(Note: This post is more technical than most stuff I write here. The intended audience here is not the general public, or even the general educated public: it’s students of geometry, broadly understood. In any case, if you don’t know what a differential form is, you’re probably not going to get much out of this.)

I’d like to show you some very nice geometry, involving some vector fields and differential forms.

Consider a surface. In fact, consider the plane, $$\mathbb{R}^2$$. That’s just the standard Euclidean plane, with coordinates $$x$$ and $$y$$.

Now let’s consider a differential 1-form on the plane; call it $$\beta$$. We’ll impose one condition on $$\beta$$: its exterior derivative $$d\beta$$ should be everywhere nonzero.

For instance, we can take $$\beta = x \; dy$$. In fact, we will take this as a running example. Its exterior derivative is $$d\beta = dx \wedge dy$$, which is just the usual Euclidean area form on the plane, and which is nowhere zero.

Now, saying that $$d\beta$$ is everywhere nonzero is the same as saying that $$d\beta$$ is an area form (although in general it might be different from the Euclidean area form $$dx \wedge dy$$); and this is also the same as saying that $$d\beta$$ is a non-degenerate 2-form. In fact, being exact, $$d\beta$$ is also closed: and hence $$d\beta$$ is a closed non-degenerate 2-form, also known as a symplectic form.

Non-degenerate 2-forms are great. When you insert a vector into one, you get a 1-form; and because of the non-degeneracy, if the vector is nonzero, then the resulting 1-form is nonzero. So you get a bijective correspondence, or duality, between 1-forms and vectors.

This means that, at each point $$p \in \mathbb{R}^2$$, the non-degenerate 2-form $$d\beta$$ provides a linear map of 2-dimensional vector spaces
$T_p \mathbb{R}^2 \rightarrow T_p^* \mathbb{R}^2,$
or in other words
$\{ \text{Vectors at p} \} \rightarrow \{ \text{1-forms at p} \},$
which sends a vector $$v$$ to the 1-form $$\iota_v d\beta$$ (i.e. the 1-form $$d\beta(v, \cdot)$$, where you have fed $$d\beta$$ one vector, but it eats two courses of vectors, and after its entree it remains a 1-form on its remaining main course). It’s a linear map and, by the non-degeneracy of $$d\beta$$, its kernel/nullspace consists solely of the zero vector. Thus it’s injective and, both vector spaces being 2-dimensional, it’s an isomorphism.

If we consider (smooth) vector fields rather than just single vectors, then we can simultaneously do this at each point of $$\mathbb{R}^2$$, and we get a map
$\{ \text{Vector fields on \mathbb{R}^2} \} \rightarrow \{ \text{1-forms on \mathbb{R}^2} \}.$

So $$d\beta$$, being a non-degenerate 2-form, gives us a way to go from 1-forms to vectors and back again. We can think of this as a duality: for each 1-form, this correspondence gives us a dual 1-form, and vice versa.

So far, we only have $$\beta$$ and $$d\beta$$. But $$\beta$$ is a 1-form! So some vector field must correspond to it. Let’s call it $$X$$. As it turns out, the 1-form $$\beta$$, the 2-form $$d\beta$$, and the vector field $$X$$, form a very nice structure.

The name of Joseph Liouville is often associated with this stuff. Often the 1-form $$\beta$$ is called a Liouville form, often the surface (or manifold in general) is called a Liouville manifold, and the whole thing is often called a Liouville structure.

In fact, we can draw pictures of such a structure.

The easiest thing to draw is the vector field $$X$$: a vector, drawn as an arrow, at each point.

