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Given a bivector, determine the decomposition of $mathbb{R}^3$ into symplectic leaves

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0

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Say I have a bivector, given by $f(z) partial_x wedge partial_y$, where $f(z)$ is a smooth function of $z$. How do I determine the decomposition of $mathbb{R}^3$ into symplectic leaves from this?

differential-geometry manifolds poisson-geometry

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asked 2 days ago

SoreySorey

600212

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add a comment | 

0

$begingroup$

Say I have a bivector, given by $f(z) partial_x wedge partial_y$, where $f(z)$ is a smooth function of $z$. How do I determine the decomposition of $mathbb{R}^3$ into symplectic leaves from this?

differential-geometry manifolds poisson-geometry

share|cite|improve this question

asked 2 days ago

SoreySorey

600212

$endgroup$

add a comment | 

$begingroup$

Say I have a bivector, given by $f(z) partial_x wedge partial_y$, where $f(z)$ is a smooth function of $z$. How do I determine the decomposition of $mathbb{R}^3$ into symplectic leaves from this?

differential-geometry manifolds poisson-geometry

share|cite|improve this question

asked 2 days ago

SoreySorey

600212

$endgroup$

share|cite|improve this question

asked 2 days ago

SoreySorey

600212

share|cite|improve this question

asked 2 days ago

SoreySorey

600212

asked 2 days ago

SoreySorey

600212

asked 2 days ago

SoreySorey

600212

1 Answer
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1

$begingroup$

Let us denote the bivector field by $$Pi=f(z)partial_{x}wedgepartial_{y}.$$
There is an associated vector bundle map $$Pi^{sharp}:T^{*}mathbb{R}^{3}rightarrow Tmathbb{R}^{3}:alphamapstoiota_{alpha}Pi,$$
which takes a one-form $alpha$ and contracts $Pi$ with it. The symplectic foliation integrates the (singular) distribution $text{Im}(Pi^{sharp})$. In this case, we have
begin{align}
text{Im}(Pi^{sharp})&=text{Span}{Pi^{sharp}(dx),Pi^{sharp}(dy),Pi^{sharp}(dz)}\
&=text{Span}{f(z)partial_{y},f(z)partial_{x}}.
end{align}
Note that this gives at each point $(x,y,z)$ either a zero-dimensional or a two-dimensional subspace, depending on whether or not $z$ is a zero of $f$.
It follows that the symplectic leaves are the following:

1) Each horizontal plane $z=c$ is a leaf whenever $f(c)neq 0$.

2) If $f(c)=0$, then all points $(x,y,c)$ are symplectic leaves.

share|cite|improve this answer

answered 2 days ago

studiosusstudiosus

2,149714

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks for the answer. What happens if instead I have a bivector field like $(x partial_x + ypartial_y)wedge partial_z$? I see that this is similar to what you have above, but it is more like an addition of two bivector fields of the form above. How does this differ?
  $endgroup$
  – Sorey
  11 hours ago
  

  
  
  
  

add a comment | 

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1

$begingroup$

Let us denote the bivector field by $$Pi=f(z)partial_{x}wedgepartial_{y}.$$
There is an associated vector bundle map $$Pi^{sharp}:T^{*}mathbb{R}^{3}rightarrow Tmathbb{R}^{3}:alphamapstoiota_{alpha}Pi,$$
which takes a one-form $alpha$ and contracts $Pi$ with it. The symplectic foliation integrates the (singular) distribution $text{Im}(Pi^{sharp})$. In this case, we have
begin{align}
text{Im}(Pi^{sharp})&=text{Span}{Pi^{sharp}(dx),Pi^{sharp}(dy),Pi^{sharp}(dz)}\
&=text{Span}{f(z)partial_{y},f(z)partial_{x}}.
end{align}
Note that this gives at each point $(x,y,z)$ either a zero-dimensional or a two-dimensional subspace, depending on whether or not $z$ is a zero of $f$.
It follows that the symplectic leaves are the following:

1) Each horizontal plane $z=c$ is a leaf whenever $f(c)neq 0$.

2) If $f(c)=0$, then all points $(x,y,c)$ are symplectic leaves.

share|cite|improve this answer

answered 2 days ago

studiosusstudiosus

2,149714

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks for the answer. What happens if instead I have a bivector field like $(x partial_x + ypartial_y)wedge partial_z$? I see that this is similar to what you have above, but it is more like an addition of two bivector fields of the form above. How does this differ?
  $endgroup$
  – Sorey
  11 hours ago
  

  
  
  
  

add a comment | 

1

$begingroup$

Let us denote the bivector field by $$Pi=f(z)partial_{x}wedgepartial_{y}.$$
There is an associated vector bundle map $$Pi^{sharp}:T^{*}mathbb{R}^{3}rightarrow Tmathbb{R}^{3}:alphamapstoiota_{alpha}Pi,$$
which takes a one-form $alpha$ and contracts $Pi$ with it. The symplectic foliation integrates the (singular) distribution $text{Im}(Pi^{sharp})$. In this case, we have
begin{align}
text{Im}(Pi^{sharp})&=text{Span}{Pi^{sharp}(dx),Pi^{sharp}(dy),Pi^{sharp}(dz)}\
&=text{Span}{f(z)partial_{y},f(z)partial_{x}}.
end{align}
Note that this gives at each point $(x,y,z)$ either a zero-dimensional or a two-dimensional subspace, depending on whether or not $z$ is a zero of $f$.
It follows that the symplectic leaves are the following:

1) Each horizontal plane $z=c$ is a leaf whenever $f(c)neq 0$.

2) If $f(c)=0$, then all points $(x,y,c)$ are symplectic leaves.

share|cite|improve this answer

answered 2 days ago

studiosusstudiosus

2,149714

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks for the answer. What happens if instead I have a bivector field like $(x partial_x + ypartial_y)wedge partial_z$? I see that this is similar to what you have above, but it is more like an addition of two bivector fields of the form above. How does this differ?
  $endgroup$
  – Sorey
  11 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

Let us denote the bivector field by $$Pi=f(z)partial_{x}wedgepartial_{y}.$$
There is an associated vector bundle map $$Pi^{sharp}:T^{*}mathbb{R}^{3}rightarrow Tmathbb{R}^{3}:alphamapstoiota_{alpha}Pi,$$
which takes a one-form $alpha$ and contracts $Pi$ with it. The symplectic foliation integrates the (singular) distribution $text{Im}(Pi^{sharp})$. In this case, we have
begin{align}
text{Im}(Pi^{sharp})&=text{Span}{Pi^{sharp}(dx),Pi^{sharp}(dy),Pi^{sharp}(dz)}\
&=text{Span}{f(z)partial_{y},f(z)partial_{x}}.
end{align}
Note that this gives at each point $(x,y,z)$ either a zero-dimensional or a two-dimensional subspace, depending on whether or not $z$ is a zero of $f$.
It follows that the symplectic leaves are the following:

1) Each horizontal plane $z=c$ is a leaf whenever $f(c)neq 0$.

2) If $f(c)=0$, then all points $(x,y,c)$ are symplectic leaves.

share|cite|improve this answer

answered 2 days ago

studiosusstudiosus

2,149714

$endgroup$

share|cite|improve this answer

answered 2 days ago

studiosusstudiosus

2,149714

answered 2 days ago

studiosusstudiosus

2,149714

answered 2 days ago

studiosusstudiosus

2,149714