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Further expounded on the main term. Referenced sharper lower bound for $L(1,\mathrm{Ad}~f)$. Was more careful with the dependence on $\epsilon$.
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2734364041
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If $f$ is a $\mathrm{GL}(2)$ newform over $\mathbb{Q}$ with prime level $p$ and $\zeta^{(p)}(s)$ denotes the Riemann zeta function $\zeta(s)$ with the Euler factor at $p$ removed, then

$$L(s,f\times \bar{f}) = \zeta^{(p)}(2s)\sum_{n=1}^{\infty} \frac{|\lambda_{f}(n)|^2}{n^s}.$$

Moreover, we have the factorization

$$L(s,f\times\bar{f}) = \zeta(s) L(s,\mathrm{Ad}~f) = \zeta(s) L(s,\mathrm{Sym}^2 f\otimes\overline{\chi}_{f}),$$

where $\chi_{f}$ is the nebentypus character. The conductor of this $L$-function is $p^2$. We assume that the spectral component $K$ (weight, Laplace eigenvalue, etc.) is absolutely bounded. By applying Corollary 1.4 in this paper (Equation 1.11, in particular), we find that if $0<\epsilon<1/4$, then

$$\sum_{n\leq x}|\lambda_{f}(n)|^2 = \frac{x}{\zeta^{(p)}(2)}\mathop{\mathrm{Res}}_{s=1}L(s,f\times\bar{f})+O((p x)^{1/2+\epsilon}),$$$$\sum_{n\leq x}|\lambda_{f}(n)|^2 = \frac{L(1,\mathrm{Ad}~f)}{\zeta^{(p)}(2)}x+O((p x)^{1/2+\epsilon}) = \frac{6L(1,\mathrm{Ad}~f)}{\pi^2}\Big(1-\frac{1}{p^2}\Big)^{-1}x+O((p x)^{1/2+\epsilon}).$$

where theThe implied constant depends on $K$ and $\epsilon$. Since the residue is $\gg p^{-\epsilon}$$L(1,\mathrm{Ad}~f)\gg 1/\log p$ (Goldfeld-Hoffstein-LockhartLieman), the asymptotic is nontrivial when $p < x^{1-\epsilon}$$p < x^{1-4\epsilon}$.

If $f$ is a $\mathrm{GL}(2)$ newform over $\mathbb{Q}$ with prime level $p$ and $\zeta^{(p)}(s)$ denotes the Riemann zeta function $\zeta(s)$ with the Euler factor at $p$ removed, then

$$L(s,f\times \bar{f}) = \zeta^{(p)}(2s)\sum_{n=1}^{\infty} \frac{|\lambda_{f}(n)|^2}{n^s}.$$

Moreover, we have the factorization

$$L(s,f\times\bar{f}) = \zeta(s) L(s,\mathrm{Ad}~f) = \zeta(s) L(s,\mathrm{Sym}^2 f\otimes\overline{\chi}_{f}),$$

where $\chi_{f}$ is the nebentypus character. The conductor of this $L$-function is $p^2$. We assume that the spectral component $K$ (weight, Laplace eigenvalue, etc.) is absolutely bounded. By applying Corollary 1.4 in this paper (Equation 1.11, in particular), we find that

$$\sum_{n\leq x}|\lambda_{f}(n)|^2 = \frac{x}{\zeta^{(p)}(2)}\mathop{\mathrm{Res}}_{s=1}L(s,f\times\bar{f})+O((p x)^{1/2+\epsilon}),$$

where the implied constant depends on $K$ and $\epsilon$. Since the residue is $\gg p^{-\epsilon}$ (Hoffstein-Lockhart), the asymptotic is nontrivial when $p < x^{1-\epsilon}$.

If $f$ is a $\mathrm{GL}(2)$ newform over $\mathbb{Q}$ with prime level $p$ and $\zeta^{(p)}(s)$ denotes the Riemann zeta function $\zeta(s)$ with the Euler factor at $p$ removed, then

$$L(s,f\times \bar{f}) = \zeta^{(p)}(2s)\sum_{n=1}^{\infty} \frac{|\lambda_{f}(n)|^2}{n^s}.$$

Moreover, we have the factorization

$$L(s,f\times\bar{f}) = \zeta(s) L(s,\mathrm{Ad}~f) = \zeta(s) L(s,\mathrm{Sym}^2 f\otimes\overline{\chi}_{f}),$$

where $\chi_{f}$ is the nebentypus character. The conductor of this $L$-function is $p^2$. We assume that the spectral component $K$ (weight, Laplace eigenvalue, etc.) is absolutely bounded. By applying Corollary 1.4 in this paper (Equation 1.11, in particular), we find that if $0<\epsilon<1/4$, then

$$\sum_{n\leq x}|\lambda_{f}(n)|^2 = \frac{L(1,\mathrm{Ad}~f)}{\zeta^{(p)}(2)}x+O((p x)^{1/2+\epsilon}) = \frac{6L(1,\mathrm{Ad}~f)}{\pi^2}\Big(1-\frac{1}{p^2}\Big)^{-1}x+O((p x)^{1/2+\epsilon}).$$

The implied constant depends on $K$ and $\epsilon$. Since $L(1,\mathrm{Ad}~f)\gg 1/\log p$ (Goldfeld-Hoffstein-Lieman), the asymptotic is nontrivial when $p < x^{1-4\epsilon}$.

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2734364041
  • 5.1k
  • 2
  • 22
  • 40

If $f$ is a $\mathrm{GL}(2)$ newform over $\mathbb{Q}$ with prime level $p$ and $\zeta^{(p)}(s)$ denotes the Riemann zeta function $\zeta(s)$ with the Euler factor at $p$ removed, then

$$L(s,f\times \bar{f}) = \zeta^{(p)}(2s)\sum_{n=1}^{\infty} \frac{|\lambda_{f}(n)|^2}{n^s}.$$

Moreover, we have the factorization

$$L(s,f\times\bar{f}) = \zeta(s) L(s,\mathrm{Ad}~f) = \zeta(s) L(s,\mathrm{Sym}^2 f\otimes\overline{\chi}_{f}),$$

where $\chi_{f}$ is the nebentypus character. The conductor of this $L$-function is $p^2$. We assume that the spectral component $K$ (weight, Laplace eigenvalue, etc.) is absolutely bounded. By applying Corollary 1.4 in this paper (Equation 1.11, in particular), we find that

$$\sum_{n\leq x}|\lambda_{f}(n)|^2 = \frac{x}{\zeta^{(p)}(2)}\mathop{\mathrm{Res}}_{s=1}L(s,f\times\bar{f})+O((p x)^{1/2+\epsilon}),$$

where the implied constant depends on $K$ and $\epsilon$. Since the residue is $\gg p^{-\epsilon}$ (Hoffstein-Lockhart), the asymptotic is nontrivial when $p < x^{1-\epsilon}$.