How do we draw $$\beta$$? It’s generally hard to draw a picture of a differential form! However for a 1-form, we can draw its kernel $$\ker \beta$$. At any point $$p$$, $$\beta_p$$ is a linear map from the tangent space $$T_p \mathbb{R^2}$$, which is a 2-dimensional vector space, to $$\mathbb{R}$$. When $$\beta_p = 0$$, $$\beta_p$$ is the zero map $$T_p \mathbb{R}^2 \rightarrow \mathbb{R}$$ and hence the kernel is the whole 2-dimensional tangent space $$T_p \mathbb{R}^2$$. But where $$\beta_p \neq 0$$, we have a nontrivial linear map from a 2-dimensional vector space to a 1-dimensional vector space. Hence $$\beta_p$$ has rank 1 and nullity 1, so the kernel is a 1-dimensional subspace of $$T_p \mathbb{R}^2$$. We then have a 1-dimensional tangent subspace at each point; in other words, $$\ker \beta$$ is a line field on $$\mathbb{R}^2$$. We can even join up the lines (i.e. integrate them) to obtain a collection of curves on $$\mathbb{R}^2$$, which become the leaves of a foliation. In this way $$\beta$$ can be drawn as a collection of curves, or singular foliation, on $$\mathbb{R}^2$$: the foliation is singular at the points where $$\beta = 0$$. True, drawing it this way only shows the kernel of $$\beta$$, i.e. in which direction you will get zero if you feed a vector into $$\beta$$, and you will not see what you get if you feed vectors in other directions. So you don’t see how “strong” $$\beta$$ is at each point. But it’s a useful way to represent $$\beta$$ nonetheless.

As for the 2-form $$d\beta$$? It’s just an area form; we won’t attempt to draw anything to represent that.

So, let’s draw what we get in our example. In our example, the 1-form is $$\beta = x \; dy$$, the 2-form is $$d\beta = dx \wedge dy$$, and the vector field $$X$$ must satisfy
$\iota_X d\beta = \beta, \quad \text{i.e.} \quad \iota_X ( dx \wedge dy ) = x \; dy.$
It’s not too difficult to calculate that $$X = x \partial_x$$ (here $$\partial_x$$ is a unit vector in the $$x$$ direction; if we’re using $$(x,y)$$ to denote coordinates, then $$\partial_x = (1,0)$$).

Being a multiple of $$\partial_x$$, $$X$$ always points in the $$x$$ direction, i.e. horizontally. When $$x$$ is positive it points to the right, when $$x$$ is negative it points to the left; and when $$x=0$$, i.e. along the $$y$$-axis, $$X$$ is zero. So $$X$$ is actually a singular vector field, in the sense that it has zeroes. And it’s zero along a whole line. (So it’s not generic.)

As for $$\beta = x \; dy$$, it is zero when $$x=0$$, so the foliation $$\ker \beta$$ is singular along the $$y$$-axis. When $$x \neq 0$$, the kernel consists of anything pointing in the $$x$$-direction. Since $$\beta$$ has only a $$dy$$, but no $$dx$$ term, if you feed it a $$\partial_x$$, or any multiple thereof, you’ll get zero. So the line field you draw is horinzontal, as is the foliation. In other words, $$\ker \beta$$ is the singular foliation consisting of horizontal lines, with singularities along the $$y$$-axis.

Note that, although we might have expected the line field of $$\ker \beta$$ and the arrows of $$X$$ to point all over the place, in fact the arrows and lines point in the same direction, i.e. horizontal! The vectors of $$X$$ point along the lines of $$\ker \beta$$. This means that, at each point, the vector $$X$$ lies in the kernel of $$\beta$$, and hence $$\beta(X) = 0$$.

Now in this example, the vector field $$X = x \partial_x$$ has a very nice property. If you flow it, then points move out horizontally, and exponentially. Indeed, if you interpret $$x \partial_x$$ as a velocity vector field, it is telling a point with horizontal coordinate $$x$$ to move to the right, with velocity $$x$$. Telling something to move as fast as where it already is, is a hallmark of exponential movement.

