Clock form and derivative
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Consider $X,Y$ two smooth vector fields of $mathbbR^2$ and
the set $C=zin mathbbR^2;, det(X,Y)(z)neq 0$.
On the set $C$, we define the form $alpha$ by :
$
alpha(X)=1,quad alpha(Y)=0.
$Given a trajectory $z(cdot)$
solution on $[0,T]$ of the system
$$
dot z(t) = X(z(t)) + u(t),Y(z(t))
$$
where $u$ is a given function and such that $z(.)in C$,
we define:
$$
int_z(cdot) alpha = T.
$$
On which set do we have $dalpha=0$ ?
differential-geometry dynamical-systems differential-forms
asked Sep 3 at 11:39
Smilia
568516
add a comment |Â
up vote
1
down vote
favorite
Consider $X,Y$ two smooth vector fields of $mathbbR^2$ and
the set $C=zin mathbbR^2;, det(X,Y)(z)neq 0$.
On the set $C$, we define the form $alpha$ by :
$
alpha(X)=1,quad alpha(Y)=0.
$Given a trajectory $z(cdot)$
solution on $[0,T]$ of the system
$$
dot z(t) = X(z(t)) + u(t),Y(z(t))
$$
where $u$ is a given function and such that $z(.)in C$,
we define:
$$
int_z(cdot) alpha = T.
$$
On which set do we have $dalpha=0$ ?
differential-geometry dynamical-systems differential-forms
asked Sep 3 at 11:39
Smilia
568516
The form $alpha$ is well-defined the minute you specify its values at $X$ and $Y$. What is the purpose of the rest of the post?
– Amitai Yuval
Sep 3 at 13:41
apparently we can have a general picture of the kernel of $dalpha$ which is the set where $Y,[X,Y]$ are collinear... but I can't see it
– Smilia
Sep 3 at 14:18
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Consider $X,Y$ two smooth vector fields of $mathbbR^2$ and
the set $C=zin mathbbR^2;, det(X,Y)(z)neq 0$.
On the set $C$, we define the form $alpha$ by :
$
alpha(X)=1,quad alpha(Y)=0.
$Given a trajectory $z(cdot)$
solution on $[0,T]$ of the system
$$
dot z(t) = X(z(t)) + u(t),Y(z(t))
$$
where $u$ is a given function and such that $z(.)in C$,
we define:
$$
int_z(cdot) alpha = T.
$$
On which set do we have $dalpha=0$ ?
differential-geometry dynamical-systems differential-forms
asked Sep 3 at 11:39
Smilia
568516
Consider $X,Y$ two smooth vector fields of $mathbbR^2$ and
the set $C=zin mathbbR^2;, det(X,Y)(z)neq 0$.
On the set $C$, we define the form $alpha$ by :
$
alpha(X)=1,quad alpha(Y)=0.
$Given a trajectory $z(cdot)$
solution on $[0,T]$ of the system
$$
dot z(t) = X(z(t)) + u(t),Y(z(t))
$$
where $u$ is a given function and such that $z(.)in C$,
we define:
$$
int_z(cdot) alpha = T.
$$
On which set do we have $dalpha=0$ ?
differential-geometry dynamical-systems differential-forms
differential-geometry dynamical-systems differential-forms
asked Sep 3 at 11:39
Smilia
568516
asked Sep 3 at 11:39
Smilia
568516
edited Sep 3 at 12:41
asked Sep 3 at 11:39
Smilia
568516
asked Sep 3 at 11:39
Smilia
568516
asked Sep 3 at 11:39
Smilia
568516
568516
The form $alpha$ is well-defined the minute you specify its values at $X$ and $Y$. What is the purpose of the rest of the post?
– Amitai Yuval
Sep 3 at 13:41
apparently we can have a general picture of the kernel of $dalpha$ which is the set where $Y,[X,Y]$ are collinear... but I can't see it
– Smilia
Sep 3 at 14:18
add a comment |Â
The form $alpha$ is well-defined the minute you specify its values at $X$ and $Y$. What is the purpose of the rest of the post?
– Amitai Yuval
Sep 3 at 13:41
apparently we can have a general picture of the kernel of $dalpha$ which is the set where $Y,[X,Y]$ are collinear... but I can't see it
– Smilia
Sep 3 at 14:18
The form $alpha$ is well-defined the minute you specify its values at $X$ and $Y$. What is the purpose of the rest of the post?
