to Kirchberg's [5, §5], but we learnt it from [1, Lemma 13.4.4]. A similar idea already appears in [2, Lemma 10] for the suspension algebra in the context of approximatively multiplicative families indexed by a continuous parameter in $(0, \infty)$, but our goals seem unrelated.

**Lemma 4.1.** Let $q : C \to B$ be a surjective $*$-homomorphism between $C^*$-algebras, with separable kernel $\mathcal{I}$. There is a pointwise bounded sequence of maps $\rho_n : C_0 \otimes C \to C_0 \otimes C$ that induces in the limit a $*$-homomorphism

$$\rho_{\infty} : C_0(C) \to \mathcal{L}(C_0(C))$$

that vanishes on $C_0(\mathcal{I})$ and for which the associated map

$$\rho_{[\infty]} : C_0(C)/C_0(\mathcal{I}) \to \mathcal{L}(C_0(C))$$

(which satisfies $\rho_{[\infty]}q_0 = \rho_{\infty}$) is an isometric $*$-homomorphism.

Moreover, we have $q_0\rho_n(x) = q_0(x)$ for all $x \in C_0 \otimes C$ and all $n$.

We could refer to the proof of Lemma 6.1 in [8] (itself based on [1, Lemma 13.4.4]), but we give the details for the reader's convenience.

**Proof.** Recall $\mathcal{I} = \ker(q)$. Let $(\sigma_n)$ be a net forming a quasicentral approximate unit of $\mathcal{I}$. This means $\sigma_n \ge 0$, $\|\sigma_n\| \le 1$, $\|\sigma_n x - x\| \to 0$ for any $x \in \mathcal{I}$ and $\|\sigma_n c - c\sigma_n\| \to 0$ 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 $f(0) = 0$. The set of polynomials $P_0 = \operatorname{span}[t^m | m > 0]$ is dense in $C_0$. Let $\rho_n : C_0 \otimes C \to C_0(C)$ be the map taking $t \mapsto f(t)c$ ($f \in C_0, c \in C$) to $t \mapsto f(t(1 - \sigma_n))c$. For instance (monomials) $\rho_n$ takes $t \mapsto t^m c$ to $t \mapsto t^m(1 - \sigma_n)^m c$. Note $\|1 - \sigma_n\| \le 1$ therefore for any $f \in C_0$ the function $t \mapsto f(t(1 - \sigma_n))$ is in $C_0(C)$ with norm $\|f\|_{C_0}$. This shows that $\sup_n \|\rho_n(y)\| < \infty$ for any $y \in C_0 \otimes C$, so that $(\rho_n)$ defines a map $\rho : C_0 \otimes C \to \ell_{\infty}(C_0(C))$.

We should warn the reader that a priori we do not know that $\rho$ is bounded on $C_0 \otimes_{\min} C = C_0(C)$, but we will show that it becomes contractive after composition with the quotient map $Q : \ell_{\infty}(C_0(C)) \to \mathcal{L}(C_0(C))$.

Since $\sigma_n \in \mathcal{I}$ we have for all $x \in P_0 \otimes C$ (and hence all $x \in C_0 \otimes C$)

$$(4.1) \qquad q_0\rho_n(x) = q_0(x).$$

Indeed, (4.1) clearly holds in the case $x = t^m \otimes c$ and hence it holds by density for any $x \in C_0 \otimes C$.
Since $\sigma_n$ is quasicentral we have

$$ \forall x, y \in C_0 \otimes \mathbf{C} \quad \limsup_n \|\rho_n(xy) - \rho_n(x)\rho_n(y)\| = 0. $$

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

$$ \limsup_n \|\rho_n(x)^* - \rho_n(x^*)\| = 0. $$

It follows that after composing $\rho$ by the quotient map $Q : \ell_\infty(C_0(\mathbf{C})) \to \mathcal{L}(C_0(\mathbf{C}))$ we obtain a map $\rho_\infty : C_0 \otimes \mathbf{C} \to \mathcal{L}(C_0(\mathbf{C}))$ which is a *-homomorphism, such that $\rho_\infty = Q\rho$ on $C_0 \otimes \mathbf{C}$. Since $C_0$ is nuclear, this extends to a *-homomorphism $\rho_\infty : C_0(\mathbf{C}) \to \mathcal{L}(C_0(\mathbf{C}))$ defined on the whole of $C_0(\mathbf{C})$ with $\|\rho_\infty\| \le 1$.

