A vector field corresponding to the complement of the tangent bundle
$begingroup$
Let $M$ be a $m$ dimensional orientable manifold, and $N$ a $m-1$ dimensional orientable submanifold in $M$, then we know at each point $x in N$, $T_{x}M = T_x N oplus$its complement. I need to produce a vector field $X$ such that $X_x$ is a vector in the complement of $T_{x}N$.
I am trying to exploit the assumption that both manifolds are orientable meaning they all have a non-vanishing top form, but what next?
UPDATE1:
Take a slice chart $(x_1, ldots, x_n)$ of $M$. Let the orientation form on $M$ be $w_M = fdx_1 wedge ldots wedge dx_n$, and the orientation form on $N$ be $w_N = fdx_1 wedge ldots wedge dx_{n-1}$. A possible approach is to find a vector $V$ such that $V lnot w_{M} = w_N$. The computation is easy. Locally $V$ should be $frac{g}{f}frac{partial}{partial x_n}$. However, how to show this expression is independent of coordinate thus can be globalized?
UPDATE2:
Trying to understand @Tsemo Aristide's answer. Locally, the choice of the sign has to be consistent. so in a neighborhood, it has to be either $u_x$ or $-u_x$ throughout, which proves the smoothness. Is this correct?
differential-topology
asked Dec 7 '18 at 23:39
KeithKeith
420317
$endgroup$
add a comment |
$begingroup$
Let $M$ be a $m$ dimensional orientable manifold, and $N$ a $m-1$ dimensional orientable submanifold in $M$, then we know at each point $x in N$, $T_{x}M = T_x N oplus$its complement. I need to produce a vector field $X$ such that $X_x$ is a vector in the complement of $T_{x}N$.
I am trying to exploit the assumption that both manifolds are orientable meaning they all have a non-vanishing top form, but what next?
UPDATE1:
Take a slice chart $(x_1, ldots, x_n)$ of $M$. Let the orientation form on $M$ be $w_M = fdx_1 wedge ldots wedge dx_n$, and the orientation form on $N$ be $w_N = fdx_1 wedge ldots wedge dx_{n-1}$. A possible approach is to find a vector $V$ such that $V lnot w_{M} = w_N$. The computation is easy. Locally $V$ should be $frac{g}{f}frac{partial}{partial x_n}$. However, how to show this expression is independent of coordinate thus can be globalized?
UPDATE2:
Trying to understand @Tsemo Aristide's answer. Locally, the choice of the sign has to be consistent. so in a neighborhood, it has to be either $u_x$ or $-u_x$ throughout, which proves the smoothness. Is this correct?
differential-topology
asked Dec 7 '18 at 23:39
KeithKeith
420317
$endgroup$
add a comment |
$begingroup$
Let $M$ be a $m$ dimensional orientable manifold, and $N$ a $m-1$ dimensional orientable submanifold in $M$, then we know at each point $x in N$, $T_{x}M = T_x N oplus$its complement. I need to produce a vector field $X$ such that $X_x$ is a vector in the complement of $T_{x}N$.
I am trying to exploit the assumption that both manifolds are orientable meaning they all have a non-vanishing top form, but what next?
UPDATE1:
Take a slice chart $(x_1, ldots, x_n)$ of $M$. Let the orientation form on $M$ be $w_M = fdx_1 wedge ldots wedge dx_n$, and the orientation form on $N$ be $w_N = fdx_1 wedge ldots wedge dx_{n-1}$. A possible approach is to find a vector $V$ such that $V lnot w_{M} = w_N$. The computation is easy. Locally $V$ should be $frac{g}{f}frac{partial}{partial x_n}$. However, how to show this expression is independent of coordinate thus can be globalized?
UPDATE2:
Trying to understand @Tsemo Aristide's answer. Locally, the choice of the sign has to be consistent. so in a neighborhood, it has to be either $u_x$ or $-u_x$ throughout, which proves the smoothness. Is this correct?
differential-topology
asked Dec 7 '18 at 23:39
KeithKeith
420317
$endgroup$
Let $M$ be a $m$ dimensional orientable manifold, and $N$ a $m-1$ dimensional orientable submanifold in $M$, then we know at each point $x in N$, $T_{x}M = T_x N oplus$its complement. I need to produce a vector field $X$ such that $X_x$ is a vector in the complement of $T_{x}N$.
I am trying to exploit the assumption that both manifolds are orientable meaning they all have a non-vanishing top form, but what next?
UPDATE1:
Take a slice chart $(x_1, ldots, x_n)$ of $M$. Let the orientation form on $M$ be $w_M = fdx_1 wedge ldots wedge dx_n$, and the orientation form on $N$ be $w_N = fdx_1 wedge ldots wedge dx_{n-1}$. A possible approach is to find a vector $V$ such that $V lnot w_{M} = w_N$. The computation is easy. Locally $V$ should be $frac{g}{f}frac{partial}{partial x_n}$. However, how to show this expression is independent of coordinate thus can be globalized?