Denoting the flow of $$X$$ for time $$t$$ by $$\phi_t$$, we have a map $$\phi_t : \mathbb{R}^2 \rightarrow \mathbb{R}^2$$. It’s an exponential function:
$\phi_t (x,y) = (x e^{t}, y).$
Indeed, you can check that $$\frac{\partial}{\partial t} \phi_t = (x e^t, 0)$$, which at time $$t=0$$ is
$\frac{\partial}{\partial t} |_{t=0} = (x,0) = X.$

Now the vector field $$X$$ (or its flow $$\phi_t$$) expands in the horizontal direction exponentially, and does nothing in the $$y$$ direction: from this it follows that $$X$$ also expands area exponentially. An infinitesimal volume $$V$$, after flowing under $$\phi_t$$ for time $$t$$, expands so that $$\frac{\partial V}{\partial t} = V$$, and hence grows exponentially: at time $$t$$, the volume has expanded from $$V$$ to $$V e^t$$. Rephrased in terms of differential forms, the Lie derivative of the area form $$d\beta = dx \wedge dy$$ under the flow of $$X$$ is the area form itself:
$L_X d\beta = d\beta.$

To see this, we use the Cartan formula $$L = d\iota + \iota d$$, which yields
$L_X d\beta = d \iota_X (d\beta) + \iota_X d (d\beta).$
From the definition of $$X$$, being dual to $$\beta$$, we have $$\iota_X (d\beta) = \beta$$; using this, and the fact that $$d^2 = 0$$, the first term becomes $$d\beta$$ and the second term is zero.

In fact, $$X$$ doesn’t just expand the area $$d\beta$$ exponentially; it also expands the Liouville 1-form $$\beta$$ exponentially. In other words,
$L_X \beta = \beta.$

To see this, we just apply the Cartan formula,
$L_X \beta = d \iota_X \beta + \iota_X d\beta = 0 + \beta.$
The first term is zero because, as we saw, $$X$$ points along $$\ker \beta$$, so $$\iota_X \beta = \beta(X) = 0$$; the second term is zero because of the definition of $$X$$ as dual to $$\beta$$.

To summarise our example so far: we started with a 1-form $$\beta$$ whose exterior derivative $$d\beta$$ was a non-degenerate 2-form. We took a vector field $$X$$ dual to $$\beta$$, using the non-degeneracy of $$d\beta$$. We have found that:

• The vector field $$X$$ points along the foliation $$\ker \beta$$; in other words, $$\beta(X) = 0$$.
• Flowing $$X$$ expands area exponentially: $$L_X d\beta = d\beta$$.
• Flowing $$X$$ in fact expands the 1-form $$\beta$$ exponentially: $$L_X \beta = \beta$$.

We proved some of these in sort-of generality, but not everything. Let’s prove it all at once in general, now.

PROPOSITION. Let $$\beta$$ be a 1-form on a surface such that $$d\beta$$ is non-degenerate. Let $$X$$ be dual to $$\beta$$, i.e. $$\iota_X d\beta = \beta$$. Then:
(i) $$\beta(X) = 0$$
(ii) $$L_X d\beta = d\beta$$
(iii) $$L_X \beta = \beta.$$

PROOF. For (i), we note $$\beta(X) = \iota_X \beta$$, and then use $$\beta = \iota_X d\beta$$ and the fact that differential forms are antisymmetric:
$\beta(X) = \iota_X \beta = \iota_X \iota_X d\beta = d\beta(X,X) = 0.$

For (ii) and (iii), we can then follow the arguments above. For (ii), we use the Cartan formula $$L_X = d \iota_X + \iota_X d$$, the fact that $$d^2 = 0$$, and the fact that $$\iota_X d\beta = \beta$$.
$L_X d\beta = d \iota_X d\beta + \iota_X d d\beta = d\beta + 0$

For (iii), we use the Cartan formula, part (i) that $$\iota_X \beta = 0$$, and the fact that $$\iota_X d\beta =\beta$$.
$L_X \beta = d \iota_X \beta + \iota_X d \beta = d 0 + \beta = \beta.$

So Liouville structures have some very nice properties.

And, this is all in fact classical physics. We can think of the plane as a phase space, and $$\beta$$ as an action $$y \; dx = p \; dq$$. Then $$d\beta$$ is the symplectic form on phase space, and the equation $$L_X d\beta = d\beta$$ shows that this symplectic form, the fundamental structure on the phase space, is expanded by the flow of $$X$$. (There is something in classical mechanics, or symplectic geometry, known as Liouville’s thoerem, which also says something about the effect of a flow of a vector field on the symplectic form.)

