2 Background and basics

We begin by recalling the definitions of the minimal and maximal $C^{*}$ -cross-norms and refer the reader to [Reference Paulsen10, Reference Pisier11] for more details. Throughout the paper, we will denote an arbitrary Hilbert space by H and by $\ell ^{2}$ a separable Hilbert space. The notion of complete positivity is shortened to c.p., while u.c.p. stands for a unital, completely positive map. Let $A_{1}$ and $A_{2}$ be unital $C^{*}$ -algebras. A $C^{*}$ -cross-norm on the algebraic tensor product $A_{1} \otimes A_{2}$ is a $C^{*}$ -norm $\gamma $ satisfying the additional condition $||a\otimes b||_{\gamma } = ||a||\cdot ||b||.$ If $\pi _{1} : A_{1} \rightarrow B(H_{1} )$ and $\pi _ 2 : A_{2}\rightarrow B(H_{2} )$ are unital $*\mbox {-homomorphisms}$ , we get a unital, $*$ -preserving homomorphism $\pi _ 1 \otimes \pi _{2} : A_{1} \otimes A_{2}\rightarrow B(H_{1}\otimes H_{2} )$ by letting $\pi _ 1 \otimes \pi _{2} (a \otimes b) = \pi _{1} (a) \otimes \pi _{2} (b). $ Thus, if for $x\in A_{1}\otimes A_{2}$ , we set $||x||_{\mathrm{min}} = \sup \{ \pi _ 1 \otimes \pi _2 (x) : \pi _ i : A _i \rightarrow B(H_ i ) ~ \mbox {unital }{*}\mbox{-homomorphism}, i = 1, 2\},$ then we obtain a $C^{*}$ -cross-norm on $A_{1}\otimes A_{2}.$ The completion of $A_ 1 \otimes A_ 2$ in this norm is denoted by $A_{1}\otimes _{\mathrm{min}} A_{2} $ and is called the minimal (or the spatial) tensor norm. It is the smallest possible $C^{*}$ -cross-norm on the algebraic tensor product $A_{1}\otimes A_{2}.$ It is not difficult to see that the minimal tensor norm is given by $||\sum _{i=1}^{n} a_{i}\otimes b_{i} ||_{\mathrm{min}} = || \sum _{i=1}^{n} \pi _{1}(a_{i})\otimes \pi _{2}(b_{i}) ||$ in $B(H)\otimes B(K)\subset B(H\otimes K),$ where $\pi _{1}:A\rightarrow B(H)$ and $\pi _{2}:B\rightarrow B(K)$ are arbitrary faithful $*$ -representations.

Let now $ \pi _ 1 : A \rightarrow B(H)$ and $\pi _{2} : B \rightarrow B(H)$ be unital $*$ -homomorphisms such that $\pi _{1} (a)\pi _{2} (b) = \pi _ 2 (b)\pi _ 1 (a) $ for all $a\in A_{1}$ and $b\in A_{2}.$ Define a unital $*\mbox{-homomorphism } \pi : A \otimes B \rightarrow B(H)$ by $\pi (x) = \sum _{i=1}^{n} \pi _{1}(a_{i})\pi _{2}(b_{i}),$ where $x = \sum ^{n}_{ i=1} a_{i}\otimes b_ i. $ Conversely, if we have a unital $*$ -homomorphism $\pi : A \otimes B \rightarrow B(H)$ and we define $\pi _{1} (a) = \pi (a \otimes I), ~ \pi _{2} (b) = \pi (I\otimes b),$ we obtain a pair of unital $*$ -homomorphisms of A and $B,$ respectively, with commuting ranges such that $\pi (a \otimes b) = \pi _ 1 (a)\pi _ 2 (b).$ We define $||x||_{ \max } = \sup \{ || \pi (x)|| : \pi : A \otimes B \rightarrow B(H) ~ \mbox { unital }*\mbox{-homomorphisms}\}.$ The completion of $A \otimes B$ in this norm, denoted by $A\otimes _{\mathrm{max}} B, $ is called the maximal tensor norm, and is the largest possible $C^{*}$ -cross-norm on the algebraic tensor product $A\otimes B.$

