Let $M$ is atomic $\sigma-$finite von Neumann algebra. Atomic means that every projection has minimal subprojection. $\sigma-$finite means that the cardinality of any set consisting of mutually orthogonal projections is no more than countable. Then there are separeble Hilbert spaces $H_n$ such that $M$ is $*-$isomorphic to the algebra $\bigoplus\limits_{n=1}^{\infty} B(H_n)$. My question is does anybody know the proof or the literature where the full proof is given?


1 Answer 1


I gave a proof earlier (still below) which I think was wrong. Here is a new proof.

Suppose that $M \subseteq \mathcal{B}(\mathcal{H})$ is separably represented. Let $(p_i)$ be a maximal family of orthogonal central projections. Note that we can assume that each $p_i$ is minimal in $Z$ (not necessarily $M$). This is because if $p_i$ is not minimal in Z, say $0 \neq q < p_i$ is minimal. Then $(p_j)_{j \neq i}\cup\{q,p_i-q\}$ is another maximal family of orthogonal central projections. We repeat this process with minimal projections of $p_i - q$. Since $M$ is $\sigma$-unital, this process will still leave us with a countable collection. We can repeat this for any $i$, which won't increase the cardinality of the collection. Now $M = \oplus_i p_iM$ (since $1 = \text{WOT-}\sum p_i$). So showing that $p_iM$ are factors will suffice; but this is true since $p_iZ = \mathbb{C}p_i$ as $p_i \in Z$ is minimal. Now $p_iM = \mathcal{B}(\mathcal{H}_i)$, where $\mathcal{H}_i = p_i\mathcal{H}$ (or abstractly, its clear that $p_iM$ are type I factors, from which it follows that they are $\mathcal{B}(\mathcal{H}_i)$).

I think the proof below is wrong since $A_i$ maximal abelian and $A_i' = \mathcal{B}(\mathcal{H}_i)$ implies that $\mathcal{H}_i$ is 1-dimensional. I leave it there in hopes that someone can tell me that either it cannot be done that way or to fix it.

We know that the center can be uniquely decomposed as $Z \simeq \oplus_i A_i$ (we will identify these), where $A_i \subseteq \mathcal{B}(\mathcal{H}_i)$ are maximal abelian (Corollary V.I.1.32 of Takesaki's "Theory of Operator Algebras I"). Since $M$, hence $Z$, is atomic (hence $A_i$ are), and this composition is unique, we must have that the unit $p_i$ of $A_i$ is minimal in $Z$ (else we could decompose the direct sum further) so that $A_i = \mathbb{C}p_i$ (note that $p_i$ is minimal in $A_i$, not $ \mathcal{B}(\mathcal{H}_i)$). Now we have that

$$M = Z' \cap M = \oplus_i A_i' \cap p_iM = \oplus (p_i\mathbb{C})' = \oplus_i \mathcal{B}(\mathcal{H}_i) $$

Note that the identities $p_i$ of $A_i$ are centrally mutually orthogonal projections, and hence there are countably many of them since $M$ is $\sigma$-unital. Since $A_i \subseteq \mathcal{B}(\mathcal{H}_i)$ is maximal abeian, $p_i = 1_{\mathcal{H}_i} \in \mathcal{B}(\mathcal{H}_i)$ and $p_iM$ is a factor (as it has trivial center), we must have that $p_iM = \mathcal{B}(\mathcal{H}_i)$. Thus $M \simeq \oplus_i \mathcal{B}(\mathcal{H}_i)$ as a countable direct sum.

Just something to note about the $p_i$'s, and how they are minimal in $Z$ (or $A_i$). Let us consider the example $M_2$. Then centre is $\mathbb{C}I$. Note that $1 \oplus 0$ is clearly smaller than $I$, but the issue is that its not central! So it is not smaller in $A = \mathbb{C}I$. This is why the dimension of $\mathcal{H}_i$ is not just 1 for all $i$.

  • $\begingroup$ What does it mean "maximal abelian von Neumann subalgebra" in this case? Because, the center $Z$ is itself abelian von Neumann subalgebra and from $A_i\subset Z$ we have $Z=A_i$. Is the $p_i$ of $A_i$ minimal projections? If they aren't then how to show $p_i p_j =0$ for $i\neq j$? If $A_i = \mathbb C p_i$ then $p_i B(H) p_i\subset A_i^{'}$ and why $A_i^{'} = p_i M p_i$? Could you be more explicit, please. $\endgroup$ Feb 26, 2020 at 8:37
  • $\begingroup$ A maximal abelian self-adjoint algebra (MASA) is a subalgebra $A \subseteq M$ such that $A' = A$. We are saying that $A_i$ is a MASA in $Z$. The $p_i$ are probably not minimal projections. $p_i \perp p_j$ for $i \neq j$ as they are in different direct summands of $Z$. The commutant of $A_i$ in $M$ (as opposed to in $Z$), is $p_iM$. $\endgroup$
    – PStheman
    Feb 26, 2020 at 13:01
  • $\begingroup$ I couldn't proof that $A_i = \mathbb C p_i$. Atomicity of $Z$ and the fact that $A_i$ is MASA in $Z$ gives us that there exists minimal projection $e_i\leq p_i$ such that $e_i\in A_i$. How to use the uniqueness of the decomposition $Z=\bigoplus\limits_{i} A_i$ to complete the proof? $\endgroup$ Feb 27, 2020 at 15:29
  • $\begingroup$ Hmm, I'm going to edit the post to make things more clear. I believe $p_i$ are actually minimal projections (in $A_i$, not $M$), so that $A_i = p_iA_i = p_i\mathbb{C}$ (one can check that $p$ minimal in $N$ is equivalent to $pNp = \mathbb{C}p$). $\endgroup$
    – PStheman
    Feb 27, 2020 at 16:06
  • $\begingroup$ I've updated the answer to make things more clear hopefully. I think I may have jumped the gun on some of the things I said initially. $\endgroup$
    – PStheman
    Feb 27, 2020 at 16:29

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