Klas Modin PRO
Mathematician at Chalmers University of Technology and the University of Gothenburg
The Structure of Groups of Diffeomorphisms
Klas Modin
Geometry of the Euler equations
\dot v + \nabla_v v = -\nabla p
\operatorname{div}v = 0
Euler's equations describe Riemannian geodesics on
\operatorname{Diff}_{\mu}(M) = \{ \varphi\in\operatorname{Diff}(M) \mid |D\varphi| = 1 \}
But how?
Riemannian metrics on Lie groups
V \in T_g G
g
\langle V,V\rangle_g =
?
G
Riemannian metrics on Lie groups
V\cdot g^{-1} \in \mathfrak{g}
e
\langle V,V\rangle_g =
\langle V\cdot g^{-1},V\cdot g^{-1}\rangle_e
G
"Right-invariant" Riemannian metric determined by inner product on \(\mathfrak{g}\)
Simple example: free rigid body
G = SO(3), \quad \mathfrak{g} = \mathfrak{so}(3)\simeq \mathbb{R}^3
Inner product: moments of inertia tensor
Arnold's example: Euler equations
G = \operatorname{Diff}_\mu(M), \quad \mathfrak{g} = \mathfrak{X}_\mu(M) = \{ v\mid \operatorname{div}v=0 \}
Inner product:
\displaystyle\langle v,v\rangle_{L^2} = \int_M \lvert v \rvert^2 \mu
Arnold's theorem: \(\gamma(t)\in \operatorname{Diff}_\mu(M)\) geodesic curve \(\Rightarrow\) vector field \(v(t) = \dot\gamma(t)\circ\gamma(t)^{-1}\) fulfills Euler's equations
But in what sense is \(\operatorname{Diff}_\mu(M)\) a Lie group?
Manifold structure of \(\operatorname{Diff}^s(M)\)
\operatorname{Diff}^s(M)
U \subset H^s
V \subset H^s
Thm [Palais, Omori, Ebin, Ebin and Marsden]
\(\operatorname{Diff}^s(M)\) is smooth Hilbert manifold if \(s>\operatorname{dim}(M)/2+1\)
\(\operatorname{Diff}^s_\mu(M)\) is a submanifold
What about Lie group structure?
Thm [Ebin]
\(\operatorname{Diff}^s(M)\) topological group if \(s>\operatorname{dim}(M)/2+1\)
Group structure:
(\varphi,\eta)\mapsto \varphi\circ \eta
\varphi\mapsto \varphi^{-1}
Failure of smoothness (illustration):
Compute derivative of left translation \(L_\varphi(\eta) = \varphi\circ\eta\)
T_\eta L_\varphi\cdot V = D\varphi\circ\eta \cdot V
H^s
What about Lie group structure?
Thm [Ebin]
\(\operatorname{Diff}^s(M)\) topological group if \(s>\operatorname{dim}(M)/2+1\)
Group structure:
(\varphi,\eta)\mapsto \varphi\circ \eta
\varphi\mapsto \varphi^{-1}
Failure of smoothness (illustration):
Compute derivative of left translation \(L_\varphi(\eta) = \varphi\circ\eta\)
T_\eta L_\varphi\cdot V = D\varphi\circ\eta \cdot V
H^s
What about Lie group structure?
Thm [Ebin]
\(\operatorname{Diff}^s(M)\) topological group if \(s>\operatorname{dim}(M)/2+1\)
Group structure:
(\varphi,\eta)\mapsto \varphi\circ \eta
\varphi\mapsto \varphi^{-1}
Failure of smoothness (illustration):
Compute derivative of left translation \(L_\varphi(\eta) = \varphi\circ\eta\)
T_\eta L_\varphi\cdot V = D\varphi\circ\eta \cdot V
H^s
What about Lie group structure?
Thm [Ebin]
\(\operatorname{Diff}^s(M)\) topological group if \(s>\operatorname{dim}(M)/2+1\)
Group structure:
(\varphi,\eta)\mapsto \varphi\circ \eta
\varphi\mapsto \varphi^{-1}
Failure of smoothness (illustration):
Compute derivative of left translation \(L_\varphi(\eta) = \varphi\circ\eta\)
T_\eta L_\varphi\cdot V = D\varphi\circ\eta \cdot V
H^{s-1}
"Fix" the Lie group structure
Option 1
\operatorname{Diff}(M)
U \subset C^\infty
V \subset C^\infty
Use Fréchet manifolds instead
- Pros: proper Lie group structure
- Cons: no Picard iterations
Option 2
Right translation \(R_\varphi(\eta) = \eta\circ\varphi\)
T_\eta R_\varphi\cdot V = V\circ\varphi
H^s
"Fix" the Lie group structure
Option 1
\operatorname{Diff}(M)
U \subset C^\infty
V \subset C^\infty
Use Fréchet manifolds instead
- Pros: proper Lie group structure
- Cons: no Picard iterations
Option 2
Right translation \(R_\varphi(\eta) = \eta\circ\varphi\)
T_\eta R_\varphi\cdot V = V\circ\varphi
H^s
"Fix" the Lie group structure
Option 1
\operatorname{Diff}(M)
U \subset C^\infty
V \subset C^\infty
Use Fréchet manifolds instead
- Pros: proper Lie group structure
- Cons: no Picard iterations
Option 2
Right translation \(R_\varphi(\eta) = \eta\circ\varphi\)
T_\eta R_\varphi\cdot V = V\circ\varphi
H^s
So, let's work only with right translation
Crazy idea, but it works!
