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 Hy\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 <y_1,\dots,y_n>. 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 Ba\mapsto a\otimes b_0,

and

\psi:B\rightarrow A\otimes_\gamma Bb\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 Bc=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 0basically 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 Ab_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 0g\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 Hs\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.

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