Show that $(f|M)(z) := (cz+d)^{-r/2}f(Mz)$ has a weight $r/2$.
0
$begingroup$
The following text is from Complex Analysis by Freitag :
For $r ∈ mathbb{Z}$ the modified Petersson notation is defined : $$(f|M)(z) := sqrt{cz+d}^{-r}f(Mz)$$ for $M ∈ SL(2, mathbb{Z})$.
In the theorem above (Proposition VI.5.11), f is a modular form of weight $r/2$ which means $f(Mz)=(cz+d)^{r/2}f(z)$ so, all $(f|M)(z)$'s must have weight $0$ because $(f|M)(z) = (cz+d)^{-r/2}f(Mz)=(cz+d)^{-r/2}(cz+d)^{r/2}f(z) =f(z).$ So how $(f|M)(z)$ has weight $r/2$ or/and consequently $F = Pi (f|M)$ weight $kr/2$?
complex-numbers proof-explanation modular-forms
asked Nov 24 '18 at 1:49
72D72D
547116
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show 1 more comment
0
$begingroup$
The following text is from Complex Analysis by Freitag :
For $r ∈ mathbb{Z}$ the modified Petersson notation is defined : $$(f|M)(z) := sqrt{cz+d}^{-r}f(Mz)$$ for $M ∈ SL(2, mathbb{Z})$.
In the theorem above (Proposition VI.5.11), f is a modular form of weight $r/2$ which means $f(Mz)=(cz+d)^{r/2}f(z)$ so, all $(f|M)(z)$'s must have weight $0$ because $(f|M)(z) = (cz+d)^{-r/2}f(Mz)=(cz+d)^{-r/2}(cz+d)^{r/2}f(z) =f(z).$ So how $(f|M)(z)$ has weight $r/2$ or/and consequently $F = Pi (f|M)$ weight $kr/2$?
complex-numbers proof-explanation modular-forms
asked Nov 24 '18 at 1:49
72D72D
547116
$endgroup$
$begingroup$
No. It meant $f^2$ is a modular form of integer weight $r$ and $F^2$ is a modular form of weight $kr$. If $r < 0$ then $lim_{Im(z) to 0} F(z)^2 = 0$ so (Schwartz reflection principle) $F^2$ is entire and bounded thus constant
$endgroup$
– reuns
Nov 24 '18 at 3:21
$begingroup$
@reuns, $F$ is not a product of $f$'s; it's a product of $f|M$'s.
$endgroup$
– 72D
Nov 24 '18 at 11:01
$begingroup$
If $f in M_k(Gamma)$ then $f|gamma in M_k(gamma^{-1}Gamma gamma)$ (when there are characters and non-integer weights it needs a little more but the idea is there)
$endgroup$
– reuns
Nov 25 '18 at 0:24
$begingroup$
@reuns, Sorry I couldn't understand! Can we analyse my argument in OP so I can know where am I doing wrong, please? I am studying from Freitag's book bc it's the easiest book I've found in MF and I am first year UG student so I will truly appreciate easy explanation.
$endgroup$
– 72D
Nov 25 '18 at 0:38
$begingroup$
$f in M_k(Gamma,chi)$ means for every $alpha in Gamma$ then $f(alpha(z)) = pmchi(alpha) (cz+d)^k f(z)$ where $(cz+d)^k = alpha'(z)^{-k/2}$. With $f_2(z) = f(gamma (z))$ then $f_2(alpha(z)) = pm chi(gammaalphagamma^{-1}) (cz+d)^k f(z)$ for $alpha in gamma^{-1}Gamma gamma$.
$endgroup$
– reuns
Nov 25 '18 at 0:44
|
show 1 more comment
0
0
0
0
$begingroup$
The following text is from Complex Analysis by Freitag :
For $r ∈ mathbb{Z}$ the modified Petersson notation is defined : $$(f|M)(z) := sqrt{cz+d}^{-r}f(Mz)$$ for $M ∈ SL(2, mathbb{Z})$.
In the theorem above (Proposition VI.5.11), f is a modular form of weight $r/2$ which means $f(Mz)=(cz+d)^{r/2}f(z)$ so, all $(f|M)(z)$'s must have weight $0$ because $(f|M)(z) = (cz+d)^{-r/2}f(Mz)=(cz+d)^{-r/2}(cz+d)^{r/2}f(z) =f(z).$ So how $(f|M)(z)$ has weight $r/2$ or/and consequently $F = Pi (f|M)$ weight $kr/2$?
complex-numbers proof-explanation modular-forms
asked Nov 24 '18 at 1:49
72D72D
547116
$endgroup$
The following text is from Complex Analysis by Freitag :
For $r ∈ mathbb{Z}$ the modified Petersson notation is defined : $$(f|M)(z) := sqrt{cz+d}^{-r}f(Mz)$$ for $M ∈ SL(2, mathbb{Z})$.