Simple example of how it works
\dot v - \dot v_{xx} + 3 v v_x - 2v_x v_{xx} - v v_{xxx} = 0
Camassa-Holm equation
A \dot v + v Av_x + 2 v_x A v = 0,
A = 1 - \partial_{xx}^2
v(t,\cdot) \in H^s(S^1), \quad s>2
Why PDE not ODE?
Idea: [Ebin and Marsden] maybe geodesic equation on \(T\operatorname{Diff}^s(S^1)\) is an ODE
H^{s-2}
H^{s-2}
H^{s-3}
Camassa-Holm as geodesic equation
\operatorname{Diff}^s(S^1)
H^s(S^1)
v(t)
H^s(S^1) \ni v(t) \mapsto (\underbrace{v(t)\circ\varphi(t)}_{\dot\varphi(t)},\varphi(t))\in T\operatorname{Diff}^s(S^1)
\dot\varphi(t) = v(t)\circ\varphi(t)
Camassa-Holm as geodesic equation
\operatorname{Diff}^s(S^1)
H^s(S^1)
\dot\varphi(t)
H^s(S^1) \ni v(t) \mapsto (\underbrace{v(t)\circ\varphi(t)}_{\dot\varphi(t)},\varphi(t))\in T\operatorname{Diff}^s(S^1)
\dot\varphi(t) = v(t)\circ\varphi(t)
\ddot\varphi = \dot v\circ\varphi + v_x\circ\varphi \, \dot\varphi
= (\dot v + v_x v)\circ\varphi = \Big( A^{-1}(-v Av_x - 2 v_x A v + A(v_x v)) \Big)\circ\varphi
= \Big( A^{-1}(-v Av_x + v A v_x + H^{s-2}\text{-terms}) \Big)\circ\varphi
Camassa-Holm as geodesic equation
\operatorname{Diff}^s(S^1)
H^s(S^1)
\dot\varphi(t)
\dot\varphi(t) = v(t)\circ\varphi(t)
\ddot\varphi = \dot v\circ\varphi + v_x\circ\varphi \, \dot\varphi
= (\dot v + v_x v)\circ\varphi = \Big( A^{-1}(-v Av_x - 2 v_x A v + A(v_x v)) \Big)\circ\varphi
= \Big( A^{-1}(-v Av_x + v A v_x + H^{s-2}\text{-terms}) \Big)\circ\varphi
Are we done?
No! RHS must be smooth as function of \( \varphi,\dot\varphi\) but \(v=\dot\varphi\circ\varphi^{-1}\)
\underbrace{\phantom{gffdffd}}_{B(v,v)}
Second piece of magic
\ddot\varphi = \tilde A^{-1}_\varphi (\tilde B_\varphi(\dot\varphi,\dot\varphi)), \quad B:H^s \times H^s\to H^{s-2}
P: H^s \to H^{s-2}, \; \tilde P_\varphi: T_\varphi\operatorname{Diff}^s(S^1)\to T^{s-2}_\varphi\operatorname{Diff}^s(S^1)
\tilde P_\varphi V = (P(V\circ\varphi^{-1}))\circ\varphi
Geodesic equation, again
Lemma: Mapping \(T\operatorname{Diff}^s(S^1)\to T^{s-1}\operatorname{Diff}^s(S^1)\) given by \[(\varphi,\dot\varphi)\mapsto (\partial_x(\dot\varphi\circ\varphi^{-1}))\circ\varphi \] is smooth
Second piece of magic
\ddot\varphi = \tilde A^{-1}_\varphi (\tilde B_\varphi(\dot\varphi,\dot\varphi)), \quad B:H^s \times H^s\to H^{s-2}
Geodesic equation, again
Lemma: Mapping \(T\operatorname{Diff}^s(S^1)\to T^{s-1}\operatorname{Diff}^s(S^1)\) given by \[(\varphi,\dot\varphi)\mapsto (\partial_x(\dot\varphi\circ\varphi^{-1}))\circ\varphi \] is smooth
Proof:
\displaystyle (\partial_x (\dot\varphi\circ\varphi^{-1}))\circ\varphi = \dot\varphi_x((\partial_x\varphi^{-1})\circ\varphi) = \frac{\dot\varphi_x}{\varphi_x}
Second piece of magic
\ddot\varphi = \tilde A^{-1}_\varphi (\tilde B_\varphi(\dot\varphi,\dot\varphi)), \quad B:H^s \times H^s\to H^{s-2}
Geodesic equation, again
Lemma: Mapping \(T\operatorname{Diff}^s(S^1)\to T^{s-1}\operatorname{Diff}^s(S^1)\) given by \[(\varphi,\dot\varphi)\mapsto (\partial_x(\dot\varphi\circ\varphi^{-1}))\circ\varphi \] is smooth
Thm: Spray \(T\operatorname{Diff}^s(S^1)\to T^{s-2}\operatorname{Diff}^s(S^1)\) given by \[(\varphi,\dot\varphi)\mapsto \tilde A^{-1}_\varphi (\tilde B_\varphi(\dot\varphi,\dot\varphi)) \] is smooth
That's it!
Examples of more recent work
- Replace \(A\) by pseudo-differential operator
[Bauer, Bruveris, Cismas, Escher, Kolev (2020)]
- Extend to semi-invariant metrics on \(\operatorname{Diff}^s(M)\) (with connections to optimal transport)
[Bauer and Modin (2020)]
- Extend to Euler equations perturbed by noise (SPDE)
[Maurelli, Modin, Schmeding (2020)]
Thank you!
Groups of diffeomorphisms
By Klas Modin
Groups of diffeomorphisms
Online-presentation given 2020-10 at Hebrew University Analysis Seminar.
