[proofplan] We compare every faithful spatial representation with the universal one and then use the standard weak-containment form of the GNS construction to obtain the reverse comparison. The key point is that every state of $A$ is obtained from a state on $B(H)$ after identifying $A$ with its faithful image under $\pi$, and analogously for $B$; this makes every GNS tensor summand norm-dominated by the faithful spatial representation. Once the two inequalities identify all faithful spatial norms, definiteness follows from an algebraic slice argument, and the $C^*$-identity is inherited from the operator norm on $B(H\otimes K)$. [/proofplan]

[step:Choose universal representations for the two factors] Let $\pi_u:A\to B(H_u)$ and $\rho_u:B\to B(K_u)$ be the universal representations of $A$ and $B$, respectively, formed as the direct sums of the GNS representations associated to all states on the corresponding unital $C^*$-algebras. Since the state spaces $S(A)\subset A^*$ and $S(B)\subset B^*$ are sets, these Hilbert direct sums are set-indexed. By [citetheorem:8566], both $\pi_u$ and $\rho_u$ are faithful unital $*$-representations. By [citetheorem:8580], the algebraic [tensor product](/page/Tensor%20Product) representation \begin{align*} \pi_u\odot\rho_u:A\odot B\to B(H_u\otimes K_u) \end{align*} is a well-defined $*$-homomorphism. Define the universal spatial seminorm \begin{align*} \|x\|_u:=\|(\pi_u\odot\rho_u)(x)\|_{B(H_u\otimes K_u)} \end{align*} for $x\in A\odot B$. [/step]

[step:Compare one faithful representation with the universal representation]Let \begin{align*} \pi:A\to B(H), \qquad \rho:B\to B(K) \end{align*} be faithful $*$-representations. We prove \begin{align*} \|(\pi\odot\rho)(x)\|_{B(H\otimes K)}\le \|x\|_u \end{align*} for every $x\in A\odot B$. Since $A$ and $B$ are unital and $\pi$ and $\rho$ are unital, $\pi(1_A)=I_H$ and $\rho(1_B)=I_K$. Hence \begin{align*} \overline{\operatorname{span}}\{\pi(a)\xi:a\in A,\ \xi\in H\}=H \end{align*} and \begin{align*} \overline{\operatorname{span}}\{\rho(b)\eta:b\in B,\ \eta\in K\}=K. \end{align*} Thus $\pi$ and $\rho$ are nondegenerate. We use the [Cyclic Decomposition Theorem](/theorems/3288) for Nondegenerate Representations: every nondegenerate $*$-representation of a $C^*$-algebra is unitarily equivalent to a Hilbert direct sum of cyclic subrepresentations. Its nondegeneracy hypothesis is satisfied by unitality, as just verified. Hence there are index sets $I$ and $J$, cyclic subrepresentations $\pi_i:A\to B(H_i)$ and $\rho_j:B\to B(K_j)$, and unitary identifications $H\cong \bigoplus_{i\in I}H_i$ and $K\cong \bigoplus_{j\in J}K_j$ under which $\pi\cong \bigoplus_{i\in I}\pi_i$ and $\rho\cong \bigoplus_{j\in J}\rho_j$. For each $i\in I$, choose a cyclic vector $\xi_i\in H_i$, and define the positive functional \begin{align*} \phi_i:A\to \mathbb C, \qquad a\mapsto (\pi_i(a)\xi_i,\xi_i)_{H_i}. \end{align*} If $\phi_i\neq 0$, define the normalized positive functional $\widehat{\phi_i}:A\to\mathbb C$ by \begin{align*} \widehat{\phi_i}(a):=\frac{\phi_i(a)}{\|\phi_i\|}. \end{align*} The cyclic representation $\pi_i$ is unitarily equivalent to the GNS representation of the state $\widehat{\phi_i}$, up to the scalar normalization of the cyclic vector. Since every state occurs as a summand of $\pi_u$, each $\pi_i$ is unitarily equivalent to a subrepresentation of $\pi_u$. For each $j\in J$, choose a cyclic vector $\eta_j\in K_j$, and define the positive functional $\psi_j:B\to\mathbb C$ by $\psi_j(b)=(\rho_j(b)\eta_j,\eta_j)_{K_j}$. If $\psi_j\neq 0$, define the normalized