Existence of a translation-invariant basis of $\ell^2$

Question

This question is heavily inspired by this other one, but is meant to be a hopefully more accessible variant of it (and I think slightly more natural).

I give four equivalent formulations of the same question in the hope of inspiring different ways to view it:

1. Let $\ell^2(\mathbb{Z};\mathbb{C})$ be the $\mathbb{C}$-vector space of square-summable $\mathbb{Z}$-indexed complex-valued sequences. Does there exist a basis $B$ of $\ell^2(\mathbb{Z};\mathbb{C})$ as a $\mathbb{C}$-vector space (not as a Hilbert space: we completely ignore topology here) which is translation-invariant in the sense that if $e \in B$ then any shift $T^r(e) : k \mapsto e(k-r)$ of $e$ is also in $B$ for any $r\in\mathbb{Z}$?

2. Is $\ell^2(\mathbb{Z};\mathbb{C})$ a free $\mathbb{C}[T^{\pm 1}]$-module, where $\mathbb{C}[T^{\pm 1}]$ is the ring of Laurent polynomials with complex coefficients, and $T$ acts on $\ell^2(\mathbb{Z};\mathbb{C})$ by shift ($T^r(e) : k \mapsto e(k-r)$)?

3. Let $L^2(\mathbb{R}/\mathbb{Z};\mathbb{C})$ be the $\mathbb{C}$-vector space of square-integrable complex-valued functions on $\mathbb{R}/\mathbb{Z}$ modulo equality a.e. Does there exist a basis $B$ of $L^2(\mathbb{R}/\mathbb{Z};\mathbb{C})$ as a $\mathbb{C}$-vector space (ignoring topology) which is stable under multiplication by $\exp(\pm 2i\pi\theta)$ (in the sense that if $f\in B$ then $\theta \mapsto \exp(2i r \pi\theta) \thinspace f(\theta)$ also belongs to $B$ for any $r\in\mathbb{Z}$)?

4. Is $L^2(\mathbb{R}/\mathbb{Z};\mathbb{C})$ a free $\mathbb{C}[T^{\pm 1}]$-module, where now $\mathbb{C}[T^{\pm 1}]$ is seen as the ring of trigonometric polynomials ($\mathbb{C}$-linear combinations of $\theta \mapsto \exp(2i k\pi\theta)$)?

(The equivalence of (1) and (2) is proved analogously to YCor's answer in the aforementioned question: the fact that $\ell^2(\mathbb{Z};\mathbb{C})$ is a torsion-free $\mathbb{C}[T^{\pm 1}]$-module follows from the fact that no linear recurrence sequence is square-integrable. The equivalence of (1) and (3) or (2) and (4) is by Fourier transform.)

A few remarks:

• $\ell^2(\mathbb{Z};\mathbb{C})$ (or, of course, $L^2(\mathbb{R}/\mathbb{Z};\mathbb{C})$, since they are the same) is a flat $\mathbb{C}[T^{\pm 1}]$-module since it is torsion-free and $\mathbb{C}[T^{\pm 1}]$ is a PID (hence a Dedekind domain). Moreover, because of a theorem of Bass (“Big Projective Modules are Free”, Ill. J. Math. 7 (1963), 24–31: theorem 4.3), to see that it is free it is enough to see that it is projective. (I don't think any of this helps, but it is still worth pointing out.)

• As I note in a partial answer to the other question, $\mathbb{C}[T^{\pm 1}]^I$ is not free for any infinite set $I$. I suspect it can probably be deduced from this that $\mathbb{C}^{\mathbb{Z}}$ (viꝫ., all complex-valued sequences) does not admit a translation-invariant vector-space basis (the module reformulation of this would be that it is a direct sum of cyclic modules: obviously, here, it is not free since it is not even torsion-free). At the other extreme, $\mathbb{C}^{(\mathbb{Z})}$ (viꝫ., complex-valued sequences with finite support) is trivially a free $\mathbb{C}[T^{\pm 1}]$-module (with rank $1$). So $\ell^2(\mathbb{Z};\mathbb{C})$ strikes me as something natural to ask “in between”. But if someone is capable of answering question (1) for other natural intermediate spaces between $\mathbb{C}^{(\mathbb{Z})}$ and $\mathbb{C}^{\mathbb{Z}}$, like $\ell^1(\mathbb{Z};\mathbb{C})$ or $c_0(\mathbb{Z};\mathbb{C})$, this also counts as an answer to this question!

• An vaguely analogous situation that may be noted is that $\mathbb{Z}^{\mathbb{N}}$ is not a free $\mathbb{Z}$-module (i.e., abelian group), $\mathbb{Z}^{(\mathbb{N})}$ trivially is one, and in between them, the set of bounded $\mathbb{Z}$-valued sequences is free (a theorem of Specker).

Answer 1

Inspired by Harry West's answer to the question about continuous functions with compact support, I can now answer my own version thusly: $\ell^2(\mathbb{Z};\mathbb{C})$ has no translation-stable $\mathbb{C}$-vector space basis.

To show this, I will use form (4) of the question and prove that $L^2(\mathbb{R}/\mathbb{Z};\mathbb{C})$ is not free as a module over the ring $\mathbb{C}[T^{\pm1}]$ of trigonometric polynomials (where $T$ is now $\theta \mapsto \exp(2i\pi\theta)$).

Indeed, let $f \in \mathbb{C}[T^{\pm1}]$ be a non-constant trigonometric polynomial which does not vanish on $\mathbb{R}/\mathbb{Z}$, e.g., $T + 3 + T^{-1}$ (that is, $3 + 2\cos(2\pi\theta)$). Now, crucially, $\frac{1}{f}$ is a well-defined continuous function $\mathbb{R}/\mathbb{Z} \to \mathbb{C}$ (in particular, it is bounded). So, for any $g \in L^2(\mathbb{R}/\mathbb{Z};\mathbb{C})$, we have $\frac{g}{f} \in L^2(\mathbb{R}/\mathbb{Z};\mathbb{C})$; in particular, $\frac{1}{f^n} \in L^2(\mathbb{R}/\mathbb{Z};\mathbb{C})$ for any $n\in\mathbb{N}$. Now if $L^2(\mathbb{R}/\mathbb{Z};\mathbb{C})$ had a $\mathbb{C}[T^{\pm1}]$-basis, expressing $\frac{1}{f^n}$ on this basis and multiplying by $f^n$, we see that the coefficients of $1$ on the basis must be multiple of $f^n$ for all $n$.

But zero is the only element of $\mathbb{C}[T^{\pm1}]$ which is divisible by $f^n$ for all $n$, because $\mathbb{C}[T^{\pm1}]$ is a principal ideal domain (as a localization of $\mathbb{C}[T]$ which is one), and $f$ is not invertible (the invertible elements of $\mathbb{C}[T^{\pm1}]$ are the $c T^r$ for $r\in\mathbb{Z}$, and $f$ is not of this form). This is a contradiction.