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.

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!

share|improve this question
1  
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

1 Answer 1

up vote 1 down vote accepted

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 $A$.

share|improve this answer
    
Ah, I see. You are saying that since I have agreed your answer is a valid expression, that I should recognize it also is exactly what I wanted. So you have here a proof that goes around our lack of a proof of the joint continuity of operator multiplication in WO topology. When I mentioned pathology, I was referring to this. Do you have an example of two nets of operators that converge weakly each, but whose product does not converge weakly? (Or at least not to the product of the limits?) Thanks. –  Jeff May 20 '12 at 1:01
    
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
    
@Jeff: By "pathology" I thought you meant you wanted examples where the weak closure of the *-algebra is not closed under multiplication, which cannot happen. I've edited my answer to clarify. You're right that it's nontrivial because of multiplication not being WOT continuous, but fortunately that isn't really a problem. Some related facts are that the SOT closure equals the WOT closure, multiplication is SOT continuous on bounded sets, and elements of the SOT closure of a *-subalgebra of B(H) are limits of bounded nets from the algebra (by Kaplansky's Density Theorem). –  Jonas Meyer May 20 '12 at 1:30

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.