Timeline for Sufficient condition for the absolute convergence of Fourier series of a function on the adele quotient $\mathbb A_k/k$
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 10, 2019 at 2:53 | vote | accept | D_S | ||
| Jan 23, 2019 at 10:18 | comment | added | Will Sawin | @D_S Yes, where $m$ is the degree of $k$ over $\mathbb Q$. In fact, we can view it as $ ( k \otimes \mathbb R) / (k \cap H)$. To do this, consider the natural map from $k \otimes \mathbb R$ to $\mathbb A / (k + H)$. The kernel clearly is $k \cap H$. Because $k+ H$ equals the finite adeles, this map is surjective. | |
| Jan 22, 2019 at 22:58 | comment | added | D_S | @WillSawin then your next claim is that $\mathbb A/(k+H)$ is actually a torus $(\mathbb R/\mathbb Z)^m$? | |
| Jan 22, 2019 at 15:07 | comment | added | reuns | Obtained from $f(x) = \sum_n \sum_{a \in \widehat{\mathbb{Z}}/n! \widehat{\mathbb{Z}}}c(a,n)\chi_{a+n! \widehat{\mathbb{Z}}}(x)$ and $\chi_{a+n! \widehat{\mathbb{Z}}}(x) = \sum_{s \in (n!)^{-1}\widehat{\mathbb{Z}}/\widehat{\mathbb{Z}}} \psi_s(-a)\psi_s(x)$ and $\psi$ the character $\mathbb{A_Q}_{fin} /\widehat{\mathbb{Z}} \to \mathbb{C}^\times$, $\psi(p^{-k}) = \exp(2i \pi p^{-k})$ | |
| Jan 22, 2019 at 15:02 | comment | added | reuns | I think I made a mistake when going from $\mathbb{Z}_p$ to $\widehat{\mathbb{Z}}$. The criterion I obtain for the absolute convergence of the Fourier series of $f : \widehat{\mathbb{Z}} \to \mathbb{C}$ should be $|f(x)-f(y)| \le C N(x-y)^{1+\epsilon}$ where $N(x) = | \widehat{\mathbb{Z}}/x \widehat{\mathbb{Z}} |= \prod_p |x_p|_p^{-1}$ @D_S | |
| Jan 22, 2019 at 14:39 | comment | added | Will Sawin | @reuns I am not aware of any such cases. | |
| Jan 22, 2019 at 6:45 | comment | added | Will Sawin | @D_S Yes, sorry it's not so clear as I typed it on my phone. | |
| Jan 22, 2019 at 4:33 | comment | added | D_S | Okay, if I understand correctly, you pull back $f$ to a function on $\mathbb A$, and claim using compactness that there is an open compact subgroup $H$ of $\mathbb A_f$ such that $f(x+h) = f(x)$ for all $x \in \mathbb A$ and $h \in H$. Then, $f$ becomes well defined on $\mathbb A/H = \mathbb A_{\infty} \times \mathbb A_f/H$. Is this correct? | |
| Jan 21, 2019 at 19:51 | history | answered | Will Sawin | CC BY-SA 4.0 |