Construction of eigenfunction for Hecke operators $T_p$
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I am reading Theorem 2 from Atkin-Lehner's Hecke operators on $Gamma_0(m)$. Let $u=(u_1(tau),u_2(tau),..,u_n(tau))$ be an orthonormal basis for the space of cusp forms on $Gamma_0(m)$ with weight $k$. Let $A_p$ denote the matrix associated with each Hecke operator $T_p$ with respect to $u$. Then we take $A$ to be a unitary matrix which can simultaneously diagonalise all $A_p.$
Then how does it follow that each element of the new basis $f=Au$ is an eigenfunction of all the operators $T_p$?
My attempt: Need $T_p(f_i) = lambda_pf_i$ for all $p$. Since $A$ can simultaneously diagonalise all $A_p,$ I have $A_p = ADA^{-1}$ ($D$ is diagonal). Then, right multiply with $Au$ leading to $A_p(Au) = ADA^{-1}Au, = ADu. $ I can't proceed further.
Will this construction of eigenforms work for any set of operators?
Let me know if any more clarifications are needed.
modular-forms
asked Nov 19 at 16:36
1.414212
234
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I am reading Theorem 2 from Atkin-Lehner's Hecke operators on $Gamma_0(m)$. Let $u=(u_1(tau),u_2(tau),..,u_n(tau))$ be an orthonormal basis for the space of cusp forms on $Gamma_0(m)$ with weight $k$. Let $A_p$ denote the matrix associated with each Hecke operator $T_p$ with respect to $u$. Then we take $A$ to be a unitary matrix which can simultaneously diagonalise all $A_p.$
Then how does it follow that each element of the new basis $f=Au$ is an eigenfunction of all the operators $T_p$?
My attempt: Need $T_p(f_i) = lambda_pf_i$ for all $p$. Since $A$ can simultaneously diagonalise all $A_p,$ I have $A_p = ADA^{-1}$ ($D$ is diagonal). Then, right multiply with $Au$ leading to $A_p(Au) = ADA^{-1}Au, = ADu. $ I can't proceed further.
Will this construction of eigenforms work for any set of operators?
Let me know if any more clarifications are needed.
modular-forms
asked Nov 19 at 16:36
1.414212
234
It is because the $T_p$ commute and $M_k(Gamma_0(m))$ is finite dimensional, so there exists a basis $U$ where they are all in Jordan normal form, and they are normal for the Peterson inner product, so they are diagonal in $U$ and $U$ is unitary for that inner product. Can you copy the article ?
– reuns
Nov 19 at 16:46
@reuns No I could not. I seem to miss a point somewhere. I will try with this insight and come back.
– 1.414212
Nov 19 at 16:49
@reuns Let $f=Au$ with $f=f_1,f_2,ldots$ and $u=u_1,u_2,ldots.$ Then $ f|T_p = Au | T_p = A(A_pu) = (AA_pA^T)f = diag(lambda_1(p),lambda_2(p),ldots)f. $ Is this true?
– 1.414212
Nov 23 at 7:10
add a comment |
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up vote
0
down vote
favorite
I am reading Theorem 2 from Atkin-Lehner's Hecke operators on $Gamma_0(m)$. Let $u=(u_1(tau),u_2(tau),..,u_n(tau))$ be an orthonormal basis for the space of cusp forms on $Gamma_0(m)$ with weight $k$. Let $A_p$ denote the matrix associated with each Hecke operator $T_p$ with respect to $u$. Then we take $A$ to be a unitary matrix which can simultaneously diagonalise all $A_p.$
Then how does it follow that each element of the new basis $f=Au$ is an eigenfunction of all the operators $T_p$?
My attempt: Need $T_p(f_i) = lambda_pf_i$ for all $p$. Since $A$ can simultaneously diagonalise all $A_p,$ I have $A_p = ADA^{-1}$ ($D$ is diagonal). Then, right multiply with $Au$ leading to $A_p(Au) = ADA^{-1}Au, = ADu. $ I can't proceed further.
Will this construction of eigenforms work for any set of operators?
Let me know if any more clarifications are needed.
modular-forms
asked Nov 19 at 16:36
1.414212
234
I am reading Theorem 2 from Atkin-Lehner's Hecke operators on $Gamma_0(m)$. Let $u=(u_1(tau),u_2(tau),..,u_n(tau))$ be an orthonormal basis for the space of cusp forms on $Gamma_0(m)$ with weight $k$. Let $A_p$ denote the matrix associated with each Hecke operator $T_p$ with respect to $u$. Then we take $A$ to be a unitary matrix which can simultaneously diagonalise all $A_p.$
Then how does it follow that each element of the new basis $f=Au$ is an eigenfunction of all the operators $T_p$?
