How do I show that $varphi(t)=(t,t^2,t^3)$ is a global chart












1












$begingroup$


Let $f_{1},f_{2}:mathbb R^{3} to mathbb R$, defined as



$f_{1}(x,y,z)=x^2+xy-y-z$



$f_{2}(x,y,z)=2x^2+3xy-2y-3z$



and $M:={(x,y,z) in mathbb R^{3}:f_{1}(x,y,z)=f_{2}(x,y,z)=0}$



I have shown that $M$ is a one-dimensional $C^{1}-$manifold. How do I go on to show that



the chart defined by: $varphi: mathbb R to mathbb R^{3}, varphi(t)=(t,t^2,t^3)$ is a global chart.



My understanding of a global chart is that $varphi$ would cover $M$ but how can I show this?






real-analysis measure-theory manifolds







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    1












    $begingroup$


    Let $f_{1},f_{2}:mathbb R^{3} to mathbb R$, defined as



    $f_{1}(x,y,z)=x^2+xy-y-z$



    $f_{2}(x,y,z)=2x^2+3xy-2y-3z$



    and $M:={(x,y,z) in mathbb R^{3}:f_{1}(x,y,z)=f_{2}(x,y,z)=0}$



    I have shown that $M$ is a one-dimensional $C^{1}-$manifold. How do I go on to show that



    the chart defined by: $varphi: mathbb R to mathbb R^{3}, varphi(t)=(t,t^2,t^3)$ is a global chart.



    My understanding of a global chart is that $varphi$ would cover $M$ but how can I show this?






    real-analysis measure-theory manifolds







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      Let $f_{1},f_{2}:mathbb R^{3} to mathbb R$, defined as



      $f_{1}(x,y,z)=x^2+xy-y-z$



      $f_{2}(x,y,z)=2x^2+3xy-2y-3z$



      and $M:={(x,y,z) in mathbb R^{3}:f_{1}(x,y,z)=f_{2}(x,y,z)=0}$



      I have shown that $M$ is a one-dimensional $C^{1}-$manifold. How do I go on to show that



      the chart defined by: $varphi: mathbb R to mathbb R^{3}, varphi(t)=(t,t^2,t^3)$ is a global chart.



      My understanding of a global chart is that $varphi$ would cover $M$ but how can I show this?






      real-analysis measure-theory manifolds







      share|cite|improve this question









      $endgroup$




      Let $f_{1},f_{2}:mathbb R^{3} to mathbb R$, defined as



      $f_{1}(x,y,z)=x^2+xy-y-z$



      $f_{2}(x,y,z)=2x^2+3xy-2y-3z$



      and $M:={(x,y,z) in mathbb R^{3}:f_{1}(x,y,z)=f_{2}(x,y,z)=0}$



      I have shown that $M$ is a one-dimensional $C^{1}-$manifold. How do I go on to show that



      the chart defined by: $varphi: mathbb R to mathbb R^{3}, varphi(t)=(t,t^2,t^3)$ is a global chart.



      My understanding of a global chart is that $varphi$ would cover $M$ but how can I show this?






      real-analysis measure-theory manifolds




      real-analysis measure-theory manifolds






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




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      asked Jan 22 at 22:50









      SABOYSABOY

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      656311






















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          $begingroup$

          The map $f:mathbb{R}rightarrowmathbb{R}^3$ defined by $f(t)=f(t,t^2,t^3)$ induces an homeomorphism between $mathbb{R}$ and $M$.



          The image of $f$ is contained in $M$.



          $t^2+tt^2-t^2-t^3=f_1(t,t^2,t^3)=0$



          $2t^2+3tt^2-2t^2-3t^3=f_2(t,t^2,t^3)=0$.



          This implies that $f(mathbb{R})subset M$.



          Define $g:mathbb{R}^3rightarrowmathbb{R}$ by $g(x,y,z)=x$, its $h$ restriction to $M$ is the inverse of $f$.



          $hcirc f(t)=g(t,t^2,t^3)=t$.



          Let $(x,y,z)in M$, $fcirc h(x,y,z)=f(x)=(x,x^2,x^3)$.



          We have $x^2+xy-y-z=0, 2x^2+3xy-2y-3z=0$ implies that:



          $-3(x^2+xy-y-z)+2x^2+3xy-2y-3z=-3x^2-3xy+3y+3z+2x^2+3xy-2y-3z$



          $=-x^2+y=0$, we deduce that $y=x^2$ and $z=x^2+xy-y=x^2+xx^2-x^2=x^3$. This shows that if $(x,y,z)in M, (x,y,z)=(x,x^2,x^3)$ and for every $(x,y,z)in M$, $fcirc h(x,y,z)=(x,y,z)$.






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            $begingroup$

            The map $f:mathbb{R}rightarrowmathbb{R}^3$ defined by $f(t)=f(t,t^2,t^3)$ induces an homeomorphism between $mathbb{R}$ and $M$.



            The image of $f$ is contained in $M$.



            $t^2+tt^2-t^2-t^3=f_1(t,t^2,t^3)=0$



            $2t^2+3tt^2-2t^2-3t^3=f_2(t,t^2,t^3)=0$.



            This implies that $f(mathbb{R})subset M$.



            Define $g:mathbb{R}^3rightarrowmathbb{R}$ by $g(x,y,z)=x$, its $h$ restriction to $M$ is the inverse of $f$.



            $hcirc f(t)=g(t,t^2,t^3)=t$.



