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Blog - fluid flows and infinite dimensional manifolds (part 1) (Rev #7)

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”:

breaking wave

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.

Fluid Flow modelled by Diffeomorphisms

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=0t_0 = 0 on a given manifold MM, the flow of every fluid package is decribed by a path on MM parameterized by time, and for every time t>t 0t \gt t_0 there is a diffeomorphism g tg^t of MM defined by the requirement that it maps the initial position yy of every fluid package to the position x=g t(y)x = g^t(y) at time tt.

schematic fluid flow

This picture is taken from the book

We will take as a model of the domain of the fluid flow a compact Riemannian manifold MM that is also a Lie group. A fluid flow, as pictured above, is then a path in the diffeomorphism group DMDM. In order to apply geometric concepts in this situation, we will have to turn DMDM or some closed subgroup of it into a manifold, which will be infinite dimensional.

The curvature of such a manifold can provide a great deal about the stability of fluid flows: On a manifold with negative curvature geodesics will diverge from each other. If we can model fluid flows as geodesics in a Riemannian manifold and calculate the curvature, we could try to infer a bound on weather forecasts. In fact, that is what Vladimir Arnold did!

If you never thought about manifolds in infinite dimensions, you may feel a little bit insecure as to how the concepts that you know from differential geometry can be generalized from finite dimensions. At least I felt this way when I first read about it. But it turns out that the part of the theory one needs to know in order to understand Arnolds insight is not that scary, so I will talk a little bit about it next.

What You Should Know about Infinite Dimensional Smooth Manifolds

The basic strategy when handling finite dimensional, smooth, real manifolds is that you have a complicated manifold MM, but also locally for every point pMp \in M a neighborhood UU and an isomorphism (a “chart”) of UU to an open subset of the vastly simpler space n\mathbb{R}^n, the “model space”. These isomorphisms can be used to transport concepts from n\mathbb{R}^n to MM. In infinite dimensions it is however not that clear what kind of model space EE should be taken in place of n\mathbb{R}^n. What structure should EE have?

Since we would like to differentiate, we should for example be able to define the differential of a curve in E:

γ:E \gamma: \mathbb{R} \to E

If we write down the differential quotent

γ(t 0):=lim t01t(γ(t 0+t)γ(t 0)) \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 EE with this structure are called topological vector spaces.

A curve that has infinite many differentials in the sense above at all points is called a smooth curve, just as in the finite dimensional case.

So EE should at least be a topological vector space. We can, of course, put more structure on EE to make it “more similar” to n\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 the space of smooth maps C ( n)C^{\infinity}(\mathbb{R}^n) of n\mathbb{R}^n \to \mathbb{R}, for example, is not a Banach space but a Fréchet space, we should not expect that we can model diffeomorphism groups on Banach spaces, but on Fréchet spaces. So we will 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:

  • 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 n\mathbb{R}^n with a Fréchet space EE, there will be certain things that won’t work out as easily as for the finite dimensional n\mathbb{R}^n, or not at all. The definition of a (topological) Fréchet manifold however is straight forward, you can keep it word for word. We will call a function between manifolds smooth if it maps smooth curves to smooth curves: In this way we get the definition of a smooth Fréchet manifold.

With tangent vectors, you may remember that there are several different ways to define them in the finite dimensional case, which turn out to be equivalent. Since there are situations in infinite dimensions where these definitions turn out to not be equivalent, I will be explicit and define tangent vectors in the “kinematical way”:

The (kinematic) tangent vector space T pMT_p M of a Fréchet manifold MM at a point pp consists of all pairs (p,c(0))(p, c'(0)) where cc is a smooth curve

c:Mwithc(0)=p c: \mathbb{R} \to M \; \text{with} \; c(0) = p

With this definition, the set of pairs (p,c(0)),pM(p, c'(0)), p \in M forms a Fréchet manifold, the tangent bundle TMTM, just as in finite dimensions.

The first serious (more or less) problem we encounter is the definition of the cotangent bundle: n\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 and promply got an answer. 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 and define it explicitly:

A differential form (a one form) α\alpha is a smooth map

α:TM \alpha: T M \to \mathbb{R}

where TMTM is the tangent bundle, such that α\alpha restricts to a linear map on every tangent space T pMT_p M.

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 n\mathbb{R}^n only, so that a lot generalizes straight forward to infinite dimensional Banach spaces. 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:

  • Richard S. Hamilton: The Inverse Function Theorem of Nash and Moser (Bulletin (New Series) of the American Mathematical Society Volume 7, Number 1, July 1982)

Reformulating the Geodesic Equation for an Invariant Metric

If MM 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 MM 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 2Gg, h_1, h_2 \in G:

d(h 1,h 2)=d(gh 1,gh 2) 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 1g,h 2g) d (h_1, h_2) = d(h_1 g, h_2 g)

How does one get a one-sided invariant metric?

