Why is this function well defined and $C^infty$? [Part 2]












0












$begingroup$


Let $Msubseteqmathbb{R}^k$ be an embedded submanifold of $mathbb{R}^k$, with dim$M=n$.



Let $(U,phi)$ be a smooth chart for $M$.



Then $phi^{-1}:phi(U)to U$ is a diffeomorphism and for each $xin phi(U)$ the map $d(phi^{-1})_x:T_xphi(U)to T_pU$ is isomorphism of $mathbb{R}$-vector spaces, with $p=phi^{-1}(x)$.



My book defines $$G:phi(U) timesmathbb{S}^{n-1} to mathbb{S}^{k-1}, quad (x,u)mapsto frac{d(phi^{-1})_x(u)}{lVert d(phi^{-1})_x(u) rVert}$$



and says that this function is $C^infty$.




Here is my understanding of the situation, and I want to know if I'm right and if there are some observations to simplify the things.




Identifying $T_pU cong T_pM$, and $T_xphi(U)cong T_x mathbb{R}^ncong mathbb{R}^n$ I can consider $d(phi^{-1})_x:mathbb{R}^nto T_pM$.



Since I can identify $T_pM$ with a subspace of $T_pmathbb{R}^k cong mathbb{R}^k$, I can consider $d(phi^{-1})_x:mathbb{R}^nto mathbb{R}^k$.



Now let $Tmathbb{R}^n$ be the tangent bundle of $mathbb{R}^n$, which is $Tmathbb{R}^n={(x,u):xin mathbb{R}^n, uin T_xmathbb{R}cong mathbb{R}^n}cong mathbb{R}^ntimesmathbb{R}^n$.



I can consider the global differential



$$d(phi^{-1}):mathbb{R}^ntimesmathbb{R}^nto Tmathbb{R}^k=mathbb{R}^ktimesmathbb{R}^k, quad (x,u)mapsto (phi^{-1}(x),d(phi^{-1})_x(u))$$ which should be $C^infty$.



Composing with the projection $mathbb{R}^ktimesmathbb{R}^ktomathbb{R}^k$ onto the last $k$ coordinates, I have now



$$d(phi^{-1}):mathbb{R}^ntimesmathbb{R}^nto mathbb{R}^k, quad (x,u)mapsto d(phi^{-1})_x(u)$$ which is $C^infty$



Now I can consider the composite map $phi(U)times mathbb{S}^{n-1}to mathbb{R}^ntimesmathbb{R}^nto mathbb{R}^k $ where the first map is the inclusion and the second is $d(phi^{-1})$ of the preceding paragraph.



So now I have $phi(U)times mathbb{S}^{n-1}to mathbb{R}^k $ which is $C^infty$.



Now I observe that if $(x,u)in phi(U)times mathbb{S}^{n-1}$, then by injectivity of $d(phi^{-1})_x$ and since $unot= 0$, I have $d(phi^{-1})_x(u)not =0$, so $phi(U)times mathbb{S}^{n-1}to mathbb{R}^ksetminus{0}, quad (x,u)mapsto d(phi^{-1})_x(u)$.



Finally, dividing by $lVert d(phi^{-1})_x(u) rVert$ I have the desired map $G$, and I have shown that it is well defined and $C^infty$.




Is my argument right? Are there some simplifications that can be made BUT without removing rigor to the argument? Is it "normal" identify all these things and proceed in this way?




N.B. In the title there is Part 2 beacuse I asked this question some time ago Why is this function well defined and $C^infty$? but without receiving any answer. I have worked completely alone on this argument so I would be happy if someone will dicuss this topic with me with a lot of details and rigor, and if you have tips or advices for me, you are welcome!!






multivariable-calculus differential-geometry smooth-manifolds smooth-functions







share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    Let $Msubseteqmathbb{R}^k$ be an embedded submanifold of $mathbb{R}^k$, with dim$M=n$.



    Let $(U,phi)$ be a smooth chart for $M$.



    Then $phi^{-1}:phi(U)to U$ is a diffeomorphism and for each $xin phi(U)$ the map $d(phi^{-1})_x:T_xphi(U)to T_pU$ is isomorphism of $mathbb{R}$-vector spaces, with $p=phi^{-1}(x)$.



    My book defines $$G:phi(U) timesmathbb{S}^{n-1} to mathbb{S}^{k-1}, quad (x,u)mapsto frac{d(phi^{-1})_x(u)}{lVert d(phi^{-1})_x(u) rVert}$$



    and says that this function is $C^infty$.




    Here is my understanding of the situation, and I want to know if I'm right and if there are some observations to simplify the things.




    Identifying $T_pU cong T_pM$, and $T_xphi(U)cong T_x mathbb{R}^ncong mathbb{R}^n$ I can consider $d(phi^{-1})_x:mathbb{R}^nto T_pM$.



    Since I can identify $T_pM$ with a subspace of $T_pmathbb{R}^k cong mathbb{R}^k$, I can consider $d(phi^{-1})_x:mathbb{R}^nto mathbb{R}^k$.



