Show that set of all cusp classes is finite.
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A cusp $κ$ of a congruence subgroup $Γ le SL_2(mathbb{Z})$ is by definition an element of $mathbb{Q}∪{{i∞}}$. In mapping $κ to frac{aκ+b}{cκ+d}$ if we take $κ=a/b in mathbb{Q}$ then $A=begin{pmatrix}x&y\-b&aend{pmatrix} in SL_2(mathbb{Z})$ maps $κ to i∞$.
On the other hand index of $Γ$ in $SL_2(mathbb{Z})$ is finite, say $SL_2(mathbb{Z}) = ΓA_1 cup dots ΓA_l$.
How to complete of the following?
Show that the set of all cusp classes $(mathbb{Q}∪{{i∞}})/Γ$ is finite.
Regarding the definition of Γ :
A subgroup $Γ ⊂ SL(2, Z)$ is called a congruence subgroup iff it contains a suitable principal congruence subgroup $Γ[q]$, i.e. $Γ[q] ⊂ Γ ⊂ Γ[1]$ in which $$ Gamma(q)=Big{M=begin{pmatrix}
a &b\
c& d
end{pmatrix}in SL(2,mathbb{Z}) | a equiv d equiv 1 bmod q text{and} b equiv c equiv 0 bmod q Big}.$$
abstract-algebra proof-verification proof-explanation modular-forms
asked Nov 23 at 19:39
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541116
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up vote
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A cusp $κ$ of a congruence subgroup $Γ le SL_2(mathbb{Z})$ is by definition an element of $mathbb{Q}∪{{i∞}}$. In mapping $κ to frac{aκ+b}{cκ+d}$ if we take $κ=a/b in mathbb{Q}$ then $A=begin{pmatrix}x&y\-b&aend{pmatrix} in SL_2(mathbb{Z})$ maps $κ to i∞$.
On the other hand index of $Γ$ in $SL_2(mathbb{Z})$ is finite, say $SL_2(mathbb{Z}) = ΓA_1 cup dots ΓA_l$.
How to complete of the following?
Show that the set of all cusp classes $(mathbb{Q}∪{{i∞}})/Γ$ is finite.
Regarding the definition of Γ :
A subgroup $Γ ⊂ SL(2, Z)$ is called a congruence subgroup iff it contains a suitable principal congruence subgroup $Γ[q]$, i.e. $Γ[q] ⊂ Γ ⊂ Γ[1]$ in which $$ Gamma(q)=Big{M=begin{pmatrix}
a &b\
c& d
end{pmatrix}in SL(2,mathbb{Z}) | a equiv d equiv 1 bmod q text{and} b equiv c equiv 0 bmod q Big}.$$
abstract-algebra proof-verification proof-explanation modular-forms
asked Nov 23 at 19:39
72D
541116
Comments are not for extended discussion; this conversation has been moved to chat.
– Aloizio Macedo♦
Nov 24 at 13:11
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
A cusp $κ$ of a congruence subgroup $Γ le SL_2(mathbb{Z})$ is by definition an element of $mathbb{Q}∪{{i∞}}$. In mapping $κ to frac{aκ+b}{cκ+d}$ if we take $κ=a/b in mathbb{Q}$ then $A=begin{pmatrix}x&y\-b&aend{pmatrix} in SL_2(mathbb{Z})$ maps $κ to i∞$.
On the other hand index of $Γ$ in $SL_2(mathbb{Z})$ is finite, say $SL_2(mathbb{Z}) = ΓA_1 cup dots ΓA_l$.
How to complete of the following?
Show that the set of all cusp classes $(mathbb{Q}∪{{i∞}})/Γ$ is finite.
Regarding the definition of Γ :
A subgroup $Γ ⊂ SL(2, Z)$ is called a congruence subgroup iff it contains a suitable principal congruence subgroup $Γ[q]$, i.e. $Γ[q] ⊂ Γ ⊂ Γ[1]$ in which $$ Gamma(q)=Big{M=begin{pmatrix}
a &b\
c& d
end{pmatrix}in SL(2,mathbb{Z}) | a equiv d equiv 1 bmod q text{and} b equiv c equiv 0 bmod q Big}.$$
abstract-algebra proof-verification proof-explanation modular-forms
asked Nov 23 at 19:39
72D
541116
A cusp $κ$ of a congruence subgroup $Γ le SL_2(mathbb{Z})$ is by definition an element of $mathbb{Q}∪{{i∞}}$. In mapping $κ to frac{aκ+b}{cκ+d}$ if we take $κ=a/b in mathbb{Q}$ then $A=begin{pmatrix}x&y\-b&aend{pmatrix} in SL_2(mathbb{Z})$ maps $κ to i∞$.
On the other hand index of $Γ$ in $SL_2(mathbb{Z})$ is finite, say $SL_2(mathbb{Z}) = ΓA_1 cup dots ΓA_l$.
How to complete of the following?
Show that the set of all cusp classes $(mathbb{Q}∪{{i∞}})/Γ$ is finite.
Regarding the definition of Γ :
A subgroup $Γ ⊂ SL(2, Z)$ is called a congruence subgroup iff it contains a suitable principal congruence subgroup $Γ[q]$, i.e. $Γ[q] ⊂ Γ ⊂ Γ[1]$ in which $$ Gamma(q)=Big{M=begin{pmatrix}
a &b\
c& d
end{pmatrix}in SL(2,mathbb{Z}) | a equiv d equiv 1 bmod q text{and} b equiv c equiv 0 bmod q Big}.$$
abstract-algebra proof-verification proof-explanation modular-forms
abstract-algebra proof-verification proof-explanation modular-forms
asked Nov 23 at 19:39
72D
541116
asked Nov 23 at 19:39
72D
541116
edited Nov 23 at 20:34
asked Nov 23 at 19:39
72D
541116
asked Nov 23 at 19:39
72D
541116
asked Nov 23 at 19:39
72D
541116
541116
Comments are not for extended discussion; this conversation has been moved to chat.
– Aloizio Macedo♦
Nov 24 at 13:11
add a comment |
Comments are not for extended discussion; this conversation has been moved to chat.
– Aloizio Macedo♦
Nov 24 at 13:11
Comments are not for extended discussion; this conversation has been moved to chat.
– Aloizio Macedo♦
Nov 24 at 13:11
Comments are not for extended discussion; this conversation has been moved to chat.
– Aloizio Macedo♦
Nov 24 at 13:11
add a comment |
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Comments are not for extended discussion; this conversation has been moved to chat.
– Aloizio Macedo♦
Nov 24 at 13:11