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I want to ask about motivations of contact topology. I mean, for example, a symplectic manifold is a generalization of phase spaces and it explains why we need a closed, non-degenerate symplectic form.

I feel like a contact manifold also has the similar explaination, so we know why a contact structure should be a smooth maximally non-integrable hyperplane field.

Thanks in advance.

a--
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An example of motivation would be optics.

e.g. consider the 3-manifold $Y = \mathbb R^2 \times S^1$ (seen as as sub-bundle of $T\mathbb R^2$). Take coordinates $(x,y,\theta)$ on $Y$ and consider the contact structure $\xi$ on $Y$ given by $$ \xi|_{x,y,\theta} = \ker (\cos (\theta) d x + \sin (\theta) d y) $$ Then $\phi_t :Y\rightarrow Y$ given by $\phi_t(x,y,\theta) = (x+t\cos(\theta), y+t\sin(\theta), \theta)$ is a family of contactomorphisms. Now, if $\gamma$ is a Legendrian curve in $(Y,\xi)$, then $\phi_t\circ\gamma$ is also Legendrian at all $t$. Consider the projection $\pi : Y\rightarrow \mathbb R^2 ; (x,y,\theta)\mapsto (x,y)$. Then $\pi\circ \phi_t\circ\gamma$ can be seen as a wavefront evoluting in $\mathbb R^2$. This can be generalised to multidimensional wavefronts using the sphere-bundle inside $T\mathbb R^n$.

A similar example easily found online is about "why it's possible to park a car using contact geometry".

Another physical motivation would be pre-quantization "à la Souriau".

Noé AC
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  • Thanks for your answer @NAC

    I want to ask a question just to make sure that I understand your answer.

    For you example, I want to go the other direction, so if we have a sphere bundel inside $T\mathbb{R}^2$, then a collection of wavefront things (I don't know what it is, but anyway) gives us a collection of Legendrians and the collection constructs a contact structure of the sphere bundle. In other words, a contact structure is a union of every possible wavefront, right?

    Sorry for bothering you with this too basic question.

     – a-- Aug 05 '17 at 23:57
  • No the contact structure here is only saying that the wavefront is propagating perpendicularly to its tangent. The advantage of dealing with a contact version on $Y$ instead of working only on $\mathbb R^2$ is that here on $\mathbb R^2$ the wavefront can make cusps (so its tangent might be ill defined somewhere). (Think of the wavefront as a singular object which is the projection of a smooth curve on $Y$ where the $\theta$ variable encodes the direction (or its perpendicular) where the wave is propagating). – Noé AC Aug 06 '17 at 01:09
  • Look p.7 in https://arxiv.org/pdf/math/0501255.pdf – Noé AC Aug 06 '17 at 01:19