UPDATE2:
Trying to understand @Tsemo Aristide's answer. Locally, the choice of the sign has to be consistent. so in a neighborhood, it has to be either $u_x$ or $-u_x$ throughout, which proves the smoothness. Is this correct?
differential-topology
differential-topology
asked Dec 7 '18 at 23:39
KeithKeith
420317
asked Dec 7 '18 at 23:39
KeithKeith
420317
edited Dec 10 '18 at 5:02
Keith
asked Dec 7 '18 at 23:39
KeithKeith
420317
asked Dec 7 '18 at 23:39
KeithKeith
420317
asked Dec 7 '18 at 23:39
KeithKeith
420317
420317
add a comment |
add a comment |
1 Answer
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Let $Omega^M$ be the volume form of $M$ and $Omega^N$ the volume form of $N$. Consider a
differentiable metric defined on $M$. For every $xin N$, there exists two vectors of norm $1$, $u_x,-u_x$ orthogonal to $T_xN$, If the restriction of $i_{u_x}{Omega^M}_x$ to $T_xN$ is $c{Omega^N}_x, c>0$ write $n(x)=u_x$ otherwise, write $n(x)=-u_x$. Show that $n(x)$ is differentiable by using local coordinates.
answered Dec 8 '18 at 0:25
Tsemo AristideTsemo Aristide
57.5k11444
$endgroup$
$begingroup$
May you be more specific about how to check the differentiability? I cannot really figure out how to write down an explicit formula locally
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– Keith
Dec 8 '18 at 2:17
add a comment |
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1 Answer
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1 Answer
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active
oldest
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$begingroup$
Let $Omega^M$ be the volume form of $M$ and $Omega^N$ the volume form of $N$. Consider a
differentiable metric defined on $M$. For every $xin N$, there exists two vectors of norm $1$, $u_x,-u_x$ orthogonal to $T_xN$, If the restriction of $i_{u_x}{Omega^M}_x$ to $T_xN$ is $c{Omega^N}_x, c>0$ write $n(x)=u_x$ otherwise, write $n(x)=-u_x$. Show that $n(x)$ is differentiable by using local coordinates.
answered Dec 8 '18 at 0:25
Tsemo AristideTsemo Aristide
57.5k11444
$endgroup$
$begingroup$
May you be more specific about how to check the differentiability? I cannot really figure out how to write down an explicit formula locally
$endgroup$
– Keith
Dec 8 '18 at 2:17
add a comment |
$begingroup$
Let $Omega^M$ be the volume form of $M$ and $Omega^N$ the volume form of $N$. Consider a
differentiable metric defined on $M$. For every $xin N$, there exists two vectors of norm $1$, $u_x,-u_x$ orthogonal to $T_xN$, If the restriction of $i_{u_x}{Omega^M}_x$ to $T_xN$ is $c{Omega^N}_x, c>0$ write $n(x)=u_x$ otherwise, write $n(x)=-u_x$. Show that $n(x)$ is differentiable by using local coordinates.
answered Dec 8 '18 at 0:25
Tsemo AristideTsemo Aristide
57.5k11444
$endgroup$
$begingroup$
May you be more specific about how to check the differentiability? I cannot really figure out how to write down an explicit formula locally
$endgroup$
– Keith
Dec 8 '18 at 2:17
add a comment |
$begingroup$
Let $Omega^M$ be the volume form of $M$ and $Omega^N$ the volume form of $N$. Consider a
differentiable metric defined on $M$. For every $xin N$, there exists two vectors of norm $1$, $u_x,-u_x$ orthogonal to $T_xN$, If the restriction of $i_{u_x}{Omega^M}_x$ to $T_xN$ is $c{Omega^N}_x, c>0$ write $n(x)=u_x$ otherwise, write $n(x)=-u_x$. Show that $n(x)$ is differentiable by using local coordinates.
answered Dec 8 '18 at 0:25
Tsemo AristideTsemo Aristide
57.5k11444
$endgroup$
Let $Omega^M$ be the volume form of $M$ and $Omega^N$ the volume form of $N$. Consider a
differentiable metric defined on $M$. For every $xin N$, there exists two vectors of norm $1$, $u_x,-u_x$ orthogonal to $T_xN$, If the restriction of $i_{u_x}{Omega^M}_x$ to $T_xN$ is $c{Omega^N}_x, c>0$ write $n(x)=u_x$ otherwise, write $n(x)=-u_x$. Show that $n(x)$ is differentiable by using local coordinates.
answered Dec 8 '18 at 0:25
Tsemo AristideTsemo Aristide
57.5k11444
answered Dec 8 '18 at 0:25
Tsemo AristideTsemo Aristide
57.5k11444
answered Dec 8 '18 at 0:25
Tsemo AristideTsemo Aristide
57.5k11444
answered Dec 8 '18 at 0:25
Tsemo AristideTsemo Aristide
57.5k11444
57.5k11444
$begingroup$
May you be more specific about how to check the differentiability? I cannot really figure out how to write down an explicit formula locally
$endgroup$
– Keith
Dec 8 '18 at 2:17
add a comment |
$begingroup$
May you be more specific about how to check the differentiability? I cannot really figure out how to write down an explicit formula locally
$endgroup$
– Keith
Dec 8 '18 at 2:17
$begingroup$
May you be more specific about how to check the differentiability? I cannot really figure out how to write down an explicit formula locally
$endgroup$
– Keith
Dec 8 '18 at 2:17
$begingroup$
May you be more specific about how to check the differentiability? I cannot really figure out how to write down an explicit formula locally
$endgroup$
– Keith
Dec 8 '18 at 2:17
add a comment |
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