In the nonunital case, both min and max tensor products are taken to be the respective subalgebras of $A^{+}\otimes _{\mathrm{min}} B^{+}$ and $A^{+}\otimes _{\mathrm{max}} B^{+},$ where $A^{+}$ and $B^{+}$ are the unitisations of A and $B.$ A $C^{*}$ -algebra is called nuclear if $A\otimes _{\mathrm{min}}B = A\otimes _{\mathrm{max}} B$ for every $C^{*}$ -algebra $B.$ It is well known that abelian $C^{*}$ -algebras, $M_{n}(\mathbb {C}),$ and the algebra of compact operators $K(H)$ are nuclear. We recall that a $C^{*}$ -algebra A is exact if the sequence

$$ \begin{align*} 0 \rightarrow A\otimes_{\mathrm{min}} B/J \rightarrow A\otimes_{\mathrm{min}} J \rightarrow A\otimes_{\mathrm{min}} B \rightarrow 0\end{align*} $$

is exact for every $C^{*}$ -algebra B and every closed, two-sided ideal $J\subset B.$

Two important and well-known results that we will use repeatedly are collected in the next proposition. We refer the reader to [Reference Pisier11], where the first is a particular case of Corollary 4.18 and the second is Proposition 7.15.

It can be proved without difficulty that if $A_{i}\subset B_{i}, ~ i=1,2,$ then $A_{1}\otimes _{\mathrm{min}} A_{2} \subset B_{1}\otimes _{\mathrm{min}} B_{2},$ but this is no longer the case for the max norm. Specifically, if $A\subset B,$ then it is possible that the inclusion $A\otimes C\subset B\otimes _{\mathrm{max}} C$ does not induce the max norm on $A\otimes C.$ If it does, we write $A\otimes _{\mathrm{max}} C\subset B\otimes _{\mathrm{max}} C.$

In [Reference Lance9], Lance observed that $A\otimes _{\mathrm{max}} C\subset A^{**}\otimes _{\mathrm{max}} C$ for all $C^{*}$ -algebras A and $C,$ and introduced the following property: given a unital inclusion $A\subset B,$ we say that A is weakly c.p. complemented in B if there exists a u.c.p. map $\psi :B\rightarrow A^{**}$ such that $\psi (a) = a$ for all $a\in A.$ If the above map $\psi $ takes values in $A,$ then A is said to be c.p. complemented in $B.$ It is easy to see that if $A\subset B\subset C$ and A is (weakly) c.p. complemented in $C,$ then A will be (weakly) c.p. complemented in B as well.

Lance proved in [Reference Lance9] that if A is weakly c.p. complemented in $B,$ then $A\otimes _{\mathrm{max}} C \subset B\otimes _{\mathrm{max}} C$ for any $C.$ The particular case when A is c.p. complemented in B will be used subsequently, so we present it as Lemma 3.1 for the benefit of the reader. A $C^{*}$ -algebra A has the WEP if, for some faithful representation on a Hilbert space $H,$ A is weakly c.p. complemented in $B(H).$ It can be seen [Reference Lance9] that this definition does not depend on the particular representation of $A.$

A u.c.p. map $\varphi :A\rightarrow B/J$ is said to be u.c.p. liftable if there exists a u.c.p. map $\psi :E\rightarrow B$ such that $\varphi |_{E}= q\circ \psi ,$ where q is the quotient map. The map $\varphi :A\rightarrow B/J$ is locally u.c.p. liftable if, for every finite dimensional operator system $E\subset A,$ there exists a u.c.p. map $\psi :E\rightarrow B$ such that $\varphi |_{E}= q\circ \psi .$ A $C^{*}$ -algebra has the lifting property (LP) (respectively the LLP) if every u.c.p. map from A to $B/J$ is u.c.p. liftable (respectively locally u.c.p. liftable) to $B.$

In [Reference Choi and Effros1], Choi and Effros proved that separable nuclear $C^{*}$ -algebras have the LP, while Kirchberg [Reference Kirchberg7] proved the same for the nonnuclear $C^{*}$ -algebra $C^{*}(\mathbb {F}_{\infty }),$ the full $C^{*}$ -algebra of the free group on countably many generators.

