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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Sign up to join this communityIt 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?
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 $\{0\}$.