It is known that $\ell^2(\mathbb{Z})$ is $\ell^1(\mathbb{Z})$-module (the module operation is the convolution).
What about the irreducible submodules? Can we characterize them?
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3$\begingroup$ When you say submodule, do you mean "norm-closed submodule"? And when you say "irreducible", do you mean "contains no non-trivial closed submodule" or "contains non non-trivial submodule in the purely algebraic sense"? $\endgroup$– Yemon ChoiCommented Dec 21, 2020 at 19:37
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2$\begingroup$ In either case, I think the right approach would be to take Fourier transforms, so that one is considering $A({\bf T})$ acting by multiplication on $L^2({\bf T})$. My instinct is that every non-zero, norm-closed, $A({\bf T})$-submodule of $L^2({\bf T})$ has the form $\chi_E \cdot L^2({\bf T})$ for some positive-measure subset $E\subseteq {\bf T})$ and hence can never be irreducible, either in the algebraic or topological sense. $\endgroup$– Yemon ChoiCommented Dec 21, 2020 at 19:40
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1$\begingroup$ I think the set of zeroes of a module $A \subset \ell^1(\mathbf{Z})$ has to be closed as a subset of $\mathbf{T}$. When $G$ is a locally compact Abelian group the ideals of $A(G)$ contain all the annihilators of closed sets but the reciprocal is false (there are examples in $A(\mathbf{R}^n)$). So in principle the question makes sense since there could be a ideals of $A(\mathbb{T})$ that is strictly smaller than the annihilator of its null set $S$ and such that no function in $A(\mathbb{T})$ with support strictly smaller than $\mathbb{T} \setminus S$ is in it (although that seems unlikely). $\endgroup$– Adrián González PérezCommented Dec 22, 2020 at 12:31
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1 Answer
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Edited because the original answer solved the problem for $\ell^1(\mathbf{Z})$ submodules not for $\ell^2(\mathbf{Z})$ ones.
An $\ell^1(\mathbf{Z})$-submodule of $\ell^2(\mathbf{Z})$ is just an invariant subspace under the left regular representation $\lambda$. If $P:\ell^2(\mathbf{Z}) \to \ell^2(\mathbf{Z})$ is the projection onto a $\lambda$-invariant subspace, then it lies in the commutant of $\lambda[\mathbf{Z}]$, which is just $L^\infty(\mathbf{T})$ after conjugating with the Fourier transform and applying the Plancherel theorem. This gives that, after taking the Fourier transform, any closed submodule is of the form $\mathbf{1}_E \cdot L^2(\mathbf{T})$, for a measurable set $E$. The reciprocal is also true. Every measurable set (up to measure zero differences) gives a submodule. Thus, the only simple module is $\{0\}$.
answered Dec 22, 2020 at 15:01
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1$\begingroup$ Maybe I missed something... but wasn't the original question about sub modules of $\ell^2(\mathbb Z)$, and not ideals in $\ell^1(\mathbb Z)$? $\endgroup$ Commented Dec 22, 2020 at 15:58
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1$\begingroup$ 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). $\endgroup$ Commented Dec 22, 2020 at 16:11
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1$\begingroup$ Ha! I just fixed the title... $\endgroup$ Commented Dec 22, 2020 at 16:14
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1$\begingroup$ 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. $\endgroup$ Commented Dec 22, 2020 at 16:34
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1$\begingroup$ 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? $\endgroup$ Commented Dec 22, 2020 at 17:03