Characteristic polynomial determines three reducible elements (line pairs) in the pencil of plane conics
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Let $eta_i,0leq ileq 4$ be 4 variables and $eta_0 x^2+4eta_1 x^3y+6eta_2 x^2y^2+4eta_3 xy^3+eta_4 y^4$ be the associated quadratic form. Consider $U=x^2,V=2xy, W=y^2$. Then the quadratic form transforms into $$eta_0 U^2+2eta_2UV+eta_2(V^2+2UW)+2eta_3VW+eta_4W^2=0quad (1)$$ and there is an additional relation $$4UW-V^2=0quad (2)$$ Denote by $Q_1$ and $Q_2$ the associated quadratic forms of $(1)$ and $(2)$.
Consider $det(Q_1+lambda Q_2)=0$ for solving $lambda$ (i.e. $(1)+lambda(2)=0$ is associated quadratic form). Then plugging $lambda$ in $(U,V,W)^T(Q_1+lambda Q_2)(U,V,W)$, one will see the polynomial splits into linear factors.
$textbf{Q:}$ Why does the polynomial splits into linear factors? I could see there is a coordinate transformation s.t. $Q_1+lambda Q_2$ is effectively (2 by 2 matrix) $oplus 0$ by $det(Q_1+lambda Q_2)=0$. That 2 by 2 is quadratic as well. Hence it reduces to quadratic of 2 variables which splits over complex number. This seems very cumbersome.
Ref. An Introduction to Invariants and Moduli, Mukai, Remark 1.26 of Chapter 1, Sec. 3(b), pg 28.
linear-algebra abstract-algebra algebraic-geometry
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$begingroup$
Let $eta_i,0leq ileq 4$ be 4 variables and $eta_0 x^2+4eta_1 x^3y+6eta_2 x^2y^2+4eta_3 xy^3+eta_4 y^4$ be the associated quadratic form. Consider $U=x^2,V=2xy, W=y^2$. Then the quadratic form transforms into $$eta_0 U^2+2eta_2UV+eta_2(V^2+2UW)+2eta_3VW+eta_4W^2=0quad (1)$$ and there is an additional relation $$4UW-V^2=0quad (2)$$ Denote by $Q_1$ and $Q_2$ the associated quadratic forms of $(1)$ and $(2)$.
Consider $det(Q_1+lambda Q_2)=0$ for solving $lambda$ (i.e. $(1)+lambda(2)=0$ is associated quadratic form). Then plugging $lambda$ in $(U,V,W)^T(Q_1+lambda Q_2)(U,V,W)$, one will see the polynomial splits into linear factors.
$textbf{Q:}$ Why does the polynomial splits into linear factors? I could see there is a coordinate transformation s.t. $Q_1+lambda Q_2$ is effectively (2 by 2 matrix) $oplus 0$ by $det(Q_1+lambda Q_2)=0$. That 2 by 2 is quadratic as well. Hence it reduces to quadratic of 2 variables which splits over complex number. This seems very cumbersome.
Ref. An Introduction to Invariants and Moduli, Mukai, Remark 1.26 of Chapter 1, Sec. 3(b), pg 28.
linear-algebra abstract-algebra algebraic-geometry
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1
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It won't show me page 27, but it does show pages 26 and 28. I recognize the 3 by 3 image of an element of the 2 by 2 modular group. Suggest you look up W. Magnus, Noneuclidean Tessellations and their groups. The relevant material goes back to a German book by Fricke and Klein, (1897), Lectures on Automorphic Forms. F+K has recently been translated into English. (I have a cheap reprint of the 1897 original).
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– Will Jagy
Dec 27 '18 at 22:08
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books.google.com/…
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– Will Jagy
Dec 27 '18 at 22:09
$begingroup$
books.google.com/…
$endgroup$
– Will Jagy
Dec 27 '18 at 22:13
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magnus; the material I meant is on page 23, not shown here: books.google.com/…
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– Will Jagy
Dec 27 '18 at 22:19
1
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@WillJagy I can check chpt 3 of Magnus's book. Thanks.
