Taken from C*-algebras and Operator Theory by Gerald Murphy.

If $H$ and $K$ are vector spaces, we denote by $H\otimes K$ their algebraic tensor product. This is linearly spanned by the elements $x\otimes y$ ($x\in H$$y\in K$).

One reason why tensor products are useful is that they turn bilinear maps (a bilinear map $\varphi$ has $\lambda\varphi(x,y)=\varphi(\lambda x,y)=\varphi(x,\lambda y)$) into linear maps ($\lambda\varphi(x,y)=\varphi(\lambda x,\lambda y)$). More precisely, if $\varphi:H\times K\rightarrow L$ is a bilinear map, where $H,\,K$ and $L$ are vector spaces, then there is a unique linear map $\varphi_1:H\otimes K\rightarrow L$ such that $\varphi_1(x\otimes y)=\varphi(x,y)$ for all $x\in H$ and $y\in K$.

If $\rho,\,\tau$ are linear functionals on the vector spaces $H,\,K$ respectively, then there is a unique linear functional $\rho\otimes\tau$ on $H\otimes K$ such that

$(\rho\otimes\tau)(x\otimes y)=\rho(x)\tau(y)$

since the function

$H\times K\rightarrow\mathbb{C}$$(x,y)\mapsto \rho(x)\tau(y)$,

is bilinear.

Suppose that the finite sum $\sum_jx_j\otimes y_j=0$, where $x_j\in H$ and $y_j\in K$. If $y_1,\dots,y_n$ are linearly independent, then $x_1=\cdot=x_n=0$. For, in this case, there exist linear functionals $\rho_j:K\rightarrow \mathbb{C}$ such that $\rho_j(y_i)=\delta_{ij}$. If $\rho:H\rightarrow\mathbb{C}$ is linear, we have

$0=(\rho\otimes \rho_j)(\sum_i x_j\otimes y_j)=\sum_i\rho(x_i)\rho_j(y_i)=\rho(x_j)$.

Thus $\rho(x_j)=0$ for arbitrary $\rho$ and this shows that all the $x_j=0$.

Similarly if the finite sum $\sum_jx_j\otimes y_j=0$ with the $x_j$ linearly independent, implies that all the $y_j$ are zero.

If $T:H_1\rightarrow H_2$ and $S:K_1\rightarrow K_2$ are linear maps between vector spaces, then by elementary linear algebra, there exists a unique linear map

$T\otimes S:H_1\otimes K_1\rightarrow H_2\otimes K_2$

such that $(T\otimes S)(x\otimes y)=Tx\otimes Sy$ for all $x\in H_1$ and $y\in K_1$. The map $(T,S)\mapsto T\otimes S$ is bilinear.

If $H$ and $K$ are normed, then there are in general many possible norms of $H\times K$ which are related in a suitable manner to the norms on $H$ and $K$, and indeed it is this very lack of uniqueness that creates the difficulties of the theory, as we shall see in the case that $H$ and $K$ are C*-algebras.

When the spaces are Hilbert spaces, however, matters are simple.

## Theorem 6.3.1

Let $H$ and $K$ be Hilbert spaces. Then there is a unique inner product $\langle\cdot,\cdot\rangle$ on $H\otimes K$ such that

$\langle x_1\otimes y_1,x_2\otimes y_2\rangle=\langle x_1,x_2\rangle\langle y_1,y_2\rangle.$

### Proof

If $\rho_1\,\rho_2$ are conjugate-linear maps from $H$ and $K$, respectively, to $\mathbb{C}$, then there is a unique conjugate-linear map $\rho_1\otimes\rho_2$ from $H\otimes K$ to $\mathbb{C}$ such that

$(\rho_1\otimes\rho_2)(x\otimes y)=\rho_1(x)\rho_2(y)$.

(Observe that $\bar{\rho_1}$ and $\bar{\rho_2}$ are linear, and set $\rho_1\otimes\rho_2=\overline{(\bar{\rho_1}\otimes\bar{\rho_2})}$.)

If $x$ is an element of a Hilbert space, let $\rho_x$ be the conjugate-linear functional defined by setting $\rho_x(y)=\langle x,y\rangle$.

Let $X$ be the vector space of all conjugate-linear functionals on $H\otimes K$. The map

$H\times K\rightarrow X$$(x,y)\mapsto \rho_x\otimes\rho_y$,

is bilinear ($\checkmark$), so there is a unique linear map $M:H\otimes K\rightarrow X$ such that

$M(x\otimes y)=\rho_x\otimes \rho_y$.

The map

$\langle\cdot,\cdot\rangle:(H\otimes K)^2\rightarrow\mathbb{C}$$(z_1,z_2)\mapsto M(z_1)(z_2)$,

is a sesquilinear form on $H\otimes K$ such that

$\langle x_1\otimes y_1,x_2\otimes y_2\rangle=\langle x_1,x_2\rangle\langle y_1,y_2\rangle.$

That it is the unique such sesquilinear form is clear ($\checkmark$).

If $z\in H\otimes K$, then $z=\sum_j x_j\otimes y_j$ (finite sum always). Let $\{e_i\}_{i\leq m}$ be an orthonormal basis for $$. Then $z=\sum^m x'_j\otimes e_j$ for some $x_j'\in H$ ($\checkmark$) and therefore,

$\langle z,z\rangle=\sum_{i,j=1}^m\langle x'_j\otimes e_i,x'_j\otimes e_j\rangle$

$=\sum_{i,j=1}^m\langle x'_i,x'_j\rangle\langle e_i,e_j\rangle$

$=\sum_{j=1}^m\|x'_j\|^2$.

