Joan Lindsay Orr



Unitary Equivalence of Nests

The theory of unitary invariants for nests was developed by Erdos. See also Davidson Chapter 7.

Prerequisites

Let $$\A$$ be an abelian von Neumann algebra on a separable Hilbert space $$\H$$. Then $$\A'$$ has a cyclic vector, which, as a consequence, is also a separating vetor for $$\A$$.

A nest is a linearly ordered set of projections in $$B(\H)$$ which is order-complete (closed under arbitrary meets and joins) and contains $$0$$ and $$I$$.

Properties of nests:

• The order topology and the weak operator topology coincide
• Nests are weak operator topology compact

The Spectral Measure

Erdos developed the theory for arbitrary Hilbert spaces, but the treatment becomes much simpler if we restrict to separable Hilberts spaces, which we shall do throughout this page.

Let $$x_0\in\H$$ be a fixed unit separating vector for $$\N''$$. The map $\alpha: N\in\N \longmapsto \ip{Nx_0}{x_0}\in[0,1]$ is continuous, increasing, one-to-one, and maps $$0$$ to $$0$$ and $$I$$ to $$1$$. Thus $$\X := \alpha(\N)$$ is a compact subset of $$[0, 1]$$ and $$\alpha^{-1}$$ is continuous. Parameterize $$\N$$ by $$N_t := \alpha^{-1}(t)$$ for $$t\in\X$$. We can easily extend the parameterization to the whole interval $$[0,1]$$ by defining $N_t := \bigvee\{N_s : s \le t, s\in\X \}$ This parameterization is monotonic and right-continuous.

Let $$x\in\H$$. The function $$g_x(t) = \ip{N_t x}{x}$$ is monotonic and right-continuous on $$\X$$. Let $$\mu_x$$ be the Lebesgue-Stieltjes measure on $$[0,1]$$ obtained from $$g_x$$. Next, for $$x, y\in\H$$ define a complex measure using the polarization identity $\mu_{x,y}(S) = \sum_{k=0}^3 i^k \mu_{x + i^ky}(S)$ It follows that for $$t\in\X$$, \begin{align} \mu_{x,y}((0,t]) &= \sum_{k=0}^3 i^k \mu_{x + i^ky}([0,t]) \\ &= \sum_{k=0}^3 i^k g_{x + i^ky}(t) - g_{x + i^ky}(0) \\ &= \sum_{k=0}^3 i^k \ip{N_t(x + i^ky)}{x + i^ky} \\ &= \ip{N_t x}{y} \end{align} Next, one verifies that for any fixed Borel set $$S$$, $$(x,y)\mapsto\mu_{x,y}(S)$$ is a bounded bilinear form. This follows because the set of $$S$$ for which this is true is a $$\sigma$$-algebra and contains all the half-open intervals $$(a, b]$$. Thus for each Borel set $$S$$ there is a bounded operator $$\E(S)$$ such that $\mu_{x,y}(S) = \ip{\E(S)x}{y}$

The map $$S\mapsto \E(S)$$ is a projection-valued measure.

Proof. Or, at least the idea of a sketch of it...

If $$\N$$ is a continuous nest then $$\E$$ is equivalent to Lebesgue measure on the Borel sets of $$[0,1]$$.

Proof. Since $$\N$$ is continuous, $$\X$$ (the range of $$\alpha$$, defined above), is equal to $$[0,1]$$ and for any $$t\in[0,1]$$ \begin{align} \ip{\E((0, t])x_0}{x_0} &= \ip{N_t x_0}{x_0} \\ &= \ip{\alpha^{-1}(t) x_0}{x_0} \\ &= \alpha(\alpha^{-1}(t)) \\ &= t \\ &= m((0, t]) \end{align} It follows that $$\ip{\E(S)x_0}{x_0} = m(S)$$ for any Borel set $$S$$ and so, since $$x_0$$ is a separating vector for $$\N''$$ (which contains all $$\E(S)$$), $$\E(S)=0$$ if and only if $$m(S)=0$$.

A nest $$\N$$ is said to be multiplicity-free if $$\N''$$ has a cyclic vector.

All multiplicity-free continuous nests are unitarily equivalent.

Proof. It suffices to show that an arbitrary multiplicity-free continuous nest $$\N$$ is unitarily equivalent to the Volterra nest on $$L^2([0,1])$$. Since $$\N$$ is multiplicity-free, $$\A := \N''$$ has a cyclic vector $$x_0$$, But then $$x_0$$ is separating for $$\A'$$, which contains $$\N''$$, and so $$x_0$$ is cyclic and separating for $$\N''$$. Let $$\E$$ be the spectral measure on $$[0,1]$$ described above, using $$x_0$$ and the parameterization of $$\N$$ derived from it.

Now for any $$f\in L^\infty([0,1])$$, \begin{align} \left\| \left(\int f(t) d\E(t)\right) x_0 \right\|^2 &= \left\langle \left(\int |f(t)|^2 d\E(t)\right) x_0, x_0\right\rangle \\ &= \int |f(t)|^2 d \langle \E(t) x_0, x_0\rangle \\ &= \int |f(t)|^2 dm(t) \\ &= \|f\|^2_2 \end{align} Thus the map $f \longmapsto \int f(t) dE(t)$ is well-defined and isometric from a dense subset of $$L^2([0,1])$$ (i.e. $$L^\infty([0,1])$$) into $$\H$$ and its range contains $$\A x_0$$, which is dense in $$\H$$. Thus it extends by continuity to a unitary $$U: L^2([0,1]) \rightarrow \H$$. Next, for $$f\in L^2([0,1])$$, any fixed $$N_t$$ is equal to $$\E((0, t])) = \int \chi_{(0,t])} d\E$$ and so \begin{align} N_t Uf &= \int \chi_{(0,t])}d\E \; \int f d\E \\ &= \int \chi_{(0,t])}f d\E \\ &= U M_{\chi_{(0,t])}} f \end{align} Thus $$U^* N_t U$$ is the multiplication operator $$M_{\chi_{(0,t])}}$$ in $$L^2([0, 1])$$.