2
$\begingroup$

My reference is Daniel Bump's book, Automorphic Forms and Representations, Chapter 3.7. Let $k$ be a number field, $G = \operatorname{GL}_2$, $B$ and $T$ the usual Borel subgroup and maximal torus of $G$. For $\chi$ an unramified character of $T(\mathbb A)/T(k)$, let $V$ be the space of "smooth" functions $f: G(\mathbb A) \rightarrow \mathbb C$ satisfying $f(bg) = \chi(b) \delta_B(b)^{\frac{1}{2}}f(g)$ which are right $K$-finite. For $f \in V$ and $g \in G(\mathbb A)$, define the Eisenstein series

$$E(g,f) = \sum\limits_{\gamma \in B(k) \backslash G(k)} f(\gamma g)$$

For suitable $\chi$, the series converges absolutely for all $g \in G(\mathbb A)$. Now $E(g,f)$ has a "Fourier expansion," which as explained by Bump is gotten as follows: the function $$\Phi: \mathbb A/k \rightarrow \mathbb C$$ $$\Phi(x) = E( \begin{pmatrix} 1 & x \\ & 1 \end{pmatrix}g,f)$$ is continuous, hence is in $L^2(\mathbb A/k)$, and therefore has a "Fourier expansion" over the characters of $\mathbb A/k$. If $\psi$ is a fixed character of $\mathbb A/k$, then $\psi_{\alpha}: x \mapsto \psi(\alpha a)$ comprise the rest of them, for $\alpha \in k$. The characters $\psi_{\alpha}$ form an orthonormal basis of $L^2(\mathbb A/k)$, so by orthogonality, $$\Phi(x) = \sum\limits_{\alpha \in k} c_{\alpha}(g,f) \psi_{\alpha}(x) \tag{1}$$

$$c_{\alpha}(g,f) = \int\limits_{\mathbb A/k} E( \begin{pmatrix} 1 & y \\ & 1 \end{pmatrix} g,f) \psi(-\alpha y) dy$$

According to Bump, we may simply set $x = 0$, giving us the Fourier expansion for the Eisenstein series

$$E(g,f) = \sum\limits_{\alpha \in k} c_{\alpha}(g,f)$$

My question: Why is this last step valid? The right hand side of equation (1) converges to $\Phi(x)$ in the $L^2$-norm. As far as I know, this is not an equation of pointwise convergence. In general, the Fourier series of a continuous function need not converge pointwise to that function everywhere (in the classical case $\mathbb R/\mathbb Z$, the Fourier series of a continuous function converges pointwise to that function almost everywhere).

$\endgroup$
4
  • 4
    $\begingroup$ When you say "usual" Borel, do you mean upper triangular matrices? And by "fixed character", do you mean fixed nontrivial character? Also, one of your sums is over $a$ instead of $\alpha$. Finally, I think the answer to your question is that absolute summability of the Fourier coefficients of a continuous function implies uniform convergence of the Fourier series. $\endgroup$ – S. Carnahan Jan 2 '19 at 7:31
  • $\begingroup$ Yes and yes. So you're saying if the Fourier coefficients are absolutely summable, then they are square-summable and the Fourier series converges uniformly to the function everywhere? $\endgroup$ – D_S Jan 2 '19 at 15:07
  • $\begingroup$ @D_S: math.stackexchange.com/questions/2918700/… $\endgroup$ – Not a grad student Jan 2 '19 at 17:54
  • $\begingroup$ Okay, I think I understand now. So the answer to my question comes down to whether $$\sum\limits_{\alpha \in k} |c_{\alpha}(g,f)| < \infty$$ $\endgroup$ – D_S Jan 2 '19 at 18:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.