This page is a blog article in progress, written by Tim van Beek.
Water waves can do a lot of things that light waves, for example, cannot, like “breaking”:
In mathematical models this difference shows up through the kind of partial differential equation (PDE) that models the waves:
Light waves are modelled by linear equations while
water waves are modelled by nonlinear equations.
Physicists like to point out that linear equations model things that do not interact, while nonlinear equations model things that interact with each other. In quantum field theory this is usually dubbed as “free fields” and “interacting fields”.
Some of these nonlinear PDE turn up as geodesic equations of infinite dimensional Riemannian manifolds that describe fluid flows. This is a fascinating observation, that is due to the Russian mathematician Vladimir Arnold. In this blog post I would like to talk a little bit about the involved concepts and show you a little toy example.
The Euler viewpoint of fluids is that there are fluid packages or “particles”. The fluid flow is described by specifying where each package or particle is at a given time t. When we start with a time $t_0 = 0$ on a given manifold $M$, the flow of every fluid package is decribed by a path on $M$ parameterized by time, and for every time $t \gt t_0$ there is a diffeomorphism $g^t$ of $M$ defined by the requirement that it maps the initial position $y$ of every fluid package to the position $x = g^t(y)$ at time $t$.
This picture is taken from the book
We will take as a model of the domain of the fluid flow a compact Riemannian manifold $M$ that is also a Lie group. A fluid flow, as pictured above, is then a path in the diffeomorphism group $DM$. In order to apply geometric concepts in this situation, we will have to turn $DM$ or some closed subgroup of it into a manifold, which will be infinite dimensional. This may look scarier at first sight as it really is, as we will see :-)
The basic strategy when handling finite dimensional, smooth, real manifolds is that you have a complicated manifold $M$, but also locally for every point $p \in M$ a neighborhood $U$ and an isomorphism (a “chart”) of $U$ to an open subset of the vastly simpler space $\mathbb{R}^n$, the “model space”. These isomorphisms can be used to transport concepts from $\mathbb{R}^n$ to $M$. In infinite dimensions it is however not that clear what kind of model space $E$ should be taken in place of $\mathbb{R}^n$. What structure should $E$ have?
Since we would like to differentiate, we should for example be able to define the differential of a curve in E:
If we write down the differential quotent
we see that we need to be able to add elements, have a scalar multiplication, and a topology such that the algebraic operations are continuous, in order to make sense of this. Sets $E$ with this structure are called topological vector spaces.
So this is the structure that $E$ should at least have. We can, of course, put more structure on $E$ to make it “more similar” to $\mathbb{R}^n$, and choose as model space in ascending order of generality:
1) A Hilbert space which has a scalar product,
2) a Banach space that does not have a scalar product, but a norm,
3) a Fréchet space that does not have a norm, but a metric,
4) a general topological vector space that need not be metrizable.
People talk accordingly of Hilbert, Banach and Fréchet manifolds.
Since our diffeomorphism groups are not Banach spaces, but at least Fréchet spaces, we are going to use the concept of Fréchet manifolds.
But if you are interested in a more general theory using locally convex topological vector spaces as model spaces, you can look it up here:
Since we replace the model space $\mathbb{R}^n$ with a Fréchet space $E$, there will be certain things that won’t work at all or at least not as easily as for the finite dimensional $\mathbb{R}^n$. The definition of a Fréchet manifold and of tangent vectors however is straight forward, so that the tangent bundle of a Fréchet manifold is canonically a Fréchet manifold. The first problem we will encounter is the definition of the cotangent bundle: $\mathbb{R}^n$ is isomorph to its dual vector space. This is still true for every Hilbert space. It fails already for Banach spaces: The dual space will still be a Banach space, but a Banach space does not need to be isomorph to its dual (those who are are called reflexive).
With Fréchet spaces things are even a little bit worse, because the dual of a Fréchet space (which is not a Banach space) is not even a Fréchet space! Since I did not know that and could not find a reference, I asked about this on mathoverflow here: Mathoverflow is a really amazing platform for this kind of question!