Here is one possibility: If you take a Lie group MM of the shelf, you get two automorphisms for free, namely the left and right multiplication by a group element gg:

L g,R g:MM L_g, R_g: M \to M
L g(h):=gh L_g(h) := gh
R g(h):=hg R_g(h) := hg

Pictorially speaking, you can use the differentials of these to transport vectors from the Lie algebra 𝔪\mathfrak{m} of MM - which is the tangential space at the group identity T idMT_{id}M - to any other tangent space T gMT_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.

To be more precise, for every tangent vectors U,VU, V of a tangent space T gMT_gM there are unique vectors X,YX, Y that are mapped to U,VU, V by the differential of the right multiplication R gR_g, that is

dR gX=UanddR gY=V dR_g X = U \; \text{and} \; dR_g Y = V

and we can define the scalar product of UU and VV to have the value of that of XX and YY:

U,V:=X,Y \langle U, V \rangle := \langle X, Y \rangle

This works for the left multiplication L gL_g, too, of course.

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 MM with tangential bundle TMTM there is a unique connection, the Levi-Civita connection, with the following properties for vector fields X,Y,ZTMX, Y, Z \in TM:

ZX,Y= ZX,Y+X, ZY(compatibility with the metric) Z \langle X, Y \rangle = \langle \nabla_Z X, Y \rangle + \langle X, \nabla_Z Y \rangle \; \text{(compatibility with the metric)}
XY YX=[X,Y](torsion freeness) \nabla_X Y - \nabla_Y X = [X, Y] \; \text{(torsion freeness)}

If we combine both formulas we get

2 XY,Z=XY,Z+YZ,XZX,Y+[X,Y],Z[Y,Z],X+[Z,X],Y 2 \langle \nabla_X Y, Z \rangle = X \langle Y, Z \rangle + Y \langle Z, X \rangle - Z \langle X, Y \rangle + \langle [X, Y], Z \rangle - \langle [Y, Z], X \rangle + \langle [Z, X], Y \rangle

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

2 XY,Z=[X,Y],Z[Y,Z],X+[Z,X],Y 2 \langle \nabla_X Y, Z \rangle = \langle [X, Y], Z \rangle - \langle [Y, Z], X \rangle + \langle [Z, X], Y \rangle

This latter formula can be written in a more succinct way if we introduce the coadjoint operator. Remeber the adjoint operator defined to be

ad XZ=[X,Z] ad_X Z = [X, Z]

With the help of the scalar product we can define the adjoint of this operator:

ad X *Y,Z:=Y,ad XZ=Y,[X,Z] \langle ad^*_X Y, Z \rangle := \langle Y, ad_X Z \rangle = \langle Y, [X, Z] \rangle

Then the formula above for the covariant derivative can be written as

2 XY,Z=ad XY,Zad Y *X,Zad X *Y,Z 2 \langle \nabla_X Y, Z \rangle = \langle ad_X Y, Z \rangle - \langle ad^*_Y X, Z \rangle - \langle ad^*_X Y, Z \rangle

Since the inner product is nondegenerate, we can eliminate ZZ and get

2 XY=ad XYad X *Yad Y *X 2 \nabla_X Y = ad_X Y - ad^*_X Y - ad^*_Y X

A geodesic curve is one whose tangent vector XX is transported parallel to itself, that is we have

XX=0 \nabla_X X = 0

Using the formula for the covariant derivative for an invariant metric above we get

XX=ad X *X=0 \nabla_X X = - ad^*_X X = 0

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 t\partial_t as an additional summand. So, in this case we get as geodesic equation (again: for an invariant metric)

XX= tXad X *X=0 \nabla_X X = \partial_t X - ad^*_X X = 0

A Simple Example: the Circle

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

[X,Y]=(uv xu xv) x [X, Y] = (u v_x - u_x v) \partial_x

where u xu_x is short for the derivative:

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

And of course we have a scalar product via

X,Y= S 1u(x)v(x)dx \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 1DS^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:

ad X *Y,Z:=Y,ad XZ=Y,[X,Z] \langle ad^*_X Y, Z \rangle := \langle Y, ad_X Z \rangle = \langle Y, [X, Z] \rangle

If we write for the vector fields X=u(x) xX = u(x) \partial_x, Y=v(x) xY = v(x) \partial_x and Z=w(x) xZ = w(x) \partial_x, this results in

ad X *Y,Z= S 1v(uw xu xw)dx= S 1(uv x+2u xv)wdx \langle ad^*_X Y, Z \rangle = \int_{S^1} v (u w_x - u_x w) d x = - \int_{S^1} (u v_x + 2 u_x v) w d x

where the last step employs integration by parts and uses the periodic boundary condition f(x+1)=f(x)f(x + 1) = f(x) for the involved functions.

So we get for the coadjoint operator

ad X *Y=(uv x+2u xv) x ad^*_X Y = - (u v_x + 2 u_x v) \partial_x

Finally, the geodesic equation

tX+ XX=0 \partial_t X + \nabla_X X = 0

turns out to be

u t+3uu x=0 u_t + 3 u u_x = 0

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