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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 $A$?

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  • $\begingroup$ 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
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    $\begingroup$ @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 $I$. $\endgroup$ – Owen Sizemore Jul 4 '12 at 6:59

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