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Dear all, I really want to know the answer of the following question. I would appreciate any help.

Assume H is a separable Hilbert space, is it possible to find two nests N1, N2 such that the intersection of Alg(N1) and Alg(N2) contains only scalars? If dimH is finite, then it is quiet easy to answer the similar question negatively, but I have no idea how to approach the question if dimH is infinite. I hope someone could provide some hint.

Thanks in advance!

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How do you do this when dim$(H)$ is finite? –  Nik Weaver Jun 26 '12 at 12:58
    
When dim(H) is finite, you can consider the dimension of the Alg(N1) and Alg(N2). –  heller Jul 2 '12 at 9:21
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2 Answers

You ask for a hint. Here is what I would try. Given any two nests $N_1$ and $N_2$, for each finite subsets $A \subseteq N_1$ and $B \subseteq N_2$ find a norm one operator in $Alg(A) \cap Alg(B)$. Then take a WOT cluster point. That is guaranteed to give you an operator in $Alg(N_1) \cap Alg(N_2)$. The interesting part is ensuring that it isn't a scalar; to do that you'll have to think about what kind of control you have over the operator in $Alg(A) \cap Alg(B)$. Most likely you either have sufficient control to easily ensure the limit operator isn't a scalar, or have so little control that you can see how to rig things so that the limit operator has to be a scalar, either of which would provide an answer.

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How about this. Let $\mathscr{A}_1$ and $\mathscr{A}_2$ be the algebras of upper and lower-triangular operators on $\ell_2$ with constant diagonals, respectively. $\mathscr{A}_1\cap \mathscr{A}_2 = \mathbb{C}I$.

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Are those algebras of the form Alg(N)? –  Yemon Choi Jun 26 '12 at 23:13
    
Rhetorical question, eh, Yemon? –  Bill Johnson Jun 26 '12 at 23:21
    
Yes, the algebra must be the reflexive algebra determined by the nest. –  heller Jun 28 '12 at 1:55
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