Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
    
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

3 Answers 3

up vote 11 down vote accepted

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 $M \sim pMp \bar{\otimes} B(\ell^2I)$.

share|improve this answer
    
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.

share|improve this answer
    
Nice explanation! –  Yemon Choi Nov 7 '10 at 2:36

could you tell me which page it is?

share|improve this answer
1  
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.