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












1












$begingroup$


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.






number-theory algebraic-geometry modular-forms arithmetic-geometry







share|cite|improve this question











$endgroup$












  • $begingroup$
    If f has rational coefficients, then what ?
    $endgroup$
    – reuns
    Jan 10 at 22:07










  • $begingroup$
    @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 $a_p=0$
    $endgroup$
    – zzy
    Jan 10 at 22:23








  • 1




    $begingroup$
    @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.
    $endgroup$
    – zzy
    Jan 10 at 22:30


















1












$begingroup$


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.






number-theory algebraic-geometry modular-forms arithmetic-geometry







share|cite|improve this question











$endgroup$












  • $begingroup$
    If f has rational coefficients, then what ?
    $endgroup$
    – reuns
    Jan 10 at 22:07










  • $begingroup$
    @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 $a_p=0$
    $endgroup$
    – zzy
    Jan 10 at 22:23








  • 1




    $begingroup$
    @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.
    $endgroup$
    – zzy
    Jan 10 at 22:30
















1












1








1


1



$begingroup$


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.






number-theory algebraic-geometry modular-forms arithmetic-geometry







share|cite|improve this question











$endgroup$




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.






number-theory algebraic-geometry modular-forms arithmetic-geometry




number-theory algebraic-geometry modular-forms arithmetic-geometry






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 11 at 15:46







zzy

















asked Jan 10 at 20:30









zzyzzy

2,4051419




2,4051419












  • $begingroup$
    If f has rational coefficients, then what ?
    $endgroup$
    – reuns
    Jan 10 at 22:07










  • $begingroup$
    @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 $a_p=0$
    $endgroup$
    – zzy
    Jan 10 at 22:23








  • 1




    $begingroup$
    @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.
    $endgroup$
    – zzy
    Jan 10 at 22:30




















  • $begingroup$
    If f has rational coefficients, then what ?
    $endgroup$
    – reuns
    Jan 10 at 22:07










  • $begingroup$
    @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 $a_p=0$
    $endgroup$
    – zzy
    Jan 10 at 22:23








  • 1




    $begingroup$
    @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.
    $endgroup$
    – zzy
    Jan 10 at 22:30


















$begingroup$
If f has rational coefficients, then what ?
$endgroup$
– reuns
Jan 10 at 22:07




$begingroup$
If f has rational coefficients, then what ?
$endgroup$
– reuns
Jan 10 at 22:07












$begingroup$
@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 $a_p=0$
$endgroup$
– zzy
Jan 10 at 22:23






$begingroup$
@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 $a_p=0$
$endgroup$
– zzy
Jan 10 at 22:23






1




1




$begingroup$
@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.
$endgroup$
– zzy
Jan 10 at 22:30






$begingroup$
@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.
$endgroup$
– zzy
Jan 10 at 22:30












1 Answer
1






active

oldest

votes


















2












$begingroup$

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 $p midmid N$ 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 $fmid_k U_p + f mid_k w_p$ 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 $f$.






share|cite|improve this answer









$endgroup$













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











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1 Answer
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active

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active

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2












$begingroup$

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 $p midmid N$ 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 $fmid_k U_p + f mid_k w_p$ 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 $f$.






share|cite|improve this answer









$endgroup$













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
















2












$begingroup$

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 $p midmid N$ 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 $fmid_k U_p + f mid_k w_p$ 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 $f$.






share|cite|improve this answer









$endgroup$













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














2












2








2





$begingroup$

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 $p midmid N$ 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 $fmid_k U_p + f mid_k w_p$ 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 $f$.






share|cite|improve this answer









$endgroup$



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 $p midmid N$ 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 $fmid_k U_p + f mid_k w_p$ 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 $f$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 11 at 9:00









David LoefflerDavid Loeffler

6,899923




6,899923












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


















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
















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




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


















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