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Take a cusp form $f$ and let $f(q) = q + a_2q^2 + q_3q^3 + \ldots$" denote its $q$-expansion (assume that the $a_k$ are integers, and that $f$ comes from an elliptic curve $E$). Of course the series $f(1) = 1 + a_2 + a_3 + \ldots$ diverges, but I wonder whether there is any work on evaluating $f(1)$ via some regularization method (I do not even know whether $\lim_{q \to 1-0} f(q)$ exists or not). I am kind of hoping that $f(1)$ is connected with the order of the Tate-Shafarevich group of the associated elliptic curve $E$ and perhaps the order of its torsion group (the actual expression would have to be invariant under isogenies).

Does this ring a bell with anyone?

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    $\begingroup$ The limit as $q \rightarrow 1$ along the real axis, say, is certainly $0$, since $f$ is a cusp form and $q=1$ corresponds to $z=0$, which is a cusp. This is easy to see numerically for small conductor; the convergence to zero is staggeringly fast. Incidentally, Euler was already aware of this phenomenon in the related case $\sum_n (-1)^n q^{n^2}$ as $q \rightarrow 1$. $\endgroup$
    – moonface
    Commented Mar 26, 2010 at 15:54
  • $\begingroup$ Maybe my answer was a bit dumb, in the case mentioned by moonface. Indeed, the value $L(f,0)$ vanishes for a cusp form $f$, due to the pole of the Gamma function at zero. So my long-winded answer is consistent with the vanishing observation of moonface. $\endgroup$
    – Marty
    Commented Mar 26, 2010 at 16:54

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I'll give a slightly uncertain answer, based somewhat on my recollection of conversations with Zagier a month ago about similar questions.

If we were to imitate Euler, we might consider $f(1)$ as $$f(1) = \sum_{n \geq 1} a_n = \sum_{n \geq 1} a_n n^{-0} = L(f,0).$$ So the analytic continuation of the L-function suggests that $f(1)$ should be identified with the value of the L-function at zero. By the functional equation, this relates to the L-function at the right edge of the critical strip.

So, for a cusp form of weight two, arising from an elliptic curve $E$ over $Q$, the value $L(f,0)$ is related to $L(E,2)$. An interpretation of this L-value, conjectured by Zagier, was proven by Goncharov and Levin, in "Zagier's conjecture on $L(E,2)$", Invent. Math. 132 (1998).

As for the analytic question, you are considering the "value" of a cusp form $f$ on the real axis, which bounds the upper half-plane. Almost by definition, there is a Sato hyperfunction $f_{bdr}$ on the real axis, which describes this boundary behavior of the holomorphic function $f$ on the upper half-plane. I am not sure if the following is published, but I have the impression that there might be a preprint now or soon which proves the following result:

At every (positive? I don't recall) rational number $q$, the hyperfunction $f_{bdr}$ is $C^\infty$ at $q$. Its value at $1$ is $L(f,0)$ as described above.

I think that saying "a hyperfunction is $C^\infty$ at $q$" means that the hyperfunction can be expressed as the distributional derivative of a continuous function -- $f = g^{(k)}$ for some $k \geq 0$ -- and $g$ happens to be $C^\infty$ at $q$. But I'm not much of an analyst.

I think that the value $f(1)$ also exists as $\lim_{z \rightarrow 1} f(z)$ limit, if $z$ approaches $1$ via a geodesic in the upper half-plane.

I don't think you'll see Sha or the torsion directly, as these appear at the central value $L(f,1)$. On the other hand, I do think you'll find $L(f,-n)$ for all $n \geq 0$ (or equivalently, $L(f,2+n)$ ), by looking at the derivatives $f^{(n)}(1)$ of the boundary hyperfunction of $f$ at $1$.

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  • $\begingroup$ Somehow it would surprise me if $f(q)$ really did tend to $L(E,0)$ in any reasonable sense, because isn't it a result of Szpiro that if $\sum_n a_n.n^{-s}$ converges as $s$ tends to 1, then it won't converge to $L(E,1)$ but rather to something like $\sqrt{2}$ times this? I forgot what he actually proved unfortunately. Somehow if you can't get it right at 1 with 'naive' methods, what chance have you of getting it right at 0? Or do you think I've got the wrong end of the stick? $\endgroup$
    – Kevin Buzzard
    Commented Mar 26, 2010 at 15:17
  • $\begingroup$ But here I'm looking at $f(z)$ as $z$ approaches $1$, via certain paths in the upper half-plane. These values have almost nothing in common with the values of the Dirichlet series as $s$ approaches zero. For cusp forms, the limits I'm talking about are in the same spirit as the convergent integral representation of the L-function. $\endgroup$
    – Marty
    Commented Mar 26, 2010 at 15:23
  • $\begingroup$ Sounds like I have indeed got the wrong end of the stick then :-) Apologies. $\endgroup$
    – Kevin Buzzard
    Commented Mar 26, 2010 at 15:52
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    $\begingroup$ Voloch is right: this $\sqrt{2}$ is related to partial Euler product behavior as $s$ tends to 1, not to the Dirichlet series representation for $L(s)$, which in fact converges for $s$ with real part greater than 5/6 (so at $s=1$). If you replace elliptic curve $L$-functions with Hecke $L$-functions for a quadratic character, you can turn the unexpected $\sqrt{2}$ into $1/\sqrt{2}$ with the Euler product as $s$ tends to $1/2$. Oh, and all such Euler product convergence implies GRH so you can numerically test this stuff and see it happen but you'll never unconditionally prove it. $\endgroup$
    – KConrad
    Commented Mar 27, 2010 at 4:01
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    $\begingroup$ Oh, and the convergence of $L(s)$ for $s$ with real part greater than 5/6 uses the elliptic modularity theorem. $\endgroup$
    – KConrad
    Commented Mar 27, 2010 at 4:02

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