(This post is a continuation of my previous one on Liouville structures, and hence is quite technical. It’s aimed at students of geometry, not the general public. It assumes you know some differential geometry. If you know what a Liouville structure is, read on.)

We’re going to take Liouville structures and move them into 3 dimensions, to obtain contact structures.

As we’ve seen, a Liouville structure on a surface $$S$$ (which necessarily has boundary, or is non-compact) is given by a 1-form $$\beta$$ such that $$d\beta$$ is non-degenerate. Then $$d\beta$$ is a symplectic form on $$S$$, and $$\beta$$ is dual to a vector field $$X$$ via the symplectic form, i.e. $$\iota_X d\beta = \beta$$. This structure has the nice property that $$X$$ points along $$\ker \beta$$, and the flow of $$X$$ expands $$\beta$$ and $$d\beta$$ exponentially. In equations, $$\beta(X) = 0$$, $$L_X \beta = \beta$$, and $$L_X d\beta = d\beta$$.

Let’s now go into the next dimension and consider a 3-dimensional space (or 3-manifold) $$M = S \times [0,1]$$. We can use $$x,y$$ as coordinates on $$S$$, and $$t$$ as a coordinate on $$[0,1]$$. So this is a thickening of $$S$$; you can think of $$S$$ as horizontal, and $$[0,1]$$ as vertical, with the coordinate $$t \in [0,1]$$ measuring height. (However $$M$$ certainly has boundary; and if $$S$$ is non-compact, then the same is true for $$M$$.)

On this 3-manifold $$M$$, let’s consider a 1-form. The 1-form $$\beta$$ is no more interesting here than it is on $$S$$, but let’s add to it a form using the third dimension. We’ll just add to $$\beta$$ the simplest possible such form, $$dt$$.

So define a 1-form $$\alpha$$ on $$M$$ by $$\alpha = \beta + dt$$.

Previously, we considered the example of a Liouville structure on $$S = \mathbb{R}^2$$, given by $$\beta = x \; dy$$. We will continue this running example. We obtain $$M = \mathbb{R}^2 \times [0,1]$$, with the 1-form $$\alpha = x \; dy + dt$$.

It’s harder to draw pictures of 1-forms in 3 dimensions, but it is possible! Just as in the 2-dimensional case, we can draw the kernel of the 1-form. But whereas the kernel of a (nowhere zero) 1-form on a surface is a line field, the kernel of a (nowhere zero) 1-form on a 3-manifold is a plane field.

Now it’s not difficult to see that, in our example, no matter where you are, the $$x$$-direction always lies in $$\ker \alpha$$. For $$\alpha$$ only contains a $$dy$$ and a $$dt$$ term; if you feed it a vector in the $$x$$-direction, you get zero. But to get a second linearly independent vector in the kernel, you need to take a more carefully chosen combination of the $$y$$ and $$t$$ directions. We can check that the combination $$\partial_y – x \partial_t$$ lies in the kernel; here $$\partial_y, \partial_t$$ denote unit vectors in the $$y$$ and $$t$$ directions.
$\alpha(\partial_y – x \partial_t) = (x \; dy + dt)(\partial_y – x \partial_t) = x \; dy(\partial_y) – dt(x \partial_t) = x-x=0.$
Thus, at each point $$(x,y,t)$$ in $$M$$, $$\ker \alpha$$ is spanned by $$\partial_x$$ and $$\partial_y – x \partial_t$$. We can write
$\ker \alpha = \langle \partial_x, \; \partial_y – x \partial_t \rangle.$

Let’s try to visualise $$\ker \alpha$$ geometrically. It’s a plane field: there is a plane at each point in $$\mathbb{R}^2 \times [0,1]$$. The plane at $$(x,y,t)$$ is spanned by $$\partial_x$$ and $$\partial_y – x \partial_t$$. When $$x=0$$, the plane is spanned by $$\partial_x$$ and $$\partial_y$$: it’s horizontal. As $$x$$ varies, $$\partial_x$$ always points along the plane, and so the planes “spin” around each line in the $$x$$-direction. As $$x$$ increases from $$0$$, the plane tilts: rather than $$\partial_y$$, the plane contains $$\partial_y -x \partial_t$$, and this negative $$\partial_t$$ component becomes larger as $$x$$ increases increases; as $$x$$ becomes large, the plane becomes close to vertical. And similarly as $$x$$ decreases from $$0$$: the plane gains a positive $$\partial_t$$ component.

