# Generation of Von Neumann Algebras

Suppose $M$ is a Von Neumann Algebra. (VNA) For me, these are subsets of some $B(H)$ that are $*$-algebras, containing the $1$ of $B(H)$, that are Weak Operator (WO) closed, or equivalently Strong Operator (SO) closed.

For a given subset $S$ of $M$ one can always refer to the VNA generated relative to $M$ by $S$. This is the smallest VNA containing $S$ that is contained in $M$, or equivalently it's an intersection over all the VNAs containing $S$ that are contained in $M$. Is there some constructive expression of this, as there is in so many other cases of similar concepts? I don't think the weak operator closure of the noncommuting-variable-polynomials evaluated at the elements of $S$ is sufficient, because operator multiplication is not continuous from the product of two WO topologies to the WO topology. (I only know that I cannot prove that it is. Does someone have an example of when this pathology arises?)

Anyway to "get my hands on" $\langle S\rangle$ would be appreciated, thanks!

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Another way to describe it is $\langle S\rangle = (S\cup S^*)''$, the double commutant of $S\cup S^*$. –  Jonas Meyer May 19 '12 at 21:08
Yeah that makes sense, but is there some sort of closure of the set of polynomials generated by S or something nice like this? What if S is just a single element? Thanks. –  Jeff May 19 '12 at 21:21

You have to also include the adjoints in your "noncommuting-variable-polynomials," but then it actually does work to just take the weak closure of the algebra. This can be seen as a consequence of von Neumann's Double Commutant Theorem, which also gives another way to describe the algebra as $\langle S\rangle = (S\cup S^*)''$, the double commutant of $S\cup S^*$. If $A$ is the unital *-subalgebra of $B(H)$ generated by $S$, then $A''=(S\cup S^*)''$ is equal to the weak closure of $A$.
Jeff: In my comment I had only mentioned that the double commutant of $S\cup S^*$ is the von Neumann algebra generated by $S$, but I waited for the answer to mention that this is also the weak closure of the *-algebra, which more directly answers your question. (I'll post another comment.) –  Jonas Meyer May 20 '12 at 1:05
Let $S$ be the unilateral shift, $S(x_0,x_1,\ldots)=(0,x_0,x_1,\ldots)$ on $\ell^2$. Then $S^n\to 0$ weakly and $(S^*)^n\to 0$ weakly but $(S^*)^nS^n=I$ for all $n$. You could make the product sequence not converge at all by interweaving $0$s, e.g. $(S,0,S^2,0,S^3,0,\ldots)\to 0$ weakly and $(S^*,0,(S^*)^2,0,(S*)^3,0,\ldots)\to 0$ weakly, but $(S^*S,0^2,(S^*)^2S^2,0^2,\ldots)=(I,0,I,0,\ldots)$ does not converge weakly. –  Jonas Meyer May 20 '12 at 1:11