Cyclic representations on $C(X)$: if $\pi_\mu\oplus\pi_\nu$ is cyclic, then $\mu\perp\nu$

Question

I'm using the definitions/conventions of $C^*$-algebras and their representations from Conway's Functional Analysis (VIII.5). If I need to supplement any definition, let me know.

Let $X$ be a compact Hausdorff space. Let $\mu$ be a positive Borel measure on $X$. Then $$ \pi_\mu:C(X)\to \mathscr B(L^2(X,\mu)),\phi\mapsto(f\mapsto \phi\cdot f) $$ defines a representation. I've shown that this representation is cyclic. I've also shown that if $\nu$ is another positive Borel measure on $X$, then $\mu\perp\nu$ implies that $\pi_\mu\oplus\pi_\nu\cong \pi_{\mu+\nu}$, which is cyclic.

I want to show that it goes the other way around too: if $\pi_\mu\oplus\pi_v\cong\pi_{\tau}$ where $\tau$ is a pos. Borel measure on $X$, then $\mu\perp\nu$.

Attempt 1:

Say $\pi_\mu\oplus\pi_n$ is cyclic with cyclic vector $(e,f)$. Arguing by contradiction, if $\pi\not\perp\mu$, then we should be able to find $(g_1,g_2)\in L^2(X,\mu)\oplus L^2(X,\nu)$ such that no net of the form $(\phi_i e,\phi_i f)$ converges to $(g_1,g_2)$. Problem is that $\mu$ and $\nu$ not being mutually singular doesn't yield the existence of an interesting set, rather it says that no interesting set exists (namely, no $A\subset X$ such that $\mu(A)=0,\nu(X-A)=0$), so I don't think I could argue here with indicator functions or something.

Attempt 2:

Since $\pi_\mu\oplus\pi_\nu$ is cyclic with cyclic vector $(e,f)$, it corresponds to the positive linear functional $f$ on $C(X)$ defined by $$ f:\phi\mapsto \langle \phi f,f\rangle+\langle \phi g,g\rangle=\int \phi f\overline f d\mu+\int \phi g\overline g d\nu. $$ By Riesz representation this functional corresponds to a positive Radon measure $\tau$: $$ f(\phi)=\int \phi d\tau. $$ If $\mu$ and $\nu$ are Radon (which I think is a fair assumption), then I believe this implies that $f\overline f \mu+g\overline g\nu=\tau$. Does this yield something interesting?

Attempt 3:

I know that any cyclic representation on $C(X)$ is of the form $\pi_\tau$ for some Radon measure $\tau$. Thus if we write $\pi_\mu\oplus\pi_\nu\cong \pi_\tau$, and we argue by contradiction, I would hope to show that the isomorphism $\Phi$ from $\pi_\tau$ to $\pi_\mu\oplus \pi_\nu$ is the same as for the case where $\mu\perp\nu$, namely $$ f\mapsto (f,f) $$ which I can then show is not surjective. If $\tau$ is a finite measure, then the constant functions are $\tau$-integrable, and since $\Phi$ is a morphism between representations, we would get $$ \Phi(f)=\Phi(f\cdot 1)=f\cdot\Phi(1) $$ for any continuous $f$. Since $\Phi=(\Phi_1,\Phi_2)$ is surjective, this shows that $\Phi_1(1)\neq 0 $ a.e. [$\mu$] and $\Phi_2(1)\neq 0$ a.e. [$\nu$], hence $\Phi_i(1)$ are cyclic vectors. Since $\tau$ is a Radon measure, $C(X)$ lies dense in $L^ 2(\tau)$ by Lusin's Theorem. This could potentially be useful?

Answer 1

I will write $H_{\tau} := L^2(X,\tau)$ and denote the representation of $C(X)$ on $L^2(X,\tau)$ by multiplication operators by $\pi_{\tau}$ whenever $\tau$ is a Borel measure on $X$. I assume you know that each $\pi_{\tau}$ is a cyclic representation of $C(X)$ and that any cyclic representation of $C(X)$ is unitarily equivalent to some $\pi_{\tau}$ (the form of your question suggests this). To focus on the core ideas I sketch details only in "footnotes." The conceptual key is that if $\mu$ and $\nu$ are not mutually singular, there is a nonzero measure $\sigma$ with $\sigma \ll \mu$ and $\sigma \ll \nu$.¹