This is in fact quite a standard geometric object, at least if you’re into the field of contact geometry. This plane field given by the kernel of $$x \; dy + dt$$ is also known (possibly after some relabllings of variables, and some sign changes) as the standard contact structure on $$\mathbb{R}^3$$. The standard contact structure on $$\mathbb{R}^3$$ (with slightly different coordinates: replace $$x,y,z$$ here with $$y,x,t$$ ). Public Domain, wikipedia

This plane field has some great properties. Try to find a surface which runs tangent to it! That is, try to find a surface in $$\mathbb{R}^2 \times [0,1]$$, which at every point is tangent to this plane field. Even if you try to find a tiny surface, you will not succeed — because this plane field has a property called non-integrability.

Frobenius’ theorem in differential geometry tells you when a plane field is integrable (i.e. tangent to a surface) or not. (In fact, it says this not just about 2-dimensional plane fields in 3 dimensions, but in general about any-dimensional plane fields in any number of dimensions.) There is a vector-field version, and a differential form version.

The vector-field version of Frobenius’ theorem says to take the Lie bracket of two vector fields spanning the plane field: if the Lie bracket points along the planes of the plane field, it’s integrable; otherwise, it is not. In our example we obtain
$[ \partial_x, \; \partial_y – x \partial_t ] = [\partial_x, \; \partial_y] – [\partial_x, \; x \partial_t] = 0 – \partial_x (x \partial_t) + x \partial_t \partial_x \\ = – \partial_t – x \partial_x \partial_t + x \partial_t \partial_x = – \partial_t.$
Here we repeatedly used the fact that the partial derivatives commute, e.g. $$\partial_x \partial_y = \partial_y \partial_x$$ or $$[\partial_x, \partial_y] = 0$$, and the product rule, in the slightly cryptic form of $$\partial_x x = 1 + x \partial_x$$. The result is $$– \partial_t$$, which is definitely not a linear combination of $$\partial_x$$ and $$\partial_y – x \partial_t$$.

The differential form version of Frobenius’ theorem says to take the wedge product $$\alpha \wedge d\alpha$$, which will be a 3-form: if it is zero, then the plane field is integrable; otherwise, it is not. In our example we obtain
$\alpha \wedge d\alpha = (x \; dy + dt) \wedge (dx \wedge dy) \\ = dt \wedge dx \wedge dy \neq 0,$
which is the standard Euclidean volume form and is definitely nowhere zero.

So, if you believe Frobenius’ theorem, or at least one or the other of these two variants, then we’ve shown that the plane field is nowhere integrable. This is exactly the definition of a contact structure: a plane field which is nowhere integrable. (This is the definition in 3 dimensions, at least.) So our plane field is a contact structure, in fact, it’s essentially the standard contact structure on $$\mathbb{R}^3$$. A 1-form whose kernel is a contact structure is called a contact form. So in this example, the 1-form $$\alpha = \beta + dt = x \; dy + dt$$ is a contact form.

Another nice feature comes from considering the surface $$S = S \times \{0\}$$, sitting inside $$S \times [0,1]$$, and how it intersects the contact planes. At every point of $$S$$ is the horizontal surface $$S$$ (which is a plane), and the plane $$\ker \alpha$$. How do these planes intersect? Well, we saw that the planes of $$\ker \alpha$$ were spanned by $$\partial_x$$ and $$\partial_y – x \partial_t$$. The first of these actually points along $$S$$, while the second does not — except when $$x = 0$$, when the planes are tangent to $$S$$. So the contact planes intersect $$S$$ along the $$x$$ direction.

Now the lines in the $$x$$ directions along $$S$$ have come up in our previous discussion — they are nothing but the kernel of the Liouville form $$\beta$$! So the planes of $$\ker \alpha$$ cut $$S$$ along the same line as $$\ker \beta$$.

Thus, the lines at each point of $$S$$, given by looking at how it intersects the contact plane $$\ker \alpha$$, form a line field on $$S$$; and if you integrate it, you get a foliation. We saw last time that $$\ker \beta$$ also forms a line field on $$S$$, which also integrates to a foliation. What we’ve found here in our example is that the foliations on $$S \times \{0\}$$ given by $$\ker \beta$$, and by intersection with $$\ker \alpha$$, coincide. (The foliations coincide when $$x \neq 0$$. When $$x = 0$$, both foliations are singular, so they are in fact equal as singular foliations.)

