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Let $H$ be an infinite dimensional separable complex Hilbert space. All C*-subalgebras of $B(H)$ are assumed to be non-degenerate.

The spectral projections of a self-adjoint element $T$ of $B(H)$ lie in the weakly closed algebra generated by $T$. In the early 1970s Pedersen proved that if a C*-subalgebra $A$ of $B(H)$ contains all of the spectral projections of each of its self-adjoint elements, then $A$ is weakly closed.1 In his words, this "characterizes von Neumann algebras (on separable Hilbert spaces) as the only C*-algebras in which the spectral theorem can be used in its full force."2

On the other hand, suppose we start with an arbitrary C*-subalgebra $A$ of $B(H)$. Must we go all the way to the weak closure to use the spectral theorem for self-adjoint elements of $A$? If not, does iterating the process of taking the smallest C*-algebra in which we can apply the spectral theorem lead to the weak closure in a finite number of steps? Pedersen asked this question over 30 years ago as a way to end a chapter on concrete C*-algebras,3 and I don't know if it has been answered. If the answer isn't known, I'd accept an answer giving a more recent reference that discusses this. For the precise question, I'll just quote Pedersen:

For any C*-subalgebra $A$ of $B(H)$ define $a(A)$ as the smallest C*-subalgebra of $B(H)$ containing all spectral projections of each self-adjoint element in $A$. It is easy to verify that $A\subset a(A)\subset A''$. If $H$ is separable is then $A''=a(A)$? This failing, is $A''=a(a...a(A)...)$ (finitely many steps)? Note that by 2.8.8 a transfinite (but countable) application of the operation $a$ will produce $A''$.

1 G. Pedersen, C*-algebras and their automorphism groups, Corollary 2.8.8, p. 38.

2 Ibid., p. 39.

3 Ibid.

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What is the reasoning in "Note that by 2.8.8 a transfinite (but countable) application of the operation a will produce $A''$." ? I see that $\omega_1$ applications of the operation $a$ produce $A''$ and also that each element of $A''$ appears at the $\alpha$-th application for some $\alpha < \omega_1$; but why is $a^{\alpha}(A)=A''$ for some $\alpha < \omega_1$? A related question: Is there some $\alpha < \omega_1$ which works for all $A \subset B(H)$? –  Andreas Thom Oct 9 '10 at 12:28
    
Andreas: That is a good question, and I do not know the answer. Perhaps I should have mentioned in the first place that I did not understand that line. I included it as part of the quote for completeness, because I hoped it would add better context. But in any case, I'm glad it inspired your question: mathoverflow.net/questions/41597/… –  Jonas Meyer Oct 20 '10 at 23:55
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