The next proposition contains several fundamental results based on Kirchberg’s work [Reference Kirchberg6].

Proof. Statements (i), (ii) and (iii) constitute Proposition 1.1 in [Reference Kirchberg6]. To prove statement (iv), suppose that $A \otimes _{\mathrm{max}} B(H) \neq A\otimes _{\mathrm{min}} B(H)$ and choose operators $a_{i} \in A, b_{i}\in B(H), ~ i=1,\ldots ,n,$ such that

$$ \begin{align*}\bigg\|\sum_{i=1}^{n} a_{i} \otimes b_{i}\bigg\|_{A\otimes_{\mathrm{min}} B(H)} < \bigg\|\sum_{i=1}^{n} a_{i} \otimes b_{i}\bigg\|_{A\otimes_{\mathrm{max}} B(H)}.\end{align*} $$

By Lemma 2.4 in [Reference Kirchberg6], there exists a separable $C^{*}$ -subalgebra $B\subset B(H)$ with the WEP such that $b_{i}\in B, ~ i=1,\ldots ,n.$ Then B is weakly c.p. complemented in $B(H)$ and

$$ \begin{align*}\bigg\|\sum_{i=1}^{n} a_{i} \otimes b_{i}\bigg\|_{A\otimes_{\mathrm{max}} B} {=} \bigg\|\sum_{i=1}^{n} a_{i} \otimes b_{i}\bigg\|_{A\otimes_{\mathrm{max}} B(H)}> \bigg\|\sum_{i=1}^{n} a_{i} \otimes b_{i}\bigg\|_{A\otimes_{\mathrm{min}} B(H)} {=} \bigg\|\sum_{i=1}^{n} a_{i} \otimes b_{i}\bigg\|_{A\otimes_{\mathrm{min}} B}, \end{align*} $$

a contradiction.

3 Preliminary results

In this section, we collect several facts that will be used subsequently in proving the main results.

Proof. By Proposition 2.1(b), the inclusion $\iota :A\rightarrow B$ extends to a u.c.p. map $\iota \otimes \mathrm {id}:A\otimes _{\mathrm{max}} C\rightarrow B\otimes _{\mathrm{max}} C$ . Therefore, for all $a_{1},\ldots ,a_{n} \in A$ and $c_{1},\ldots ,c_{n} \in C$ ,

$$ \begin{align*} \bigg\|\sum_{i=1}^{n} a_{i} \otimes c_{i}\bigg\|_{B\otimes_{\mathrm{max}} C} = \bigg\|\sum_{i=1}^{n} \iota(a_{i}) \otimes c_{i}\bigg\|_{B\otimes_{\mathrm{max}} C} & = \bigg\|(\iota\otimes \mathrm{id})\bigg(\sum_{i=1}^{n} a_{i} \otimes c_{i}\bigg) \bigg\|_{B\otimes_{\mathrm{max}} C} \\ & \leq \bigg\|\sum_{i=1}^{n} a_{i}\otimes c_{i}\bigg\|_{A\otimes_{\mathrm{max}}C}. \end{align*} $$

Similarly, $\theta $ extends to a u.c.p. map $\theta \otimes \mathrm {id}:B\otimes _{\mathrm{max}}C \rightarrow A\otimes _{\mathrm{max}} C,$ so

$$ \begin{align*} \bigg\|\sum_{i=1}^{n} a_{i}\otimes c_{i}\bigg\|_{A\otimes_{\mathrm{max}}C} = \bigg\| \sum_{i=1}^{n} \theta(a_{i})\otimes c_{i}\bigg\|_{A\otimes_{\mathrm{max}}C} & = \bigg\|(\theta\otimes \mathrm{id})\bigg(\sum_{i=1}^{n} a_{i} \otimes c_{i}\bigg) \bigg\|_{A\otimes_{\mathrm{max}} C} \\ & \leq \bigg\|\sum_{i=1}^{n} a_{i}\otimes c_{i}\bigg\|_{B\otimes_{\mathrm{max}}C}. \\[-3.7pc] \end{align*} $$