Anyway, above we saw one example of a Liouville structure on the plane. Here’s another one, which is more “radial”.

Take $$\beta = \frac{1}{2}(x dy – y dx)$$. Then $$d\beta = \frac{1}{2}(dx \wedge dy – dy \wedge dx) = dx \wedge dy$$. The vector field $$X$$ dual to $$\beta$$ is then $$\frac{1}{2}(x \; dx + y \; dy)$$, which is a radial vector field. The kernel of $$\beta$$ is also the radial direction: $$\beta(X) = (xdy – ydx)(x \partial_x + y \partial_y) = xy – yx = 0$$. The flow of $$X$$ then looks like it should expand area exponentially, and it does.

Liouville structures can exist in other places too, and not just on surfaces: they exist in higher dimensions too. Notice that the fact that $$\beta$$ was on a surface was never actually used in the proof above: it could have been any manifold, with any 1-form $$\beta$$ such that $$d\beta$$ is non-degenerate.

However, not every manifold has Liouville structures. In fact, there are many surfaces on which no Liouville structures exist. Any compact surface (without boundary) has no Liouville structure.

Why is this? The idea is pretty simple. If you have a closed and bounded surface, it’s pretty hard to have a smooth vector field $$X$$ which expands the area! Your surface has a given finite area, and then you flow it along $$X$$ — moving the points of the surface around by a diffeomorphism — and now it has exponentially larger area! This is pretty paradoxical, and indeed it’s a contradiction.

The plane escapes this paradox, because it’s not bounded. You can indeed expand the plane by any factor you like, and it’s still the same plane. Surfaces with boundary also escape the paradox, because the flow of $$X$$ will not be defined for all time: eventually points will be pushed off the edge.

But a sphere torus does not escape the paradox. Nor does a torus, or any higher genus compact surface without boundary.

Further, Liouville structures can only exist in even dimensions. So you can’t have one on a 3-dimensional space. But you can have one on a 4-dimensional space. Why is this? It can be seen from linear algebra. You simply can’t have non-degenerate 2-forms in odd dimensions. (The easiest way I know to see this is as follows. Let $$\omega$$ be a non-degenerate 2-form on an $$n$$-dimensional vector space. Choose a basis $$e_1, \ldots, e_n$$ and write $$\omega(e_i, e_j)$$ as the $$(i,j)$$ term of the $$n \times n$$ matrix $$A$$ for $$\omega$$. The facts that $$\omega$$ is antisymmetric and non-degenerate mean that $$A^T= -A$$ and $$\det A \neq 0$$ respectively. But then $$\det A = \det A^T = \det (-A) = (-1)^n \det A$$, so $$(-1)^n = 1$$, and $$n$$ is even.

Just as a Liouville structure cannot exist on a compact surface (without boundary), it can’t exist on any compact manifold (without boundary). The argument is similar: because $$X$$ expands the 2-form $$d\beta$$, it also expands all its exterior powers, and hence the volume form of the manifold, whatever (even) dimension it may be.

So, we’ve seen that a Liouville structure can only exist on a manifold with even dimension, and which is not compact, or has boundary. (If you know about de Rham cohomology, it’s not difficult to see why the manifold must have $$H^2 \neq 0$$.)

But when it does, we have a wonderful little geometric triplet $$\beta, d\beta$$ and $$X$$.

Lovely Liouville geometry
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### 2 thoughts on “Lovely Liouville geometry”

• 2020-09-20 at 6:52 am

“When βp=0, β is the zero map TpR2→R2 and hence the kernel is the whole 2-dimensional tangent space TpR2.”
Is there a typo here? Shouldn’t it be “the zero map TpR2→0”?

• 2020-09-21 at 10:57 am

There is a typo, thanks. βp is a map from TpR2 -> R, not R2. The codomain is R (all differential forms map into R, at least in this setting) but the image is just zero. It’s updated now.