Question about proof of n-1 form inducing normal unit vector field












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Suppose we have a $n-1$ dimensional manifold $M subset mathbb{R}^n$ and a non-vanishing $n-1$ form $omega$ on $M$. This implies the existence of a normal unit vector field on $M$.



The proof of this goes as follows:



"Let $xin M$, there exists a neighborhood $U$ of $x$ in $mathbb{R}^n$, a submersion $f:Urightarrow mathbb{R}$ such that $Mcap U=f^{-1}(0)$, $T_xM={uin T_xmathbb{R}^n:df_x(u)=0}$. Write $df_x=(partial f_1(x),...,partial f_n(x))$. You can identify $(partial f_1(x),...,partial f_n(x))$ with a vector $u_x$ of $T_xmathbb{R}^n$ such that $Vect(u_x)$ is a supplementary space to $T_xM$. Let $Omega$ be the canonical volume form $dx_1wedge...wedge dx_n$, $(partial f_1(x)dx_1+...+partial f_n(x)dx_n)wedge omega =cOmega$, if $c>0$, define $n(x)={1over{|u_x|}}u_x$ if $c<0$, define $n(x)=-{1over{|u_x|}}u_x$. Remark that $n(x)$ is well defined in a neighborhood of $x$ since it does not depend of the choice of $f$. In fact $u_x$ is a unit vector orthogonal to $T_xM$ relatively to the usual scalar product and it continuously depends of $x$."



Anyways, I have a few questions about the proof
The first one is: Why is the wedge "$(partial f_1(x)dx_1+...+partial f_n(x)dx_n)wedge omega = cOmega $" important, how is fundamentally linked to $n(x)$, I don't see the importance, and okay $omega$ is non vanishing and there is an $i$ such that $partial f_i(x)$ is not zero, but that's no reason to say that the wedge product is not zero, so $c$ could still be zero.










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    0












    $begingroup$


    Suppose we have a $n-1$ dimensional manifold $M subset mathbb{R}^n$ and a non-vanishing $n-1$ form $omega$ on $M$. This implies the existence of a normal unit vector field on $M$.



    The proof of this goes as follows:



    "Let $xin M$, there exists a neighborhood $U$ of $x$ in $mathbb{R}^n$, a submersion $f:Urightarrow mathbb{R}$ such that $Mcap U=f^{-1}(0)$, $T_xM={uin T_xmathbb{R}^n:df_x(u)=0}$. Write $df_x=(partial f_1(x),...,partial f_n(x))$. You can identify $(partial f_1(x),...,partial f_n(x))$ with a vector $u_x$ of $T_xmathbb{R}^n$ such that $Vect(u_x)$ is a supplementary space to $T_xM$. Let $Omega$ be the canonical volume form $dx_1wedge...wedge dx_n$, $(partial f_1(x)dx_1+...+partial f_n(x)dx_n)wedge omega =cOmega$, if $c>0$, define $n(x)={1over{|u_x|}}u_x$ if $c<0$, define $n(x)=-{1over{|u_x|}}u_x$. Remark that $n(x)$ is well defined in a neighborhood of $x$ since it does not depend of the choice of $f$. In fact $u_x$ is a unit vector orthogonal to $T_xM$ relatively to the usual scalar product and it continuously depends of $x$."



    Anyways, I have a few questions about the proof
    The first one is: Why is the wedge "$(partial f_1(x)dx_1+...+partial f_n(x)dx_n)wedge omega = cOmega $" important, how is fundamentally linked to $n(x)$, I don't see the importance, and okay $omega$ is non vanishing and there is an $i$ such that $partial f_i(x)$ is not zero, but that's no reason to say that the wedge product is not zero, so $c$ could still be zero.










    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      Suppose we have a $n-1$ dimensional manifold $M subset mathbb{R}^n$ and a non-vanishing $n-1$ form $omega$ on $M$. This implies the existence of a normal unit vector field on $M$.



      The proof of this goes as follows:



      "Let $xin M$, there exists a neighborhood $U$ of $x$ in $mathbb{R}^n$, a submersion $f:Urightarrow mathbb{R}$ such that $Mcap U=f^{-1}(0)$, $T_xM={uin T_xmathbb{R}^n:df_x(u)=0}$. Write $df_x=(partial f_1(x),...,partial f_n(x))$. You can identify $(partial f_1(x),...,partial f_n(x))$ with a vector $u_x$ of $T_xmathbb{R}^n$ such that $Vect(u_x)$ is a supplementary space to $T_xM$. Let $Omega$ be the canonical volume form $dx_1wedge...wedge dx_n$, $(partial f_1(x)dx_1+...+partial f_n(x)dx_n)wedge omega =cOmega$, if $c>0$, define $n(x)={1over{|u_x|}}u_x$ if $c<0$, define $n(x)=-{1over{|u_x|}}u_x$. Remark that $n(x)$ is well defined in a neighborhood of $x$ since it does not depend of the choice of $f$. In fact $u_x$ is a unit vector orthogonal to $T_xM$ relatively to the usual scalar product and it continuously depends of $x$."



      Anyways, I have a few questions about the proof
      The first one is: Why is the wedge "$(partial f_1(x)dx_1+...+partial f_n(x)dx_n)wedge omega = cOmega $" important, how is fundamentally linked to $n(x)$, I don't see the importance, and okay $omega$ is non vanishing and there is an $i$ such that $partial f_i(x)$ is not zero, but that's no reason to say that the wedge product is not zero, so $c$ could still be zero.










      share|cite|improve this question









      $endgroup$




      Suppose we have a $n-1$ dimensional manifold $M subset mathbb{R}^n$ and a non-vanishing $n-1$ form $omega$ on $M$. This implies the existence of a normal unit vector field on $M$.



      The proof of this goes as follows:



      "Let $xin M$, there exists a neighborhood $U$ of $x$ in $mathbb{R}^n$, a submersion $f:Urightarrow mathbb{R}$ such that $Mcap U=f^{-1}(0)$, $T_xM={uin T_xmathbb{R}^n:df_x(u)=0}$. Write $df_x=(partial f_1(x),...,partial f_n(x))$. You can identify $(partial f_1(x),...,partial f_n(x))$ with a vector $u_x$ of $T_xmathbb{R}^n$ such that $Vect(u_x)$ is a supplementary space to $T_xM$. Let $Omega$ be the canonical volume form $dx_1wedge...wedge dx_n$, $(partial f_1(x)dx_1+...+partial f_n(x)dx_n)wedge omega =cOmega$, if $c>0$, define $n(x)={1over{|u_x|}}u_x$ if $c<0$, define $n(x)=-{1over{|u_x|}}u_x$. Remark that $n(x)$ is well defined in a neighborhood of $x$ since it does not depend of the choice of $f$. In fact $u_x$ is a unit vector orthogonal to $T_xM$ relatively to the usual scalar product and it continuously depends of $x$."



      Anyways, I have a few questions about the proof
      The first one is: Why is the wedge "$(partial f_1(x)dx_1+...+partial f_n(x)dx_n)wedge omega = cOmega $" important, how is fundamentally linked to $n(x)$, I don't see the importance, and okay $omega$ is non vanishing and there is an $i$ such that $partial f_i(x)$ is not zero, but that's no reason to say that the wedge product is not zero, so $c$ could still be zero.







      differential-geometry manifolds differential-forms vector-fields tangent-spaces






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      asked Dec 4 '18 at 15:41









      AkatsukiMalikiAkatsukiMaliki

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