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I just watched Kannan Soundararajan's talk on the distributions of valus of zeta and $L$-functions at virtual ICM 2022. In his talk, he introduced a theorem on Ramanujan's ternary form $\phi_{1}: x^{2} + y^{2} + 10z^{2}$ - on which numbers can be represented by such a form. The proof introduces another form $\phi_{2}: 2x^{2} + 2y^{2} + 3z^{2} - 2xz$ and he said that the number of integer solutions $R_{i}(N, \phi_{i}) = \#\{(x, y, z)\,:\, \phi_{i}(x, y, z) = N\}$ is related to certain modular form of weight $3/2$ - $a(N):= (1/4)(R(N, \phi_{1}) - R(N, \phi_{2}))$ becomes a coefficient of such a form $f(z) = \sum a(n)q^{n}$. Using the Shimura correspondence, it is related to some weight 2 modular form $F(z) = \sum A(n)q^{n}$. Finally, using Waldspurger's formula $$a(n)^{2} = C\sqrt{n}L\left(\frac{1}{2}, F\times \chi_{-40n}\right) = C\sqrt{n} \sum_{k \geq 1} \frac{A(k)}{k}\chi_{-40n}(k)$$ he concludes that the Lindelof hypothesis for $L(s, F\times \chi_{-40n})$ (which is implied by Riemann hypothesis) implies (a weaker version of) Ramanujan's conjecture for $f$, $$|a(n)|^{2} \leq C(\epsilon) n^{1/4 + \epsilon}.$$

I wonder if such a strategy can be applied to prove (weaker version of) Ramanujan's conjecture for other modular forms. For example, given half-weight modular form $f$ and corresponding integral weight modular form $F$, does the Lindelof hypothesis on $L(s, F)$ implies (a weak version of) Ramanujan's conjecture on $f$? Also, is it possible to prove Ramanujan's conjecture for integral weight modular forms with a similar strategy (e.g. assuming the Riemann hypothesis or the Lindelof hypothesis for certain $L$-functions)?

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    $\begingroup$ (1) Yes, if I remember correctly, Waldspurger's formula, and thus the relation, works for every modular form of half-integral weight. (2) There's an appraoch, but a completely different one, based on the symmetric power $L$-functions. Theoretically holomorphic continuation and Riemann hypothesis for all the symmetric power $L$-functions of the modular form would do the trick, but analytic continuation on the half-space $s>1$ would work just as well if you could do it for every single symmetric power. $\endgroup$
    – Will Sawin
    Commented Jul 15, 2022 at 1:53
  • $\begingroup$ the thing I don't see is the fact that the two quadratic forms are in the same genus. Thus, for example, one may consider Siegel's description of numbers of representation by the genus. And, you see, every eligible number is represented by one or both of the forms; all excpet $4^k (16m+6)$ See zakuski.math.utsa.edu/~kap $\endgroup$
    – Will Jagy
    Commented Jul 15, 2022 at 2:25

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