    Now let $Tmathbb{R}^n$ be the tangent bundle of $mathbb{R}^n$, which is $Tmathbb{R}^n={(x,u):xin mathbb{R}^n, uin T_xmathbb{R}cong mathbb{R}^n}cong mathbb{R}^ntimesmathbb{R}^n$.



    I can consider the global differential



    $$d(phi^{-1}):mathbb{R}^ntimesmathbb{R}^nto Tmathbb{R}^k=mathbb{R}^ktimesmathbb{R}^k, quad (x,u)mapsto (phi^{-1}(x),d(phi^{-1})_x(u))$$ which should be $C^infty$.



    Composing with the projection $mathbb{R}^ktimesmathbb{R}^ktomathbb{R}^k$ onto the last $k$ coordinates, I have now



    $$d(phi^{-1}):mathbb{R}^ntimesmathbb{R}^nto mathbb{R}^k, quad (x,u)mapsto d(phi^{-1})_x(u)$$ which is $C^infty$



    Now I can consider the composite map $phi(U)times mathbb{S}^{n-1}to mathbb{R}^ntimesmathbb{R}^nto mathbb{R}^k $ where the first map is the inclusion and the second is $d(phi^{-1})$ of the preceding paragraph.



    So now I have $phi(U)times mathbb{S}^{n-1}to mathbb{R}^k $ which is $C^infty$.



    Now I observe that if $(x,u)in phi(U)times mathbb{S}^{n-1}$, then by injectivity of $d(phi^{-1})_x$ and since $unot= 0$, I have $d(phi^{-1})_x(u)not =0$, so $phi(U)times mathbb{S}^{n-1}to mathbb{R}^ksetminus{0}, quad (x,u)mapsto d(phi^{-1})_x(u)$.



    Finally, dividing by $lVert d(phi^{-1})_x(u) rVert$ I have the desired map $G$, and I have shown that it is well defined and $C^infty$.




    Is my argument right? Are there some simplifications that can be made BUT without removing rigor to the argument? Is it "normal" identify all these things and proceed in this way?




    N.B. In the title there is Part 2 beacuse I asked this question some time ago Why is this function well defined and $C^infty$? but without receiving any answer. I have worked completely alone on this argument so I would be happy if someone will dicuss this topic with me with a lot of details and rigor, and if you have tips or advices for me, you are welcome!!






    multivariable-calculus differential-geometry smooth-manifolds smooth-functions







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      Let $Msubseteqmathbb{R}^k$ be an embedded submanifold of $mathbb{R}^k$, with dim$M=n$.



      Let $(U,phi)$ be a smooth chart for $M$.



      Then $phi^{-1}:phi(U)to U$ is a diffeomorphism and for each $xin phi(U)$ the map $d(phi^{-1})_x:T_xphi(U)to T_pU$ is isomorphism of $mathbb{R}$-vector spaces, with $p=phi^{-1}(x)$.



      My book defines $$G:phi(U) timesmathbb{S}^{n-1} to mathbb{S}^{k-1}, quad (x,u)mapsto frac{d(phi^{-1})_x(u)}{lVert d(phi^{-1})_x(u) rVert}$$



      and says that this function is $C^infty$.




      Here is my understanding of the situation, and I want to know if I'm right and if there are some observations to simplify the things.




      Identifying $T_pU cong T_pM$, and $T_xphi(U)cong T_x mathbb{R}^ncong mathbb{R}^n$ I can consider $d(phi^{-1})_x:mathbb{R}^nto T_pM$.



      Since I can identify $T_pM$ with a subspace of $T_pmathbb{R}^k cong mathbb{R}^k$, I can consider $d(phi^{-1})_x:mathbb{R}^nto mathbb{R}^k$.



      Now let $Tmathbb{R}^n$ be the tangent bundle of $mathbb{R}^n$, which is $Tmathbb{R}^n={(x,u):xin mathbb{R}^n, uin T_xmathbb{R}cong mathbb{R}^n}cong mathbb{R}^ntimesmathbb{R}^n$.



      I can consider the global differential



      $$d(phi^{-1}):mathbb{R}^ntimesmathbb{R}^nto Tmathbb{R}^k=mathbb{R}^ktimesmathbb{R}^k, quad (x,u)mapsto (phi^{-1}(x),d(phi^{-1})_x(u))$$ which should be $C^infty$.



      Composing with the projection $mathbb{R}^ktimesmathbb{R}^ktomathbb{R}^k$ onto the last $k$ coordinates, I have now



      $$d(phi^{-1}):mathbb{R}^ntimesmathbb{R}^nto mathbb{R}^k, quad (x,u)mapsto d(phi^{-1})_x(u)$$ which is $C^infty$



      Now I can consider the composite map $phi(U)times mathbb{S}^{n-1}to mathbb{R}^ntimesmathbb{R}^nto mathbb{R}^k $ where the first map is the inclusion and the second is $d(phi^{-1})$ of the preceding paragraph.



      So now I have $phi(U)times mathbb{S}^{n-1}to mathbb{R}^k $ which is $C^infty$.



