n-1 form inducing normal unit vector field












2












$begingroup$


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






differential-geometry manifolds differential-forms vector-fields







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    2












    $begingroup$


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






    differential-geometry manifolds differential-forms vector-fields







    share|cite|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$


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






      differential-geometry manifolds differential-forms vector-fields







      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 $M$. How would this imply the existence of a normal unit vector field on $M$?






      differential-geometry manifolds differential-forms vector-fields




      differential-geometry manifolds differential-forms vector-fields






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 3 '18 at 23:35







      AkatsukiMaliki

















      asked Dec 3 '18 at 23:24









      AkatsukiMalikiAkatsukiMaliki

      313110




      313110






















          1 Answer
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          $begingroup$

          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 $x$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Where do you use the fact that $omega$ is non-vanishing? Is it to say that c is either positive or negative? Moreover, I don't see how you relate the form to $n(x)$, how does $n$ kind of depend on c. And is there a reason why you only added dx1?
            $endgroup$
            – AkatsukiMaliki
            Dec 4 '18 at 8:44










          • $begingroup$
            The fact that $omega$ does not vanish is used to remark that $cneq 0$.
            $endgroup$
            – Tsemo Aristide
            Dec 4 '18 at 9:33












          • $begingroup$
            But you wrote obly dx1, why not all up to dxn?
            $endgroup$
            – AkatsukiMaliki
            Dec 4 '18 at 10:09










          • $begingroup$
            Only* . But i dont see the relation betwenen c and n(x).
            $endgroup$
            – AkatsukiMaliki
            Dec 4 '18 at 10:12










          • $begingroup$
            I think c can be zero in this case, take $omega = dx + dy$ and you could in fact get after doing the wedge : $dxdy + dydx = 0$.
            $endgroup$
            – AkatsukiMaliki
            Dec 4 '18 at 15:06











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          $begingroup$

          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 $x$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Where do you use the fact that $omega$ is non-vanishing? Is it to say that c is either positive or negative? Moreover, I don't see how you relate the form to $n(x)$, how does $n$ kind of depend on c. And is there a reason why you only added dx1?
            $endgroup$
            – AkatsukiMaliki
            Dec 4 '18 at 8:44










          • $begingroup$
            The fact that $omega$ does not vanish is used to remark that $cneq 0$.
            $endgroup$
            – Tsemo Aristide
            Dec 4 '18 at 9:33












          • $begingroup$
            But you wrote obly dx1, why not all up to dxn?
            $endgroup$
            – AkatsukiMaliki
            Dec 4 '18 at 10:09










          • $begingroup$
            Only* . But i dont see the relation betwenen c and n(x).
            $endgroup$
            – AkatsukiMaliki
            Dec 4 '18 at 10:12










          • $begingroup$
            I think c can be zero in this case, take $omega = dx + dy$ and you could in fact get after doing the wedge : $dxdy + dydx = 0$.
            $endgroup$
            – AkatsukiMaliki
            Dec 4 '18 at 15:06
















          1












          $begingroup$

          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 $x$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Where do you use the fact that $omega$ is non-vanishing? Is it to say that c is either positive or negative? Moreover, I don't see how you relate the form to $n(x)$, how does $n$ kind of depend on c. And is there a reason why you only added dx1?
            $endgroup$
            – AkatsukiMaliki
            Dec 4 '18 at 8:44










          • $begingroup$
            The fact that $omega$ does not vanish is used to remark that $cneq 0$.
            $endgroup$
            – Tsemo Aristide
            Dec 4 '18 at 9:33












          • $begingroup$
            But you wrote obly dx1, why not all up to dxn?
            $endgroup$
            – AkatsukiMaliki
            Dec 4 '18 at 10:09










          • $begingroup$
            Only* . But i dont see the relation betwenen c and n(x).
            $endgroup$
            – AkatsukiMaliki
            Dec 4 '18 at 10:12










          • $begingroup$
            I think c can be zero in this case, take $omega = dx + dy$ and you could in fact get after doing the wedge : $dxdy + dydx = 0$.
            $endgroup$
            – AkatsukiMaliki
            Dec 4 '18 at 15:06














          1












          1








          1





          $begingroup$

          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 $x$.






