Covering Spaces
of a Robot Arm's Flat Torus C-Space
Consider a double-pendulum robot arm with c-space a 2D torus
\(T^2\).
My CAD model of a 2D pendulum (click for OnShape link)
My 3D print of my 2D CAD model
The 2D c-space of my double pendulum CAD model with a (not closed)
geodesic traced out on it
A video of Prof. Jim Yorke
displaying chaotic properties of the double pendulum
at a Mardem lecture at UWM.
A video of Prof. Steve
Brunton displaying control engineering of the chaos on the double
pendulum.
A video of a Boston Dynamics
robot dancing.
(When on can "make the system dance" [usually metaprhorical, but
literal in this case], one generally has the desired level of control
of the system.)
We shall identify \(T^2\) with \([0,2\pi]^2\), with units of
radians, via the standard coordinate patch \(\phi: [0, 2\pi]^2 \to
T^2\) given by
\(\phi(\theta_1, \theta_2) = ((\cos(\theta_1), \sin(\theta_1)),
(\cos(\theta_2), \sin(\theta_1)))\) and therefore endow it with the
units of radians, although it is technically unitless. The Riemannian
metric (r-metric) \(g(\theta_1, \theta_2)\) (written as \(M(\theta)\)
in the
following graphic from https://youtu.be/BjD-pL819LA?si=q8_It-jJFEq0AmEu&t=212;
\(g(\theta)\) in the following graphic is the gravity term) may be read
off the equation of motion (EoM):
The
Riemannian metric in the standard coordinate patch for this robot arm
is given by
Note that the entries in the Riemannian metric
have dimensions of \(\left[\text{Length}^2 \cdot
\text{Mass}\right]\). (It should be noted that there are
dimensions/units used for the
describing the robot arm and hence, perhaps bizarrely, for describing
the
Riemannian metric. Points in the
manifold and tangent vectors to the manifold are part of the intrinsic
topology of the underlying smooth manifold of the c-space and have only
dimensions of [Angular Measure], while inner products/norms of tangent
vectors and
lengths of curves/distances between points in the manifold depend on
Riemannian metric, which is extrinsic to the topology of the underlying
manifold; inner products of tangent vectors have dimensions of
\(\displaystyle\left[\text{Angular Measure}^2\cdot\text{Length}^2 \cdot
\text{Mass}\right]\) while norms of tangent vectors and lengths of
curves/distances between points have units of
\(\displaystyle\left[\text{Angular Measure}\cdot\text{Length} \cdot
\text{Mass}^{\frac{1}{2}}\right]\).)
This link I think I computes the sectional curvature correctly in
MATLAB, and it is identically 0. This link (shamelessly pirated from [1]) I think
confirms the
computation of the sectional curvature of my Riemannian manifold in
Mathematica (although it heavily uses the fact we are in dimension 2;
I'll need to work on that). This link
shows that the volume of the torus is 693.121
\(\left[\text{rad}^2\cdot\text{m}^2\cdot\text{kg}\right]\)
First, let us consider the infimum of path lengths metric
\(\left(\text{in
}\displaystyle\left[\text{rad}\cdot\text{m} \cdot
\text{kg}^{\frac{1}{2}}\right]\right)\) on \((T^2,g)\) itself. (NB:
This is per a discussion with Boris Okun on
08/10/2023.)
Next, let us consider the "standard" loop based at (0,0) that goes
around the "first" circle of the torus rel endpoints. This loop may be
homotoped rel its endpoints to the shortest path in its homotopy class
rel endpoints; by the argument in [4] and its second comment, this
second loop should be the unique geodesic in this homotopy class.
Consider the tangent
vector at the origin, \(\vec{v}_1\), with the property that the
geodesic associated to \(\vec{v}_1\) goes around this second loop, a
geodesic, from \(t=0\) to \(t=1\). In a similar way, form a second
tangent vector \(\vec{v}_2\) at the origin so that the geodesic based
at (0,0) that is homotopic to the "second" circle of
the torus rel endpoints associated to this tangent vector goes around
once from \(t=0\) to \(t=1\). Call this first
lattice \(\Lambda_1 = (\vec{v}_1, \vec{v}_2)\)
Continuing, consider the tangent space at \((0,0)\) to \((\mathbb{R}^2,
i)\), with its standard Riemannian metric. Form the lattice
\(\Lambda_2 = (\vec{w}_1, \vec{w}_2)\) having
\(\vec{w}_1 = ||\vec{v}_1||\hat{i}\) and having \(\vec{w}_2\) with the
same
norm as \(\vec{v}_2\), the same
unordered angle with \(\vec{w}_1\) that \(\vec{v}_1\) has with
\(\vec{v}_2\), and pointing in the upper-half plane (UHP).
A Straightening-Up Lemma Of
Sorts:
By [6], according to the Cartan-Hadamard Theorem, each tangent space at
the identity, together with the exponential map, is the universal cover
of its
associated torus. But, of course, any two simply-connected covering
spaces (with covering
maps) are canonically homeomorphic (diffeomorphic, indeed, isometric,
in fact, in this
case)! In particular, by [3] and [6], there is an isometry
\(f\) which goes from \((T^2, g)\) to a "straight" flat torus using a
lattice \(\Lambda_2\) and a parallelogram in \(\mathbb{R}^2\) formed by
using the two exponential maps, sending \(\exp_1(t_1\vec{v}_1
+t_2\vec{v}_2)\) to \(\exp_2(t_2\vec{w}_1 + t_2\vec{w}_2)\).
