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Let $\mathbb{T}$ be the unit circle and consider the convolution group algebra $L^1(\mathbb{T})$. Let $I_n$ be the closed ideal generated by the polynomial $p_n(z)=z^n-1$ in $L^1(\mathbb{T})$. Let $I=\bigcap_{n\geq1}I_n$.

Q. What is the character space of the quotient $\frac{L^1(\mathbb{T})}{I}$?

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  • $\begingroup$ I think $I$ is just the constant functions. The dual of $L^1$ is $\mathbb Z$ and the dual of $L^1({\mathbb T})/I$ is ${\mathbb Z}\smallsetminus \{0\}$. The projection $L^1\to L^1/I$ dualises to the inclusion. $\endgroup$
    – user473423
    Commented Apr 5, 2022 at 12:57
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    $\begingroup$ @Echo why only constant functions? Take an open set $A\subset \mathbb{T}$ which contains all roots of unity but measure of $A$ is less than 1. Any summable function supported outside $A$ belongs to $I$. $\endgroup$
    – Fedor Petrov
    Commented Apr 5, 2022 at 17:45
  • $\begingroup$ We are talking about convolution here. Note that $z^n*z^m$ is zero if $n\ne m$ and equals $z^n$ if $n=m$. This means that the ideal generated by $p_n$, $n\ne 0$, is 2-dimensional. It is spanned by $1$ and $z^n$. $\endgroup$
    – user473423
    Commented Apr 6, 2022 at 5:22
  • $\begingroup$ I need to ask just to make sure if I get the notation right: is $I_n=\{f*g_n: f\in L^1(\mathbb{T})\}$ where $g_n(t) = p_n(e^{2\pi i t}) = e^{2\pi i nt} - 1$ ? $\endgroup$
    – Onur Oktay
    Commented Apr 6, 2022 at 21:26
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    $\begingroup$ Well, in that case, $f\in I_n$ iff $\hat{f}(k) = 0$ for all $k\notin\{0,n\}$. So $f\in I$ iff $\hat{f}(k) = 0$ for all $k\neq 0$ iff $f$ is constant. Here $(\hat{f}(n))_{n\in\mathbb{Z}}$ denote the Fourier series coefficients of $f\in L^1(\mathbb{T})$. Second, every character $\tilde{h}$ of $L^1(\mathbb{T})/I$ is of the form $\tilde{h}(f+I)=h(f)$ for some character $h$ of $L^1(\mathbb{T})$ such that $I\subset\ker{h}$. Consequently, the character space of $L^1(\mathbb{T})/I$ is isomorphic to $\mathbb{Z}\backslash\{0\}$. $\endgroup$
    – Onur Oktay
    Commented Apr 10, 2022 at 15:27

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