In this post, without further mention all C*-algebras are assumed to have an identity element and subalgebras inherit the identity.

Question: Let $\mathcal{C}$ be a C*-subalgebra of $\mathcal{B}$. Suppose that for every C*-algebra $\mathcal{A}$ containing $\mathcal{C}$ as a C*-subalgebra and every $\mathcal{C}$-bimodule map $\phi:\mathcal{A}\to\mathcal{B}$, $\|\phi\| = \|\phi\|_{cb}$. Does it follow that $\mathcal{C}$ is matricially norming for $\mathcal{B}$?


  • Let $\mathcal{A}$ and $\mathcal{B}$ be C*-algebras and let $\phi:\mathcal{A}\to\mathcal{B}$ be a linear map. Define, for each $n\in\mathbb{N}$, $\phi_n:M_n(\mathcal{A})\to M_n(\mathcal{B})$ by $\phi_n((a_{ij}))=(\phi(a_{ij}))$. The cb norm of $\phi$ is defined by $\|\phi\|_{cb}=\sup_{n\in\mathbb{N}}\|\phi_n\|$. (The "cb" stands for completely bounded, and $\phi$ is called completely bounded when its cb norm is finite.)

  • A C*-subalgebra $\mathcal{C}$ of a C*-algebra $\mathcal{B}$ is called matricially norming for $\mathcal{B}$ if for every $n\in\mathbb{N}$ and every $n$-by-$n$ matrix $(b_{ij})$ in the C*-algebra $M_n(\mathcal{B})$, $$\|(b_{ij})\|=\sup\left\{ \left\| \sum_{i,j=1}^n x_i b_{ij} y_j \right\|: x_i,y_j\in\mathcal{C}, \left\| \sum_{i=1}^n x_i x_i^* \right\|\leq1, \left\| \sum_{j=1}^n y_j^* y_j \right\|\leq1\right\}.$$


  • The matricially norming condition says that matrices over $\mathcal{B}$ are "normed" by contractive row and column matrices over $\mathcal{C}$. That is, for $B\in M_n(\mathcal{B})$, $\|B\|=\sup\|RBC\|$, where the sup is taken over all rows $R$ and columns $C$ of length $n$ of elements of $\mathcal{C}$ such that $\|RR^*\|\leq1$ and $\|C^*C\|\leq1$. (So that, for instance, it is readily seen that $\mathbb{C}$ is matricially norming for $\mathbb{C}$.) Sometimes just "norming" is used instead of "matricially norming", but I am using the more descriptive term found in Paulsen's book.

  • The converse to the question holds. That is, if $\mathcal{C}$ is a matricialy norming C*-subalgebra of $\mathcal{B}$, then for every C*-algebra $\mathcal{A}$ containing $\mathcal{C}$ as a C*-subalgebra and every $\mathcal{C}$-bimodule map $\phi:\mathcal{A}\to\mathcal{B}$, $\|\phi\| = \|\phi\|_{cb}$. This is what inspired the question, brought to me by a fellow graduate student after we had studied chapter 8 in Paulsen's book.

  • The subalgebra of scalars is matricially norming for a commutative C*-algebra, so if there is a counterexample it is with a noncommutative $\mathcal{B}$.

  • I believe that results in this paper by Smith can be used to show that there is no counterexample when $\mathcal{C}=\mathbb{C}$. It is shown there that if all bounded linear maps from all C*-algebras into $\mathcal{B}$ are completely bounded, then $\mathcal{B}$ must be isomorphic to a subalgebra of $M_n(C(X))$ for some $n\in\mathbb{N}$ and some compact Hausdorff space $X$. But then if $\mathcal{B}$ is noncommutative, the transpose map should give an example of a map whose cb norm is bigger than its operator norm.

  • For more on matricially norming C*-subalgebras, the paper by Pop, Sinclair, and Smith is a good place to start.


1 Answer 1


Dear Jonas, My colleague Bill Johnson has drawn this to my attention. Your question has a positive answer after noting that a $C$-bimodule map lifts to a $C\otimes M_n$-bimodule map on $A\otimes M_n$ If $X$ is an $n\times n$ matrix over $A$ of norm 1 then for any row and column contractions $R$, $C$ over $C$ we have $$\|\phi_n(X)\|=sup\{\|R\phi_n(X)C\|: \|R\|,\|C\|\leq 1\} =sup\{\|\phi(RXC)\|: \|R\|,\|C\|\leq 1\} \leq \|\phi\|$$ showing that $\|\phi_n\|\leq \|\phi\|. Of course, the reverse inequality holds and we have proved that the norm and cb-norm coincide. Best of luck with your studies, Roger Smith

Sorry, I (mis)read your question rather quickly so the answer above is not an answer at all. Will think about it.


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