(This post is the third in a series on geometry. (A geometric series, har har har.) They all assume you know about differential forms and such things. The first was on Liouville geometry, also known as exact symplectic geometry, on surfaces. The second went from them to contact geometry. So I’m assuming you know what those are.)

We’ve seen that if you take a Liouville 1-form $$\beta$$ on a surface $$S$$ (i.e. such that $$d\beta$$ is nondegenerate, hence a symplectic form), then the 1-form $$\alpha = \beta + dt$$ on the 3-manifold $$M = S \times [0,1]$$ obtained by thickening $$S$$ is a contact form. (Here $$t$$ is the coordinate on $$[0,1]$$.)

Moreover, we’ve seen that on each slice $$S \times \{t\}$$ of this thickening, the characteristic foliation (i.e. the pattern of how the slice intersects the contact planes) $$\mathcal{F}$$ coincides with $$\ker \beta$$.

We’ve also noted that this contact form $$\alpha$$ is a vertically invariant contact form on $$M$$: it has no dependence on $$t$$. Indeed, the flow of the vertical vector field $$\partial_t$$ preserves $$\alpha$$, and hence is a contact vector field. Thus each slice $$S \times \{t\}$$ is transverse to a contact vector field, and hence is a convex surface.

Thus, starting from the simple but elegant structure of a Liouville 1-form on a surface, we have been led to 3-dimensional contact geometry, and convex surfaces.

What we’re going to do now is go in the other direction, and start from a convex surface.

We’re going to make a clear distinction now between a contact structure and a contact form. A contact form is a 1-form $$\alpha$$ such that $$\alpha \wedge d\alpha$$ is non-degenerate, i.e. so that $$\ker \alpha$$ is a non-integrable plane field. A contact structure $$\xi$$ is a non-integrable plane field. So any contact form $$\alpha$$ defines a contact structure $$\xi$$ by $$\xi = \ker \alpha$$, but a contact structure $$\xi$$ has many 1-forms defining it (at least locally). Given any contact form $$\alpha$$ such that $$\ker \alpha = \xi$$, we can multiply $$\alpha$$ by any smooth nonzero real-valued function $$f$$, and $$f\alpha$$ is then another contact 1-form, with $$\ker(f\alpha) = \ker \alpha = \xi$$.

Well, let’s return to the definition of a convex surface: it’s an embedded surface $$S$$ in a contact 3-manifold for which there is a vector field $$X$$ transverse to $$S$$. Said tersely, a convex surface is a surface with a transverse contact vector field.

Now, given a convex surface, we can introduce coordinates as we please. Let us define a coordinate $$t$$ by the transverse vector field $$X$$. So let $$X = \partial_t$$. We can then let $$t=0$$ on the surface $$S$$, and flowing along $$X = \partial_t$$, we obtain a coordinate $$t$$ which measures how far from $$S$$ we have flowed along $$X$$. Using this coordinate, we can describe a neighbourhood of $$S$$ as $$S \times [-\varepsilon, \varepsilon]$$, for some sufficiently small $$\varepsilon$$, where $$S$$ appears as $$S \times \{0\}$$ and the coordinate on the $$[-\varepsilon, \varepsilon]$$ factor is precisely $$t$$. For simplicitly, we can take $$\varepsilon = 1$$; by slowing down the vector field $$X$$ we can in fact fit this $$S \times [-1,1]$$ inside the previous $$S \times [-\varepsilon, \varepsilon]$$.

So now we have a neighbourhood of $$S$$ given as $$S \times [-1,1]$$, and the transverse contact vector field is $$X = \partial_t$$.

If we further denote by $$x,y$$ some local coordinates on $$S$$, then $$x,y,t$$ form some local coordinates on $$S \times [-1,1]$$. So the contact form $$\alpha$$ (or indeed any 1-form) can be written in the form
$\alpha = f \; dx + g \; dy + u \; dt,$
where $$f,g,u$$ are real-valued functions on $$S \times [-1,1]$$. Now the functions $$f,g,u : S \times [-1,1] \rightarrow \mathbb{R}$$ might in general depend on $$x,y,t$$. But as $$X = \partial_t$$ s a contact vector field, the contact planes given by $$\ker \alpha$$ don’t depend on the $$t$$ coordinate at all. And hence we can take the contact form $$\alpha$$ not to depend on $$t$$ either. (Possibly $$\alpha$$ might depend on $$t$$, since multiplying $$\alpha$$ by any nonzero real-valued function produces a 1-form with the same kernel; but for such an $$\alpha$$, we can “normalise” it, multiplying by a nonzero function, to make it independent of $$t$$. Or indeed replacing $$f(x,y,t), g(x,y,t), u(x,y,t)$$ with $$f(x,y,0), g(x,y,0), u(x,y,0)$$ would have the same effect.)

In other words, since $$S$$ is a convex surface, there is a contact form $$\alpha$$ where $$f,g,u$$ only depend on $$x,y$$, and not $$t$$. We can write
$\alpha = f(x,y) \; dx + g(x,y) \; dy + u(x,y) \; dt.$
Written in this way, the first two terms $$f(x,y) \; dx + g(x,y) \; dy$$ denote a 1-form purely on the surface $$S$$. Indeed, any 1-form on $$S$$ can be written this way. So let’s call it $$\beta$$. In a similar way, $$u(x,y)$$ can be regarded as a function purely on the surface $$S$$. We then have
$\alpha = \beta + u \; dt$
where $$\beta$$ is a 1-form on $$S$$, and $$u$$ is a real-valued function on $$S$$.