In the theorem above (Proposition VI.5.11), f is a modular form of weight $r/2$ which means $f(Mz)=(cz+d)^{r/2}f(z)$ so, all $(f|M)(z)$'s must have weight $0$ because $(f|M)(z) = (cz+d)^{-r/2}f(Mz)=(cz+d)^{-r/2}(cz+d)^{r/2}f(z) =f(z).$ So how $(f|M)(z)$ has weight $r/2$ or/and consequently $F = Pi (f|M)$ weight $kr/2$?
complex-numbers proof-explanation modular-forms
complex-numbers proof-explanation modular-forms
asked Nov 24 '18 at 1:49
72D72D
547116
asked Nov 24 '18 at 1:49
72D72D
547116
edited Nov 24 '18 at 11:21
72D
asked Nov 24 '18 at 1:49
72D72D
547116
asked Nov 24 '18 at 1:49
72D72D
547116
asked Nov 24 '18 at 1:49
72D72D
547116
547116
$begingroup$
No. It meant $f^2$ is a modular form of integer weight $r$ and $F^2$ is a modular form of weight $kr$. If $r < 0$ then $lim_{Im(z) to 0} F(z)^2 = 0$ so (Schwartz reflection principle) $F^2$ is entire and bounded thus constant
$endgroup$
– reuns
Nov 24 '18 at 3:21
$begingroup$
@reuns, $F$ is not a product of $f$'s; it's a product of $f|M$'s.
$endgroup$
– 72D
Nov 24 '18 at 11:01
$begingroup$
If $f in M_k(Gamma)$ then $f|gamma in M_k(gamma^{-1}Gamma gamma)$ (when there are characters and non-integer weights it needs a little more but the idea is there)
$endgroup$
– reuns
Nov 25 '18 at 0:24
$begingroup$
@reuns, Sorry I couldn't understand! Can we analyse my argument in OP so I can know where am I doing wrong, please? I am studying from Freitag's book bc it's the easiest book I've found in MF and I am first year UG student so I will truly appreciate easy explanation.
$endgroup$
– 72D
Nov 25 '18 at 0:38
$begingroup$
$f in M_k(Gamma,chi)$ means for every $alpha in Gamma$ then $f(alpha(z)) = pmchi(alpha) (cz+d)^k f(z)$ where $(cz+d)^k = alpha'(z)^{-k/2}$. With $f_2(z) = f(gamma (z))$ then $f_2(alpha(z)) = pm chi(gammaalphagamma^{-1}) (cz+d)^k f(z)$ for $alpha in gamma^{-1}Gamma gamma$.
$endgroup$
– reuns
Nov 25 '18 at 0:44
|
show 1 more comment
$begingroup$
No. It meant $f^2$ is a modular form of integer weight $r$ and $F^2$ is a modular form of weight $kr$. If $r < 0$ then $lim_{Im(z) to 0} F(z)^2 = 0$ so (Schwartz reflection principle) $F^2$ is entire and bounded thus constant
$endgroup$
– reuns
Nov 24 '18 at 3:21
$begingroup$
@reuns, $F$ is not a product of $f$'s; it's a product of $f|M$'s.
$endgroup$
– 72D
Nov 24 '18 at 11:01
$begingroup$
If $f in M_k(Gamma)$ then $f|gamma in M_k(gamma^{-1}Gamma gamma)$ (when there are characters and non-integer weights it needs a little more but the idea is there)
$endgroup$
– reuns
Nov 25 '18 at 0:24
$begingroup$
@reuns, Sorry I couldn't understand! Can we analyse my argument in OP so I can know where am I doing wrong, please? I am studying from Freitag's book bc it's the easiest book I've found in MF and I am first year UG student so I will truly appreciate easy explanation.
$endgroup$
– 72D
Nov 25 '18 at 0:38
$begingroup$
$f in M_k(Gamma,chi)$ means for every $alpha in Gamma$ then $f(alpha(z)) = pmchi(alpha) (cz+d)^k f(z)$ where $(cz+d)^k = alpha'(z)^{-k/2}$. With $f_2(z) = f(gamma (z))$ then $f_2(alpha(z)) = pm chi(gammaalphagamma^{-1}) (cz+d)^k f(z)$ for $alpha in gamma^{-1}Gamma gamma$.