When you have a contact form $$\alpha$$ on a 3-manifold, and a surface $$S$$ in that 3-manifold, you can always consider, in a similar way, how the contact planes intersect the tangent plane to $$S$$, at each point of $$S$$. The result is a (singular) line field on $$S$$, which integrates to a (singular) foliation: this is called the characteristic foliation $$\mathcal{F}$$ of $$S$$. In fancy language, at a point $$p \in S$$, the characteristic foliation is given by the intersection of the tangent plane $$T_p S$$ to $$S$$, with the contact plane $$\ker \alpha_p$$ there:
$\mathcal{F}_p = T_p S \cap \ker \alpha_p.$
In simple language, the characteristic foliation of a surface is the pattern of how it intersects the contact planes.

What we’ve found, in our example, is that the foliation $$\ker \beta$$ coincides with the characteristic foliation on $$S \times \{0\}$$. And in fact the same occurs not just on $$S \times \{0\}$$, but on any slice $$S \times \{t\}$$.

Let’s now consider our second example from last time: $$\beta = \frac{1}{2} (x \; dy – y \; dx)$$ on the surface $$S = \mathbb{R}^2$$. This was the “radial” example where $$X$$ and $$\ker \beta$$ both pointed out directly out along lines from the origin.

In this example, we obtain on $$M = \mathbb{R^2} \times [0,1]$$ the 1-form $$\alpha = \beta + dt = \frac{1}{2} (x \; dy – y \; dx) + dt$$. You can check that along the $$t$$-axis $$x=y=0$$, $$\ker \alpha$$ is a horizontal plane, spanned by $$\partial_x$$ and $$\partial_y$$. At other points, the kernel is spanned by $$x \partial_x + y \partial_y$$, which points horizontally radially outward from the $$t$$-axis, and $$y \partial_x – x \partial_y + \frac{x^2+y^2}{2} \; dt$$, which has an angular component $$y \partial x – x \partial y$$ (which is in the $$\theta$$ direction in cyclindrical coordinates), and a $$dt$$ component. As $$x^2 + y^2$$ becomes larger, i.e. further from the $$t$$-axis, the $$dt$$ term becomes larger and the planes become more vertical. The result is a plane field known as the standard cylindrically symmetric contact structure, because it’s cylindrically symmetric, and it’s a contact structure. From John Etnyre’s lecture notes on open book decompostions and contact structures. The picture is by Stephan Schonenberger.

It’s perhaps not surprising that if you take something on the plane which is radially symmetric, and then just make a 3-d thing out of it, which doesn’t change in the new ($$t$$) dimension, then you get something which is cylindrically symmetric.

Let’s check that we actually have a contact form, by applying the differential form version of Frobenius’ theorem:
$\alpha \wedge d\alpha = \left( \frac{1}{2} x \; dy – \frac{1}{2} y \; dx + dt \right) \wedge \left( dx \wedge dy \right) \\ = dt \wedge dx \wedge dy.$
Most of the terms immediately cancel out in the wedge product because of the anti-symmetry: all that’s left is the standard 3-dimensional Euclidean volume form, which is definitely nowhere zero.

You can also check that on $$S \times \{0\}$$, or in fact on any slice $$S \times \{t\}$$, the characteristic foliation points radially outward from the origin. (Indeed, we just saw that the radial vector field $$x \partial_x + y \partial_y$$ lies in $$\ker \alpha$$.) This again coincides with the foliation $$\ker \beta$$ from the Liouville 1-form.

In fact, this argument works generally, not just on this example. The differential form version of Frobenius’ theorem can always easily be applied to show $$\alpha$$ is a contact form. Moreover, it’s not difficult to show that the characteristic foliation coincides with the foliation from the Liouville 1-form.

PROPOSITION. Starting from any Liouville 1-form $$\beta$$ on any surface $$S$$, the 1-form $$\alpha = \beta + dt$$ is a contact form on $$M = S \times [0,1]$$.

Moreover, for each $$t \in [0,1]$$, the characteristic foliation on $$S \times \{t\}$$ coincides with the foliation of $$\ker \beta$$.

