Euler–Poincaré and Lagrange–Poincaré Reduction
This entry builds the Euler–Poincaré equation one step at a time, for an arbitrary Lie group. It then adds forces, closes the system, and introduces the momentum map. It ends with Lagrange–Poincaré reduction, which handles systems that are a group plus internal shape, and a few worked examples.
Conventions. Repeated indices are summed. Vectors and covectors are bold. Points of a manifold are not bold. For a covector $\bfalpha$ and a vector $\bfv$, the pairing is $\left(\bfalpha, \bfv\right)$. A hat marks a unit vector.
Reaction Wheel Applet
The group and its Lie algebra
Let $G$ be a Lie group of dimension $n$, with identity $e$. A configuration is a point $g \in G$, and a motion is a curve $g(t)$.
The Lie algebra is the tangent space at the identity:
\[\frakg = T_e G.\]It is an $n$-dimensional vector space. Its bracket is defined in the section on the basis, bracket, and structure constants.
Left translation and body velocity
For each $g \in G$, left translation is the map $L_g : G \to G$ given by $L_g(h) = gh$. It is a diffeomorphism with inverse $L_{g^{-1}}$. Its pushforward at $h$ is a linear isomorphism
\[(L_g)_\ast : T_h G \to T_{gh} G.\]The body velocity is the velocity $\dot{g} \in T_g G$ translated back to the identity:
\[\bfxi = (L_{g^{-1}})_\ast\, \dot{g} \in \frakg.\]The reverse map gives the reconstruction equation:
\[\dot{g} = (L_g)_\ast\, \bfxi.\]For a matrix group, $(L_g)_\ast$ is left multiplication by $g$, so $\bfxi = g^{-1}\dot{g}$.
Basis, bracket, and structure constants
Let $\bfe_1, \dots, \bfe_n$ be any basis of $\frakg$. The body velocity has components $\xi^i$, with $\bfxi = \xi^i \bfe_i$.
Each $\bfu \in \frakg$ defines a left-invariant vector field $\bfu^L$ on $G$ by $\bfu^L(g) = (L_g)_\ast\, \bfu$. The bracket on $\frakg$ is the vector-field bracket of these fields, evaluated at the identity:
\[[\bfu, \bfv] = [\bfu^L, \bfv^L](e).\]The structure constants $c^k_{ij}$ are the components of the bracket of basis vectors:
\[[\bfe_i, \bfe_j] = c^k_{ij}\, \bfe_k.\]They are antisymmetric, $c^k_{ij} = -c^k_{ji}$. They satisfy the Jacobi identity, $c^l_{ij} c^m_{lk} + c^l_{jk} c^m_{li} + c^l_{ki} c^m_{lj} = 0$. They depend on the basis.
The reduced Lagrangian
Let $L : TG \to \bbR$ be the Lagrangian. Assume it is left-invariant. That is, for every fixed $h \in G$,
\[L\big(hg, (L_h)_\ast\, \dot{g}\big) = L(g, \dot{g}).\]Define the reduced Lagrangian $l : \frakg \to \bbR$ by $l(\bfxi) = L(e, \bfxi)$. Taking $h = g^{-1}$ in the invariance condition shows that
\[L(g, \dot{g}) = l(\bfxi).\]So $L$ depends only on the body velocity. In components, $l$ is a function of the $n$ numbers $\xi^1, \dots, \xi^n$.
The dual space and the momentum
The dual space $\frakg^\ast$ is the space of linear maps $\frakg \to \bbR$. Its dual basis $\bfe^1, \dots, \bfe^n$ is defined by $\left(\bfe^k, \bfe_i\right) = \delta^k_i$, where $\delta^k_i$ is the Kronecker delta. A covector has lower-index components, $\bfalpha = \alpha_k \bfe^k$, and $\left(\bfalpha, \bfv\right) = \alpha_k v^k$.
The body momentum $\bfmu \in \frakg^\ast$ is the derivative of $l$ at the current body velocity. Its components are
\[\mu_k = \frac{\partial l}{\partial \xi^k},\]evaluated at $\bfxi(t)$. Coordinate-free, $\left(\bfmu, \bfv\right)$ is the derivative of $l(\bfxi + s\bfv)$ in $s$ at $s = 0$. We write this as $\bfmu = \delta l / \delta \bfxi$.
