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 $\phi\_2$?



Is independence meaningful for commutative $C^*$-algebras?
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).