When $$u \neq 0$$, we can do even better. We can then divide the whole 1-form $$\alpha$$ by $$u$$ — and remember, multiplying the contact form by a nonzero function results in another contact form defining the same contact structure. So this allows us effectively to assume that $$u=1$$, and that $$\alpha$$ is of the form $$\beta + dt$$, where again $$\beta$$ is a 1-form and $$u$$ a real-valued function on $$S$$.

Now if $$\alpha = \beta + dt$$ is a contact form, then it must satisfy the contact condition of being non-integrable, i.e. $$\alpha \wedge d\alpha$$ must be non-degenerate. Not every possible 1-form $$\beta$$ on $$S$$ and every function $$u$$ on $$S$$ will make $$\beta + dt$$ a contact form. Which possible 1-forms $$\beta$$ make a contact form? We can compute $$\alpha \wedge d\alpha$$ to find out:
$\alpha \wedge d\alpha = (\beta + dt) \wedge d\beta = \beta \wedge d\beta + dt \wedge d\beta = dt \wedge d\beta.$
This is a contact form if and only if $$d\beta$$ is a non-degenerate 2-form on $$S$$ — that is, if $$\beta$$ is a Liouville 1-form.

Thus we have proved the following.

PROPOSITION. Let $$S$$ be a convex surface in a 3-manfold with a contact structure $$\xi$$. Defining a transverse coordinate $$t$$ via the transverse contact vector field, $$S$$ has a neighbourhood on which $$\xi$$ has a contact form $$\beta + u \; dt$$, where $$\beta$$ is a 1-form and $$u$$ is a real-valued function on $$S$$.

If further $$u$$ is nowhere zero, then $$S$$ has a neighbourhood on which $$\xi$$ has a contact form $$\beta + dt$$, where $$\beta$$ is a Liouville 1-form on $$S$$.

In our previous episode, starting from Liouville structures on surfaces, we were led to convex surfaces in contact 3-manifolds. And now, we have gone back, from convex surfaces to Liouville structures.

Now we know that not every surface has a Liouville structure: we saw previously that there can’t be one if $$S$$ is compact without boundary. And so a convex surface also can’t have a local contact form $$\beta + dt$$ if $$S$$ is compact without boundary.

But, amazingly enough, Giroux proved that, in a certain sense, almost any embedded surface in a contact manifold is convex — including almost any embedded compact surface without boundary.

Such a convex surface $$S$$, compact without boundary, has a local contact form of the type $$\beta + u \; dt$$, as we’ve discussed. And remember we said that if $$u$$ is nowhere zero, then we could divide out by $$u$$ and obtain a local contact form of the type $$\beta + dt$$. But they can’t have a local contact form of the type $$\beta + dt$$. Hence for any convex surface $$S$$, compact without boundary, the contact form $$\beta + u \; dt$$ must have some zeroes of $$u$$.

And as it turns out, the zeroes of $$u$$ are very interesting and important.

What happens at the zeroes of $$u$$? They are precisely where the contact planes are vertical, i.e. where $$\partial_t$$, or the contact vector field $$X$$, lies in $$\xi = \ker \alpha$$. Indeed,
$\alpha(X) = \alpha(\partial_t) = \beta(\partial_t) + u dt (\partial_t) = u.$
Here we used the fact that $$\beta(\partial_t) = 0$$, since $$\beta$$ is a 1-form on $$S$$, which is independent of the $$t$$ coordinate. So $$\alpha(X) = 0$$ precisely when $$u=0$$.

The set of points where $$u=0$$ is called the dividing set (or decoupage, in the original French). It turns out that it’s a curve on $$S$$ and it splits $$S$$ into pieces where $$u>0$$ and where $$u<0$$. (This was proved by Giroux.)

Note that when $$u>0$$, we have $$\alpha(X)>0$$; and when $$u<0$$, we have $$\alpha(X)<0$$. Suppose we paint one side of the contact planes white, and the other side black. We think of the black side as “positive”, and the white side as “negative”, in the following sense. Given any vector $$V$$, we will have $$\alpha(V) > 0$$ when $$V$$ points out of the white side, $$\alpha(V) = 0$$ when $$V$$ points along the plane, and $$\alpha(V) < 0$$ when $$V$$ points out of the black side.

Thus, the contact planes are white side up when $$u<0$$, they become vertical along the dividing set $$u=0$$, where they flip over to be black side up when $$u>0$$. A convex disc. The dividing set is drawn in red. The dividing set is usually drawn in red.

The standard notation is that the dividing set (i.e. where $$u=0$$) is denoted $$\Gamma$$; the region of $$S$$ where $$u>0$$ is denoted $$R_+$$; and the region of $$S$$ where $$u<0$$ is denoted $$R_-$$.

The best thing is that, if you just consider the subset of $$S \times [-1,1]$$ where $$u>0$$ (say), i.e. $$R_+ \times [-1,1]$$, you can divide $$\alpha$$ out by $$u$$, and obtain a contact form of the type $$\beta + dt$$, where $$\beta$$ is Liouville. So the characteristic foliation on $$R_+$$ is a Liouville foliation, and there is a flow tangent to it which exponentially expands an area form. The same applies to $$R_-$$.

So in fact a convex surface can be regarded as made up of two Liouville structures pieced together along a dividing set, where the contact planes flip over.