          share|cite|improve this answer











          $endgroup$



          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 $x$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 4 '18 at 9:31

























          answered Dec 4 '18 at 1:25









          Tsemo AristideTsemo Aristide

          57k11444




          57k11444












          • $begingroup$
            Where do you use the fact that $omega$ is non-vanishing? Is it to say that c is either positive or negative? Moreover, I don't see how you relate the form to $n(x)$, how does $n$ kind of depend on c. And is there a reason why you only added dx1?
            $endgroup$
            – AkatsukiMaliki
            Dec 4 '18 at 8:44










          • $begingroup$
            The fact that $omega$ does not vanish is used to remark that $cneq 0$.
            $endgroup$
            – Tsemo Aristide
            Dec 4 '18 at 9:33












          • $begingroup$
            But you wrote obly dx1, why not all up to dxn?
            $endgroup$
            – AkatsukiMaliki
            Dec 4 '18 at 10:09










          • $begingroup$
            Only* . But i dont see the relation betwenen c and n(x).
            $endgroup$
            – AkatsukiMaliki
            Dec 4 '18 at 10:12










          • $begingroup$
            I think c can be zero in this case, take $omega = dx + dy$ and you could in fact get after doing the wedge : $dxdy + dydx = 0$.
            $endgroup$
            – AkatsukiMaliki
            Dec 4 '18 at 15:06


















          • $begingroup$
            Where do you use the fact that $omega$ is non-vanishing? Is it to say that c is either positive or negative? Moreover, I don't see how you relate the form to $n(x)$, how does $n$ kind of depend on c. And is there a reason why you only added dx1?
            $endgroup$
            – AkatsukiMaliki
            Dec 4 '18 at 8:44










          • $begingroup$
            The fact that $omega$ does not vanish is used to remark that $cneq 0$.
            $endgroup$
            – Tsemo Aristide
            Dec 4 '18 at 9:33












          • $begingroup$
            But you wrote obly dx1, why not all up to dxn?
            $endgroup$
            – AkatsukiMaliki
            Dec 4 '18 at 10:09










          • $begingroup$
            Only* . But i dont see the relation betwenen c and n(x).
            $endgroup$
            – AkatsukiMaliki
            Dec 4 '18 at 10:12










          • $begingroup$
            I think c can be zero in this case, take $omega = dx + dy$ and you could in fact get after doing the wedge : $dxdy + dydx = 0$.
            $endgroup$
            – AkatsukiMaliki
            Dec 4 '18 at 15:06
















          $begingroup$
          Where do you use the fact that $omega$ is non-vanishing? Is it to say that c is either positive or negative? Moreover, I don't see how you relate the form to $n(x)$, how does $n$ kind of depend on c. And is there a reason why you only added dx1?
          $endgroup$
          – AkatsukiMaliki
          Dec 4 '18 at 8:44




          $begingroup$
          Where do you use the fact that $omega$ is non-vanishing? Is it to say that c is either positive or negative? Moreover, I don't see how you relate the form to $n(x)$, how does $n$ kind of depend on c. And is there a reason why you only added dx1?
          $endgroup$
          – AkatsukiMaliki
          Dec 4 '18 at 8:44












          $begingroup$
          The fact that $omega$ does not vanish is used to remark that $cneq 0$.
          $endgroup$
          – Tsemo Aristide
          Dec 4 '18 at 9:33






          $begingroup$
          The fact that $omega$ does not vanish is used to remark that $cneq 0$.
          $endgroup$
          – Tsemo Aristide
          Dec 4 '18 at 9:33














          $begingroup$
          But you wrote obly dx1, why not all up to dxn?
          $endgroup$
          – AkatsukiMaliki
          Dec 4 '18 at 10:09




          $begingroup$
          But you wrote obly dx1, why not all up to dxn?
          $endgroup$
          – AkatsukiMaliki
          Dec 4 '18 at 10:09












          $begingroup$
          Only* . But i dont see the relation betwenen c and n(x).
          $endgroup$
          – AkatsukiMaliki
          Dec 4 '18 at 10:12




          $begingroup$
          Only* . But i dont see the relation betwenen c and n(x).
          $endgroup$
          – AkatsukiMaliki
          Dec 4 '18 at 10:12












          $begingroup$
          I think c can be zero in this case, take $omega = dx + dy$ and you could in fact get after doing the wedge : $dxdy + dydx = 0$.
          $endgroup$
          – AkatsukiMaliki
          Dec 4 '18 at 15:06




          $begingroup$
          I think c can be zero in this case, take $omega = dx + dy$ and you could in fact get after doing the wedge : $dxdy + dydx = 0$.
          $endgroup$
          – AkatsukiMaliki
          Dec 4 '18 at 15:06


















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