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While studying, I recently came across the following interesting problem.

Let's say that the (level one) weight $k$ modular forms $M_k(\Gamma(1))$ have dimension $d$. We know by the ring structure of $M_*(\Gamma(1))$ that the map $M_k(\Gamma(1)) \rightarrow \mathbb{C}^d$, sending a form to its first $d$ Fourier coefficients, is an isomorphism. This suggests a natural basis for $M_k(\Gamma(1))$: the pre-images of the standard basis vectors $(e_1, \ldots, e_n)$ for $\mathbb{C}^d$. Let's call this basis $g_1, \ldots, g_d$.

The claim is that if $T_n$ is the $n$th Hecke operator, then

$$T_n = \sum_{j=1}^{d-1} a_n(g_j) T_j$$

acting on $S_k(\Gamma(1))$.

How can you prove this identity?

I had thought to approach this by using the formula for the Fourier coefficients of the Hecke operators. We know that it suffices to check that both sides agree on the basis vectors $g_k$, which we can check by comparing the first $d$ Fourier coefficients. In principle, we can write down a formula for $b_i(T_n g_k)$ and $\sum_{j=1}^{d-1} a_n(g_j) b_i(T_j(g_k))$, but I tried doing this and it seemed really messy.

EDIT: I originally accepted the answer below, but now I don't know that it's correct - see my comment.

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The space of cusp forms $S_k(\Gamma(1))$ admits a basis of normalized ($a_1=1$) eigenforms, so it suffices to prove the identify for normalized eigenforms.

Let $f = \alpha_1 g_1 + \cdots \alpha_{d-1} g_{d-1}$ be a normalized eigenform in $S_k(\Gamma(1))$. Then, $$T_n(f) = a_n(f)f = [\alpha_1a_n(g_1) + \cdots + \alpha_{d-1}a_n(g_{d-1})]f,$$ where $a_n(\cdot)$ denotes the $n$th Fourier coefficient of the modular form.

On the other hand, since $a_j(f) = \alpha_j$ for $0 < j < d$, we have $$\sum_{j=1}^{d-1}a_n(g_j)T_j(f) = \sum_{j=1}^{d-1}a_n(g_j)a_j(f)f = \sum_{j=1}^{d-1}a_n(g_j)\alpha_jf.$$

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  • $\begingroup$ Apologies, but when I revisited this I realized that you are assuming something significant - namely, $S_k(\gamma(1))$ has a basis of eigenforms with nonvanishing $a_1$. Actually, I don't think this is true, is it? $\endgroup$
    – Tony
    Jun 4, 2014 at 12:05
  • $\begingroup$ There is a theorem of Hecke (see e.g. Iwaniec, Topics in Classical Automorphic Forms, Thm 6.15) that the space $S_k(\Gamma(1))$ admits a basis by eigenforms, essentially because the Hecke operators commute, so we can simultaneously diagonalize for them. The $a_1$ do not vanish, because otherwise all the Fourier coefficients would vanish, because $a_n = \lambda(n) a_1$. So in fact we can take our basis to be of normalized eigenforms, i.e. $a_1=1$. My answer was poorly worded and I will clarify. Thanks! $\endgroup$
    – John M
    Jun 4, 2014 at 15:43
  • $\begingroup$ Very silly of me - I knew these facts, but I was simultaneously thinking that the square of an eigenform is an eigenform and thought that it was possible for $a_1 = 0$. $\endgroup$
    – Tony
    Jun 4, 2014 at 15:52
  • $\begingroup$ Thanks though for pointing it out. As originally worded (taking $f$ as just any cusp form), what I wrote was incorrect. $\endgroup$
    – John M
    Jun 4, 2014 at 15:56

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