What this article covers

Part 2 handed us one vertical pendulum and two reconstructions: free SR extremals with generically cuspidal projections, and smooth pinned Euler elastica in closed Jacobi-elliptic form. This part asks when symmetry produces equal-cost competitors. A curve can be perfectly taut locally and still be beaten by a completely different curve of the same length. The set of endpoints where that happens — where two distinct shortest geodesics tie — is the Maxwell stratum. We show it is forced by the reflection group hiding inside the pendulum equation. We compute the mirror-pair tie exactly on the smooth elastica family — it lands at arc length $4K(k^2)$, the very period that controlled the curvature in Part 2 — and set up the sharper result of the free sub-Riemannian problem, whose cut fires at *half* that (Part 4).

Two ways to stop being optimal

Recall the model from Part 1. One proposed completion rule selects a shortest horizontal path in $\mathrm{SE}(2)$ between two oriented points — where “shortest” always means the sub-Riemannian length, the only length the contact geometry defines. Part 2 solved the local equations: one pendulum governs every candidate, and its smooth face — the elastica family, curvature $\kappa(s) = 2k\,\mathrm{cn}(s\mid k^2)$ for the generic inflectional case — is the family we compute with below (the free SR geodesics are its cuspidal siblings; Part 2, Appendix A3).

But solving the geodesic equation only gives candidates. A geodesic is guaranteed to be the shortest path only for a while. As you extend it, one of two things eventually goes wrong.

The first is local failure. Past its first conjugate point, a geodesic is no longer even a local minimum — some tiny perturbation beats it. The conjugate time $t_{\mathrm{conj}}$ is when this first happens.

The second is global failure, and it can strike much earlier. A geodesic can be a perfect local minimum — taut, no shortcut in any thin tube around it — while, somewhere else in $\mathrm{SE}(2)$, an entirely different geodesic of exactly the same length reaches the same endpoint. At that endpoint neither curve is uniquely shortest; they tie. The moment of the first tie is the cut time $t_{\mathrm{cut}}$, and it is what we actually care about: beyond it, the “completed contour” the cortex would draw is no longer well-defined by minimality alone.

The two clocks

Along a geodesic from the origin, optimality can fail two ways: $$t_{\mathrm{cut}} \;\le\; t_{\mathrm{conj}}.$$ Conjugate ($t_{\mathrm{conj}}$): loss of local minimality, a fold of the geodesic family. Cut ($t_{\mathrm{cut}}$): loss of global minimality, the first endpoint reachable equally fast by a different geodesic. This article is about the second clock — and the symmetric mechanism that sets it.

The Maxwell mechanism

Why would two different geodesics ever reach the same point with the same length? For a generic geometry, they might not — you would have to solve transcendental equations and hope for a coincidence. But $\mathrm{SE}(2)$ is not generic. It is loaded with symmetry, and symmetry manufactures coincidences on purpose.

Here is the mechanism. Suppose the problem has a symmetry — a transformation $\varepsilon$ of the geodesics that (i) preserves length and the starting point, but (ii) sends a geodesic $\gamma$ to a genuinely different geodesic $\varepsilon(\gamma)$. If, at some time $t$, the two happen to arrive at the same endpoint,

\[\gamma(t) \;=\; \varepsilon(\gamma)(t), \qquad \gamma \neq \varepsilon(\gamma),\]

then that endpoint is a Maxwell point by construction — two distinct equal-length geodesics tie there. No luck required; the symmetry forces it. So the whole problem of locating the cut time reduces to a much more tractable one: find the symmetries, then find where a geodesic first meets its own symmetric image.

The four symmetries of the pendulum

The symmetries live not in the plane but in the pendulum that Part 2 derived. Recall the reduction: the costate angle obeys

\[\ddot\varphi + \sin\varphi = 0, \qquad E = \tfrac12\dot\varphi^2 - \cos\varphi,\]

and the inflectional family is the librating pendulum, $-1 < E < 1$, oscillating between turning points $\pm\varphi_{\max}$. This pendulum has two obvious discrete symmetries, and they generate a third.

• Time reversal $\varepsilon^1 : s \mapsto -s$. Run the pendulum backwards. Because the equation has no friction, a reversed solution is still a solution.
• Reflection $\varepsilon^2 : \varphi \mapsto -\varphi$. Flip the pendulum left-right. Since $\sin(-\varphi) = -\sin\varphi$, the equation is unchanged, so this too maps solutions to solutions. Downstairs in the plane it is the mirror $y \mapsto -y$, which flips the sign of the curvature: $\kappa \mapsto -\kappa$.
• The composite $\varepsilon^3 = \varepsilon^1!\circ\varepsilon^2$: reverse time and reflect.

Together with the identity, these close up into a group under composition: each element undoes itself, and any two of them compose to the third. That is precisely the Klein four-group $\mathbb{Z}_2 \times \mathbb{Z}_2$ — the working core of the full reflection group $(\mathbb{Z}_2)^3$ that Moiseev–Sachkov (2010) identified for this problem.

