### Geodesics of left invariant metrics on matrix Lie groups – Part 2 Conservation laws

In the last post, Geodesics of left invariant metrics on matrix Lie groups – Part 1,we have derived Arnold’s equation – that is a half of the problem of finding geodesics on a Lie group endowed with left-invariant metric.
[latexpage]
Suppose $G$ is a Lie group, and $g(\xi,\eta)$ is a scalar product (i.e. a nondegenerate bilinear form) on its Lie algebra $Lie(G)$. Then, using left translations $g$ defines a left invariant (Riemannian or pseudo-Riemannian) metric on the whole group $G.$ If $t\mapsto a(t)\in G$ is a path in $G,$ we use left translations to define the image $\omega(t)$ of the tangent vector $\dot{a}(t)$ in $Lie(G)$
$$\omega(t)=a(t)^{-1}\dot{a}(t).$$
Usually it is written more carefully, using $L_{a^{-1}}$ instead of $a^{-1}$, but I am using a simplified notation, well adapted to dealing with problems for matrix groups.

If $a(t)$ is a geodesic for the metric $g$, then $\omega(t)$ satisfies the Arnold’s equation
$$\dot{\omega}=B(\omega,\omega),\label{eq:ar1}$$
where
$B:Lie(G)\times Lie(G)\rightarrow Lie(G)$
is defined as
$$B^k_{ij}=g^{kl}B_{ij,l}=g^{kl}C_{jl,i}=g^{kl}g_{im}C_{jl}^m,$$
and $C_{jl}^m$ are the structure constants of $G$, that is
$$[\xi_i,\xi_j]=C_{ij}^k\, \xi_k$$
where $\xi_i$ form a basis for $Lie(G)$ and
$$B_{ij,l}=g_{lk}B_{ij}^k,\,C_{ij,l}=g_{lk}C_{ij}^k.$$

Equation (\ref{eq:ar1}) is a system of nonlinear ordinary differential equations with constant coefficients. In this form it can be found in Arnold’s 1966 paper “Sur la g\’eom\’etrie differentielle des groupes de Lie de dimension inﬁnie et ses applications \`a l’hydrodynamique des ﬂuides parfaits”, Ann. Inst.
Fourier (Grenoble) 16 (1966), 319-361.

In his blog post “The Euler-Arnold equation” Terrence Tao mentions that some people call this equation the “Euler-Arnold” equation, while some other prefer to skip Arnold’s name completely and call it “Euler-Poincare equation”. Go-figure!

Solving equations (\ref{eq:ar1}) is just one half of the whole problem of finding a geodesic. Once $\omega^{i}(t)$ are known, we need to solve the linear differential equation with variable coefficients that results from the definition of $\omega$ (\ref{eq:do}):
$$\dot{a}(t)=a(t)\omega(t),$$

For the case of the rotation group and free rigid body we did it in Taming the T-handle continued.

B. Kolev in his paper “Lie groups and mechanics. An introduction” shows that for the rotation group O(n) the equations of motion for the free rigid body are completely integrable. We will not need this result, but is useful to know that in general case there are always two quadratic constants of motion corresponding to “kinetic energy” and “square of the angular momentum”. We were discussing these constants of motion for the group O(3) in
“Asymmetric Spinning Top – The Hardest Concept To Grasp In Physics” – they are used in so-called Poinsot construction. We will discuss a version of it for the group O(2,1) in the future.

The first observation is that the “kinetic energy” $T(t)=g_{ij}\omega(t)^i\omega(t)^j$ is, in fact, a constant, independent of $t.$ To see that this is the case, we differentiate:
$$\frac{dT(t)}{dt}=g_{kl}\dot{\omega}^{k}\omega^l+g_{kl}\omega^{k}\dot{\omega^l}=2g_{kl}\omega^k\dot{\omega}^l.\label{eq:dt}$$

In the previous post we wrote Eq. (\ref{eq:ar1}) as

$$g_{kl}\,\dot{\omega}^l=C_{jk,i}\,\omega^{i}\omega^{j}.\label{eq:mk}$$

Substituting into Eq. (\ref{eq:dt}) we get

$$\frac{dT(t)}{dt}=2C_{jk,i}\omega^{i}\omega^j \omega^{k}=0.$$

The result is zero because $C_{jk,i}$ is antisymmetric in $j,k$ while $\omega^j\omega^k$ is symmetric. That is an often used property: if $A^{ij}=-A^{ji}$ and $B_{ij}=B_{ji},$ then the contraction $A^{ij}B_{ij}=0.$ Indeed $A^{ij}B_{ij}=-A^{ji}B_{ij}=-A^{ji}B_{ji}=-A^{ij}B_{ij},$ where in the last equality we have exchanged dummy indices names $i\leftrightarrow j$. If a number is equal to its negative, it must be zero.

