Transfinite induction, a theorem of Pedersen, and chains of subalgebras of $B(H)$ - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T17:46:04Z http://mathoverflow.net/feeds/question/41597 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/41597/transfinite-induction-a-theorem-of-pedersen-and-chains-of-subalgebras-of-bh Transfinite induction, a theorem of Pedersen, and chains of subalgebras of $B(H)$ Andreas Thom 2010-10-09T14:30:11Z 2011-02-03T19:54:33Z <p>This post is closely related to <a href="http://mathoverflow.net/questions/24440/must-we-close-weakly-to-apply-the-spectral-theorem" rel="nofollow">this one</a>. (In fact I copied some of its content.)</p> <p>Let $H$ be an infinite dimensional separable complex Hilbert space. All $C^{\star}$-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 <a href="http://blms.oxfordjournals.org/cgi/reprint/4/2/171" rel="nofollow">proved</a> that if a $C^{\star}$-subalgebra $A$ of $B(H)$ contains all of the spectral projections of each of its self-adjoint elements, then $A$ is weakly closed, i.e. $A=A''$. Now, concerning the consequences of this result, Pedersen says in his book:</p> <p>"For any $C^\star$-subalgebra $A$ of $B(H)$ define $a(A)$ as the smallest $C^{\star}$-subalgebra of $B(H)$ containing all spectral projections of each self-adjoint element in $A$. [...] Note that [...] a transfinite (but countable) application of the operation $a$ will produce $A''$."</p> <blockquote> <p><strong>Question 1:</strong> What is the reasoning here? </p> </blockquote> <p>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 $\alpha &lt; \omega_1$ (This is due to the fact that the closure in the norm-topology is a sequential closure and hence, only countably many elements play a role.); but why is $a^{\alpha}(A)=A''$ for some $\alpha &lt; \omega_1$?</p> <p>Maybe there is something deeper behind:</p> <blockquote> <p><strong>Question 2:</strong> Is there an $\omega_1$-chain of unital subalgebras of $B(H)$?</p> </blockquote> <p>Here $\omega_1$-chain means an $\omega_1$-index family $(A_{\alpha})_{\alpha&lt; \omega_1}$ of subalgebras of $B(H)$ such that for all $\beta &lt; \omega_1$ we have $$\overline{\cup_{\alpha&lt; \beta}A_{\alpha}}^{\|.\|} \subsetneq A_{\beta}.$$</p> http://mathoverflow.net/questions/41597/transfinite-induction-a-theorem-of-pedersen-and-chains-of-subalgebras-of-bh/41617#41617 Answer by Eric Wofsey for Transfinite induction, a theorem of Pedersen, and chains of subalgebras of $B(H)$ Eric Wofsey 2010-10-09T18:01:19Z 2010-10-09T18:11:28Z <p>The answer to Question 2 is yes; such a chain can be found using only diagonal operators with respect to some fixed basis $(e_n)_{n\in{\mathbb N}}$ for $H$. As a starting point, you can find an $\omega_1$-sequence of subsets $S_\alpha\subset{\mathbb N}$ such that if $\alpha&lt;\beta$, $S_\beta\setminus S_\alpha$ is finite and $S_\alpha\setminus S_\beta$ is infinite (that is, the sequence $(S_\alpha)$ is strictly decreasing "modulo finite sets"). This is easy to do because by diagonalizing, given a countable decreasing sequence of infinite subsets of ${\mathbb N}$, you can always find another infinite set which is strictly contained in all of them modulo finite sets. Now let $A_\alpha$ denote the algebra of diagonal operators such that the eigenvalues of the eigenvectors $e_n$ for $n\in S_\alpha$ form a convergent sequence. These are closed because a uniform limit of convergent sequences is convergent. These are nested because if $\alpha&lt;\beta$, then all but finitely many elements of $S_\beta$ are contained in $S_\alpha$ so if the $S_\alpha$-eigenvalues converge, so do the $S_\beta$-eigenvalues.</p> <p>Here's a more conceptual explanation of this construction. The algebra of diagonal operators is naturally isomorphic to the algebra $C_b({\mathbb N})$ of bounded continuous functions on $\mathbb N$, which is in turn naturally isomorphic to the algebra $C(\beta{\mathbb N})$ of all continuous functions on the Stone-Cech compactification of $\mathbb N$. Each set $S_\alpha\subset{\mathbb N}$ has a closure $\overline{S_\alpha}\subset\beta{\mathbb N}$, and the statement that the $S_\alpha$ are decreasing modulo finite sets says exactly that the sequence of closed sets $C_\alpha=\overline{S_\alpha}\setminus S_\alpha$ is actually decreasing. The subalgebra $A_\alpha$ is then exactly the set of continuous functions on $\beta{\mathbb N}$ which are constant on the set $C_\alpha$.</p>