positive functional $\widehat{\psi_j}:B\to\mathbb C$ by \begin{align*} \widehat{\psi_j}(b):=\frac{\psi_j(b)}{\|\psi_j\|}. \end{align*} The cyclic representation $\rho_j$ is unitarily equivalent to the GNS representation of the state $\widehat{\psi_j}$, up to the scalar normalization of the cyclic vector. Since every state occurs as a summand of $\rho_u$, each $\rho_j$ is unitarily equivalent to a subrepresentation of $\rho_u$. Therefore each tensor product representation \begin{align*} \pi_i\odot\rho_j:A\odot B\to B(H_i\otimes K_j) \end{align*} is unitarily equivalent to a subrepresentation of $\pi_u\odot\rho_u$. Since \begin{align*} H\otimes K\cong \bigoplus_{(i,j)\in I\times J}(H_i\otimes K_j) \end{align*} and, under this identification, \begin{align*} \pi\odot\rho\cong \bigoplus_{(i,j)\in I\times J}(\pi_i\odot\rho_j), \end{align*} we obtain, for every $x\in A\odot B$, \begin{align*} \|(\pi\odot\rho)(x)\|_{B(H\otimes K)} = \sup_{(i,j)\in I\times J}\|(\pi_i\odot\rho_j)(x)\|_{B(H_i\otimes K_j)}. \end{align*} Each summand is a compression of $(\pi_u\odot\rho_u)(x)$ to an invariant subspace, so \begin{align*} \|(\pi_i\odot\rho_j)(x)\|_{B(H_i\otimes K_j)} \le \|(\pi_u\odot\rho_u)(x)\|_{B(H_u\otimes K_u)}. \end{align*} Taking the supremum over $(i,j)\in I\times J$ gives \begin{align*} \|(\pi\odot\rho)(x)\|_{B(H\otimes K)}\le \|x\|_u. \end{align*}[/step]

[guided]We want to compare an arbitrary faithful spatial representation with the universal one. The universal representation is built from all states, so the useful bridge is the cyclic decomposition of representations. First remove the part on which the representation acts as zero. Define \begin{align*} H_0:=\overline{\operatorname{span}}\{\pi(a)\xi:a\in A,\ \xi\in H\}. \end{align*} If $\zeta\in H_0^\perp$, then for all $a\in A$ and $\xi\in H$ we have \begin{align*} (\pi(a)^*\zeta,\xi)_H=(\zeta,\pi(a)\xi)_H=0. \end{align*} Since $\pi(a)^*=\pi(a^*)$, this gives $\pi(a^*)\zeta=0$ for every $a\in A$, and hence $\pi(c)\zeta=0$ for every $c\in A$. Thus $H_0^\perp$ is annihilated by $\pi(A)$. The same construction for $\rho$ gives \begin{align*} K_0:=\overline{\operatorname{span}}\{\rho(b)\eta:b\in B,\ \eta\in K\}, \end{align*} and $K_0^\perp$ is annihilated by $\rho(B)$. Consequently $(\pi\odot\rho)(x)$ acts nontrivially only on $H_0\otimes K_0$, so its operator norm is unchanged if we replace $H$ and $K$ by $H_0$ and $K_0$. This reduces the proof to nondegenerate representations. For a nondegenerate representation, the standard cyclic decomposition theorem says that the representation is a Hilbert direct sum of cyclic subrepresentations. Thus there are index sets $I$ and $J$ and cyclic representations \begin{align*} \pi_i:A\to B(H_i), \qquad \rho_j:B\to B(K_j) \end{align*} such that \begin{align*} H\cong \bigoplus_{i\in I}H_i, \qquad K\cong \bigoplus_{j\in J}K_j, \end{align*} and \begin{align*} \pi\cong \bigoplus_{i\in I}\pi_i, \qquad \rho\cong \bigoplus_{j\in J}\rho_j. \end{align*} Now fix $i\in I$ and choose a cyclic vector $\xi_i\in H_i$. Define \begin{align*} \phi_i:A\to \mathbb C, \qquad a\mapsto (\pi_i(a)\xi_i,\xi_i)_{H_i}. \end{align*} This is a positive linear functional because for every $a\in A$, \begin{align*} \phi_i(a^*a)=(\pi_i(a^*a)\xi_i,\xi_i)_{H_i}=(\pi_i(a)\xi_i,\pi_i(a)\xi_i)_{H_i}\ge 0. \end{align*} If $\phi_i\neq 0$, then $\widehat{\phi_i}:=\phi_i/\|\phi_i\|$ is a state. The GNS representation associated to $\widehat{\phi_i}$ is unitarily equivalent to the cyclic representation generated by $\xi_i$, with only the cyclic vector normalized. Since the universal representation $\pi_u$ is the direct sum of the GNS representations associated to all states on $A$, the representation $\pi_i$ occurs as a subrepresentation of $\pi_u$. Repeating this argument for the cyclic summands of $\rho$ shows that each $\rho_j$ occurs as a subrepresentation of $\rho_u$. Tensoring the cyclic summands gives \begin{align*} H\otimes K\cong \bigoplus_{(i,j)\in I\times J}(H_i\otimes K_j), \end{align*} and, under this unitary identification, \begin{align*} \pi\odot\rho\cong \bigoplus_{(i,j)\in I\times J}(\pi_i\odot\rho_j). \end{align*} For a direct sum of bounded operators, the operator norm is the supremum of the norms of the summands. Hence, for every $x\in A\odot B$, \begin{align*} \|(\pi\odot\rho)(x)\|_{B(H\otimes K)} = \sup_{(i,j)\in I\times J}\|(\pi_i\odot\rho_j)(x)\|_{B(H_i\otimes K_j)}. \end{align*} But each $\pi_i\odot\rho_j$ is a subrepresentation of $\pi_u\odot\rho_u$, so each operator $(\pi_i\odot\rho_j)(x)$ is obtained by restricting $(\pi_u\odot\rho_u)(x)$ to an invariant subspace. Restriction to an invariant subspace cannot increase operator norm, so \begin{align*} \|(\pi_i\odot\rho_j)(x)\|_{B(H_i\otimes K_j)} \le \|(\pi_u\odot\rho_u)(x)\|_{B(H_u\otimes K_u)}. \end{align*} Taking the supremum over all pairs $(i,j)$ proves \begin{align*} \|(\pi\odot\rho)(x)\|_{B(H\otimes K)}\le \|x\|_u. \end{align*}[/guided]

[step:Prove the reverse domination using Tomiyama's extension theorem]Let $\pi:A\to B(H)$ and $\rho:B\to B(K)$ be faithful unital $*$-representations. We prove \begin{align*} \|x\|_u\le \|(\pi\odot\rho)(x)\|_{B(H\otimes K)} \end{align*} for every $x\in A\odot B$. We use Tomiyama's weak-containment theorem for the spatial tensor product: if $C\subset B(H)$ and $D\subset B(K)$ are unital $C^*$-subalgebras, then for every pair of states $\alpha\in S(C)$ and $\beta\in S(D)$, the tensor product of the GNS representations associated to $\alpha$ and $\beta$ is weakly contained in the concrete spatial representation \begin{align*} C\odot D\to B(H\otimes K), \qquad c\otimes d\mapsto c\otimes d. \end{align*} Equivalently, for every $y\in C\odot D$, \begin{align*} \|(\sigma_\alpha\odot\tau_\beta)(y)\|_{B(H_\alpha\otimes K_\beta)}\le \|y\|_{B(H\otimes K)}, \end{align*} where $\sigma_\alpha:C\to B(H_\alpha)$ and $\tau_\beta:D\to B(K_\beta)$ are the corresponding GNS representations. Apply this theorem to the unital inclusions $\pi(A)\subset B(H)$ and $\rho(B)\subset B(K)$. Let $\phi\in S(A)$ and $\psi\in S(B)$. Define states $\alpha_\phi:\pi(A)\to\mathbb C$ and $\beta_\psi:\rho(B)\to\mathbb C$ by \begin{align*} \alpha_\phi(\pi(a)):=\phi(a) \end{align*} for $a\in A$, and \begin{align*} \beta_\psi(\rho(b)):=\psi(b) \end{align*} for $b\in B$. These definitions are well-defined because $\pi$ and $\rho$ are injective. Let \begin{align*} \pi_\phi:A\to B(H_\phi) \end{align*} and \begin{align*} \rho_\psi:B\to B(K_\psi) \end{align*} be the corresponding GNS representations. The GNS representation of $\alpha_\phi$ on $\pi(A)$ is unitarily equivalent to $\pi_\phi\circ\pi^{-1}:\pi(A)\to B(H_\phi)$, and the GNS representation of $\beta_\psi$ on $\rho(B)$ is unitarily equivalent to $\rho_\psi\circ\rho^{-1}:\rho(B)\to B(K_\psi)$. Tomiyama's theorem therefore gives weak containment of $\pi_\phi\odot\rho_\psi$ in $\pi\odot\rho$ after transporting along the $*$-isomorphisms $\pi:A\to\pi(A)$ and $\rho:B\to\rho(B)$. Hence, for every $x\in A\odot B$, \begin{align*} \|(\pi_\phi\odot\rho_\psi)(x)\|_{B(H_\phi\otimes K_\psi)}\le \|(\pi\odot\rho)(x)\|_{B(H\otimes K)}. \end{align*} The hypotheses match exactly: $\pi(A)$ and $\rho(B)$ are unital $C^*$-subalgebras because $\pi$ and $\rho$ are unital $*$-homomorphisms, and the identifications with $A$ and $B$ are injective because $\pi$ and $\rho$ are faithful. Taking the supremum over all states $\phi\in S(A)$ and $\psi\in S(B)$, which are precisely the tensor GNS summands appearing in $\pi_u\odot\rho_u$, gives \begin{align*} \|x\|_u\le \|(\pi\odot\rho)(x)\|_{B(H\otimes K)}. \end{align*} Combining this inequality with the previous step gives, for every faithful unital pair $\pi,\rho$, \begin{align*} \|(\pi\odot\rho)(x)\|_{B(H\otimes K)}=\|x\|_u. \end{align*} Thus any two faithful unital pairs give the same value on $x$, and we denote the common value by $\|x\|_{\min}$.[/step]

[guided]The previous step proved only one inequality: every faithful spatial representation is bounded above by the universal representation. To get equality, we need a genuinely external input. The needed input is Tomiyama's weak-containment theorem for the spatial tensor product: for unital concrete $C^*$-subalgebras $C\subset B(H)$ and $D\subset B(K)$, each tensor GNS representation coming from states on $C$ and $D$ is weakly contained in the concrete representation on $H\otimes K$. We verify the hypotheses before applying it. Because $\pi:A\to B(H)$ is a faithful unital $*$-representation, $\pi(A)$ is a unital $C^*$-subalgebra of $B(H)$ and $\pi:A\to\pi(A)$ is a $*$-isomorphism. Hence each state $\phi\in S(A)$ defines a state $\alpha_\phi:\pi(A)\to\mathbb C$ by the rule $\alpha_\phi(\pi(a))=\phi(a)$; this rule is well-defined because $\pi$ is injective. Similarly, $\rho(B)$ is a unital $C^*$-subalgebra of $B(K)$ and each $\psi\in S(B)$ defines a state $\beta_\psi:\rho(B)\to\mathbb C$ by $\beta_\psi(\rho(b))=\psi(b)$. Let \begin{align*} \pi_\phi:A\to B(H_\phi) \end{align*} and \begin{align*} \rho_\psi:B\to B(K_\psi) \end{align*} be the GNS representations associated to $\phi$ and $\psi$. Tomiyama's theorem says that the tensor GNS representation $\pi_\phi\odot\rho_\psi$ is weakly contained in the concrete spatial representation $\pi\odot\rho$. In norm terms, weak containment gives exactly \begin{align*} \|(\pi_\phi\odot\rho_\psi)(x)\|_{B(H_\phi\otimes K_\psi)}\le \|(\pi\odot\rho)(x)\|_{B(H\otimes K)} \end{align*} for every $x\in A\odot B$. Why is this the right replacement for the tempting product-state argument? The product functional obtained from arbitrary state extensions to $B(H)$ and $B(K)$ is not automatically known to be continuous for the concrete spatial norm without precisely this theorem. Tomiyama's theorem supplies that missing continuity and avoids circularly assuming representation independence. The universal representation $\pi_u\odot\rho_u$ is the Hilbert direct sum of the tensor GNS representations $\pi_\phi\odot\rho_\psi$ over all state pairs $\phi\in S(A)$ and $\psi\in S(B)$. The norm of a direct sum of bounded operators is the supremum of the norms of the summands, so taking the supremum over all state pairs gives \begin{align*} \|x\|_u\le \|(\pi\odot\rho)(x)\|_{B(H\otimes K)}. \end{align*} Together with the opposite inequality from the previous step, this proves equality with the universal spatial norm for every faithful unital pair. Hence two faithful unital pairs necessarily give the same norm value on $x$.[/guided]

