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 $A$?
Asked
Viewed
135 times

$\begingroup$ Can you add more detail? Have you thought about it? What if $A$ is a factor, does that simplify it? $\endgroup$ – MTS Jul 3 '12 at 22:10

$\begingroup$ 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. $\endgroup$ – Owen Sizemore Jul 4 '12 at 0:06

$\begingroup$ @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. $\endgroup$ – MTS Jul 4 '12 at 0:38

$\begingroup$ @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 $\endgroup$ – Owen Sizemore Jul 4 '12 at 1:59

1$\begingroup$ @MTS. Yes this was what I was trying to say in my first comment. As soon as there is any noncommutativity 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 $I$. $\endgroup$ – Owen Sizemore Jul 4 '12 at 6:59