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This question was asked in a slightly different way here, and then I realized that it could be of higher interest in the physics community.

Let $\mathcal{P}(\mathcal{H})$ be an (atomic) lattice of projections on a separable Hilbert space $\mathcal{H}$. Let $\mathcal{B}$ be a maximal Boolean subalgebra (block) of $\mathcal{P}(\mathcal{H})$. Then, by spectral theorem, there exists a measure space $(X,\mu)$ and unitary operator $U:\mathcal{H}\rightarrow L^2(X,\mu)$ such that for each $P\in\mathcal{B}$, it holds that $UPU^{-1}$ is a multiplication by a measurable function $f_P:X\rightarrow\mathbb{R}$.

The interesting implication is that if $\mathcal{B}$ is a complete** block, then $\mathcal{B}\simeq\mathcal{B_0}$, where $\mathcal{B_0}$ is a measure algebra of $(X,\mu)$. Obviously, whenever $\mathcal{B}$ is an atomic BA, $\mathcal{B_0}$ is also atomic. (Details can be found e.g. in Takeuti's "Two applications of logic to mathematics" pg. 63.)

I have several questions to that issue:

  • minor one: to what extent $(X,\mu)$ is unique in the spectral theorem?
  • major one 1: my intuition is that if $\mathcal{B}$ is a complete block generated (and maximized by Zorn's lemma) by spectral measure of a self-adjoint operator with a continuous spectrum, then $\mathcal{B}$ is atomless. On the other hand, if a complete $\mathcal{B}$ comes similarly from a self-adjoint operator with a pure point spectrum, then $\mathcal{B}$ is atomic (must $\mu$ be discrete then?) - is this correct?*
  • major one 2: what can be said about atomicity of a complete block $\mathcal{B}$ generated by a self-adjoint operator with mixed (both continuous and point) spectrum?
  • aside: could this be related to the presence of minimal projections in von Neumann algebras? I suspect the answer is no, since from the very beginning every block in $\mathcal{P}(\mathcal{H})$ being a BA is commutative, hence not factor.

*This would come from the fact that projections $P$ under the isomorphism are mapped to characteristic functions $\chi_P$; if a projection's range is a one-dimensional subspace of $\mathcal{H}$, then $\chi_P=\chi_{\{p\}}$ for some $p\in X$.

**Edit: I realized that the completeness is a subtle point here: a complete BA of projections is not only complete as a BA (i.e. containing all the sup's), but also in the following, stronger sense: if $P=\mathrm{sup}(P_a)$, then $P(\mathcal{H})$ (the range of $P$) is the closure of the linear space spanned by $\bigcup P_a(\mathcal{H})$.

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First of all I am not sure to understand well your terminology, but I think that maximality implies completeness (the $\sup$ of a set of projectors, viewed as the strong limit of a class of projectors, commutes with all the projectors of the block using the strong operator topology and so it belongs to the block as it is maximal). So your blocks are complete.

Regarding your first question actually I do not know, I think it is unique but I should check my conjecture and I do not have much time now. Sorry.

Regarding your second question. I assume that atomic means that for every $P\neq 0$ in the block there is an atom $Q$ with $Q \leq P$. If the spectrum has only continuous component the lattice not only is not atomic, but it does not contain atoms because the atoms are associated to the single eigenvalues. If the spectrum of $A$ is a pure point spectrum, then $P_{\sigma_p(A)}=I$ by definition (notice that $\sigma_c(A)\neq \emptyset$ is still possible, think of a self-adjoint compact operator whose eigenvectors are accumulated by $0$ which, in turn, is not an eigenvalue). By difference, $P_{\sigma_c(A)}=0$. Therefore, every Borel set $E\subset \sigma(A)$ either intersects $\sigma_p(A)$ or produces the zero projector. In the first case an atom $Q \leq P_E$ exists since it is associated to some eigenvalue of $A$ included in $E$.

Regarding your third question, as the atoms are all of the form $P_{\{\lambda\}}$ with $\lambda \in \sigma_p(A)$, then the lattice may be atomic or not. As I pointed out above, pure point spectrum does not mean that $\sigma_c(A)$ is empty, but only that $P_{\sigma_p(A)}=I$, in this case however the lattice is atomic. In more complicated cases where the continuous part of the spectrum is full open segment for instance and the point spectrum includes also some isolated points, the lattice is not atomic.

I do not understand well your last question. Think of a Hermitian matrix $A$ so that its commutative lattice is generated by the atoms $P_\lambda$, where $\lambda$ is every eigenvalue of $A$. The commutative von Neumann algebra generated by the lattice coincides to the algebra of all complex functions $f(A)$ spectrally defined. It is clear that, even if the algebra is not a factor, the projectors $P_\lambda$ are minimal projectors of that algebra.

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  • $\begingroup$ Well, I had to edit the question to explain completeness, it is not obvious that every block is complete also in the stronger sense. You elaborated well on the spectra. However, I have trouble to think of an example of atomless BA of projections that in the process of maximizing and completing remains atomless (i.e. 1-dim projections are not included). This is the only unclear point to me. Last point: lack of atoms in some BA of projections and lack of minimal projections in type II, III factors have to be accidental, since BA of projections by definition can not by a factor of any type. $\endgroup$ – krzysiekb Jan 4 '17 at 16:13
  • $\begingroup$ " I have trouble to think of an example of atomless BA of projections that in the process of maximizing and completing remains atomless (i.e. 1-dim projections are not included)" Could you be more explicit? In particular what do you mean by "process of maximizing and completing"? $\endgroup$ – Valter Moretti Jan 4 '17 at 16:32
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    $\begingroup$ Regarding completeness of a maximal block, I was actually referring to $\sigma$-completeness since the Hilbert space is supposed to be sparable. In that case, given an at most countable sequence of elements $P_n$ in the block, $\sup_n P_n$ can always be written as a strong limit sum $P= \sum_m Q_n$ where $Q_n Q_m =0$ if $n\neq m$ and each $Q_n$ is in the block (essentially by means of Gramm-Schmidt procedure). Therefore $P$ commutes with every projector commuting with each $Q_m$, that is with every element of the block. By maximality $P$ belongs to the block. $\endgroup$ – Valter Moretti Jan 4 '17 at 16:47
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    $\begingroup$ An example in $H= L^2(\mathbb R, dx)$ is the spectral measure of the position operator. It is known that if an orthogonal projector $Q$ (actually a bounded self-adjoint operator) commutes with every element of that spectral measure then it is a function of it. The only possibility for a projector it is that $Q$ belongs to the spectral measure (in other words we have a maximal and also $\sigma$ complete block). Since the spectrum of the operator is continuous, the block is atomless. $\endgroup$ – Valter Moretti Jan 4 '17 at 18:02
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    $\begingroup$ Another example is the joint spectral measure of $P_1,P_2,P_3$ in $L^2(\mathbb R^3, d^3p)$. Those three operators define a maximal set of compatible observables: a bounded self-adjoint operator commuting with their joint spectral measure is a bounded measurable function of it (an element of the von Neumann algebra generated by that spectral measure)...the lattice is atomless $\endgroup$ – Valter Moretti Jan 4 '17 at 18:12

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