Thus $\langle\cdot,\cdot\rangle$ is positive and if $\langle z,z\rangle=0$, then the $x'_j=0$ and $z=0$. Therefore $\langle\cdot,\cdot\rangle$ is an inner product $\bullet$

If $H$ and $K$ are as in Theorem 6.3.1, we shall always regard $H\otimes K$ as a pre-Hilbert space with the above inner product. The Hilbert space completion of $H\otimes K$ is denoted by $H\hat{\otimes} K$, and called the Hilbert space tensor product of $H$ and $K$. Note that ($\checkmark$)

$\|x\otimes y\|=\|x\|\|y\|.$

It is an elementary exercise to show that if $E$ and $F$ are orthonormal bases for $H$ and $K$, respectively, then $E\otimes F=\{e_i\otimes f_j:e_i\in E\,f_j\in F\}$ is an orthonormal basis for $H\hat{\otimes}K$ ($\checkmark$).

If $H_0\,K_0$ are closed vector subspaces of $H,\,K$, respectively, then the inclusion map $H_0\otimes K_0\rightarrow H\hat{\otimes}K$ is isometric when $H_0\otimes K_o$ has its canonical inner product. It follows that we may regard $H_0\hat{\otimes}K_0$ as a closed vector subspace $H\hat{\otimes}K$.

## Lemma 6.3.2

Let $H,\,K$ be Hilbert spaces and suppose that $T\in B(H)$ and $s\in B(K)$. Then there is a unique $T\hat{\otimes} S\in B(H\hat{\otimes} K)$ such that

$(T\hat{\otimes}S)(x\otimes y)=Tx\otimes Sy$.

Moreover, $\|T\hat{\otimes}S\|=\|T\|\|S\|$.

### Proof

The map $(T,S)\mapsto T\otimes S$ is bilinear, so to show that $T\otimes S:H\otimes K\rightarrow H\otimes K$ is bounded, we may assume that $T$ and $S$ are unitaries, since the unitaries span the C*-algebras $B(H)$ and $B(K)$. If $z\in H\otimes K$, then we may write $z=\sum_j x_j\otimes y_j$ where the $y_j$ are orthogonal. Hence

$\|(T\otimes S)(z)\|^2=\left\|\sum_{j=1}^n Tx_j\otimes Sy_j\right\|^2,$

$=\sum_{j=1}^n\|Tx_j\otimes Sy_j\|^2,$

$=\sum_{j=1}^n\|Tx_j\|^2\|Sy_j\|^2,$

$=\sum_{j=1}^n\|x^j\|^2\|y_j\|^2,$

$=\|z\|^2$.

Consequently, $\|T\otimes S\|=1$.

Thus, for all operators $T,\,S$ on $H,\, K$, respectively, the linear map $T\otimes S$ is bounded on $H\otimes K$ and hence has an (isometric) extension to a bounded linear map $T\hat{\otimes}S$ on $H\hat{\otimes}K$.

It is easily verified that the maps

$B(H)\rightarrow B(H\hat{\otimes} K)$$T\mapsto T\hat{\otimes}I_K$,

and

$B(K)\rightarrow B(H\hat{\otimes}K)$$S\mapsto I_H\hat{\otimes} S$,

are injective $\star$-homomorphisms and therefore isometric ($\checkmark$). Hence, $\|T\hat{\otimes}I_K\|=\|T\|$ and $\|I_H\hat{\otimes} S\|=\|S\|$, so

$\|(T\hat{\otimes}S)\|=\|(T\hat{\otimes}I_K)(I_H\hat{\otimes}S)\|\leq\|T\|\|S\|$.

If $\varepsilon$ is a sufficiently small positive number, and if $T,\,S\neq0$, then there are unit vectors $x,\,y$ such that

$\|Tx\|>\|T\|-\varepsilon$ and $\|Sy\|>\|S\|-\varepsilon$.

Hence,

$\|(T\hat{\otimes}S)(x\otimes y)\|=\|Tx\|\|Sy\|$,

$>(\|T\|-\varepsilon)(\|S\|-\varepsilon)$,

so $\|T\hat{\otimes}S\|>(\|T\|-\varepsilon)(\|S\|-\varepsilon)$. Letting $\varepsilon\rightarrow0$, we get $\|T\hat{\otimes}S\|\geq\|T\|\|S\|$ $\bullet$

### Remark

Let $H$ and $K$ be Hilbert spaces and suppose that $T_1,\,T_2\in B(H)$ and $S_1,\,S_2\in B(K)$. It is routine to show that

$(T_1\hat{\otimes}S_1)(T_2\hat{\otimes}S_2)=T_1T_2\hat{\otimes}S_1S_2$

and

$(T\hat{\otimes}S)^*=T^*\hat{\otimes}S^*$.

(I’m happy with the second and the first is done below I think.)

If $T_1,\dots,T_n$ are operators on $H$ and $S_1,\dots,S_n$ are linearly independent operators on $K$ such that (the finite sum) $\sum_j T_j\hat{\otimes}S_j=0$, then all of the $T_j$ are zero. For if $x\in H$, choose orthonormal vectors $e_1,\dots,e_n\in H$ such that

$\mathbb{C}T_1x+\cdots+\mathbb{C}T_nx\subset \mathbb{C}e_1\cdots+\mathbb{C}e_m$.