To get the formula for the second quadratic invariant we need to return to the Ad-invariant scalar product that we have denoted $\mathring{g}$ in Killing vectors, geodesics, and “Noether’s theorem”:
$$\mathring{g}(\eta_1,\eta_2)=\mbox{const}\, \frac{1}{2}Re(\mbox{Tr}(\eta_1\eta_2))).$$
The fact that $\mathring{g}$ is Ad-invariant implies an important relation between the matrix $\mathring{g}_{ij}$ and the structure constants $C_{ij}^k$
that we are going to use. Ad-invariance means that:
$$\mathring{g}(e^{t\eta}\xi_1 e^{-t\eta},e^{t\eta}\xi_2 e^{-t\eta})=\mathring{g}(\xi_1,\xi_2)$$
for all $t\in\mathbf{R}$ and $\xi_1,\xi_2,\eta\in Lie(G).$
Differentiating at $t=0$ we get
$$\mathring{g}([\eta,\xi_1],\xi_2)+\mathring{g}(\xi_1,[\eta,\xi_2])=0.$$
Setting $\xi_1=\xi_i,\xi_2=\xi_j,\eta=\xi_k$ we get
$$C_{ki}^l\mathring{g}_{lj}+C_{kj}^l\mathring{g}_{li}=0.$$
Multiplying both sides by $g^{im}g^{jn}$ we obtain
$$C_{ki}^n\,g^{im}+C_{kj}^m\,g^{jn}=0\label{eq:adj}$$
We can now derive the second conservation law. The angular momentum $m_i$ is defined as
$$m_i=g_{ij}\omega^{j}.\label{eq:m}$$
Notice that $m$ is a covector, a one-form on $Lie(G)$, it is in the dual $Lie(G)^*$ of $Lie(G).$ It is the metric that connects the space to its dual. While vectors in $Lie(G)$ play an active role, they generate transformations, elements in the dual, one-forms from $Lie(G)^*$, are “passive”, they evaluate vectors to numbers. It is the metric that is the third element here, that allows the active principle to connect to the passive principle. The metric depends on the mass distribution. In application to rigid bodies the inertia tensor is encoded in the metric on the rotation group.

The second conservation law states that the square of the angular momentum evaluated with the Ad-invariant metric $\mathring{g}$ is constant:

m_0^2(t)=\mathring{g}^{ij}m_i(t)m_j(t)=\mbox{const}.
To verify we differentiate and use Eq. (\ref{eq:mk}) rewritten as
$$\dot{m}_k=C_{jk}^{i}\,m_{i}\omega^{j}.\label{eq:mk1}$$
$$\frac{dm_0^2(t)}{dt}=2\mathring{g}^{kl}\dot{m}_k m_l=2\mathring{g}^{kl}C_{jk}^{i}m_im_l\omega^j.$$
Now, according to Eq. (\ref{eq:adj}) $\mathring{g}^{kl}C_{jk}^{i}$ is antisymmetric in $(i,l)$, while the product $m_im_l$ is symmetric, therefore we get zero:
$$\frac{dm_0^2(t)}{dt}=0.$$

In the following posts we will first return to the case of the rotation group in three dimensions and the rigid body, and then try to apply a similar reasoning to the case of the Lorentz group O(2,1) in 2+1 dimensions.

### Without goals everything is a propos of nothing

We are still traveling in the land of 4D butterflies made of remarkable circles such as those introduced in the note Meeting with remarkable circles three days ago. We did not travel too far during these three days, but we managed to familiarize ourselves with one particular family of these peculiar creatures in the last post Being a 4D butterfly – how does it feel? It was supposed to be an illustration of More than one path, which I illustrated with the wisdom of Miyamoto Musashi, expert Japanese swordsman and rōnin:

But, in fact, my many geodesics 4D butterfly picture of yesterday did not fit the Buddhist philosophy of many paths to the top of the mountain:

What we see are many paths, all of the same type, but each leading to similar, but a different mountain. It is, perhaps, a proper illustration for the saying that:

It sounds amusing, but, as you can read in Don’t think of a black cat” by Peter Wright,

Every scenario or paradigm and situation should have a goal or a collection of goals, outcomes that are worked towards. Without goals everything is a propos of nothing (as with Dirk and Deke … or maybe All I Wanna Do by Sheryl Crow).