            Let $(x,y,z)in M$, $fcirc h(x,y,z)=f(x)=(x,x^2,x^3)$.



            We have $x^2+xy-y-z=0, 2x^2+3xy-2y-3z=0$ implies that:



            $-3(x^2+xy-y-z)+2x^2+3xy-2y-3z=-3x^2-3xy+3y+3z+2x^2+3xy-2y-3z$



            $=-x^2+y=0$, we deduce that $y=x^2$ and $z=x^2+xy-y=x^2+xx^2-x^2=x^3$. This shows that if $(x,y,z)in M, (x,y,z)=(x,x^2,x^3)$ and for every $(x,y,z)in M$, $fcirc h(x,y,z)=(x,y,z)$.






            share|cite|improve this answer









            $endgroup$


















              2












              $begingroup$

              The map $f:mathbb{R}rightarrowmathbb{R}^3$ defined by $f(t)=f(t,t^2,t^3)$ induces an homeomorphism between $mathbb{R}$ and $M$.



              The image of $f$ is contained in $M$.



              $t^2+tt^2-t^2-t^3=f_1(t,t^2,t^3)=0$



              $2t^2+3tt^2-2t^2-3t^3=f_2(t,t^2,t^3)=0$.



              This implies that $f(mathbb{R})subset M$.



              Define $g:mathbb{R}^3rightarrowmathbb{R}$ by $g(x,y,z)=x$, its $h$ restriction to $M$ is the inverse of $f$.



              $hcirc f(t)=g(t,t^2,t^3)=t$.



              Let $(x,y,z)in M$, $fcirc h(x,y,z)=f(x)=(x,x^2,x^3)$.



              We have $x^2+xy-y-z=0, 2x^2+3xy-2y-3z=0$ implies that:



              $-3(x^2+xy-y-z)+2x^2+3xy-2y-3z=-3x^2-3xy+3y+3z+2x^2+3xy-2y-3z$



              $=-x^2+y=0$, we deduce that $y=x^2$ and $z=x^2+xy-y=x^2+xx^2-x^2=x^3$. This shows that if $(x,y,z)in M, (x,y,z)=(x,x^2,x^3)$ and for every $(x,y,z)in M$, $fcirc h(x,y,z)=(x,y,z)$.






              share|cite|improve this answer









              $endgroup$
















                2












                2








                2





                $begingroup$

                The map $f:mathbb{R}rightarrowmathbb{R}^3$ defined by $f(t)=f(t,t^2,t^3)$ induces an homeomorphism between $mathbb{R}$ and $M$.



                The image of $f$ is contained in $M$.



                $t^2+tt^2-t^2-t^3=f_1(t,t^2,t^3)=0$



                $2t^2+3tt^2-2t^2-3t^3=f_2(t,t^2,t^3)=0$.



                This implies that $f(mathbb{R})subset M$.



                Define $g:mathbb{R}^3rightarrowmathbb{R}$ by $g(x,y,z)=x$, its $h$ restriction to $M$ is the inverse of $f$.



                $hcirc f(t)=g(t,t^2,t^3)=t$.



                Let $(x,y,z)in M$, $fcirc h(x,y,z)=f(x)=(x,x^2,x^3)$.



                We have $x^2+xy-y-z=0, 2x^2+3xy-2y-3z=0$ implies that:



                $-3(x^2+xy-y-z)+2x^2+3xy-2y-3z=-3x^2-3xy+3y+3z+2x^2+3xy-2y-3z$



                $=-x^2+y=0$, we deduce that $y=x^2$ and $z=x^2+xy-y=x^2+xx^2-x^2=x^3$. This shows that if $(x,y,z)in M, (x,y,z)=(x,x^2,x^3)$ and for every $(x,y,z)in M$, $fcirc h(x,y,z)=(x,y,z)$.






                share|cite|improve this answer









                $endgroup$



                The map $f:mathbb{R}rightarrowmathbb{R}^3$ defined by $f(t)=f(t,t^2,t^3)$ induces an homeomorphism between $mathbb{R}$ and $M$.



                The image of $f$ is contained in $M$.



                $t^2+tt^2-t^2-t^3=f_1(t,t^2,t^3)=0$



                $2t^2+3tt^2-2t^2-3t^3=f_2(t,t^2,t^3)=0$.



                This implies that $f(mathbb{R})subset M$.



                Define $g:mathbb{R}^3rightarrowmathbb{R}$ by $g(x,y,z)=x$, its $h$ restriction to $M$ is the inverse of $f$.



                $hcirc f(t)=g(t,t^2,t^3)=t$.



                Let $(x,y,z)in M$, $fcirc h(x,y,z)=f(x)=(x,x^2,x^3)$.



                We have $x^2+xy-y-z=0, 2x^2+3xy-2y-3z=0$ implies that:



                $-3(x^2+xy-y-z)+2x^2+3xy-2y-3z=-3x^2-3xy+3y+3z+2x^2+3xy-2y-3z$



                $=-x^2+y=0$, we deduce that $y=x^2$ and $z=x^2+xy-y=x^2+xx^2-x^2=x^3$. This shows that if $(x,y,z)in M, (x,y,z)=(x,x^2,x^3)$ and for every $(x,y,z)in M$, $fcirc h(x,y,z)=(x,y,z)$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 23 at 0:04









                Tsemo AristideTsemo Aristide

                59.2k11445




                59.2k11445






























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