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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?

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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.

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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

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