Proof. (a) We apply Lemma 3.1 for B replaced by $A\otimes _{\mathrm{min}} B$ (respectively $B\otimes _{\mathrm{min}} A$ ) and identify A with the subalgebra $A\otimes I$ (respectively $I\otimes A$ ) of $ A\otimes _{\mathrm{min}} B.$ As far as $\theta $ is concerned, choose an arbitrary state $\varphi $ on B and define $\theta :A\otimes _{\mathrm{min}} B\rightarrow A$ by

$$ \begin{align*}\theta\bigg(\sum_{i=1}^{n} a_{i}\otimes b_{i} \bigg) = \sum_{i=1}^{n} \varphi(b_{i}) a_{i}. \end{align*} $$

Respectively, define $\theta :B\otimes _{\mathrm{min}} A\rightarrow A$ by

$$ \begin{align*}\theta\bigg(\sum_{i=1}^{n} b_{i}\otimes a_{i} \bigg) = \sum_{i=1}^{n} \varphi(b_{i}) a_{i}. \end{align*} $$

(b) We use induction. The case $n=2$ is part (a). Suppose now that the property is true for n and consider $n+1$ algebras $A_{1}, \ldots , A_{n+1}.$ For $i=1$ ,

$$ \begin{align*}A_{1}\otimes_{\mathrm{max}} B \subset [A_{1}\otimes_{\mathrm{min}} (A_{2}\otimes_{\mathrm{min}}\cdots\otimes_{\mathrm{min}} A_{n+1})] \otimes_{\mathrm{max}} B\end{align*} $$

by part (a), while for $2\leq i \leq n+1$ ,

$$ \begin{align*}A_{i}\otimes_{\mathrm{max}} B \subset (A_{2}\otimes_{\mathrm{min}}\cdots\otimes_{\mathrm{min}} A_{n+1}) \otimes_{\mathrm{max}} B\end{align*} $$

by the inductive hypothesis and

$$ \begin{align*} (A_{2}\otimes_{\mathrm{min}}\cdots\otimes_{\mathrm{min}} A_{n+1}) \otimes_{\mathrm{max}} B \subset [A_{1}\otimes_{\mathrm{min}} (A_{2}\otimes_{\mathrm{min}}\cdots\otimes_{\mathrm{min}} A_{n+1})] \otimes_{\mathrm{max}} B\end{align*} $$

by part (a). The conclusion follows.

4 Tensor products with the Calkin algebra

In this section, we prove that a $C^{*}$ -algebra whose max and min tensor products with the Calkin algebra are equal must be exact and have the LLP. This result, which represents the core technical ingredient needed in the proof of the main result, has also been folklore for a while. It first appeared in print recently as Corollary 10.10 in [Reference Pisier11]. The proof we present here is different and shorter than the one in [Reference Pisier11].

Proof. $\Leftarrow $ The sequence

$$ \begin{align*} 0 \rightarrow A\otimes_{\mathrm{max}} B(\ell^{2})/K(\ell^{2}) \rightarrow A\otimes_{\mathrm{max}} K(\ell^{2}) \rightarrow A\otimes_{\mathrm{max}} B(\ell^{2}) \rightarrow 0\end{align*} $$

is exact and, since A is exact, so is the sequence

$$ \begin{align*} 0 \rightarrow A\otimes_{\mathrm{min}} B(\ell^{2})/K(\ell^{2}) \rightarrow A\otimes_{\mathrm{min}} K(\ell^{2}) \rightarrow A\otimes_{\mathrm{min}} B(\ell^{2}) \rightarrow 0.\end{align*} $$

Since A has the LLP, $A\otimes _{\mathrm{min}} B(\ell ^{2}) = A\otimes _{\mathrm{max}} B(\ell ^{2}).$ This, together with the fact that $A\otimes _{\mathrm{min}} K(\ell ^{2}) = A\otimes _{\mathrm{max}} K(\ell ^{2})$ (since $K(\ell ^{2})$ is nuclear), leads to the conclusion.

