Uniqueness of $L$-series of cusp forms
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For a cusp form $f$, one gets an $L$-series by taking the Mellin transform as we have
$$ tilde{f}(s) = (2pi)^{-s} Gamma(s) L(s,f). $$
My question is: is this operation injective? It seems to me that this should be the case and that one should be able to recover $f$ using the inverse transform, but I could not find anything on the subject.
I would also assume that if the $L$-series admits an Euler product, then it determines $f$ completely by determining its Fourier coefficients. Is that correct?
modular-forms mellin-transform l-functions
asked Nov 18 at 13:57
hrt
457310
add a comment |
up vote
0
down vote
favorite
For a cusp form $f$, one gets an $L$-series by taking the Mellin transform as we have
$$ tilde{f}(s) = (2pi)^{-s} Gamma(s) L(s,f). $$
My question is: is this operation injective? It seems to me that this should be the case and that one should be able to recover $f$ using the inverse transform, but I could not find anything on the subject.
I would also assume that if the $L$-series admits an Euler product, then it determines $f$ completely by determining its Fourier coefficients. Is that correct?
modular-forms mellin-transform l-functions
asked Nov 18 at 13:57
hrt
457310
1
$(2pi)^{-s} Gamma(s) L(s,f)$ is the Mellin transform of $f(iy)$, that is the Fourier transform of $f(i e^{-u}) e^{-sigma u}$, which is inversible. And $f(z)$ is the analytic continuation of $f(iy)$. For non-holomorphic modular forms it is a little less obvious. A theorem I don't fully understand is that any eigensystem for the Hecke algebra is a modular form (so that $f^sigma, sigma in Gal(overline{mathbb{Q}},mathbb{Q})$ is a modular form if $f$ is an eigenform)
– reuns
Nov 18 at 15:36
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
For a cusp form $f$, one gets an $L$-series by taking the Mellin transform as we have
$$ tilde{f}(s) = (2pi)^{-s} Gamma(s) L(s,f). $$
My question is: is this operation injective? It seems to me that this should be the case and that one should be able to recover $f$ using the inverse transform, but I could not find anything on the subject.
I would also assume that if the $L$-series admits an Euler product, then it determines $f$ completely by determining its Fourier coefficients. Is that correct?
modular-forms mellin-transform l-functions
asked Nov 18 at 13:57
hrt
457310
For a cusp form $f$, one gets an $L$-series by taking the Mellin transform as we have
$$ tilde{f}(s) = (2pi)^{-s} Gamma(s) L(s,f). $$
My question is: is this operation injective? It seems to me that this should be the case and that one should be able to recover $f$ using the inverse transform, but I could not find anything on the subject.
I would also assume that if the $L$-series admits an Euler product, then it determines $f$ completely by determining its Fourier coefficients. Is that correct?
modular-forms mellin-transform l-functions
modular-forms mellin-transform l-functions
asked Nov 18 at 13:57
hrt
457310
asked Nov 18 at 13:57
hrt
457310
asked Nov 18 at 13:57
hrt
457310
asked Nov 18 at 13:57
hrt
457310
asked Nov 18 at 13:57
hrt
457310
457310
1
$(2pi)^{-s} Gamma(s) L(s,f)$ is the Mellin transform of $f(iy)$, that is the Fourier transform of $f(i e^{-u}) e^{-sigma u}$, which is inversible. And $f(z)$ is the analytic continuation of $f(iy)$. For non-holomorphic modular forms it is a little less obvious. A theorem I don't fully understand is that any eigensystem for the Hecke algebra is a modular form (so that $f^sigma, sigma in Gal(overline{mathbb{Q}},mathbb{Q})$ is a modular form if $f$ is an eigenform)
– reuns
Nov 18 at 15:36
add a comment |
1
$(2pi)^{-s} Gamma(s) L(s,f)$ is the Mellin transform of $f(iy)$, that is the Fourier transform of $f(i e^{-u}) e^{-sigma u}$, which is inversible. And $f(z)$ is the analytic continuation of $f(iy)$. For non-holomorphic modular forms it is a little less obvious. A theorem I don't fully understand is that any eigensystem for the Hecke algebra is a modular form (so that $f^sigma, sigma in Gal(overline{mathbb{Q}},mathbb{Q})$ is a modular form if $f$ is an eigenform)
– reuns
Nov 18 at 15:36
1
1
$(2pi)^{-s} Gamma(s) L(s,f)$ is the Mellin transform of $f(iy)$, that is the Fourier transform of $f(i e^{-u}) e^{-sigma u}$, which is inversible. And $f(z)$ is the analytic continuation of $f(iy)$. For non-holomorphic modular forms it is a little less obvious. A theorem I don't fully understand is that any eigensystem for the Hecke algebra is a modular form (so that $f^sigma, sigma in Gal(overline{mathbb{Q}},mathbb{Q})$ is a modular form if $f$ is an eigenform)
– reuns
Nov 18 at 15:36
$(2pi)^{-s} Gamma(s) L(s,f)$ is the Mellin transform of $f(iy)$, that is the Fourier transform of $f(i e^{-u}) e^{-sigma u}$, which is inversible. And $f(z)$ is the analytic continuation of $f(iy)$. For non-holomorphic modular forms it is a little less obvious. A theorem I don't fully understand is that any eigensystem for the Hecke algebra is a modular form (so that $f^sigma, sigma in Gal(overline{mathbb{Q}},mathbb{Q})$ is a modular form if $f$ is an eigenform)
– reuns
Nov 18 at 15:36
add a comment |
1 Answer
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If $f$ and $g$ have the same Mellin transform, then $L(s,f)=L(s,g)$. These are two Dirichlet series, with coefficients being respectively the Fourier coefficients $a_n(f)$ and $a_n(g)$ of $f$ and $g$ at the cusp $i infty$.