      Now I observe that if $(x,u)in phi(U)times mathbb{S}^{n-1}$, then by injectivity of $d(phi^{-1})_x$ and since $unot= 0$, I have $d(phi^{-1})_x(u)not =0$, so $phi(U)times mathbb{S}^{n-1}to mathbb{R}^ksetminus{0}, quad (x,u)mapsto d(phi^{-1})_x(u)$.



      Finally, dividing by $lVert d(phi^{-1})_x(u) rVert$ I have the desired map $G$, and I have shown that it is well defined and $C^infty$.




      Is my argument right? Are there some simplifications that can be made BUT without removing rigor to the argument? Is it "normal" identify all these things and proceed in this way?




      N.B. In the title there is Part 2 beacuse I asked this question some time ago Why is this function well defined and $C^infty$? but without receiving any answer. I have worked completely alone on this argument so I would be happy if someone will dicuss this topic with me with a lot of details and rigor, and if you have tips or advices for me, you are welcome!!






      multivariable-calculus differential-geometry smooth-manifolds smooth-functions







      share|cite|improve this question











      $endgroup$




      Let $Msubseteqmathbb{R}^k$ be an embedded submanifold of $mathbb{R}^k$, with dim$M=n$.



      Let $(U,phi)$ be a smooth chart for $M$.



      Then $phi^{-1}:phi(U)to U$ is a diffeomorphism and for each $xin phi(U)$ the map $d(phi^{-1})_x:T_xphi(U)to T_pU$ is isomorphism of $mathbb{R}$-vector spaces, with $p=phi^{-1}(x)$.



      My book defines $$G:phi(U) timesmathbb{S}^{n-1} to mathbb{S}^{k-1}, quad (x,u)mapsto frac{d(phi^{-1})_x(u)}{lVert d(phi^{-1})_x(u) rVert}$$



      and says that this function is $C^infty$.




      Here is my understanding of the situation, and I want to know if I'm right and if there are some observations to simplify the things.




      Identifying $T_pU cong T_pM$, and $T_xphi(U)cong T_x mathbb{R}^ncong mathbb{R}^n$ I can consider $d(phi^{-1})_x:mathbb{R}^nto T_pM$.



      Since I can identify $T_pM$ with a subspace of $T_pmathbb{R}^k cong mathbb{R}^k$, I can consider $d(phi^{-1})_x:mathbb{R}^nto mathbb{R}^k$.



      Now let $Tmathbb{R}^n$ be the tangent bundle of $mathbb{R}^n$, which is $Tmathbb{R}^n={(x,u):xin mathbb{R}^n, uin T_xmathbb{R}cong mathbb{R}^n}cong mathbb{R}^ntimesmathbb{R}^n$.



      I can consider the global differential



      $$d(phi^{-1}):mathbb{R}^ntimesmathbb{R}^nto Tmathbb{R}^k=mathbb{R}^ktimesmathbb{R}^k, quad (x,u)mapsto (phi^{-1}(x),d(phi^{-1})_x(u))$$ which should be $C^infty$.



      Composing with the projection $mathbb{R}^ktimesmathbb{R}^ktomathbb{R}^k$ onto the last $k$ coordinates, I have now



      $$d(phi^{-1}):mathbb{R}^ntimesmathbb{R}^nto mathbb{R}^k, quad (x,u)mapsto d(phi^{-1})_x(u)$$ which is $C^infty$



      Now I can consider the composite map $phi(U)times mathbb{S}^{n-1}to mathbb{R}^ntimesmathbb{R}^nto mathbb{R}^k $ where the first map is the inclusion and the second is $d(phi^{-1})$ of the preceding paragraph.



      So now I have $phi(U)times mathbb{S}^{n-1}to mathbb{R}^k $ which is $C^infty$.



      Now I observe that if $(x,u)in phi(U)times mathbb{S}^{n-1}$, then by injectivity of $d(phi^{-1})_x$ and since $unot= 0$, I have $d(phi^{-1})_x(u)not =0$, so $phi(U)times mathbb{S}^{n-1}to mathbb{R}^ksetminus{0}, quad (x,u)mapsto d(phi^{-1})_x(u)$.



      Finally, dividing by $lVert d(phi^{-1})_x(u) rVert$ I have the desired map $G$, and I have shown that it is well defined and $C^infty$.




      Is my argument right? Are there some simplifications that can be made BUT without removing rigor to the argument? Is it "normal" identify all these things and proceed in this way?




      N.B. In the title there is Part 2 beacuse I asked this question some time ago Why is this function well defined and $C^infty$? but without receiving any answer. I have worked completely alone on this argument so I would be happy if someone will dicuss this topic with me with a lot of details and rigor, and if you have tips or advices for me, you are welcome!!






      multivariable-calculus differential-geometry smooth-manifolds smooth-functions




      multivariable-calculus differential-geometry smooth-manifolds smooth-functions






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




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      edited Feb 12 at 19:27







      Minato

















      asked Jan 22 at 11:57









      MinatoMinato

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