7
$\begingroup$

Let $k\geq 1$ and let $f=\sum_{n\geq 1}a(n)q^{n}$, $a(n)\in\mathbb{R}$, be a normalised cuspidal Hecke eigenform of weight $2k$ for $\Gamma_{0}(N)$ without complex multiplication. So the result of Barnet-Lamb et al. tells us that the numbers $\frac{a(p)}{2p^{k-1/2}}$ are $\mu$-equidistributed in $[-1,1]$, where $\mu$ is the probability measure on the interval $[-1,1]$ defined by $\frac{2}{\pi}\sqrt{1-t^{2}}\,dt$, and $p$ runs through the primes not dividing $N$.

I am looking for any generalisation of Sato-Tate conjecture when Fourier coefficients $a(n)$ are complex numbers, $a(n)\in\mathbb{C}$?

$\endgroup$

1 Answer 1

6
$\begingroup$

First, note that if $f = \sum_{n \geq 1} a(n) q^{n}$ is a cuspidal Hecke eigenform lying in the new subspace of $S_{k}(\Gamma_{0}(N), \chi)$, where $\chi$ is a Dirichlet character modulo $n$, then one can determine the argument of $a(n)$ in terms of the values of $\chi$. In particular, for $\gcd(n,N) = 1$, then $$ \overline{a(n)} = \chi(n) a(n). $$ (A proof of this relation can be found in Iwaniec's "Topics in Classical Automorphic Forms" text.) This is the reason that if $\chi$ is trivial, all the Fourier coefficient of $f$ are real. In general, if $\zeta$ is a root of unity and $\chi(p) = \zeta^{2}$, then $\frac{a(p)}{\zeta}$ is real.

A generalization of the Sato-Tate conjecture is stated and proved in the paper of Barnet-Lamb et. al. (which can be found online here). In particular, Theorem B of that paper states that if $\zeta$ is a root of unity with $\zeta^{2}$ in the image of $\chi$, then the numbers $$ \left\{ \frac{a(p)}{2p^{(k-1)/2} \zeta} : p~\text{prime}, \chi(p) = \zeta^{2} \right\}$$ are equidistributed in $[-1,1]$ with respect to $\frac{2}{\pi} \sqrt{1-t^{2}}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.