Timeline for Shift invariant subspaces of $l^1$
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
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| Jun 30, 2017 at 21:12 | comment | added | Pietro Majer | @Yemon Choi: The idea was that the function $f_E:=\mathrm{dist}(\cdot,E)$ is lipschitz, hence it is in $A(\mathbb{T})$; and the hull of the ideal generated by $f_E$ is $E$ | |
| Jun 30, 2017 at 20:33 | comment | added | Nik Weaver | (Actually, my original problem was on the 2-torus and dealt with an analogous question relative to quantum tori. Then I proved that the answer is independent of the rotation parameter, and hence we can set is to zero and reduce to the commutative case ...) | |
| Jun 30, 2017 at 20:32 | comment | added | Nik Weaver | @YemonChoi: funny you should say that, because I was led to this question by a problem in operator algebras. Let $L^\infty(\mathbb{T})$ act by multiplication on $L^2(\mathbb{T})$ and let $\mathcal{V} \subseteq B(L^2(\mathbb{T}))$ be a weak* closed bimodule over $L^\infty(\mathbb{T})$ which is invariant for the action of $\mathbb{T}$ on itself. Can we characterize $\mathcal{V}$? This reduces to the problem I asked. | |
| Jun 30, 2017 at 16:42 | vote | accept | Nik Weaver | ||
| Jun 30, 2017 at 16:32 | comment | added | Yemon Choi | @PietroMajer It is true that for every proper closed subset $E \subset {\bf T}$ there is a closed ideal of ${\rm A}({\bf T})$ whose hull is $E$, but I'm not sure that this is as immediate as your comment seems to suggest. I think this requires something like the fact that ${\rm A}({\bf T})$ is a regular Banach algebra of functions on ${\bf T}$... but perhaps I am overlooking something simpler | |
| Jun 30, 2017 at 16:21 | comment | added | Yemon Choi | Nik: as Matt's answer indicates below, the ideals of A(T) are not well understood in full generality, which is why "Banach algebras bad, operator algebras trendy" or some such. Pietro: the closed ideals of C^1(T) are well understood but they are NOT merely the ones given by the zero-set construction; in the language of Matt's answer, points of T are not sets of synthesis for C^1(T). It's a nice exercise to work out exactly which ideals have "hull" equal to a given singleton... | |
| Jun 30, 2017 at 15:40 | answer | added | Matthew Daws | timeline score: 5 | |
| Jun 30, 2017 at 15:04 | comment | added | Pietro Majer | If $I$ is a closed ideal of $A(\mathbb{T})$, its uniform closure is a closed ideal of $C^0(\mathbb{T})$, and this correspondence is certainly surjective. If $J$ is a closed ideal of $C^0(\mathbb{T})$, its trace on $A(\mathbb{T})$ is a closed ideal of $A(\mathbb{T})$. Saying that these maps are inverse of each other means that each closed ideal of $A(\mathbb{T})$ is also closed in $A(\mathbb{T})$ wrto the uniform topology... is it true? | |
| Jun 30, 2017 at 13:49 | comment | added | Pietro Majer | Given that $C^1(\mathbb{T})\subset A(\mathbb{T})\subset C^0(\mathbb{T})$, I would say closed ideals of $A(\mathbb{T})$ are in bijection with closed ideals of $C^0(\mathbb{T})$, and in bijection with closed subsets of $\mathbb{T}$ as zero-sets (this is true for maximal ideals, by Gelfand theorem). | |
| Jun 30, 2017 at 13:23 | comment | added | Nik Weaver | These are both great insights ... surely the ideals of $A(\mathbb{T})$ must be well understood? If they all have the desired form, I think these two comments solve the problem completely. | |
| Jun 30, 2017 at 10:14 | comment | added | Pietro Majer | Shift-invariant closed subspaces of $\ell^1$ are ideals wrto convolution, so the question equivalently asks for a characterization of the closed ideals of the Wiener algebra $A(\mathbb{T})$, doesn't it? | |
| Jun 30, 2017 at 8:16 | comment | added | YCor | To be picky, to have a complete description in the $\ell^2$ case, one has to say for which pairs $S_1,S_2$ of measurable subsets of the circle we have $F_{S_1}=F_{S_2}$, where $F_S$ is the set of elements in $L^2$ of the circle that are zero outside $S$; I guess it holds iff $S\triangle T$ has measure zero. In the $\ell^1$ case, Fourier transforms are continuous and hence the discussion will be different: letting $G_S$ be the set of $\ell^1$ elements whose Fourier transform is zero outside $S$, I guess $G_{S_1}=G_{S_2}$ iff $S_1$ and $S_2$ have equal interiors. | |
| Jun 30, 2017 at 3:15 | history | asked | Nik Weaver | CC BY-SA 3.0 |