So, if we naively define the cotangent space as in finite dimensions by taking the dual space of every tangent space, then the cotangent bundle won’t be a Fréchet manifold.
We will therefore have to be careful with the definition of differential forms for Fréchet manifolds: Since the general definition in finite dimensions does not generalize, we will need to define differential forms for every example at hand.
Another pitfall is that theorems from multivariate calculus may fail in Fréchet spaces, like the existence and uniqueness theorem of Picard-Lindelöf for ordinary differential equations. Things are much easier in Banach spaces: If you take a closer look at multivariate calculus, you will notice that a lot of definitions and theorems actually make use of the Banach space structure of $mathbb{R}^n$ only, so that a lot generalizes straight forward to infinite dimensional Banach space. But that is less so for Fréchet spaces.
It is nevertheless possible to define both integrals and differentials that behave much in the expected way. You can find a nice exposition of how this can be done in this paper:
If $M$ is both a Riemannian manifold and a Lie group, it is possible to define the concept of left or right invariant metric. A left or right invariant metric d on $M$ is one that does not change if we multiply the arguments with a group element:
A metric d is left invariant iff for all $g, h_1, h_2 \in G$:
Similarly, d is right invariant iff:
How does one get a one-sided invariant metric?
Here is one possibility: If you take a Lie group $M$ of the shelf, you get two automorphisms for free, namely the left and right multiplication by a group element $g$:
You can use the differentials of these to transport vectors from the Lie algebra $\mathfrak{m}$ of $M$ - which is the tangential space at the group identity $T_{id}M$ - to any other tangent space $T_g M$. If you can define a scalar product on the Lie algebra, you can use this trick to transport the scalar product to all the other tangential spaces by left or right multiplication, which will get you a left or right invariant metric.
For a one-sided invariant metric, the geodesic equation looks somewhat simpler than for general metrics. Let us take a look at that:
On a Riemannian manifold $M$ with tangential bundle $TM$ there is a unique connection, the Levi-Civita connection, with the following properties for vector fields $X, Y, Z \in TM$:
If we combine both formulas we get
If the scalar products are constant along every flow, i.e. the metric is (left or right) invariant, then the first three terms on the right hand side vanish, so that we get
This latter formula can be written in a more succinct way if we introduce the coadjoint operator. Remeber the adjoint operator defined to be
With the help of the scalar product we can define the adjoint of this operator:
Then the formula above for the covariant derivative can be written as
Since the inner product is nondegenerate, we can eliminate $Z$ and get
A geodesic curve is one whose tangent vector $X$ is transported parallel to itself, that is we have
Using the formula for the covariant derivative for an invariant metric above we get
as a reformulation of the geodesic equation.
For time dependent dynamical systems, we have the time axis as an additional dimension and every vector field has $\partial_t$ as an additional summand. So, in this case we get as geodesic equation (again: for an invariant metric)
As a simple example we will look at the circle $S^1$ and its diffeomorphism group $DS^1$. The Lie algebra $Vec(S^1)$ of $DS^1$ can be identified with the space of all vector fields on $S^1$. If we sloppily identify $S^1$ with $\mathbb{R}/mathbb{Z}$ with coordinate $x$, then we can write for vector fields $X = u(x) \partial_x$ and $Y = v(x) \partial_x$ the commutator
where $u_x$ is short for the derivative:
And of course we have a scalar product via
which we can use to define either a left or a right invariant metric on $DS^1$, by transporting it via left or right multiplication to every tangent space.
Let us evaluate the geodesic equation for this example. We have to calculate the effect of the coadjoint operator:
If we write for the vector fields $X = u(x) \partial_x$, $Y = v(x) \partial_x$ and $Z = w(x) \partial_x$, this results in
where the last step employs integration by parts and uses the periodic boundary condition $f(x + 1) = f(x)$ for the involved functions.
So we get for the coadjoint operator
Finally, the geodesic equation
turns out to be