Constrained variations
Fix an interval $[a, b]$. A variation of $g(t)$ is a family $g(t, s)$ with $g(t, 0) = g(t)$ and fixed endpoints. Define
\[\bfeta = (L_{g^{-1}})_\ast\, \frac{\partial g}{\partial s} \in \frakg.\]It vanishes at $t = a$ and $t = b$. The key fact is that $\bfxi$ cannot be varied freely. Its variation is
\[\delta \xi^k = \dot{\eta}^k + c^k_{ij}\, \xi^i \eta^j, \qquad \delta \bfxi = \dot{\bfeta} + [\bfxi, \bfeta].\]Why. The fields $\bfe_i^L$ form a frame on $G$, and $[\bfe_i^L, \bfe_j^L] = c^k_{ij}\, \bfe_k^L$ everywhere, since a bracket of left-invariant fields is left-invariant. Let $\bftheta^k$ be the dual one-forms. The identity $d\bftheta(\bfx, \bfy) = \bfx(\bftheta(\bfy)) - \bfy(\bftheta(\bfx)) - \bftheta([\bfx, \bfy])$ gives
\[d\bftheta^k(\bfe_i^L, \bfe_j^L) = -c^k_{ij}.\]Pull back along $\sigma(t, s) = g(t, s)$. Then $\sigma^\ast \bftheta^k = \xi^k\, dt + \eta^k\, ds$. Equating $d(\sigma^\ast \bftheta^k)$ with $\sigma^\ast(d\bftheta^k)$ gives
\[\frac{\partial \eta^k}{\partial t} - \frac{\partial \xi^k}{\partial s} = -c^k_{ij}\, \xi^i \eta^j.\]Setting $s = 0$ gives the result.
The Euler–Poincaré equation
Hamilton’s principle on $G$ is equivalent to $\delta \int_a^b l(\bfxi)\, dt = 0$ under the constrained variations from the section above. Substituting and integrating by parts,
\[\delta \int_a^b l\, dt = \int_a^b \mu_k \left( \dot{\eta}^k + c^k_{ij}\, \xi^i \eta^j \right) dt = \int_a^b \left( -\dot{\mu}_j + c^k_{ij}\, \xi^i \mu_k \right) \eta^j\, dt.\]The boundary terms vanish because $\bfeta$ does. Since $\bfeta$ is otherwise arbitrary, we get the Euler–Poincaré equation:
\[\dot{\mu}_k = c^j_{ik}\, \xi^i \mu_j.\]Coordinate-free form. For $\bfxi \in \frakg$, define $\mathrm{ad}^\ast_{\bfxi} : \frakg^\ast \to \frakg^\ast$ by $\left(\mathrm{ad}^\ast_{\bfxi}\, \bfalpha, \bfv\right) = \left(\bfalpha, [\bfxi, \bfv]\right)$. Pairing with $\bfe_k$ gives $(\mathrm{ad}^\ast_{\bfxi}\, \bfalpha)_k = c^j_{ik}\, \xi^i \alpha_j$. So the equation is
\[\dot{\bfmu} = \mathrm{ad}^\ast_{\bfxi}\, \bfmu.\]The index form does not avoid $\mathrm{ad}^\ast$. It spells out what $\mathrm{ad}^\ast$ does to components.
External forces
A force on the group is a covector $F \in T^\ast_g G$. Its body-frame version $\bff \in \frakg^\ast$ is defined by
\[\left(\bff, \bfv\right) = \left(F, (L_g)_\ast\, \bfv\right)\]for all $\bfv \in \frakg$. Hamilton’s principle is replaced by the Lagrange–d’Alembert principle, $\delta \int l\, dt + \int \left(\bff, \bfeta\right)\, dt = 0$. This adds one term:
\[\dot{\mu}_k = c^j_{ik}\, \xi^i \mu_j + f_k, \qquad \dot{\bfmu} = \mathrm{ad}^\ast_{\bfxi}\, \bfmu + \bff.\]Potentials. Left multiplication moves any point of $G$ to any other. So a left-invariant potential $V(g)$ must be constant, and $l$ cannot contain a nontrivial one. A potential that breaks the symmetry enters as a force instead, with $\left(\bff, \bfv\right) = -dV_g\big((L_g)_\ast\, \bfv\big)$. Such a force depends on $g$, so the momentum equation no longer closes on $\frakg^\ast$ by itself.