– Amitai Yuval
Sep 3 at 13:41
The form $alpha$ is well-defined the minute you specify its values at $X$ and $Y$. What is the purpose of the rest of the post?
– Amitai Yuval
Sep 3 at 13:41
apparently we can have a general picture of the kernel of $dalpha$ which is the set where $Y,[X,Y]$ are collinear... but I can't see it
– Smilia
Sep 3 at 14:18
apparently we can have a general picture of the kernel of $dalpha$ which is the set where $Y,[X,Y]$ are collinear... but I can't see it
– Smilia
Sep 3 at 14:18
add a comment |Â
1 Answer
1
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oldest
votes
up vote
2
down vote
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By the intrinsic expression for the exterior derivative, we have $$dalpha(X,Y)=X(alpha(Y))-Y(alpha(X))-alpha([X,Y])=-alpha([X,Y]).$$ So, $alpha$ is closed at every point where $alpha([X,Y])=0$. But, by construction, the kernel of $alpha$ at a point $p$ is the line spanned by $Y(p)$. Hence, $dalpha$ vanishes at $p$ if and only if $Y$ and $[X,Y]$ are colinear at $p$.
answered Sep 3 at 15:07
Amitai Yuval
14.6k11026
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
By the intrinsic expression for the exterior derivative, we have $$dalpha(X,Y)=X(alpha(Y))-Y(alpha(X))-alpha([X,Y])=-alpha([X,Y]).$$ So, $alpha$ is closed at every point where $alpha([X,Y])=0$. But, by construction, the kernel of $alpha$ at a point $p$ is the line spanned by $Y(p)$. Hence, $dalpha$ vanishes at $p$ if and only if $Y$ and $[X,Y]$ are colinear at $p$.
answered Sep 3 at 15:07
Amitai Yuval
14.6k11026
add a comment |Â
up vote
2
down vote
accepted
By the intrinsic expression for the exterior derivative, we have $$dalpha(X,Y)=X(alpha(Y))-Y(alpha(X))-alpha([X,Y])=-alpha([X,Y]).$$ So, $alpha$ is closed at every point where $alpha([X,Y])=0$. But, by construction, the kernel of $alpha$ at a point $p$ is the line spanned by $Y(p)$. Hence, $dalpha$ vanishes at $p$ if and only if $Y$ and $[X,Y]$ are colinear at $p$.
answered Sep 3 at 15:07
Amitai Yuval
14.6k11026
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
By the intrinsic expression for the exterior derivative, we have $$dalpha(X,Y)=X(alpha(Y))-Y(alpha(X))-alpha([X,Y])=-alpha([X,Y]).$$ So, $alpha$ is closed at every point where $alpha([X,Y])=0$. But, by construction, the kernel of $alpha$ at a point $p$ is the line spanned by $Y(p)$. Hence, $dalpha$ vanishes at $p$ if and only if $Y$ and $[X,Y]$ are colinear at $p$.
answered Sep 3 at 15:07
Amitai Yuval
14.6k11026
By the intrinsic expression for the exterior derivative, we have $$dalpha(X,Y)=X(alpha(Y))-Y(alpha(X))-alpha([X,Y])=-alpha([X,Y]).$$ So, $alpha$ is closed at every point where $alpha([X,Y])=0$. But, by construction, the kernel of $alpha$ at a point $p$ is the line spanned by $Y(p)$. Hence, $dalpha$ vanishes at $p$ if and only if $Y$ and $[X,Y]$ are colinear at $p$.
answered Sep 3 at 15:07
Amitai Yuval
14.6k11026
answered Sep 3 at 15:07
Amitai Yuval
14.6k11026
answered Sep 3 at 15:07
Amitai Yuval
14.6k11026
answered Sep 3 at 15:07
Amitai Yuval
14.6k11026
14.6k11026
add a comment |Â
add a comment |Â
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The form $alpha$ is well-defined the minute you specify its values at $X$ and $Y$. What is the purpose of the rest of the post?
– Amitai Yuval
Sep 3 at 13:41
apparently we can have a general picture of the kernel of $dalpha$ which is the set where $Y,[X,Y]$ are collinear... but I can't see it
– Smilia
Sep 3 at 14:18