Suppose we have a nonholonomic mechanical system, say Lagrangian, for example the Chaplygin sleigh is a model of a knife in the plane. Its configuation space is $Q = S^1\times \mathbb{R}^2$ with local coordinates $q = (\theta,x,y)$. The nonholonomic constraint is "no admissible velocities are perpendicular to the blade" which is specified by a one-form

$$\omega_q = \sin\theta\,\mathrm dx + \cos\theta\,\mathrm dy.$$ When evaulating on a velocity $\dot{q}$ we specify the velocity constraint $$v = \omega_q\cdot \dot{q} = -\dot{x}\sin\theta + \dot{y}\cos\theta = 0.$$

What is the mathematical/physical meaning of the exterior derivative of the constraint one-form? Since $\omega_q$ is a one-form, in physics, we typically interpret it as a force. Mathematically, we integrate this form over a 'line' to get the work due to the force over the 'line'. The exterior derivative is a 2-form, (does this have any physical significance??)

$$\mathrm d\omega = -\cos\theta\,\mathrm d\theta \wedge\,\mathrm dx - \sin\theta\,\mathrm d\theta \wedge\,\mathrm dy$$

If we evaluate this 2-form along a tangent (velocity) vector $\dot{q}$, we find the 'force'

$$\alpha_q = \mathrm d\omega(\dot{q},.) = \left(\dot{x}\cos\theta + \dot{y}\sin\theta \right)\,\mathrm d\theta - \cos\theta\dot{\theta}\,\mathrm dx - \sin\theta \dot{\theta}\,\mathrm dy.$$

Does this 'force', $\alpha_q \in T^*Q$, have any direct/obvious physical/mathematical physics/differential geometric interpretation??


  • $\begingroup$ This is an interesting question (maybe better suited for physics SE). My first thoughts were as follows: Your form $\alpha_q$ is just $i_{\dot{q}} d \omega$. By the Cartan formula, $$ i_{\dot{q}} d \omega = \mathcal{L}_{\dot{q}} \omega - d i_{\dot{q}} \omega $$ But we've already said that $$ i_{\dot{q}} \omega = 0 $$ Thus $$ i_{\dot{q}} d\omega = \mathcal{L}_{\dot{q}} \omega $$ Your form is precisely the Lie derivative of $\omega$ along $\dot{q}$. That is, it measures how $\omega$ changes as it is dragged along a curve satisfying the constraint. $\endgroup$ Commented Jul 18, 2021 at 22:51
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    $\begingroup$ But this doesn't work because $\dot{q}$ is not a vector field, so $d i_{\dot{q}} \omega$ doesn't make sense. We have to extend $\dot{q}$ to a vector field $X$ which also satisfies $\omega(X) = 0$ for the above analysis to go through. There must be a better way of looking at it. $\endgroup$ Commented Jul 18, 2021 at 22:52

1 Answer 1


Note that, in a sense, the constraint one-form $\omega$ is not itself "physically meaningful": What matters the distribution $\mathcal{D}$ of "allowed directions", and there are many possible choices of (local) generating one-forms, and any interpretation of $d\omega$ will have to take this redundancy into account. Still there is a useful interpretation of $(d\omega)|_\mathcal{D}\neq 0$ as saying that the constraint $\omega(\dot{q})=0$ can be effectively circumvented by local motion.

Throughout, let $\mathcal{D}$ be a smooth distribution of constant rank on a manifold $M$, and $\Omega^kM$ denote the set of differential $k$-forms on $M$.

1.An aside on the Frobenius theorem

Given a differential form $\omega$. we may restrict it to $\mathcal{D}$ to obtain a smoothly varying alternating multilinear form $\omega|_\mathcal{D}$ on $\mathcal{D}$. The key property of one-forms that define the constraint is that they vanish when restricted in this way. Instead, we can define the set of all differential forms (of any order) which vanish when restricted to $\mathcal{D}$, denoted by $I_\mathcal{D}=\{\omega\in\Omega^kM:\omega|_\mathcal{D}=0\}$, which has a few useful properties.