$\Rightarrow $ First we prove that A has the LLP. To get a contradiction, suppose, by using Proposition 2.2(iv), that there exists a separable WEP $C^{*}$ -algebra B with the property $A\otimes _{\mathrm{min}} B \neq A\otimes _{\mathrm{max}} B.$ As a consequence, there exist operators $a_{i}\in A, b_{i}\in B, i=1,\ldots ,n,$ and $\varepsilon>0$ such that

$$ \begin{align*}\bigg\|\sum_{i=1}^{n} a_{i} \otimes b_{i}\bigg\|_{A\otimes_{\mathrm{max}}B} =1 \quad\mbox{and}\quad \bigg\|\sum_{i=1}^{n} a_{i} \otimes b_{i}\bigg\|_{A\otimes_{\mathrm{min}}B} < 1-\varepsilon. \end{align*} $$

We may assume without loss of generality that B is faithfully represented on a separable Hilbert space, viewed as $\ell ^{2},$ in such a way that $B \cap K(\ell ^{2}) = \{0\}.$ Then the quotient map $\pi : B(\ell ^{2}) \rightarrow B(\ell ^{2})/K(\ell ^{2})$ induces a $*$ -isomorphism between B and $\pi (B) \subset B(\ell ^{2})/K(\ell ^{2}).$ If we view the Calkin algebra represented faithfully on some (nonseparable) Hilbert space $H,$ then

$$ \begin{align*}\pi(B) \subset B(\ell^{2})/K(\ell^{2}) \subset B(H).\end{align*} $$

Since $\pi (B)$ has the WEP, it is weakly c.p. complemented in $B(H);$ therefore, also weakly c.p. complemented in $B(\ell ^{2})/K(\ell ^{2}),$ as remarked in Section 2. In particular, $ A\otimes _{\mathrm{max}} \pi (B) \subset A \otimes _{\mathrm{max}} B(\ell ^{2})/K(\ell ^{2}).$ It follows that

$$ \begin{align*} \bigg\|\sum_{i=1}^{n} a_{i}\otimes b_{i} \bigg\|_{A\otimes_{\mathrm{max}} B} = \bigg\|\sum_{i=1}^{n} a_{i}\otimes \pi(b_{i}) \bigg\| _{A\otimes_{\mathrm{max}} \pi(B)} = \bigg\|\sum_{i=1}^{n} a_{i}\otimes \pi(b_{i}) \bigg\| _{A\otimes_{\mathrm{max}} B(\ell^{2})/K(\ell^{2})}. \end{align*} $$

However, by using the hypothesis, we obtain

$$ \begin{align*} \bigg\| \sum_{i=1}^{n} a_{i}\otimes \pi(b_{i}) \bigg\| _{A\otimes_{\mathrm{max}} B(\ell^{2})/K(\ell^{2})} & = \bigg\|\sum_{i=1}^{n} a_{i}\otimes \pi(b_{i}) \bigg\| _{A\otimes_{\mathrm{min}} B(\ell^{2})/K(\ell^{2})} \\ = \bigg\|\sum_{i=1}^{n} a_{i}\otimes \pi(b_{i}) \bigg\| _{A\otimes_{\mathrm{min}} \pi(B)} & = \bigg\|\sum_{i=1}^{n} a_{i}\otimes b_{i} \bigg\| _{A\otimes_{\mathrm{min}} B} < 1 - \varepsilon. \end{align*} $$

It follows that ${ ||\sum _{i=1}^{n} a_{i}\otimes b_{i} ||_{A\otimes _{\mathrm{max}} B} < 1 - \varepsilon ,}$ which contradicts the assumption that ${ ||\sum _{i=1}^{n} a_{i}\otimes b_{i} ||_{A\otimes _{\mathrm{max}} B} = 1.}$ We conclude that A has the LLP.

