find the general integral manifolds of a given distribution
$begingroup$
Let $D$ be the distribution on $M={(x,y,z),x,y,z>0}$ generated by $X=yfrac{partial}{partial z}-zfrac{partial}{partial y}$ and $Y=zfrac{partial}{partial x}-xfrac{partial}{partial z}$, show $D$ is involutive and find the general integral manifolds of $D$.
I compute the flows of $X$ and $Y$, there are a lot of $sin$ and $cos$ in the form of flows of $X$ and $Y$ , I got $phi_t(x,y,z)=(x,-zsin t+ycos t,ysin t+zcos t)$ and $psi_s(x,y,z)=(zsin s+xcos s,y,-xsin s+zcos s)$, but to find the integral manifolds at last, I can not eliminate $s$ and $t$ to get a general form
differential-geometry
asked Dec 4 '18 at 23:23
PoorMathStudentPoorMathStudent
82
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$begingroup$
Let $D$ be the distribution on $M={(x,y,z),x,y,z>0}$ generated by $X=yfrac{partial}{partial z}-zfrac{partial}{partial y}$ and $Y=zfrac{partial}{partial x}-xfrac{partial}{partial z}$, show $D$ is involutive and find the general integral manifolds of $D$.
I compute the flows of $X$ and $Y$, there are a lot of $sin$ and $cos$ in the form of flows of $X$ and $Y$ , I got $phi_t(x,y,z)=(x,-zsin t+ycos t,ysin t+zcos t)$ and $psi_s(x,y,z)=(zsin s+xcos s,y,-xsin s+zcos s)$, but to find the integral manifolds at last, I can not eliminate $s$ and $t$ to get a general form
differential-geometry
asked Dec 4 '18 at 23:23
PoorMathStudentPoorMathStudent
82
$endgroup$
add a comment |
$begingroup$
Let $D$ be the distribution on $M={(x,y,z),x,y,z>0}$ generated by $X=yfrac{partial}{partial z}-zfrac{partial}{partial y}$ and $Y=zfrac{partial}{partial x}-xfrac{partial}{partial z}$, show $D$ is involutive and find the general integral manifolds of $D$.
I compute the flows of $X$ and $Y$, there are a lot of $sin$ and $cos$ in the form of flows of $X$ and $Y$ , I got $phi_t(x,y,z)=(x,-zsin t+ycos t,ysin t+zcos t)$ and $psi_s(x,y,z)=(zsin s+xcos s,y,-xsin s+zcos s)$, but to find the integral manifolds at last, I can not eliminate $s$ and $t$ to get a general form
differential-geometry
asked Dec 4 '18 at 23:23
PoorMathStudentPoorMathStudent
82
$endgroup$
Let $D$ be the distribution on $M={(x,y,z),x,y,z>0}$ generated by $X=yfrac{partial}{partial z}-zfrac{partial}{partial y}$ and $Y=zfrac{partial}{partial x}-xfrac{partial}{partial z}$, show $D$ is involutive and find the general integral manifolds of $D$.
I compute the flows of $X$ and $Y$, there are a lot of $sin$ and $cos$ in the form of flows of $X$ and $Y$ , I got $phi_t(x,y,z)=(x,-zsin t+ycos t,ysin t+zcos t)$ and $psi_s(x,y,z)=(zsin s+xcos s,y,-xsin s+zcos s)$, but to find the integral manifolds at last, I can not eliminate $s$ and $t$ to get a general form
differential-geometry
differential-geometry
asked Dec 4 '18 at 23:23
PoorMathStudentPoorMathStudent
82
asked Dec 4 '18 at 23:23
PoorMathStudentPoorMathStudent
82
asked Dec 4 '18 at 23:23
PoorMathStudentPoorMathStudent
82
asked Dec 4 '18 at 23:23
PoorMathStudentPoorMathStudent
82
asked Dec 4 '18 at 23:23
PoorMathStudentPoorMathStudent
82
82
add a comment |
add a comment |
1 Answer
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$begingroup$
Differential forms for the win. Note that if you take the $1$-form
$$omega = x,dx+y,dy+z,dz,$$
then $omega(X)=omega(Y)=0$. So the distribution is given by the kernel of $omega$. Of course, $omega = dbig(frac12(x^2+y^2+z^2)big)$, so integral manifolds of $D$ are the level sets $x^2+y^2+z^2=c$ for $c>0$.
answered Dec 5 '18 at 19:41
Ted ShifrinTed Shifrin
63.6k44591
$endgroup$
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$begingroup$
Differential forms for the win. Note that if you take the $1$-form
$$omega = x,dx+y,dy+z,dz,$$
then $omega(X)=omega(Y)=0$. So the distribution is given by the kernel of $omega$. Of course, $omega = dbig(frac12(x^2+y^2+z^2)big)$, so integral manifolds of $D$ are the level sets $x^2+y^2+z^2=c$ for $c>0$.
answered Dec 5 '18 at 19:41
Ted ShifrinTed Shifrin
63.6k44591
$endgroup$
add a comment |
$begingroup$
Differential forms for the win. Note that if you take the $1$-form
$$omega = x,dx+y,dy+z,dz,$$
then $omega(X)=omega(Y)=0$. So the distribution is given by the kernel of $omega$. Of course, $omega = dbig(frac12(x^2+y^2+z^2)big)$, so integral manifolds of $D$ are the level sets $x^2+y^2+z^2=c$ for $c>0$.
answered Dec 5 '18 at 19:41
Ted ShifrinTed Shifrin
63.6k44591
$endgroup$
add a comment |
$begingroup$
Differential forms for the win. Note that if you take the $1$-form
$$omega = x,dx+y,dy+z,dz,$$
then $omega(X)=omega(Y)=0$. So the distribution is given by the kernel of $omega$. Of course, $omega = dbig(frac12(x^2+y^2+z^2)big)$, so integral manifolds of $D$ are the level sets $x^2+y^2+z^2=c$ for $c>0$.
answered Dec 5 '18 at 19:41
Ted ShifrinTed Shifrin
63.6k44591
$endgroup$
Differential forms for the win. Note that if you take the $1$-form
$$omega = x,dx+y,dy+z,dz,$$
then $omega(X)=omega(Y)=0$. So the distribution is given by the kernel of $omega$. Of course, $omega = dbig(frac12(x^2+y^2+z^2)big)$, so integral manifolds of $D$ are the level sets $x^2+y^2+z^2=c$ for $c>0$.
answered Dec 5 '18 at 19:41
Ted ShifrinTed Shifrin
63.6k44591
answered Dec 5 '18 at 19:41
Ted ShifrinTed Shifrin
63.6k44591
answered Dec 5 '18 at 19:41
Ted ShifrinTed Shifrin
63.6k44591
answered Dec 5 '18 at 19:41
Ted ShifrinTed Shifrin
63.6k44591
63.6k44591
add a comment |
add a comment |
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