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What is the order of growth of coefficients of half-integral weight modular forms on congruence subgroup with character ? There is the usual Hecke trick to compute bounds on the coefficients of integral weights on $SL_2(\mathbb{Z})$ but I don't know how to adapt the argument. I've looked online at many articles but didn't find anything. My guess is that it should be $O(n^{k/2})$ where $k$ is the half-integral weight.

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I assume you're referring to cuspforms. My answer assumes cuspforms throughout. The Hecke bound still applies. But one can say better.

It seems that we should have essentially the same bound as in the full integral weight case. That is, we should have that $$ a(n) \ll n^{\frac{k-1}{2} + \epsilon}. $$ In the full integral weight case, one can say a bit more about the $\epsilon$ factor in terms of the divisor function, but that might not be true for half-integral weight.

One reason to suspect that this is the truth is because it holds on average. The standard Rankin--Selberg technique of studying the Dirichlet series $$ D(s) = \sum_{n \geq 1} \frac{\lvert a(n) \rvert^2}{n^{s + k - 1}} $$ applies. One can show that $D(s)$ has meromorphic continuation to $\mathbb{C}$ with polynomial growth in vertical strips, and that the rightmost pole is $s = 1$. This implies that $$ \sum_{n \leq X} \lvert a(n)/n^{\frac{k-1}{2}} \rvert^2 = c X + o(X), $$ and thus that $\lvert a(n) \rvert \approx n^{\frac{k-1}{2}}$ on average.

But unlike the full integral weight case, the results of Deligne don't apply and we do not know how to prove this.

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    $\begingroup$ And of course by Waldspurger's theorem, if we knew such bounds for $n$ a fundamental discriminant, then we would know the Lindelof hypothesis for certain $L$-functions. So this is much harder than Deligne's theorem. $\endgroup$ Commented Sep 20, 2022 at 20:36
  • $\begingroup$ Thank you guys ! Great insights from your part. $\endgroup$ Commented Sep 21, 2022 at 14:40

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