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Is there an analog of Fourier series in the function field setting based on the Carlitz exponential? I mean, something like:

Let $\Omega$ be the completion of an algebraic closure of $\mathbb F_q\left(\left(\frac1T\right)\right)$ and $\exp_{\mathcal C}(z)=\sum_{n\ge0}\frac{z^{q^n}}{D_n}$ be the Carlitz exponential on $\Omega$ where $$D_n=\prod_{\substack{a\in\mathbb F_q[T]\\ \deg a= n\\ a\text{ monic}}}a.$$ One denotes by $\xi\in\Omega$ the smallest (relatively to degree) period of $\exp_{\mathcal C}$. Let $f$ be an entire function on $\Omega$ such that $f(z+a)=f(z)$ for all $z\in\Omega$ and $a\in\mathbb F_q[T]$. Can we expand $f$ as $$f(z)=\sum_{a\in\mathbb F_q[T]}b_a\exp_{\mathcal C}(\xi a z),\text{ where }b_a\in\Omega.$$

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Yes, there is and is called $t$-expansion rather than $q$-expansion - may be, though, you might want to replace your $\Omega$ by $$ \widehat{\bar{K_\infty}}\setminus K_\infty $$ where I denote by $K_\infty$ your completion $\mathbb{F}_q\big(\big(\frac{1}{T}\big)\big)$. This is the content of the paper On the coefficients of Drinfel'd modular forms by E-U. Gekeler, Inventiones Math. 93 (1988).

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  • $\begingroup$ Are there conditions for which f admits such an expansion? ANd when there exists a Fourier expansions, is there a formula giving the coefficients $b_a$ $\endgroup$
    – joaopa
    Commented Apr 25, 2013 at 4:47
  • $\begingroup$ Well, as in the "classical" case, you need some condition at the cusps if you want to insist that the expansion is "integral" rather than rational. Similarly, assuming your periodic function to be a modular form, you have a notion of eigenform for which the coefficients get related to Hecke eigenvalues. In Section 6. of the paper Gekeler describes also in some detail the coefficients of Eisenstein series, but I would advise you to have a direct look at his paper. $\endgroup$
    – Filippo Alberto Edoardo
    Commented Apr 25, 2013 at 5:42
  • $\begingroup$ I thought more about a more general criteria. For example, in the classical casse, any periodic entire function on $\mathbb C$ admits a Fourier expansion. Is this the case in the function field setting? $\endgroup$
    – joaopa
    Commented Apr 25, 2013 at 5:55
  • $\begingroup$ Well, yes, periodicity plus some convergence at cusps (to guarantee convergence at $0$ in the disk) is enough. But, again, I advise you to look at Gekeler's Inventiones paper (for this, Def. 5.6 might be what you want). $\endgroup$
    – Filippo Alberto Edoardo
    Commented Apr 25, 2013 at 7:15

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