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Let $\mathcal{A}$ be a $C^*$-algebra. Consider vector space of matrices of size $n\times n$ whose entries in $\mathcal{A}$. Denote this vector space $M_{n,n}(\mathcal{A})$. We can define involution on $M_{n,n}(\mathcal{A})$ by equality $$ [a_{ij}]^*=[a_{ji}^*],\qquad\text{where}\quad [a_{ij}]\in M_{n,n}(\mathcal{A}). $$ Thus we have an involutive algebra $M_{n,n}(\mathcal{A})$. It is well known that there exist at most one norm on $M_{n,n}(\mathcal{A})$ making it a $C^*$-algebra. This norm does exist. Indeed take universal representation $\pi:\mathcal{A}\to\mathcal{B}(H)$ and define linear injective $^*$-homomorphism $$ \Pi:M_{n,n}(\mathcal{A})\to\mathcal{B}\left(\bigoplus\limits_{k=1}^n H\right):[a_{ij}]\mapsto\left((x_1,\ldots,x_n)\mapsto\left(\sum\limits_{j=1}^n\pi(a_{1j})x_j,\ldots,\sum\limits_{j=1}^n\pi(a_{nj})x_j\right)\right) $$ Hence we can define norm on $M_{n,n}(\mathcal{A})$ as $\left\Vert[a_{ij}]\right\Vert_{M_{n,n}(\mathcal{A})}=\Vert\Pi([a_{ij}])\Vert$. At first sight this definition depends on the choice of representation, but in fact it does not.

My question This norm on $M_{n,n}(\mathcal{A})$ can be defined internally. Namely $$ \Vert[a_{ij}]\Vert_{M_{n,n}(\mathcal{A})}=\sup\left\Vert\sum\limits_{i=1}^n\sum\limits_{j=1}^n x_i a_{ij}y_j^*\right\Vert $$ where supremum is taken over all tuples $\{x_i\}_{i=1}^n\subset\mathcal{A}$, $\{y_i\}_{i=1}^n\subset\mathcal{A}$ such that $\left\Vert\sum\limits_{i=1}^n x_i x_i^*\right\Vert\leq 1$, $\left\Vert\sum\limits_{i=1}^n y_i y_i^*\right\Vert\leq 1$. Is there proof of this fact without usage of structural theorem for $C^*$-algebras, a straightforward proof which can be made by simple checking axioms of $C^*$-algebras?


P.S. There is another answer on this question on mathoverflow.net

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This doesn't answer the question, but for further reference this fact is proved (using irreducible representations) as Lemma 2.3 (i) in "Norming C*-algebras by C*-subalgebras" by Pop, Sinclair, and Smith. The norm also has the internal characterization $\|A\|=r(\sqrt{A^*A})$, where $r$ denotes the spectral radius, as for any C*-algebra. –  Jonas Meyer Jan 31 '12 at 23:00
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I think it would be good manners for you to now link to the copy of the question that you have posted on MathOverflow –  user16299 Feb 1 '12 at 16:22
    
In fact, since the question has now been answered on both sites, I suggest that it be closed here. –  user16299 Feb 1 '12 at 16:46
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Actually, I don't think either should be closed. In many ways, Jonas's answer below is complimentary to mine over at MO, so why not leave them both? –  Matthew Daws Feb 1 '12 at 17:08
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@Yemon: I do see the logic. I don't feel like I know the community views here as I do over at MO, so I think I won't make any further comments... –  Matthew Daws Feb 1 '12 at 20:48

1 Answer 1

up vote 6 down vote accepted

This norm comes from considering $M_n(\mathcal A)$ as acting as operators on the Hilbert C*-module $\mathcal A^n$, and no Hilbert space representation is required. I will try to give a fairly minimal overview of the situation in this special case, and more details can be found in the first chapter of Lance's Hilbert C*-modules: a toolkit for operator algebraists. How straightforward it is depends on your familiarity with these objects. Please feel free to ask for elaboration.

Define on the direct sum $\mathcal A^n$ the $\mathcal A$-valued inner product $\langle\cdot,\cdot\rangle:\mathcal A^n\times\mathcal A^n\to\mathcal A$, given by $$\langle (x_i),(y_i)\rangle=\sum_{i=1}^n x_i^*y_i.$$ The norm on $\mathcal A^n$ is $\|(x_i)\|=\sqrt{\|\langle(x_i),(x_i)\rangle\|}$ (the Cauchy-Schwarz inequality for Hilbert C*-modules, Proposition 1.1 on page 3 of Lance, gives one way to see that this is in fact a norm). Let $\mathcal L(\mathcal A^n)$ denote the set of adjointable operators on $\mathcal A^n$. These are the maps $T:\mathcal A^n\to\mathcal A^n$ such that there exists a map $T^*:\mathcal A^n\to\mathcal A^n$ satisfying $\langle T(x_i),(y_i)\rangle=\langle(x_i),T^*(y_i)\rangle$ for all $(x_i),(y_i)\in\mathcal A^n$. With the operator norm, $\mathcal L(\mathcal A^n)$ is a closed subalgebra of the Banach algebra of all bounded operators on the Banach space $\mathcal A^n$, so $\mathcal L(\mathcal A^n)$ is a Banach algebra. With the conjugate linear involutive anti-automorphism $T\mapsto T^*$, it is also a $*$-algebra. A straightforward computation shows that $\|T^*T\|=\|T\|^2$ for all $T$, so $\mathcal L(\mathcal A^n)$ is a C*-algebra.

Let $\pi:M_n(\mathcal A)\to\mathcal L(\mathcal A^n)$ be defined by $\pi[a_{ij}](x_i)=\left(\sum\limits_{j=1}^n a_{ij}x_j\right)$; that is, $\pi$ is the action of $M_n(\mathcal A)$ on $\mathcal A^n$ by multiplying matrices with column vectors. The fact that $\pi([a_{ij}]^*)=(\pi[a_{ij}])^*$ shows that the codomain of $\pi$ is appropriate. Since $\pi$ is an injective $*$-homomorphism between $C^*$-algebras, it is isometric (alternatively, this could be used to define the unique C*-norm on $M_n(\mathcal A)$). (Incidentally, $\pi$ is surjective if and only if $\mathcal A$ is unital.)

Let's see how this gives the characterization in question of the norm. Let $[a_{ij}]\in M_n(\mathcal A)$. From the definition of the norm on $\mathcal A^n$ and the Cauchy-Schwarz inequality for Hilbert C*-modules, $$\sup\limits_{\|(x_i)\|,\|(y_i)\|\leq 1}\left\Vert\sum\limits_{i=1}^n\sum\limits_{j=1}^n x_i^* a_{ij}y_j\right\Vert=\sup\limits_{\|(x_i)\|,\|(y_i)\|\leq 1}\|\langle (x_i),\pi[a_{ij}](y_i)\rangle\|=\|\pi[a_{ij}]\|=\|[a_{ij}]\|.$$ This is what you have, but with a slightly different appearance due to the convention I used (following Lance) for how the inner product on $\mathcal A^n$ is defined.

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Cf. Matt's answer on MathOverflow –  user16299 Feb 1 '12 at 16:44
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@Yemon: Thanks. I saw your comment on the question after I posted, and hadn't seen the MO version before. Funny, I ended up posting only 1 minute before Matthew, so I can't really blame the unannounced crossposting for the redundancy. –  Jonas Meyer Feb 1 '12 at 16:47

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