$endgroup$
– reuns
Nov 25 '18 at 0:44
$begingroup$
No. It meant $f^2$ is a modular form of integer weight $r$ and $F^2$ is a modular form of weight $kr$. If $r < 0$ then $lim_{Im(z) to 0} F(z)^2 = 0$ so (Schwartz reflection principle) $F^2$ is entire and bounded thus constant
$endgroup$
– reuns
Nov 24 '18 at 3:21
$begingroup$
No. It meant $f^2$ is a modular form of integer weight $r$ and $F^2$ is a modular form of weight $kr$. If $r < 0$ then $lim_{Im(z) to 0} F(z)^2 = 0$ so (Schwartz reflection principle) $F^2$ is entire and bounded thus constant
$endgroup$
– reuns
Nov 24 '18 at 3:21
$begingroup$
@reuns, $F$ is not a product of $f$'s; it's a product of $f|M$'s.
$endgroup$
– 72D
Nov 24 '18 at 11:01
$begingroup$
@reuns, $F$ is not a product of $f$'s; it's a product of $f|M$'s.
$endgroup$
– 72D
Nov 24 '18 at 11:01
$begingroup$
If $f in M_k(Gamma)$ then $f|gamma in M_k(gamma^{-1}Gamma gamma)$ (when there are characters and non-integer weights it needs a little more but the idea is there)
$endgroup$
– reuns
Nov 25 '18 at 0:24
$begingroup$
If $f in M_k(Gamma)$ then $f|gamma in M_k(gamma^{-1}Gamma gamma)$ (when there are characters and non-integer weights it needs a little more but the idea is there)
$endgroup$
– reuns
Nov 25 '18 at 0:24
$begingroup$
@reuns, Sorry I couldn't understand! Can we analyse my argument in OP so I can know where am I doing wrong, please? I am studying from Freitag's book bc it's the easiest book I've found in MF and I am first year UG student so I will truly appreciate easy explanation.
$endgroup$
– 72D
Nov 25 '18 at 0:38
$begingroup$
@reuns, Sorry I couldn't understand! Can we analyse my argument in OP so I can know where am I doing wrong, please? I am studying from Freitag's book bc it's the easiest book I've found in MF and I am first year UG student so I will truly appreciate easy explanation.
$endgroup$
– 72D
Nov 25 '18 at 0:38
$begingroup$
$f in M_k(Gamma,chi)$ means for every $alpha in Gamma$ then $f(alpha(z)) = pmchi(alpha) (cz+d)^k f(z)$ where $(cz+d)^k = alpha'(z)^{-k/2}$. With $f_2(z) = f(gamma (z))$ then $f_2(alpha(z)) = pm chi(gammaalphagamma^{-1}) (cz+d)^k f(z)$ for $alpha in gamma^{-1}Gamma gamma$.
$endgroup$
– reuns
Nov 25 '18 at 0:44
$begingroup$
$f in M_k(Gamma,chi)$ means for every $alpha in Gamma$ then $f(alpha(z)) = pmchi(alpha) (cz+d)^k f(z)$ where $(cz+d)^k = alpha'(z)^{-k/2}$. With $f_2(z) = f(gamma (z))$ then $f_2(alpha(z)) = pm chi(gammaalphagamma^{-1}) (cz+d)^k f(z)$ for $alpha in gamma^{-1}Gamma gamma$.
$endgroup$
– reuns
Nov 25 '18 at 0:44
|
show 1 more comment
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$begingroup$
No. It meant $f^2$ is a modular form of integer weight $r$ and $F^2$ is a modular form of weight $kr$. If $r < 0$ then $lim_{Im(z) to 0} F(z)^2 = 0$ so (Schwartz reflection principle) $F^2$ is entire and bounded thus constant
$endgroup$
– reuns
Nov 24 '18 at 3:21
$begingroup$
@reuns, $F$ is not a product of $f$'s; it's a product of $f|M$'s.
$endgroup$
– 72D
Nov 24 '18 at 11:01
$begingroup$
If $f in M_k(Gamma)$ then $f|gamma in M_k(gamma^{-1}Gamma gamma)$ (when there are characters and non-integer weights it needs a little more but the idea is there)
$endgroup$
– reuns
Nov 25 '18 at 0:24
$begingroup$
@reuns, Sorry I couldn't understand! Can we analyse my argument in OP so I can know where am I doing wrong, please? I am studying from Freitag's book bc it's the easiest book I've found in MF and I am first year UG student so I will truly appreciate easy explanation.
$endgroup$
– 72D
Nov 25 '18 at 0:38
$begingroup$
$f in M_k(Gamma,chi)$ means for every $alpha in Gamma$ then $f(alpha(z)) = pmchi(alpha) (cz+d)^k f(z)$ where $(cz+d)^k = alpha'(z)^{-k/2}$. With $f_2(z) = f(gamma (z))$ then $f_2(alpha(z)) = pm chi(gammaalphagamma^{-1}) (cz+d)^k f(z)$ for $alpha in gamma^{-1}Gamma gamma$.
$endgroup$
– reuns
Nov 25 '18 at 0:44