One approach is by analyzing commutants. In general, $\tau$ is any measure and $\tau' \ll \tau$, the $L^2$ space $H_{\tau}$ contains a subspace isomorphic to $H_{\tau'}$,² such that an isometry implementing this isomorphism also implements a unitary equivalence between $\pi_{\tau'}$ and a subrepresentation of $\pi_{\tau}$.³ So if $\mu$ and $\nu$ are not mutually singular, there is thus a nonzero $\sigma$ such that $\pi := \pi_{\mu} \oplus \pi_{\nu}$ contains a subrepresentation unitarily equivalent to $\pi_{\sigma} \oplus \pi_{\sigma}$,⁴ and it suffices to show that $\pi_{\sigma} \oplus \pi_{\sigma}$ is not cyclic.⁵ But the commutant of any cyclic representation of $C(X)$ is abelian⁶ while the commutant of $\pi(C(X))$ in $\mathcal{B}(H_{\sigma} \oplus H_{\sigma})$ clearly is not.⁷ End of proof.

Some language from representation theory is helpful here. A representation $\tau$ of a $C^*$-algebra is said to be "multiplicity-free" if it contains no subrepresentation unitarily equivalent to a representation of the form $\tau' \oplus \tau'$, where $\tau'$ is a nonzero representation. While the representations $\pi_{\tau}$ of $C(X)$ are all multiplicity-free⁸, the argument above shows that when $\mu$ and $\nu$ are not mutually singular, the $\pi_{\mu} \oplus \pi_{\nu}$ is not multiplicity-free. (I am not sure if this perspective or language appears in Conway, but see e.g. Arveson's Invitation to $C^*$-algebras for more from this perspective.)

As an alternative to analyzing commutants, one can also simply exhibit nonzero vectors in the orthocomplements of the spaces $\pi(C(X)) \xi$ when the measures $\mu$ and $\nu$ are not mutually singular. As before, we will use the existence of a nonzero measure $\sigma$ with $\sigma \ll \mu$ and $\sigma \ll \nu$.

For motivation, consider the example of $X$ finite and discrete, with $\mu$ and $\nu$ counting measures on subsets $E$ and $F$ of $X$ respectively⁹; such representations $\mu$ and $\nu$ are plainly mutually singular iff $E$ and $F$ are disjoint. By indexing the elements of $E$ and $F$, one can identify $H_{\mu}$ with $\mathbb{C}^{|E|}$ and $H_{\nu}$ with $\mathbb{C}^{|F|}$ and the action of $\pi$ on $H := \mathbb{C}^{|E|+|F|}$ is by diagonal matrices, where the diagonal entries of $f \in C(X)$ list the values of $f$ first on $E$ and then on $F$. If $E$ and $F$ are disjoint (i.e., if $\mu$ and $\nu$ are mutually singular), the image of $\pi$ is easily seen to be the set of all diagonal matrices, and as you already know, this representation is cyclic.¹⁰ If $E$ and $F$ are not disjoint (i.e., if $\mu$ and $\nu$ fail to be mutually singular), equally clear is that the the image of $\pi$ is not the set of all diagonal matrices, but a subspace of diagonal matrices having some fixed pair of entries equal to one another.¹¹ Fixing indices $1 \leq m \leq |E| < n \leq |E| + |F|$ with the property that the $m$th and $n$th diagonal entries of each matrix $\pi(f)$ are equal, let $\xi \in H$ be given. If $\xi_m$ (resp. $\xi_n$ ) happens to be zero, then the $m$th standard basis vector $e_m$ (resp. the $n$th standard basis vector $e_n$) is nonzero and in $[\pi(C(X))\xi]^{\perp}$,¹² while if both $\xi_m$ and $\xi_n$ are nonzero, then the nonzero vector $\bar{\xi_m}^{-1} e_m - \bar{\xi_n}^{-1} e_n$ is in $[\pi(C(X))\xi]^{\perp}$.¹³ So we can see very concretely how $\pi$ fails to be cyclic when $E$ and $F$ are not mutually singular in this case.

