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Let $G$ be a compact abelian group. The unitary characters of $G$ form an orthonormal basis of $L^2(G)$, so every square integrable function $f: G \rightarrow \mathbb C$ admits a Fourier expansion

$$f(x) = \sum\limits_{\chi \in \hat{G}} c_{\chi} \chi(x) \tag{1}$$

where the $c_{\chi}$ are uniquely determined complex numbers satisfying $\sum\limits |c_{\chi}|^2 < \infty$, and the right hand side converges to $f$ in the $L^2$-norm.

If moreover $$\sum\limits |c_{\chi}| < \infty \tag{2}$$ then (1) is actually a pointwise limit (and in fact a uniform limit).

When $G = \mathbb R/\mathbb Z$, it is well known that a sufficient condition for (2) is that $f$ be smooth (even just $C^1$).

What about when $G = \mathbb A_k/k$ for $k$ a number field, and $\mathbb A_k$ the adeles of $k$? There is a notion of a smooth function on $\mathbb A_k$ (being smooth in the archimedean argument, and locally constant in the nonarchimedean). Does the Fourier series of a smooth function $f$ on $\mathbb A_k/k$ satisfy (2)? Or if not, is there a well known sufficient condition on $f$ for (2) to hold?

The example I have in mind is the Fourier expansion of Eisenstein series, which I've asked about in a previous question here.

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Observe that, because we can cover by open sets, on each of which the function is constant under translation by some open in the finite adeles, by compactness we can take a finite subcover, do the function is translation-invariant by the intersection, another open subgroup. Modding our the whole adele group by this, and quotienting by the field, we just get avtorus.

So the problem reduces to the Torus case, which is well-known.

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  • $\begingroup$ 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? $\endgroup$ – D_S Jan 22 '19 at 4:33
  • $\begingroup$ @D_S Yes, sorry it's not so clear as I typed it on my phone. $\endgroup$ – Will Sawin Jan 22 '19 at 6:45
  • $\begingroup$ @reuns I am not aware of any such cases. $\endgroup$ – Will Sawin Jan 22 '19 at 14:39
  • $\begingroup$ 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 $\endgroup$ – reuns Jan 22 '19 at 15:02
  • $\begingroup$ 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 $\mathbb{A_Q}_{fin} /\widehat{\mathbb{Z}} \to \mathbb{C}^\times$, $\psi(p^{-k}) = \exp(2i \pi p^{-k})$ $\endgroup$ – reuns Jan 22 '19 at 15:07

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