All I Wanna DoSheryl Crow
This ain’t no country club
And it ain’t no disco
This is New York City
1, 2, 1, 2
“All I wanna do is have a little fun before I die, ”
Says the man next to me out of nowhere
It’s apropos of nothin’
He says, “His name is William”
But I’m sure he’s Bill or Billy or Mac or Buddy
And he’s plain ugly to me
And I wonder if he’s ever had a day of fun in his whole life

[latexpage]
I promised to explain in detail how exactly the 4D butterfly is created. If only to have some fun. Yesterday I was not completely sure that my idea was good. Today I think it is OK, therefore worth sharing.

The key is the algorithm described in Meeting with remarkable circles. It produces a path $q(t)$ in the 3D sphere in 4D space, the 3-sphere of quaternions of unit length. The function $q(t)$ is a very particular solution of the evolution equation for a free rigid body rotating in zero-gravity space about its center of mass – like in Dzhanibekov effect. It is a very special solution, with just one flip, and also with time origin $t=0$ arranged so, that it is exactly in the middle of the flip, between one story of almost uniform rotation one way in the past, and another story of almost uniform rotations, another way, in the future. For our particular path we have
$$q(0)=(\frac{\sqrt{3}}{2},0,\frac{1}{2},0),\quad\mbox{i.e. } q(0)=\frac{\sqrt{3}}{2}+\frac{1}{2}\,\mathbf{j}.$$

The function $q(t)$ is a solution of the quaternion evolution equation (cf. Quaternion evolution)
$$\dot{q}(t)=\frac{1}{2}q(t)\widehat{\boldsymbol{\Omega} (t)},\label{eq:qd}$$
where $\Omega_i(t)=L_i(t)/I_i,$ and $\mathbf{L}$ is a solution of Euler’s equations:
\begin{eqnarray}
\frac{dL_1}{dt}+(\frac{1}{I_2}-\frac{1}{I_3})L_2L_3&=&0,\\
\frac{dL_2}{dt}+(\frac{1}{I_3}-\frac{1}{I_1})L_3L_1&=&0,\\
\frac{dL_3}{dt}+(\frac{1}{I_1}-\frac{1}{I_2})L_1L_2&=&0.
\end{eqnarray}

There are two simple ways in which we get, from our particular solution $q(t),$ other solutions. The first way is to multiply $q(t)$ by a constant unit quaternion $u.$ If $q(t)$ is a solution, then $uq(t)$ is another solution. This follows by multiplying both sides of Eq. (\ref{eq:qd}) by $u$ form the left, and by noticing that $u,$ being constant, enters under the symbol of differentiation $d/dt,$ indicated by the dot over $q.$

The second method is by shifting the origin of time. For any fixed $s$ the function $t\mapsto q(t+s)$ is also a solution of Eq. (\ref{eq:qd}) if $q(t)$ is. Formally it follows from the rule of differentiation of a composite function (the Chain rule) and from the fact that $\frac{d}{dt}(t+s)=1.$

For producing the family of solutions used in the 4D butterfly image we use both methods as follows. Let
$q(t,s)$ be defined as follows:

$$q(t,s)=q(0)q(s)^*q(s+t).$$

Then, for any fixed $s,$ the function $t\mapsto q(t,s)$ is a solution of Eq. (\ref{eq:qd}) with the property that $q(0,s)=q(0).$ that is all trajectories have the same origin, namely $q(0).$ In verifying this last property we use the fact, that $q(s)$ are unit quaternions, therefore $q(s)^*q(s)=1.$

To produce the butterfly on the picture above I used $s$ from $s=-4$ to $s=4,$ with step $0.4.$ Today I went further on, with $s$ between $-10$ and $10,$ step $0,5$, to produce this image:

An interesting symmetry shows up when looking from far away, between circles that are “bridges”and circles that are “attractors”.

Some mystery is hidden there, waiting for being discovered in the future.

But, all of the above was in the spirit of “Happiness is a journey, not a destination” philosophy. What about many paths to the top of the mountain?

That would be keeping the attractor circles fixed (these are the two tops of mountain, one in the past, one in the future), and changing only the bridges connecting these limit circles.

This is, in fact, much easier. We just shift the time origin, and skip the part of multiplication from the left. Here are images produced this way.

The two eternal return circles are the same, only the bridges connecting them are shifting.

This last image resembles images from more realistic Dzhanibekov’s many-flips histories. It is time now to study them… if only to have some fun.

### Circles of eternal return

Let me recall the particular trajectory that we are discussing. In fact the particular part of this particular trajectory where the flip occurs.

Click on the image to open gif animation. It will take time to load as it is 2.5 MB.

[latexpage]
The flip is much like one of these flips that Russian cosmonaut Dzhanibekov observed in zero gravity, and was very much perplexed by the phenomenon.