With some care, the approach just outlined extends to the general case. Fix a nonzero $\sigma$ with $\sigma \ll \mu$ and $\sigma \ll \nu$ and let $h_{\sigma,\mu}$ and $h_{\sigma,\nu}$ respectively denote the Radon-Nikodym derivatives of $\sigma$ with respect to $\mu$ and $\nu$, and let $\xi = (\xi_1, \xi_2) \in H := H_{\mu} \oplus H_{\nu}$ be given. If $\xi_1$ (resp. $\xi_2$) vanishes on a set of positive $\sigma$-measure, $\xi_1$ (resp. $\xi_2$) then also vanishes on a set of positive measure $P$ for which $1_P h_{\sigma,\mu}$ is in $L^2(\mu)$, and the vector $(1_P h_{\sigma,\mu},0)$ (resp. $(0,1_P h_{\sigma,\nu})$) is then seen to be a nonzero vector in $[\pi(C(X)) \xi]^{\perp}$.¹⁴ Similarly, if both $\xi_1$ and $\xi_2$ are nonzero almost everywhere, there is $c > 0$ for which the set $\{x \in X: \text{$|\xi_1| \geq c$ and $|\xi_2| \geq c$}\}$ has positive $\sigma$-measure, and a subset $P$ of this set, also of positive $\sigma$-measure, for which $(1_P \bar{\xi_1}^{-1} h_{\sigma,\mu}, -1_P \bar{\xi_2}^{-1} h_{\sigma,\nu})$ is in $H$, where it is seen to be a nonzero vector¹⁵ in $[\pi(C(X)) \xi]^{\perp}$.¹⁶

