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In the proof of this theorem, which says all of the type 1 factors (factors with minimal projections) are isomorphic to $B(\ell^2(I))$ for some $I$, I want to know a few things:

  1. The supposed isomorphism is usually obtained as follows: if $M$ is a factor with a minimal projection, then after some argument one obtains a mutually orthogonal set of minimal projections $p_i$ (with index set $I$) such that the strongly convergent sum over all the $i\in I$ is $1$. I.e. the ranges of the projection give an orthogonal decomposition of the Hilbert space on which $M$ acts. Also, this decomposition is such that every $x$ in $M$ has the property that $e_ixe_j=x_{ij}e_{ij}$ for some matrix $x_{ij}$ over the index set $I$. (Here $e_{ij}$ are the partial isometries that take the range of $p_i$ unitarily to the range of $p_j$, since all minimal projections are equivalent like this.)

The suggested isomorphism is the map that takes $x$ to its $x_{ij}$ which is clearly well defined. With some analysis, one finds that it actually maps into $B(\ell^2(I))$ It is clearly injective and a homomorphism. It also preserves involution, so that we know it's isometric. But why is it surjective? If you decide to construct the inverse, please remember to prove that its image consists entirely of bounded operators in $M$. Bounded is important.

2.I don't feel right calling this an isomorphism when the subject matter is Von Neumann algebras unless the isomorphism is both a homeomorphism between weak topologies and also between strong topologies. Otherwise it's as if I'm just dealing with unital $C^*$- algebras. Are either of these the case? If so, please prove it. Thanks.

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What do you mean by the weak and strong topologies on a von Neumann algebra? –  Qiaochu Yuan May 27 '12 at 5:10
The notion of weak and strong topologies exist for $B(H)$. $M$ is a subset, hence can have the induced topology. –  Jeff May 27 '12 at 5:16
Yes, but it doesn't exist for an abstract von Neumann algebra, does it? –  Qiaochu Yuan May 27 '12 at 5:17
I did not realize there was any such thing as an abstract VNA. To me they are weakly closed unital C* subalgebras of B(H). Let me know what your definition is, but my question is wrt my definition. –  Jeff May 27 '12 at 5:19
An abstract von Neumann algebra is a $C^{\ast}$-algebra which, as a Banach space, has a predual. You can find this definition on the Wikipedia page, for example. So it is just a $C^{\ast}$-algebra with an extra property and the correct notion of isomorphism of von Neumann algebras is the obvious one. –  Qiaochu Yuan May 27 '12 at 5:31

1 Answer 1

up vote 1 down vote accepted

It's possible to give "formal" proofs of this result that don't actually deal with the issue you're talking about. So it helps to pay attention to the details of that argument, and to handle them in the "right" way.

[I will assume we are dealing with von Neumann algebras on separable Hilbert spaces only--- mainly because I haven't ever really thought about the more general case.]

Let $M$ be a factor with a minimal projection on a Hilbert space $H$. Let $\{p_i: i \in I\}$ be a maximal family of mutually orthogonal minimal projections in $M$. By hypothesis the set $I$ is either finite or countable. So I will assume $I$ is either the set of positive integers, or the set $\{1,2,\dots,n\}$ for some finite $n$.

Arguments you are probably familiar with show that all of the projections $p_i$, $i \in I$, are Murray-von Neumann equivalent to one another, and the partial isometries implementing these equivalences give rise to a system of matrix units in $M$, allowing you to "think of elements of $M$ as block matrices" and thus get a map from $M$ to $\mathcal{B}(\ell^2(I))$...

But it's simpler from a technical point of view to complete the following program:

  1. Prove that the partial isometries implementing the equivalences among the $p_i$, $i \in I$, generate $M$ as a von Neumann algebra.

  2. Observe (or prove) that the space $H$ decomposes as an orthogonal direct sum $$ H = \bigoplus_{i \in I} p_i H, $$ so there is unitary $U$ from the Hilbert space $K = \ell^2(I) \otimes p_1 H$ (the Hilbert space tensor product) to $H$ satisfying $$ U (c_i \otimes \eta) = e_{i1} \eta, \qquad i \in I, \quad \eta \in p_1 H, $$ where here for each $i \in I$, I let $c_i$ denote the element of $\ell^2(I)$ that is $1$ at $i$ and $0$ elsewhere (this is sort of the "usual" orthonormal basis of $\ell^2(I)$) and $e_{i1}$ denotes the matrix unit in $M$ with initial space $p_1 H$ and final space $p_i H$.