Then there are scalars $\{\lambda_{ij}\}$ (which aren’t chosen uniquely — this representation is not unique.) such that $T_jx=\sum_{i}\lambda_{ij}e_i$ for any of the $T_j$. If $y\in K$, we have

$0=\sum_{j=1}^n\left(\sum_{i=1}^m\lambda_{ij}e_i\right)\otimes S_jy=\sum_{i=1}^m\left(e_i\otimes \left(\sum_{j=1}^n\lambda_{ij}S_jy\right)\right)$,

so $\sum_j\lambda_{ij}S_jy=0$. This shows that $\sum_j\lambda_{ij}S_j=0$, and therefore, by linear dependence of the $S_j$, we get that the $\lambda_{ij}=0$ for all $i,\,j$. It follows that $T_1x=\cdots T_nx=0$ so all of the $T_j$ are zero.

If $A$ and $B$ are algebras, there is a unique multiplication on $A\otimes B$ such that

$(a_1\otimes b_1)(a_2\otimes b_2)=a_1a_2\otimes b_1b_2$.

for all $a_1,\,a_2\in A$ and $b_1,\,b_2\in B$: Let $L_a$ denote left multiplication by $a$ and let $X$ be the vector space of all linear maps on $A\otimes B$. If $a\in A$ and $b\in B$, then $L_a\otimes L_b\in X$, and the map

$A\times B\rightarrow X$$(a,b)\mapsto L_a\otimes L_b$,

is bilinear. Hence, there is a unique map $M:A\otimes B\rightarrow X$ such that $M(a\times b)=L_a\otimes L_b$ for all $a$ and $b$. The bilinear map

$(A\otimes B)^2\rightarrow A\otimes B$$(c,d)\mapsto cd=M(c)(d)$,

is readily seen to be the required unique multiplication on $A\otimes B$. We call $A\otimes B$ endowed with this multiplication the algebra tensor product of the algebras $A$ and $B$.

If $A$ and $B$ are *-algebras, then there is a unique involution on $A\otimes B$ such that $(a\otimes b)^*=a^*\otimes b^*$ for all $a$ and $b$. The existence of such an involution is easily seen if we show that

$\sum_{j=1}^na_j\otimes b_j=0\Rightarrow \sum_{j=1}^na_j^*\otimes b_j^*$.

If $A,\,B$ are *-algebras, then we want to make $A\otimes B$ into a *-algebra. We want to define $i:A\otimes B\rightarrow A\otimes B$ by

$i\left(\sum_ia_i\otimes b_i\right)=\sum_ia_i^*\otimes b_i^*$.

If $\sum_ia_i\otimes b_i=\sum_ja'_j\otimes b'_j$ then $\sum_ka_k''\otimes b_k''=0$ for some reindexing, then if $\sum_ia_i\otimes b_i=0\Rightarrow \sum_ia_i^*\otimes b_i^*=0$, then we get that $\sum_ka_k''^*\otimes b_k''^*=0$ which implies that $\sum_ia_i^*\otimes b_i^*=\sum_ja_j'^*\otimes b_j'^*$.

Suppose, therefore that $\sum_i a_i\otimes b_i=0$, and assume without loss of generality that the $b_i$ are linearly independent. Now if $a_i=0$, this implies that $a_i^*=0$ which means that $\sum_ia_i^*\otimes b_i^*=0$ as required.

We call $A\otimes B$ with the above involution the *-algebra tensor product of $A$ and $B$.

If $A_0,\,B_0$ are *-subalgebras of $A,\,B$, respectively, we may clearly regard $A\otimes B$ as a *-subalgebra of $A\otimes B$.

### Remark 6.3.2

Let $\varphi:A\rightarrow C$ and $\psi:B\rightarrow C$ be *-homomorphisms from *-algebras $A$ and $B$ into a *-algebra $C$ such that every element of $\varphi(A)$ commutes with every element of $\psi(B)$. Then there is a unique *-homomorphism $\pi:A\otimes B\rightarrow C$ such that

$\pi(a\otimes b)=\varphi(a)\psi(b)$.

This follows from the observation that the map

$A\times B\rightarrow C$$(a,b)\mapsto \varphi(a)\psi(b)$,

is bilinear and so induces a linear map $\pi:A\otimes B\rightarrow C$, which is easily seen to have the required properties (The commutativity is needed to show that $\pi((a\otimes b)^*)=[\pi(a\times b)]^*$).

If $A_1,\,A_2,\,B_1,\,B_2$ are *-algebras and $\varphi:A_1\rightarrow A_2$ and $\psi:B_1\rightarrow B_2$ are *-homomorphisms, then $\varphi\otimes \psi:A_1\otimes B_1\rightarrow A_2\otimes B_2$ is a *-homomorphism. The proof is a routine exercise ($\checkmark$).

We shall use the next result to show that there is at least one C*-norm on $A\otimes B$ if $A$ and $B$ are C*-algebras.

## Theorem 6.3.3

Suppose that $(H,\varphi)$ and $(K,\psi)$ are representations of the C*-algebras $A$ and $B$ respectively. Then there exists a unique *-homomorphism $\pi:A\otimes B \rightarrow B(H\hat{o\times}K)$ such that

$\pi(a\otimes b)=\varphi(a)\hat{\otimes}\psi(b)$.

Moreover, if $\varphi$ and $\psi$ are injective, so is $\pi$.

### Proof

The maps

$\varphi_0:A\rightarrow B(H\hat{\otimes}K)$$a\mapsto \varphi(a)\hat{\otimes}I_K$,

and

$\psi_0:B\rightarrow B(H\hat{\otimes}K)$$b\mapsto I_H\otimes \psi(b)$,

are *-homomorphisms, and the elements of $\varphi_0(A)$ commute with the elements of $\psi_0(B)$ ($\checkmark$). Hence, by the previous remark, there is a unique *-homomorphism $\pi:A\otimes B\rightarrow B(H\hat{\otimes}K)$ such that $\pi(a\otimes b)=\varphi_0(a)\psi_0(b)=\varphi(a)\hat{\otimes}\psi(b)$.

