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Like the title: is it true that a self-adjoint unital subalgebra of $L(H)$ closed in the weak operator topology (a Von Neumann algebra) has a trivial null space? Why? Thank you.

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  • $\begingroup$ What would the null space be? $\endgroup$ Mar 10 '14 at 1:11
  • $\begingroup$ A null space of a set $S$ (in my case the set is the self-adjoint unital subalgebra of $L(H)$) is the set of the elements $\xi \in H$ such that $A\xi=0$ for all $A \in S$. $\endgroup$
    – Benzio
    Mar 10 '14 at 10:45
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If the unit of the subalgebra is the identity operator (that is, the algebra is non-degenerate), then the answer is yes: if $A\xi=0$ for all $A$, then in particular $\xi=I\,\xi=0$.

If the unit of the subalgebra is a projection $P\ne I$, then the null space is $(I-P)L(H) $.

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  • $\begingroup$ Very clear! Thanks! $\endgroup$
    – Benzio
    Mar 10 '14 at 12:55
  • $\begingroup$ You are welcome. $\endgroup$ Mar 10 '14 at 14:55

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