We have

$$ (4.2) \qquad \forall f \in C_0 \otimes \mathcal{I} \quad \limsup_n \|\rho_n(f)\| = 0, $$

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

$$ \rho_{[\infty]} : C_0(\mathbf{C})/C_0(\mathcal{I}) \to \mathcal{L}(C_0(\mathbf{C})). $$

We have clearly $\rho_{[\infty]}q_0 = Q\rho$ on $C_0 \otimes \mathbf{C}$. To check that $\rho_{[\infty]}$ is isometric it suffices to show that for any $x \in C_0 \otimes \mathbf{C}$ we have $\|\rho_{[\infty]}q_0(x)\| \ge \|q_0(x)\|$. By (4.1) we have

$$ (4.3) \qquad \forall x \in C_0 \otimes \mathbf{C} \quad \forall n \quad \|q_0(x)\| \le \|\rho_n(x)\|. $$

This implies $\|q_0(x)\| \le \limsup \|\rho_n(x)\| = \|Q\rho(x)\| = \|\rho_{[\infty]}q_0(x)\|$. Thus $\rho_{[\infty]}$ is isometric. $\square$

**Remark 4.2.** [On unitizations] We start by some standard background. Recall that the unitization of a $C^*$-algebra $A$ is the unique unital $C^*$-algebra $\hat{A} \supset A$ such that any unital *-homomorphism $u : A \to B$ into a unital $C^*$-algebra $B$ uniquely extends to a unital *-homomorphism. The vector space $\hat{A}$ is spanned by $A$ and the unit of $\hat{A}$, and $A$ is an ideal in $\hat{A}$ such that $\hat{A}/A \simeq \mathbb{C}$. The Banach spaces $\hat{A}$ and $A \oplus \mathbb{C}$ are isomorphic with equivalent norms. Given a linear map $u : A \to B$ from a $C^*$-algebra $A$ to a unital one, let $\hat{u} : \hat{A} \to B$ denote its unital extension to the unitization of $A$. When $u$ is a *-homomorphism (resp. an isometric one), so is $\hat{u}$.

Recall that $C(\mathbf{C})$ (resp. $C(B)$) denotes the unitization of $C_0(\mathbf{C})$ (resp. $C_0(B)$). Using the identifications $C_0(B) \simeq C_0(\mathbf{C})/C_0(\mathcal{I}) \simeq C_0(\mathbf{C}/\mathcal{I})$ we have an isometric *-homomorphism

$$ \rho_{[\infty]} : C_0(\mathbf{C}/\mathcal{I}) \to \mathcal{L}(C_0(\mathbf{C})). $$

Clearly $C_0(\mathcal{I})$ can be viewed as a subalgebra (in fact an ideal) in $C(\mathbf{C})$. Using the characteristic property of unitizations, it is easy to verify that $C(\mathbf{C})/C_0(\mathcal{I})$ can be identified with $C(B) = C(\mathbf{C}/\mathcal{I})$. We have a canonical isometric embedding

$$ i_{\mathcal{L}} : \mathcal{L}(C_0(\mathbf{C})) \subset \mathcal{L}(C(\mathbf{C})), $$

derived from the inclusion $C_0(\mathbf{C}) \subset C(\mathbf{C})$. Since $\mathcal{L}(C(\mathbf{C}))$ is unital we have a unital *-homomorphism $\tilde{\rho}_{[\infty]} : C(\mathbf{C}/\mathcal{I}) \to \mathcal{L}(C(\mathbf{C}))$ extending $i_{\mathcal{L}}\rho_{[\infty]}$. Clearly, $\tilde{\rho}_{[\infty]}$ remains isometric. Moreover, denoting

$$ V = \mathbb{C}1_{C(\mathbf{C})} + C_0 \otimes \mathbf{C} \subset C(\mathbf{C}) $$

the map $\rho = (\rho_n) : C_0 \otimes \mathbf{C} \to L(C_0(\mathbf{C})) \subset L(C(\mathbf{C}))$ has a unital extension $\tilde{\rho} : V \to L(C(\mathbf{C}))$ and $V$ is a dense subspace of $C(\mathbf{C})$. Let $Q' : L(C(\mathbf{C})) \to \mathcal{L}(C(\mathbf{C}))$ denote the quotient map. It is then easy to check that $Q'\tilde{\rho}$ is the restriction to $V$ of $\tilde{\rho}_{[\infty]}$.