Now suppose that the representations $(H,\varphi)$ and $(K,\psi)$ are faithful (that is $\varphi$ and $\psi$ are injective), and let $c\in\text{ker }\pi$. We can write $c$ in the (finite sum) form $c=\sum_ja_j\otimes b_j$, where the $b_j$ are linearly independent. Then the $\psi(b_j)$ are linearly independent, because $\psi$ is injective and $\sum_j\varphi(a_j)\hat{\otimes}\psi(b_j)=0$ ($\checkmark$). Hence, all the $\varphi(a_j)$ are zero. Since $\varphi$ is injective, all the $a_j$ are zero also, so $c=0$. Thus $\text{ker }\pi=0$ $\bullet$

Let $A$ and $B$ be C*-algebras with universal representations $(H,\varphi)$ and $(K,\psi)$ respectively. By Theorem 6.3.3 there is a unique injective *-homomorphism $\pi:A\otimes B\rightarrow B(H\hat{\otimes}K)$ such that $\pi(a\otimes b)=\varphi(a)\hat{\otimes} \psi(b)$ for all $a$ and $b$. Hence the function

$\|\cdot\|_*:A\otimes B\rightarrow \mathbb{R}^+$$c\mapsto \|\pi(c)\|$,

is a C*-norm on $A\otimes B$, called the spatial C*-norm. Note that$\|a\otimes b\|_*=\|a\|\|b\|$ ($\pi$ is an injective *-homomorphism given by $\pi(a\otimes b)=\varphi(a)\otimes\psi(b)$. Now

$\|a\otimes b\|_*=\|\pi(a\otimes b)\|=\|\varphi(a)\otimes\psi\|$

$=\|\varphi(a)\|\|\psi(b)\|=\|a\|\|b\|$,

since $\varphi,\,\psi$ are injective C*-isomorphisms.).

We call the C*-completion of $A\otimes B$ with respect to $\|\cdot\|_*$ the spatial tensor product of $A$ and $B$, and denote it by $A\otimes_*B$.

In general, there may be more than one C*-norm on $A\otimes B$. If $\gamma$ is a C*-norm on $A\otimes B$, we denote the C*-completion of $A\times B$ with respect to $A\otimes_\gamma B$ with respect to $\gamma$ by $A\otimes_\gamma B$.

## Lemma 6.3.4

Let $A,\, B$ be C*-algebras and let $\gamma$ be a C*-norm on $A\otimes B$. Then for each $a_0\in A$ and $b_0\in B$ the maps

$\varphi:A\rightarrow A\otimes_\gamma B$$a\mapsto a\otimes b_0$,

and

$\psi:B\rightarrow A\otimes_\gamma B$$b\mapsto a_0\otimes b$,

are continuous.

### Proof

Since $\varphi$ is a linear map between Banach spaces, we may invoke the closed graph theorem (the domain of $\varphi$ so we have the ‘converse’ $\checkmark$). Thus, to show that $\varphi$ is continuous we need only to show that if a sequence $\{a_n\}$ converges to $0$ in $A$ and the sequence $\{\varphi(a_n)\}$ converges to $c\in A\otimes_\gamma B$, then $c=0$ (also continuity at $0$ will imply continuity as $\varphi$ is linear). We may suppose that $a_n$ and $b_0$ are positive (by replacing $a_n$ by $a_n^*a_n$ and $b_0$ by $b_0^*b_0$ if necessary). Hence $c\geq 0$ ($\checkmark$). If $\rho$ is a positive linear functional on $A\otimes_\gamma B$, then the linear functional

$\tau:A\rightarrow\mathbb{C}$$a\mapsto \rho(a\otimes b_0)$,

is positive, hence continuous. Consequently, $\rho(c)=\lim_{n\rightarrow\infty}\rho(a_n\otimes b_0)=\lim_{n\rightarrow \infty}\tau(a_n)=0$, since $\lim_{n\rightarrow\infty}a_n=0$. Since $\rho$ is an arbitrary positive linear functional on $A\otimes_\gamma B$$c=0$. Therefore, $\varphi$ is continuous. Similarly, $\psi$ is continuous $\bullet$

## Theorem 6.3.5

Let $A,\,B$ be non-zero C*-algebras and suppose that $\gamma$ is a C*-norm on $A\otimes B$. Let $(H,\pi)$ be a non-degenerate representation ($\forall\,x\in H$$\exists\,a\in A$$\pi(a)x\neq 0$basically everything isn’t sent to zero) of $A\otimes_\gamma B$. Then there exist unique *-homomorphisms $\varphi:A\rightarrow B(H)$ and $\psi:B\rightarrow B(H)$ such that

$\pi(a\otimes b)=\varphi(a)\psi(b)=\psi(b)\varphi(a)$.

Moreover, the representations $(H,\varphi)$ and $(H,\psi)$ are non-degenerate.

### Proof

Let $H_0=\pi(A\otimes B)H$. If $z\in H_0$, it can be written in the form

$z=\sum_{i=1}^n\pi(a_i\otimes b_i)(x_i)$.

