It is well known that a separable Hilbert space $H$ decomposes as a direct integral in the presence of an abelian von Neumann algebra $\mathscr A\subseteq B(H)$.

More precisely, and quoting from Kadison & Ringrose, Theorem (14.2.1), there is a (locally compact, complete, separable metric) measure space $(X, \mu )$ such that $H$ is the direct integral of Hilbert spaces $\{H_p\}$ over $(X, \mu )$ and $\mathscr A$ is the algebra of diagonalizable operators relative to this decomposition.

In the study of von Neumann algebras (on separable Hilbert spaces), only a few measure spaces ever need to be considered, such as discrete spaces, the unit interval, and unions of these. In fact the above integral decomposition Theorem still holds even if we require $X$ to be among these few examples.

However, there are instances in which some pre-determined spaces are relevant to one's intentions. For example, suppose that $Y$ is a nice (locally compact, complete, separable metrizable) topological space and that $C_0(Y)$ is faithfully represented on some separable Hilbert space $H$ through a non-degenerate representation. Then the double commutant $$\mathscr A = C(Y)''$$ qualifies as a commutative von Neumann algebra on $H$, and hence $H$ decomposes as a direct integral over some OTHER measurable space $(X, \mu )$, as seen above.

Question 1. Can we always take $X$ to be $Y$ with its Borel structure? Of course one would then like such a decomposition to respect the original representation of $C_0(Y)$ in the sense that each $f$ in $C_0(Y)$ acts on $H_y$ as scalar multiplication by the complex number $f(y)$, for almost all $y$.

Question 2. In case of an affirmative answer to the above, is there a concrete description of each fiber $H_y$? Could $H_y$ perhaps be described in terms of spectral projections for smaller and smaller neighborhoods of $y$?


1 Answer 1


The answer to question 1 is "yes" and this can for instance be proven by the following concrete construction that also somehow answers question 2.

I will make the (nonrestrictive) assumption that the given representation of $C_0(Y)$ on $H$ is nondegenerate, in the sense that $C_0(Y) H$ is total in $H$.

Choose a countable family of unit vectors $(\xi_i)_{i \in I}$ in $H$ such that the subspaces $C_0(Y)\xi_i$ are mutually orthogonal and densely span $H$. Once this choice is made, everything will be canonical.

For every $i \in I$, denote by $\mu_i$ the probability measure on $Y$ given by $\int_Y F d\mu_i = \langle F \xi_i,\xi_i\rangle$ for all $F \in C_0(Y)$. Choose $\alpha_i > 0$ with $\sum_{i \in I} \alpha_i = 1$ and define the probability measure $\mu$ on $Y$ by $\mu = \sum_{i \in I} \alpha_i \mu_i$. Since $\mu_i \prec \mu$, take Borel functions $F_i \geq 0$ on $Y$ such that $d\mu_i / d\mu = F_i$. Consider the Hilbert space $K = \ell^2(I)$ with its canonical orthonormal basis $(e_i)_{i \in I}$. Define the measurable field of Hilbert spaces $(H_y)_{y \in Y}$ by defining $H_y$ as the closed linear span of the vectors $e_i$ with $F_i(y) > 0$. Define the measurable sections $s_i$ of this field by $s_i(y) = F_i(y)^{1/2} e_i$.

There now is a unique unitary operator $U : \int_Y^\oplus H_y d\mu(y) \to H$ satisfying $UF = FU$ for all $F \in C_0(Y)$ and $U(s_i) = \xi_i$ for every $i \in I$. This provides the required direct integral decomposition over the space $Y$ and also gives a quite concrete description of $H_y$.

  • $\begingroup$ Thanks Stefaan for this very nice answer! This is very much what I was hoping to be true! After the choice of the $\xi_i$, I think there is still a bit of non-uniqueness since the $F_i$ are only determined up to null sets but the precise values of $F_i(y)$ affect the choice of $H_y$. Nevertheless, of course, no one expects much uniqueness in a construction like this. $\endgroup$
    – Black
    May 13, 2021 at 14:09
  • $\begingroup$ @Black : as you write, I do not think that one can expect a canonical construction defining $H_y$ for every $y \in Y$. There will always appear a measure class on $Y$ and a construction of $H_y$ for almost every $y \in Y$. $\endgroup$ May 13, 2021 at 14:27

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