The Dual Maass form for $SL(3,mathbb{Z})$
$begingroup$
Let $phi(z)$ be a Maass form of type $(v_1,v_2)in mathbb{C}^2$ for $SL(3,mathbb{Z})$. Then the dual Maass form $$ tilde{phi}(z):= phi(w.(z^{-1})^{intercal}.w),,,,,,,,,,,,w=begin{pmatrix}
& & 1 \
& -1& \
1 &
end{pmatrix}$$ is a Maass form of type $(v_2,v_1)$ .
I am having difficulty showing the part $$Delta_1tilde{phi}=lambda_1tilde{phi}\Delta_2tilde{phi}=lambda_2tilde{phi} $$ where $Delta_1, Delta_2$ are the generators of the centre of the ring of differential operators and $lambda_1$ and $lambda_2$ are the eigenvalues of $Delta_1$ and $Delta_2$ respectively corresponding to the eigenfunction $I_{v_2,v_1}(z)$ .
Note that if
$$z=begin{pmatrix}
1 & x_2 & x_3\
& 1 & x_1\
& & 1
end{pmatrix}.begin{pmatrix}
y_1y_2 & & \
& y_1 & \
& & 1
end{pmatrix}$$ then
$$w.(z^{-1})^{intercal}.w=begin{pmatrix}
1 & x_1 & x_1x_2-x_3\
& 1 & x_2\
& & 1
end{pmatrix}.begin{pmatrix}
y_2y_1 & & \
& y_2 & \
& & 1
end{pmatrix}$$ and hence we have
$$I_{v_1,v_2}(z)={y_1}^{v_1+2v_2}{y_2}^{2v_1+v_2}=I_{v_2,v_1}(w.(z^{-1})^{intercal}.w)$$ I am not able to see the 'chain rule' the text is then suggesting to use for the conclusion.
modular-forms automorphic-forms
asked Dec 5 '18 at 9:39
pkspks
18411
$endgroup$
add a comment |
$begingroup$
Let $phi(z)$ be a Maass form of type $(v_1,v_2)in mathbb{C}^2$ for $SL(3,mathbb{Z})$. Then the dual Maass form $$ tilde{phi}(z):= phi(w.(z^{-1})^{intercal}.w),,,,,,,,,,,,w=begin{pmatrix}
& & 1 \
& -1& \
1 &
end{pmatrix}$$ is a Maass form of type $(v_2,v_1)$ .
I am having difficulty showing the part $$Delta_1tilde{phi}=lambda_1tilde{phi}\Delta_2tilde{phi}=lambda_2tilde{phi} $$ where $Delta_1, Delta_2$ are the generators of the centre of the ring of differential operators and $lambda_1$ and $lambda_2$ are the eigenvalues of $Delta_1$ and $Delta_2$ respectively corresponding to the eigenfunction $I_{v_2,v_1}(z)$ .
Note that if
$$z=begin{pmatrix}
1 & x_2 & x_3\
& 1 & x_1\
& & 1
end{pmatrix}.begin{pmatrix}
y_1y_2 & & \
& y_1 & \
& & 1
end{pmatrix}$$ then
$$w.(z^{-1})^{intercal}.w=begin{pmatrix}
1 & x_1 & x_1x_2-x_3\
& 1 & x_2\
& & 1
end{pmatrix}.begin{pmatrix}
y_2y_1 & & \
& y_2 & \
& & 1
end{pmatrix}$$ and hence we have
$$I_{v_1,v_2}(z)={y_1}^{v_1+2v_2}{y_2}^{2v_1+v_2}=I_{v_2,v_1}(w.(z^{-1})^{intercal}.w)$$ I am not able to see the 'chain rule' the text is then suggesting to use for the conclusion.
modular-forms automorphic-forms
asked Dec 5 '18 at 9:39
pkspks
18411
$endgroup$
1
$begingroup$
Isn't $S[f](z) = T [f(rho(z))](rho^{-1}(z))$ a differential operator when $T$ is ? (here $rho(z) = w z^{-top} w$) Since $T mapsto S$ is an isomorphism of the ring of differential operators, if $T$ is in center then so is $S$. So at least you know $phi(rho(z))$ is a Maass form when $phi(z)$ is ?