  3. It immediately follows from 1. and 2. that $U^* M U$ ($ = \{U^* m U: m \in M\}$) is a von Neumann subalgebra of $\mathcal{B}(K)$ generated by the operators $U^* e_{ij} U$, $i, j \in I$.

  4. Do a calculation that shows for each $i, j \in I$ that $U^* e_{ij} U$ is just the operator $E_{ij} \otimes \operatorname{id}$ on $\mathcal{B}(K)$, where $\{E_{ij}: i, j \in I\}$ is the usual system of matrix units in $\mathcal{B}(\ell^2(I))$ coming from the orthonormal basis $\{c_i: i \in I\}$ and $\operatorname{id}$ denotes the identity operator on $p_1 H$.

    You know (as a special case of 1.) that the operators $E_{ij} \otimes \operatorname{id}$, $i, j \in I$, generate the von Neumann subalgebra $\mathcal{B}(\ell^2(I)) \otimes \operatorname{id}$ of $\mathcal{B}(K)$.

So not only is $M$ "isomorphic to" some $\mathcal{B}(\ell^2(I))$, but we can write down a specific unitary operator between the Hilbert space on which $M$ acts and the Hilbert space $\ell^2(I) \otimes p_1 H$ that implements the isomorphism.

Note that the unitary implementing the isomorphism is not, generally, coming from a unitary between $H$ and $\ell^2(I)$. This makes intuitive sense, though: for example, the von Neumann subalgebra of $\mathcal{B}(H)$ (with $H$ infinite dimensional) generated by two mutually orthogonal equivalent projections summing to $1$ is clearly isomorphic (as an complex algebra) to $\mathcal{B}(\mathbb{C}^2)$, but there is no hope of finding a unitary from $H$ to $\mathbb{C}^2$. This isomorphism exists only in the world of abstract algebra: it is not spatial. And in some contexts, you only want to consider spatial isomorphisms (ones implemented by unitaries in the above way) as your "isomorphisms."

You are right to find it sort of "fishy" when people write down maps between von Neumann algebras that using "matrix theoretic" formulas on generators and don't comment on issues like these. But, when algebraic formulas work, very often it is because there is a unitary between some underlying Hilbert spaces that implements the formulas (and, to find the all-details-included proof, you often just need to identify or construct the relevant Hilbert spaces). It is trivially true that maps of the form $x \mapsto U^* x U$, with $U$ unitary, map von Neumann algebras to von Neumann algebras, and play nice with all the topologies. If you do it right, that is what is going on here.

I hope this helped.

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This was very helpful. I got some good practice working on the details of step 4 because I used the Kaplansky density theorem together with some theory of nets in order to show that the $B(l^2(I))$ copy in B(K) is a Von Neuman algebra over K, i.e. showing it was weakly closed. –  Jeff May 27 '12 at 8:50
@Jeff it is very good practice to get sensitive to details (and know multiple ways of checking them). A surprising lot of people never take the time, which is OK if you only use the most "algebraic" aspects of von Neumann algebras, but very risky for analysis. (A good example of the subtlety: if $A \subseteq \mathcal{B}(H)$ and $B \subseteq \mathcal{B}(K)$ are von Neumann algebras, one can show without too much hassle that $A \otimes B$ is a von Neumann subalgebra of $\mathcal{B}(H \otimes K)$. What is its commutant? [...] –  leslie townes May 28 '12 at 1:32
"Obviously" it's the tensor product $A' \otimes B'$ of the commutants of $A$ and $B$. But a proof of this in general (ie, independent of special hypotheses on the algebras $A,B$) was not known until I think the 1960s. (I think Dixmier discusses this a bit in one of the later editions of his book on von Neumann algebras.) So even the "algebra" part of von Neumann algebras is subtle, and demands close attention. –  leslie townes May 28 '12 at 1:34
Yes actually the desire to prove a similar result came up yesterday. I was unaware it was so difficult. I ended up just showing it was a Von Neumann algebra directly without using Bicommutants. –  Jeff May 28 '12 at 2:17

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