All in all this shows that there is a version of Lemma 4.1 valid for the unital cone algebras as follows.
**Lemma 4.3.** Let $\mathbf{C} : \mathbf{C} \to \mathcal{B}$ be a surjective $\mathcal{C}$-algebra, with separable kernel $\mathcal{I}$. Let $i : C_0(\mathbf{C})/C_0(\mathcal{I}) \to C(\mathbf{C})/C_0(\mathcal{I})$ be the natural inclusion. Let $\tilde{q}_0 : C(\mathbf{C}) \to C(\mathbf{C})/C_0(\mathcal{I})$ be the unital map associated to $iq_0$. There is a pointwise bounded sequence of unital maps $\tilde{\rho}_n : \mathcal{V} \to \mathcal{V}$ that induces in the limit a unital $\mathcal{C}$-homomorphism

$$ \tilde{\rho}_{\infty} : C(\mathbf{C}) \to \mathcal{L}(C(\mathbf{C})) $$

that vanishes on $C_0(\mathcal{I})$ and for which the associated map

$$ \tilde{\rho}_{[\infty]} : C(\mathbf{C})/C_0(\mathcal{I}) \to \mathcal{L}(C(\mathbf{C})) $$

(which satisfies $\tilde{\rho}_{[\infty]}\tilde{q}_0 = \tilde{\rho}_{\infty}$) is a unital isometric $\mathcal{C}$-homomorphism. Moreover, we have $\tilde{q}_0\tilde{\rho}_n(x) = \tilde{q}_0(x)$ for all $x \in \mathcal{V}$ and all $n$.

**Remark 4.4.** Alternatively, we could denote the unitization of $\mathcal{L}(C_0(\mathbf{C}))$ by $\hat{\mathcal{L}}$ and by $\alpha : \mathcal{L}(C_0(\mathbf{C})) \to \hat{\mathcal{L}}$ the natural inclusion. Then $\tilde{\rho}_{[\infty]} = \widehat{\alpha\rho_{[\infty]}}\widehat{i_{\mathcal{L}}}$.

Now let $j : \mathbf{C} \to \mathcal{B}$ be an embedding so that $j_0 : C_0(\mathbf{C}) \to C_0(\mathcal{B})$ and $F = \rho_{[\infty]}j_0$ are clearly also embeddings. Let $\tilde{j}_0 : C(\mathbf{C}) \to C(\mathcal{B})$ be the canonical extension of $j_0$ to the unitizations (again an isometric $\mathcal{C}$-homomorphism). Recall $\mathcal{B} = \mathbf{C}/\mathcal{I}$. Using the identification

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

we may set

$$ \tilde{F} = \tilde{\rho}_{[\infty]}\tilde{j}_0 : C(\mathbf{C}) \to \mathcal{L}(C(\mathbf{C})). $$

Note that $\tilde{F}$ is but the canonical unital extension of the map $i_{\mathcal{L}}F$. We refer the reader to the diagrams below.

**Lemma 4.5.** Let $j$ be as above and let

$$ F : C_0(\mathbf{C}) \to \mathcal{L}(C_0(\mathbf{C})) \quad (\text{resp. } \tilde{F} : C(\mathbf{C}) \to \mathcal{L}(C(\mathbf{C}))) $$

be as above. Let $f = (f_n)$ be a c.c. lifting for $F$ (resp. $\tilde{F}$). Then the condition (1.1) holds.

