**Background:** If $\mathcal{H}$ is a Hilbert space and $U:\mathcal{H}\rightarrow\mathcal{H}$ is a unitary operator, then for each $f \in \mathcal{H}$ the sequence $(\langle U^nf,f\rangle)\_{n = 1}^\infty$ is positive definite, so it can be viewed as the Fourier coefficients of some measure $\mu\_f$ on the torus $\mathbb{T}$. If $f,g \in \mathcal{H}$ are such that $\mu\_f\perp\mu\_g$, i.e., $\mu\_f$ and $\mu\_g$ are mutually singular, then $\langle f,g\rangle = 0$. The most prominent example of this is when $\mu\_f$ is a discrete measure and $\mu\_g$ is a continuous measure, as this corresponds to the compact-weak mixing decomposition of $\mathcal{H}$.


My goal is to find a proof or reference of the generalization of this fact to the setting of second countable locally compact type I groups $G$. In particular, we recall that for such groups $G$, the dual space $\widehat{G}$ of unitary equivalence classes of irreducible representations of $G$ has the structure of a standard measure space with respect to the Mackey-Borel structure. Now if $\pi$ is a representation of $G$ on some separable Hilbert space $\mathcal{H}$, then there exists a measure $\mu$ on $\widehat{G}$ so that we can write $\pi = \int^{\oplus}\_{\widehat{G}}\pi\_pd\mu(p)$. This is essentially Theorem 7.40 in 'A course in abstract harmonic analysis' by G. Folland, or Theorem 8.6.6 in '$C^\*$-Algebras' by J. Dixmier. We say that a set $A \subseteq \widehat{G}$ *carries* $\pi$ if $\mu(\widehat{G}\setminus A) = 0$ (Definition 8.6.9 from Dixmier).


In conclusion, I would like a proof or a reference for the following conjecture.


**Conjecture:** Let $\pi$ be a representation of $G$ on $\mathcal{H}$, and let $A$ and $B$ be closed subspaces of $\mathcal{H}$ that are invariant under the action of $\pi$. If $\pi|\_A$ and $\pi|\_B$ are carried by disjoint sets, then $A\perp B$.


P.S. This feels like it should be an easy consequence of the uniqueness portion of Theorem 7.40 of Folland's book, but I have been having difficulty making the details work.



Spectral disjointness of unitary representations of Type I groups and orthogonality
A friend provided an answer to my question in a private communication. I have added some details and will present it below in my own words.


**Additional notation:** Below I use $\mu\_{\pi}$ to refer to the measure $\mu$ from my initial post with respect to a particular representation $\pi$. I also use $[\mu]$ to denote the equivalence class of measures that are mutually absolutely continuous with $\mu$.


Let us assume for the sake of contradiction that $A\not\perp B$ and let $P:A\rightarrow B$ denote the unique orthogonal projection. Since $A$ and $B$ are $\pi$-invariant, we see that for any $g \in G$, the map $P\_g := \pi(g)P\pi(g)^{-1}:A\rightarrow B$ is also an orthogonal projection, so $P$ and $\pi$ commute. Letting $A = \text{ker}(P)\oplus N\_1$ and $B = \text{Im}(P)\oplus N\_2$, we see that $P:N\_1\rightarrow N\_2$ is a bijection between two $\pi$-invariant subspaces. Recalling that $P$ and $\pi$ commute, one can easily verify that $P':N\_1\rightarrow N\_2$ given by $P'f = \frac{||f||}{||Pf||}Pf$ is a unitary operator that commutes with $\pi$, so $\pi\_1 := \pi|\_{N\_1}$ and $\pi\_2 := \pi|\_{N\_2}$ are unitarily equivalent representations of $G$, so $[\mu\_{\pi\_1}] = [\mu\_{\pi\_2}]$. Letting $\pi\_k := \pi|\_{\text{ker}(P)}$ and $\pi\_I := \pi|\_{\text{Im}(P)}$, we see that $[\mu\_{f\_1}] = [\mu\_{\pi\_k}]+[\mu\_{\pi\_1}]$ and $[\mu\_{f\_2}] = [\mu\_{\pi\_I}]+[\mu\_{\pi\_2}]$, which yields the desired contradiction.