[step:Verify that the common function is a norm] Choose faithful unital representations \begin{align*} \pi:A\to B(H) \end{align*} and \begin{align*} \rho:B\to B(K). \end{align*} Define the map $N_{\pi,\rho}:A\odot B\to[0,\infty)$ by \begin{align*} N_{\pi,\rho}(x):=\|(\pi\odot\rho)(x)\|_{B(H\otimes K)}. \end{align*} The map $N_{\pi,\rho}$ is homogeneous and satisfies the triangle inequality because it is the composition of the [linear map](/page/Linear%20Map) $\pi\odot\rho$ with the operator norm on $B(H\otimes K)$. It remains only to prove definiteness. If $\|x\|_{\min}=0$, then \begin{align*} (\pi_u\odot\rho_u)(x)=0. \end{align*} By representation independence, every faithful unital spatial representation has the same norm as the universal one. In particular, choose faithful unital concrete realizations of $A$ and $B$; these exist by [citetheorem:8566]. We use a bounded vector-functional slice argument to prove algebraic faithfulness. Write $x=\sum_{m=1}^n a_m\otimes b_m$ with $b_1,\dots,b_n$ linearly independent. Define the finite-dimensional subspace \begin{align*} E:=\operatorname{span}\{\rho(b_1),\dots,\rho(b_n)\}\subset B(K). \end{align*} Because $\rho$ is faithful, the operators $\rho(b_1),\dots,\rho(b_n)$ are linearly independent. For each $m$, choose the coordinate functional $\theta_m:E\to\mathbb C$ satisfying $\theta_m(\rho(b_l))=\delta_{ml}$. Vector functionals on $B(K)$ separate points of $E$: if $R\in E$ and $R\ne 0$, then there is $\eta\in K$ with $R\eta\ne 0$, and choosing $\zeta:=R\eta$ gives $(R\eta,\zeta)_K\ne 0$. Since $E$ is finite-dimensional, the restrictions to $E$ of vector functionals span $E^*$. Hence for each $m$ there are an integer $q_m\in\mathbb N$, vectors $\eta_{m,r},\zeta_{m,r}\in K$, and scalars $c_{m,r}\in\mathbb C$ for $1\le r\le q_m$ such that \begin{align*} \theta_m(R)=\sum_{r=1}^{q_m}c_{m,r}(R\eta_{m,r},\zeta_{m,r})_K \end{align*} for every $R\in E$. Now use the operator equality $(\pi\odot\rho)(x)=0$ in $B(H\otimes K)$. For arbitrary $\xi,\xi'\in H$, applying this zero operator to $\xi\otimes\eta_{m,r}$ and pairing with $\xi'\otimes\zeta_{m,r}$ gives \begin{align*} 0=\sum_{l=1}^n (\pi(a_l)\xi,\xi')_H(\rho(b_l)\eta_{m,r},\zeta_{m,r})_K \end{align*} for every $1\le r\le q_m$. Multiplying by $c_{m,r}$, summing over $r$, and using the definition of $\theta_m$ gives \begin{align*} 0=\sum_{l=1}^n (\pi(a_l)\xi,\xi')_H\theta_m(\rho(b_l))=(\pi(a_m)\xi,\xi')_H. \end{align*} Since $\xi,\xi'\in H$ were arbitrary, $\pi(a_m)=0$. Faithfulness of $\pi$ gives $a_m=0$ for every $m$. Since $b_1,\dots,b_n$ are linearly independent, $x=0$. Thus $\|\cdot\|_{\min}$ is a norm on $A\odot B$. [/step]

[step:Inherit the $C^*$-identity from the operator norm] Let $x\in A\odot B$, and choose faithful unital representations \begin{align*} \pi:A\to B(H) \end{align*} and \begin{align*} \rho:B\to B(K). \end{align*} By [citetheorem:8580], $\pi\odot\rho$ is a $*$-homomorphism. Therefore \begin{align*} (\pi\odot\rho)(x^*x)=(\pi\odot\rho)(x)^*(\pi\odot\rho)(x). \end{align*} Using the $C^*$-identity in the concrete operator algebra $B(H\otimes K)$, we obtain \begin{align*} \|x^*x\|_{\min} = \|(\pi\odot\rho)(x^*x)\|_{B(H\otimes K)}. \end{align*} Hence \begin{align*} \|x^*x\|_{\min} = \|(\pi\odot\rho)(x)^*(\pi\odot\rho)(x)\|_{B(H\otimes K)}. \end{align*} The operator-norm $C^*$-identity gives \begin{align*} \|(\pi\odot\rho)(x)^*(\pi\odot\rho)(x)\|_{B(H\otimes K)} = \|(\pi\odot\rho)(x)\|_{B(H\otimes K)}^2. \end{align*} By the definition of the common minimal norm, \begin{align*} \|(\pi\odot\rho)(x)\|_{B(H\otimes K)}^2=\|x\|_{\min}^2. \end{align*} Therefore \begin{align*} \|x^*x\|_{\min}=\|x\|_{\min}^2. \end{align*} This proves that $\|\cdot\|_{\min}$ is a $C^*$-norm on the algebraic $*$-algebra $A\odot B$. No completeness assertion is made at this algebraic stage; completeness is obtained only after taking the norm completion. [/step]