# The Azimuth Project Blog - fluid flows and infinite dimensional manifolds (part 1) (Rev #1)

## Or: how fluid flows can be modelled on infinite dimensional Riemannian manifolds

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”:

Tim van Beek: picture of breaking waves.

In mathematical models this difference shows up through the kind of partial differential equation 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”.

#### What You Should Know about Infinite Dimensional Smooth Manifolds

The basic strategy when handling smooth (real) manifolds is that you have a complicated manifold $M$, but also locally for every point $p$ 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:

$\gamma: \mathbb{R} \to E$

If we write down the differential quotent

$\gamma'(t_0) := \lim_{t \to 0} \frac{1}{t} (\gamma(t_0 +t) - \gamma(t_0))$

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 get “closer” to the finite dimensional case, 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, this is the concept what we are going to use.

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:

• Andreas Kriegl and Peter W. Michor: “The convenient setting of global analysis”, AMS 1999 (it is free for download 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. The definition of a Frèchet manifold and of tangent vectors however is straight forward, so that the tangent bundle of a Frèchet manifold is canonically a Frèchet 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 Frèchet spaces things are even a little bit worse, because the dual of a Frèchet space (which is not a Banach space) is not even a Fréchet space! 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.

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, we will 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.

#### Reformulating the Geodesic Equation for an Invariant Metric

If you take a Lie group $G$ of the shelf, you get two automorphisms for free, left and right multiplication by a group element $g$:

$L_g, R_g: G \to G$
$L_g(h) := gh$
$R_g(h) := hg$

You can use the differentials of these to transport vectors from the Lie algebra $\mathfrak{g}$ of $G$ to any other tangent space $T_g G$. If we have a scalar product on $\mathfrak{g}$, we can in this way define a left or right invariant Riemannian metric on $G$.

A left or right invariant metric d on G is one that does not change if we multiply the arguments with a group element:

A metric d is left invariant iff for all $g, h_1, h_2 \in G$:

$d (h_1, h_2) = d(g h_1, g h_2)$

Similarly, d is right invariant iff:

$d (h_1, h_2) = d(h_1 g, h_2 g)$

#### A Simple Example: the Circle

As a simple example we will look at the circle $S^1$ and its diffeomorphism group $DS^1$. The Lie algebra \mathfrak{DS^1} can be identified with the space of all vector field on $\mathfrak{DS^S^1$. If we sloppily identify $S^1$ with \mathbb{R}\\\mathbb{Z} with coordinate $\mathbb{R}\\x$, then we can write for vector fields $X = u(x) \partial_x$ and $Y = v(x) \partial_x$ the commutator

$[X, Y] = (u_x v + u v_x) \partial_x$

where $u_x$ is short for the derivative:

$u_x := \frac{d u}{d x}$

And of course we have a scalar product via

$\langle X, Y \rangle = \int_{S^1} u(x) v(x) d x$

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.

category: blog, climate