Contact manifold and Darboux's theorem

[cianfa72]

TL;DR Summary

Contact manifold and Darboux's theorem for one-form $\theta$ such that $d\theta$ is a 2-form with constant rank 0

Hi, I'm studying the concept of contact manifold -- Contact geometry
A related theorem is Darboux's theorem for one-forms -- Darboux theorem

In the particular case of one-form $\theta \neq 0$ such that $d\theta$ has constant rank 0 then if $\theta \wedge (d\theta)^0 \neq 0$ there exists a local coordinate chart such that $\theta=dx_1$.

My question is: what does it mean $d\theta$ has rank 0 ? Thanks.

 

[fresh_42]

I suspect that it means to be a scalar.

 

[jbergman]

TL;DR Summary: Contact manifold and Darboux's theorem for one-form $\theta$ such that $d\theta$ is a 2-form with constant rank 0

Hi, I'm studying the concept of contact manifold -- Contact geometry
A related theorem is Darboux's theorem for one-forms -- Darboux theorem

In the particular case of one-form $\theta \neq 0$ such that $d\theta$ has constant rank 0 then if $\theta \wedge (d\theta)^0 \neq 0$ there exists a local coordinate chart such that $\theta=dx_1$.

My question is: what does it mean $d\theta$ has rank 0 ? Thanks.

My understanding is that a rank 0 form is just 0. We know that the exterior derivative of closed forms is 0, and in general $d^2 \omega =0$ for all forms.

Also, looking at wikipedia, the $\neq$ case only holds for $p\gt 0$.

 

[cianfa72]

My understanding is that a rank 0 form is just 0.

Yes, me too. On the other hand to me a writing of type $(d\theta)^0$ is actually indefinite. What should that mean ? Do the wedge product with itself 0 times...

Also, looking at wikipedia, the $\neq$ case only holds for $p\gt 0$.

Yes, I suspect there is a typo in Wikipedia entry in the next Frobenius's theorem section. It should hold for $p=1$.

 

[cianfa72]

From Wikipedia entry on Contact geometry

each point in $\mathbf R^3$ has a plane associated to it by the contact structure, in this case as the kernel of the one-form dz − y dx. These planes appear to twist along the y-axis. It is not integrable, as can be verified by drawing an infinitesimal square in the x-y plane, and follow the path along the one-forms. The path would not return to the same z-coordinate after one circuit.

I'm in trouble to grasp the sentence in bold. What does it mean follow a path along one circuit ? Thanks.

 

[jbergman]

From Wikipedia entry on Contact geometry

View attachment 335689I'm in trouble to grasp the sentence in bold. What does it mean follow a path along one circuit ? Thanks.

I think the idea is that those planes represent tangent subspaces at each point defined by tangent vectors in the kernel of $dz -ydx$. For instance, at $(0,0,0)$ the form would be just $dz$ so the tangent subspace lies in the $x,y$ plane.

Now draw a rectangle in the x,y plane but now when we traverse it we follow along a tangent vector in the subspace at each point. So if the plane is slanted upwards we would move up as we followed are path.

What this is saying is that if you followed a such a closed path you would end up at a different z coordinate then you started at.

 

[fresh_42]

It also says ...

In mathematics, contact geometry is the study of a geometric structure on smooth manifolds given by a hyperplane distribution in the tangent bundle satisfying a condition called 'complete non-integrability'.

And, $(d\alpha)^0=0.$

 

[cianfa72]

And, $(d\alpha)^0=0.$

So the above is just a definition. I.e. $(d\alpha)^0=0$ by definition for every form $d\alpha$ ?

 

[cianfa72]

What this is saying is that if you followed a such a closed path you would end up at a different z coordinate then you started at.

As closed path you mean follow the perimeter of the rectangle that is the projection on x-y plane of the path followed along a tangent vector picked at each point in the tangent (slanted) subspaces ?

 

[jbergman]

As closed path you mean follow the perimeter of the rectangle that is the projection on x-y plane of the path followed along a tangent vector picked at each point in the tangent (slanted) subspaces ?

Yes. Basically you have a parametrized path, $\gamma(t)$ whose projection onto the x,y plane is a rectangle and where $\frac{d\gamma}{dt}$ is a tangent vector in the tangent subspace at each point $\gamma(t)$.

You can't do that with a curve that stays in the xy plane. And such a curve that starts there would end up above or below it.

 

[fresh_42]

So the above is just a definition. I.e. $(d\alpha)^0=0$ by definition for every form $d\alpha$ ?

I saw $(d\alpha)^k=\underbrace{d\alpha \wedge \ldots \wedge d\alpha}_{k \;times}$. So $\displaystyle{(d\alpha)^0=\wedge_{k\in \emptyset}}\; d\alpha =\text{ neutral element }$ which is zero in a an algebra.

 

[cianfa72]

So $\displaystyle{(d\alpha)^0=\wedge_{k\in \emptyset}}\; d\alpha =\text{ neutral element }$ which is zero in an algebra.

On the field of reals the rank of a 2-form $d\alpha$ is $p$ if and only if $(d\alpha)^p \neq 0$ and $(d\alpha)^{p+1}=0$.

So what is the real content of Darboux's theorem for a one-form $\alpha$ such that the 2-form $d\alpha$ has rank $p=0$ ?

 

[cianfa72]

BTW Contact manifolds can be used as a model for thermodynamic systems. In particular if we consider two couples of conjugate variables we get a 5D contact manifold -- see how-exactly-is-the-formalism-of-thermodynamics-based-on-contact-geometry

On that manifold is defined a one-form $\alpha \neq 0$ such that $d\alpha$ has rank $2$, indeed $2\cdot 2 + 1 = 5$. Thanks to Darboux' theorem we get $$\alpha=dU -TdS + pdV$$
The equation of state of a substance/system is actually represented by a Legendrian submanifold of dimension 2. So there are 3 equations between the 5 state variables $(U,T,S,p,V)$ in order to define a 2d (immersed/embedded) submanifold.

Is the above correct ? Thanks.

 

[jbergman]

BTW Contact manifolds can be used as a model for thermodynamic systems. In particular if we consider two couples of conjugate variables we get a 5D contact manifold -- see how-exactly-is-the-formalism-of-thermodynamics-based-on-contact-geometry

On that manifold is defined a one-form $\alpha \neq 0$ such that $d\alpha$ has rank $2$, indeed $2\cdot 2 + 1 = 5$. Thanks to Darboux' theorem we get $$\alpha=dU -TdS + pdV$$
The equation of state of a substance/system is actually represented by a Legendrian submanifold of dimension 2. So there are 3 equations between the 5 state variables $(U,T,S,p,V)$ in order to define a 2d (immersed/embedded) submanifold.

Is the above correct ? Thanks.

No idea, but the linked reference looks interesting. Thanks for sharing.

 

[cianfa72]

See also this thread.