Suppose we have two such expressions; that is, $z$ can also be written as

$z=\sum_{i=1}^n\pi(a_i\otimes b_i)(x_i)=\sum_{j=1}^m\pi(c_j\otimes d_j)(y_j)$,

where $a_i,\,c_j\in A$$b_i,\,d_j\in B$ and $x_i,\,y_j\in H$. If $\{u_\lambda\}_{\lambda\in\Lambda}$ is an approximate unit for $B$ and $a\in A$, then

$\pi(a\otimes u_\lambda)(z)=\sum_{i=1}^n\pi(aa_i\otimes u_\lambda b_i)(x_i)=\sum_{j=1}^m\pi(ac_j\otimes u_\lambda d_j)(y_j)$,

so in the limit (using Lemma 6.3.4)

$\lim_{\lambda}\pi(a\otimes u_\lambda)(z)=\sum_{i=1}^n\pi(aa_i\otimes b_i)(x_i)=\sum_{j=1}^m\pi(ac_j\otimes d_j)(y_j)$.

We can therefore well-define a map $\varphi(a):H_0\rightarrow H_0$ by setting $\varphi(a)(z)=\sum_{i=1}^n\pi(aa_i\otimes b_i)(x_i)$, if $z$ is in the ‘$x_i$‘ form ($\checkmark$). Since $\varphi(a)z=\lim_\lambda \pi(a\otimes u_\lambda)(z)$, it is clear that $\varphi(a)$ is linear ($\checkmark$). By Lemma 6.3.4, there exists a positive number $M$ (depending on $a$) such that $\|\pi(a\otimes b)\|\leq M\|b\|$, so $\|\pi(a)(z)\|\leq M\|z\|$. Hence, $\varphi(a)$ is bounded. Since $H_0$ is dense in $H$ (because $(H,\pi)$ is non-degenerate), we can therefore extend $\varphi(a)$ uniquely to a bounded linear map on $H$, also denoted by $\varphi(a)$.

Suppose now that $\{v_\mu\}_{\mu\in M}$ is an approximate unit for $A$. Reasoning as before, for all $b\in B$, we have that $\lim_\mu \pi(v_\mu\otimes b)(z)=\sum_i\pi(a_i\otimes bb_i)(x_i)$, if $z$ is in that representation. We can therefore well-define a map $\psi(b):H_0\rightarrow H_0$ by setting $\psi(b):H_0\rightarrow H_0$ by $\psi(b)(z)=\sum_i\pi(a\otimes bb_i)(x_i)=\lim_\mu\pi(v_\mu\otimes b)(z)$. The linear map $\psi(b)$ is bounded and extends uniquely to a bounded linear map on $H$, also denoted by $\psi(b)$.

A routine verification shows that the maps

$\varphi:A\rightarrow B(A)$$a\mapsto \varphi(a)$

and

$\psi:B\rightarrow B(H)$$b\mapsto \psi(b)$,

are *-homomorphisms ($\checkmark$), and that $\pi(a\otimes b)=\varphi(a)\psi(b)=\psi(b)\varphi(a)$.

Now suppose that $\varphi_1:A\rightarrow B(H)$ and $\psi_1:B\rightarrow B(H)$ are another pair of *-homomorphisms such that $\pi(a\otimes b)=\varphi_1(a)\psi_1(b)=\psi_1(b)\varphi_1(a)$. Suppose that $z\in H$ is such that $\varphi_1(a)z=0$. Then $\pi(a\otimes b)z=0$ fir all $a\in A$ and $b\in B$, and therefore $\pi(c)z=0$ for all $c\in A\otimes_\gamma B$. By non-degeneracy of $(H,\pi)$, we have $z=0$. Thus, $\varphi_1$ is non-degenerate. Similarly, $\psi_1$ is non-degenerate.

In particular, $\varphi$ and $\psi$ are degenerate.

If $\{v_\mu\}$ and $\{u_\lambda\}$ are approximate units as above, then the nets $\{\varphi_1(v_\mu)\}$ and $\{\psi_1(u_\lambda)\}$ converge strongly to $I_H$ (by non-degeneracy of  $\varphi_1$ and $\psi_1$). Hence, $\{\pi(a\otimes u_\lambda)\}_\lambda$ converges strongly to both $\varphi_1(a)$ and $\varphi(a)$ for all $a\in A$, hence $\varphi_1=\varphi$. Similarly, $\{\pi(v_\mu\otimes b)\}_\mu$ converges strongly to both $\psi_1(b)$ and $\psi(b)$ for all $b\in B$ so $\psi_1=\psi$ $\bullet$

We denote the maps $\varphi$ and $\psi$ in the preceding theorem by $\pi_A$ and $\pi_B$, respectively.

## Corollary 6.3.6

Let $A$ and $B$ be C*-algebras and let $\gamma$ be a C*-seminorm on $A\otimes B$. Then

$\gamma(a\otimes b)\leq \|a\|\|b\|$.

### Proof

Let $\delta:=\max(\gamma,\|\cdot\|_*)$, so $\delta$ is a C*-norm on $A\otimes B$. Let $(H,\varphi)$ be the universal representation of $A\otimes_\delta B$. This is faithful and non-degenerate, so Theorem 6.3.5 applies. If $a\in A$ and $b\in B$, then $\pi(a\otimes b)=\pi_A(a)\pi_B(b)$.

