I was reading the section on the structure of type I von Neumann algebras in John B. Conway's "A course in operator theory" and a few questions about certain definitions and references arose, I was just seeking some clarification :).

Let $A$ be a $C^*$-algebra, $M_n(A)$ is the algebra of $n \times n$ matrices with entries in $A$, $A^{(n)}$ is the direct sum of $A$ with itself $n$ times.

On page 285 he defines:

"50.10 Definition. If $\mathcal{A}$ is a von Neumann algebra and $n$ is some cardinal number, let $M_n(\mathcal{A}) = \{ (\mathcal{A}')^{(n)} \}'$."

1 - What does this $M_n(\mathcal{A})$ have to do with the algebra of $n \times n$ matrices defined above?

Then on the next page (page 286) he writes:

"Another way to define $M_n(\mathcal{A})$ is to define the tensor product of von Neumann algebras. In this setting $M_n(\mathcal{A})$ is precisely $\mathcal{A}\otimes \mathcal{B}(\mathcal{K})$, where $\mathcal{K}$ is a Hilbert space with dimension $n$. In fact the Hilbert space $\mathcal{H} \otimes \mathcal{K}$ can be identified with $\mathcal{H}^{(n)}$. With this identification , $\mathcal{A}' \otimes \mathbb{C} \subseteq \mathcal{B}(\mathcal{H} \otimes \mathcal{K})$ is identified with $\mathcal{A}'^{(n)}$ and $M_n(\mathcal{A})$ with $\mathcal{A} \otimes \mathcal{B}(\mathcal{K})$. The reader can see the references for an exposition of the tensor product of von Neumann algebras."

I couldn't find the reference for this so I wanted to ask:

2 - In this paragraph, is this supposed to apply only to type I algebras?

3 - What does it mean for $\mathcal{A}' \otimes \mathbb{C} \subseteq \mathcal{B}(\mathcal{H} \otimes \mathcal{K})$ to be "identified" with $\mathcal{A}'^{(n)}$ and $M_n(\mathcal{A})$ with $\mathcal{A} \otimes \mathcal{B}(\mathcal{K})$? Does that mean that $\mathcal{A}' \otimes \mathbb{C}$ is isomorphic to $\mathcal{A}'^{(n)}$ and $M_n(\mathcal{A})$ is isomorphic to $\mathcal{A} \otimes \mathcal{B}(\mathcal{K})$?

Does anyone know? Or know where I can find a reference for that last paragraph?


1 Answer 1


I believe the problem here is the notation. Conway uses the notation

  • $H^{(n)}$ to mean the $n$-fold direct sum of $H$, and
  • $A^{(n)}$ for the $n$-fold direct sum of the operator $A\in B(H)$.
  • $\mathcal{A}$ for the $C^*$ or von Neumann algebra.

(I didn't have Conway's book in my office, but I looked at the preview on Google Books, and searched for "cardinal number," and I found Definition 6.3.(a). I don't know if he also uses the notation $\mathcal{A}^{(n)}$ for the $n$-fold direct sum of $\mathcal{A}$, but I can see how this could be confusing!)

In what you've written, $\mathcal{A}^{(n)}$ should be the $n$-fold amplification of $\mathcal{A}$, i.e., the image of the map $$ A\mapsto \begin{pmatrix} A & & \\ & \ddots & \\ & & A \end{pmatrix} \in B(H^{(n)}). $$ So $M_n(\mathcal{A})$ is the operators on $H^{(n)}$ commuting with the diagonal action of the algebra $\mathcal{A}'$, as expected. I hope this clears up the confusion.

I think it would be more clear to use lower case letters for operators and upper case letters for algebras, e.g., $a\in \mathcal{A}$.

  • $\begingroup$ Thank you! So this is for all algebras not just type $I$, and just to be clear "to be identified" means that $M_n(\mathcal{A})$ is isomorphic to the tensor product of von Neumann algebras $\mathcal{A} \otimes \mathcal{B}(\mathcal{K})$? $\endgroup$ Commented Oct 4, 2013 at 13:58
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    $\begingroup$ Yes. For any von Neumann algebra $\mathcal{A}$ acting on $\mathcal{H}$, $(\mathcal{A}'\otimes 1_{\mathcal{K}})'=\mathcal{A}\otimes B(\mathcal{K})$ is an example of the commutation theorem. See Theorem 5.9 in Takesaki's Theory of operator algebra I. $\endgroup$ Commented Oct 4, 2013 at 14:31
  • $\begingroup$ Will do, thank you for your answer and your patience Dave. $\endgroup$ Commented Oct 8, 2013 at 6:53
  • $\begingroup$ Dear Dave just a question, you're saying that $A$ is $^*$-isomorphic to $A \bar{\otimes} B(\mathcal{K})$ for any Hilbert space $\mathcal{K}$ of any dimension $n$? What is the algebra $A' \bar{\otimes} 1_\mathcal{K}$? Is this algebra supposed to be $^*$-isomorphic to $A'$? If so, are their commutants also $^*$-isomorphic so that we get $A \cong A \bar{\otimes} B(\mathcal{K})$? I'm a bit confused. $\endgroup$ Commented Aug 20, 2014 at 22:44
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    $\begingroup$ @Mario, No, I am not saying $\mathcal{A}$ is $*$-isomorphic to $\mathcal{A}\otimes B(\mathcal{K})$. This is false. Just take $\mathcal{A}=\mathbb{C}$, and $\mathcal{K}\cong \mathbb{C}^2$. Rather, $\mathcal{A}$ is $*$-isomorphic to $\mathcal{A}\otimes 1_{\mathcal{K}}\subset \mathcal{A}\otimes B(\mathcal{K})$, which consists of the matrices over $\mathcal{A}$ which are diagonal and constant along the diagonal. This should hopefully answer your second question - just replace $\mathcal{A}$ with $\mathcal{A}'$. $\endgroup$ Commented Aug 20, 2014 at 23:55

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