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Operator algebras

In analogy to concrete operator spaces we define (cf. [BRS90, Def. 1.1]): An operator algebra is a closed, not necessarily self-adjoint subalgebra $ X$ of $ B({\mathcal{H}})$ ( $ {\mathcal{H}}$ a Hilbert space).
Example: For selfadjoint $ X$ we have the theory of $ C^*$-algebras.
As in the operator space situation, one can also adopt an abstract point of view: here, this leads to considering Banach algebras which are operator spaces and are equipped with a multiplication compatible with the operator space structure. These provide an abstract characterization of the (concrete) operator algebras ( cf. below : analogue of Ruan's theorem for operator algebras).

If $ X$, $ Y$, $ Z$ are operator spaces, and if $ \Phi: X \times Y \rightarrow Z$ is bilinear, we can define another bilinear map in the following way (cf.: Amplification of bilinear mappings):

$\displaystyle \Phi^{(n,l)}:M_{n,l}(X) \times M_{l,n}(Y)$ $\displaystyle \rightarrow$ $\displaystyle M_n(Z)$  
$\displaystyle \left(\left[x_{ij}\right],\left[y_{jk}\right]\right)$ $\displaystyle \mapsto$ $\displaystyle \left[ \sum_{j=1}^{l}
\Phi(x_{ij},y_{jk})\right] ~ (l, n \in {\mathbb{N}}).$  

This is called the bilinear amplification25 of $ \Phi$.

$ \Phi$ is called completely bounded if $ \Vert\Phi\Vert _{\mathrm{cb}} :=
{\sup}_n \Vert\Phi^{(n,n)}\Vert< \infty$, and completely contractive if $ \Vert\Phi\Vert _{\mathrm{cb}} \le 1$.26 Compare this definition with the approach presented in Completely bounded bilinear Mappings .

[In the sequel, for Banach algebras with unit $ e$, we will require $ \Vert e\Vert=1$.] An operator space $ (X,\Vert\cdot\Vert _n)$ with a bilinear, associative and completely contractive map $ m: ~
X \times X \rightarrow X$, the multiplication, is called an abstract operator algebra (cf. [BRS90, Def. 1.4]). Here, the multiplication on $ M_n(X)$ is just the matrix multiplication $ m_n$.
In the unital case $ m$ is automatically associative [BRS90, Cor. 2.4].

We have an analogue of Ruan's theorem ([Ble95, Thm. 2.1], cf. also [BRS90, Thm. 3.1]): Let $ A$ be a unital Banach algebra and an operator space. Then $ A$ is completely isometrically isomorphic to an operator algebra if and only if the multiplication on $ A$ is completely contractive.
This yields the following stability result:
1.) The quotient of an operator algebra with a closed ideal is again an operator algebra [BRS90, Cor. 3.2].
With this at hand, one deduces another important theorem on hereditary properties of operator algebras:
2.) The class of operator algebras is stable under complex interpolation [BLM95, (1.12), p. 320].

Adopting a more general point of view than in the Ruan type Representation Theorem above, one obtains the following [Ble95, Thm. 2.2]: Let $ A$ be a Banach algebra and an operator space. Then $ A$ is completely isomorphic to an operator algebra if and only if the multiplication on $ A$ is completely bounded. (cf. the chapter Examples !)

Basic examples of operator algebras are provided by the completely bounded maps on some suitable operator spaces. More precisely, for an operator space $ X$, one obtains the following [Ble95, Thm. 3.4]: $ \mathit{CB}(X)$ with the composition as multiplication, is completely isomorphic to an operator algebra if and only if $ X$ is completely isomorphic to a column Hilbert space . - An analogue statement holds for the isometric case.

In the following result, for operator algebras $ A$ and $ B$, the assumption that $ A$ and $ B$ be (norm-) closed, is essential (in contrast to the whole rest) [ER90b, Prop. 3.1]:
A unital complete isometry $ \varphi$ between $ A \subset
B({\mathcal{H}})$ and $ B \subset B(\mathcal{K})$ ( $ {\mathcal{H}}$, $ \mathcal{K}$ Hilbert spaces), where $ \mathrm{1\!\!\!\:l}_{B({\mathcal{H}})} \in A$, $ \mathrm{1\!\!\!\:l}_{B(\mathcal{K})} \in B$, is already an algebra homomorphism.



Footnotes

... amplification25
In the literature, e.g. in [BRS90, p. 190], the bilinear amplification $ \Phi^{(n,n)}$ is often referred to as the amplification and is noted by $ \Phi^{(n)}$.
....26
In order to define the notion of complete boundedness of bilinear mappings, it suffices to consider only the $ \Phi^{(n,n)}$ instead of all the $ \Phi^{(n,l)}$; this definition is usually chosen in the literature about completely bounded bi- and, analogously, multilinear maps [BRS90, p. 190], [CES87, p. 281].


Subsections
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Next: Examples Up: Multiplicative Structures Previous: Completely bounded module homomorphisms   Contents   Index
Prof. Gerd Wittstock 2001-01-07