Why is this implying a normal vector field?
$begingroup$
Suppose $omega$ is a $n-1$ form on a $n-1$ dimensional manifold and $(a_1(x)dx_1 + ... + a_n(x)dx_n )wedge omega = cOmega $, with $c neq 0$ and $Omega =dx_1wedge...wedge dx_n$. Moreover $(a_1(x),...,a_n(x))= a(x)$ is a normal vector on the manifold, with a nonzero norm. Apparently with this you can say there exists a normal unit vector field on $M$, where you say $n(x) = frac{a(x)}{| a(x)| }$ if $c >0$ and $n(x) = -frac{a(x)}{| a(x)| }$ if $c < 0$. Why can you say this, how is from the wedge product seen that the normal vector fields is smooth, as in it "doesn't keep switching" .
differential-geometry manifolds vector-fields exterior-algebra
asked Dec 4 '18 at 18:20
AkatsukiMalikiAkatsukiMaliki
313110
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add a comment |
$begingroup$
Suppose $omega$ is a $n-1$ form on a $n-1$ dimensional manifold and $(a_1(x)dx_1 + ... + a_n(x)dx_n )wedge omega = cOmega $, with $c neq 0$ and $Omega =dx_1wedge...wedge dx_n$. Moreover $(a_1(x),...,a_n(x))= a(x)$ is a normal vector on the manifold, with a nonzero norm. Apparently with this you can say there exists a normal unit vector field on $M$, where you say $n(x) = frac{a(x)}{| a(x)| }$ if $c >0$ and $n(x) = -frac{a(x)}{| a(x)| }$ if $c < 0$. Why can you say this, how is from the wedge product seen that the normal vector fields is smooth, as in it "doesn't keep switching" .
differential-geometry manifolds vector-fields exterior-algebra
asked Dec 4 '18 at 18:20
AkatsukiMalikiAkatsukiMaliki
313110
$endgroup$
$begingroup$
Is the manifold embedded in $Bbb R^n$?
$endgroup$
– edm
Dec 5 '18 at 4:03
$begingroup$
Yeeah it is in Rn
$endgroup$
– AkatsukiMaliki
Dec 5 '18 at 7:06
$begingroup$
And is $a_1(x)dx_1 + cdots + a_n(x)dx_n$ defined on the manifold or on the whole $Bbb R^n$?
$endgroup$
– edm
Dec 5 '18 at 11:07
add a comment |
$begingroup$
Suppose $omega$ is a $n-1$ form on a $n-1$ dimensional manifold and $(a_1(x)dx_1 + ... + a_n(x)dx_n )wedge omega = cOmega $, with $c neq 0$ and $Omega =dx_1wedge...wedge dx_n$. Moreover $(a_1(x),...,a_n(x))= a(x)$ is a normal vector on the manifold, with a nonzero norm. Apparently with this you can say there exists a normal unit vector field on $M$, where you say $n(x) = frac{a(x)}{| a(x)| }$ if $c >0$ and $n(x) = -frac{a(x)}{| a(x)| }$ if $c < 0$. Why can you say this, how is from the wedge product seen that the normal vector fields is smooth, as in it "doesn't keep switching" .
differential-geometry manifolds vector-fields exterior-algebra
asked Dec 4 '18 at 18:20
AkatsukiMalikiAkatsukiMaliki
313110
$endgroup$
Suppose $omega$ is a $n-1$ form on a $n-1$ dimensional manifold and $(a_1(x)dx_1 + ... + a_n(x)dx_n )wedge omega = cOmega $, with $c neq 0$ and $Omega =dx_1wedge...wedge dx_n$. Moreover $(a_1(x),...,a_n(x))= a(x)$ is a normal vector on the manifold, with a nonzero norm. Apparently with this you can say there exists a normal unit vector field on $M$, where you say $n(x) = frac{a(x)}{| a(x)| }$ if $c >0$ and $n(x) = -frac{a(x)}{| a(x)| }$ if $c < 0$. Why can you say this, how is from the wedge product seen that the normal vector fields is smooth, as in it "doesn't keep switching" .
differential-geometry manifolds vector-fields exterior-algebra
differential-geometry manifolds vector-fields exterior-algebra
asked Dec 4 '18 at 18:20
AkatsukiMalikiAkatsukiMaliki
313110
asked Dec 4 '18 at 18:20
AkatsukiMalikiAkatsukiMaliki
313110
asked Dec 4 '18 at 18:20
AkatsukiMalikiAkatsukiMaliki
313110
asked Dec 4 '18 at 18:20
AkatsukiMalikiAkatsukiMaliki
313110
asked Dec 4 '18 at 18:20
AkatsukiMalikiAkatsukiMaliki
313110
313110
$begingroup$
Is the manifold embedded in $Bbb R^n$?
$endgroup$
– edm
Dec 5 '18 at 4:03
$begingroup$
Yeeah it is in Rn
$endgroup$
– AkatsukiMaliki
Dec 5 '18 at 7:06
$begingroup$
And is $a_1(x)dx_1 + cdots + a_n(x)dx_n$ defined on the manifold or on the whole $Bbb R^n$?
$endgroup$
– edm
Dec 5 '18 at 11:07
add a comment |
$begingroup$
Is the manifold embedded in $Bbb R^n$?
$endgroup$
– edm
Dec 5 '18 at 4:03
$begingroup$
Yeeah it is in Rn
$endgroup$
– AkatsukiMaliki
Dec 5 '18 at 7:06
$begingroup$
And is $a_1(x)dx_1 + cdots + a_n(x)dx_n$ defined on the manifold or on the whole $Bbb R^n$?
$endgroup$
– edm
Dec 5 '18 at 11:07
$begingroup$
Is the manifold embedded in $Bbb R^n$?
$endgroup$
– edm
Dec 5 '18 at 4:03
$begingroup$
Is the manifold embedded in $Bbb R^n$?
$endgroup$
– edm
Dec 5 '18 at 4:03
$begingroup$
Yeeah it is in Rn
$endgroup$
– AkatsukiMaliki
Dec 5 '18 at 7:06
$begingroup$
Yeeah it is in Rn
$endgroup$
– AkatsukiMaliki
Dec 5 '18 at 7:06
$begingroup$
And is $a_1(x)dx_1 + cdots + a_n(x)dx_n$ defined on the manifold or on the whole $Bbb R^n$?
$endgroup$
– edm
Dec 5 '18 at 11:07
$begingroup$
And is $a_1(x)dx_1 + cdots + a_n(x)dx_n$ defined on the manifold or on the whole $Bbb R^n$?
$endgroup$
– edm
Dec 5 '18 at 11:07
add a comment |
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$begingroup$
Is the manifold embedded in $Bbb R^n$?
$endgroup$
– edm
Dec 5 '18 at 4:03
$begingroup$
Yeeah it is in Rn
$endgroup$
– AkatsukiMaliki
Dec 5 '18 at 7:06
$begingroup$
And is $a_1(x)dx_1 + cdots + a_n(x)dx_n$ defined on the manifold or on the whole $Bbb R^n$?
$endgroup$
– edm
Dec 5 '18 at 11:07