# $p$-th coefficient of weight $2$ new form with $p | N$ must be $1$ or $0$ or $-1$?

Let $$f= \sum_{n=1}^{\infty}a_nq^n \in S_2^{new}(N)$$ be a normalized new form of weight $$2$$ with respect to $$\Gamma_0(N)$$ and assume $$p|N$$ is a prime. Then must $$a_p=0$$ if $$p^2|N$$ and belongs to $$\{-1,1\}$$ if $$p|N$$?

If $$f$$ has rational coefficients, then this can be seen from the corresponding modular elliptic curve as $$N$$ is the conductor.

So here I don't assume $$f$$ has rational coefficients so then the definition field $$K_f$$ may be large than $$\mathbb Q$$. Then again there is a modular abelian variety, but it's subtle because the $$L$$-function of the abelian variety matches the product of $$L$$-functions of Galois conjugation of $$f$$ rather than $$f$$, and I don't know much about Neron model.

Can it be proved only using knowledge of modular forms? I am also happy with a proof involving algebraic geometry of abelian varieties.

• If f has rational coefficients, then what ? – reuns Jan 10 at 22:07
• @reuns Then $a_p$ is equal to $p-\#E^{ns}(F_p)$ for bad primes, if $p||N$ then the reduction is multiplicative so $a_p=1$ or $-1$, if $p^2|N$ then the reduction is addictive so $a_p=0$ – zzy Jan 10 at 22:23
• @reuns I find such relations by playing with datas in LMFDB, for example if you check this new form lmfdb.org/ModularForm/GL2/Q/holomorphic/88/2/1/b of level 88, then $a_8=0, a_11=-1$. But this modular form is not with rational coefficients. – zzy Jan 10 at 22:30

Yes, this is a standard property of modular forms which you can prove "by hand" using a computation with double cosets; suitably stated, the property holds for all weights, including weight 1 (whereas modular abelian varieties are a weight 2 thing).

In the $$p^2 \mid N$$ case, the idea is to check that the double coset $$\Gamma_0(N) \begin{pmatrix} 1 & 0 \\ 0 & p \end{pmatrix} \Gamma_0(N)$$ giving the Hecke operator $$U_p$$ is actually stable under right-multiplication by $$\Gamma_0(N/p)$$, so if $$f \in S_k(N)$$, then $$f \mid_k U_p$$ is actually in $$S_k(N/p)$$. However, $$f \mid_k U_p$$ also has the same Hecke eigenvalues away from $$p$$ as $$f$$, so if $$f$$ is new of level $$N$$, it had better be 0.

For $$p \mid\mid N$$ this needs to be modified slightly because the index of $$\Gamma_0(N)$$ in $$\Gamma_0(N/p)$$ is $$p+1$$ instead of $$p$$. This gives an extra term in the formula, and you end up deducing that $$f\mid_k U_p + f \mid_k w_p$$ is zero, where $$w_p$$ is the Atkin-Lehner operator. Since $$w_p$$ is an involution, its eigenvalue had better be $$\pm 1$$ and this gives the result. (For general weights $$k$$ there is an extra normalisation factor coming out here, and you get that the $$U_p$$ eigenvalue is $$\pm p^{(k-2)/2}$$ instead.)

For newforms of $$\Gamma_1(N)$$ levels instead of $$\Gamma_0(N)$$ levels, there is a slightly more complicated statement, where you have to keep track of the powers of $$p$$ dividing both $$N$$ and the conductor of the character of $$f$$.

• Thank you! Is there an explanation using the corresponding Abelian variety? – zzy Jan 11 at 15:45