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Let us suppose that M is a von Neumann algebra on some hilbert space $H$ such that $M = A''$ for some $C^*$-algebra $A\subseteq B(H)$.

I am wondering when $A$ will be dense in $M$ and by which topology. I also want to know that if I have two functional $S,T \in M^*$ such that $S|_A=T|_A$, can I say that $T=S$?

Thank you very much.

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up vote 2 down vote accepted

Let $K\subseteq H$ be the closed subspace generated by $AH$. Then von Neumann's double commutant theorem says that if $K=H$, then $M$ is the closure of $A$ in the weak operator, strong operator, ultraweak, or ultrastrong topologies. If $K$ is a proper subspace of $H$, then $A=A|_K\times \{0\}|_{K^\perp}$, $A''=(A|_K)''\times \mathbb C I_{K^\perp}$, and $A$ is weakly dense in $(A|_K)''\times \{0\}$ but not $M$. The condition $K=H$ is also known as $A$ acting nondegenerately on $H$. In particular, it holds if $A$ contains the identity operator on $H$.

So assuming that $A$ acts nondegenerately, $A''=M$ is equivalent to density of $A$ in $M$ in various topologies that are coarser than the norm topology. If $M^*$ is the set of norm-continuous linear functionals on $M$, then it is possible in general for distinct functionals in $M^*$ to agree on $A$, if at least one of them is not ultraweakly continuous. $A$ is a norm-closed subspace of $M$, so if $A\neq M$ then there are nonzero continuous functionals on the Banach space $M/A$, hence by composing with the quotient map, there are nonzero continuous functionals on $M$ that vanish on $A$.

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If two elements of $M^*$ are in the form of $f(\cdot)=\langle \cdot \xi,\eta\rangle$ and $g(\cdot)=\langle \cdot \xi',\eta'\rangle$ for some $\xi', \eta', \xi', \eta'\in H$. Then what can we say about $f$ and $g$, if $f|_A=g|_A$? – unknown is my last name Jul 5 '12 at 6:25
@unknownismylastname: Those functionals are continuous in the weak operator topology, so WOT density of $A$ in $M''$ says that $f=g$. – Jonas Meyer Jul 5 '12 at 14:47

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