In other words, at each $$(p,t) \in S \times [0,1]$$, the intersection of $$\ker \alpha$$ with the tangent space to $$S \times \{t\}$$ is equal to $$\ker \beta$$.
$\ker \alpha_{(p,t)} \cap T_{(p,t)} (S \times \{t\}) = \ker \beta_{(p,t)}$

PROOF. Consider $$\alpha \wedge d\alpha$$; we’ll show it is nowhere zero.
$\alpha \wedge d\alpha = (\beta + dt) \wedge d\beta = \beta \wedge d\beta + dt \wedge d\beta$
Now $$\beta \wedge d\beta$$ is a 3-form, but $$\beta$$ is a 1-form on the surface $$S$$ only: it has no $$t$$-component. So $$\beta \wedge d\beta$$ is a 3-form… on the 2-dimensional space $$S$$! Consequently it must be zero.

Thus $$\alpha \wedge d\alpha = dt \wedge d\beta$$. Now we use the fact that $$\beta$$ is a Liouville 1-form: this means that $$d\beta$$ is a non-degenerate 2-form on $$S$$. When we wedge it with $$dt$$ then we must get a non-degenerate 3-form on $$S \times [0,1]$$. This means $$\alpha$$ is a contact form.

Now consider a point $$(p,t) \in S \times [0,1]$$, and a tangent vector $$V$$ to $$S \times \{t\}$$ there. So $$V$$ is horizontal: it has no $$t$$ component, and $$dt(V) = 0$$. So $$\beta(V)= 0$$ if and only if $$\alpha(V) = \beta(V) + dt(V) = 0$$. Consequently for a vector $$V \in T_{(p,t)} S \times \{t\}$$, we have $$V$$ lies in $$\ker \alpha$$ iff it lies in $$\ker \beta$$. In other words, $$\ker \alpha \cap T_{(p,t)} S \times \{t\} = \ker \beta$$. QED

Furthermore, this proof works in higher dimensions too. As long as you have a Liouville structure on a manifold $$S$$ of any dimension $$2n$$ (Liouville structures only exist in even dimension; this was discussed in the prequel), you’ll get a contact form on $$S \times [0,1]$$ this way. (The only change is that $$\beta \wedge d\beta$$ is a $$(2n+1)$$-form on a $$(2n)$$-manifold.)

In any case, this means that Liouville geometry leads naturally to contact geometry: we just add another dimension!

Indeed, sometimes this construction is called a contactisation!

(See, for example, section 2.3 of this 1998 paper of Yasha Eliashberg, Helmut Hofer and Dietmar Salamon.)

Liouville geometry is another name for symplectic geometry where the symplectic form is exact, i.e. the symplectic form is $$d\beta$$. So contactisation upgrades symplectic geometry — the mathematics of classical Hamiltonian mechanics — to an odd-dimensional counterpart.

Another interesting thing to note is that in this “contactisation” construction, the 1-form $$\alpha = \beta + dt$$ doesn’t actually depend on $$t$$; it’s invariant under translations in the $$t$$ direction. We saw in the second example that a radially symmetric Liouville form gave rise to a cylindrically symmetric contact structure.

And indeed, when you take $$\alpha = \beta + dt$$, the 1-form $$\beta$$ on the surface $$S$$ doesn’t depend on $$t$$; and $$dt$$ is just the same $$dt$$ regardless of what $$t$$ is!

Such a contact form is sometimes called “vertically invariant”.

In fancy language, it means that the flow of the vertical vector field $$\partial_t$$ preserves the contact form. In even fancier language, it means that the Lie derivative of the contact form $$\alpha$$ in the direction of $$\partial_t$$ is zero:
$L_{\partial_t} \alpha = 0.$

A vector field whose flow preserves a contact structure is often called a contact vector field. So $$\partial_t$$ is a contact vector field.

Our 3-manifold $$M$$, the thickened surface $$S \times [0,1]$$, can be considered as a family of surfaces $$S_t = S \times {t}$$, for $$t \in [0,1]$$. And there is a vector field $$\partial_t$$, transverse to all of these surfaces.

Emmanuel Giroux discovered in the 1990s that, when you have a surface in a contact 3-manifold with a contact vector field transverse to it, very nice stuff happens. He called such a surface convex.

So Liouville geometry leads naturally not just to contact geometry, but to convex surfaces in contact geometry. More on that, another day.

From Liouville geometry to contact geometry
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