Show that $Delta ne 0$ and LI of $G_4^3$ and $G_6^2$ are equivalent.
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In Complex Analysis by Freitag it is claimed that $Delta ne 0$ and LI of $G_4^3$ and $G_6^2$ are equivalent; that is, if $Delta = (60G_4)^3 - 27(140G_6)^2 ne 0$ then $G_4^3$ and $G_6^2$ are linearly independent and conversely. But $Delta = (60G_4)^3 - 27(140G_6)^2 ne 0$ means that only one (up to a multiplication of $Delta$ by a nonzero number) linear combination of $G_4^3$ and $G_6^2$ is nonzero, so how that implies that all linear combination of $G_4^3$ and $G_6^2$ is nonzero? Same question goes for the converse.
complex-analysis modular-forms
asked Nov 15 at 18:43
72D
52516
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In Complex Analysis by Freitag it is claimed that $Delta ne 0$ and LI of $G_4^3$ and $G_6^2$ are equivalent; that is, if $Delta = (60G_4)^3 - 27(140G_6)^2 ne 0$ then $G_4^3$ and $G_6^2$ are linearly independent and conversely. But $Delta = (60G_4)^3 - 27(140G_6)^2 ne 0$ means that only one (up to a multiplication of $Delta$ by a nonzero number) linear combination of $G_4^3$ and $G_6^2$ is nonzero, so how that implies that all linear combination of $G_4^3$ and $G_6^2$ is nonzero? Same question goes for the converse.
complex-analysis modular-forms
asked Nov 15 at 18:43
72D
52516
1
Consider the values of $G_4^3$ and $G_6^2$ at the cusp $iinfty$.
– Lord Shark the Unknown
Nov 15 at 18:49
@LordSharktheUnknown, both are zero at $iinfty$, am I right? if so it can come from $Delta = 0$ at $iinfty$. However, I can't relate this to the problem of their equivalence.
– 72D
Nov 15 at 18:55
No, the point is that neither is zero at infinity, but the only linear combination that is is $Delta$ (up to a a constant factor).
– Lord Shark the Unknown
Nov 15 at 19:14
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
In Complex Analysis by Freitag it is claimed that $Delta ne 0$ and LI of $G_4^3$ and $G_6^2$ are equivalent; that is, if $Delta = (60G_4)^3 - 27(140G_6)^2 ne 0$ then $G_4^3$ and $G_6^2$ are linearly independent and conversely. But $Delta = (60G_4)^3 - 27(140G_6)^2 ne 0$ means that only one (up to a multiplication of $Delta$ by a nonzero number) linear combination of $G_4^3$ and $G_6^2$ is nonzero, so how that implies that all linear combination of $G_4^3$ and $G_6^2$ is nonzero? Same question goes for the converse.
complex-analysis modular-forms
asked Nov 15 at 18:43
72D
52516
In Complex Analysis by Freitag it is claimed that $Delta ne 0$ and LI of $G_4^3$ and $G_6^2$ are equivalent; that is, if $Delta = (60G_4)^3 - 27(140G_6)^2 ne 0$ then $G_4^3$ and $G_6^2$ are linearly independent and conversely. But $Delta = (60G_4)^3 - 27(140G_6)^2 ne 0$ means that only one (up to a multiplication of $Delta$ by a nonzero number) linear combination of $G_4^3$ and $G_6^2$ is nonzero, so how that implies that all linear combination of $G_4^3$ and $G_6^2$ is nonzero? Same question goes for the converse.
complex-analysis modular-forms
complex-analysis modular-forms
asked Nov 15 at 18:43
72D
52516
asked Nov 15 at 18:43
72D
52516
edited Nov 15 at 19:08
asked Nov 15 at 18:43
72D
52516
asked Nov 15 at 18:43
72D
52516
asked Nov 15 at 18:43
72D
52516
52516
1
Consider the values of $G_4^3$ and $G_6^2$ at the cusp $iinfty$.
– Lord Shark the Unknown
Nov 15 at 18:49
@LordSharktheUnknown, both are zero at $iinfty$, am I right? if so it can come from $Delta = 0$ at $iinfty$. However, I can't relate this to the problem of their equivalence.
– 72D
Nov 15 at 18:55
No, the point is that neither is zero at infinity, but the only linear combination that is is $Delta$ (up to a a constant factor).
– Lord Shark the Unknown
Nov 15 at 19:14
add a comment |
1
Consider the values of $G_4^3$ and $G_6^2$ at the cusp $iinfty$.
– Lord Shark the Unknown
Nov 15 at 18:49
@LordSharktheUnknown, both are zero at $iinfty$, am I right? if so it can come from $Delta = 0$ at $iinfty$. However, I can't relate this to the problem of their equivalence.
– 72D
Nov 15 at 18:55
No, the point is that neither is zero at infinity, but the only linear combination that is is $Delta$ (up to a a constant factor).
– Lord Shark the Unknown
Nov 15 at 19:14
1
1
Consider the values of $G_4^3$ and $G_6^2$ at the cusp $iinfty$.
– Lord Shark the Unknown
Nov 15 at 18:49
Consider the values of $G_4^3$ and $G_6^2$ at the cusp $iinfty$.
– Lord Shark the Unknown
Nov 15 at 18:49
@LordSharktheUnknown, both are zero at $iinfty$, am I right? if so it can come from $Delta = 0$ at $iinfty$. However, I can't relate this to the problem of their equivalence.
– 72D
Nov 15 at 18:55
@LordSharktheUnknown, both are zero at $iinfty$, am I right? if so it can come from $Delta = 0$ at $iinfty$. However, I can't relate this to the problem of their equivalence.
– 72D
Nov 15 at 18:55
No, the point is that neither is zero at infinity, but the only linear combination that is is $Delta$ (up to a a constant factor).
– Lord Shark the Unknown
Nov 15 at 19:14
No, the point is that neither is zero at infinity, but the only linear combination that is is $Delta$ (up to a a constant factor).
– Lord Shark the Unknown
Nov 15 at 19:14
add a comment |
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Consider the values of $G_4^3$ and $G_6^2$ at the cusp $iinfty$.
– Lord Shark the Unknown
Nov 15 at 18:49
@LordSharktheUnknown, both are zero at $iinfty$, am I right? if so it can come from $Delta = 0$ at $iinfty$. However, I can't relate this to the problem of their equivalence.
– 72D
Nov 15 at 18:55
No, the point is that neither is zero at infinity, but the only linear combination that is is $Delta$ (up to a a constant factor).
– Lord Shark the Unknown
Nov 15 at 19:14