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While reading the following theorem for Apostol's modular functions and dirichlet series in number theory, I have a question:

(Theorem 6.14, page 130)

Assume that k is even and $k\geq 4$. If the space $M_k$ contains a simultaneous eigenform $f$ with Fourier expansion $f(\tau)= \sum_{m=0}^{\infty} c(m) x^m$ ,where $x= e^{2πi\tau}$, then $c(1)\neq 0$.

Proof: The coefficient of $x$ in the Fourier expansion of $T_n(f)$ is $\gamma_n(1)= c(n)$. Since f is a simultaneous eigenform this coefficient is also equal to $\lambda(n) c(1)$, so $c(n) = \lambda(n) c(1)$ for all $ n\geq 1$.

I am not able to deduce how does f being a simultaneous eigenform implies that this coefficient will also be equal to $\lambda(n) c(1)$.

If f is an eigenform for every Hecke operator $T_n, n\geq 1$ then f is called a simultaneous eigenform.

If $ f\in M_k$ and has a Fourier expansion $f(\tau)= \sum_{m=0}^{\infty} c(m) x^m$ where $x= e^{2πi\tau} $, then $T_n f$ has the Fourier expansion $T_n(f)(\tau) = \sum_{m=0}^{\infty} \gamma_{n} (m) x^m $, where $\gamma_n(m) = \sum_{d |(m,n)} d^{k-1} c(\frac{m n}{d^2})$

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  • $\begingroup$ Is "will also not equal" a typo for "will also be equal"? $\endgroup$ Feb 19, 2023 at 22:43
  • $\begingroup$ i don't know this stuff that well, but as far as i would guess: $T_nf(\tau )=\lambda (n)f(\tau )$ because it's an eigenform, so comparing the fourier coefficients for both sides of this equality gives $\gamma _n(m)=\lambda (n)c(m)$ and in particular $\gamma _n(1)=\lambda (n)c(1)$ $\endgroup$
    – tomos
    Feb 20, 2023 at 12:58
  • $\begingroup$ @DavidLoeffler you are right. Sorry! $\endgroup$
    – Jack
    Feb 20, 2023 at 19:30
  • $\begingroup$ @tomos That is exactly what's going on, you should repost your comment as an answer. $\endgroup$ Feb 20, 2023 at 19:54
  • $\begingroup$ ah right, ye sure $\endgroup$
    – tomos
    Feb 20, 2023 at 19:57

1 Answer 1

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Let $n\in \mathbb N$ be given. As $f$ is an eigenform for $T_n$, then $T_nf=\lambda (n)f$, and in particular the ceofficients of their Fourier series must be equal. In your notation, the Fourier series of the LHS is $$\sum _{m}\gamma _n(m)x^m$$ whilst the Fourier series of the RHS is $$\lambda (n)\sum _mc(m)x^m$$ so $\lambda (n)c(m)=\gamma _n(m)$ and in particular $\lambda (n)c(1)=\gamma _n(1)=c(n)$.

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