Element	Action on pendulum	Action on plane curve	Fixed set in the phase plane $(\varphi, \dot\varphi)$
$e$	identity	identity	everything
$\varepsilon^1$	$s \mapsto -s$ (time reversal)	reverse traversal	the axis $\dot\varphi = 0$ — the turning points
$\varepsilon^2$	$\varphi \mapsto -\varphi$ (mirror)	mirror $y \mapsto -y$, $\kappa \mapsto -\kappa$	the origin alone — it pairs instants, pins none
$\varepsilon^3$	both	mirrored, traversed backwards	the axis $\varphi = 0$ — the bottom crossings

(In the phase plane, $\varepsilon^1$ is the reflection across the horizontal axis; $\varepsilon^3$ — which flips $\varphi$ but, reversing time too, preserves $\dot\varphi$ — is the reflection across the vertical axis; and $\varepsilon^2$, flipping both, is the half-turn about the origin.)

What makes a symmetry useful is its fixed set downstairs, in $\mathrm{SE}(2)$ itself. If the geodesic $\gamma$ crosses a configuration that $\varepsilon$ leaves fixed, then at that instant its partner $\varepsilon(\gamma)$ passes through the same configuration — a tie, provided the partner is a genuinely different curve. For the mirror $\varepsilon^2$, acting downstairs by $(x, y, \theta) \mapsto (x, -y, -\theta)$, that fixed set is the launch axis with heading along it: ${\,y = 0,\ \theta \in {0, \pi}\,}$. The instants where a geodesic crosses this set are its candidate Maxwell times — and the next section computes the first one exactly.

highlight symmetry — none — ε¹ : time reversal (s → −s) ε² : mirror (φ → −φ) ε³ : both composed one libration orbit (blue) and the chosen symmetry's geometry (orange)

Figure 1. The reflection core acting on the pendulum. Phase plane of $\ddot\varphi + \sin\varphi = 0$: horizontal axis is the pendulum angle $\varphi$ (radians), vertical axis its rate $\dot\varphi$. The bold blue closed loop is one libration orbit (one inflectional geodesic, energy $E = 2k^2 - 1$); faint loops are neighbouring orbits. Choose a group element to see its geometry (orange): $\varepsilon^1$ reflects the phase plane across the horizontal axis $\dot\varphi = 0$, pinning the two turning points; $\varepsilon^3$ reflects across the vertical axis $\varphi = 0$, pinning the two bottom-crossings; $\varepsilon^2$ flips both signs — the half-turn about the origin — pinning no instant but pairing every phase point with its antipode (an example pair is marked). The libration orbit is carried to itself by all three: that is what makes them symmetries of the geodesic family.

The first fork, computed exactly

Take the mirror $\varepsilon^2$ and first illustrate the mechanism where everything is smooth and explicit: on the elastica family of Part 2. The free SR problem has a related reflection action on its costate cylinder, but different reconstruction equations and a different first-event formula. Apply the planar mirror to an inflectional elastica $\gamma_A$ with curvature $\kappa_A(s) = +2k\,\mathrm{cn}(s\mid k^2)$. The image $\gamma_B = \varepsilon^2(\gamma_A)$ is the curve with the opposite curvature, $\kappa_B(s) = -2k\,\mathrm{cn}(s\mid k^2)$ — a different curve (it bends the other way) but with identical length at every arc length $s$. This is the $\sigma$-symmetric pair: two mirror-image curves leaving the origin.

They start together. When do they first meet again? In the plane, the reflection acts by $y \mapsto -y$, so $\gamma_B(s) = \bigl(x_A(s),\, -y_A(s),\, -\theta_A(s)\bigr)$. As points of $\mathrm{SE}(2)$ — position and heading — the two coincide exactly when

\[y_A(s) = 0 \quad\text{and}\quad \theta_A(s) \equiv 0 \pmod{2\pi}.\]

Part 2 gave the closed form for this inflectional elastica:

\[y_A(s) = 2k\bigl(1 - \mathrm{cn}(s\mid k^2)\bigr) \;\ge\; 0.\]

This is the whole calculation in one line. Since $\mathrm{cn}(s\mid k^2) \le 1$ with equality only at $s = 0, 4K(k^2), 8K(k^2), \dots$, the height $y_A(s)$ returns to zero first at

\[s = 4K(k^2),\]

exactly one spatial period of the curvature. The heading returns with it: Part 2 gave $\theta_A(s) = 2\arcsin!\bigl(k\,\mathrm{sn}(s\mid k^2)\bigr)$, which vanishes wherever $\mathrm{sn}$ does — at $s = 0,\, 2K,\, 4K, \dots$ (it never winds; the heading just oscillates within $\pm 2\arcsin k$). At $s = 2K(k^2)$ the heading is zero but the height is maximal, $y_A = 4k$; the first instant both conditions hold together is $s = 4K(k^2)$. So the mirror pair re-coincides — position and heading at once — for the first time at $s = 4K(k^2)$, for every modulus $k$.

