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Oct 7, 2023 at 14:29 comment added Christian Remling 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.
Oct 7, 2023 at 12:36 comment added bm3253 @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.
Oct 7, 2023 at 7:13 history edited YCor CC BY-SA 4.0
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Oct 7, 2023 at 4:42 review Close votes
Oct 13, 2023 at 4:15
Oct 7, 2023 at 4:19 comment added Christian Remling 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.
Oct 7, 2023 at 2:30 comment added Nik Weaver 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.
S Oct 7, 2023 at 1:38 review First questions
Oct 7, 2023 at 5:41
S Oct 7, 2023 at 1:38 history asked bm3253 CC BY-SA 4.0