$endgroup$
– user45765
Dec 27 '18 at 22:29
add a comment |
$begingroup$
Let $eta_i,0leq ileq 4$ be 4 variables and $eta_0 x^2+4eta_1 x^3y+6eta_2 x^2y^2+4eta_3 xy^3+eta_4 y^4$ be the associated quadratic form. Consider $U=x^2,V=2xy, W=y^2$. Then the quadratic form transforms into $$eta_0 U^2+2eta_2UV+eta_2(V^2+2UW)+2eta_3VW+eta_4W^2=0quad (1)$$ and there is an additional relation $$4UW-V^2=0quad (2)$$ Denote by $Q_1$ and $Q_2$ the associated quadratic forms of $(1)$ and $(2)$.
Consider $det(Q_1+lambda Q_2)=0$ for solving $lambda$ (i.e. $(1)+lambda(2)=0$ is associated quadratic form). Then plugging $lambda$ in $(U,V,W)^T(Q_1+lambda Q_2)(U,V,W)$, one will see the polynomial splits into linear factors.
$textbf{Q:}$ Why does the polynomial splits into linear factors? I could see there is a coordinate transformation s.t. $Q_1+lambda Q_2$ is effectively (2 by 2 matrix) $oplus 0$ by $det(Q_1+lambda Q_2)=0$. That 2 by 2 is quadratic as well. Hence it reduces to quadratic of 2 variables which splits over complex number. This seems very cumbersome.
Ref. An Introduction to Invariants and Moduli, Mukai, Remark 1.26 of Chapter 1, Sec. 3(b), pg 28.
linear-algebra abstract-algebra algebraic-geometry
$endgroup$
Let $eta_i,0leq ileq 4$ be 4 variables and $eta_0 x^2+4eta_1 x^3y+6eta_2 x^2y^2+4eta_3 xy^3+eta_4 y^4$ be the associated quadratic form. Consider $U=x^2,V=2xy, W=y^2$. Then the quadratic form transforms into $$eta_0 U^2+2eta_2UV+eta_2(V^2+2UW)+2eta_3VW+eta_4W^2=0quad (1)$$ and there is an additional relation $$4UW-V^2=0quad (2)$$ Denote by $Q_1$ and $Q_2$ the associated quadratic forms of $(1)$ and $(2)$.
Consider $det(Q_1+lambda Q_2)=0$ for solving $lambda$ (i.e. $(1)+lambda(2)=0$ is associated quadratic form). Then plugging $lambda$ in $(U,V,W)^T(Q_1+lambda Q_2)(U,V,W)$, one will see the polynomial splits into linear factors.
$textbf{Q:}$ Why does the polynomial splits into linear factors? I could see there is a coordinate transformation s.t. $Q_1+lambda Q_2$ is effectively (2 by 2 matrix) $oplus 0$ by $det(Q_1+lambda Q_2)=0$. That 2 by 2 is quadratic as well. Hence it reduces to quadratic of 2 variables which splits over complex number. This seems very cumbersome.
Ref. An Introduction to Invariants and Moduli, Mukai, Remark 1.26 of Chapter 1, Sec. 3(b), pg 28.
linear-algebra abstract-algebra algebraic-geometry
linear-algebra abstract-algebra algebraic-geometry
edited Dec 27 '18 at 22:28
user26857
39.5k124284
39.5k124284
asked Dec 27 '18 at 19:22
user45765user45765
2,7232724
2,7232724
1
$begingroup$
It won't show me page 27, but it does show pages 26 and 28. I recognize the 3 by 3 image of an element of the 2 by 2 modular group. Suggest you look up W. Magnus, Noneuclidean Tessellations and their groups. The relevant material goes back to a German book by Fricke and Klein, (1897), Lectures on Automorphic Forms. F+K has recently been translated into English. (I have a cheap reprint of the 1897 original).
$endgroup$
– Will Jagy
Dec 27 '18 at 22:08
$begingroup$
books.google.com/…
$endgroup$
– Will Jagy
Dec 27 '18 at 22:09
$begingroup$
books.google.com/…
$endgroup$
– Will Jagy
Dec 27 '18 at 22:13
$begingroup$
magnus; the material I meant is on page 23, not shown here: books.google.com/…
$endgroup$
– Will Jagy
Dec 27 '18 at 22:19
1
$begingroup$
@WillJagy I can check chpt 3 of Magnus's book. Thanks.