By theorem 11.3 in Apostol, Introduction to Analytic Number Theory, one gets $a_n(f) = a_n(g)$ for every $n geq 1$. This implies that $f=g$ (having the same Laurent series).
answered Nov 18 at 15:33
Watson
15.8k92870
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
If $f$ and $g$ have the same Mellin transform, then $L(s,f)=L(s,g)$. These are two Dirichlet series, with coefficients being respectively the Fourier coefficients $a_n(f)$ and $a_n(g)$ of $f$ and $g$ at the cusp $i infty$.
By theorem 11.3 in Apostol, Introduction to Analytic Number Theory, one gets $a_n(f) = a_n(g)$ for every $n geq 1$. This implies that $f=g$ (having the same Laurent series).
answered Nov 18 at 15:33
Watson
15.8k92870
add a comment |
up vote
1
down vote
accepted
If $f$ and $g$ have the same Mellin transform, then $L(s,f)=L(s,g)$. These are two Dirichlet series, with coefficients being respectively the Fourier coefficients $a_n(f)$ and $a_n(g)$ of $f$ and $g$ at the cusp $i infty$.
By theorem 11.3 in Apostol, Introduction to Analytic Number Theory, one gets $a_n(f) = a_n(g)$ for every $n geq 1$. This implies that $f=g$ (having the same Laurent series).
answered Nov 18 at 15:33
Watson
15.8k92870
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
If $f$ and $g$ have the same Mellin transform, then $L(s,f)=L(s,g)$. These are two Dirichlet series, with coefficients being respectively the Fourier coefficients $a_n(f)$ and $a_n(g)$ of $f$ and $g$ at the cusp $i infty$.
By theorem 11.3 in Apostol, Introduction to Analytic Number Theory, one gets $a_n(f) = a_n(g)$ for every $n geq 1$. This implies that $f=g$ (having the same Laurent series).
answered Nov 18 at 15:33
Watson
15.8k92870
If $f$ and $g$ have the same Mellin transform, then $L(s,f)=L(s,g)$. These are two Dirichlet series, with coefficients being respectively the Fourier coefficients $a_n(f)$ and $a_n(g)$ of $f$ and $g$ at the cusp $i infty$.
By theorem 11.3 in Apostol, Introduction to Analytic Number Theory, one gets $a_n(f) = a_n(g)$ for every $n geq 1$. This implies that $f=g$ (having the same Laurent series).
answered Nov 18 at 15:33
Watson
15.8k92870
answered Nov 18 at 15:33
Watson
15.8k92870
answered Nov 18 at 15:33
Watson
15.8k92870
answered Nov 18 at 15:33
Watson
15.8k92870
15.8k92870
add a comment |
add a comment |
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$(2pi)^{-s} Gamma(s) L(s,f)$ is the Mellin transform of $f(iy)$, that is the Fourier transform of $f(i e^{-u}) e^{-sigma u}$, which is inversible. And $f(z)$ is the analytic continuation of $f(iy)$. For non-holomorphic modular forms it is a little less obvious. A theorem I don't fully understand is that any eigensystem for the Hecke algebra is a modular form (so that $f^sigma, sigma in Gal(overline{mathbb{Q}},mathbb{Q})$ is a modular form if $f$ is an eigenform)
– reuns
Nov 18 at 15:36