This
leads to a commutative diagram of covering spaces
The problem now is to identify \(\Lambda_1\), that is, to identify
the two smallest-length closed geodesics
emanating from \((0,0)\) [rad] in \((T^2,g)\) with winding numbers
(1,0) and (0,1) respectively, their unit tangent
vectors, and the geodesics' lengths.
Since \(g\) is independent of \(\theta_1\), \(\gamma_1:\ [0,2\pi] \to
T^2\) given by \(\gamma_1(t) = (t,0)\) should be a geodesic (\(t\) is a
unitless parameter). Hence, if
\(C_1\) is the closed curve corresponding to \(\gamma_1\), we should
have \(\displaystyle L(C_1) = \oint\limits_{C_1}\ 1\ ds
= \int\limits_0^{2\pi} 1||\hat{\theta}_1||\ dt = \int\limits_0^{1}
2\pi||\hat{\theta}_1||\ dt' = \int\limits_0^{1}
2\pi\left(\sqrt{\hat{\theta}_1^T
g\hat{\theta}_1}\right)\ dt' =
\int\limits_0^{1} 2\pi(8.887)\ dt' = 55.842\
\left[\text{rad}\cdot\text{m} \cdot
\text{kg}^{\frac{1}{2}}\right]\). So, we
should have \(\displaystyle \vec{v}_1 = 2\pi\hat{\theta}_1\
\left(\text{norm has units of }\left[\text{rad}\cdot\text{m} \cdot
\text{kg}^{\frac{1}{2}}\right]\right)\), and \(\displaystyle\vec{w}_1 =
55.842\hat{i}\
\left(\text{norm has units of }\left[\text{rad}\cdot\text{m} \cdot
\text{kg}^{\frac{1}{2}}\right]\right)\).
Finding the other tangent vector \(\vec{v}_2\) would involve finding
the second
geodesic. This would involve starting with the path \(\gamma_4:\ [0,
2\pi] \to T^2\) given by \(\gamma_4(t) = ((1,0),(\\cos(t), \sin(t))\)
and use the Calculus
of Variations to find the variation on its homotopy class rel
endpoints, \(\gamma_5(t) = (\cos[\eta(t)], \sin[\eta(t)]]), (\cos(t),
\sin(t))\) minimizing the length in MATLAB or Python (see [7]),
integrating the unitless scalar function 1 over the closed curve
\(C_4\) associated to \(\gamma_5\) with respect to arc length to get
the geodesic's length, and reparameterizing \(\gamma_5\) to
\(\gamma_6\) whose domain is [0, 1]; then \(\displaystyle\vec{v}_2 =
\left.\frac{d\gamma_6}{dt}\right|^{t=0}\).
The Calculus of Variations problem is to minimize \(\displaystyle
J[\eta(t)] = \int\limits_0^{2\pi}
\left|\left|\frac{d\gamma_5}{dt}\right|\right|\ dt =
\int\limits_0^{2\pi} ||\eta'(t)\hat{\eta(t)}_1 + 1\hat{t}_2||\ dt =
\int\limits_0^{2\pi} \left(\sqrt{[\eta'(t)\hat{\eta(t)}_1 +
1\hat{t}_2]^T\begin{bmatrix} 28.927\cos(t) + 50.06
& 14.463\cos(t) + 10.567 \\ 14.463\cos(t) + 10.567 &
10.567\end{bmatrix}[\eta'(t)\hat{\eta(t)}_1 + 1\hat{t}_2]}\right)\ dt\)
subject
to \(\eta(0) = \eta(2\pi) = 0\).
The Calculus of Variations problem is solved by setting \(\displaystyle
L(t, s, s') = \sqrt{[s'\hat{\eta(t)}_1 +
1\hat{t}_2]^T\begin{bmatrix} 28.927\cos(t) + 50.06
& 14.463\cos(t) + 10.567 \\ 14.463\cos(t) + 10.567 &
10.567\end{bmatrix}[s'\hat{\eta(t)}_1 + 1\hat{t}_2]}\) and solving
\(\displaystyle \frac{\partial}{\partial t}\left\{\frac{\partial
L}{\partial s'}\right\} = 0\).
This
MATLAB script
appears to show there is no (real, nonzero) solution to the Calculus of
Variations problem. If this is the case, then \(\eta(t) \equiv 0\) and
\(\displaystyle \vec{v}_2 = 2\pi\hat{\theta}_1\
\left(\text{norm has units of }\left[\text{rad}\cdot\text{m} \cdot
\text{kg}^{\frac{1}{2}}\right]\right)\), so that \(||\vec{w}_2||\) is
20.425 \(\left[\text{rad}\cdot\text{m} \cdot
\text{kg}^{\frac{1}{2}}\right]\) and the angle between
\(\vec{w}_1\) and \(\vec{w}_2\) is 0.523 [rad], so
\(\displaystyle\vec{w}_2 =
17.691\hat{i} + 10.209\hat{j}\
\left(\text{norm has units of }\left[\text{rad}\cdot\text{m} \cdot
\text{kg}^{\frac{1}{2}}\right]\right)\). It remain to validate the
result by showing the surface area of of the torus
\((\mathbb{R}^2/\Lambda_2, i)\) is 693.121
\(\left[\text{rad}^2\cdot\text{m}^2\cdot\text{kg}\right]\).
\(\blacksquare\).
