Every quadratic polynomial is conjugate to one of the polynomial $z^2+t$
$begingroup$
Every quadratic polynomial is conjugate to one of the polynomial $z^2+t$
where conjugation means conjugation by a linear fractional transformation $alpha(z)=frac{az+b}{cz+d}$ and $f^{alpha}={alpha}^{-1}f{alpha}$.
Now if $p$ is said to be a fixed point by $f$ then ${alpha}^{-1}(p)$ is said to be a fixed point by $f^{alpha}$.
Again any quadratic polynomial can have two distinct or same root in $Bbb C$. I think I have to connect this but I can't! Even in mathstack I got this but not what I want.
I think I got one idea if we just consider the affine linear transformation ${alpha}(z)=az+b$ then
If $f$ has two distinct roots say $x,yin Bbb C$ then we choose $tneq 0$ then we can choose ${alpha}(z)$ s.t ${alpha}(x)=t$ and ${alpha}(y)=-t$ then $fcirc {alpha}^{-1}(t)=0$ and $fcirc {alpha}^{-1}(-t)=0$. Moreover, $fcirc {alpha}^{-1}(z)to infty$ as $z to infty$. Can we say something using these?
algebraic-geometry polynomials dynamical-systems moduli-space arithmetic-dynamics
asked Dec 23 '18 at 6:12
GimgimGimgim
27113
$endgroup$
add a comment |
$begingroup$
Every quadratic polynomial is conjugate to one of the polynomial $z^2+t$
where conjugation means conjugation by a linear fractional transformation $alpha(z)=frac{az+b}{cz+d}$ and $f^{alpha}={alpha}^{-1}f{alpha}$.
Now if $p$ is said to be a fixed point by $f$ then ${alpha}^{-1}(p)$ is said to be a fixed point by $f^{alpha}$.
Again any quadratic polynomial can have two distinct or same root in $Bbb C$. I think I have to connect this but I can't! Even in mathstack I got this but not what I want.
I think I got one idea if we just consider the affine linear transformation ${alpha}(z)=az+b$ then
If $f$ has two distinct roots say $x,yin Bbb C$ then we choose $tneq 0$ then we can choose ${alpha}(z)$ s.t ${alpha}(x)=t$ and ${alpha}(y)=-t$ then $fcirc {alpha}^{-1}(t)=0$ and $fcirc {alpha}^{-1}(-t)=0$. Moreover, $fcirc {alpha}^{-1}(z)to infty$ as $z to infty$. Can we say something using these?
algebraic-geometry polynomials dynamical-systems moduli-space arithmetic-dynamics
asked Dec 23 '18 at 6:12
GimgimGimgim
27113
$endgroup$
add a comment |
$begingroup$
Every quadratic polynomial is conjugate to one of the polynomial $z^2+t$
where conjugation means conjugation by a linear fractional transformation $alpha(z)=frac{az+b}{cz+d}$ and $f^{alpha}={alpha}^{-1}f{alpha}$.
Now if $p$ is said to be a fixed point by $f$ then ${alpha}^{-1}(p)$ is said to be a fixed point by $f^{alpha}$.
Again any quadratic polynomial can have two distinct or same root in $Bbb C$. I think I have to connect this but I can't! Even in mathstack I got this but not what I want.
I think I got one idea if we just consider the affine linear transformation ${alpha}(z)=az+b$ then
If $f$ has two distinct roots say $x,yin Bbb C$ then we choose $tneq 0$ then we can choose ${alpha}(z)$ s.t ${alpha}(x)=t$ and ${alpha}(y)=-t$ then $fcirc {alpha}^{-1}(t)=0$ and $fcirc {alpha}^{-1}(-t)=0$. Moreover, $fcirc {alpha}^{-1}(z)to infty$ as $z to infty$. Can we say something using these?
algebraic-geometry polynomials dynamical-systems moduli-space arithmetic-dynamics
asked Dec 23 '18 at 6:12
GimgimGimgim
27113
$endgroup$
Every quadratic polynomial is conjugate to one of the polynomial $z^2+t$
where conjugation means conjugation by a linear fractional transformation $alpha(z)=frac{az+b}{cz+d}$ and $f^{alpha}={alpha}^{-1}f{alpha}$.
