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Let $U$ be a unitary operator on a Hilbert space $\mathcal{H}$. We can assume $\mathcal{H}$ is an $L^2$ space and $U$ acts as multiplication by a function $u$ with $|u(x)| = 1$ a.e (by the spectral theorem). A subspace $W \subset \mathcal{H}$ is reducing for $U$ if and only if the projection $P_W$ commutes with $U$. Given that these projections are self-adjoint, it's tempting to wonder whether one can use the spectral theorem for commuting normal operators to classify these projections, and hence the reducing subspaces of $U$.

Can this be done even with the (bilateral) shift operator $S$ on $L^2(T)$? Usually one classifies the reducing subspaces for $S$ by first establishing that the only operators commuting with $S$ are the multiplication operators, and of those the only idempotents are multiplications by characteristic functions. I'm wondering if there's a more general argument behind this, using spectral theory, which generalizes to other unitary operators.

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  • $\begingroup$ There's a degeneracy issue you need to watch out for. For instance, if $U$ is the identity then every projection commutes with it, not just those which are multiplication operators. So you need to worry about multiplicity, but that is the only subtlety here. $\endgroup$
    – Nik Weaver
    Commented Oct 7, 2023 at 2:30
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    $\begingroup$ That's cracking a nut with a sledgehammer. Just observe that the reducing subspaces are the ones that are invariant under $U$ and $U^*$ (or better yet, define them this way). If $U$ is multiplication by $z$ in $L^2(T,\mu)$, then, since $p(z,\overline{z})$ is dense in $C(T)$, these subspaces are exactly the spaces $L^2(A)$, $A\subseteq T$. In general, there is also the issue mentioned by Nik. $\endgroup$
    – Christian Remling
    Commented Oct 7, 2023 at 4:19
  • $\begingroup$ @ChristianRemling: I know that using spectral theory is overkill for the specific example of the shift. But, this is not the only case I care about. In reality, I have two noncommuting unitary operators and need to classify those subspaces which are reducing for both. I don't know much spectral theory, so I asked this simpler question to gauge whether this is the appropriate tool to use here. As for the other comment, I don't expect the only operators to commute with U to be multiplication operators. In fact, I know this isn't the case for the operators I have in mind. $\endgroup$
    – bm3253
    Commented Oct 7, 2023 at 12:36
  • $\begingroup$ I don't think one can say much about two operators without further assumptions. There could be no common reducing subspaces, already for $2\times 2$ matrices, or there could be many. $\endgroup$
    – Christian Remling
    Commented Oct 7, 2023 at 14:29

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