  • $I_\mathcal{D}$ is uniquely determined by the distribution $\mathcal{D}$.
  • $I_\mathcal{D}$ is an ideal of of the algebra of differential forms, in the sense that for any $\omega\in I_\mathcal{D}$ and $\mu\in\Omega^kM$, we have $\omega\wedge\mu\in I_\mathcal{D}$.
  • If defining one-forms $\omega_1,\cdots,\omega_m$ generate $\mathcal{D}$, then they also generate the ideal $I_\mathcal{D}$, in the sense that $$ I_\mathcal{D}=\left\{\sum_{i=1}^m\omega_i\wedge\mu_i:\mu_1,\cdots,\mu_m\in\Omega^k M\right\}\cup\{0\} $$

A beautiful result known as the Frobenius theorem states that a distribution $\mathcal{D}$ is integrable (i.e. holonomic) if and only if the corresponding ideal is closed under exterior differentiation, i.e. $\omega|_\mathcal{D}=0\implies (d\omega)|_\mathcal{D}=0$. This implies we should look at the restriction $(d\omega)|_\mathcal{D}$ rather than $d\omega$ itself, and that this restriction should somehow indicate the extent to which the constraint imposed be $\omega$ fails to be integrable.

2. "Violating" constraints with local motion

Let $\omega\in I_\mathcal{D}$ be a constraint one form, and $X,Y\in\Gamma(D)$ be two vector fields compatible with the constraints. Starting at a point $p$ we can define a constraint-satisfying path with a time parameter $\tau$ and a speed parameter $v$:

  • Flow along an integral curve of $vX$ for time $\tau$,
  • flow along an integral curve of $vY$ for time $\tau$,
  • flow along an integral curve of $-vX$ for time $\tau$,
  • flow along an integral curve of $-vY$ for time $\tau$,
  • repeat ad infinitum.

One can show that in the limit $\tau\to 0,v\to\infty$ with $v^2\tau=1$, this path approaches an integral curve of the Lie bracket vector field $[X,Y]$, so $[X,Y]$ is a kind of "average velocity" from alternately moving back and forth along $X$ and $Y$. We can then ask if this average velocity satisfies the constraint imposed by $\omega$: $$\begin{align*} \omega([X,Y])=&\omega([X,Y])+0-0 \\ =&\omega([X,Y])+X(\omega(Y))-Y(\omega(X)) \\ =&d\omega(X,Y) \end{align*}$$ We see that it satisfies the constraint precisely when the vector fields satisfy $d\omega(X,Y)=0$. Conversely, if $(d\omega)|_\mathcal{D}\neq 0$ at $p$, we can find vector fields $X,Y$ which satisfy the constraints such that their alternating flow in the manner described above, starting at $p$, "violates" the constraint.

3.Your example, and parallel parking

If you're familiar with parallel parking, you have already seen this principle in action. Most automobiles are described more or less accurately by your model, in that they have a "heading" direction and can only move forward/backward along that direction or rotate their heading. However, by alternating small clockwise/counterclockwise rotations and forward/backward motion, they can effectively move perpendicular to their heading.

You can explicitly compute this: At the point $(x,y,\theta)=(0,0,0)$, the constraint $\omega_{(0,0,0)}=dy$ forbids motion in the $y$ direction. However, choosing vector fields $X(x,y,\theta)=\cos\theta\partial_x-\sin\theta\partial_y$ and $\Theta(x,y,\theta)=\partial_\theta$, we have $[X,\Theta]_{(0,0,0)}=\partial_y$ and $d\omega(X,\Theta)_{(0,0,0)}=1$, so alternately moving along these vector fields will result in effective motion in the $y$ direction.

  • $\begingroup$ Do constraints which you can/can't effectively violate through this sort of "wiggling" have names? $\endgroup$ Commented Jul 20, 2021 at 19:47
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    $\begingroup$ @CharlesHudgins Constraints which cannot be violated this way are exactly integrable (i.e. holonomic) constraints. If if $\omega\neq0$ and $d\omega=0$ on a neighborhood, then you can find a local coordinate $y$ such that the constraint $\omega(\dot{q})=0$ is equivalent to $y(q)=\text{constant}$. $\endgroup$
    – Kajelad
    Commented Jul 20, 2021 at 22:32

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