$endgroup$
– reuns
Dec 5 '18 at 10:51
$begingroup$
Yes got it. The point is S is also in the centre and hence is a polynomial in $Delta_1$ and $Delta_2$ The eigenvalues comes out from applying S to the last equation i wrote. Thanks
$endgroup$
– pks
Dec 5 '18 at 10:56
add a comment |
$begingroup$
Let $phi(z)$ be a Maass form of type $(v_1,v_2)in mathbb{C}^2$ for $SL(3,mathbb{Z})$. Then the dual Maass form $$ tilde{phi}(z):= phi(w.(z^{-1})^{intercal}.w),,,,,,,,,,,,w=begin{pmatrix}
& & 1 \
& -1& \
1 &
end{pmatrix}$$ is a Maass form of type $(v_2,v_1)$ .
I am having difficulty showing the part $$Delta_1tilde{phi}=lambda_1tilde{phi}\Delta_2tilde{phi}=lambda_2tilde{phi} $$ where $Delta_1, Delta_2$ are the generators of the centre of the ring of differential operators and $lambda_1$ and $lambda_2$ are the eigenvalues of $Delta_1$ and $Delta_2$ respectively corresponding to the eigenfunction $I_{v_2,v_1}(z)$ .
Note that if
$$z=begin{pmatrix}
1 & x_2 & x_3\
& 1 & x_1\
& & 1
end{pmatrix}.begin{pmatrix}
y_1y_2 & & \
& y_1 & \
& & 1
end{pmatrix}$$ then
$$w.(z^{-1})^{intercal}.w=begin{pmatrix}
1 & x_1 & x_1x_2-x_3\
& 1 & x_2\
& & 1
end{pmatrix}.begin{pmatrix}
y_2y_1 & & \
& y_2 & \
& & 1
end{pmatrix}$$ and hence we have
$$I_{v_1,v_2}(z)={y_1}^{v_1+2v_2}{y_2}^{2v_1+v_2}=I_{v_2,v_1}(w.(z^{-1})^{intercal}.w)$$ I am not able to see the 'chain rule' the text is then suggesting to use for the conclusion.
modular-forms automorphic-forms
asked Dec 5 '18 at 9:39
pkspks
18411
$endgroup$
Let $phi(z)$ be a Maass form of type $(v_1,v_2)in mathbb{C}^2$ for $SL(3,mathbb{Z})$. Then the dual Maass form $$ tilde{phi}(z):= phi(w.(z^{-1})^{intercal}.w),,,,,,,,,,,,w=begin{pmatrix}
& & 1 \
& -1& \
1 &
end{pmatrix}$$ is a Maass form of type $(v_2,v_1)$ .
I am having difficulty showing the part $$Delta_1tilde{phi}=lambda_1tilde{phi}\Delta_2tilde{phi}=lambda_2tilde{phi} $$ where $Delta_1, Delta_2$ are the generators of the centre of the ring of differential operators and $lambda_1$ and $lambda_2$ are the eigenvalues of $Delta_1$ and $Delta_2$ respectively corresponding to the eigenfunction $I_{v_2,v_1}(z)$ .
Note that if
$$z=begin{pmatrix}
1 & x_2 & x_3\
& 1 & x_1\
& & 1
end{pmatrix}.begin{pmatrix}
y_1y_2 & & \
& y_1 & \
& & 1
end{pmatrix}$$ then
$$w.(z^{-1})^{intercal}.w=begin{pmatrix}
1 & x_1 & x_1x_2-x_3\
& 1 & x_2\
& & 1
end{pmatrix}.begin{pmatrix}
y_2y_1 & & \
& y_2 & \
& & 1
end{pmatrix}$$ and hence we have
$$I_{v_1,v_2}(z)={y_1}^{v_1+2v_2}{y_2}^{2v_1+v_2}=I_{v_2,v_1}(w.(z^{-1})^{intercal}.w)$$ I am not able to see the 'chain rule' the text is then suggesting to use for the conclusion.