We are investigating the jungle of math that we have to apply in order to simulate such a behavior of an ideal Platonic rigid body. At present we are analyzing the special conditions when only one flip happens. Apart of a rather short (as compared to eternity) period of time when the flip occurs, the body rotates uniformly about its middle axis. Its trajectory in the rotation group, represented by quaternions of unit norm, follows a circle. One circle before flip, another circle after the flip.

In this note I am taking a close look at these two circles. Can we find their exact coordinate representation in 3D after stereographic projection?
Yes, we can, and we will do it now.
We use the formulas from Meeting with remarkable circles.
For the body of our choice, with moments of inertia $I_1=1, I_2=2,I_3=3,$ and on the trajectory where $d=1/I_2,$ the constants that enter the solution have values
\begin{eqnarray}
A_1&=&\sqrt{\frac{I_1(I_3-I_2)}{I_2(I_3-I_1)}}=1/2,\\
A_2&=&1,\\
A_3&=&\sqrt{\frac{I_3(I_2-I_1)}{I_2(I_3-I_1)}}=\frac{\sqrt{3}}{2},\\
B&=&\frac{1}{I_2}\,\sqrt{\frac{(I_2-I_1)(I_3-I_2)}{I_1I_3}}=\frac{1}{2\sqrt{3}},\\
\delta&=&\frac{\sqrt{I_2(I_3-I_1)}-\sqrt{I_1(I_3-I_2)}}{\sqrt{I_3(I_2-I_1)}}=\frac{1}{\sqrt{3}}.
\end{eqnarray}
In the solution of the Euler’s equations we have functions $\tanh (Bt)$ and $\mathrm{sech} (Bt)$
\begin{eqnarray}
l_1(t)&=& A_1\,\mathrm{sech }(B t),\\
l_2(t)&=&A_2\,\tanh( B t),\\
l_3(t)&=&A_3\,\mathrm{sech }(B t),
\end{eqnarray}
with $\mathrm{sech }(x)=1/\cosh(x).$
Here are the plots of these functions:

For $|t|>100$ we have $l_1(t)$ smaller than $10^{-12},$ and for $|t|>200$ smaller than $10^{-25}.$ We can consider it being zero for all practical purposes.
As for $l_2(t),$ it becomes practically constant and equal 1 for $t>100,$ and $-1$ for $t<-100.$ We then have the function $\psi(t)$ $$\psi(t)=\frac{t}{I_2}+2\arctan\left(\delta \tanh(B t/2)\right).$$ Again for all practical purposes for $t>100$ we have
$$\psi(t)\approx t/2+\pi/3,$$
and for $t<-100,$ $$\psi(t)\approx t/2-\pi/3.$$ Let us substitute these formulas into the solution for $q(t)$ \begin{eqnarray} q_0(t)&=&\frac{\sqrt{1+l_1(t)}\cos\frac{\psi(t)}{2}}{\sqrt{2}},\\ q_1(t)&=&\frac{\sqrt{1+l_1(t)}\sin\frac{\psi(t)}{2}}{\sqrt{2}},\\ q_2(t)&=&\frac{l_3(t)\cos\frac{\psi(t)}{2}+l_2(t)\sin\frac{\psi(t)}{2}}{\sqrt{2(1+l_1(t))}},\\ q_3(t)&=&\frac{-l_2(t)\cos\frac{\psi(t)}{2}+l_3(t)\sin\frac{\psi(t)}{2}}{\sqrt{2(1+l_1(t))}}. \end{eqnarray} We obtain \begin{eqnarray} q_0(t)&=&\frac{\cos\frac{\psi(t)}{2}}{\sqrt{2}},\\ q_1(t)&=&\frac{\sin\frac{\psi(t)}{2}}{\sqrt{2}},\\ q_2(t)&=&\pm\frac{\sin\frac{\psi(t)}{2}}{\sqrt{2}},\\ q_3(t)&=&\frac{\mp\cos\frac{\psi(t)}{2}}{\sqrt{2}}. \end{eqnarray} The upper signs concern $t>100$ the lower signs concern $t<-100.$ In stereographic projection we are plotting $(q_1,q_2,q_3)$ divided by $1-q_0.$ We see that the "future circle" is in the plane $x=y,$ while the past circle in the orthogonal plane $x=-y.$
I did my geometric exercises and I have found that “the future circle” has origin at $(0,0,-1),$ while “the past circle” has the origin at $(0,0,1).$ Both circles have radius $\sqrt{2}.$

So, we have complete information about these “circles of eternal return”.
One more comment: in Another geodesic line I was showing what I thought was a subtle structure of the trajectory approaching the circle:

But now I know that it was a numerical artefact. As noticed in today’s post, for $|t|>100$ computer will see only one line, no fine structure. To be sure I asked my computer to take more points on the plot (100000 instead of 10000) and all the fine structure disappeared.