Let $S_k(\Gamma_1(N))^{\textrm{new}}$ be the new subspace of weight $k$ cusp forms on $\Gamma_1(N)$. Section 5.8 of Diamond and Shurman's book claims that every $f \in S_k(\Gamma_1(N))^{\textrm{new}}$ is an eigenform for all of the Hecke operators $T_n$. Clearly I am confused about something here, because I don't see how this is possible. If $f,g \in S_k(\Gamma_1(N))^{\textrm{new}}$ are linearly independent then for some $T_n$ the corresponding eigenvalues $\lambda_n$ for $f$ and $\mu_n$ for $g$ must be different. By the claim $f + g \in S_k(\Gamma_1(N))^{\textrm{new}}$ is also an eigenform for $T_n$. Let $\rho_n$ be the corresponding eigenvalue. Then $$\lambda_n + \mu_ng = T_n(f + g) = \rho_n(f + g).$$ But by linear independence this implies $\rho_n = \lambda_n$ and $\rho_n = \mu_n$, thus $\lambda_n = \mu_n$, contradicting that these two eigenvalues are different.


1 Answer 1


I think you need to include the previous statement to get the intended meaning the full paragraph is (emphasis mine):

Recall from Corollary 5.6.3 that the spaces $S_k (Γ_1 (N ))^ {old}$ and $S_k (Γ_1 (N ))^{new}$ have orthogonal bases of eigenforms for the Hecke operators $\{T_n , n : (n, N ) = 1\}$. Let f be such an eigenform. Using the Main Lemma, this section will show that if $f ∈ S_k (Γ_1 (N ))^{ new}$ then in fact $f$ is an eigenform for all $T_n$ and $\langle n\rangle$.

So Diamond and Shurman are assuming that $f$ is an element of the orthonormal eigenbasis of some old or newspace and in addition it is new then the conclusion is that $f$ is an eigenform not just for a subset of $T_n$'s (those with $(n,N)=1$) but all of them.

  • $\begingroup$ Ah ok. I obviously didn't read this part carefully enough. Thanks! $\endgroup$ Oct 28, 2019 at 5:39

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