Hence

$\delta(a\otimes b)=\|\pi(a\otimes b)\|\leq\|\pi_A(a)\|\|\pi_B(b)\|\leq\|a\|\|b\|$,

so

$\gamma(a\otimes b)\leq \delta(a\otimes b)\leq\|a\|\|b\|$ $\bullet$

Let $A$ and $B$ be C*-algebras. Denote by $\Gamma$ the set of all C*-norms $\gamma$ on $A\otimes B$. We define $\|c\|_{\max}:=\sup_{\gamma\in\Gamma}\gamma(c)$ for each $c\in A\otimes B$. If $c=\sum_j a_j\otimes b_j$, then for any $\gamma\in\Gamma$ we have

$\gamma(c)\leq \sum_{j=1}^n\gamma(a_j\otimes b_j)\leq \sum_{j=1}^n\|a_j\|\|b_j\|$

by Corollary 6.3.6. Hence, $\|c\|_{\max}<\infty$.

It is readily verified that

$\|\cdot\|_{\max}:A\otimes B\rightarrow \mathbb{R}^+$$c\mapsto\|c\|_{\max}$,

is a C*-norm, called the maximal C*-norm ($\checkmark$). We denote by $A\otimes_{\max} B$ the C*-completion of $A\otimes B$ under this norm, and call $A\otimes_{\max} B$ the maximal tensor product of $A$ and $B$.

If $\gamma$ is a C*-seminorm, then $\gamma\leq\|\cdot\|_{\max}$. The maximal tensor product has a very useful universal property.

## Theorem 6.3.7

Let $A,\,B$ and $C$ be C*-algebras and suppose that $\varphi:A\rightarrow C$ and $\psi:B\rightarrow C$ are *-homomorphisms such that every element of $\varphi(A)$ commutes with every element of $\psi(B)$. Then there is a unique *-homomorphism $\pi:A\otimes_{\max} B\rightarrow C$ such that

$\pi(a\otimes b)=\varphi(a)\psi(b)$.

### Proof

Uniqueness is clear ($\checkmark$). By Remark 6.3.2 there is a *-homomorphism $\pi:A\otimes B\rightarrow C$ satisfying the equation in the statement of the theorem. The function

$\gamma:A\otimes B\rightarrow \mathbb{R}^+$$c\mapsto \|\pi(c)\|$,

is a C*-seminorm. Hence $\gamma(c)\leq\|c\|_{\max}$ for all $c\in A\otimes B$. Therefore, $\pi$ is a norm-decreasing *-homomorphism, and so extends to a norm-decreasing *-homomorphism  on $A\otimes_{\max} B$ $\bullet$

We say that a C*-algebra is nuclear if, for each C*-algebra $B$, there is only one C*-norm on $A\otimes B$.

### Remark

If a *-algebra $A$ admits a complete C*-norm $\|\cdot\|$, then it is the only C*-norm on $A$. For if $\gamma$ is another C*-norm on $A$, and $B$ denotes the C*-completion of $A$ with respect to $A$, then the inclusion $(A,\|\cdot\|)\rightarrow (B,\gamma)$ is an injective *-homomorphism and therefore isometric, so $\gamma=\|\cdot\|$.

### Example

For each $n\geq 1$, the C*-algebra $M_n(\mathbb{C})$ is nuclear. The reason is that for each C*-algebra $A$, the *-algebra $M_n(\mathbb{C})\otimes A$ admits a complete C*-norm. This is seen by showing that the unique linear map $\pi:M_n(\mathbb{C})\otimes A\rightarrow M_n(A)$, such that $\pi([\lambda_{ij}]_{ij}\otimes a)=[\lambda_{ij}a]_{ij}$ for $[\lambda_{ij}]_{ij}\in M_n(\mathbb{C})$ and $a\in A$, is a *-isomorphism (all fine except $\pi(c_1+c_2)=\pi(c_1)+\pi(c_2)$. I had calculated this directly but there is a slicker way

Fix $=\Lambda\in M_n(\mathbb{C})$ and $a\in A$. Now the linear maps $M\mapsto [M^i_ja]$ and $x\mapstp [\lambda^i_jx]$ are linear maps $M_n(\mathbb{C})\rightarrow M_n(A)$ and $A\rightarrow M_n(A)$ hence there exists a unique linear map $\pi:M_n(\mathbb{C}\otimes A)\rightarrow M_n(A)$.).

We are going to show that all finite-dimensional C*-algebras are nuclear and for this we shall need to determine the structure theory for such algebras.

## Theorem 6.3.8

If $A$ is a non-zero finite-dimensional C*-algebra, it is *-isomorphic to $\bigoplus_{i=1}^k M_{n_i}(\mathbb{C})$.

### Proof

If $A$ is simple, the result is immediate from Remark 6.2.1 (have glossed over this section). We prove the general result by induction on the dimension $m$ of $A$. The case $m=1$ is obvious. Suppose the result holds for all dimensions less than $m$. We may suppose that $A$ is not simple, and so contains a non-zero proper closed ideal $I$, and we may take $I$ to be of minimum dimension. In this case, $I$ has no non-trivial ideals, so $I$ is *-isomorphic to $M_{n_1}(\mathbb{C})$ for some $n_1\in\mathbb{N}$. Hence, $I$ has a unit $I_I$, so $I=AI_I$ and $I_I$ commutes with all the elements of $A$ ($\checkmark$). Also $A(I_A-I_I)$ is a C*-subalgebra of $A$ ($\checkmark$) and the map

$A\rightarrow Ap\oplus A(I_A-I_I)$$a\mapsto (ap,a(I_A-I_I))$,

is a *-isomorphism. Since the algebra $A(I_A-I_I)$  has dimension less than $A$,it is *-isomorphic to $\bigoplus_{i=2}^kM_{n_i}(\mathbb{C})$. Hence, by the Axiom of Induction, $A$ is *-isomorphic to $\bigoplus_{i=1}^kM_{n_i}(\mathbb{C})$ $\bullet$

## Theorem 6.3.9

A finite-dimensional C*-algebra is nuclear.

