Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $H$ be a Hilbert space, and $A$ a $C^*$-subalgebra of $B(H)$ (the bounded operators on $H$). Let $B$ be the strong-operator closure of $A$, so that in particular, $B$ is a von-Neumann-algebra.

According to the Kaplansky-Theorem:

The … in the unit ball of $A$ is s-o dense in the … in unit ball of $B$, where … is:

  • unitaries
  • self adjoints

Does it hold, that $\operatorname{Proj}(A)$ is strongly dense in $\operatorname{Proj}(B)$ (or even weakly dense).

I.e. does the Kaplansky-Theorem hold for projections? (Or a weaker version of it.)

If not, why not?

share|improve this question

1 Answer 1

No. E.g., let $A=C[0,1]$ acting as multiplication operators on $L^2[0,1]$. The only projections in $A$ are $0$ and $1$. The strong closure contains multiplications by the characteristic functions of all measurable subsets of $[0,1]$.

More generally, infinite-dimensional C*-algebras need not have projections, while von Neumann algebras are spanned by projections.

share|improve this answer
    
What about if $B$ is, say, the hyperfinite $II_1$ factor? –  RS8 Nov 21 '12 at 17:46
    
(See edit: I meant B.) –  RS8 Nov 21 '12 at 17:48
    
I don't know if it is true in that case. –  Jonas Meyer Nov 21 '12 at 17:51
    
Could one obtain a bridge result, say: Kaplansky holds for partial isometries? That would also be useful to me. –  RS8 Nov 21 '12 at 17:57
    
RS8: In the example in my answer, the set of partial isometries in $A$ is not strongly dense in the set of partial isometries in $B$. I don't know about weak density. –  Jonas Meyer Nov 21 '12 at 21:50

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.