# Why are factors the natural building stones of a von Neumann algebra?

If $\mathcal{H}$ is a separable Hilbert space then we may decompose a von Neumann algebra $\mathcal{M}\subset B(\mathcal{H})$ into a direct integral of factors, therefore one considers factors as the building block of von Neumann algebras and focuses on factors when classifying them.

My question is: Why are factors the natural building block of a von Neumann algebra (despite that it finally works)?

• From what I've read, the proofs are not very straightforward. I'm almost certain the results are in "On Rings of Operators: Reduction Theory" written by Von Neumann himself, published in 1949. – Aweygan Jan 26 '17 at 15:28
• I don't need a proof, I am more interested on a "birds view" and I am sorry when I didn't make this clear enough in my question. – Sebastian Bechtel Jan 26 '17 at 15:31
• I see. Well I wish I could provide such a view, but alas I am working my way towards such an understanding myself. Good luck! – Aweygan Jan 26 '17 at 15:48

## 2 Answers

To understand any category of algebraic objects (groups, rings, etc.) one of the key goals is to understand the simple objects (the ones, $M$, for which all morphisms $M\rightarrow N$ for any object $N$ is injective). For von Neuman algebras and normal homomorphisms the simple objects are the factors.

• Thank you for this answer! In my lecture notes there is a corollary that all normal *-homomorphisms from a factor are injective and I added as a note that I should figure out why this is interesting :P Now I can replace this reminder with your answer. – Sebastian Bechtel Jan 26 '17 at 20:25

What Murray-von Neumann did was to show that there is an infinite-dimensional generalization of the following fact.

If $\mathcal H$ is finite-dimensional and $\mathcal M\subset\mathcal B(\mathcal H)$ is a von Neumann algebra, it is a basic exercise that we can see $\mathcal B(\mathcal H)$ as $M_n(\mathbb C)$ for $n=\dim\mathcal H$. And in that situation, $\mathcal M$ is isomorphic to $$\tag{*}\bigoplus_{k=1}^{\ell} M_{n(k)}(\mathbb C),$$ where the blocks are given by the minimal central projections (i.e., each block is $P\mathcal M P$, with $P$ a minimal central projection.

When $\mathcal H$ is infinite-dimensional, the same idea works. Thing is, now the centre may not have minimal projections, but what they proved is that there exists a Borel space $X$ and a Borel measure $\mu$ such that $$\mathcal M=\int_X^{\oplus} \mathcal M_\lambda\,d\mu(\lambda),$$ where the function $\lambda\longmapsto \mathcal M_\lambda$ is factor-valued a.e.

In the particular case when $X$ is finite, $\mu$ is the counting measure and the $\mathcal M_\lambda$ are finite-dimensional, one recovers $(*)$.

• If I recall correctly, finding a generalization to the Artin-Wedderburn theorem along with the corresponding fact about decomposing measurable actions into a direct integral of ergodic actions was a significant part of von Neumann's motivation in studying von Neumann algebras. – Owen Sizemore Jan 26 '17 at 17:59
• I cannot comment on that, really. I do know that von Neumann was very interested in setting the mathematical framework for quantum mechanics. I also read somewhere (maybe an article by Murray) that the "Rings of Operators" papers grew out of the fact that they initially tried to show that any factor was isomorphic to $B(H)$. Maybe this was a second step after what you suggest. – Martin Argerami Jan 26 '17 at 20:15
• My lecturer pointed out (as a note when we have classified finite c*-algebras) that this was von Neumanns attempt to extend the Artin-Wedderburn theorem to the von Neumann setting, so your remark seems reasonable to me. – Sebastian Bechtel Jan 26 '17 at 20:29
• @MartinArgerami do I understand you correctly that using factors was not the starting point of the idea to decompose a vNA but rather that one had a decomposition into $pMp$ with $p$ a minimal central projection (in the case of finite VNAs) and then realized that $pMp$ are factors? – Sebastian Bechtel Jan 26 '17 at 20:43
• No, the analogy with the finite-dimensional case is too evident, so Murray and von Neumann must have been expecting factors. What they also expected, as in the finite-dimensional case, was for all factors to be type I. – Martin Argerami Jan 26 '17 at 22:17