### Proof

Let $A$ be a finite-dimensional C*-algebra, which we may suppose to be the direct sum $A=\bigoplus_{i=1}^k M_{n_i}(\mathbb{C})$. Let $B$ be an arbitrary C*-algebra. A routine verification shows that the unique linear map

$\pi:A\otimes B\rightarrow \bigoplus_{i=1}^k(M_{n_i}(\mathbb{C})\otimes B)$,

such that

$\pi((a_1,\dots,a_k)\otimes b)=(a_1\otimes b,\dots,a_k\otimes b)$

for all $a_j\in M_{n_j}(\mathbb{C})$ and $b\in B$, is a *-isomorphism. Hence, $A\otimes B$ admits a complete C*-norm ($\checkmark$), so it admits only one C*-norm. This shows that $A$ is nuclear $\bullet$

The next result suggests that nuclear algebras exist in abundance.

## Theorem 6.3.10

Let $S$ be a non-empty set of C*-subalgebras of a C*-algebra $A$ which is upwards directed (if $B,\,C\in S$, then there exists a $D\in S$ such that $B,\,C\subset D$). Suppose that $\cup S$ is dense in $A$ and that all the algebras are nuclear. Then $A$ is nuclear.

### Proof

Let $B$ be an arbitrary C*-algebra and suppose $\beta,\,\gamma$ are C*-norms on $A\otimes B$. Set $C=\cup_{D\in S}D\otimes B$ (we may regard $D\otimes B$ as a *-subalgebra of $A\otimes B$ for each $D\in S$). Then $C$ is a *-subalgebra of $A\otimes B$ and it is clear that $C$ is dense in the C*-algebras $A\otimes_\beta B$ and $A\otimes_\gamma B$. Now $\beta=\gamma$ on $D\otimes B$ for each $D\in S$, by nuclearity of $D$, so $\beta=\gamma$ on $C$, and therefore the identity map on $C$ extends to a *-isomorphism $\pi:A\otimes_\beta B\rightarrow A\otimes_\gamma B$. If $a\in A$ and $b\in B$, then there exists a sequence $\{a_i\}$ in $\cup S$ such that $a=\lim_{n\rightarrow a_n}$ in $A$, so $a\otimes b=\lim_{n\rightarrow \infty}a_n\otimes b$ in $A\otimes_\beta B$ and in $A\otimes_\gamma B$, and therefore

$\pi(a\otimes b)=\lim_{n\rightarrow\infty}\pi(a_n\otimes b)=\lim_{n\rightarrow\infty}a_n\otimes b=a\otimes b$,

where convergence is with respect to $\gamma$. Therefore, $\pi=I_{A\otimes B}$. Hence for any $c\in A\otimes B$, we have $\gamma(c)=\gamma(\pi(c))=\beta(c)$, so $\gamma=\beta$ on $A\otimes B$. Therefore, the algebra $A$ is nuclear $\bullet$

### Example

If $H$ is a Hilbert space, $K(H)$ is a nuclear C*-algebra. To see this, suppose that $e_1,\dots,e_n$ are orthonormal vectors in $H$. If $P=\sum_{j=1}^n|e_j\rangle\langle e_j|$, then $P$ is a projection, $P\in K(H)$, and the map

$\varphi:M_n(\mathbb{C})\rightarrow PK(H)P$$[\lambda_{ij}]\mapsto\sum_{i,j=1}^n$\lambda|e_i\rangle\langle e_j|\$,

is a *-isomorphism. We show surjectivity only: If $T\in PK(H)P$, then

$T=PTP$

$=\sum_{i,j=1}^n|e_i\rangle\langle e_i|T|e_j\rangle\langle e_j|$

$=\sum_{i,j=1}^n\langle Te_j,e_i\rangle|e_i\rangle\langle e_j|$

$\varphi([\langle Te_j,e_i\rangle])$.

Since $PK(H)P$ is finite-dimensional, it is nuclear (by Theorem 6.3.9).

If $E$ is an orthonormal basis for $H$, let $I$ be the set of all finite non-empty subsets of $E$ made into an upwards-directed poset by setting $i\leq j$ if $i\subset j$. For $i\in I$, let $P_i=\sum_{x\in i}|x\rangle\langle x|$ and $A_i=P_iK(H)P_i$. Each $A_i$ us therefore a nuclear C*-algebra, and the set $S=\{A_i:i\in I\}$ is upwards-directed.

If $T$ is a finite-rank operator on $H$, we can write $T=\sum_{k=1}^m|x_k \rangle\langle y_k|$ for some $x_k,\,y_k\in H$. Therefore,

$TP_i=\sum_{k=1}^m|x_k\rangle\langle P_iy_k|$,

so $\lim_{i\in I}TP_i=\sum_{k=1}^m|x_k\rangle\langle y_k|=T$, since $\lim_{i}P_iy=y$ for $y\in H$. From this it follows by norm-density of the finite-rank operations in $K(H)$ that $\lim_iTP_i=T$ for all $T\in K(H)$. Therefore $\lim_iP_iTP_i=T$. Hence, $\cup S$ is dense in $K(H)$. By Theorem 6.3.10, $K(H)$ is a nuclear $C*-algebra$.

An AF-algebra is a C*-algebra that contains an increasing sequence $\{A_n\}_{n\in\mathbb{N}}$ of finite-dimensional C*-subalgebras such that $\bigcup_{n\in\mathbb{N}}A_n$ is dense in $A$.