My attempt: Need $T_p(f_i) = lambda_pf_i$ for all $p$. Since $A$ can simultaneously diagonalise all $A_p,$ I have $A_p = ADA^{-1}$ ($D$ is diagonal). Then, right multiply with $Au$ leading to $A_p(Au) = ADA^{-1}Au, = ADu. $ I can't proceed further.
Will this construction of eigenforms work for any set of operators?
Let me know if any more clarifications are needed.
modular-forms
modular-forms
asked Nov 19 at 16:36
1.414212
234
asked Nov 19 at 16:36
1.414212
234
asked Nov 19 at 16:36
1.414212
234
asked Nov 19 at 16:36
1.414212
234
asked Nov 19 at 16:36
1.414212
234
234
It is because the $T_p$ commute and $M_k(Gamma_0(m))$ is finite dimensional, so there exists a basis $U$ where they are all in Jordan normal form, and they are normal for the Peterson inner product, so they are diagonal in $U$ and $U$ is unitary for that inner product. Can you copy the article ?
– reuns
Nov 19 at 16:46
@reuns No I could not. I seem to miss a point somewhere. I will try with this insight and come back.
– 1.414212
Nov 19 at 16:49
@reuns Let $f=Au$ with $f=f_1,f_2,ldots$ and $u=u_1,u_2,ldots.$ Then $ f|T_p = Au | T_p = A(A_pu) = (AA_pA^T)f = diag(lambda_1(p),lambda_2(p),ldots)f. $ Is this true?
– 1.414212
Nov 23 at 7:10
add a comment |
It is because the $T_p$ commute and $M_k(Gamma_0(m))$ is finite dimensional, so there exists a basis $U$ where they are all in Jordan normal form, and they are normal for the Peterson inner product, so they are diagonal in $U$ and $U$ is unitary for that inner product. Can you copy the article ?
– reuns
Nov 19 at 16:46
@reuns No I could not. I seem to miss a point somewhere. I will try with this insight and come back.
– 1.414212
Nov 19 at 16:49
@reuns Let $f=Au$ with $f=f_1,f_2,ldots$ and $u=u_1,u_2,ldots.$ Then $ f|T_p = Au | T_p = A(A_pu) = (AA_pA^T)f = diag(lambda_1(p),lambda_2(p),ldots)f. $ Is this true?
– 1.414212
Nov 23 at 7:10
It is because the $T_p$ commute and $M_k(Gamma_0(m))$ is finite dimensional, so there exists a basis $U$ where they are all in Jordan normal form, and they are normal for the Peterson inner product, so they are diagonal in $U$ and $U$ is unitary for that inner product. Can you copy the article ?
– reuns
Nov 19 at 16:46
It is because the $T_p$ commute and $M_k(Gamma_0(m))$ is finite dimensional, so there exists a basis $U$ where they are all in Jordan normal form, and they are normal for the Peterson inner product, so they are diagonal in $U$ and $U$ is unitary for that inner product. Can you copy the article ?
– reuns
Nov 19 at 16:46
@reuns No I could not. I seem to miss a point somewhere. I will try with this insight and come back.
– 1.414212
Nov 19 at 16:49
@reuns No I could not. I seem to miss a point somewhere. I will try with this insight and come back.
– 1.414212
Nov 19 at 16:49
@reuns Let $f=Au$ with $f=f_1,f_2,ldots$ and $u=u_1,u_2,ldots.$ Then $ f|T_p = Au | T_p = A(A_pu) = (AA_pA^T)f = diag(lambda_1(p),lambda_2(p),ldots)f. $ Is this true?
– 1.414212
Nov 23 at 7:10
@reuns Let $f=Au$ with $f=f_1,f_2,ldots$ and $u=u_1,u_2,ldots.$ Then $ f|T_p = Au | T_p = A(A_pu) = (AA_pA^T)f = diag(lambda_1(p),lambda_2(p),ldots)f. $ Is this true?
– 1.414212
Nov 23 at 7:10
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It is because the $T_p$ commute and $M_k(Gamma_0(m))$ is finite dimensional, so there exists a basis $U$ where they are all in Jordan normal form, and they are normal for the Peterson inner product, so they are diagonal in $U$ and $U$ is unitary for that inner product. Can you copy the article ?
– reuns
Nov 19 at 16:46
@reuns No I could not. I seem to miss a point somewhere. I will try with this insight and come back.
– 1.414212
Nov 19 at 16:49
@reuns Let $f=Au$ with $f=f_1,f_2,ldots$ and $u=u_1,u_2,ldots.$ Then $ f|T_p = Au | T_p = A(A_pu) = (AA_pA^T)f = diag(lambda_1(p),lambda_2(p),ldots)f. $ Is this true?
– 1.414212
Nov 23 at 7:10