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

Suppose that a von Neumann algebra $A$ acts on $\ell_2(I)$ and $I$ has the minimal cardinality for which it holds. Can we caluclate (or at least give a reasonable lower bound) for the cardinality of the set of projections in $A$?

share|improve this question
    
Can you add more detail? Have you thought about it? What if $A$ is a factor, does that simplify it? –  MTS Jul 3 '12 at 22:10
    
You should think about this more. Just think of the example where $|I|=2$. So you are asking about unital subalgebras of $M_2(\mathbb{C})$. There depending on which algebra A is there are either the minimal possible for a von Neumann algebra, 2, or as many as uncountably many (which is the maximum if $|I|$ is at most countable. –  Owen Sizemore Jul 4 '12 at 0:06
    
@Owen But if the algebra is just the scalars, so that there are only two projections, then 2 is not the minimal cardinality of a basis of a Hilbert space that the algebra can act on. –  MTS Jul 4 '12 at 0:38
    
@MTS. Yes I was not interpreting the questions that way but now that you mention it I think that is what is meant by "$I$ has the minimal cardinality for which it holds". In that case the cardinality of Proj(A) should be at least cardinality of $I + 2$. I haven't thought through the details, but I'm pretty sure that should be true –  Owen Sizemore Jul 4 '12 at 1:59
1  
@MTS. Yes this was what I was trying to say in my first comment. As soon as there is any non-commutativity there are going to be lots (uncountably many...?) projections. So the least of amount of projections would occur in an abelian algebra. My statement about $I+2$ was stupid. Now that I think it should be that the least amount of projections occurs in $l^\infty(I)$, which should have number of projections equal to the cardinality of the power set of $I$. –  Owen Sizemore Jul 4 '12 at 6:59
show 1 more comment

Know someone who can answer? Share a link to this question via email, Google+, Twitter, or Facebook.

Your Answer

 
discard

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

Browse other questions tagged or ask your own question.