Question about proof of n-1 form inducing normal unit vector field
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
asked Dec 4 '18 at 15:41
AkatsukiMaliki
308110
add a comment |
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
asked Dec 4 '18 at 15:41
AkatsukiMaliki
308110
add a comment |
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
asked Dec 4 '18 at 15:41
AkatsukiMaliki
308110
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
differential-geometry manifolds differential-forms vector-fields tangent-spaces
asked Dec 4 '18 at 15:41
AkatsukiMaliki
308110
asked Dec 4 '18 at 15:41
AkatsukiMaliki
308110
asked Dec 4 '18 at 15:41
AkatsukiMaliki
308110
asked Dec 4 '18 at 15:41
AkatsukiMaliki
308110
asked Dec 4 '18 at 15:41
AkatsukiMaliki
308110
308110
add a comment |
add a comment |
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