Now if $p$ is said to be a fixed point by $f$ then ${alpha}^{-1}(p)$ is said to be a fixed point by $f^{alpha}$.
Again any quadratic polynomial can have two distinct or same root in $Bbb C$. I think I have to connect this but I can't! Even in mathstack I got this but not what I want.
I think I got one idea if we just consider the affine linear transformation ${alpha}(z)=az+b$ then
If $f$ has two distinct roots say $x,yin Bbb C$ then we choose $tneq 0$ then we can choose ${alpha}(z)$ s.t ${alpha}(x)=t$ and ${alpha}(y)=-t$ then $fcirc {alpha}^{-1}(t)=0$ and $fcirc {alpha}^{-1}(-t)=0$. Moreover, $fcirc {alpha}^{-1}(z)to infty$ as $z to infty$. Can we say something using these?
algebraic-geometry polynomials dynamical-systems moduli-space arithmetic-dynamics
algebraic-geometry polynomials dynamical-systems moduli-space arithmetic-dynamics
asked Dec 23 '18 at 6:12
GimgimGimgim
27113
asked Dec 23 '18 at 6:12
GimgimGimgim
27113
edited Dec 23 '18 at 6:17
Gimgim
asked Dec 23 '18 at 6:12
GimgimGimgim
27113
asked Dec 23 '18 at 6:12
GimgimGimgim
27113
asked Dec 23 '18 at 6:12
GimgimGimgim
27113
27113
add a comment |
add a comment |
1 Answer
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$begingroup$
This is basically just completing the square. Every quadratic has the form $a(z+b)^2+c$ for some $a,b,cinmathbb{C}$ with $a$ nonzero. Conjugating by $zmapsto z-b$ you get $az^2+d$ for some $d$ and then conjugating by $zmapsto z/a$ you get $z^2+t$ for some $t$.
answered Dec 23 '18 at 6:25
Eric WofseyEric Wofsey
187k14215344
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1 Answer
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1 Answer
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active
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votes
$begingroup$
This is basically just completing the square. Every quadratic has the form $a(z+b)^2+c$ for some $a,b,cinmathbb{C}$ with $a$ nonzero. Conjugating by $zmapsto z-b$ you get $az^2+d$ for some $d$ and then conjugating by $zmapsto z/a$ you get $z^2+t$ for some $t$.
answered Dec 23 '18 at 6:25
Eric WofseyEric Wofsey
187k14215344
$endgroup$
add a comment |
$begingroup$
This is basically just completing the square. Every quadratic has the form $a(z+b)^2+c$ for some $a,b,cinmathbb{C}$ with $a$ nonzero. Conjugating by $zmapsto z-b$ you get $az^2+d$ for some $d$ and then conjugating by $zmapsto z/a$ you get $z^2+t$ for some $t$.
answered Dec 23 '18 at 6:25
Eric WofseyEric Wofsey
187k14215344
$endgroup$
add a comment |
$begingroup$
This is basically just completing the square. Every quadratic has the form $a(z+b)^2+c$ for some $a,b,cinmathbb{C}$ with $a$ nonzero. Conjugating by $zmapsto z-b$ you get $az^2+d$ for some $d$ and then conjugating by $zmapsto z/a$ you get $z^2+t$ for some $t$.
answered Dec 23 '18 at 6:25
Eric WofseyEric Wofsey
187k14215344
$endgroup$
This is basically just completing the square. Every quadratic has the form $a(z+b)^2+c$ for some $a,b,cinmathbb{C}$ with $a$ nonzero. Conjugating by $zmapsto z-b$ you get $az^2+d$ for some $d$ and then conjugating by $zmapsto z/a$ you get $z^2+t$ for some $t$.
answered Dec 23 '18 at 6:25
Eric WofseyEric Wofsey
187k14215344
answered Dec 23 '18 at 6:25
Eric WofseyEric Wofsey
187k14215344
answered Dec 23 '18 at 6:25
Eric WofseyEric Wofsey
187k14215344
answered Dec 23 '18 at 6:25
Eric WofseyEric Wofsey
187k14215344
187k14215344
add a comment |
add a comment |
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