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I am referring for instance to this question about coefficients of automorphic forms on $GL(3)$. I know that the Ramanujan on average bound is known and gives $$\sum_{n^2 m < x} |\lambda(n,m)|^2 \ll x^{1+\varepsilon}.$$

Is there anything known (in terms of upper bounds, with explicit dependence in the fixed $m$ or $n$) about the partial sums (except trivially bounding by the above) : $$\sum_{n^2< x} |\lambda(n,m)|^2 \qquad \text{and} \qquad \sum_{m < x} |\lambda(n,m)|^2 ?$$

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On the Ramanujan conjecture $\lambda(m,n) \ll 1$. As there is no cancellation in the second sum, essentially (upto $x^\epsilon$) the best upper bound which one may expect for that is $x^{1+\epsilon}$.

For the first sum for the same reason the best possible bound would be $x^{1/2+\epsilon}$. To prove that note that $\lambda(n,m)=\overline{\lambda(m,n)}$. So $$\sum_{n^2<x}|\lambda(n,m)|^2=\sum_{n<\sqrt{x}}|\lambda(m,n)|^2\ll x^{1/2+\epsilon},$$ using the bound from the second sum.

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  • $\begingroup$ But how do you get the bound on the second sum? (for now it seems a bit circular) $\endgroup$
    – Amomentum
    Commented Mar 26, 2020 at 1:48
  • $\begingroup$ I thought that $n$ in the second sum is fixed, say $n_0$. Then the second sum is bounded by $\sum_{n^2m<n_0^2x} |\lambda(n,m)|^2\ll_{n_0} x^{1+\epsilon}$ from the first estimate in your question. $\endgroup$
    – Subhajit Jana
    Commented Mar 26, 2020 at 8:22
  • $\begingroup$ But this is wasteful since we have a rough $n^2$ bound on the right then, which is a loss compared to Ramanujan conjecture. Is there any way to do better? $\endgroup$
    – Amomentum
    Commented Mar 28, 2020 at 4:55
  • $\begingroup$ If you are asking for Ramanujan on average uniformly in $n$, to prove that it will be as hard as proving Ramanujan: Let $n$ be any large prime bigger than $x$ so that $(m,n)=1$ for all $m<x$ so that $$\lambda(n,m)=\lambda(n,1)\lambda(1.m).$$ If we have $$\sum_{m<x} |\lambda(n,m)|^2\ll x^{1+\epsilon}$$ uniformly in $n$ then using the fact that $$\sum_{m<x}|\lambda(1,m)|^2\gg x^{1-\epsilon}$$ we obtain $$\lambda(n,1)\ll x^\epsilon < n^\epsilon.,$$ i.e. Ramanujan. $\endgroup$
    – Subhajit Jana
    Commented Mar 31, 2020 at 6:51

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