Timeline for Irreducible sub-modules of $\ell^2(\mathbb{Z})$
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 23, 2020 at 16:10 | vote | accept | ABB | ||
| Dec 23, 2020 at 12:18 | history | edited | Adrián González Pérez | CC BY-SA 4.0 |
deleted 269 characters in body
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| Dec 22, 2020 at 19:27 | comment | added | Matthew Daws | I second that Adrian: write your comment up as an answer... | |
| Dec 22, 2020 at 17:03 | comment | added | Yemon Choi | Adrian: yes, I mentally corrected the title and was answering that question. Perhaps you would like to replace this answer with an expanded version of your comment? | |
| Dec 22, 2020 at 16:34 | comment | added | Adrián González Pérez | In that case the answer is even easier (I think). An $\ell^1(\mathbf{Z})$-submodule is invariant under the left regular representation by taking Dirac deltas. The projection onto an invariant subset belongs to the commutant von Neumann algebra, which in this case is just $L^\infty(\mathbf{T})$ itself. Any such projection is given by a measurable set up to measure equivalence. Reciprocally every measurable set gives an invariant subspace. So, Yemon's idea should work fine. | |
| Dec 22, 2020 at 16:14 | comment | added | Matthew Daws | Ha! I just fixed the title... | |
| Dec 22, 2020 at 16:11 | comment | added | Adrián González Pérez | My bad. I just read $\ell^1$ instead of $\ell^2$ (It didn't help that there is a typo in the title and it said $\ell^1$ there). | |
| Dec 22, 2020 at 15:58 | comment | added | Matthew Daws | Maybe I missed something... but wasn't the original question about sub modules of $\ell^2(\mathbb Z)$, and not ideals in $\ell^1(\mathbb Z)$? | |
| Dec 22, 2020 at 15:01 | history | answered | Adrián González Pérez | CC BY-SA 4.0 |