Closing the system
The full system is three relations:
\[\mu_k = \frac{\partial l}{\partial \xi^k}, \qquad \dot{\mu}_k = c^j_{ik}\, \xi^i \mu_j + f_k, \qquad \dot{g} = (L_g)_\ast\, \bfxi.\]The first must be inverted to recover $\bfxi$ from $\bfmu$. This works locally when the Hessian $M_{jk} = \partial^2 l / \partial \xi^j \partial \xi^k$ is invertible. In the common quadratic case $l = \frac{1}{2} M_{ij}\, \xi^i \xi^j$, with $M_{ij}$ constant, symmetric, and positive-definite, we have $\mu_k = M_{kj}\, \xi^j$. Then
\[\xi^i = M^{ik} \mu_k,\]where $M^{ik}$ is the inverse matrix of $M_{ik}$.
To integrate, advance $\bfmu$ with the Euler–Poincaré equation, recover $\bfxi$ at each instant, and advance $g$ with the reconstruction equation. If $\bff$ does not depend on $g$, the first two steps close on their own, and $g$ can be reconstructed afterward.
The momentum map
Let $R_g(h) = hg$ be right translation. The adjoint map $\mathrm{Ad}_g : \frakg \to \frakg$ is
\[\mathrm{Ad}_g\, \bfzeta = (L_g)_\ast\, (R_{g^{-1}})_\ast\, \bfzeta.\]Its dual $\mathrm{Ad}^\ast_g$ is defined by $\left(\mathrm{Ad}^\ast_g\, \bfalpha, \bfzeta\right) = \left(\bfalpha, \mathrm{Ad}_g\, \bfzeta\right)$.
The symmetry of $L$ is left multiplication. Its momentum map is the function $\bfJ : TG \to \frakg^\ast$ given by
\[\bfJ(g, \dot{g}) = \mathrm{Ad}^\ast_{g^{-1}}\, \bfmu.\]In components, write $\mathrm{Ad}_{g^{-1}}\, \bfe_k = A^i_k(g)\, \bfe_i$. Then $J_k = A^i_k(g)\, \mu_i$.
Meaning. Each $\bfzeta \in \frakg$ generates the symmetry $g \mapsto \exp(s\bfzeta)\, g$, whose velocity field is $(R_g)_\ast\, \bfzeta$. The number $\left(\bfJ, \bfzeta\right)$ is the Noether momentum conjugate to that symmetry. So $\bfJ$ packages all the conserved quantities into one covector. Both $\bfmu$ and $\bfJ$ describe the same momentum. $\bfmu$ sees it from the body frame, and $\bfJ$ from the spatial frame.
Evolution. Along a motion, $\frac{d}{dt}\, \mathrm{Ad}_{g^{-1}}\, \bfzeta = -[\bfxi, \mathrm{Ad}_{g^{-1}}\, \bfzeta]$. Therefore
\[\frac{d}{dt} \left(\bfJ, \bfzeta\right) = \left(\dot{\bfmu} - \mathrm{ad}^\ast_{\bfxi}\, \bfmu,\, \mathrm{Ad}_{g^{-1}}\, \bfzeta\right).\]By the forced Euler–Poincaré equation, this gives
\[\dot{\bfJ} = \mathrm{Ad}^\ast_{g^{-1}}\, \bff.\]With no force, $\bfJ$ is constant, even though $\bfmu$ generally is not.
Lagrange–Poincaré reduction
Many systems are not a pure group. They have a group part and an internal shape. Lagrange–Poincaré reduction handles this case.
Setup. Let $Q$ be the configuration space, and let $G$ act freely on $Q$ from the left. Assume $L : TQ \to \bbR$ is $G$-invariant. The shape space is the quotient $S = Q/G$. Its points are configurations with the symmetry removed.
The product case. Suppose $Q = G \times S$, with $G$ acting only on the first factor, $h \cdot (g, s) = (hg, s)$. Let $s^a$ be coordinates on $S$. The reduction then proceeds in four steps.