**Proof.** Let $V \subset \mathcal{C}$ be a f.d. subspace and let $\Delta = \mathcal{P}_N \otimes V \subset C_0 \otimes \mathbf{C}$ where $\mathcal{P}_N = \text{span}[t, \dots, t^N]$. Recall $jV \subset \mathcal{B} = \mathbf{C}/\mathcal{I}$. Let $\psi : jV \to \mathbf{C}$ be any linear lifting, i.e. we have $q\psi j(x) = j(x)$ for any $x \in \mathcal{V}$, and hence $q_0\psi_0j_0(x) = j_0(x)$ for any $x \in \Delta$. Let $g_n = \rho_n\psi_0j_0|_{\Delta} : \Delta \to C_0(\mathbf{C})$ (actually $g_n$ takes values in $\mathcal{P}_N \otimes \mathbf{C}$). We first claim that $g_n$ is asymptotic to $f_{n|\Delta}$ by which we mean

$$ \forall x \in \Delta \quad \|g_n(x) - f_n(x)\| \to 0. $$

Indeed, we have $Q(f(x)) = F(x) = \rho_{[\infty]}j_0(x)$ and also $Q(g(x)) = Q\rho\psi_0j_0(x) = \rho_{\infty}\psi_0j_0(x) = \rho_{[\infty]}q_0\psi_0j_0(x) = \rho_{[\infty]}j_0(x)$. Thus $Q(f(x)) = Q(g(x))$ which is equivalent to our first claim. Our second claim is that

$$ \forall n \quad \|g_n^{-1}|_{g_n(\Delta)}\|_{cb} \le 1. $$

Indeed, since (see Lemma 4.1) $q_0\rho_n = q_0$ on $C_0 \otimes \mathbf{C}$ we have $q_0g_n = q_0\psi_0j_0 = j_0$, from which we derive $\|x\| = \|j_0(x)\| \le \|g_n(x)\|$. Equivalently $\|g_n^{-1}|_{g_n(\Delta)}\| \le 1$. Similarly since $q_0$ is c.c. and $j_0$ completely isometric we obtain our second claim.

To conclude, an elementary perturbation argument shows that the two claims together with the complete contractivity of the $f_n$'s imply

$$ (4.4) \qquad \limsup \|f_n^{-1}|_{f_n(E)}\|_{cb} \le 1. $$
Indeed, by [6, Lemma 2.13.2 p. 69] for any $\varepsilon > 0$ and any $E \subset C_0(\mathbf{C})$ there are $\Delta \subset \mathcal{P} \otimes \mathbf{C}$ and a complete isomorphism $h : C_0(\mathbf{C}) \to C_0(\mathbf{C})$ such that $h(\Delta) = E$ and $\|h - Id\|_{cb} < \varepsilon$ and also for convenience $\|h^{-1} - Id\|_{cb} < \varepsilon$. Since $\Delta$ is f.d. $\limsup \|(f_n - g_n)_{|\Delta}\|_{cb} = 0$ by the first claim. By the two claims for any $0 < \varepsilon < 1$ there is $N = N_\varepsilon$ such that $\sup_{n \ge N} \|(f_n - g_n)_{|\Delta}\|_{cb} \le \varepsilon$ and also $\|g_n^{-1}\|_{|g_n(\Delta)}\|_{cb} \le (1 - \varepsilon)^{-1}$. Fix $n \ge N$. Let $x \in E$ with $\|x\| = 1$. Let $y = h^{-1}(x) \in \Delta$. Then $1 - \varepsilon \le \|y\| \le 1 + \varepsilon$. We have $\|g_n(y)\| \ge (1 - \varepsilon)\|y\|$ and $\|f_n(y) - f_n(x)\| \le \|y - x\| \le \varepsilon$. It follows

$$ \|f_n(x)\| \ge \|f_n(y)\| - \varepsilon \ge \|g_n(y)\| - \varepsilon\|y\| - \varepsilon \ge (1 - \varepsilon)\|y\| - \varepsilon\|y\| - \varepsilon \ge (1 - 2\varepsilon)(1 - \varepsilon) - \varepsilon. $$

This shows that $\|f_n^{-1}\|_{|f_n(E)} \le [(1 - 2\varepsilon)(1 - \varepsilon) - \varepsilon]^{-1}$.

An entirely similar reasoning with $x$ in the unit sphere of (say) $\mathcal{B} \otimes_{\min} E$ shows the same upper bound for $\|f_n^{-1}\|_{|f_n(E)}\|_{cb}$. Since this holds for any $n \ge N$ and any $\varepsilon > 0$ (4.4) alias (1.1) follows.

