Every quadratic polynomial is conjugate to one of the polynomial $z^2+t$












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$begingroup$


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







share|cite|improve this question











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    0












    $begingroup$


    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







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      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







      share|cite|improve this question











      $endgroup$




      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






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 23 '18 at 6:17







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      asked Dec 23 '18 at 6:12









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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 $t$.






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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 $t$.






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              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 $t$.






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                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 $t$.






                share|cite|improve this answer









                $endgroup$



                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 $t$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 23 '18 at 6:25









                Eric WofseyEric Wofsey

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