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I recently learned why the Tate's thesis, especially Poisson summation formula, over a function field $F = \mathbb{F}_q(X)$ of a smooth projective curve $X_{/ \mathbb{F}_q}$ implies Riemann-Roch theorem. This essentially follows from (a version of) Poisson summation formula

$$ \sum_{a \in F} f(ax) = \frac{1}{|x|} \sum_{a \in F} \widehat{f}\left(\frac{a}{x}\right) $$

and apply this to $f = \mathbf{1}_{\mathcal{O}_{\mathbb{A}_F}}$ & interpret both sides as cardinalities of linear systems of certain divisors (I followed Ramakrishnan-Valenza, Chapter 7.2). Here we choose $x = x(D)$ an idele with $v(x(D)_v) = n_v$ for $D = \sum_v n_v v$. After reading this, I have two question arose in my mind:

  1. We also have a similar Poisson summation formula for $\mathrm{GL}_n$ that is used by Godement-Jacquet for their theory of automorphic $L$-functions of $\mathrm{GL}_n$. If we read the formula for a function field again, does it give any Riemann-Roch-like theorem for vector bundles over $X$? (I'm not sure which one is the right direction for generalization - increasing the dimension of $X$ or the rank of bundle on $X$).

  2. As Tate did in his thesis, we can choose a different test function $f$ other than the simplest one $\mathbf{1}_{\mathcal{O}_{\mathbb{A}_F}}$. Especially, we can choose $f = \otimes_v f_v$ with $f_v = \psi_v$ an additive character with conductor strictly larger than $0$, for all but finitely many $v$. In this case, both sides might count something else other than just $\#\mathcal{L}(D)$ or $\#\mathcal{L}(K - D)$. Can we explicitly describe what those object would be? (I think it could be just some translation $D' = D + D_0$ for some explicit $D_0$ given in terms of $\psi_v$'s, so may not give anything new..)

If 1 and 2 are both nontrivial, then we may mix these two.

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    $\begingroup$ Surely it's not right to say Tate's thesis is the Riemann-Roch theorem but Tate's thesis gives the Riemann-Roch theorem? From Riemann-Roch we can derive functional equations of L-functions of everywhere unramified idele class group characters but there is the ramified case as well. $\endgroup$
    – Will Sawin
    Commented Feb 5 at 19:25
  • $\begingroup$ @WillSawin Thanks, I totally agree with you that the wording is not correct - I may misheard in the past. $\endgroup$
    – Seewoo Lee
    Commented Feb 5 at 19:53
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    $\begingroup$ Or, perhaps it's that Poisson summation is Riemann-Roch? (In the function-field case.) $\endgroup$ Commented Feb 5 at 20:26

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