## Corollary 6.3.11

All AF-algebras are nuclear.

## Appendix: Non-negative Definite Kernels

Suppose that $S$ is a set. We say that $k:S\times S\rightarrow\mathbb{C}$ is a non-negative kernel if for all $n\in\mathbb{N}$$(s_1,\dots,s_n)\in S^n$ and $(\alpha_1,\dots,\alpha_n)\in\mathbb{C}^n$, we have that

$\sum_{i,j=1}^n\overline{\alpha_i}\alpha_jk(s_i,s_j)\geq 0$

$\Leftrightarrow \left\langle \left(\begin{array}{c}\alpha_1\\\vdots\\\alpha_n\end{array}\right)\left(\begin{array}{ccc}k(s_1,s_1)&\cdots&k(s_1,s_n)\\\vdots&\ddots&\vdots\\k(s_n,s_1)&\cdots&k(s_n,s_n)\end{array}\right) \left(\begin{array}{c}\alpha_1\\\vdots\\\alpha_n\end{array}\right)\right\rangle\geq 0$

$\Leftrightarrow [k(s_i,s_j)]\in M_n(\mathbb{C})_+$.

Let $F(S)=\{\text{functions }S\rightarrow\mathbb{C}\text{ of finite support}\}$. Define

$\langle f,g\rangle_0:=\sum_{s,t\in S}\overline{f(s)}g(t)k(s,t)$

which is defined because of finite support. Then $\langle f,f\rangle_0\geq 0$$g\mapsto \langle f,g\rangle_0$ is linear and

$\overline{\langle f,g\rangle_0}=\langle g,f\rangle_0$.

$\langle,\rangle_0$ is a semi-inner product, so the Cauchy-Schwarz Inequality holds. Define $N=\{f\in F(S):\langle f,f\rangle_0=0\}$. Now $N$ is a subspace by Cauchy-Schwarz, and now set

$\langle f+N,g+N\rangle:=\langle f,g\rangle_0$.

This defines an inner product on $F(S)/N$. Now complete to get a Hilbert space $H$, together with a map $\chi:S\rightarrow H$$s\mapsto \delta^s+N$. Now $\overline{\text{Lin}(\chi(s):s\in S)}=H$ since $\{\delta^s\}_{s\in S}$ is a basis of $F(S)$, with

$\langle \chi(s),\chi(t)\rangle=\langle \delta^s+N,\delta^t+N\rangle=\langle\delta^s,\delta^t\rangle_0=k(s,t)$.

Moreover, if $K$ is any Hilbert space and $\varphi:S\rightarrow K$ a map such that $\langle \varphi(s),\varphi(t)\rangle=k(s,t)$m then consider the map $\chi(s)\mapsto\varphi(s)$. Since

$\langle \chi(s),\chi(t)\rangle=\langle \varphi(s),\varphi(t)\rangle$,

there exists a unique isometry $U:H\rightarrow K$ such that $U\chi(s)=\varphi(s)$. In particular, if $\overline{\text{Lin}(\varphi(s))}=K$, then $U$ is unitary.

Now let $S=H\times K$, and define

$k((h_1,k_1),(h_2,k_2))=\langle h_1,h_2\rangle\langle k_1,k_2\rangle$.

$k$ is a non-negative definite kernel. Let $((h_1,k_1),\dots,(h_n,k_n))\in S^n$ and then

$\left(\begin{array}{cc}k((h_1,k_1),(h_2,k_2))&\cdots\\\vdots&\ddots\end{array}\right)$

$=\left(\langle h_i,h_j\rangle\langle k_i,k_j\rangle\right)=\left(\langle h_i,h_j\rangle\right)\circ\left(\langle k_i,k_j\rangle\right)$

where $A\circ B$ is the Schur product. Moreover,

$\sum_{i,j}\overline{\alpha_i}\alpha_j\langle h_i,h_j\rangle=\left\|\sum_i\alpha_ih_i\right\|^2\geq0$.

In other words, $\left(\langle h_i,h_j\rangle\right)\geq 0$ and $\left(\langle k_i,k_j\rangle\right)\geq 0$ — and the Schur product of positive matrices is positive (Theorem 12).

There exists a Hilbert space $H\otimes K$ and a map $\chi:S\rightarrow H\otimes K$ such that if $\chi(h,k):=h\otimes k$, then

$\langle h_1\otimes k_1,h_2\otimes k_2\rangle=k((h_1,k_1),(h_2,k_2))=\langle h_1,h_2\rangle\langle k_1,k_2\rangle$,

and also $\overline{\text{Lin}(h\otimes k)}=H\otimes K$.

Pick unitaries $U\in B(H)$$V\in B(K)$. Define $\chi^{U,V}:S\rightarrow H\otimes K$ by

$\chi^{U,V}(h,k)=Uh\otimes Vk$.

Then

$\langle\chi^{U,V}(h_1,k_1),\chi^{UV}(h_2,k_2)\rangle=\langle Uh_1\otimes Vk_1,Uh_2\otimes Vk_2\rangle$

$=\langle Uh_1,Uk_2\rangle\langle Vk_1,Vk_2\rangle=\langle\chi(h_1,k_1),\chi(h_2,k_2)\rangle$.

Also $\text{Ran }\chi^{U,V}=\text{Ran }\chi$, so total. Thus, by ‘uniqueness’ of $(H\otimes K,\chi)$, there exists a unique unitary on $H\otimes K$ mapping $h\otimes k\mapsto Uh\otimes Vk$ — call this $U\otimes V$.