First, form the body velocity $\bfxi = (L_{g^{-1}})_\ast\, \dot{g}$ exactly as before. Second, use invariance to write $L$ as a reduced Lagrangian
\[l(\bfxi, s, \dot{s}) = L(g, \dot{g}, s, \dot{s}).\]Third, vary. The variation $\delta \bfxi$ is constrained as in the section on constrained variations, while $\delta s$ is free with fixed endpoints. Fourth, collect the terms multiplying $\bfeta$ and those multiplying $\delta s$. This gives
\[\mu_k = \frac{\partial l}{\partial \xi^k}, \qquad \dot{\mu}_k = c^j_{ik}\, \xi^i \mu_j + f_k,\] \[\frac{d}{dt}\, \frac{\partial l}{\partial \dot{s}^a} - \frac{\partial l}{\partial s^a} = u_a,\] \[\dot{g} = (L_g)_\ast\, \bfxi.\]Here $\bff \in \frakg^\ast$ is the external force in the body frame, and $u_a$ are generalized forces on the shape, such as internal actuators. Note that $\mu_k$ is evaluated at $(\bfxi, s, \dot{s})$, so it depends on the shape. That is how the two parts couple.
Internal forces do not change $\bfJ$. The $u_a$ appear only in the shape equation. So with $\bff = 0$, the momentum map $\bfJ = \mathrm{Ad}^\ast_{g^{-1}}\, \bfmu$ stays constant, whatever the actuators do. Internal motion can change the attitude, but only by trading momentum between the parts.
The general case. Usually $Q$ is a principal $G$-bundle over $S$, not a global product. Then a principal connection is needed to split each velocity into a group part and a shape part. The reduced equations keep the same two-part form. They gain extra terms involving the curvature of the connection, which couple shape motion to group motion. These terms are what allow a body to reorient itself at zero angular momentum.
Special cases. If $S$ is a single point, then $Q = G$ and the equations reduce to Euler–Poincaré. If $G$ is trivial, then $Q = S$ and they reduce to Euler–Lagrange.
Example: a rigid body with a reaction wheel
Let $g \in SO(3)$ be the attitude of the body. Let $\phi$ be the angle of an axisymmetric wheel relative to the body, spinning about the body-fixed unit axis $\hat{\bfa}$. Then
\[Q = SO(3) \times S^1, \qquad G = SO(3), \qquad S = S^1.\]Let $\bfomega$ be the body angular velocity, $I$ the locked inertia matrix of body plus wheel, and $I_w$ the wheel’s axial inertia. The reduced Lagrangian is
\[l(\bfomega, \dot{\phi}) = \frac{1}{2}\, \bfomega \cdot I\bfomega + I_w\, \dot{\phi}\, (\hat{\bfa} \cdot \bfomega) + \frac{1}{2} I_w\, \dot{\phi}^2.\]It does not depend on $\phi$, because spinning an axisymmetric wheel does not change the mass distribution. The momenta are
\[\bfmu = \frac{\partial l}{\partial \bfomega} = I\bfomega + I_w\, \dot{\phi}\, \hat{\bfa}, \qquad p = \frac{\partial l}{\partial \dot{\phi}} = I_w \left( \hat{\bfa} \cdot \bfomega + \dot{\phi} \right).\]Let $u$ be the motor torque and $\bftau$ an external body-frame torque. Using the $SO(3)$ structure constants from the rigid body example below, the equations are
\[\dot{\bfmu} = \bfmu \times \bfomega + \bftau, \qquad \dot{p} = u, \qquad \dot{g} = g\, \bfomega^{\wedge}.\]The velocities are recovered by
\[\bfomega = \left( I - I_w\, \hat{\bfa} \hat{\bfa}^{\mathsf{T}} \right)^{-1} \left( \bfmu - p\, \hat{\bfa} \right), \qquad \dot{\phi} = \frac{p}{I_w} - \hat{\bfa} \cdot \bfomega.\]With $\bftau = 0$, the spatial angular momentum $\bfJ = g\bfmu$ is constant for every motor torque $u$. Starting from rest, $\bfJ = 0$. Spinning the wheel up turns the body the other way. Stopping the wheel stops the body, in a new attitude.
Examples
The rigid body on $SO(3)$
The configuration is $g \in SO(3)$. The Lie algebra $\mathfrak{so}(3)$ is the space of skew-symmetric $3 \times 3$ matrices. Identify it with $\bbR^3$ by the wedge map: for $\bfv \in \bbR^3$, the matrix $\bfv^{\wedge}$ satisfies $\bfv^{\wedge}\, \bfw = \bfv \times \bfw$ for all $\bfw$. Under this map the bracket becomes the cross product.
With the standard basis of $\bbR^3$, the structure constants are $c^k_{ij} = \varepsilon_{ijk}$, the Levi-Civita symbol. The body angular velocity $\bfomega$ is defined by $\bfomega^{\wedge} = g^{-1}\dot{g}$.