This completes the proof for $F$, the case of $\tilde{F}$ is identical. $\square$

The situation is summarized by the following diagrams.

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

By Remark 4.2 we also have a unital isometric $*$-homomorphism $\tilde{F} : C(\mathbf{C}) \to \mathcal{L}(C(\mathbf{C}))$ that factors through $C(B)$.

Intuitively, we would like to apply Lemma 4.1 to the case of a surjective $*$-homomorphism $q : C^*(\mathbb{F}) \to \mathcal{B}$ ($\mathbb{F}$ some large enough free group) together with an isometric $*$-homomorphism $j : C^*(\mathbb{F}) \to \mathcal{B}$, and to use the fact that $Id_{\mathcal{B}}$ is $\mathcal{C}$-nuclear but since we wish to remain in the separable realm, we need to make some adjustments. There is clearly an embedding $\mathcal{C} \subset \mathcal{B}$, and it is easy to show that there is a separable $C^*$-subalgebra $B \subset \mathcal{B}$ with WEP that contains $\mathcal{C}$. Let $j : \mathcal{C} \to B$ be the inclusion map. Since $B$ is separable we also have a surjective $*$-homomorphism $q : \mathcal{C} \to B$, whence $F = \rho_{[\infty]}j_0 : C_0(\mathcal{C}) \to \mathcal{L}(C_0(\mathcal{C}))$. The passage to the cone algebras is harmless: $C_0(\mathcal{C})$ has the LP and $C_0(B)$ the WEP, i.e. its identity map is $\mathcal{C}$-nuclear. A fortiori the map $F$ is $\mathcal{C}$-nuclear.

Thus we obtain:

**Theorem 4.6.** There is an isometric $*$-homomorphism $F : C_0(\mathcal{C}) \to \mathcal{L}(C_0(\mathcal{C}))$ (and a unital one $\tilde{F} : C(\mathcal{C}) \to \mathcal{L}(C(\mathcal{C}))$) that is also $\mathcal{C}$-nuclear.

*Remark 4.7* (Unital variant). Let $f = (f_n)$ be a unital c.c. lifting of $\tilde{F}$. Then if we start by a unital subspace $E_0$, the construction produces unital algebras $Z$ and $A$ (see Remark 2.6).

We are thus in a position to apply Theorem 3.12.

**Theorem 4.8.** In the situation just described there is an inductive system associated to the map $F : C_0(\mathcal{C}) \to \mathcal{L}(C_0(\mathcal{C}))$ (resp. $\tilde{F} : C(\mathcal{C}) \to \mathcal{L}(C(\mathcal{C}))$) for which the associated (resp. unital) $C^*$-algebra $A$ has the WEP and the LLP but is not nuclear. Moreover, $A$ locally embeds in $\mathcal{C}$.

*Proof.* We apply Theorem 3.12 with $C = C_0(\mathcal{C})$ to the map $F = \rho_{[\infty]}j_0 : C_0(\mathcal{C}) \to \mathcal{L}(C_0(\mathcal{C}))$ derived above from the coexistence of $j$ and $q$ (or of $j_0$ and $q_0$) on $\mathcal{C}$. By Lemma 4.5, the condition (1.1) holds. By Proposition 2.8 the algebra $A$ locally embeds in $C_0(\mathcal{C})$ which itself locally embeds
in $\mathcal{C}$ (see Remark 2.9). By Proposition 3.10 A has the LLP. We use $D = \mathcal{C}$ in Theorem 3.12. Then A is $\mathcal{C}$-nuclear which means it has the WEP.