$endgroup$
– user45765
Dec 27 '18 at 22:29
add a comment |
1
$begingroup$
It won't show me page 27, but it does show pages 26 and 28. I recognize the 3 by 3 image of an element of the 2 by 2 modular group. Suggest you look up W. Magnus, Noneuclidean Tessellations and their groups. The relevant material goes back to a German book by Fricke and Klein, (1897), Lectures on Automorphic Forms. F+K has recently been translated into English. (I have a cheap reprint of the 1897 original).
$endgroup$
– Will Jagy
Dec 27 '18 at 22:08
$begingroup$
books.google.com/…
$endgroup$
– Will Jagy
Dec 27 '18 at 22:09
$begingroup$
books.google.com/…
$endgroup$
– Will Jagy
Dec 27 '18 at 22:13
$begingroup$
magnus; the material I meant is on page 23, not shown here: books.google.com/…
$endgroup$
– Will Jagy
Dec 27 '18 at 22:19
1
$begingroup$
@WillJagy I can check chpt 3 of Magnus's book. Thanks.
$endgroup$
– user45765
Dec 27 '18 at 22:29
1
1
$begingroup$
It won't show me page 27, but it does show pages 26 and 28. I recognize the 3 by 3 image of an element of the 2 by 2 modular group. Suggest you look up W. Magnus, Noneuclidean Tessellations and their groups. The relevant material goes back to a German book by Fricke and Klein, (1897), Lectures on Automorphic Forms. F+K has recently been translated into English. (I have a cheap reprint of the 1897 original).
$endgroup$
– Will Jagy
Dec 27 '18 at 22:08
$begingroup$
It won't show me page 27, but it does show pages 26 and 28. I recognize the 3 by 3 image of an element of the 2 by 2 modular group. Suggest you look up W. Magnus, Noneuclidean Tessellations and their groups. The relevant material goes back to a German book by Fricke and Klein, (1897), Lectures on Automorphic Forms. F+K has recently been translated into English. (I have a cheap reprint of the 1897 original).
$endgroup$
– Will Jagy
Dec 27 '18 at 22:08
$begingroup$
books.google.com/…
$endgroup$
– Will Jagy
Dec 27 '18 at 22:09
$begingroup$
books.google.com/…
$endgroup$
– Will Jagy
Dec 27 '18 at 22:09
$begingroup$
books.google.com/…
$endgroup$
– Will Jagy
Dec 27 '18 at 22:13
$begingroup$
books.google.com/…
$endgroup$
– Will Jagy
Dec 27 '18 at 22:13
$begingroup$
magnus; the material I meant is on page 23, not shown here: books.google.com/…
$endgroup$
– Will Jagy
Dec 27 '18 at 22:19
$begingroup$
magnus; the material I meant is on page 23, not shown here: books.google.com/…
$endgroup$
– Will Jagy
Dec 27 '18 at 22:19
1
1
$begingroup$
@WillJagy I can check chpt 3 of Magnus's book. Thanks.
$endgroup$
– user45765
Dec 27 '18 at 22:29
$begingroup$
@WillJagy I can check chpt 3 of Magnus's book. Thanks.
$endgroup$
– user45765
Dec 27 '18 at 22:29
add a comment |
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$begingroup$
It won't show me page 27, but it does show pages 26 and 28. I recognize the 3 by 3 image of an element of the 2 by 2 modular group. Suggest you look up W. Magnus, Noneuclidean Tessellations and their groups. The relevant material goes back to a German book by Fricke and Klein, (1897), Lectures on Automorphic Forms. F+K has recently been translated into English. (I have a cheap reprint of the 1897 original).
$endgroup$
– Will Jagy
Dec 27 '18 at 22:08
$begingroup$
books.google.com/…
$endgroup$
– Will Jagy
Dec 27 '18 at 22:09
$begingroup$
books.google.com/…
$endgroup$
– Will Jagy
Dec 27 '18 at 22:13
$begingroup$
magnus; the material I meant is on page 23, not shown here: books.google.com/…
$endgroup$
– Will Jagy
Dec 27 '18 at 22:19
1
$begingroup$
@WillJagy I can check chpt 3 of Magnus's book. Thanks.
$endgroup$
– user45765
Dec 27 '18 at 22:29