To prove exactness, we will show that in the short exact sequence

$$ \begin{align*} 0 \rightarrow A\otimes_{\mathrm{max}} B(\ell^{2})/K(\ell^{2}) \rightarrow A\otimes_{\mathrm{max}} K(\ell^{2}) \rightarrow A\otimes_{\mathrm{max}} B(\ell^{2}) \rightarrow 0,\end{align*} $$

we can replace ‘max’ by ‘min’. Since A has the LLP, it follows that $A\otimes _{\mathrm{max}} B(\ell ^{2}) = A\otimes _{\mathrm{min}} B(\ell ^{2}).$ By hypothesis, we also have $A\otimes _{\mathrm{max}} B(\ell ^{2})/K(\ell ^{2}) = A\otimes _{\mathrm{min}} B(\ell ^{2})/K(\ell ^{2}) $ which, together with $A\otimes _{\mathrm{max}} K(\ell ^{2}) = A\otimes _{\mathrm{min}} K(\ell ^{2}),$ leads to

$$ \begin{align*} 0 \rightarrow A\otimes_{\mathrm{min}} B(\ell^{2})/K(\ell^{2}) \rightarrow A\otimes_{\mathrm{min}} K(\ell^{2}) \rightarrow A\otimes_{\mathrm{min}} B(\ell^{2}) \rightarrow 0.\end{align*} $$

This implies that A is exact by Theorem 1.1 in [Reference Kirchberg5].

5 Detecting nuclearity

This section is devoted to the main results.

The connection between the two notions in the above definition is given in the next proposition.

Proof. Suppose that B satisfies

$$ \begin{align*} (A_{1}\otimes_{\mathrm{min}} A_{2}\otimes_{\mathrm{min}} \cdots \otimes_{\mathrm{min}} A_{n}) \otimes_{\mathrm{min}} B = (A_{1}\otimes_{\mathrm{min}} A_{2}\otimes_{\mathrm{min}} \cdots \otimes_{\mathrm{min}} A_{n})\otimes_{\mathrm{max}} B. \end{align*} $$

By Corollary 3.2(b),

$$ \begin{align*} A_{i}\otimes_{\mathrm{max}} B &\subset (A_{1}\otimes_{\mathrm{min}} A_{2}\otimes_{\mathrm{min}} \cdots \otimes_{\mathrm{min}} A_{n})\otimes_{\mathrm{max}} B\\ & = (A_{1}\otimes_{\mathrm{min}} A_{2}\otimes_{\mathrm{min}} \cdots \otimes_{\mathrm{min}} A_{n})\otimes_{\mathrm{min}} B; \end{align*} $$

therefore, $A_{i}\otimes _{\mathrm{min}} B = A_{i}\otimes _{\mathrm{max}} B$ for $1\leq i \leq n.$ Since $\mathbb {S}$ detects nuclearity, the conclusion follows.

We arrive at the main result of this section.

6 Other tensor product characterisations of nuclearity

In the same circle of ideas as nuclearity detection, we mention the issue of characterising nuclearity by the equality of the min and max norms on certain tensor products.

To illustrate this idea, we recall that a $C^{*}$ -algebra is nuclear if and only if any Hilbert space representation generates an injective von Neumann algebra. Equivalently, A is nuclear if and only if $A^{**}$ is injective [Reference Pisier11, 8.16]. Since, for von Neumann algebras, injectivity and the WEP are equivalent [Reference Pisier11, 9.26], we obtain the following characterisation of nuclearity [Reference Pisier11, 9.27].

Proposition 6.1. A $C^{*}$ -algebra A is nuclear if and only if

$$ \begin{align*} C^{*}(\mathbb{F}_{\infty})\otimes_{\mathrm{min}} A^{**} = C^{*}(\mathbb{F}_{\infty})\otimes_{\mathrm{max}} A^{**}.\end{align*} $$

Recall that, given a $C^{*}$ -algebra $A,$ the opposite $C^{*}$ -algebra, denoted by $A^{\mathrm{op}},$ is the same algebra with the same involution but with reversed product, so that the product of a and b in $A^{\mathrm{op}}$ is defined as $ba.$ We conclude with one more characterisation of nuclearity.

Proposition 6.2. A $C^{*}$ -algebra A is nuclear if and only if

$$ \begin{align*}A\otimes_{\mathrm{min}} (A^{\mathrm{op}}\otimes_{\mathrm{min}} B(\ell^{2})/K(\ell^{2})) = A\otimes_{\mathrm{max}} (A^{\mathrm{op}}\otimes_{\mathrm{min}} B(\ell^{2})/K(\ell^{2}) ).\end{align*} $$