modular-forms automorphic-forms
modular-forms automorphic-forms
asked Dec 5 '18 at 9:39
pkspks
18411
asked Dec 5 '18 at 9:39
pkspks
18411
asked Dec 5 '18 at 9:39
pkspks
18411
asked Dec 5 '18 at 9:39
pkspks
18411
asked Dec 5 '18 at 9:39
pkspks
18411
18411
1
$begingroup$
Isn't $S[f](z) = T [f(rho(z))](rho^{-1}(z))$ a differential operator when $T$ is ? (here $rho(z) = w z^{-top} w$) Since $T mapsto S$ is an isomorphism of the ring of differential operators, if $T$ is in center then so is $S$. So at least you know $phi(rho(z))$ is a Maass form when $phi(z)$ is ?
$endgroup$
– reuns
Dec 5 '18 at 10:51
$begingroup$
Yes got it. The point is S is also in the centre and hence is a polynomial in $Delta_1$ and $Delta_2$ The eigenvalues comes out from applying S to the last equation i wrote. Thanks
$endgroup$
– pks
Dec 5 '18 at 10:56
add a comment |
1
$begingroup$
Isn't $S[f](z) = T [f(rho(z))](rho^{-1}(z))$ a differential operator when $T$ is ? (here $rho(z) = w z^{-top} w$) Since $T mapsto S$ is an isomorphism of the ring of differential operators, if $T$ is in center then so is $S$. So at least you know $phi(rho(z))$ is a Maass form when $phi(z)$ is ?
$endgroup$
– reuns
Dec 5 '18 at 10:51
$begingroup$
Yes got it. The point is S is also in the centre and hence is a polynomial in $Delta_1$ and $Delta_2$ The eigenvalues comes out from applying S to the last equation i wrote. Thanks
$endgroup$
– pks
Dec 5 '18 at 10:56
1
1
$begingroup$
Isn't $S[f](z) = T [f(rho(z))](rho^{-1}(z))$ a differential operator when $T$ is ? (here $rho(z) = w z^{-top} w$) Since $T mapsto S$ is an isomorphism of the ring of differential operators, if $T$ is in center then so is $S$. So at least you know $phi(rho(z))$ is a Maass form when $phi(z)$ is ?
$endgroup$
– reuns
Dec 5 '18 at 10:51
$begingroup$
Isn't $S[f](z) = T [f(rho(z))](rho^{-1}(z))$ a differential operator when $T$ is ? (here $rho(z) = w z^{-top} w$) Since $T mapsto S$ is an isomorphism of the ring of differential operators, if $T$ is in center then so is $S$. So at least you know $phi(rho(z))$ is a Maass form when $phi(z)$ is ?
$endgroup$
– reuns
Dec 5 '18 at 10:51
$begingroup$
Yes got it. The point is S is also in the centre and hence is a polynomial in $Delta_1$ and $Delta_2$ The eigenvalues comes out from applying S to the last equation i wrote. Thanks
$endgroup$
– pks
Dec 5 '18 at 10:56
$begingroup$
Yes got it. The point is S is also in the centre and hence is a polynomial in $Delta_1$ and $Delta_2$ The eigenvalues comes out from applying S to the last equation i wrote. Thanks
$endgroup$
– pks
Dec 5 '18 at 10:56
add a comment |
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1
$begingroup$
Isn't $S[f](z) = T [f(rho(z))](rho^{-1}(z))$ a differential operator when $T$ is ? (here $rho(z) = w z^{-top} w$) Since $T mapsto S$ is an isomorphism of the ring of differential operators, if $T$ is in center then so is $S$. So at least you know $phi(rho(z))$ is a Maass form when $phi(z)$ is ?
$endgroup$
– reuns
Dec 5 '18 at 10:51
$begingroup$
Yes got it. The point is S is also in the centre and hence is a polynomial in $Delta_1$ and $Delta_2$ The eigenvalues comes out from applying S to the last equation i wrote. Thanks
$endgroup$
– pks
Dec 5 '18 at 10:56