Let us now show how to make sure that A is not nuclear. For that purpose we choose for $E_0$ a suitable subspace of $C_0(\mathcal{C})$, for instance we can select $E_0$ linearly spanned by $(t \otimes U_1, \cdots, t \otimes U_d)$ where $(U_1, \cdots, U_d)$ are a $d$-tuple of free unitary generators of $\mathbb{F}_\infty$ viewed as unitaries in $\mathcal{C}$ (here $t$ means abusively the function $t \mapsto t$ on $[0, 1]$). It is known (see [6, Th. 21.5, p. 336]) that for any subspace $S$ of a nuclear $C^*$-algebra we have

$$d_{cb}(E_0, S) \geq d/2\sqrt{d-1}.$$

When $d \geq 3$ we have $d/2\sqrt{d-1} > 1$. Thus by (2.14) to show that A is not nuclear with this choice of $E_0$ it suffices to choose our sequence $(\varepsilon_n)$ so that $(1 + \eta_1)^2 < d/2\sqrt{d-1}$. See Remark 4.9 below for an alternate argument. $\square$

*Remark 4.9* (How to ensure that $\mathcal{C}$ locally embeds in A). In the construction of the inductive system we may always enlarge $E_n$ at any step and then the next steps must of course be modified accordingly. Thus, if $S_n \subset C_0(\mathcal{C})$ is any sequence of f.d.s.a. subspaces that is dense for the $d_{cb}$ distance in the space of f.d.s.a. subspaces of $C_0(\mathcal{C})$, we may define $E_n = T_n(E_{n-1}) + T_n(E_{n-1})T_n(E_{n-1}) + S_n$ for each $n \geq 1$. Then (2.14) shows that we have a subspace $A'_n \subset A_n$ corresponding to $S_n \subset E_n$ such that $d_{cb}(S_n, A'_n) \leq (1 + \eta_{n+1})^2$. If we use a sequence $(S_n)$ where each of the original $S_n$'s is repeated infinitely many times, this shows that for the resulting system $C_0(\mathcal{C})$ and hence $\mathcal{C}$ itself locally embeds in A.

# 5 Appendix

**Lemma 5.1.** Let A be a $C^*$-algebra that locally embeds in $\mathcal{C}$. Then A has the LLP if and only if A is $j$-nuclear, where $j: \mathcal{C} \to B$ is any given isometric $*$-homomorphism from $\mathcal{C}$ to a $C^*$-algebra B with WEP.

*Proof.* The only if part is obvious. To show the if part, assume that A is $j$-nuclear. Let $u: A \to C/\mathcal{I}$ be a c.p.c.c. map. Since $Id_A$ is $j$-nuclear, the map $j \otimes u$ is a contraction from $\mathcal{C} \otimes_{\min} A \to B \otimes_{\max} C/\mathcal{I}$. Let $E \subset A$ be a f.d. subspace. Let $t \in E^* \otimes A$ be the tensor associated to the inclusion map $E \to A$. We have $\|t\|_{\min} = 1$. Recall that $E^*$ always locally embeds in $\mathcal{C}$ when E does (see [6, Lem. 21.10 p. 339]). Therefore there is a c.i. embedding $k: E^* \to \mathcal{C}$. We have $[k \otimes Id_A](t) \in \mathcal{C} \otimes A$ with $\|[k \otimes Id_A](t)\|_{\min} = 1$. After composing with $j$, we find $\|[jk \otimes Id_A](t)\|_{\max} \leq 1$, and since c.p.c.c. maps between $C^*$-algebras preserve the max norm, we have $\|[jk \otimes u](t)\|_{\max} \leq 1$. Here $t' = [jk \otimes u](t) \in jk(E^*) \otimes_{\max} C/\mathcal{I} \subset B \otimes_{\max} C/\mathcal{I}$. By the projectivity of the max-tensor product, there is an element $s$ in the unit ball of $jk(E^*) \otimes_{\max} C \subset B \otimes_{\max} C$ lifting $t'$, meaning, denoting by $q: C \to C/\mathcal{I}$ the quotient map, that $t' = [Id \otimes q](s)$. A fortiori $\|s\|_{\min} \leq 1$. Since $jk$ is c.i. we recover from $s$ an element $\hat{t}$ in the unit ball of $E^* \otimes_{\min} C$, such that $[Id \otimes q](\hat{t}) = t$. Let $v: E \to C$ be the linear map associated to $\hat{t}$. Then $v$ lifts the inclusion $u|_E$ and $\|v\|_{cb} = 1$. This yields the LLP for A. $\square$

**Acknowledgement.** I am grateful to Jean Roydor for useful conversations and the simpler proof of Proposition 3.8.
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