I don't know very much about spectral theory so probably the answer to my question has a basic reference which I would appreciate.

Let's say I have two self-adjoint operators on a Hilbert space and that they commute. I know that, since they generate a commutative $C^*$ algebra, on the one hand they can be viewed as continuous functions on some compact Hausdorff space. On the other hand, individually, they can be viewed as multiplication operators $L^2$ on some measure space.

Now, one (extremely basic) question I have is whether it can be arranged for them to be viewed as multiplication operators on the same measure space. I am guessing the answer is yes, but I would like a reference for it or a counterexample since I don't really know.

Now I am sort of assuming the answer to question 1 is yes, but here's question 2. Let's say I already have two multiplication operators in hand, $\phi_1 : X \to {\mathbb R}$ and $\phi_2 : X \to {\mathbb R}$ where $X$ has measure $\mu$. I would say that $\phi_1$ and $\phi_2$ are independent if the pushforward of $\mu$ by $\phi_1 \times \phi_2 : X \to {\mathbb R}^2$ is the product measure of the pushforwards of $\mu$ by the maps $\phi_1$ and $\phi_2$ individually. Notice that if I compose $\phi_1 \circ T$ and $\phi_2 \circ T$ with any measure-preserving map, the resulting maps are still independent since the pushed-forward measures do not change at all.

My second question is whether the notion of independence makes sense for such operators (or more generally for members of a commutative $C^*$ algebra) without having a measure at all. For example, is being independent an operator-theoretic property of $\phi_1$ and $\phi_2$?


For your first question, yes, you can view two commuting normal operators as functions on the same measure space. Let us for simplicity assume that $T_i : i=1,2$ are self-adjoint acting on a Hilbert space $H$. Then there exists a measure $\mu$ defined on $\mathbb{R}^2$ with values in projections oh $H$, so that: $\mu(A)$ and $\mu(B)$ commute for any Borel subsets of $\mathbb{R}$; $\mu(A)$ and $\mu(B)$ are perpendicular whenever $A$ and $B$ are disjoint; and $\mu$ is $\sigma$-additive (where you use the strong operator topology to define limits of sums of perpendicular projections). Moreover, you have:

Given $c_1, c_2$ and $C_1, C_2$ with $c_i < C_i$, $\mu( [c_1, C_1] \times [c_2,C_2])$ is the orthogonal projection onto the set $\{\xi \in H: c_i \Vert \xi \Vert \leq \Vert T_i \xi \Vert \leq C_i \Vert \xi \Vert\}$ and $T_i = \int x_i d\mu(x_1,x_2)$ (here $x_1,x_2$ are the two coordinates on $\mathbb{R}^2$ and the convergence of the integral is in the strong sense, i.e., at every vector). This means in particular that for any $\xi$, $$\langle T_i \xi ,\xi \rangle = \int x_i \langle d\mu (x_1,x_2)\xi ,\xi \rangle $$ (note that $\langle \xi ,\mu (\cdot)\xi\rangle $ is a positive scalar-valued measure).

Now you would prefer to have scalar-valued measures. To do so, denote by $M$ the von Neumann algebra generated by $T_1,T_2$, and find a sequence of vectors $\xi_i$ in your Hilbert space so that $M\xi_i \perp M\xi_j$ for all $i\neq j$ and $\textrm{span}(M\xi_i : i=1,2,\dots)$ is dense in $H$. Let us arrange that $\sum \Vert\xi_i\Vert^2 =1$. Let $H_i$ be the closure of $M\xi_i$. Then if you denote $\mu_i(B) = \langle \xi_i \mu (B),\xi_i\rangle$, you find that $M\xi_i \cong L^2(\mathbb{R}^2,\mu_i$ in a way that takes $T_i$ to the multiplication operator by $x_i$. The original action of $T$ on $H$ can then be viewed as multiplication by $x_i$ on the disjoint union of measure spaces $\sqcup_i (\mathbb{R}^2,\mu_i)$ with the probability measure given by the sum of the $\mu_i$'s.

Any book that does the spectral theorem will have an equivalent statement. Probably you can find it in Conway's functional analysis book.

Regarding your second question: it does make sense to talk about independence of operators, provided that you have chosen a measure (i.e., a state on your $C^*$-algebra). The definition for multiplication operators is exactly the one for random variables: $\phi_i: (X,\mu)\to \mathbb{R}$ are independent if the push-forward of $\mu$ is the product measure. But this of course depends on the measure $\mu$ in a crucial way. For example, if you take $x_1,x_2: (\mathbb{R^2},\mu)\to \mathbb{R}$ and you take $\mu$ to be the Lebesgue measure on $[0,1]^2$, then the $x_i$'s will be independent. But take instead the $\delta$-measure concentrated on the diagonal $\{(x_1,x_2):x_1=x_2,0\leq x_j\leq 1/\sqrt{2}\}$ (i.e., $\mu(B)$ is the Lebesgue measure on the diagonal of the intersection of $B$ with a diagonal line segment from the origin to $(1/\sqrt{2},1/\sqrt{2})$). In that case the push-forwards of $\mu$ by $x_1$ and $x_2$ are equal and are not the product of the individual push-forwards. Thus it is absolutely essentially that you talk about independence of functions on measure spaces (=self-adjoint operators in an abelian $C^*$-algebra, =random variables) with respect to the chosen measure (=state, =expectation).


If $X$ is a separable compact Hausdorff space then for any Radon measure $\mu$ with dense support we have $C(X) \subset L^\infty(X, \mu)$, so the answer to your first question is yes.

Independence will depend on the measure you take, consider for instance the difference between taking Lebesgue measure on $[0, 1] \times [0,1]$ versus taking a measure which gives most of its weight to the diagonal. In the former case functions of $x$ will be independent from functions of $y$, while this is not the case in the latter.

These types of results would be covered in any introductory book on operator algebras. For instance R. Douglas "Banach algebra techniques in operator theory" (http://www.ams.org/mathscinet-getitem?mr=1634900).


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