Mirror-pair tie (inflectional elastica)

$$\boxed{\; s^{\ast}(k) \;=\; 4K(k^2), \;}$$ one full curvature period, with $K$ the complete elliptic integral of the first kind. This is the same $4K(k^2)$ that set the curvature's spatial period in Part 2 — now reappearing as a meeting time: the first instant an elastica ties with its mirror twin. (For the free SR problem the reflection strata fire earlier — at half a pendulum period — which is exactly the cut-time story of Part 4.)

The identity is worth pausing on. In Part 2, $4K(k^2)$ was a fact about one curve — how far you travel before its curvature pattern repeats. Here it is a fact about two curves — how far you travel before an elastica and its mirror partner arrive at the same place at the same time. The two roles of the elliptic period coincide on the elastica family; the same reflection machinery, run in the costate coordinates of the free problem, is the technical heart of the Maxwell-strata theorem of Moiseev–Sachkov (2010).

$k$ 0.70 arc length $s$ — blue: $\kappa = +2k\,\mathrm{cn}$  ·  red: mirror $\kappa = -2k\,\mathrm{cn}$

Figure 2. The σ-symmetric pair forks and re-meets. The blue pinned elastica ($\kappa = +2k\,\mathrm{cn}$) and its mirror image (red, dashed, $\kappa = -2k\,\mathrm{cn}$) leave the origin (black dot) together. Drag $s$ to extend them. The vertical guide on the height plot (right) shows $y_A(s) = 2k(1-\mathrm{cn})$, the gap to the mirror axis; its first return to zero — marked, at $s = 4K(k^2)$ — is the pair's elastica Maxwell coincidence, where the two equal-length extremals arrive at the same $\mathrm{SE}(2)$ point (orange). The slider's arc-length readout turns orange once you pass it. Horizontal axis of the left panel: plane $x$; right panel: arc length $s$ against height $y_A$ (elastica arc-length units).

From the first tie to the cut time

The mirror pair showed the mechanism in its cleanest form. Now the honest accounting. For the free sub-Riemannian problem, Moiseev–Sachkov (2010) run this same reflection machinery through all seven $\varepsilon$’s, in the costate’s elliptic coordinates, and collect the earliest coincidence of each stratum into one function on the cotangent space — the first Maxwell time $\mathfrak t(\lambda)$.

Maxwell strata bound the cut time (Moiseev–Sachkov 2010)

A tie destroys uniqueness of the minimiser, so along every SR geodesic $$t_{\mathrm{cut}}(\lambda) \;\le\; \mathfrak t(\lambda),$$ and for the inflectional family the strata fire strikingly early: $\mathfrak t = 2K(k^2)$ — half a pendulum period, half the elastica mirror-tie value computed above.

That inequality is the punchline of Part 3 — and a cliffhanger. The reflection strata say the free geodesic cannot remain globally shortest past $2K(k^2)$. What this part does not settle is whether the cut happens exactly there or strictly earlier — whether the bound is tight.

It is: $t_{\mathrm{cut}} = \mathfrak t(\lambda)$ exactly, for every family — with the extra surprise that along the whole inflectional family local optimality never fails at all. Proving that, and confronting what remains genuinely open beyond $\mathrm{SE}(2)$, is Part 4.

Summary

• A geodesic loses optimality two ways: locally at its conjugate point, globally at its cut point, with $t_{\mathrm{cut}} \le t_{\mathrm{conj}}$.
• Global optimality dies at a Maxwell point — where two distinct equal-length geodesics tie — and ties are forced by symmetry, not luck.
• The pendulum carries the reflection group $(\mathbb{Z}_2)^3$ of Moiseev–Sachkov; the working core on one orbit is a Klein four-group (time reversal, mirror, composite).
• The mirror’s $\sigma$-symmetric pair, computed on the smooth elastica family, first re-meets at $s = 4K(k^2)$ — one full curvature period, the same elliptic clock as Part 2.
• For the free SR problem the collected strata give the first Maxwell time $\mathfrak t(\lambda)$, with $t_{\mathrm{cut}} \le \mathfrak t = 2K(k^2)$ on the inflectional family — half a pendulum period. Whether equality holds is Part 4.

References

1. I. Moiseev & Yu. L. Sachkov (2010). "Maxwell strata in sub-Riemannian problem on the group of motions of a plane." ESAIM: COCV 16(2): 380–399. arXiv:0807.4731
2. Yu. L. Sachkov (2010). "Conjugate and cut time in the sub-Riemannian problem on the group of motions of a plane." ESAIM: COCV 16(4): 1018–1039.
3. Yu. L. Sachkov (2011). "Cut locus and optimal synthesis in the sub-Riemannian problem on the group of motions of a plane." ESAIM: COCV 17(2): 293–321. arXiv:0903.0727
4. A. A. Agrachev & Yu. L. Sachkov (2004). Control Theory from the Geometric Viewpoint. Springer. Chapter 17 — symmetries of the exponential map and Maxwell strata.
5. Yu. L. Sachkov (2008). "Maxwell strata in the Euler elastic problem." Journal of Dynamical and Control Systems 14(2): 169–234 — the same reflection-symmetry method applied to Euler's elastica.