Let $I_{ij}$ be the body-frame inertia matrix. Then
\[l(\bfomega) = \frac{1}{2} I_{ij}\, \omega^i \omega^j, \qquad \mu_k = I_{kj}\, \omega^j.\]The forced Euler–Poincaré equation, with body torque $\bftau$, is
\[\dot{\mu}_k = \varepsilon_{ikj}\, \omega^i \mu_j + \tau_k, \qquad \dot{\bfmu} = \bfmu \times \bfomega + \bftau.\]Substituting $\bfmu = I\bfomega$ gives Euler’s equations, $I\dot{\bfomega} = (I\bfomega) \times \bfomega + \bftau$. The system closes with $\bfomega = I^{-1}\bfmu$ and $\dot{g} = g\, \bfomega^{\wedge}$.
Here $\mathrm{Ad}_{g^{-1}}\, \bfzeta = g^{\mathsf{T}} \bfzeta$, so the momentum map is $\bfJ = g\bfmu$. This is the spatial angular momentum. It obeys $\dot{\bfJ} = g\bftau$, so it is constant for a free body.
A planar body on $SE(2)$
The configuration is $g \in SE(2)$, the position and heading of a body in the plane. Choose a basis of $\mathfrak{se}(2)$ in which $\bfe_1$ is rotation and $\bfe_2, \bfe_3$ are translations along the two body axes. The nonzero brackets are
\[[\bfe_1, \bfe_2] = \bfe_3, \qquad [\bfe_1, \bfe_3] = -\bfe_2.\]So $c^3_{12} = 1$ and $c^2_{13} = -1$, with their antisymmetric partners. The body velocity $\bfxi = \xi^i \bfe_i$ has angular rate $\xi^1$ and body-frame velocities $\xi^2, \xi^3$.
Take $l = \frac{1}{2} \left( I_0 (\xi^1)^2 + m_2 (\xi^2)^2 + m_3 (\xi^3)^2 \right)$. Here $I_0$ is the rotational inertia, and $m_2, m_3$ are the effective masses along the body axes. They differ for a body in an ideal fluid, because of added mass. Then $\mu_1 = I_0\, \xi^1$, $\mu_2 = m_2\, \xi^2$, and $\mu_3 = m_3\, \xi^3$. The Euler–Poincaré equation gives
\[\dot{\mu}_1 = (m_2 - m_3)\, \xi^2 \xi^3, \qquad \dot{\mu}_2 = m_3\, \xi^1 \xi^3, \qquad \dot{\mu}_3 = -m_2\, \xi^1 \xi^2.\]When $m_2 = m_3$, the first equation says the spin is constant. The other two say the body-frame velocity rotates opposite to the body, so the spatial velocity is constant. When $m_2 \neq m_3$, translation drives rotation. This is the Munk moment.
Summary
| Index form | Coordinate-free form | |
|---|---|---|
| Body velocity | $\xi^i$, with $\bfxi = \xi^i \bfe_i$ | $\bfxi = (L_{g^{-1}})_\ast\, \dot{g}$ |
| Momentum | $\mu_k = \partial l / \partial \xi^k$ | $\bfmu = \delta l / \delta \bfxi$ |
| Constrained variation | $\delta \xi^k = \dot{\eta}^k + c^k_{ij}\, \xi^i \eta^j$ | $\delta \bfxi = \dot{\bfeta} + [\bfxi, \bfeta]$ |
| Dynamics | $\dot{\mu}_k = c^j_{ik}\, \xi^i \mu_j + f_k$ | $\dot{\bfmu} = \mathrm{ad}^\ast_{\bfxi}\, \bfmu + \bff$ |
| Reconstruction | $\dot{g} = (L_g)_\ast\, \xi^i \bfe_i$ | $\dot{g} = (L_g)_\ast\, \bfxi$ |
| Momentum map | $J_k = A^i_k(g)\, \mu_i$ | $\bfJ = \mathrm{Ad}^\ast_{g^{-1}}\, \bfmu$ |
Further reading. Marsden and Ratiu, Introduction to Mechanics and Symmetry. Holm, Schmah, and Stoica, Geometric Mechanics and Symmetry. Cendra, Marsden, and Ratiu, Lagrangian Reduction by Stages, Memoirs of the AMS, 2001.