# A property of simultaneous eigenforms

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 $$\gamma_n(m) = \sum_{d |(m,n)} d^{k-1} c(\frac{m n}{d^2})$$

• Is "will also not equal" a typo for "will also be equal"? Feb 19, 2023 at 22:43
• 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 $\gamma _n(1)=\lambda (n)c(1)$ Feb 20, 2023 at 12:58
• @DavidLoeffler you are right. Sorry!
– Jack
Feb 20, 2023 at 19:30
• @tomos That is exactly what's going on, you should repost your comment as an answer. Feb 20, 2023 at 19:54
• ah right, ye sure Feb 20, 2023 at 19:57

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 $$\lambda (n)c(1)=\gamma _n(1)=c(n)$$.