How is the bijection obtained?
$begingroup$
I am reading in Serge Lang's book " Introduction to Modular Forms " .
On page 8 there is written that there is a bijection between functions of lattices , homogenous of degree -k and functions g on H ( the upper half plane ) satisfying the condition $$ g(alpha(z))=(cz+d)^kg(z) (1) , $$
but I do not understand how it is obtained . I think the second part says :
Given a function g satisfying (1) .Define $$ G(z,1)=Gbegin{pmatrix}
z\
1\
end{pmatrix}=g(z) , $$
$$ where we consider the lattice [z,1] . Then for any lattice L=[omega_1,omega_2] we have \ g(frac{omega_1}{omega_2})=Gbegin{pmatrix}
frac{omega_1}{omega_2}\
1\
end{pmatrix}=omega_2^kGbegin{pmatrix}
omega_1\
omega_2\
end{pmatrix}=omega_2^kG(L)
Longrightarrow G(L)=omega_2^{-k}g(frac{omega_1}{omega_2})Longrightarrow G(lambda L)=lambda^{-k}G(L)
$$
But the first part I do not know how to show .
Thanks for helping !
modular-forms modular-function
asked Dec 5 '18 at 21:26
MatilloMatillo
147
$endgroup$
add a comment |
$begingroup$
I am reading in Serge Lang's book " Introduction to Modular Forms " .
On page 8 there is written that there is a bijection between functions of lattices , homogenous of degree -k and functions g on H ( the upper half plane ) satisfying the condition $$ g(alpha(z))=(cz+d)^kg(z) (1) , $$
but I do not understand how it is obtained . I think the second part says :
Given a function g satisfying (1) .Define $$ G(z,1)=Gbegin{pmatrix}
z\
1\
end{pmatrix}=g(z) , $$
$$ where we consider the lattice [z,1] . Then for any lattice L=[omega_1,omega_2] we have \ g(frac{omega_1}{omega_2})=Gbegin{pmatrix}
frac{omega_1}{omega_2}\
1\
end{pmatrix}=omega_2^kGbegin{pmatrix}
omega_1\
omega_2\
end{pmatrix}=omega_2^kG(L)
Longrightarrow G(L)=omega_2^{-k}g(frac{omega_1}{omega_2})Longrightarrow G(lambda L)=lambda^{-k}G(L)
$$
But the first part I do not know how to show .
Thanks for helping !
modular-forms modular-function
asked Dec 5 '18 at 21:26
MatilloMatillo
147
$endgroup$
add a comment |
$begingroup$
I am reading in Serge Lang's book " Introduction to Modular Forms " .
On page 8 there is written that there is a bijection between functions of lattices , homogenous of degree -k and functions g on H ( the upper half plane ) satisfying the condition $$ g(alpha(z))=(cz+d)^kg(z) (1) , $$
but I do not understand how it is obtained . I think the second part says :
Given a function g satisfying (1) .Define $$ G(z,1)=Gbegin{pmatrix}
z\
1\
end{pmatrix}=g(z) , $$
$$ where we consider the lattice [z,1] . Then for any lattice L=[omega_1,omega_2] we have \ g(frac{omega_1}{omega_2})=Gbegin{pmatrix}
frac{omega_1}{omega_2}\
1\
end{pmatrix}=omega_2^kGbegin{pmatrix}
omega_1\
omega_2\
end{pmatrix}=omega_2^kG(L)
Longrightarrow G(L)=omega_2^{-k}g(frac{omega_1}{omega_2})Longrightarrow G(lambda L)=lambda^{-k}G(L)
$$
But the first part I do not know how to show .
Thanks for helping !
modular-forms modular-function
asked Dec 5 '18 at 21:26
MatilloMatillo
147
$endgroup$
I am reading in Serge Lang's book " Introduction to Modular Forms " .
On page 8 there is written that there is a bijection between functions of lattices , homogenous of degree -k and functions g on H ( the upper half plane ) satisfying the condition $$ g(alpha(z))=(cz+d)^kg(z) (1) , $$
but I do not understand how it is obtained . I think the second part says :
Given a function g satisfying (1) .Define $$ G(z,1)=Gbegin{pmatrix}
z\
1\
end{pmatrix}=g(z) , $$
$$ where we consider the lattice [z,1] . Then for any lattice L=[omega_1,omega_2] we have \ g(frac{omega_1}{omega_2})=Gbegin{pmatrix}
frac{omega_1}{omega_2}\
1\
end{pmatrix}=omega_2^kGbegin{pmatrix}
omega_1\
omega_2\
end{pmatrix}=omega_2^kG(L)
Longrightarrow G(L)=omega_2^{-k}g(frac{omega_1}{omega_2})Longrightarrow G(lambda L)=lambda^{-k}G(L)
$$
But the first part I do not know how to show .
Thanks for helping !
modular-forms modular-function
modular-forms modular-function
asked Dec 5 '18 at 21:26
MatilloMatillo
147
asked Dec 5 '18 at 21:26
MatilloMatillo
147
asked Dec 5 '18 at 21:26
MatilloMatillo
147
asked Dec 5 '18 at 21:26
MatilloMatillo
147
asked Dec 5 '18 at 21:26
MatilloMatillo
147
147
add a comment |
add a comment |
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