1. This follows, for example, from a "Lebesgue decomposition"-type theorem, where I use quotes and the indefinite article because textbooks formulate these slightly differently and sometimes use different names for them. Examples of such are this MO post (the "Background" statement), this MSE post (discussing a proof of one aspect of this type of theorem and referencing two other MSE posts about Lebesgue decompositions).
2. If $h$ denotes the Radon-Nikodym derivative of $\tau'$ with respect to $\tau$, the relevant isometry $H_{\tau'} \to H_{\tau}$ is given by the recipe $\xi \mapsto \sqrt{h} \xi$ (short exercise to check that this is well defined).
3. The most substantive part of this might be the fact that the range of the isometry in (2) is a reducing subspace for the operators in $\pi_{\tau}(C(X))$. One way of seeing this is to note that if $A$ is any Borel subset of $X$, the range of "multiplication by the indicator function of $A$" on $H_{\tau}$ is clearly reducing for $\pi_{\tau}(C(X))$, and to note/verify that the range of the isometry from (2) has this form, where $A$ can be taken to be the subset $\{x \in X: h(x) > 0\}$ (where $h$ is the Radon-Nikodym derivative from (2); $h$ and $A$ are defined only up to a set of $\tau$-measure zero, but because of the identifications made in the space $H_{\tau}$, any choice will do).
4. Follows from (1) and two uses of (3).
5. It is immediate from the definition of "cyclic" that if $\pi'$ on $K \subseteq H$ is a subrepresentation of $\pi$ on $H$, and $\pi$ is cyclic, then $\pi'$ is also cyclic (the orthogonal projection onto $K$ of any cyclic vector for $\pi$ is a cyclic vector for $\pi'$).
6. A cyclic representation of $C(X)$ has the form $\pi_{\tau}$ for some measure $\tau$, and the commutant of $\pi_{\tau}(C(X))$ on $H_{\tau}$ is $L^{\infty}(X,\tau)$. The closest I can find on MSE is this post (identifying the commutant of $L^{\infty}(\mathbb{R})$ in $\mathcal{B}(L^2(\mathbb{R})$). A similar argument shows that any operator on $L^2(X,\tau)$ that commutes with all multiplication operators in $C(X)$ is in $L^{\infty}(X,\tau)$. It is also possible to see that the commutant of $\pi_{\tau}(C(X))$ must be abelian in a way that does not directly identify what it is. See this MSE post (showing that a unital abelian $C^*$-subalgebra of $\mathcal{B}(H)$ with a cyclic vector necessarily has an abelian commutant).
7. For example, a short calculations show that the operator on $H_{\sigma} \oplus H_{\sigma}$ given by $(\xi_1, \xi_2) \mapsto (\xi_2,0)$ and its adjoint $(\xi_1,\xi_2) \mapsto (0, \xi_1)$, which do not commute, are in the commutant of $\pi_{\sigma} \oplus \pi_{\sigma}$. (More generally, thinking of elements of $H_{\sigma}$ as $2 \times 1$ "column vectors," one can locate a copy of $M_2(\mathbb{C})$ in the commutant of $\pi_{\sigma} \oplus \pi_{\sigma}$.)
8. Conway hopefully proves this, or at least gives some idea of why it is true. Being multiplicity-free is equivalent to having an image whose commutant is abelian, so this statement amounts to an identification of $\pi_{\tau}(C(X))'$ in $H_{\tau}$ (see point (6)). A book might naturally do this near a discussion of the GNS representation, or in a discussion of abelian von Neumann algebras.
9. There is actually no loss of generality (within the world of finite and discrete $X$) in assuming that $\mu$ and $\nu$ are counting measures, because any measure on a finite and discrete $X$ is mutually absolutely continuous with respect to a unique counting measure. [In general, when one replaces a measure $\tau$ with another that is mutually absolutely continuous with respect to it, the unitary equivalence class of $\pi_{\tau}$ does not change. The details of this are similar to points (1), (2) above.]
10. Any vector in $H$ with all entries nonzero will be cyclic for $\pi$ in this case, as also holds in general. See generally e.g. this MSE post (identifying all cyclic vectors for $L^{\infty}(\mu)$ acting on $L^2(\mu)$).
11. If $E$ and $F$ are not disjoint, when listing the values of any $f \in C(X)$ first on $E$ and then on $F$, a value will repeat in that list for each element of $E$ that is also an element of $F$, and these values will repeat in the same places for every $f \in C(X)$.
12. The desired orthogonality comes from the fact that the pointwise product of any $f\xi$ with the conjugate of the given vector is zero.
13. The desired orthogonality comes from the fact that the pointwise product of any $f\xi$ with the conjugate of the given vector has $m$th entry $f(p)$, $n$th entry $-f(p)$, and all other entries $0$ (here $p$ is the point of $E \cap F$ corresponding to the indices $m$ and $n$).
14. Starting with any set $P$ of positive $\sigma$-measure on which $|\xi_j| = 0$, we can additionally assume that $1_P h_{\sigma,\mu}$ is in $H_{\mu}$ (when a priori it is only, like $h_{\sigma,\mu}$ itself, in $L^1(X,\mu)$) by replacing $P$ with $\{p \in P: h_{\sigma,\mu} \leq n\}$ for $n$ sufficiently large (any such has the property that the vector constructed from it will be in $L^2(\mu)$, and these sets have positive $\sigma$-measure for $n$ sufficiently large). The element of $H$ thus constructed is nonzero because its norm in $H$ is $\sigma(P)$. The desired orthogonality follows from the same facts as in the discrete case (12).
15. The point of "$c$" is to ensure that the functions $\xi_1^{-1}$ and $\xi_2^{-1}$ are bounded on $P$. The subset $P$ can be chosen of the form $\{x \in X: |\xi_1| \geq c, |\xi_2| \geq c, h_{\sigma,\mu} \leq n, h_{\sigma,\nu} \leq n\}$ (any such set has the property that the vector constructed from $P$ is in both $L^2(\mu)$ and $L^2(\nu)$, and $P$ will have positive $\sigma$-measure for $n$ sufficiently large). The vector thus constructed is nonzero in $H_{\mu}$ e.g. because the inner product of its $H_{\mu}$ component with $\xi_1$ is $\sigma(P) \neq 0$.
16. The desired orthogonality comes from the fact that the pointwise product of any $f \xi_1$ with the conjugate of the $H_{\mu}$ component of the given vector is $f 1_P h_{\sigma,\mu}$, while the pointwise product of $\xi_2$ with the conjugate of the $H_{\nu}$ component of the given vector is $f 1_P h_{\sigma,\nu}$. It follows the inner product of any $\pi(f)(\xi)$ with the constructed vector is $\int_P f h_{\sigma,\mu} \, d\mu - \int_P f h_{\sigma,\nu} \, d\nu = \int_P f \, d\sigma - \int_P f \, d\sigma = 0$. This is the same idea as the discrete case (13), with the two copies of $L^2(X,\sigma)$ inside $L^2(X,\mu)$ and $L^2(X,\nu)$ playing the roles of the indices $m$ and $n$ in the finite case. Very roughly speaking, they are distinct "locations" within $H$ where we can leverage the fact that the corresponding "diagonal entries" of the operators $\pi(C(X))$ are constrained to be equal to one another.