n-1 form inducing normal unit vector field
$begingroup$
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
asked Dec 3 '18 at 23:24
AkatsukiMalikiAkatsukiMaliki
313110
$endgroup$
add a comment |
$begingroup$
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
asked Dec 3 '18 at 23:24
AkatsukiMalikiAkatsukiMaliki
313110
$endgroup$
add a comment |
$begingroup$
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
asked Dec 3 '18 at 23:24
AkatsukiMalikiAkatsukiMaliki
313110
$endgroup$
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
asked Dec 3 '18 at 23:24
AkatsukiMalikiAkatsukiMaliki
313110
asked Dec 3 '18 at 23:24
AkatsukiMalikiAkatsukiMaliki
313110
edited Dec 3 '18 at 23:35
AkatsukiMaliki
asked Dec 3 '18 at 23:24
AkatsukiMalikiAkatsukiMaliki
313110
asked Dec 3 '18 at 23:24
AkatsukiMalikiAkatsukiMaliki
313110
asked Dec 3 '18 at 23:24
AkatsukiMalikiAkatsukiMaliki
313110
313110
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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 $x$.
answered Dec 4 '18 at 1:25
Tsemo AristideTsemo Aristide
57k11444
$endgroup$
$begingroup$
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
|
show 2 more comments
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 $x$.
answered Dec 4 '18 at 1:25
Tsemo AristideTsemo Aristide
57k11444
$endgroup$
$begingroup$
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
|
show 2 more comments
$begingroup$
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 $x$.
answered Dec 4 '18 at 1:25
Tsemo AristideTsemo Aristide
57k11444
$endgroup$
$begingroup$
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
|
show 2 more comments
$begingroup$
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 $x$.
answered Dec 4 '18 at 1:25
Tsemo AristideTsemo Aristide
57k11444
$endgroup$
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 $x$.
answered Dec 4 '18 at 1:25
Tsemo AristideTsemo Aristide
57k11444
edited Dec 4 '18 at 9:31
answered Dec 4 '18 at 1:25
Tsemo AristideTsemo Aristide
57k11444
answered Dec 4 '18 at 1:25
Tsemo AristideTsemo Aristide
57k11444
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
|
show 2 more comments
$begingroup$
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
|
show 2 more comments
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