# Expressing a normalized eigenform in basis of newforms

Let $$S_k(\Gamma_1(N))$$ denote the space of weight $$k$$ cusp forms on $$\Gamma_1(N)$$. The set $$\mathcal{B} = \{f(n\tau) : f \textit{ is a newform of level M with nM \mid N}\}$$ a basis of $$S_k(\Gamma_1(N))$$. Let $$g \in S_k(\Gamma_1(N))$$ be a normalized eigenform. Then there exists a unique newform $$f$$ of level $$M \mid N$$ such that $$a_p(f) = a_p(g)$$ (here $$a_n(h)$$ denotes the $$n$$-th Fourier coefficient of a form $$h$$) for all primes $$p \not\mid N/M$$. I want to show that $$g(\tau) = \sum_{d \mid N/M}c_df(d\tau)$$ for some $$c_d \in \mathbb{C}$$.

Expression $$g$$ in the basis $$\mathcal{B}$$, we have $$g(\tau) = \sum_{i,j}c_{i,j}f_i(n_{i,j}\tau), \hspace{.5cm} c_{i,j} \in \mathbb{C}$$ where each $$f_i$$ is a newform of level $$M_i \mid N$$ and each $$n_{i,j} \mid N/M_i$$. Let $$f_k$$ be the newform such that $$a_p(g) = a_p(f_k)$$ for all $$p \not\mid N/M_i$$. Given a prime such that $$p \not\mid N$$, consider the linear operator $$T = \prod_{t \neq k}(T_p - a_p(f_t)),$$ where $$T_p$$ is the $$p$$-th Hecke operator $$T_p$$ and $$a_p(f) = a_p(f) \cdot \textrm{id}$$. Since $$p \not \mid N$$, the Hecke operator $$T_p$$ is the same at all levels dividing $$N$$. Applying $$T$$ to both sides of our expression for $$g$$ in the basis $$\mathcal{B}$$ and recalling that $$T_p(h) = a_p(h)h$$ for a normalized eigenform $$h$$, we obtain

$$\prod_{t \neq k}(a_p(g) - a_p(f_t)) \cdot g = \sum_{i,j} \prod_{t \neq k}(a_p(f_t) - a_p(f_j)) \cdot c_{i,j}f_i(n_{i,j}\tau)$$

The terms with $$i \neq k$$ on the right-hand side vanish and $$a_p(f_k) = a_p(g)$$, thus $$\prod_{t \neq k}(a_p(g) - a_p(f_t)) \cdot g = \prod_{t \neq k}(a_p(g) - a_p(f_t)) \cdot \sum_{j} c_{k,j}f_k(n_{k,j}\tau)$$ We are thus done if we can choose a prime $$p$$ such that $$p \not\mid N$$ and $$a_p(f_k) \neq a_p(f_t)$$ for all $$t$$. I know that each fixed $$t$$ we can choose such a prime by strong multiplicity one, but can we choose a prime that works for all $$t$$ simultaneously?

I realized we can just pick primes $$p_t \not\mid N$$ such that $$a_{p_t}(g) \neq a_{p_t}(f_t)$$ for $$t \neq k$$ by strong multiplicity one and the uniqueness of $$f_k$$. Then the proof goes through if we define the operator $$T$$ by $$T = \prod_{t \neq k} (T_{p_t} - a_{p_t}(f_t)).$$