# Why is every factor a tensor product of a $\sigma$-finite factor and a factor of type I?

Here I am not assuming the factor is represented on a separable Hilbert space. This is quoted on page 370 of Takesaki II, then a bit later on page 381, and I haven't been able to find a proof prior to this point in the book or in Takesaki I.

-
If the definition of "factor" is "von Neumann algebra with trivial commutant" then I am rather puzzled by the claimed result (but I don't have access to a copy of Takesaki vol. 2 at present to check) – Yemon Choi Nov 6 '10 at 6:43
For me, a factor is a von Neumann algebra with trivial center. – Kevin Ventullo Nov 6 '10 at 7:37
Sorry, I meant trivial centre! and my previous remark was wrong-headed, please ignore. – Yemon Choi Nov 6 '10 at 8:44
@Kevin: it is usually nicer to include the complete question in the body of the question, repeating the title in your case. – Mariano Suárez-Alvarez Nov 6 '10 at 16:29
@Mariano: Ok, I will do that in the future. – Kevin Ventullo Nov 7 '10 at 8:30

It is because a von Neumann algebra is $\sigma$-finite if it has a faithful normal state, there is a partion of unity $1 = \sum_{i\in I} p_i$ by mutually orthogonal projections equivalent to any given projection $p$ in an infinite factor, and such a decomposition induces the isomorphism $M \sim pMp \bar{\otimes} B(\ell^2I)$.

-
Great answer. Thanks! – Kevin Ventullo Nov 7 '10 at 8:32

Note that "$\sigma$-finite" is a tricky notion. For example, any ${\rm\ II}_1$ factor, even non-separable, is $\sigma$-finite, because the finite trace is a faithful state.

In any case, the argument one needs is that of Makoto Yamashita, with a few clarifications. Note that for type I and type II factors, the assertion is trivial: for type I, the $\sigma$-finite factor can be taken to be $\mathbb{C}$. For type ${\rm\ II}_1$, it is always $\sigma$-finite. And any type ${\rm\ II}_\infty$ is the tensor of a ${\rm\ II}_1$ with a ${\rm\ I}_\infty$, so again the assertion holds.

This leaves us then with a factor of type III. The question is why does there exist a projection $p$ with $pMp$ $\sigma$-finite. It is well-known that any von Neumann algebra has a faithful semifinite normal weight. From this one can deduce that there has to exist a projection $p$ where the weight is finite. And then one deduces that, restricted to $pMp$, the weight is a faithful normal state. As $M$ is type III, one can construct a family of projections $\{p_j\}_{j\in J}$ with $p_j$ equivalent to $p$ for all $j$. This equivalences can be used to construct a system of "matrix units", from where the isomorphism $$M\simeq(pMp)\otimes B(\ell^2(J))$$ follows.

-
Nice explanation! – Yemon Choi Nov 7 '10 at 2:36

could you tell me which page it is?

-
To whoever voted this down: Please remember that a rep 1 user can't comment yet. – Harald Hanche-Olsen Nov 6 '10 at 9:42