Geodesics on \({SL}(n)\) with the Hilbert-Schmidt metric, and one application to fluids

Lefschetz seminar, Clark University — 2021.11.12

Geodesics on \({SL}(n)\) with the Hilbert-Schmidt metric, and one application to fluids

Willie Wai Yeung Wong
Michigan State University

Slides available :

Credits page

Audrey will present a poster on this at JMM2022; do stop by and say hi!


  1. Set-up
  2. Physics
  3. Geometry; prior results
  4. Our results
  5. Application; what's next?

1. Set-up

Dramatis personae

Hilbert-Schmidt metric on $SL(n)$

$M(n)$ with $\langle A,B\rangle$ $\qquad\cong\qquad$ $\mathbb{R}^{n\times n}$ with standard metric.

$SL(n)$ is a co-dimension 1;
induced Riemannian metric — the HS metric.

Not bi-invariant!
(Non-compact, but semi-simple, so Killing form is pseudo-Riemannian.)

The Question

What is the geometry of this Riemannian manifold?

Why emphasis on geodesics?

2. Physics

Rigid Body


Rigid Body

For write-up with more details: click here.

Action Principle

\[ S = \int \left[ \mathrm{vol}(\Omega) |x'(t)|^2 + \mathrm{tr}\left( A'(t) I_\Omega A'(t)^T \right) \right] ~dt\]

The integrand defines a positive definite quadratic form on the tangent space of $\mathbb{R}^n\times SO(n)$

Solutions are geodesics w.r.t. this Riemannian metric.
Note that the metric given by $I_\Omega$ is not the bi-invariant metric on $SO(n)$.

Motion on $\mathbb{R}^n$ decouples from motion on $SO(n)$.

Affine Motion

Why do we only care when $I_\Omega = \mathrm{Id}_n$?

Incompressible Fluids

Relating affine motion to fluid motion

3. Geometry; prior results

Discrete symmetry

\(A \to A^T\) is isometry of \(M(n)\), and fixes \(SL(n)\), so is isometry on \(SL(n)\).

Lemma. $S(n)\cap SL(n)$ is totally geodesic in \(SL(n)\).

Corollary. $\mathring{S}_+(n)$ is totally geodesic in \(SL(n)\).

Continuous symmetries

Polar decomposition

Every $A\in SL(n)$ has a unique factorization as $OP$ (or $PO$): $ \begin{cases} O\in SO(n) \newline P\in \mathring{S}_+(n)\end{cases}$

$SL(n)$ foliated by cosets

So the geodesics are easy to describe, right?

Problem: Left actions generate right-invariant vector fields and vice versa

... so even with our conserved quantities, the motions on the two factors do not split.

$n = 2$ is special: analysis

Will return to this classification a bit later.

$n = 2$ is special: geometry



Call $SO(n)\subsetneq SL(n)$ the "throat": it describes the points closest to the origin in $M(n)$.

Second fundamental form

Sign convention: $II(X,X) \gt 0$ if it curves away from the origin.

Second f.f. defined via normal pointing away from origin of \(M(n)\).

Geodesic equation

Let $X,Y\in T_A SL(n)$, then \[ II(X,Y) = \frac{\mathrm{tr}(A^{-1}X A^{-1}Y)}{|A^{-1}|} \] and the geodesic equation reads \[ \ddot{A} = \frac{ \mathrm{tr}( A^{-1} \dot{A} A^{-1} \dot{A})}{\mathrm{tr}(A^{-1} A^{-T})} A^{-T}.\]

$n = 2$ geodesic classification

How much of these survive in higher dimensions?

$n = 3$ prior results

Analyses of some special explicit solutions and their asymptotics

Sideris (ARMA 2017)

$n$-independent prior results

Sideris (ARMA 2017)

4. Our results

Higher dimensions are more curvy

Linear solutions are linearly stable

Bounded geodesics abound in $n \geq 4$

Plot of pulsating motion
Numerical simulation for $n = 6$, for generic data when axes are paired; plots are the three semi-major axial lengths.

Refinement of Virial argument

\[ \frac{d^2}{dt^2} |A|^2 = \frac{|\dot{A}|^2}{|A|} + \frac{n}{|A|^2} II(\dot{A}, \dot{A}).\]

Lemma. When $n = 2$, we have $\frac{d^2}{dt^2} |A(t)|^2 \leq 0 \implies |A(t)| = |\mathrm{Id}|$.

Corollary. All geodesics except for the throat rotation are unbounded.

Emphatically false when $n \geq 3$.

Virial revisited

Theorem. [sufficient conditions for unboundedness]

  1. Solution is tangent to $\mathring{S}_+(n)$ (or another coset)
  2. Solution has vanishing total angular momentum
  3. Solution has vanishing total vorticity

New formulation of equations

Given solution $A(t)$, define \[ \beta = A^{-1}A^{-T}, \quad \omega = A^T \dot{A} + \dot{A}^T A, \quad \zeta = A^T \dot{A} - \dot{A}^T A \]

Above is the vorticity version; angular momentum version moves the transpose to the other factor.

Cannot do both simultaneously.

New formulation of equations

Solves problem with Killing vector field not pointing in the fibre direction.

\begin{align*} \frac{d}{dt} \beta & = -\beta \omega \beta \newline \frac{d}{dt}\omega & = \frac12(\omega + \zeta)^T\beta(\omega + \zeta) + \underbrace{\frac{\mathrm{tr}\omega\beta\omega\beta}{2\mathrm{tr}\beta}}_{\geq 0} I + \underbrace{\frac{\mathrm{tr}\zeta\beta\zeta\beta}{2\mathrm{tr}\beta}}_{\leq 0} I \newline \frac{d}{dt}\zeta &= 0 \end{align*}

Compatibility condition: $\quad \mathrm{tr}\beta\omega = 0$.

Preserves "block diagonal structure"

Block diagonal solutions

Assume $\beta, \omega,\zeta$ decompose into $2\times 2$ blocks (with one $1\times 1$ block when $n$ is odd) along the diagonal.

A boundedness criterion

Theorem. Let $n$ be even. If the initial data of $\beta,\omega$ are built from $2\times 2$ blocks that are pure-trace, and if $\zeta$ has no vanishing bocks, then the solution is bounded.

When $n = 2$, hypothesis only possible with rotation at throat.

"Swirling and shear flows"

Theorem. If the initial data has the form \[ \beta_0 = \begin{pmatrix} b_1 \mathrm{Id}_{2m} \newline & b_2 \mathrm{Id}_{n - 2m}\end{pmatrix}, \quad \omega_0 = \begin{pmatrix} w_1 \mathrm{Id}_{2m} \newline & w_2 \mathrm{Id}_{n-2m} \end{pmatrix} \] and writing $\varepsilon$ for the antisymmetric $2\times 2$ matrix \[ \zeta_0 = \begin{pmatrix} z (\underbrace{\varepsilon \oplus \cdots \oplus \varepsilon}_{m \text{ copies}}) \newline & 0 \end{pmatrix};\] then the corresponding solution is unbounded, and asymptotically linear.

Sideris [ARMA 2017] described $n = 3$ and $m = 1$ with explicit integration

Further generalization?

Plot generalized swirling flow.
Numerical simulation of a generalized swirling and shear flow with three sets of axes, instead of two.

5. Application; what's next?

Asymptotically linear solutions can be unstable

Connection to fluids

Further questions

Thank you!